ELEMENTS ALGEBRA MAJOR D. H. HILL, PEOFESSOK OP MATHEMATICS AND CIVIL ENQINEEHnfO IN DAVIDSON COLLEGE, N. C. LATE FROFESSOa OF MATHEMATICS IN WASHINGTON COLLBOE, VA. PHILADELPHIA: J. B. LIPPINCOTT & CO. 1857. Entered, according to the Act of Congress, in the year 1857, by J. B. LIPPINCOTT & CO., in the Clerk's Office of the District Court of the United States for the Eastern District of Pennsylvania. STEREOTYPED BY J. FAGAN. TESTIMONIALS. From T. J. Jackson, Professor of Natural and Experimental Philo- sophy, Virginia Military Institute. " Froin an examination of various portions of Major D. H. Hill's Algebra, in manuscript, I regard it as superior to any other work with which I am acquainted on the same branch of science." From a Teacher of Mathematics. " Having also examined several chapters of Major Hill's Algebra, I have no hesitation in concurring in the above opinion of Professor Jackson." WILLIAM Mclaughlin. From J. L. Campbell, Professor of Natural Science, Washington College, Virginia. " While I fully concur with Professor Jackson and Mr. McLaughlin in the opinion they express of Major Hill's Algebra, I will add, that I regard the method adopted by the author, of incorporating into the work some of the elementary principles of the Calculus, as giving it peculiar value as a college text-book." From William Gilham, Professor of Chemistry and Geology, Virginia Military Institute, and late Professor at West Point, iV. Y. " Having read the greater portion of Mnjor Hill's Algebra, I consider it as better adapted for use as a college text-book than any work on the subject with which I am acquainted." From Professor J. A. Leland, late Professor of Mathematics in the State Military Institute of S. C. " I have examined with care, many of the proof sheets of Major Hill's Algebra, and it affords me pleasure to concur in the favorable opinions above expressed." 105971 IV TESTIMONIALS. This work of Professor Hill's is the product of a mind intensely in love with Algebra. It bears the marks of unremitting and intelligent toil. It is exhaus- tive on the subjects it treats ; and, in the abundance and aptness of its illustra- tious, reminds one of the richness and simphcity of Euler. CHARLES PHILLIPS, Prof. CM Engineering, University, N. C. CONTENTS TAQB DEFINITIONS AND PRELimNARY REMARKS 13 ADDITION 18 Case L When the Terms are like, and have like Signs 18 XL When the Quantities are like, but aflFected with unlike Signs.. 19 III. When the Quantities are similar and dissimilar, and have like and unlike Signs 20 SUBTRACTION 21 MULTIPLICATION 24 Case I. When both Multiplicand and Multiplier are Simple Quantities 24 n. When the Multiplicand is a Compound Quantity, and the Multiplier a Monomial 27 III. When the Multiplicand and Multiplier are both Compound Quantities 29 Theorems 33 Factoring Folynomials 35 DIVISION '. 38 Case I. When the Dividend and Divisor are both Monomials 38 II, When the Dividend is a Compound Quantity, and the Divisor a Monomial 41 III. When the Dividend and Divisor are both Polynomials 42 Principles in Division 48 ALGEBRAIC FRACTIONS 51 Reduction of Fractions 54 Case I. To reduce a Simple Fraction to its Simplest Form 54 II. To reduce a Mixed Quantity to the Form of a Fraction 55 III. To reduce a Fraction to an Entire or j\Iixed Quantity 56 1* (V) Yl CONTENTS. _, , ^ PAGE Reduction of Fractions (continued). ^ Ca^e IV. To develop a Fraction into a Series 57 V. To reduce Fractious having Diiferent Denominators to Equi- valent Fractions having the same Denominator 69 VI. To add Fractional Quantities together 61 VII. To Subtract one or more Fractional Quantities from one or more Fractional Quantities 62 VIII. To multiply Fractional Quantities together 64 IX. To divide Fractional Quantities by each other 65 X. To reduce a Compound Fraction to its Lowest Terms 68 Least Common Multiple 76 Corollary 78 Least Common Multiple of Fractions 79 Greatest Common Divisor of Fractions 80 EQUATIONS OF THE FIRST DEGREE 82 Solution of Equations of the First Degree 85 First Transformation — clearing an Equation of Fractions 85 Second Transformation — transposing Terms 86 Third Transformation — clearing the Unknown Quantity of Coefficients 86 Solutions of Problems producing Equations of the First Degree with an Unknown Quantity 91 GEOMETRICAL PROPORTION 95 Theorems 96 General Remarks 103 Problems in Geometrical Proportion 105 NEGATIVE QUANTITIES 107 Problem of the Couriers ; 110 General Problems '. 116 ELIMINATION BETWEEN SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE 129 Elimination by Comparison 130 Elimination by Addition and Subtraction 131 Elimination by the Greatest Common Divisor 133 Examples in Elimination between two Simple Equations of the First Degree involving two Unknown Quantities 135 Elimination between any Number of Simultaneous Equations 140 General Remarks 149 Problems producing Simultaneous Equations of the First Degree 148 CONTENTS. VU PAGE VANISHING FRACTIONS 159 FORMATION OF THE POWERS AND EXTRACTION OF ROOTS 163 Of Incommensurable Numbers 175 Extraction of the Square Root of Fractions 177 Mixed Numbers 184 Roots of Numbers entirely Decimal 186 Square Root of Fractions expressed decimnlly 187 Extraction of the Cube Root of Numbers 188 Approximate Roots of Incommensurable Numbers 19G Cube Root of Fractions 197 Approximate Root to within a certain Decimal 199 Cote I. Approximate Root of Whole Numbers to within a certain Decimal 199 II. Approximate Roots of Mixed Numbers to within a cer- tain Decimal 200 Approximate Root of Decimal Fractions to within a certain Decimal.... 202 Approximate Root of Vulgar Fractions to within a certain Decimal 203 General Remarks on the Extraction of the Cube Root 204 Extraction of the Square Root of Algebraic Quantitiet 206 Square Root of Monomials 206 Square Root of Polynomials 207 Square Root of a Polynomial involving Negative Exponents 214 Square Root of Incommensurable Polynomials 216 Square Root of Polynomials containing Terms affected with Fractional Exponents 217 Cube Root of Polynomials 217 Cube Root of Incommensurable Polynomials 222 Cube Root of Polynomials involving Fractional Exponents 223 Cube Root of Polynomials containing one or more Terms affected with Negative Exponents 224 Quantities affected with Fractional Exponents 225 Multiplication of Quantities affected with Entire or Fractional Ex- ponents 227 Division of Quantities affected with Fractional and Entire Exponents — Monomials 228 Raising to Powers Quantities affected with Fractional and Entire Ex- ponents — Monomials 230 VIU CONTENTS. PAGE Quantities affected with Fractional Exponents (continued). Extraction of Roots of Quantities aifected with Entire or Fractional Exponents — Monomials 232 Promiscuous Examples 233 Calculus of Radicals 236 First Principle 238 Second Principle 240 Third Principle 241 Fourth Principle 242 Fifth Principle 244 Sixth Principal 247 SeTenth Principle 248 Eighth Principle 250 To make Surds rational by Multiplication 253 Case I. Monomial Surds 253 Corollary 254 II. To find a Multiplier that will make rational an Expression consisting of a Monomial Surd connected with Rational Terms, or consisting of two Monomial Surds 256 Corollary 261 III. To make rational an Expression containing three or more Terms of the Square Root 262 Corollary .". 264 Extraction of the Square Root of a Monomial Surd connected with a Rational Term, or of two Monomial Surds 265 First Principle 265 Second Principle 265 Third Principle 266 Imaginary Quantities 270 EQUATIONS OF THE SECOND DEGREE 276 Incomplete Equations 276 Properties of Incomplete Equations 284 Binomial Equations 286 General Problems in Binomial Equations 287 Complete Equations of the Second Degree 289 First Property. Every Complete Equation of the Second Degree has two Values, and but two, for the Unknown Quantity .., 295 CONTENTS. IX EQUATIONS OF THE SECOND DEGREE — Complete Equations of the Second Degi'ee (continued). Second Property. The First Member of every Equation of the Second Degree can be decomposed into two Factors of the First Degree with respect to x, the First Factor being the Algebraic Sum of x and the First Value of x with its Sign changed ; and the Second Factor being the Algebraic Sum of x and the Second Value with its Sign changed. The Second Member of the Equation after this Decomposition will be Zero 296 Third Property. The Algebraic Sum of the Values of every Com- plete Equation of the Second Degree is equal to the Coefficient of the First Power of the Unknown Quantity, with its Sign changed.. 298 Fourth Property. The Product of the Values in every Complete Equation of the Form, x^ -\- px t= q, is equal to the Second Mem- ber or Absolute Term, with its Sign changed 298 Fifth Property. The Value of x, in every Complete Equation of the Second Degree, is half the Coefficient of the First Power of X, plus or minus the Square Root of the Square of half this Co- efficient increased by the Absolute Term 299 General Examples 300 Discussion of Complete Equations of the Second Degree 304 Irrational, Imaginary, and Equal Values 307 Suppositions made upon the Constants 308 Explanation of Imaginary Values 309 Explanation of Negative Solutions 311 General Problems involving Complete and Incomplete Equations of the Second Degree 311 Trinomial Equations 326 Problem of the Lights 329 UNDETERMINED COEFFICIENTS 335 Failing Cases 341 Particular Cases 345 DERIVATION 347 Theorems 351 X CONTENTS. PAGE BINOMIAL FORMULA 356 Demonstration 357 Formation of Powers by the Rule 360 Demonstration of the Binomial Formula for any Exponent 364 Development of Binomials aflFected witli Negative and Fractional Ex- ponents 865 Consequences of the Binomial Formula — Square of any Polynomial... 367 Cube of any Polynomial 368 Extraction of the nth Root of Whole Numbers and Polynomials 369 Approximate Root of an Irrational Number to within a certain Vulgar Fraction 372 Approximate Root of Whole Numbere to within a certain Decimal 373 Approximate wth Root of a Mixed Number to within a certain De- cimal , 373 Approximate wth Root of Numbers entirely decimal 374 PERMUTATIONS AND COMBINATIONS 374 Combinations 377 Permutations in which Letters are repeated 385 Partial Permutation 386 General Examples in Permutations and Combinations 387 LOGARITHMS , 390 First Property. The Logarithm of the Product of any Number of Factors taken in the same System, is equal t-o the Sum of the Logarithms of those Factors 392 Second Property. The Logarithm of the Quotient arising from divi- ding one Quantity by another, is equal to the Logarithm of the Dividend minus the Logarithm of the Divisor 393 Third Property. The Logarithm of the Power of a Number is equal to the Exponent of the Power into the Logarithm of the Number.. 394 Fourth Property. The Logarithm of the Root of a Number is equal to the Logarithm of the Number divided by the Index of the Root. 395 Fifth Property. The Logarithm of the Reciprocal of a Number is equal to the Logarithm of the Number taken negatively 396 CONTENTS. Xl PAGE LOGARITHMS (continued). Sixth Property. The Logarithm of any Base, taken in its own Sys- tem, is Unity 397 Seventh Property. If we have a Table of Logarithms calculated to a Particular Base, the Logarithms of the Numbers in this Table, divided by the Logarithm of a Second Base, taken in the First System, will give, as Quotients, the Logarithms of the same Num- bers in the Second System 397 Differential of ax 398 Logarithmic Series 399 Common and other Logarithms found from Napierian 401 Measure of any Modulus 401 Table of Napierian Logarithms 401 Advantages of the Common System 404 Application of Common Logarithms 406 To find the Logarithm of a given Number below 10000 407 To find the Logarithm of a Number above 10-000 408 To 'find a Number corresponding to a given Logarithm 409 Solution of Exponential Equations by means of Logarithms 411 ARITHMETICAL PROGRESSION 412 Formula for the nth Term 413 Formula for the First Term 414 Formula for the Number of Terms , 415 Formula for the Sum of the Terms 41G To find the Arithmetical Mean 418 Insertion of Means between the Extremes of a Proportion 420 GEOMETRICAL PROGRESSION 422 Formula for the nth Term 423 Formula for the Ratio of the Progression 424 Formula for the Sum of the Series .' 425 An Infinite decreasing Progression 427 INEQUALITIES 431 Xll CONTENTS. FAOE GENERAL THEORY OF EQUATIONS 436 Equal Values 459 Derived Polynomials 462 Equation of Differences 467 Irrational Values 470 Newton's Method of Approximation 474 Lagrange's Method of Approximation 475 General Solution of Numerical Equations 477 STURM'S THEOREM 478 DESCARTES' RULE 491 ELIMINATION BETWEEN TWO EQUATIONS OF ANY DEGREE 495 ELEMENTS OF ALGEBRA Article 1. — A mathematical principle is a truth admitted as self- evident, or proved by a course of reasoning called a demonstration. Thus, it is a self-evident principle, that if equals be added to equals, the results will be equal. But it requires a demonstration to show that if the sura of two quantities be added to the difference of those quanti- ties, the result will be equal to twice the greater quantity. 2. Science is knowledge gained by theory or experiment, employed in the investigation of principles. Art is the application of acquired principles to the practical purposes of life. 3. Quantity is anything that can be measured or numbered. A thing is said to be measured when its magnitude is expressed in terms of the unit of measure, and is said to be numbered when this unit is unknown or indefinite. Thus, 12 bushels of corn is a definite measure, the unit of measure being one bushel of corn. But a quantity, expressed by the number 12, or even by 12 bushels, would only be numbered, for it might be 12 bushels of gold, or 12 of flour, or 12 of anything else. Lines are quantities, becaiise they can be measured in terms of yards, feet, inches, &c. Time is quantity, being measured by hours, minutes, seconds, &c. The unit of measure for time is generally the second. The operations of the mind, such as hope, fear, joy, grief, &c., are not quantities. For, although we speak of a great hope and a small hope, there is no definite unit of comparison by which to measure its magnitude. Numbers are not quantities, but simply the agents by which to express the abstract relations between quantities of the same kind; that is to say, every number may be regarded as the quotient arising 2 13 14 ELEMENTS OP ALGEBRA. from dividing one quantity by another of the same kind, independently of tlie species to which they belong. Thus, the number 12 may express the abstract relation between 12 inches and 1 inch, or between 12 months and 1 month; and may, there- fore, represent the quotient arising from the division of the foot by the inch, or the year by the month. And so it may express the quotient between any two quantities whatever, provided they are of the same kind. 4. Mathematics is the science of quantity. Arithmetic is that branch of mathematics in which the quantities considered are represented by numbers. 5. The word Algebra is of Arabic origin, and signifies to consolidate. Algebra enables us to investigate arithmetical principles in a consoli- dated, and, at the same time, general manner, and may be regarded as a compendious, and also universal arithmetic. 6. The quantities considered in Algebra are represented by numbers and letters, and the operations to be performed are indicated by signs. The numbers, letters, and signs are generally called symbols. The numbers and letters that represent quantities are, for conve- nience, most usually called quantities themselves. The student, how- ever, should remember that they are only the representatives of quantities. 7. The sign + is called plus, i. e., more, and when prefixed to a quantity, signifies that it is to be added to some other quantity expressed or understood. Thus, a + 6 is read a plus h, and signifies that h is to be added to a. The expression + c signifies that c is to be added to some quantity not expressed. When no sign is written before a quan- tity, the sign -f is always understood. In the expression a -f &, a is understood to be afiiected with the plus sign. 8. A horizontal line, thus, — is called minus, i. e. less, and when prefixed to a quantity, signifies that it is to be substracted from some other quantity, expressed or understood. Thus, a — 6 is read a minus h, and signifies that h is to be taken from a. The expression — c signifies that c is to be taken from some quantity not expressed. 9. A Greek cross x is called the sign of multiplication, and when placed between quantities, indicates that they are to be multiplied together. Thus, a X h is read a multiplied by h, or simply a into h, the sign indicating a multiplication to be performed. ELEMENTS OE ALaEBRA. 15 10 The multiplication of quantities is sometimes indicated by a point. Thus, a . h indicates that a and h are to be multiplied together. 11. The multiplication of literal factors is usually indicated by wri- ting them one after another, thu^ a i c is the same as a X 6 X c. This notation cannot be employed when numbers are used, for the product thus expressed would be confounded with some other number. The multiplication of 2 by 4, for instance, cannot be indicated by writing the one after the other, because the product would be mistaken for 42 or 24. 12. When several terms connected by the sign +, or — , arc to be multiplied by a single term, the multiplication is indicated by means of parentheses. Thus, (a -\- h -{■ c) m, signifies that the sum of a, b, and c is to be multiplied by m. "When the multiplier itself is composed of more than one term, it is also enclosed in parenthesis: Thus, (a + b) (m — c) indicates that the sum of a and b is to be multiplied by the diflFerence of m and c. A horizontal or vertical line is also used to col- lect terms for multiplication. Thus, a x to + « -f c, indicates that the sum of m, n, and c is to be multiplied by a. The same thing may be indicated by a vertical bar, thus 4-TO I + n + c 13. The coefficient of a quantity is a number or letter prefixed to a quantity, showing how often it has been added to itself. Thus, instead of writing a + a -r a, which denotes the addition of a three times, we abridge the notation by writing 3a. So also, 10.ry, signifies the addition of xj/ ten times. In like manner, mx signifies the addition of X, m times. The coefficient serves as a brief mode of indicating the addition of a quantity to itself. When no coefficient is written, 1 is always to be understood. 14. The exponent is a small number or letter written a little above and to the right of a number or letter, and indicates the number of times it enters into itself as a factor. Thus, we write a^ instead of aa, and call the result a square. a^ " " aaa, " " " " a cube. a* " " aaaa, " " " " a to the fourth power. The exponent enables us to abbreviate the manner of indicating the 16 ELEMENTS OF ALGEBRA. multiplication of a quantity by itself. When no exponent is wiitten, 1 is always to be understood. Division is denoted by three signs. The division of a by & may be indicated hj a-i-h, or — , or a\b. 15. The sign = is called the sign of equality, and is read " equal to." When placed between two quantities, it indicates that they are equal to each other. Thus, a = b, is read a equal to b. In like man- ner, 2 + 4 = 6 is read 2 plus 4 equal to 6. 16. The sign ^ is called the sign of inequality, and is read "greater than," when the opening is toward the left; and "less than," when opening is toward the right. Thus, a ^ Z> is read, a greater than b ; and c <^ m is read, c less than m. 17. A root of a quantity is a quantity which, multiplied by itself a certain number of times, will produce the given quantity. To indicate the extraction of a root, we use the sign V , called the radical sign, placing a number or letter to the left and over the sign to indicate what root is to be extracted. Thus, ^'a denotes the square root of a; %/~a the cube root of a ; V~a the i\>'^ root of a. 18. The number or letter placed over the radical sign is called the index of the radical. When no index is written, we always understand that the square root is to be extracted. Thus, s/ a means that the square root of a is to be taken. 19. A simple quantity is one in which the letters and numbers of which it is composed are not connected by the sign plus or minus. Thus, a, aZ>, — and c are simple quantities or monomials. All quan- tities not simple are compound, and called binomials when composed of two terms; trinomials when composed of three; and polynomials when composed of more than three. Each of the literal factors which enter into a term is called a dimension of the term. The degree of a term is the number of its dimensions. When a factor enters more than once, its dimension is denoted by the exponent. Thus, a^ is of the second dimension. 20. When several factors are multiplied together, the sum of the exponents denotes the dimension of the term. Thus, ab is of the second dimension, af-bc, of the fourth, aWcd, of the sixth, &c. If the term is a fraction, its degree is denoted by the difference between ELEMENTS OF ALGEBRA. 17 the sums of the exponents of the numerator and denominator. Thus is of the second degree, — ~ of the first degree, &c. 21. A monomial is said to be liomogcneoxis, ^hen all of its literal factors are of the same dimension. A polynomial is said to be homo- geneous, when all its terms arc of the same degree. Thus ah, a?h^, a!^L^, are each separately homogeneous monomials ; 4a*Z/ -f 2Z/V — m^, is a homogeneous polynomial. 22. Like quantities are those which are composed of the same letters, and which have their corresponding letters affected with the same ex- ponents ; SaP + lOah^ — 3a6^ are like quantities. But Sai- -f lOah — Sa^b'^ are unlike ; for, though the letters are like, the exponents of these letters are different. p. 17. 23. The reciprocal of a quantity is 1 divided by that quantity. The .1 1 2 13 reciprocal of 2 is -- ; of a, - ; of r ■ o o^ o> •-'*^'^- 2 a 3 2 2 3 24. Quantities, affected with the plus i?ign, are called iwsitive quanti- ties, and those affected with the minus sign, are called ncfjative quan- tities. The student, however, should bear in mind that quantities can- not be positive or negative in themselves, and that by these phrases, we wish merely to signify additive and subtractivc quantities. 25. To familiarize the student with the foregoing symbols, we sub- join a few examples fur practice : — 1. Express in algebraic language that twice the product of x into y, divided by t^iree times the product of z into to, shall be equal to G. 2. Express that the sum of a and b, when added to their difference is equal to twice the greater quantity a. 3. That one-third of a quantity ??i, multiplied by ith of a second quantity ?), and that product increased by 100, the result will be a hun- dred thousand. 4. Find the value of this expression, , when b = G-i:, f = 144, m.n m=10, n=l. Find the value of the expression (x a), when m = 4, ?i = 3, x = 2, a = l. Find the value of the expression —,-, when 7?i = 4, n = S, c:^=6, nrii h = 10, 1 = 100. 2* B 18 ADDITION. ADDITION. 26. Addition is the connecting of several terms together by the sign plus or minus, so that they nuiy be reduced into a single expres- sion. There are three distinct cases : — CASE I. 27. When the terms are like and have like signs. Add the coefficients of the several terms together, prefix the common sign to this sum, and write after it the common letter or letters, with their primitive exponents. Thus, + 2a + 3a are like quantities referred to the same unit, and may therefore be added, just as two whole numbers are added in Arith- metic. The -1- 2a indicates that a is to be added twice to some quan- tity not expressed, and + 3a that a is again to be added three times to the same quantity. This addition can obviously be indicated at once by writing + 5a instead of + 2a 4- 3a. Suppose, for example, that a represented one dollar, then 2a would represent two dollars, and 3a three dollars, and the sum, five dollars, would, of course, be represented by 5a. In like manner, — 36 — 46 is equal to — 76, because the minus signs indicate that the quantities represented by 36 and 46 are to be taken from some quantity not expressed, and subtracting 76 at once is obviously the same as first subtracting 36 and then subtracting 46. We may then write — 36 — 46 = — 76, the minus sign before the 76 indi- cating that it is to be taken from some quantity not expressed, and the whole expression denoting that the subtraction of 76 is the same as the successive subtraction of 36 and 46. This will be made plainer to the beginner by attributing to 6 some known value, as a pound or an ounce. EXAMPLES. -f a — a + &— y + m — z — 4,-2 + 2f + 2a — 2a + 26- 2^ + 5„i_ 2s — 5z' + 5/ -f- 3a — 3a -f 56 — 4y + 8??!,— 3z — 9,r + 7f -1- 4a — 4a + 96— 8y + 10m— 7z — 12,-2 + sy +"io^" ~10^' + 176—15^ + 2im — ldz — W^ + 22/ ADDITION. 19 CASE II. 28, When the quantities are like, but affected with unlike signs. RULE. Add the quantities affected tcith the positive siyn hy the last ride, then add those affected with the negative sign in like manner. Subtract the smaller of the coefficients of those sums from the greater, annex the common letter or letters to the difference, and prefix the sign of the greater sum. If we were required to add -{- ba — 60 + 4a — a together, we could, for the reasons already given, write + 9a for the positive terms, and 7a for the negative terms. By this, we must understand that 9a has to be added to some quantity not expressed, and that 7a has to be subtracted from the sum of 9a, and this unexpressed quantity. Now to take 7a from 9a, and add the remainder to the unexpressed quantity is plainly the same as taking 7a from the sum of 9a, and the unexpressed quan- tity. But the difference between 9a and 7a is 2a, hence the sum of -f 5a — 6a + 4a — a is + 2a. The plus 2a denotes that 2a has to be added to an unexpressed quantity. If we were required to add — 5a + Ga — 4a -f- a together, we could, as has been shown, collect the negative terms into a single term, — 9o, and the positive terms into a single term -|-7a. Wc would then be required to take 9a from the sum of 7a and some unexpressed quantity, or, which would be the same thing, to take 9a from 7a, and add the remainder to the unexpressed quantity. But 9a is made up of 7a added to 2a, and when, therefore, we subtract 9a from 7a, the 7a's will strike each other out, and there will still remain 2a to be sub- tracted. The required and unexecuted subtraction is indicated by the minus sign before the 2a, and the true sum of — 5a -f 6a — 4a -f a is therefore — 2a. EXAMPLES. 1. Add 2 1 — Gh + ah — 27j together. Ans. +ah — 6i or (+ a — 6) J. 2. Add 4c -{- 3c — mc -{- no together. Ans. (7 -\- 71 — w) c. 3. Add xi/ -{- 2xy + hxy — pxy together. Ans. (3 -f & — p) xy. 20 ADDITION. 4. Add 4a — 7a — 3a — 10a — 12a together. Ans. — 28a. 5. Add 4ax — 7ax — dax — Wax — 12acc together. A71S. — 2Sax. CASE III. 29. When the quantities arc similar and dissimilar, and have like and unlike signs. RULE. Add all (he sets of similar terms hy the. last two rules, and write their sums one after another, and connect with them all the single terms with their oion signs. Since the letters are always the representatives of quantities, it is obvious that the quantities represented by dissimilar terms cannot be added into one sum. Thus, 4a + 46 neither make 8a nor 86. The quantity a might represent a year, and the quantity h a pound ; then 4a would represent 4 years, and Ah would represent 4 pounds, and the addition ought neither to give 8 years nor 8 pounds. Hence, we can only reduce the similar terms in sets, and connect the results with their appropriate signs. 3jn^ — n^ xy + Ix bax + y 2nx+ w 3x^ + ni' z -\-y x^ -\- Ay 2w -\- bnx 2m^ -\- x^ xy +ox Ax} + % 6».x4-s Qm^ + Ax^ — n"" 2xy + Wx+z^y bx^-\-bax^lAy V6nx + ^io-\-s 30. It often facilitates an algebraic operation to arrange the terms in a certain order. In addition, this arrangement is effected by placing all the like terms beneath one another. Ta^ 4 Qxy + 5^ + Aw la" + Aw + Qxy + bz w -\-Az + 8a-y+ 12a^ may be written 12a- + ?r + 'ixy + Az Sz + 9xy+ bw + 100a- 100a" + bw + 9xy + Sz llda'+10w + 2nxy + Viz 31. Sometimes the addition of a compound quantity to a single quan- tity, or to another compound quantity, is not actually performed, but indicated by the parenthesis (), or vinculum. Thus, 46 + (6 — c), or 4t ^ Ij — c, indicate that h — c is to be added to Ah. SUBTRACTION. 21 Remarks. 32. It will be noticed tliat the term addition in Algebra, is used in an extended sense, the operation being often arithmetical substraction. The addition of a negative quantity is, in fact, the same as the subtrac- tion of the same quantity regarded as positive. The use of negative quantities in Algebra constitutes one of its marked drfiferences from Arithmetic, in which the numbers employed are always supposed to be positive. By reference to the examples, another remarkable distinction between Algebra and Arithmetic will be observed ; each sum indicates Avhat (|uantities were added together, whilst an arithmetical sura con- tains no trace of the numbers employed. Thus, the sum of 10 and 5 is 15, but the result does not point out the numbers that were added, for 15 might proceed from the addition of 14 and 1, 12 and o, 13 and 2, &c. It will be seen hereafter, that in all algebraical operations, the result contains some, if not all, the quantities employed in the inves- tij^atious. SUBTRACTION. 33. Subtraction is taking the dlfR'rence between two or more quantities, and may be regarded as the undoing of a previous addition. If we were required to subtract + G — 4, or 2 from 12, the result plainly ought to be 10. But if we write G — 4 beneath the 12, and perform the subtraction of each term separately, the subtraction of 6 from 12 will give a remainder G, which is too small by 4, because we were not re^juired to take 6, but 2 from 12. We can only correct the error by adding + 4, and, hence, we have 12 — ( + G — 4) = 12 — G 4- 4 = 10, as before. We observe, in the last result, that the signs of G and 4 have both been changed. Again, let it be required to subtract + i — c from a. The sub- traction of h from a will be indicated by writing a — h, but tliis re- mainder is too small by c, because we were not required to subtract h itself from a, but what remained of h after it was reduced by c. The error can only be corrected by adding plus -: to the result. Hence, a — (+Z> — c)= a — 6 + c. SUBTRACTION, In this general example, we see that all the signs of the subtrahend have been changed, and then, it has been added, as in addition. Hence, we have the following RULE. Conceive the signs of all the terms of the subtrahend to he changed, and then add up these terms as in addition. EXAMPLES. — 6a + 46 Sy/~X + 5a2 7x^ -r m 2 V a + ax + 4a— i Q^'x— Sa^ — oVx + loa- 7x' + 2m — m 4s/of + Qax — 10a + 5b — 2Va — bax From y/x + b — 10 +Sa^+7Qi/ +18a+ 19x + lira Take d —20 + dx + bn—2^x+ 36+ 4aH.:; + 2s 3v/x- — 26 + 10* +4a^ + lSa+Qn—d—z—2s +7Qi/ + 10 In the third example we have written zei'o as the difference between 7x^ and 7x^, but it is more usual to denote zero by a blank. From iixt/ + 14 v/y + s + 2n — w Take Sxi/ + 2 — 11 + 11 v/y— 2n + w 11 + SV7+ 4n — 2m; From 12a^ + Ux — cx^ + m^ + 46- — c^x^ + 71s Take 46^ — cV — Qbx + ex'' — m!" — 3«s + 106^ — io>/ + 12a- 106x — 2cx^ + 2m^ + 4ns — 10b- + wi/ " 34. The remainder added to the subtrahend ought to be equal to the minuend, and the result, therefore, can be verified, as in Arithmetic. It will be seen that an Algebraic subtraction does not necessarily im- ply diminution, and that the remainder may not only exceed the sub- trahend and minuend separately, but may be equal to their arithmetical sum. In general, the subtraction of a negative quantity is equivalent to the addition of the same quantity taken positively. Thus, — 6 taken from a gives a + 6 for a remainder. This subject may be illustrated by a simple example : Suppose a to represent the value of an estate exclusive of its liabilities, and 6 a mortgage upon that estate, then a — b SUBTRACTION. 23 will represent the actual value of the estate. Now, suppose some friend of the owner should determine to cancel his debt ; he could do this, either by removing the mortgage (in algebraical language, taking away — h), or by giving him a sum of money equal to the debt, that is, a sum represented by + h. So we see that, taking a-^vay — 6 is equi- valent to adding -f- h. 35. Quantities arc considered negative, when opposed in character or direction to other quantities of the same kind, that are assumed to be positive. Thus, if a ship leave port with the intention of sailing due north, but encounters adverse winds, and is driven south part of the time; to get the distance passed over .north, we must obviously subtract the distance sailed over south. If then the direction north be considered positive, the direction south must be considered negative. Suppose the vessel sails first day 100 miles north to 10 south, second day 80 miles north and 30 south, the entire distance passed over in a northern direction will be expressed by 100 + 80 — 10 — 30 = 140. If a man agree to labor for a dollar per day, and to forfeit half a dollar for every idle day, and he labor 4 days and is idle 2, his wages will be 4x 1 — 2x J, or4x l + 2( — *) = 3. We see that we have re- garded as negative, cither the forfeiture as opposed to gain, or the idle days as opposed to the working days. If a man's age be now thirty j-ears, his age four years hence will be expressed by 30 + 4 ; his age four years ago by 30 — 4. Here future time being positive, past time is negative. If a steamboat sail with a velocity of 10 miles per hour, and encounter a head wind that would carry it back at the rate of 8 miles per hour, then its rate of advance will be expressed by 10 — 8, or 2 miles per hour. But if it be carried back at the rate of 12 miles per hour, then its rate of advance will be expressed by 10 — 12, or — 2 miles per hour. It will then plainly be carried back, and the minus indicates a change of direction. Distance, when estimated as positive in one direction, is considered nega- tive in the opposite direction. Thus, let the distance AB = n and BC = w, C '- i , the distance BC being considered positive on the A B ' ' ° ^ right of B. Then AC ^ wi + n. Now suppose the distance BC be esti- mated on the left of B, the point C falling between B and A, then AC = m — n. We see that the distance BC, which was regarded as positive on the right of B, became negative when estimated on the left. 24 MULTIPLICATION. In general, the minus sign may be regarded as always indicating either that a quantity has changed its character or direction, or that it is just the opposite of something else of the same kind that is con- sidered positive. MULTIPLICATION. 36. Multiplication is a short method of adding the multiplicand to itself as many times as there are units in the multiplier. Thus, to multiply a by Z) is to add a to itself h times, and since the addition of a to itself twice is 2a, three times 3a, &c., the addition of a to itself h times will be ha. Hence the product of a by & will be la or ah, for it plainly matters not in what order we write the factors. There are three cases : — CASE I. 37. When both multiplicand and multiplier are simple quantities. Before giving a rule for the multiplication, it will be necessary to show that the product of simple quantities having like signs, both plus or both minus, is always positive, and that the product is always nega- tive when they have unlike signs. The product of + a by -{-h is plainly + ah, because plus a added to itself h times m.ust, of course, be positive. And — a by -\- h must be negative, because — a added to itself any number of times must, of course, retain its sign. But — a by — h will give + ah, for the h having a minus sign before it, indicates the reverse of what it did before, and therefore denotes that — a is to be subtracted from itself h times. But we have seen that the subtraction of — a is the same as the addition of H- a ; in like manner, the subtrac- tion of — a twice is the same as the addition of + 2o ; the subtraction of — a tliroe times the same as the addition of + 3a ; and the subtraction of — a, h times, the same as the addition of + a, + h times, or + ah. Hence, the product — o by — & is + ah, and we see that like signs produce plus, and unlike signs, minus. Let it now be required to multiply a^ by a". The exponents indicate that a enters twice as a factor in a^, and four times as a factor in a, hence, it will enter 6 times as a factor in the result. MULTIPLICATION. 25 Hence, a^ x a'' := aaaaaa = a^, and we see that the multiplication is effected by adding the exponents of the same letter. To multiply a^ by ab is the same as multiplying a^ by a, and then multiplying the result by h. But to multiply a^ by a, we have only to add the exponents of a^ and a, hence the result is a'. Now multiply by b, and we have a'^b. The multiplication is effected by adding the exponents of like letters, and writing after the result the letter which is not common to the mul- tiplicand and multiplier, with its primitive exponent. In like manner a^ X ah^ will be a^b^. To multiply a\- by ab, is the same as multiplying a^c by a, and then the result by b. But a'^c by a, as we have seen, gives a^c, and that product by b, will plainly be a\b. And the multiplica- tion is again effected by adding the exponents of like letters, and writing after the result the letters, which are not common, affected with their primitive exponents. The same reasoning can be extended to any num- ber of factors, the multiplication being effected in every instance, by adding the exponents of like letters, and writing after the result, the letters not common, with their primitive exponents. Thus, a^cd by ab gives aW6, and ah-d by abm gives a\dbni. We have taken mono- mials whose coefficients were unity. If we were required to multiply a by 2b, the product of a by 6 has to be nuiltiplied by 2, but the pro- duct a by 6 is ab, hence the result is 2ab. If we were required to multiply 2a by b, then 2a has to be added to itself as many times as there are units in b, but to add 2a to itself twice gives 2a x 2 or 4a, to add it three times gives 2a X 3, or Ga, and to add it b times gives 2a X b, or 2ab. In like manner, if we were required to multiply 2rt by 4i, then 2a has to be added to itself -ib times, and the result will plainly be 2a x 46, or Sab : the same result that we would get by multiplying the literal factors together, and prefixing the product of the coefficient for a new coefficient. Hence we have for the multipli- cation of monomials, the following RULE. 38. Multqjlr/ tJie coefficients to. 12. 4«-'"i"c by — Sa^i-'c-^ ^72s. — 32a''^"-='c-3. 13. 4a^6V by — Sa^i^C Jws.— 32a^Z;V. 14. 6a-='i-''c--3 by 7a-^i-V-''. Ans. 42a-^i-^c-l 15. 3a«Z;V by 9a-='i-'^c-^ ^rjs. 27a^Z-V. 16. a^^V by Axyzh. Ans. Ax^ Y+'zf+'h. 17. a^^^-rc by ac. A71S. + x^yzac^. 18. x^fc by xy. Ans. x^ yc. 19. rr^/c by a-7/^ Ans. x^'^\ifc^ 20. x^-\jc by:.y. Ans. x^^Yc 21. rr-'-yc-^ by cV. Ans. x^^yc. Remnrlcs. 39. We see, from the last four examples, that we can make any change in the position of the exponents, provided we retain the factors, and get the same result. We see, from examples 13, 14, and 15, that when the factors of the multiplicand and multiplier are homogeneous, the factors of the result will also be homogeneous. Example 11 shows that this will not be true when the multiplicand is alone made up of homogeneous factors ; and Example 1 shows that it will not be true when the multiplier alone is homogeneous. 40. Any number of monomials may be multiplied together, in accord- ance with the preceding principles. MULTIPLICATIOy 27 EXAMPLES. 1. o\" X abX hdXmn. Ans. a?l?dcmn. 2. — ah X — ah X Id X nui. Ans. a^U^dcmn. 3. — a'c X — ah X — hd X mn. Ans. — a%\Jcmn. 4. — a-c X — ah X hd X — mn. Ans. — a^h^dcmn. We see, that when the monomials are all positive, the result will be positive ; and when the negative monomials are even in number, the result will also be positive. But when an odd number of negative mo- nomials are multiplied together, as in the fourth Example, or when an odd number of negative monomials are multiplied by a positive mono- mial, as in Example 3, the result will be negative. The result would also be negative, if an odd number of negative monomial factors was multiplied into any number of positive monomial factors. CASE II. 41. When tlae multiplicand is a compound quantity and the multi- plier a monomial. The multiplicand is to be repeated as many times as there are units in the multiplier; each term of the multiplicand is then to be multiplied by the multiplier, and the partial products to be connected with their appropriate signs. We might a.ssumc, what has already been proved for monomials, that the product of a positive quantity by a positive, or of a negative quantity by a negative, gives a positive result; and that the product of a positive quantity by a negative gives a nega- tive i-esult. But we can demonstrate this rule more rigorously : Let it be required to multiply a — a hy + c. We know that a — a is zero, and that zero repeated c times must still be zero. Hence, the product of a — a by + c must be zero. But + ahy + c gives -|- ac for a product, and, therefore, the product of — a by -|- c must be — ac, in order to cancel the first product. Hence, the product of a negative quantity by a positive gives a negative result. Or, to express the whole algebraically, a — a = + c ■j-ac — ac = o 28 M U Ti T I P L I C A T I N . Let it now be re((uired to multiply a — a by — c. The multipli- cand being zero, tlie product must be zero. But, from wliat has just been shown, the product of + a and — c is — ac, hence, the product of — a by — c must be + ac, to destroy the first product. Or, in other words, the product of a negative quantity by a negative must be effected with the positive sign. Algebraically, a — a = o — c We conclude that the product of a positive quantity by a positive, and of a negative quantity by a negative, is positive ; and that the product of a positive quantity by a negative is negative. Briefly, we say like signs give a positive result, and unlike, a negative result. RULE. Multiply each term of the vudtiplicaud hy tlie midliplicr, and con- nect the res^dts ivith their appropriate signs. EXAMPLES. a^ -{-h by a. Ans. a"" -f ah. a^ — h by a. A71S. a^~ah. — a^ -\-h by a. Ans. — a^ + ah. — d' — h by — a. Ans. «' -f o&. — fr -f Z» by — a. Ans. a^' — ah. a^t + c -f d by a. Ans. a^h -f ac + ad. a^h — c 4- d^ by a. Ans. a^h — ac -f ad. a% -{- c — d by a. Ans. a^h + ac — ad. — d'h-^c + d by a. Ans. — a^'h -f ac + ad. — a'^h — c -\- d by a. Ans. — a^h — ac -f ad. — a^h — c — d by a. Ans. — a^h — ac — ad. — d'h — c — d by — a. Ans. a^h -\- ac -\- ad. — a^h — c-\-d by — a. Ans. a^h -\- ac — ad. — a^b-\-c-\-d by — a. Ans. a^h — ac — ad. a^b + c-j- d by — a. Ans. — aNj — ac — • ad. m^n + a -f 2c by s. Ans. m"ns -f- as -j- 2c,9. m^n + a + 2c by 3c Ans. Scm'n -f Sac -f Gc^ m^n -f a + 2c by 4wn. Ans. 4??i^;i'^ + 4amn -f Smnc. 8m — 6a + llcby 2c;= Ans. 16mc'' — 12ac='-f 220". MULTIPLICATION. 29 CASE III. 42. When tlie multiplicand and multiplier are both compound quan- tities. The multiplicand, as in the two preceding cases, is to be repeated as many times as there are units in the multiplier. The multiplicand and multiplier will, in genei*al, be made up of some positive and some ncg:'- tive terms. Let a denote the sum of all the positive terms in the mul- tiplicand, and b the sum of all the negative terms. Let c denote the sum of the positive terms in the multiplier, and d the sum of all th(> negative terms. We write the multiplier beneath the multiplicand and multiply all the terms of the one by all the terms of the other. Thus, a c — b — d ac — be — ad + bd ac — be — ad -f- bd. To explain this result, let us drop fur a moment the consideration of — d, then a — b must be repeated e times. From what has been shown, the result of this multiplication will be ac — be. But we were not required to multiply a — i by c alone, but by c after it had been diminished by d ; hence, the result, ac — be, is too great by a — b taken d times. To correct the error, then, we must subtract the pro- duet of a — bhj d from the first product ac — be. The product of a — b by -f d, from what has been shown, will be + ad — bd, and to indicate that this must be subtracted, we write it in parenthesis with the minus sign before it. The whole result will be oc — be — (-f ad — bd) = ac — be — ad + bd, since the signs of the quantities sub- tracted must be changed. By examining the result, we will observe, as before, that the product of quantities affected with like signs is positive, and that of quantities affected with unlike signs is minus. 43. It is found most convenient to arrange both polynomials with reference to the highest or lowest exponent of the same letter. Thus, x^ -{- x^ + X -\- a is arranged with reference to the highest exponent of X ; and a + x + x^ + x^ is arranged with reference to the lowest exponent of the same letter. In these expressions, x is supposed to enter to the zero power in the term a, as will be explained under the head of Division. 3* MULTIPLICATION. RULE. 44. Arrange the miiUipUcand and multiplier tvith reference to the highest or loicest exponent of the same letter (if tliey have a common letter'), and then multiply each te^m of the one polynomial by each term of the other polynomial, leg inning on the left. Set down the result of the midtiplication of the second term of the multiplier under that of the first term, only removed one place further to the right, and the result of the multiplication of the third term of the multiplier %inder that of the second, and contimie the operation until the multiplication is com- plete. Then reduce the lohole result to the simplest form. 1. X + a X? + ax — ax — a? ■:i? + — d'- =■ x^ — a^. . X^ + O.T^ X x^ -\- ax -\- \ xb + ax'^ — x^ + ax'^ + ax^ — ax^ + a;' + ax^ — x cc5 + 2ax'^ + ax^ + o — x 2. x^ -{- xy -\- a X +y x^ + x'y + ax -\- x^y + ar/+ ay X* + Ix^y + ax + xy^ + ay. a? + 2xy + f X — y x' + 2x^y + y'^x — x^y — 2y'^x — y^ *' + x'y — y'^x — / 45. It will be observed in the above examples, that by arranging with reference to a certain letter, and by removing each result one place further to the right than the preceding, the terms which reduce fall immediately the one below the other. And it is to save the trouble of looking for the terms which reduce, that this arrangement and sys- tem are made. It will also be noticed that in every product there are two terms which do not reduce with other terms, viz., those which result from the multiplication of the quantities affected with the highest and lowest exponents of the arranged letter. In the first example, x^ and — a^ are the irreducible terms, in the second, x? and ay. MULTIPLICATIOX. 31 EXAMPLES. 6. Multiply X -\- 7/hj X -{- 1/. Ans. x' + 2xi/+7/\ 7. Multiply X + y by cc — i/. Ans. 7? — 1)^. 8. Multiply x^ + Ixy + ^^^ by + cc + y. .4ns. x^ + Zx^ij + ^n'^x + y^. 9. Multiply X' — 2xy -\- y^\>^ x + y. sins. X? — x-y — xy'^ -\- y^. 10. Multiplyx^ — T/^byx^+y. Ans. x^ + y^x" — x'y' — /. 11. Multiply a- + lax + a;^ by x -^r a. Ans. a;' + Sux'^ + 3alt+ a'. 12. Multiply .t' + a^" + 3aa; + 3«.7;2 by x + a. .'1/)S. x" + 4ax' + 6al<;2 + \c{'x + a^ 13. Multiply 7x-^ + 2« by %y'-- Ans. 2.\oh/ + 6^7/^. 14. Multiply 4a;* + y^ by 4.c* — y*. .4«s. 16x^— /. 15. Multiply x^ H- o.c^ + a^x + a' by 2r/ + 2.f. ^ns. 2a;'' + hix^ + 4(/^.r + 4a='a: + 2^". 16. Multiply x^ + xy + / by 1x + 7y. ^«s. 7.<' + 14a-y + 14a-y^ 4- ly^. 17. Multiply a;' + ax' + a^x^ + a'x + a< by 4a; + 4a. ylns. 4x* + Sax* + Sa^x' + 8aV + Sa^x + 4al 18. Multiply 2x' + 2ax + 2a' by 3x + 3y. Ans. ^x^ + Gax' + Gx^y + Ga'x + G«xy + Ga^- 19. Multiply 2a2 — 2^^ by 2a + 26. .4«s. 4a3 + 4a^6 — \aV — \h\ 20. Multiply 2x + 3a + 46 by y + z. Ans. 2xy + Zay + 46y + 2x3 + 3a2 + Ahz. 21. Multiply 2x + 3a + 46 by 2x — 2y. Ans. 4x' — 4.ry + 6ax + 86x — Qay — 8iy. 22. Multiply 4/ + 4x^ + 4m?i by 2x + 2y + 2z. Ans. 8y=' + Syx' + Sywin + 8x' + Sxy' + ^xmn + Sy''^ + Sx's + 8m?tz. MULTIPLICATION. RemarJis. 46. It -will be seen, by inspecting tlie above results, that when the two polynomials are both homogeneous, the product is also homoge- neous, and of a degree equal to the sum of the dimensions of the mul- tiplicand and multiplier. Thus, in Example 7th, the first polynomial is homogeneous, and of the first degree ; and the second is also homo- geneous of the first degree, the product is homogeneous of the second degree. In Example 8th, the multiplier is homogeneous of the second degree, and the multiplier is homogeneous of the first degree; the product is homogeneous of the third degree. The same may be noticed in several other examples. 2d. We notice also that when the coefl&cients are the same in each term of the multiplicand, and the same in each term of the multiplier, and all the terms of both polynomials are positive, that the sum of the coefiicients in the product will be equal to the product arising from mul- tiplying the sum of the coefiScients in one polynomial by the sum of the coefiicients in the other. Thus, in Example 16, each coefficient of the multiplicand is 1. and each coefficient of the multiplier 7. The sum of •the coefficients of the multiplicand is 3, and that of the multiplier 14 : the sum of the coefficients in the product is 3 X 14 = 42. The same law may be noticed in Examples 15, 18, and 22. 47. Examples 10 and 19 show. that when the multiplicand is com- posed of the diff"erence of two terms, whose coefficients are equal, the algebraic sum of the coefficients in the product is zero. Examples 7 and 21 show that this sum is also zero when the multiplier is composed of two terms with contrary signs and equal coefficients. 3d. It has already been remarked that there are always two terms, which do not reduce with any other terms. We can only reduce similar terms, and when the two polynomials have been arranged with respect to a certain letter, the products of the extreme terms are dissimilar to the other partial products. The whole process of division of polynomials is based upon this fact, and it ought to be remembered. By attending to the above laws in regard to the product, we can often by a simple inspection detect errors in the multiplication. There are three theorems of great importance, which must be com- mitted to memory. MULTIPLICATION. 33 Theorem I. 48. The square of the sum of two quantities is equal to the square of the first, plus the double product of the first by the second, plus the square of the second. Let X denote the firi/ + 2ay + //. Required the square of 2^ -f x + «t + n. Let 2}/ -{- x = z, and m + '* = s. Then (2y + x + m + nf = (z + sf = z' + 2zs + s' = (2y + xf + 2 (2>/ + X) (m + n) + (m + nf == 4/ + 4yx + x' + 4>/m + iyn -f 2xm + 2.rn + ni^ + 2»i/( + n\ Required the square of oy + 4x + m + 2n-f 5. Represent the first three terms by a single letter, and the last two also by a single letter, and proceed as before. TlIEOHEM II. 49. The square of the diiTorcnce of two quantities is equal to the square of the first, minus the double product of the first by the second, plus the square of the second. Let X and a denote the quantities, then x — o = difference. And (x — a)^ = (x — a) (x — a) = x^ — 2ax + a^, as enunciated, So(10 — 5)^ = 10^— 20x 5+ (5)2=^100 — 100 + 25 = 25. And(40 — 6)^ = (40)^ — 80 X 6 + 6^ = 1600—480 + 36 = 1156. 34 . MULTIPLICATION. In like manner (a;" — a"")^ = x^'" — 2x"'(r + a^'". llequircd tlie square of 2x — a. Ans. 4x'^ — 4ax + a^. Required the square of 3a; + h — a. Ans. (3x + hy — 2 (Sx + h)a+ a^ = dx^ + Q>hx + 1/— Qax — 2ah + aK llequircd square of a;^ — a^. Ans. x^ — 2x^a? + a^. Theorem III. 50. The sum of two quantities multiplied by the difference of the same quantities, is equal to the difference of their squares. Let X and a denote their quantities. Then x -{• a= sum, and x — a = difference, and (.« + o) {x — a) == x^ — a^, by performing the multiplication indicated. So (10 + 5) (10 — 5) = (10/ — (5)^ = 100 — 25 == 75. And (40 + 6) (40 — 6) = (40)'' — (6)* = 1600 — 36 = 1564. Multiply (4« + 6) by (4a — 6). Ans. 16a^ — 6^ Multiply la + /j + f by 7a + 6 — c. Ans. (7a + ly — c' = 49a2 + Uab + J/ — c\ Multiply x -\- y + z + m\)j .K -\- y + z — m. Ans. (,x +1J + zj — m'' = x' + f+ z" + 2xy + 2xz + 2ijz — m". Multiply X -\- y ^r z -{- m\)^ X ■\- y — z — m. Ans. {x + yf — {z + mj- = a-^ -f 2xy -\- y'^ — ^^ — 2mz — m'- ■V Remarks. 51. If we omit the exponents of the extreme terms in the expres- sion, x^ + 2«x + a^, and connect these terms with the exponents so omitted, by the sign of the middle term 2oa-, we have a; + a, the bino- mial, which squared gave x? -f 2aa; + a^- In like manner, if we omit the exponents of x^ and o^ in the expression, a? — 2aa; -f o?, and con- nect them with their exponents omitted by the sign of the middle term, we have x — a, the binomial, which squared gave a;** — 2ax + a;l In like manner, if we have given the difference of two squares, we can readily determine the quantities which, by being multiplied together, gave this difference. Thus n^ — ?j^ = (m + n) (m — ?i), and r- — .s^-:= (r + s) (r — s). Omit the exponents, connect the terms by the sign plus, and we have the sum of the two quantities ; omit and connect by the .sign minus, and FACTORING POLYNOMIALS. WO have the difference. And the product of the sum by the difference, is the difference of the squares. These remai-ks are preliminary to an important subject. FACTORING POLYXOMIALS. 52. It is often a matter of great importance to resolve a polynomial into its factors. The reduction of expressions can often be effected in this way, and in no other. Practice alone can make the student expert in the resolution of an algebraic expression into its factors. A few rules may, however, assist the beginner. 53. 1st. Look out for the terms which have a commoa factor, and write them- in parenthesis, as a multiplier of that factor j next look for another set of terms also having a common factor, and write them in like manner; proceed thus until all the terms are taken up. Observe whether the parenthetical expressions are the same, if so, multiply the common parenthesis into the ahjchraic sum of the terms to which it serves as a coefficient. 1. hx + ha->r CX + ca = h (./- + a) + c (.c + o) = (x + a) {h + c). 2. hx + ha — cx — ca = h{x -\- a) + (x^ a) (— c) = (x + a) {h -c). 3. — hx — ha -f ex + ca = (— h) (.r + a) + (.c + a) c = (x + a) (r 4. hx + ex -\- ha + dx + da -f ca = (.c -{. a)h -{■ (x -\- a) c-\- (^x-'r a) d = (x+a) (h+c-ird). 5. hx-\- ex -\- ha — dx — d(t -f- ca = (x + o) (h -\- c — d). 6. a^ +ax + hx-{- ah = {x + a) x +(x- -{-a')h = (x+ a) (x + 6). 7. x^ — ax -{■ hx — ah = (x — «) (x + />). 8. x^ — ax + hx — ah + x^ — ax' = (x — a)(x+h + x"). 9. x^ — ax + hx — ah -f mx^ — ajnx^ = (x — o) (x + h + mx^). 10. x^ — ax + ix — cd) + amx^ — mx^ = (x — a)(x -\- h — mx^). 54. 2d. After having found a common factor or parenthesis, see whether that factor may not admit of farther reduction. (Articles 48, 49, and 50.) 3G F A C T U 11 1 N (i P U L Y N tJ M I A J. S , EXAMPLES. 1. W + Ihn + 6 + cn^ + 2c??. + c = h (n^ + 2w + 1) + c (u' + 2« + !) = (]> + (■) («2 + 2u + 1) = (?, + c) (n + 1)1 (Article 48.) 2. bn'' — 2/';h + Z> + C7i' — 2ai + c = {b + c) (« — 1)''. (Art. 49). 3. am" + hm^ — an' — hit^ = (a + h) (vi' —ir) = (a + b) (m + 7i)(^m — n). (xVrticle 50). 4. am^ — bm^ — an^ + bn^ = (a — b') (vi + n) (m — «). 5. «=> + 2n' + 7i = n (n + \)\ 6. n^ -f 2;i^ = n^ (ji + 2). Admits of no lower reduction. 7. «='— 2n:2 + 7i = n (n — If. 8. ii^ — 2n^ + n + 07111^ — 2mn + m = (n + m) {ii — 1)1 9. n^ — 271^ -\- n — wi/t^ + 2mn — m = (ii — vi) (n — 1)^. 10. — n^ + 2?t^ — n + mii' — 2mn + m = (m — n) (n — 1)1 11. _ „3 + 2»2— ?i — m/i" + 2??j?i — wi = — (/I + m) (?i — 1)1 12. m^ — n' — 2c7i — c^ = m^ — (h + cf. We have now the differences of two squares, and to apply the formula x" — a^ = (x + a) (x- — a). (Art. 50). We see that m^ = .r^, or on = .r, and (n + cf = ry^, or w + c = a. Hence (x -\- a) = {m + ?i + c), and (.r — ff) = hn — n — c). Therefore m^ — (?i + cf = (vu + n + c) (?n — n — c). 13. m^ — 11^ + 2cn. — c^ = Hi^ — (n — cf = (m + n — c) (in + c — ??). In this Example m = x, and n — c = a. 14. m- + 2Z;??i + b^— n"" — 2cn ^ n" = (m + bf — (n + cf = (m + b + n -\- c) X (m + b — n — c). 55. 3d. Expressions may sometimes be thrown into factors by an artifice, when they do not at first glance appear to admit of resolution into factors. One of the simplest contrivances to effect a decomposition into factors is the addition to, and subtraction of the same quantity from the given expression, which operation docs not, of course, alter its value. EXAMPLES. 1. x^^2x — l'i = x- — 2x + 1 — 15— l = a;^ — 2a: -r 1 — 16 = (.r — 1)^ — lG = (a' — 1)^ — (4)=^^(x — 1 + 4) (a;— 1 — 4). Art. 50. ={x + o) (x. — b). FACTORING POLYNOMIALS. 37 In this example, the decomposition was effected by adding + 1, and subtracting the same from the given expression. The first three terms of tlie equivalent expression were thus made a perfect square = (x — 1^), and the whole was made the difference between two squares. 2. x" + -ix — 12 = x' + -ix + 4 — 12 — ,4 = (x + 2)^ — (4)^ = (x + 0) (X — 2). 3. x' + 2x — S =x' + 2x + l — S — l= (x + lf — (Sy = (., + 4) (x — 2). 4. x' + 4x — 21 = x' + 4x + 4 — 21 — 4 = (.f + 2)^ — (5/ = (x + 7) (x — n). 5. x^ + 8x — 48 = x' + Sx + IG — 48 — IG = (x + 4/ — (8)^ = (x + 12)(x — 4). 6. x' ^ 10.r + 24 = x" 4- 10a: + 25 + 24 — 2-5 = (x + 5)^ — (1/ = (x -h 6) (X + 4). 7. ;y2 _u 8x + 12 = X' + 8x +1G -f 12 — IG = (x + 4)= — (2/ = (x + G) (X + 2). 8. x^ + 20x + 84 = x^ + 20x + 100 + 84 — 100 = (x + 10/ — (4y = (x + 14) (re + G). 9. x' 4- 12x + 2 = x' + 12x + 30 + 2 — 3G = (x + G)^ — V (34/ = (x + G + V34) (x + 6— v/34). ' 10. x' + 3x + 5= x' + 3x 4- » + 5— 9 = C.r 4. |)2 _ ^"(=^17/ = (-*^ + 1 + V— y) (-r + 5 — v/— '.,')• 11. x^ — lOx 4- 24 = x^ — lOx 4- 25 4- 24 — 25 = (.r _5/ — (1/ = (,;_4)(x-6). 12. x^ — 8x 4- 12 = x^ — 8x + ] G + 1 2 — IG = (x — 4)^ — (2/ = (,c_2)(x-G). 13. x^ — 1 Ox — 24 = x' — lOx + 25 — 24 — 25 = (x — 5)- — (7/ = (x + 2)(x— 12). 14. x' — 10./- — 24 = ./•' — 10./;2 4- 25 — 24 — 25 = ^x^ — of — (7/ = (.r^4-2)0-^ — 12). 15. ^>-6 _j. 4,.3_ 12 = .,/• + 4.f3 4- 4 — 12 — 4 = (.,;' 4- 2)^ — (4)" = (x'4-G)(.f^ — 2). IG. (f- 4- 2aIj — SI/ = >r + -lab -\- Ir — W = (a -f If — (2hf = (a + 3i) (a — h-). 17. a^ 4- 2((h — m^ — Ihm = rt^ 4- 2ah + // — m^ — 2hm — 1" = {a + ly — (m + h'f = (a 4- hi + 2Z/) (a — m). 18. a' + 2ab — TO^ + 2bm = a" + 2ab 4- &' — m^ 4 2bm — b^ = (a + bf—(in — bf =^{a + m){a — m-\- 2b'). 4 D I ^' I S I N . DIVISION. 56. Division consists iu finding how many times one quantity is contained in another. The quantity divided is called the dividend, that by which it is divided, the divisor, and the result obtained the quotient. 57. It follows from this that the quotient multiplied by the divisor must give the dividend. The quotient is said to be exact when the dividend contains the divisor an exact number of times. When this is not so, the quotient is called imperfect. It will be seen that the object of division is to find a quantity called the quotient, which, when multiplied by the divisor, will give the dividend. The result of the division of 4rtx by 2a is plainly 2x, because 2a X 2a; = 4aa", the dividend. So, ■ — - = la, because 7a X ac = 7a^<:. ac But — ; — — gives — 1x for a quotient, because 2a X ( — 2x) = — 4ax. i , — 7a^c ^ . ^ „ 2 And = — ia, because — i a x ac = — ia'^c. ac 58. And, in general, since the quotient into the divisor must give the dividend ; when the sign of the quantity to be divided is unlike that by which it is divided, the result will be negative, and when the sign of the quantity to be divided is like that by which it is divided, the result will have the positive sign. Thus, J- = + a, because + a X — h = — ah. 59. In Division, then, as in multiplication, like signs produce +, and unlike signs — . There are three cases iu Division. CASE I. Whe7i the dividend and divisor are hoth monomials. 60. Divide a^ by a ; we are to find a quantity, which, multiplied by a, will give aj^. This quantity is plainly a^, because a^ X a = a^. DIVISION. 39 And, since, in multiplication, we add the exponents of like letters, Ave must, in division, subtract the exponents of like letters. Divide 6a' by 2«, the result is obviously 3a^ ; because 2a X oa^ = Qa". And since, in multiplication, we multiply the coefficients together for a new coefficient, we must, in division, divide the coefficients for a new coefficient. Divide a'^b by a, the result is plainly ah, because oh X a = ci^h. Divide a^6V by a, the result is aiV. And, in general, if there arc letters in the dividend not common to the divisor, they will enter into the quotient with their primitive exponents. RULE. 61. Divide the coefficient «f the dividend hi/ the coefficient of the divisor for the coefficient of the quotient. Write after this neio coeffi- cient all the letters common to dividend and divisor affected icith expo- nents equal to the excels (f the exponents of the dividend over those of the same letters in divisor, and icrite the letters common to the dividend only with their primitive exponents. Give the quotient the sijn -{■ , where the monomials have likf siijns, and the si(/n — , ichcre they have unlike signs. F.X.VMPLES. 1. Divide x^y by x. 2. Divide — x^y by x. 8. Divide — .r'y by — .r. 4. Divide x^y by — x. 5. Divide 4a"'Z>Pc by 2tl 6. Divide — AOx^'yz^ by — 20x>'"2-''. 7. Divide a-?/ by x'y. 8. Divide X y by a;,?/. 9. Divide a;V by x'lf. 10. Divide x\if by a^V- 11. Divide .^s' by 2V. 12. Divide z's^ by zV. 13. Divide lOOOa^Z-" by SOOa-^i-". 14. Divide 1000a-"Z;-" by — SOOa-^i". 1 15. Divide a; ° ?/" by .t^". Ans. x'y. Ans. -xV- Ans. ^% Ans. -xhj. Ans. 2a^U'c- Ans. 2x-y Ans. x-y. Ans. X ~y~\ Ans. x-^y-'. Ans. xy. Ans. z~^s. Ans. zs~\ Ans. 2a'"'h'". Ans. — 2a~' 40 DIVISION. 02. It will be seen that the division of monomials is impossible when the coefficient of the dividend is not divisible by that of the divisor, and when the divisor contains one or more letters than the dividend. 63. In such cases, the quotient appears in the form of a fraction, which may admit of farther reduction by striking out the common fac- tors. Thus, the quotient arising from the division of 7a^ by 'la^b, is -r— rr, because — ^r- X la^b is plainly equal to the dividend 7«^. But la^b Za^b -r-^ may be reduced to -— by striking out the common factor, a^. 'lo?b '' 'lb ^ "^ ' In like manner, la^b, divided by 2«//, is -—- = -— . 64. We will now demonstrate two principles which will enable us to reduce such expressions still lower. 1st. Any quantity raised to the zero power is equal to unity, that is, a° = 1. For — by the rule for the exponents in division is equal to ara-m _ f,o^ j^j^^^ ^ny quantity divided by itself is also equal to one. Hence, — = a" is also equal to 1. Therefore, rt" = 1. In like manner, 2" = 1, and . (1000)" = 1, and (a + by = 1, &c. 65. 2d. A quantity may be transferred from the numerator to the denominator, or from the denominator to the numerator, by changing the sign of its exponent. For a~'° may be multiplied by — = 1, without altering its value. „ a" a" 1 Hence, a~ "■ = a""" x — = — = — a™ «™ o™ But we know that «-"> = —-— which we have seen, is also eqiial to — . 1 ri™ The quantity a"" has then gone from the numerator to the denominator, by changing the sign of its exponent. In like manner, a"*""" may be multiplied by -3- = 1, without altering its value. Hence, So, o™ Cf-" a" 1 a™ T - = a"" X a"" — a~" , — rt~' 1 1 z-p z-p z-f X ~ p ZP zv 3— p z" T DIVISION. 41 The quantity z^ has passed from the denominator to the numerator by changing the sign of its exponent. 1 1 Z^ Z^ 2P ' z~'^ z~'' z^ z° 1 66. By the first principle, the quotient in Example 15 may be _L-2 _L-2 changed into x" . 1 = x- ■> By the second principle, the quotient of Icrb by 2a^b may be changed into -;j^l>~\ and the quotient of la'^b by 2«Z/^ may be changed into —ah~\ Divide x^t/ by x^i/^c. Ans. a~'^~'c~', or 67. In this example, and in all similar examples, when the divisor contains a letter or letters not contained in the dividend, we may con- ceive this letter, or these letters, also to enter into dividend raised to the zero power, and to execute the division, we have only to subtract the exponents of like letters, as before. Thus, x^y may be written x^i/c°, and we have only to subtract 1, the exponent of c- in the divisor, from o, the exponent of c in the dividend. The result will be o — 1, or — 1. Divide x^y by a?b\* ; then x^y = x''ya''b°c'', and the result will be x'ya-'b-'c-". 68. Strictly speaking, then, there is but one case in which the divi- sion of monomials will not give an entire quotient, and that is, when the coefficient of the dividend is not exactly divisible by the coefficient of the divisor. CASE II. 69. When the dividend is a compound quantity, and the divisor a monomial. The dividend may be regarded as the product of each term of the quotient sought by the monomial divisor; hence, to find this quotient, we must divide each term of the dividend by the divisor. 4* 42 DIVISION. 70. Divide each term of the dioidcnd separately hy each term of the divisor, as in Case I., and conned the partial quotients hi/ their ap- propriate signs. EXAMPLES. 1. Divide 6x-/ — ^xhjz + 802 by 2.z. Ans. ^xhfz-' — 1x?ij^\a. 2. Divide oc^ + 2a.c + a^ by x. Ans. ic + 2a + o^.»~'. 3. Divide r? + .r"+' + a-''+^ + x^"^ by a:?. 71h,s. i + x + :r- + .7/ 4. Divide x-^ + x-p+' + rc-P+'' + x-'^"^^ by a;-p. ^??.s. 1 + .T + x^ + x' 5. Divide X- -P-! .T-P- -2_ - X' -p-3 .a:- p-4 by .TP. yl is. .X-2P-' — .T" -2r-2 X" -2p-3 6. Divide X~ -p-1 — a^-P- -2_ -X' -;^3 Ans p-4 X by -1 _ ir-p. X 7. Divide S.r'' + 2y -f 2^ by 2;^. Ans. -^xh-' + yrr^ ^~ 3 z" 8. Divide 3x' + 2j/ + .^'^ by 2^-'. ^?2S. — xH ^ yz ■\- — 9. Divide \^a^l? — \^ahx + 15o?>.Ty by 5a6. ^4?(s. SaZ* — 2.T -(- 3a;?/. 10. Divide A^a% — lOa^^x- H- 15aZ>,ry by — 5a?;. Ans. — 8a + 2a; — 3x7/. CASE III. 71. When the dividend and divisor are both polynomials. We must keep in view that the dividend is the collection, after addi- tion and subtraction, of the partial products arising- from multiplying each term of the quotient, when found, by each term of the divisor. If, then, we can find a true term of the quotient and multiply it into each term of the divisor, we will form so much of the dividend as was composed of the product of this term by the whole divisor. And when we have subtracted this product from the dividend, the remainder will be made up of the partial products of the remaining terms of the quotient not yet found by each term of the divisor. Now, if we can find another DIVISION. 43 true term of the quotient and multiply it into the divisor, the product will be so much of the dividend as is made up by the multiplication of this second term of the quotient by each term of the divisor. And if ■we subtract this product from the first remainder, the new remainder, if there be any, will be so much of the dividend as is made up of the product of the remaining terms of the quotient not yet found by each term of the divisor. We can thus regard each remainder in succession as a new dividend until all the terms of the quotient are found. 72. The whole difficulty then consi.^ts in finding, in succession, true terras of the quotient. Now, we have seen that in the product of poly- nomials there are always two terms which are dissimilar from the other terms, and, consequently, irreducible with them. They arc the terms arising from the multiplication of the two terms in the multiplicand and multiplier affected with the highest and lowest exponent of the same letter. If, then, we divide that term of the dividend which contains the highest exponent of a certain letter by that term of the divisor which contains the highest exponent of the same letter, we are sure of getting a true term of the quotient. For this terra of the dividend has been formed without reduction from the multiplication of the corres- ponding term in the divisor by the quotient found. 73. In like manner, if wc divide the terms afiected with the lowest exponent of the same letter, the one by the other, we are sure of get- ting a true term of the quotient sought. For this reason, the dividend and divisor are arranged with reference to the highest or lowest exponent of the same letter, generally with reference to the highest. AVhen so arranged, the first term of each successive remainder will con- tain the highest or lowest exponent of the letter according to which the polynomials are arranged, and will, when divided by the fir^t term of the divisor, give a true terui of the quotient. The divisor in Algebra is written on the right of the dividend, and the quotient imraediately under the divisor. Divide a^ + 2ah + h^ hj a + h. Dividend. Divisor. Dividpud. Divisor. or i^ -f 2ha + a- \ h + a h^ + ha \ h -u a 2ab + h^ a + h a -f h + ah -f V' ah + h' 1 Quotinit. ha + a^ \ Quotient. ha +_a^ + [) Rrinainder. -f Remainder. 44 DIVISION. 74. We see, that by arranging dividend and divisor with reference to the ascending or descending powers of the same letter, and dividing the first term of the dividend and the first of the remainder by the first term of the divisor, we necessarily get true terms of the quotient. But if the polynomials are not arranged we will not get true terms of the quotient. Write the dividend and divisor thus, Dividend. Divisor. V + h' h a — — &c., Quotient. a We get — for the first term of the quotient, which is not a true result. It is also plain that the division would never end. 75. Since each successive remainder may be regarded as a new poly- nomial, to be divided by the divisor, we can change the arrangement of the remainders at pleasure, provided we make a corresponding change in the divisor. lu the example above we might arrange the remainder ah + h"^ with reference to h, and write it V^ + ah, provided we change the arrangement of the divisor, and write it 6 + a. ^ RULE. 76. Arrange the dividend and divisor with reference to the ascend- ing or descending powers of the same letter, and then divide the first term on the left of the dividend hy the first term on the left of the divi- sor. This will give the first term of the quotient. Multijyly this term into each term of the divisor, and suhtract the product from the dividend. Divide the first term of the remainder hy the first term of the divisor for the second term of the quotient. Multi- ply this term into the divisor, as hefore, and suhtract the pjruduct froin the first remainder. Continue this process, dividing the first term of each remainder hy the first term of the divisor until we get a remainder, zero, when the division is said to he exact. But, if the exponents of the letter, according to which the arrange- ment is made, are all positive, and the first term of any remainder is DIVISION. 45 not divisible hy the first term of the divisor, the quotient is incomplete ; or, -if some of the exjwnents are nerjative, and the letter, according to which the arrangement has been made, has disappeared from any of the successive remainders, the quotient is also incomplete. EXAMPLES. 1. Divide x=' + Sax^ + Sa^x + a"" hy x + a. Ans. x'^ + 2ax + a*. 2. Divide x* + 4«j;' + Gr/^x^ + Aa^x + x' by x + a. Ans. a;» + 3ax* + 3«^x + a\ 3. Divide x^ — ^x^y + 'ixy^ — y by x — y. Ans. x^ — 'l.ry + y'^. 4. Divide 4a== — I'' by 2a + h. Ans. 2a — b. 5. Divide x" — y"" by x — y. X —y G. Divide x' — y^ by x^ — y^. Ans. x« + xY + ./.y + /. 7. Divide x" — y'^ by x- — /. Ans. x'" + x^^^ + xy + xy + ^y + ^"'• 8. Divide x' — y by x^ — y\ Ans. x^ -\- xy + xy* + xy* — ;/ 9. Divide x' — ?/^ by x' — y^. Ans. X* + xy 4- y^. 10. Divide x^ + / by x" + y\ Ans. x« — xV^ + xy — / + 2^y^ •T^ + f. 11. Divide x'^' + y'^ by x" + /. Ans. x'" ~ xy + xy — xy + xy — y'" + 2//'^ 12. Divide x' + y by x" + /. -4??s. x^ — xy + xy* — xy* + ?/ x^ + /. 13. Divide x® + y^ by x' + y\ J.ns. X* — x'^' + y^- 46 DIVISION. 14. Divide x'' + if^ by x" -\- f. Ans. .t'" — x'l/ + y"», 15. Divide .r'^ + y'^ by x"^ + if. Ans. x'^ — .ry + a;y — xV + y'\ 16. Divide 7ff.r — lai/ + 7«6 + Z>x — hy -\- Ir by Ta + h. Ans. X — y -{- h. 17. Divide x'^y-^ + a; ' + :^-y + ?/-' by a;^ + if. Ans. x-^ + ?/-^. 18. Divide 14.x» — 14/ by Ix^ — If. Ans. 2x« + 2xy + 2xy + 2/. 19. Divide a^ — 2nx + a;^ + 2f/y — Zxy by 2y + a — x. Ans. a — X. 20. Divide ax^ — 2>arx-\-ar^+hx'^ — '2hi\x-\-hr^-\-cr'^ — '2crx-\-cx?- hyx — r. Ans. {x — r) (a + i + c). 21. Divide s'^ — 2sx -[■ x^ -\- rs — 2xr -\- r^ -\- rshj s — x -\- r. Ans. s — X -\- r. 22. Divide 50x« + 50/ by 25^' + 25/. Ans. 2x' — 2^^/' + 2f. 2.3. Divide (a -{- h -\- c) (x — rf by (a J^ h -\- c) {x — r). Ans. X — r. 24. Divide ^V' + ^'^ + oc" + y + fh + x'-f by a;' + /. Ans. y~^ -\- Jj -{■ x~'^. 25. Divide A^x^ — 64^^^^ by Ix^ + 8^y. ^Hs. 7icy — 8r^. 26. Divide a« — h^ by a^ + V. Ans. a' — oH;' + h' — 2^ a' + //. 27. Divide a'" — ?/° by a^ + h\ Alls, a' — a'l/ + a'h* — a%' + h' — 2h^ a^ + lA 28. Divide a' — I' by n^ + ?>'• Ans. a" — a'Jr + a'b' — fA 29. Divide a'' — V by «=' + h\ Ans. a^ — a'^I/ + a'Z*« — 6'. 30. Divide a'° — l'° by a' + ?/. 31. Divide 4 by 1 + :r. Ans. 4 — 4.r + 4^2 — 4j-' + 4a;'' ^1 + X. DIVISION. 47 32. Divide 4 by x + 1. Ans. 4./;-' — 4.r-^ + 4.r-^ — 4.r-^ a; + 1. 33. Divide m — ?ix + px^ — ^x^ by 1 + a-. * Ans. m — (m + n) x + (??^ + n -\- p) x^ — otlier terms. 34. Divide 1 + x by 1 — :r. Ans. 1 + 2x + 2.f2 + 2.6=' + &c. 35. Divide 1 — x by 1 + x. Ans. 1 — 2.r + 2.v;^ — 2..=' + &c. 30. Divide x'= + «^ + ax + 1 + xi/ by x + o. Ans. .r -\- y -^ 1 X -\- a. 37. Divide x^ + x^ + 5 by x=' + 2. J»S. .T + 1 + 3 — 2.T i?f?na?-/is. 77. I. When the exponents of the arranged letter in dividend and divisor are all positive, the division will not be exact if the first terra of any of the reinaiudcrs docs not contain the arranijed letter to a higher power than it is contained in the divisor. "When, therefore, we got a remainder in which the first terra contains the arranged letter to a lower degree than it is contained in the divisor, we need proceed no further, for we can never arrive at an exact quotient. If some of the exponents of the letter, according to which the poly- nomials are arranged, are negative, the last rule will not hold good, for the result of the division will be a term affected with a negative expo- nent, and there ought to be one or more such terms in the quotient, [f, however, the letter, according to which the ai'rangement has been made, has disappeared from any of the successive remainders, we need proceed no further. II. Examples 6, 7, and 9, show that the difference between the like powers of two quantities is divisible by the like powers of a lower degree of those quantities, when the common exponent of the dividend, divided by the common exponent of the divisor, gives an exact quotient. And they show that the number of terms in the quotient is expressed by the quotient of the exponents. Thus, in example 6 there are four terms 8 in the quotient, the same as— the quotient of the exponents. In ex- 48 DIVISION. ample 9, the quotient of the exponents is three, and the division is exact, with three terms in the result. III. Examples 13, 14, and 15, show that the sum of the like powers of two quantities is divisible by the sum of the like powers of the same quantities when the result of the division of the exponents is odd. IV. Examples 28, 29, and 30, show that the diiFerence between the like powers of two quantities is divisible by the sum of their like powers when the quotient of the exponents is even. In the last two cases, as in the first, the quotient of the exponents gives the nuuibor of terms in the result. V. If all the exponents of the dividend and divisor are positive, we can tell, by a simple inspection, in many examples, whether the divi- sion is possible; when the extreme terms of the arranged polynomials are not divisible by each other, the quotient will not be exact. Take ^ -f ^Ixy + 'if', to be divided by ^ -^ tf ; the division is impossible, because the exponents of the result must be positive ; and oi^ by x* will give a;"'. It, of course, does not follow that the division will give a complete quotient when the extreme terms are divisible by each other. Take a;^ -f 4 + if, to be divided by .-r ■\- y ) the extreme terms are divisible by each other, but the quotient of the polynomials is not exact. PRINCIPLES IN DIVISION. 78. I. We will now show that the difference between the like powers of two quantities will be divisible by the difference of their like powers of a lower degree, whenever the quotient of the common exponent of the dividend, by the common exponent of the divisor, is exact. That is, that a"' — If" is divisible by a" — ^ i", when m is exactly divisible by n. „m _ ^^m I „"— ^;". a'"~"h" — b"' = h" (a™ " — h"' "). Performing the division, we get a"'~" for a quotient, and l>" (a'"-" — 6"—°) for a remainder. If this remainder is divisible by a" — i", the dividend will also be divisible by a" — l" ; for, represent the dividend by D, the divisor by d, the quotient by q, and the remainder by R. Then a" — h"", or J) = qd + B,. Now, qd is plainly divisible by d, and if R be also divisible by d, the first member must also be divisible by d, otherwise we would have the sum of two entire quantities equal to a DIVISION. 49 fraction. Kence, in general, if the remainder is divisible by the divi- sor, the dividend will also be divisible by it. Then, if we can prove that the remainder, b" (a""" — b'"~°), can be divided by a" — Z>", we can prove that the dividend can also be divided by a' — 1°. But if the factor, a'"~° — i"""", is divisible by the divisor, h" times that factor will also be divisible by the divisor ; and the remainder giving an exact quo- tient, the dividend will also give an exact quotient. That is, a"" — J" will be divisible by a" — b" whenever a"""" — i""" is divisible by a" — b"; or, in other words, if the difference of the like power of two quanti- ties is divisible by the difference of the like power of the n*"* degree, the difference of the like powers of a degree higher by n will also be divisible by the difference of the n**" degree. But we know that a" — b" is divisible by itself, a° — b" ; hence, by the principle just demon- strated, a^° — b^" mu.st be divisible by a" — b". And since a* — i*" is divisible by a" — b", a^" — i"" must also be divisible by a" — b", and so on, for powers of a degree greater by n, until it finally reaches and divides a™ — b'". This power must eventually be reached, because m is supposed to be a multiple of n. 79. In Example 8, x'' — y, divided by j:^ — i/', the quotient was not exact, because m or 7 was not a multiple of n or 2. In Examples 6, 7, and 9, the quotients were exact, because in each case m was a multiple of n. 80. It is plain thatj^a"" — pb"" can be divided by qa" — qb" whenjj and m are multiples of q and n. For we can put the expressions under the form of p {a"' — b""), and q (a" — i°), and the quotient will, of course, be exact when j^ will divide q, and a" — b"' will divide a" — b\ Thus, in Example 18 we found Wx^ — 14// divisible by Ix^ — ly'^ : the dividend can be written, 14 (x^ — y), and the divisor 7 (.c^ — y^, and since 14 will divide 7, and x^ — y* will divide .'-^ — ?/^, the quo- tient is exact. 81. The demonstration holds good whenever n, added to itself a cer- tain number of times, will produce m, and is therefore true for quanti- ties affected with negative and fractional exponents. Divide a"' — b-' by cr^ — b-\ Ans. a'^ 4- b-\ Divide o-« — i-6 by a-' — b-\ Ans. a'^ + i"^. Divide «-« — b'' by a~^ — b'". Avs. a"* + a'^'b-^+b-^. 5 D 50 DIVISION Divide a- — h^' by J — tl Ans. J + h\ Divide a^ — l^ by a^ — 5 - Ans. a + a't^ + b. Dividea® — i'sby a* — 6*. ^«s. J + Jl^ + Jb^ + a%^ + b^- 82. II. The difference of the like powers of two quantities is divisi- Dle by the sum of their like powers of a lower degree, when the result of the division of the common exponent of the dividend by the common exponent of the divisor is an even number; which is evident from the principle of Article 50. 83. III. The sum of the like powers of two quantities is divisible by the sum of their like powers of a lower degree, when the quotient arising from dividing the common exponent of dividend by the common exponent of the divisor is an odd number. That is, «"■ + b"" is divisible by a" + V", when m, divided by rt, is an odd number. Performing the division, we will have — Z*" (a"""" — J""") for a re- mainder, and whenever the factor, a""" — i"*""", is divisible by a" + 6", the dividend a™ + b'^ will be divisible by the same divisor. But we have just shown that d""" — i""" can be divided by a" + Z)° when m — n . , in — n . m IS an even number, and smce is the same as 1, n n 71 „ 'ni , , , . 1 . „ ???- — n . the quotient of — must be odd in case the quotient of is even. Hence, the truth of the proposition. Thus, in Examples 13, 14, and 15, the quotients were exact, because for each of these examples — was an odd number. So, in like manner, x^ + a", x^ + a^, x} + a', &c., are divisible by a; + a, and give 3, 5, 7, &c., terms in the quo- tient, with signs alternately plus and minus. 84. IV. It is plain that the sum of the like powers of two quantities cannot be divided by the difference of their like powers ; that is, a"' + i" cannot be divided by a" — Z»", because it is not possible to decom- pose the sum of two quantities into factors, one of which will be a dif- ference between the quantities. AIC FRACTIONS. 51 ALGEBRAIC FRACTIONS. 85. A fraction is a hrohcn part of unity. The denominator denotes the number of equal parts into which the unit has been broken or divided, and the numerator expresses the number of these equal parts taken. Thus, the fraction f indicates that the unit has been divided into three equal parts, and that two of these parts have been taken. In like manner, the fraction — indicates that the unit has been divided b into h equal parts, and that a of these parts have been taken. 8G. Every quantity not expressed under a fractional form is called an entire quantity. 87. An expresssion, made up in part of an entire quantity, and in part of a fraction, is called a mixed quantity. Thus, 4 + J, and a — — are mixed quantities. c 88. A proper fraction is one in which the numerator is less than the denominator. An improper fraction is one in which the numerator is greater than the denominator. 89. A simple fraction is one whose numerator and denominator arc simple quantities. Thus, — is a simple fraction. 90. A compound fraction is one which has a compound expression in the numerator or denominator, or in both. Thus, , , ' c ^ c -\- d and -j are compound fractions. 91. The minus sign before a fraction changes the signs of all the terms in the numerator. Thus, is equivalent to . b b 92. A few of the principles involved in operations upon fractions will now be demonstrated. I. The multiplication of a fraction by an entire quantity is effected by multiplying the numerator, or dividing the denominator by the entire quantity. For to multiply — by r, is to repeat — , c times, and since — taken 02 ALGEBRAIC FRACTIONS. twice, is V> tlivce times is — , &c, the result of multiplyin<2; — , c times oc will plainly ba — . The multiplication can also be performed by divid- ing the denominator by c ; for, if we divide the denominator by f, and q is the quotient of the division, then the unit will be divided into c times fewer parts than before, and, of course, each part is c times as great as before. Now, if we write a over q, we will have taken as many parts of the unit as before ; and since each part is c times greater, the fraction — is plainly c times greater than the fraction — . The multiplication of the numerator by a whole number may also be demonstrated in another way. — is taken to be increased c fold, and if the denominator remains unchanged, while the numerator is multi- plied by c, the number of parts into which the unit is divided remains the same, but the number of those parts taken is increased c fold, the oc new fraction — must then be c fold greater than the old. b 93. II. The division of a fraction by an entire quantity is effected by dividing the numerator, or multiplying the denominator by the entire quantity. For, to divide -- by c, is to diminish the value of the fraction c fold, and if the result of the division of a by c is q ; then g' is c times smaller than a, and y- is c times smaller than — . Because, while the parts of the unit have remained unchanged, c fold fewer of these parts have been taken. The division can also be effected by multiplying the denominator by c, for then the parts into which the unit is divided being increased c fold, the value of each part must be decreased c fold, and, of course, when the same number of these c fold diminished parts are taken, as at first, the result must be c times smaller than before. 94. III. The value of a fraction is not altered by multiplying the numerator and denominator by the same quantity. The fraction -- is not altered in value by multiplying the numerator and denominator by c. For, to multiply the numerator by c is, from what has just been ALGEBRAIC FRACTIONS. 5H shown, to increase the value of the fraction c fold, and to multiply the denominator by c is to diminish the value of the fraction c fold, and, of course, the fraction has undergone no change of value since it has been increased and decreased equally. 95. IV. The viilue of a fraction is not changed by dividing the numerator and denominator by the same quantity. Because the two divisions cancel each other, and leave the fraction in its primitive condition. 90. Y. If the same quantity be added to the numerator and denomi- nator of a proper fraction, the value of the fraction will be increased. To show this, it will be necessary to show that if the same quantity be added to two quantities differing in magnitude, the smaller of these will be increased more, proportionally, than the larger. Take the num- bers 1 and 2, add 1 to both, the first will be doubled, but the second will not be ; add 3 to both, the first will be increased four fold, while the second Avill only be 2'] times greater than before. In general, let a and b represent the two quantities, a being less than h- add a to both, the first will be doubled, but the second M'ill not be, because a -\- h is loss than 2/j. Now, when the same quantity is added to the numerator and denominator of a proper fraction, the numerator is increased more proportionally than the denominator ; the number of parts taken, then, is increased without their being an equal decrease in the size of those parts. Hence, the new fraction must be greater than the old. 97. VI. By adding large quantities to the numerator and denomi- nator, we can make the value of the fraction approximate indefinitely near, though it can never become unity. Thus, add 1000 to the nume- rator and denominator of i, the new fraction is j^o^' al"^os*^^> though not quite, unity. Now, add a million to both terms of the fraction, and the result will be indefinitely near, without being altogether unity. 98. VII. If the same quantity is added to the numerator and deno- minator of an im]m)per fraction, the value of the fraction will be decreased. For, the denominator being smaller than the numerator, will be increased more proportionally, and the value of the fraction must be diminished. Take | as an example, add 10, and the fraction becomes |4 ajcm 36aa; + 42aV Ans. ac + m 6 +7aV ■ 6. Reduce 7. Reduce \. Reduce ALaEBRAIC FRACTIONS. 55 4m^ — \r? 8 (m + n) ac + 10 + 0" + dc a + h +c + d • Zxy + Q^y - - llxYt Ans. 1 + 3x?/ — ^xhjh, Ans. ^bxy + oaxy + Vlx^f 1 »;.) hex' (m — n) hex' (m — n) hex' (m — n) hex' By reducing the above results by Case I., we will get back the ori- ginal fractions. We can thus verify the correctness of the results obtained. ALGEBRAIC FRACTIONS. 61 CASE VI. 106. To add fractional quantities top^ether. RULE. Reduce the fractions^ to a common denominator, if necessary , and over this common denominator write the algebraic sum of the new numerators. Fractions, which have different denominators, cannot be added to- gether previous to reduction to the same denominator, because they represent different things. Tims, J and 1 neither make i nor | ; the first indicates that one of the three parts into which the unit has been been divided has been taken ; the second indicates that Ave have taken one of the four parts into which unity has been divided. Since, therefore, the parts taken are different iu magnitude, they cannot be added together. It would, obviously, be just as proper to add a peck (the fourth of a bushel) and a quart (the thirty-second part of a bushel), and call the sum two pecks, or two quarts, as to add two fractions with diflcreut denominators. EXAMPLES. -f»ii=^ /> , — C . , X + h — C 1. Add — H , and together. - a a a ^ 2. Add -, -f - and — -. a a a 3. Add —V- to . 4. Add X — a -\ ^1 ^, a A ns. X- + ha — ac a' Ans. 4c : + ex — 2b 2c 3..;^ ■ — 2ax + a"" r — 2a 2 (x — a) _ . ..2x 3x 4x C)x . 164a; 6. Mi'—l+J^to'-+J-'^. X ^ + y c n nc (x'-' — ;/) -|- bcxn -f- nx {x -f y)^ — mcx (x -\- y) ncx {x 4- y) 6 02- ALGEBRAIC FRACTIONS. cc — y m ic + y c ^^^^ 2mc (x' + f) + (c" + m') (a-'' — .y') mc (.x^ — ?/^) 8. Add 4 + 4.x to J + ^. ^n.. lL^+1). 9. Add 4 — 4a: to 1 - -|-. ^«5. mLlZ^). X 17 Tx — 1) 10. Add 4.r — 4 to ^ — }. Ans. — ^ '-. 4 4 11. Add -^^ to .T — a H . 6 X -j- a X — a (x + a + bx — ha) (a;^ — a^) + 2al^ Ans^ b (x' — a^) Rcmarh. 107. When none of the terms have been reduced with each other, the result divided by the common denominator ought to give back the original series of fractions. This may be noticed in the first three examples. CASE VII. 108. To subtract one or more fractional quantities from one or more fractional quantities. RULE. Mahe all the denominators the same, if the fractions have not a com- mon denominator, then write into one sum the numerators of the minu- end, and from this sum take the algebraic sum of the numerator, or numerators of the subtrahend. Write the difference over the common denominator. The same reasons which show that we cannot add fractions, with different denominators, prove that we cannot subtract fractions with unlike denominators. The fractions represent different things until reduced to a common denominator. ALGEBRAIC FRACTIONS. 1. Subtract -|- from y. ^ns. — — . 2. Subtract -J from — . AnAns. ^. 6 2 t» 3. Required the diflference between x and -;^. X X Ans. -^ OT -^. 4. Take the difference between x + -— and x — -r-. Ans. + o, or — a. ^ „ ■, a 4- h X „ a — h 5. Subtract — 1- -;y from — - — i ^> ^^' + -^O /7 , N Ans. — 2 ;- — , or — (^ + ^)- 6. Subtract — f- x from — - — Ans. — 2 ^—.— , or — (h + 2x) 7. Subtract 2x + ^ -f 4 from '^-. 3(yH-6) — 12a; — 2a — 24 Ans. --^^^^— ! — . 8. Subtract ^^-^ from 2x4--^ + "l- 2 o Ans. 12-^ + 2« + 24-8(y + &) ^ A , o -f .r - a — X 0. Subtract — - — from — ^ — . 10. Subtract —^- from —f-. 11. Subtract x — — from ' . 6 7>l 6 Alls. (a + 5.--) 6 a -\- 5x Ans. Ti — . , , (x + c) — mhx + mx Ans. b 1 • vib 6Ei A L Ci E B 15 A 1 C 1' R A C T I O N S . CASE VIII. 109. To multiply fractional quantities together. RULE. ]f the quantities are mixed, reduce them to a fractional form , then mxdtiphj their numerators together for the numerator of the product required, and their denominators together for the denominator of this j^roduct. Let it be required to multiply -j- by — . To multiply — by c, is to repeat — , c times ; the result will then plainly be — . Because, while the size of the parts into which the unit has been divided has remained the same, the number of parts taken has been increased c fold. — is then, obviously, c times 2;reater than—-. But we were not required to multiply — by c, but by the quotient ari- sing from the division of c by d; our multiplier has then been d times too great, and, of course, the product is d times too great. The result, then, must be corrected by dividing by d. Hence, the true product of — by — is — „ in accordance with the rule. b d bd 1. Multiply 20^ + 1 by I-. Ans. ^J^J^, . 2. Multiply ^^tf by ""-^ . Ans. ''—^. >. 3lultip]y ^— by --. Ans. a ac 4. Multiply -~^^ by ^. Ans. —. V- X -\- 1/ x—y x^—y' ALGEBRAIC FRACTIONS. 65 5. Multiply ^44-— 'by^- f-*- A-hs ^~^ "*" ^~'^~^ ~ ^~^^~' ~ ^~^ r^ 4- 7/3 6. Multiply '^ ^'^ by x'+U' {a 2 + h) {x- Ans. 1. Ans. — . A 7. Multiply 7 + ^^^ by ^ ^, ' — j/ '^ 'y +i/■ ^ ,, , . , ex , 1 // 8. Multiply c + — by — + — . y -^ c r,/; 9. Multiply L±i:!L-L/ byl^=:i:^-±^^.- c (rt^ — b^) _-,.,.- a + 6 , ac — he 10. Multiply ., , 7 by.j ; . 1— lGx^ + %xy — f 11. Multiply 8 + ^ — c by ./: — ^ + ^. Ans. 4 G4.r + 4a.7- — 8r.r — 32a — 2^^ + 5ao + l(3c — 2c 12. Multiply ^_ibyj+^, 9*2 — 4/r :>;^ V ^"^- -^1^*^^ 4-9 CASE IX. 110. To divide fraetioual quaiitities by each other. RULE. If tlie quantity to Ic divided is not in the form of a simple fraction, reduce it to that form ; and. if the quantity to he used as a divisor is not already a simple fraction, make it so. Then invert the terms of the divisor and proceed as in the last case. G* E m ALGEBRAIC FRACTIONS. Take, as an example, — - to be divided by — . The result of the division of — by c must be c times smaller than—. It will then plainly be — , because, wliile tlic number of parts taken has remained unaltered, tlie size of tlicsc parts has been diminished c fold, since their number has been increased c fold. But we were not re- quired to divide — by c, but by the quotient of c by d. Our divisor, then, has been d times too lar2;e, and the result, — , of course, d times be too small. The error, then, must be corrected by multiplying by d. Hence, ^ — : — - ■= ^— , in accordance with the rule. h d be 111. The demonstration is general, and is applicable when the divi- dend and divisor are mixed quantities, or made up of several fractions ; or, when either dividend or divisor is a mixed quantity, or composed of more than one fraction. For the fractions, if not already under the form of -J- and -— , can be put under these forms without difficulty. The demonstration in the last case is general, for the same reasons. EXAMPLES. 1. Divide — by -^. Ans. %x. 2. Ax Divide ~ by 5. Ans. -^. 3. Divide -^ by i. Ans. 4x. 4. Divide " — -j- — by - ■ + 1 8 Ans. 2(a; — 1). 5. Divide ^'' J !: + Hv^^ 1 1' ^^ X — Ans. 1. G. Divide ^^^'^ by - a +6 ( y a' — b^' Ans. (x—y) (ci — h). ALGEBRAIC FRACTION'S. 67 r. Divide m by — ~ — y. Am. -ir'^-^^7^—. II a' — Ixy ^. Divide !!^^ by ^^-^V2x. Am. "^"^^ y ■ n {x^ + y 9. Divide 1 + ^ — 4:c by — 1 — 4-' + 4.r. o o ^l«s. — 1. 10. Divide 1 + 4^ _ 4x by + 1 — 4 + 4.t. + a;2 — 12x J^??s. 3 — a;^ + Vlx ^^ ^. .. 2(x+y-)+a?j - , , Cr — 7/)« + fl2^ 11. Divide — ^^ — h + xhx -T-. — -^ — — ; — . a -^ 4{x + y) + 2ax Ans. 2^±^^^±^. cr\x — y + ax 10. Divide 'i(^l+^'-l + .bj U:r + y) + 2a. X — y + ax Ans. ^ . 2a + X + 2 x m + 71 x lo. Divide ^ 2o + - by -^, -. . Ah{a + x — ch + 1^ Ans. — '-. m -{- n — 2bx Rcmarlis. 112. The quotient, multiplied by the divisor, ought to give the divi- dend ; and in the last case, the product, divided by one of the factors, ought to give the other factor. "We are thus enabled to verify our results. We have seen that both the numerator and denominator of a fraction can be multiplied by the same quantity without altering the value of the fraction. Hence, by multiplying the term of a fraction by minus X 1 iinity, the signs of these terms may be altered. Thus, " — - — may be •xi 1 — x -\- a — a ., _ — a a written — ; may be written =-, cvc. Hence, j = — . — — +6 — bo That is, the quotient of two negative quantities is positive. So, - — T, or, J = -, read minus the fraction — . + O — Ob ALGEBRAIC FRACTIONS. CASE X. 113. To reduce a compound fraction to its lowest terms. Wc have seen tliat a simple fraction can often be reduced by remov- ing the factors common to the numerator and denominator. When the common factors of a compound fraction can be detected by inspection, we have but to remove them, and the fraction is reduced to its lowest terms. Thus, a'^ + ah — ac um -\- an can be reduced to its lowest terms by the removal of the common factor a. But the common factor, or common factors, of a compound fraction cannot always be detected by inspection, and some process becomes necessary to discover them. This process is called finding the (jrealest common divisor. 114. The greatest quantity that will divide two or more quantities, is their greatest common divisor. When the greatest common divisor of the numerator and denomina- tor of a compound fractions is found, it can be reduced to its lowest terms by dividing by this divisor. 115. The greatest common divisor, though most usually obtained in order to reduce compound fractions, is also frequently found between (quantities not written in the fractional form. IIG. We will then explain the method of finding the greatest com- mon divisor between two polynomials, without regarding them as nume- rator and denominator of a compound fraction. 117. The determination of the greatest common divisor of two poly- nomials depends upon two principles. 1st. The common divisor of two polynomials contains, as factors, all the common factors of the two polynomials, and does not contain any other factors. For, let A and B be the two polynomials, and D their greatest common divisor. Now, it is plain that any quantity C, made up of a part of the common factors of A and B, would divide both quantities, and would, therefore, be a common divisor ; but C, multiplied by the remaining factors com- mon to A and B, would still divide them. Hence, C would be a divi- sor, but not the greatest divisor of A and B. Again, any quantity M, made up of all, or a part, of the common fac- ALGEBRAIC FRACTIONS. 69 tors of A and B, and containing also a factor, d, not common to the two polynomials, will not divide them. For, if we proceed to the divi- sion, the common factors of A, B, and M will strike out, and leave the factor (1 as a denominator of the reduced polynomials. Thus, ad will not divide a^ -f ah and ac + am ; for, though it contains their common factor, a, it also contains a factor d, not common. The result of the division will be , and - — - — . The greatest common divisor, D. a d then, contains all the common factors of A and B, and contains nu other factors. 118. 2d. The greatest common divisor of two polynomials, A and B, will enter into the successive remainders which arise from dividing A by B, and B by the remainder, and the second remainder by the first, aud so on, until there is no remainder. For, denote by A' and B' the quotients arising from dividing A and B by the greatest common divisor, D ; then A = A'D, and B = B'D. Divide A by B ; or, what is the same thing, divide A'D by B'D, aud call the quotient Q. Thus, A'D I B'D QB'D Q (A' — QB') D = 1st Bemaindcr = MD. by making A' — QB' = IM. We see that the first remainder contains the greatest common divisor; and as it is also in the divisor, wc have a right to seek it between these polynomials. Now, divide B'D by the remainder, and call the new quotient (7, and we get B'D I MD Q':md q ' (B' — Q'M) D = 2d Kemaindcr = XD, representing B' — Q'M by it. Wc see that the greatest common divisor enters also into the second remainder; and as it is also in the first remainder, we have a right to seek it between these two remainders. Divide MD byND; the remainder, if any, will still contain the greatest common divisor. 119. Let us apply these principles in finding the greatest common divisor of the polynomials, A and B. First arrange them with respect to a certain letter, and regard that polynomial which contains the highest power of this letter as the divi- 70 ALGEBRAIC FRACTIONS. dend, and the other polynomial as the divisor. Next, examine if A (which we suppose the dividend) contains a factor common to all its terms, but not common to all the terms of B. By the first principle, this factor can constitute no part of the common divisor, and may be suppressed. It is not absolutely necessary to suppress it, because it will disappear in division, and not appear in the remainder, and, there- fore, by the second principle, cannot make part of the greatest common divisor. But if there is a factor common to all the terms of B, and not common to those of A, it must be suppressed. For A would not be, divisible by B until it had been multiplied by the common factor of B, and the multiplication would make this factor common to A and B ; and, hence, by the first principle, it must be common to the divisor. The greatest common divisor would then contain a factor which did not originally belong to A. 120. If there is a factor common to A and B which can be detected by inspection, it may be divided out and set aside, as making part of the greatest common divisor. 121. The next step, after setting aside, or suppressing factors seen by inspection, is to divide A by B. If the coefiicient of the first term of A is not divisible by the coefiicient of the first term of B, it may be made divisible by multiplying all the terms of A by the coefiicient of the first term of B, or by any other quantity that will make the division possible. This multiplication will not efi"ect the result, because the factor introduced into it will disappear in the division. 122. The remainder, after division of A by B, will contain the great- est common divisor ; and, as B also contains it by hypothesis, we have a right to seek it between B and the remainder. The second remain- der, if any, contains D also, but its coefficient or multiplier is smaller than in the first remainder; and so tlie coefiicient of J), in each re- mainder, is smaller than in the preceding, until it finally becomes unity. "When this final remainder is used as a divisor, it will go as many times in the preceding divisor as there are units in the coefficient of D in that divisor, and there will be no remainder. 123. When, therefore, wc get a remainder zero, we conclude that the last divisor is D X 1, or D itself. 124. If the given polynomials have no common divisor, we can dis- cover the fact by the same tests that show when one polynomial is not divisible by another. We will find either that the letter, according to which the arrangement has been made, has disappeared from some of ALGEBRAIC FRACTIONS. 71 the remainders, or it enters to a higher power in that remainder which is used as a divisor than in the one used as a dividend. The preceding principles and demonstrations for finding the greatest common divisor lead to the following RULE. 125. Arraivje the two polynomials with reference to a certain letter, as in division; use that one tchich contains the highest j^oicer of this letter as the dividend. 2d. Next, set aside the factors, if any, common to dividend and divisor, as part of the greatest common divisor, and suj^ress those fac- tors tohich are common to the one and not to the other. 3c7. Prepare the dividend, if necessary, for division, hy multiplying hy any quantity that loill make its first term divisible hy the first term of the divisor ; divide the one polynomial hy the other, and suppress in the remainder any factor that may he common to cdl its terms and not common to those^of the divisor. Aih. Use the remainder so reduced as a divisor, and the last divisor as a dividend, and proceed c(s before; and continue in tliis manner, using eadh successive remainder as a divisor, and the last divisor as a dividend, until tliere is no remainder, or until it is evident that the polynomials have no common divisor. „ , a'^cb — a^c)/ + a^b.r — a^i/.r . lieduce — '^ '-^— to its lowest terms. mcb + mcb + mbx + myx a^ is common to the numerator and not to the denominator, m is com- mon to the denominator and not to the numerator. These factors must then be suppressed, as constituting no part of the greatest common divisor. Then, ch — cy + hx — yx \ cb -f cy + hx -\- yx cb -\- cy -f hx + yx 1 Quotient. — 'Icy — "lyx = —2y{c-\- x). Suppress — 2y, a factor common to the remainder and not to the divisor. Tlien, cb + cy -\- hx -\- yx \ c -\- x cb + bx ^~-\- y 2d Quotient. cy + yx cy + yx + 0. 72 ALGEBRAIC FRACTIONS. Hence, c -}- x is tlic greatest common divisor, and the reduced frac- tion obtained by dividing both terms of the fraction by the G, C, D, is «' (&—;/) m (6 + v/)' The factor suppressed in the first remainder might have been + 2y, instead of — 2?/. The greatest common divisor then would have been — c — X. And, in general, the two polynomials have two greatest common divisors, differing in the signs of all their terms. The reason „ , . . , . . Ad — Ad . ,. . , 01 this IS obvious, since ^f— = ^ , the common divisor to the two JJd — lj(t terms of the fraction, may be either + (7, or — d. 2. Find the greatest common divisor of x^ + 4x^ + bx + 2, and 2x^-{- Sx -f 1. Prepare for division by multiplying the first polyno- mial by 4, the square of the coefficient of the first term of the second polynomial. Then, Ax'' + ICr^ + 20:c -f 8 I 2:c" -f 3x + 1 4.x=^ + Cx^ + 2x I 2x 4^5 Quotient. 10^2 + 18a; + 8 lO.x^' + 15x + 5 3:c + 3 = 3 (x + 1). Remainder. Suppress the common factor, 3, of the remainder, and continue the division. 2^2 + Sx4-1 \ x + 1 2x^ + 2x 2x + 1. 2d Quotient, a' -f 1 = G.C.D. X + 1 X + 1 126. If we had multiplied, to prepare for division, by the first in- stead of the second power of the cocfiicient of the first terra, we would only have found one term in the quotient by the first division, and it would have been necessary to multiply the remainder by the coefficient of the first term to obtain a second term of the quotient. It is easier to prepare the polynomial by multiplying by the second power of the coefficient of the first term. 3. Reduce ,--^ ; = to its lowest terms. 6x^ — 4x- + 1 Multiply the numerator by 3' = 27, to prepare for division. ALGEBRAIC FRACTIONS. 73 27x^ + WSz' — Slx' — 'ZlOx + 210 | Hx^ — ix + 1 21x'— 36x«+ 9x' 9x^ + 48x + 34 144a;'- 144a;'- — 90x' — 192.r^ — 270x + 216 '+ 48a; 102x2 102a;^ — 318.r + 216 — 136.C + 34 Suppress — 182. — 182a; + 182: = - -182 (a- - ■!)• Then, 3a-2 — 4x + 1 1 a; • -1 3a;^ — 3x Sx- -1 — a; + l — a; + l . + 5x' -^ ox — 1 127. In the above examples, the difference of the exponents of the arranged letter being two, we multiply by the square of the coefficient of the first term of the divisor to prepare the dividend for division. It could have been prepared by multiplying by 3, but we would only have gotten one term in the quotient, and there Avould have been two re- mainders to prepare for division. In general, when preparation for division is necessary, we save time by multiplying the dividend by the coefficient of the first term of the divisor, raised to a power one greater than the difference of exponents of the arranged letter in the first terms of the two polynomials. 4. Find the greatest common divisor of «'' — a^x + ax^ — x^, and a' — rrx + ax- — x^. Ans. a — X. _ _ - cx^ -f ex* + or — x^ — x^ — xi/^ 5. Reduce -^ , '^ , ^—, — . ^,. ■ fa-' — ex + CI/ — X -\- x — Xi/ Ans. GCD c — x. And reduced fraction, -= s —„. sr — x^ + 1/^ 7 74 ALGEBRAIC FRACTIONS. 6. llcduce '- ^^, ^'^„ '^"^, !^, to its lowest terms. Ans. GCDx^ — ?/l Reduced fraction, ^"^ + ^^ + f ' + < c + 2 - -r, ■, '"'f' — 2ama; + mx^ + a^ — 2ax + x^ . . 7. Ecduco 5 — -r 5 to its lowest terms. na^ — Zanx + nx^ Reduced fraction, Ans. GCB a^ — 2ax + m + 1 8. Find the GCD of 3.^*= — 3/ and 40;=* + 4/. Ans. x^ + y^. 9. Find tlie GCD of x" ■\- x" — ax'' + a-x — «j; + a^ and x'' — x^ — ax} + a^x; + ax — cr. J-jis. £c^ — ax -\- a-. 10. Find the GCD of a:'« — ^'^ and x^ + x^ + xif + y\ Ans. .c^ + ,y^. 11. licduce ^r — - — -— — to its lowest terms. X? + \ax^ + Soil- + 2a^ Reduced fraction, Ans. GCD 3? + 2ax + a\ x^ + 2o.r + a" X + 2a 12. Reduce -5 r ; , , to its lowest terms. x- — X + x//~^ — ,y + •'*■ — -^ Ans. GCD .r-2 + 3/-' 4- X. Reduced fraction, =. a; — 1 13. Find the GCD of .7;^ + x^_// + 1+^ + 1/ + yx-^ and x^ + yx^ + ^'^' + y + J/= + ^• d[?!S. a;'' + ^. ^ , ^ . :^V + 2*" + 3a; + o-y + 1 ^ . . 14. Reduce - — -^ = -. — \ , ,^ to its lowest tenus. a;^+ 2x + 1 +yx-' + ?/' + 2y ^Ihs. GCD a;-' + y + 2. Reduced fraction, ^— — ^. ALGEBRAIC FRACTIONS. 75 Eemainder in this example, 2x — y + 1 — ary — jfx = 2a; + 1 ->r xy — 2xy — y — y'x = X (x"' + y -\'T)—yx (a;-' + 2 + y) = (x—yx)(x-' + 2+y-). Suppress factor, cc — yx. 15. Reduce :; ^— — ; ~^—, ^ to its lowest terms. 1 + jy-' + Reduced fraction, ' a;-' + y- Ans. GCD x + 7f. Eemarhs. 128. The last two examples show the importance of suppressing tho factor common to the remainder and not common to the divisor. Greatest common divisor of three or more polynomials. 129. The foregoing principles can be readily extended to finding the greatest common divisor of three or more polynomials. It is evident, that if we use one of the polynomials as a divisor, and divide all the others by it, that the remainders, if any, must contain the common divisor of all the polynomials. Suppress, in each of these remainders, the factors not common to the polynomial used as a divisor, and take that remainder which contains the lowest power of the arranged letter as the new divisor, and divide the other remainders by it; and con- tinue in this way until we get an exact divisor; it will be the GCD sought. To illustrate by an example, find the GCD of a-2 + ax — 2.V — 2r(, :r^ — .>• — 2, and ./'^ — 2./-' — 4.r + 8. Use x^ — X — 2 as the divisor, the remainder will be ax — X— 2a + 2, or a (x — 2) — {x — 2) = (./ — !) (x — 2), and — a;- — 2a; + 8 Suppress the factor, a — 1, in the first remainder, and the result, X — 2, will exactly divide the other two remainders ; hence, x — 2 is the greatest common divisor of the given polynomials. Again, take x''+ax + x + a, x'' + Sx^ + ?jx^l, and x^ + 2x'' — ax^ — 2ax -f .x — a. 76 ALGEBRAIC FRACTIONS. Use the first as the divisor; tlie three remainders will he 0, — 2« (,r + 1) -f :f + 1 + d' {x + 1), aud 1a^ {x + 1) — 2a{x-\- 1), or 0, and (..• + 1) (1 + a^ — 2a), and (.« + 1) C-«' — 2fl). Hence, x -\- I = GCD. 3d. Find the GCD of a'' — x\ a? + 2ax + a' aud (c" + 3«lr + 3ax2 4- .-r^ J.HS. a + X. 4th. Find the GCD oix^ — -f, x^ — f, x'' — i/'% and .r^' —,?/=•. ^Ins. .x^ — 1/^. 5th. Find the GCD of .r- — 1, :r.^ — l,x'' — 2x + 1, and x'' — Sx' + nx — i. Alls. X — 1. 130. It will be seen that the GCD of any number of polynomials, as well as for two, may have its sign changed. Thus, the last GCD may he X — 1, or 1 — X, and so for the others. LEAST COMxMON MULTIPLE. 131. The least common multiple of two or more quantities, is the least quantity that they will exactly divide. Thus, the least common multiple of 2, 4, and 6, is 12; of a^, a, and h, is a-h; of a^ + ah, I and ^/, is a^h^ + «Z/, &c. It is plain that the product of all the quantities will be a common multiple of these quantities; that is, it will be exactly divisible by them. Thus, 2x4x6 = 24 is a common multiple of 2, 4, and 6, but it is not their least common multiple. In like manner, a'^h is a common multiple of a^, a, and h, but not their least common multiple. Multiply a common multiple by anything whatever, the pro- duct will still be a common multiple. Hence, there may bo an infinite number of common multiples, but there can be but one least common multiple. 132. No quantity will be divisible by another, unless it contains all the factors of the second quantity. So, no quantity will be divisible by two or more quantities, unless it is divisible by each, and by all the factors of these quantities. To be the least common multiple of the given quantities, it must contain no more factors than they contain; and these factors must not enter to higher powers than in the given quan- tities. Thus, abc is not the least common multiple of a and h, because ALGEBRAIC FRACTIONS. 77 it contains a factor, c, that they do not contain. Neither is a^b the least common muhiple^ because the factor a enters to a higher power than in the given quantities. The expressions abc, and a^b, are multi- ples, but not hast common multiples. 133. If the given quantities contain a common factor, this must enter into the least common multiple raised to the highest power ti) which it is raised in the expressions to be divided, but it must not be repeated. 134. It must enter to the highest power, else the multiple which we formed would not be divisible by the expression containing the highest power of the common factor, and it must must not be repeated, else the multiple would not be the least common multiple. 135. From these principles we derive the following RULE. Decompose the given quantities into their prime factors, form a pro- duct composed of all the factors not common, and of the highest powers of the common factors, talcinfj care to let no factor enter more than once. The product so formed will le the least common multiple required. Form the least common multiple of ax^, a^x, bx*, and a'. Decomposing, we have a.x^, a^.x, h.x*, and a'. Hence, least common multiple, a'^.x*.b = aVjx*. It is plain that the common factor, a, must enter to the highest power (the third) to which it enters in one of the given quantities, else the multiple would not be divisible by that quantity. It is also plain that x must enter to the highest power, and that the factor b, not common, must also form part of the least common multiple, other- wise, the expression coutaiuiug b would not be a divisor. Form the least common multiple of 2, 8, 3, 9, a, a^, and 5b. Decomposing into factors we have 2, 2', 3, 3^, a, o?, h.b. Hence, 2'.3^.alZ».5 = 360a^i is the least common multiple required. By inspecting the result, 2^, 3^, ha^.b, it is plain that it is the least common multiple. It is divisible by 2, because it contains a factor 2^ ; 7* 78 ALGEBRAIC FRACTIONS. by 8, because it contains 2^; by 3, because it contains 3^, &c. More- over, it is the least product that will divide the given quantities, for it is made up of the least factors that will fulfil the required conditions. The number 2 need not be raised to the third power to divide 2, but it must be to divide 8 ; so 3 need not be raised to the second power to divide 3, but it must be to divide 9. We see, too, that no factor has been taken more than once. 3d. Find the least common multiple of (a^ + ax^^, 9a'', 21, and 7ah. A}is. 3^7 (« + .t') a^h = 63 (a + x") a^h. 4th. Find the least common multiple of ,t^ — ?/, x -\- y, 7x — 7y, and x'^ — 7/\r. Ans. 1x (.c^ — ?/^). 5th. Find the least common multiple of ?Hi^h\ 4 Qx + If, IQcc^lr, and 40 {x + l)/>. Ans. 3.2^5 (.r + Yfa^J? = 240 (x + 1)V/A 6th. Find the least common multiple of 2a, oa^, 4i, 5//, Gc, 7c^, 8d\ dJ, and lOahcd. Ans. 2\B'.b.7a%'c\I' = 2520a'b\M\ CoraUarj/. 136. The least common multiple of two or more quantities can also be found by dividing their product by the greatest common divisor. For, in the division of the product of the quantities by their greatest common divisor, the lowest powers of their common factors alone are divided out, and the quotient is made up of the factors not common, multiplied by those that are common, raised to the highest powers to which they enter in any of the given quantities, which, as we have seen, is the composition of the least common multiple. 137. Conversely, if we know the least common multiple of two or more quantities, we can find their greatest common divisor by multi- plying the quantities together, and dividing their product by the least common multiple. For, let A, B, and C denote the quantities, D, their greatest common divisor, and L their least common multiple. Then, since = L, -^ — = D. This relation between the greatest com- mon divisor and the least common multiple, enables us to verify our results iu finding either. ALGEBRAIC FRACTIONS. LEAST COMMON MULTIPLE OF FRACTIONS. 138. It is often important to find the least common multiple of frac- tions, and as the rule for finding it for entire quantities fails in this case, it becomes necessary to demonstrate another nxle. Let us take the fractions "We arc required to find the least quantity that they will exactly divide, and give entire quotients. Now, to divide by a fraction, is to divide by the numerator, and multiply by the denominator. The quantity, then, that will be divisible by one of the fractions, as -— , must be divi- de sible by a^, or it will not be divisible after it has been multiplied by hc^. And, as the same reasoning may be extended to the other fractions, the quantity sought must evidently be divisible by each of the numerators ; and, in order to be the least quantity that will fulfil this condition, it must be the least common multiple of the numerators. But, since the denominators are to be multiplied by this least common multiple of the numerators, it is plain that a^h cannot be the quantity sought. For, rt^i, the least common multiple of the numerators, divided by any quan- tity that will exactly divide the denominators, will be a smaller quantity than aH> itself. For instance, — will be divisible by the given quan- tities, and be a smaller quantity than a^b. It is plain, too, that a^b, divided by the greatest quantity that will divide the denominators, will still be smaller than -— , and will be the smallest quantity that will be exactly divisible by the given quantities. Hence, -j. is the least com- mon multiple of the fractions 6?' P?^"'^^- RULE. 139. Find the least common multiple of tlie mimerators, and divide it by the greatest common divisor of the denominators ; the fraction so formed tcill be the least common multiple of the (jiven fractions. 80 A li G E B R A 1 C F R A C T I N S . 2. Find the least comiuou multiple of -f^, 4^, and K-. 3. Find tlic least common multiple of Arts. V- 8 (a' + ax') -^3—,—-^—, and-. Ans. 4. Find the least common multiple of -«, oa, y ' 9 ' 5 ^l?is. II0? (a- — rc^). 5. Find the least common multiple of 1 (x + ff 49 (x + 9/) 21a;2 a: + .y —3 — > —9—' -2r' ^^^^ -i^r- Ans. tm^L±J^^49(x + ,yx^. o 6. Find the least common multiple of 8a- IBah X -f 3"' "2"' "5 73a- 7Sah X -\- 1/ ^ a ' and — Ans. loa'h (x + ^). The demonstration being founded upon the hypothesis, that the frac- tions are reduced to their lowest terms, the rule is, of course, only applicable to such fractions. GREATEST COMMON DIVISOR OF FRACTIONS. 140. The greatest common divisor of two or more fractions is the greatest quantity that will exactlj"^ divide them, giving entire quotients. This quantity must be a fraction ; for nothing but a fraction will divide a fraction reduced to its lowest terms, and give an entire quo- tient. The greatest common divisor of several fractions will then be a fraction itself; and since, when we divide a fraction by another frac- tion, we divide the numerator of the fraction assumed as the dividend by the numerator of the fraction taken as the divisor, and divide the denominator of the divisor by the denominator of the dividend, it fol- lows, that the divisor sought must have a numerator that will divide each of the numerators of the given fractions, and a denominator that ALGEBRAIC FRACTlbXS. 81 will be divisible by each of the given denominators. But, since the value of a fraction increases with the increase of its numerator and the decrease of its denominator, it is plain that a fraction, whose numerator would divide all the given numerators, and whose denominator would be divisible by all the given denominators, would not be the greatest fraction that will divide the given fractions, unless it has the greatest numerator and the least denominator that will fulfil the required conditions. Or, in other words, unless its numerator is the greatest common divisor of the given numerators, and its denominator the least common multiple of the given denominators. RULE. 141. Find the greatest common divisor of the numerators, and di- vide it hij the least common multiple of the denominators ; the fraction so formed will he the least common multiple of the given fractions. EXAMPLES. 2a' 8a' 4a^ '^^^ 1. Find the GCD of ~, ~, J?-, and 0.(^ 2. Find the GCD of 7.r' a.^- (a + .r) 12 ' ;i ' -4-—' ^"'^ 2- 3. Find the GCD of ^, -y, *f-„ and ^. d 4 12 db 4. Find the GCD ot^,^,—, and — . za 6 X X 5. Find the GCD of -^, -, ^^~, and ^. 7 21 3 6. Find the GCD of —-, -y, ^— t— :„ and -„. _x- X ax -f- X' X Ans. Remarks. Ans. 27* Ans. 12" A71S. 1 3(j' 1 ins. Qax Ans X I2x' (a' + x) 142. The greatest common divisor of entire quantities may be unity, and the quantities are then said to be prime with respect to each other. But it is evident that fractions can never be prime with respect to each F 82 EQUATIONS OF THE FIRST DEGREE. other; for, though their nuniertators may be prime, as in the 3d, 4th, and 6th examples, the least common multiple of their denominators can always be formed. 143. We may also remark that, as the least common multiple of entire quantities can always be formed ; so, likewise, all fractions what- ever have a least common multiple. 144. Knowing the least common multiple of any number of frac- tions, we can find their greatest common divisor by multiplying the fractious together, and dividing their product by their least common multiple. 145. Conversely, knowing their greatest common divisor, we can find their least common multiple by multiplying the fractions together, and dividing the product by the greatest common divisor. EQUATIONS OF THE FIRST DEGREE. 146. An Equation is an expression containing two equal quantities, with the sign of equality between them. Thus, cc = a — & is an equa- tion, and expresses, that the quantity represented by x is equal to the dilTerence of the quantities represented by a and b. 147. The part on the left of the sign of equality is called the y?rs^ member of the equation, and that on the right the second member. 148. Problems can be stated or expressed, and their solutions ob- tained by means of equations ; that is, we can express, in algebraic lan- guage, the relation between an unknown quantity to be found, and one or more known quantities, and, by certain opcratioas, can find the value of the unknown quantity. 149. The expressing the relation is called the statement of the pro- blem; and the operation performed after the statement, to find the value of the unknown quantity, is called the sohUion of the problem. Thus, let it bo required to find a quantity, which, being added once to itself, will give a sum equal to h. Let x be the unknown quantity; then, by the conditions, x + x = b, or, 2x = b. Then, if twice x is equal to b, x itself must be equal to half of b, raid x = — is the final result. In tiiis, making the equation, .r -\- x = & is tlie statement, EQUATIO-NS OF THE FIRST DEGREE. S3 and the subsequent operation is the solution. If the vakie found for the unknown quantitity ( — j be substituted in the equation x -\-x = h, we will have —+ — =/>, a true eqviation. When the value found for the unknown quantity substituted in the equation of the problem makes the two members equal to each other, the equation is said to be satisfied, and we conclude that the solution is true. 150. The unknown quantity, the thing to be found, is usually repre- sented by one of the final letters of the alphabet, x, i/, and z. Known quantities are generally represented by the first letters of the alphabet, a, I), c, and cL 151. An equation with one unknown quantity is of the first degree, when the highest exponent of the unknown quantity in any term is unity; of the second, when the highest exponent of the unknown quantity in any term is two, &e. X + a = 5 is an equation of the first degree. x^ -\- X = a is an equation of the second degree. x'^ -\- x^ -]- X = a is an equation of the third degree. X* -\- px^ + mx^ + nx = a is an ef[uation of the fourth degree. 152. Equations which contain the unknown quantity from the high- est to the first power inclusive, are called complete equations. The above are all complete equations with one unknown quantity. 153. Equations in which some of the powers of the unknown quan- tity are missing, are called incomplete equations, x^ = a is an incom- plete equation of the second degree, x^ + x^ = a is an incomplete equation of the third degree. 154. A numerical equation is one in which all the known quantities are represented by numbers. Thus, .v^ + 2x = 4 is a numerical equation. 155. A literal equation is one in which all the quantities, known and unknown, are represented by letters. Thus, x'^ + ax = h is a literal equation. 156. An equation in which the known quantities are partly repre- sented by letters, and partly by numbers, is a mixed equation. Thus, x^ + ax = h -\- 2 is a mixed equation. 84 EQUATIONS OF THE FlUST DEGREE, 157. An identical cfjuatiou is one in wliich the two members differ only in form, if they differ at aU. Thus, 2x = — , ^^^^ — = x + —, and X = X, arc identical equations. 158. An iiiJctcrminatc equation is one in which the value of the unknown quantity is indeterminate, 159. A single equation with two unknown quantities is necessarily indeterminate, for we must assume the value of one before we can determine that of the other. Thus, a: -f y = 10 is an indeterminate C(]uation, because, by assuming ?/ = 1, 5, 4, &,c., we find x = 9, 5, 6, &c. And, by attributing an infinite number of arbitrary values to i/, there will be an infinite number of values for x. IGO. All identical equations are necessarily indeterminate. Thus, the equation x = x will be true, or satisfied, when x is 1, 10, 1000, or anything whatever. 161. The oivn sign of a quantity is the sign with which the quantity is affected previous to any operation being performed upon it. The essential sign is the sign with which it is affected after the operation. Thus, multiply + « by — 1 ; the own sign of a is plus, and its essen- tial sign minus. Eut, multiply + a by + 1, and the own and essen- tial signs are the same. Subtract + ^ from a, then a — (-(-&) = a — h. Here, h appears in the second member with the essential sign. 162. Since a positive quantity cannot be equal to a negative, and, conversely, it is evident that the essential sign of the two members of an equation must be the same. 163. Since quantities can only be equal to quantities of the same kind, it is plain that the two members of an equation must be composed of quantities of the same kind. Thus, if one member represent time, the other must represent time also. These two principles are important, and ought to be remembered. 164. Several axioms, or self-evident propositions, are assumed as the basis of the principles by which equations are solved. 1. If equals be added to or subtracted from equals, the results will be equal. 2. If equals be multiplied or divided by equals, the results will be equal. EQUATIONS OF THE FIRST DEGREE. R'^ SOLUTION OF EQUATIOXS OF THE FIRST DEGREE. 165. The transformation of an equation consists in changing its form, without destroying the equality of the two members. There are three transformations of equations. First TraitKjurmatlon. IGG. The first transformation consists in clearing an equation of its fractions. By the second axiom, we have a right to multiply botli nicmhers of an equation by the same quantity ; and it is evident that if we multi- ply the two members by the least common multij^lc of the denominators, the resulting equation will be free of fractions. For, since each deno- minator will divide the least common multiple, it is plain that the pro- duct of each numerator by the least common multiple will be divisible by each denominator. Thus, take the equation — + '-+2=4, mul- tiply each member by 0, the least common multiple of the denomina- tors, and the resulting equation is 3x + a^ + 12 = 24. It is plain, that if we multiply the two members by 12, the product of the denominators, the resulting equation will also be free from frac- tions. For each denominator will divide the product of 12 by its numerator, inasmuch as it will divide 12 itself. Multiplying by 12, we get G.T + 2ic + 24 = 48. Now, divide both members by 2, which we have a right to do by the second axiom, and we get 3x -f- a; = 12, the same equation as before. Again, take the equation — + — + ^ + 2 = 5. By the first method (the least common multiple being 12), we get i\x + 2x + ox + 24 = GO. By the second method, we get 24a; + 8x + 12.r + 96 = 240. Divide both members by 4, and the results are the same. It willbe seen that the second method is the same as mul- tiplying each numerator into all the denominators except its own, and each entire quantity by ttie product of all the denominators. r.ULE. Multiplif hotli memhcrii hi/ the least common mulliple, or hrj the 'pro- duct of the denominators. 8 86 EQUATIONS OF T II K i'lRST DEGRKE. When tlu^ lonst coininon multiple cau be found without difficulty, the first method is preferable. S'.-cond Tramfvnnaiion. 167. This consists in transposing known terms to the second mem- ber, and unknown terras to the first member. Take the equation 2x — 2 = x -f- a. (1) By the first axiom, we have a right to add + 2 to both members of the equation. Adding, we have 2a; — 2 + 2 = a; + a-r2; or, since the — 2 and + 2 destroy each other in the first member, the equation becomes 2x = x -\- a -\- 2. By comparing this with the equation marked (1), we see that 2 has passed into the second member, by changing its sign. Resume the equation 2x = x + a + 2, and add minus x to both members. AYe get 2x — x = x — x + a -{- 2 ; or, since + x and — X destroy each other in the second member, 2x — x = a + 2. Hence, x has passed into the first member by changing its sign. We have demonstrated that quantities can be transposed from one member to another, provided we change their signs, since transiDosition is nothing more than adding to both members the quantities to be transposed, with their signs changed. The demonstration can be as readily made by subtracting the quantities with their appropriate signs. Hence, the second transformation is equivalent to adding to both mem- bers the quantities to be transposed, with their signs changed, or sub- tracting these quantities with their own signs from both members. RULE. Change the &!x is equal to 36 ; hence, obviously, llx = 36. Now, by the second axiom, we have a right to divide both EQUATIONS OF THE FIRST DEGREE. 87 members by the same quantity. Divide by 11, and we have x = |f . The unknown quantity now stands unconnected with known terms, and we have its trae value. Take ax -^-hx — ex =■ m. By the rules for factorinir, we have (a + h — c) x = m; and by the second axiom, x = ; . a -\- b — c These illustrations show that the third transformation is used when the equation has been cleared of its fractions, if it contained any, and when all the known quantities in the first member have been removed to the second, and when all the terms involving the unknown quan- tity have been placed in the first member, if not already there. RULE. Collect into a single algebraic sum all the corfficicnfs of the vnhioicn quantity, and divide the two memhers hij this sum; if there is hut one coefficient, divide both members by it. EXAMPLES. 1. Clear the unknown quantity of its coefficient in the equation, 2.1: = 6 + c Ans ,- = i±^' •> 2. Clear the unknown quantity of its coefficient in the equation, bx + 3x — ex = a. Ans. x = . b -\-o — c 3. Clear the unknown quantity of its coefficient in the equation, 6x -\- ax — 2x == n. n Ans. b -\- a 4. Clear the unknown quantity of its coefficient in the equation, 7x + 3x + 2x = 12. Ans. X = 1. 5. Clear the unknown quantity of its coefficient in the equation, '2x — 7x = a. a — a Ans. X = r = — p . — o 88 EQUATIONS OF THE FIRST DEGREE. 1G9. It will be seen tliat the object of the three traiisforniatious is to free the unknown quantity of its connection with known terms, and make it stand alone in the first member of the equation. When so situated, its true value is known, being expressed by the second mem- ber of the equation. 170. The various steps in the solution of an equation, can be best illustrated by taking an example which will require all three trans- formations. Take • |-f ^+ fl = 6_2:r. First step. To clear the equation of its fractions. Result. 5x + ^x + 10a = 10Z> — 20a:. Second step. To get the known terms to second member, and un- known to first member. Result bx -\- ^x + 20a) = lOi — 10a. Third step. To clear the unknown quantity of its coefficient. Result. X = - — ^ = ~ \ 171. Frequently, only two transformations are necessary in the solu- tion of an equation ; sometimes only one, but always at least one. Take ax -\- h = c. r — h By second transformation, ax ^ c — h ; and by third, x = . Take 2x — ox = m} then, by third transformation, — x = m. In this the unknown quantity appears with the negative sign ; but we were required to find the value of + a; and not of — x. Multiply both members by minus unity, which we have a right to do by the second axiom, and — x becomes positive. Hence, x = — m is the required value. 172. In general, when the sign of the unknown quantity is negative in the final result, and the conditions of the problem require it to be aflfected with the positive sign, both members must be multiplied by minus unity. 173. We observe, also, that the signs of all the terms of an equation EQUATIONS OF THE FIRST DEGREE. 89 can be changed without altering the equality of the two members, since we have a right to multiply these members by minus unity. 174. From the foregoing principles we derive, for the solution of equations of the first degree, the following RULE. I. If the equation contain fractions, dear it of them hi/ vmltiplying hath members hi/ the product of the denominators, or their least common mtdtiple. ^ II. Bring all the terms involving the unJcnown quantity/ to the first memher {if not already there'), and all the known terms to the second. III. Collect into a single algebraic sum all the coefficients of the un- knotcn quantity, and divide hoth members hi/ this sum. IV. If the nnl-nown quantiti/ appear in the final result icith a nega- tive sign, midiiplg both members hi/ minus unitij. EXAMPLES. 1. Solve the equation x + 3.r — \x = 14. Ans. x = 4. 2. Solve the equation 4x -f -J^ — '—-{-—' = 50. .Ins. X = 10. 3. Solve the equation ^ + 3.c + :^ + ^ — GO = — G. Ans. X = 15. 4. Solve the equation 2x -\- 3x + ix — 7 = — |. Ans. X = J. (. CI 1 ii .' ^ X , X X X 24a; 176 5. Solve the equation _+-+_+ - - + _ _= —. J d 4 12 7 < Ans. X = 12. 6. Solve tho equation 2x + 3.x + 43- — 7 = — 4. Ans. X = i. 7. Solve the equation ~ \- \- ox = — 4- oc + 1. a c a Ans. x = c. 90 EQUATIONS OF THE FIRST DEGREE. 8. Solve tlie equation ».r -\- —^ -{■ 4x — 5 = n'' -f- 4 («- — 1). Ans. X = ?i^ X 9. Solve the equation — — -\- -ix -{- h — o = 56 + 3a -f 1. Ans. x = a -\- h. X X X X 10. Solve the equation \- -r- -\ -, = ('Z*rc7. a b c (I (ahcdf Ans. X = Led + acd -j- al)d — ahc 11. Solve the equation -^ — -r + — + - = + 2 4 ^> 9;t a (P^ -\- pin -f- pa -f 4/^ + 4m -{- 4«) p + « T/' + m) 4p 9?l (? Ans. cc = p -f- ?n + a. z 12. Solve the equation z A \- ab + a:: = m. a , a (in — ab) Ans. z = --^ -!. a' + « + 1 13. Solve the equation t/ — -^ + ^+ a + b^ = a -}- -^ — c. to o — 105c ^--^-uT+sEb- 14. Solve the equation -^^ + ^ — ^ + a + 3 — 67j = 10. Sab (7 4- GZ. — a) Ans. y = Ga '-—- ~^. Ida -f 6 175. A few examples follow of equations in whixih the unknown quantity is affected with negative exi^onents, and in which the degree of the equation is apparently higher than the first. 2.r-' -4- X — 1 2i' + 4 1. Solve the equation -; = — . A71S. X = 2. 2. Solve the equation p 1-1 = 2. ox " Ans. X = ^. 3. Solve the equation ^ + -— = -^—- :'. 5 5 5 Ans. x = 5. EQUATIONS OF THE FIRST DEGREE. 91 , ^ , , . ,' + 7x' — Sx 12.;;^ + 14.7;^ 4. Solve the equation = -: x. Ans. rr = 6. a;-i X 2x~^ 1 5. Solve the equation zr-r + '— — '-^^r— = ;; — ;• . , G. Solve the equation x~^ -\ h — = , 1 _ h a + 1 + ah ax Ans. .T = — . o 7. Solve the equation 1 ■{■ I + 1 + o'j + i = SOx"'. Ans. X = 40. X-* h c 8. Solve the Cfiuatiou ax~' ■\ = f- — ,• ax X a {c — a) Ans. X = — H T-- 1 — ab q 1 Q 9. Solve the equation 2x-^ + — 2 + -2 = ^^~^ + — 2- Ans. a; = 1 . Gax-^ 2m 10. Solve the equation 2x~^ = — ^r— H . o X Ans. 0: = a -{■ 7n. 170. Solufions of prohhms produciuij equations of the first di^yree with an unknoicn quantity. It has been remarked, that the solution of a problem consist.s of two distinct parts, the statement and the solution. The examples just uiven, illustrate the manner of solving the equa- tion after the statement has been made. No general rules can be given for making the statement, which de- pends mainly upon the ingenuity of the student. Sometimes there is but a single thing to be determined, and in that case, by representing it by one of the final letters of the alphabet, and following the conditions of the problem, we get the equation or state- ment required. Often, however, two or more things are to bo determined, then the undetermined quantity, upon which the other quantities depend, is re- presented by one of the final letters of the alphabet. As an example of the first class of problems, let it be required to find a quantity, which, added to half itself, will give a sum equal to 6. 92 EQUATIONS or Tin: first degree. Let X represent the quantity, then half of the quantity will be repre- sented by l.r, and by the conditions we liavc x -\- Jx = G. Solving the equation, we get x = i. As an example of the second class of problems, take the following : A man invests half of his money in State stocks, and a third of the remainder in a Savings' Institution, and has four hundred dollars left. Required the entire amount of his capital, and the amount of each of his two investments. Since his investments depend upon his capital, let x = capital. placed in Savings' Institution. The remainder, after these investments, is plainly x — (Jx -f i^) ) but this remainder is, by the conditions of the problem, equal to four hundred dollars. Hence, x — (•].« -|- \x^ = 400. Solving, we get x = 1200. Then, i^x = 600, Investment in State stock. ix = 200, " in Savings' Institution. 400, The amount uninvested. 1200, Original capital. The solution has been verified by adding the two investments to what was left, and getting a sum equal to the original capital. The solutions of all problems ought to be verified in a similar manner. We have seen that an equation is said to be satisfied when the value found for the unknown quantity substituted in the given equation, makes the two members equal to each other But the value of the unknown quantity in the solution of a problem must not only satisfy the equation of the problem (the statement), but must also fulfil the required conditions. In consequence of the use of negative quantities in Algebra, the equation of the problem (the state- ment), is often satisfied, when the required conditions are not fulfilled in a strict, arithmetical sense. As an illustration, let it be required to find a quantity, which, when added to 4, will give a sum equal to 2. Let x be the quantity, then •r + 4=2. Hence, x = —2. Now, — 2 will satisfy the equation of the problem, but will not fulfil the conditions in an arithmetical sense; for we were required to find a quantity, which, when added to 4, would give a sum equal to 2, and the quantity found is really to be taken from 4. EQUATIONS or THE FIRST DEGREE. 93 It will be seen hereafter, that there are several cases in which the equation may be satisfied, and the conditions not fulfilled. A farmer has two kinds of oats ; the one kind worth 80 cents per bushel; the other 20 cents per bushel. He mixes 50 bushels of the superior article with a certain amount of the inferior, and sells the mix- ture at 22 cents per bushel. How much of the second quality did he put with the first in order to make the rate of sale the same? Let X represent the amount of the inferior article ; then, by the con- dition, 50 . 30 -f 2Qx = (50 -H x) 22. Hence, x-= 200. In this example, the verification of the equation and the fulfilment of the con- ditions are the same. 2. A farmer has 50 bushels of oats, worth 30 cents per bushel, and 200 bushels, worth 20 cents per bushel. He mixes the two lots to- gether. At what price ought he to sell the mixture ? Let X represent the required price of the mixture per bushel; then, 50 . 30 -f 20 . 200 = 250x. Hence, x = 22. 3. What number must be subtracted from the numerator and deno- minator of the fraction J, to make the new fraction the reciprocal of the old ? Ans. x = 7. 4. What number must be subtracted from the numerator and denomi- nator of any fraction — to make the new fraction, the reciprocal of the old. Ans. X = m + n. The solution is general, and is true for any fraction whatever. 5. What number, added to the numerator and denominator of the fraction s, will make the new fraction 4 times as great as the old ? Ans. X = — 3^. What does this value satisfy ? 6. What number, added to the numerator and denominator of an_\ fraction — , will make the new fraction 4 times as great as the old ? 11 ° Ans. X = •4m 7. What number, added to any fraction — , will give a sum equal to the numerator. ^ m (n — 1) Ans. X = — ^^ . 94 EQUATIONS OF THE FIRST DEGREE. 8. What number, added to any fraction — , will give a sum equal to the denominator ? . n' — m Ans. X = . n In what case will this result fall to satisfy the conditions of the problem ? 9. What number, multiplied by any fraction — , will give a product equal to the sum of the numerator and denominator? _ (m 4- n) n m How may the last four results be made applicable to any fraction whatever ? 10. In a square floor of a college building there is a certain number of brick ; if one more brick is added to each side of the floor, and the square form be preserved, there will be 61 more brick than at first. How many brick does the floor contain? Ans. x^ = 900. 11. Two pipes lead into a reservoir capable of containing 160 hogs- heads of water. The first pipe can fill it in 16 hours, and the two to- gether in 12 hours. In what time can the second alone fill it ? Ans. X = 48 hours. 12. Two travellers travel at the same rate, the one starting before the other J at 12 o'clock the first has travelled 4 times as far as the second, but when they had each gone 15 miles further, the entire dis- tance passed over by the first was only double that of the second. Eequired these distances. Ans. 45 and 22 J miles. 12. The sum of two numbers is a, and their diiFerence h. Required the two numbers. Ans. The greater, 7^+7^, the smaller — — — . AVe see from this result that, knowing the sum and difference of two quantities, we get the greater by adding the half sum to the half diffe- rence, and the less, by subtracting the half difference from the half sum. This formula is of extensive application, and ought to be remem- bered. 13. A Californian gold digger wishes to sell a vessel full of gold mixed with sand. The vessel, when filled with gold, will weigh ten pounds, and when filled with sand, one pound; the mixture weighs seven pounds. How much gold is in it ? Ans. Q-^\ lbs. GEOMETRICAL PROPORTION. 95 14. The railing around the altar of the Cathedral in the City of Mexico is a composition of gold and silver. Assuming that 279 pounds of the composition loses 20 pounds when immersed in water, and that 19 i pounds of gold lose one pound in water, and 10^ pounds of silver lose one pound in water. Required the proportion of gold and silver in the alloy. Ans. Gold : Silver : : 156 : 123. 15. Two men purchase together a barrel of flour, weighing 196 pounds, for 5 dollars ; the first pays half a dollar more than the second. How ought they to divide the flour ? .4ns. The first oudit to have 1074 lbs, the second 881. GEOMETRICAL PROPORTION. 177. Ratio is the quotient arising from dividing one quantity by another of the same kind. Thus, if M and N represent quantities of the same kind, then — expresses the ratio of M to X. Four quantities, M, N, P, and Q, are said to be proportional when the ratio of the first to the second is the same as that of the third to the fourth. Thus, 04, 8, 10, and 2 constitute a proportion, because the first, divided by the second, is equal to the third divided by the fourth. The proportionality of four quantities, M, N, P, and Q, is expressed thus, M : N : : P : Q, and is read, M is to N as P is to Q. The first and last terms of a proportion are called the extremes ; the second and third the means. Of four proportional quantities, the first and third are called antecedents ; the other two, consequents. Of three quantities, M, N, and P ; when M : X : : N : P, then X is said to be a mean proportional between M and P; and X is, at the same time, antecedent and consequent. Two quantities, M and X, are said to be reciprocally proportional, when the one increases as fast as the other diminishes. One of the quantities must be equal to a fraction with a constant numerator, and with the other quantity for its denominator. 4 X tional. 96 Ci K V M i; T l\ I C A J. 1' R P R T I N . Equi-multiples of two quantities are the products wliicli arise from multiplying tlicm by the same quantity. Thus, aJI and oN are equi- multiples of M and N. the common factor being a. Theorem I. If four quantities are in proportion, the product of the extremes ■will equal the product of the means. ]M P Foi', when M : N : : P : Q, we know that — = — ; and by clearing of fractions, MQ = PN.. This theorem furnishes an important test of the proportionality of four quantities. Whenever the product of the extremes is equal to that of the means, the proportion is true ; and when that is not the case, it is a false proportion. Theorem II, 178. When the product of two quantities is equal to the product of two other quantities, two of them may be taken as the extremes, and two as the means of a proportion. For, suppose ]MQ = NP, divide both members by QN, the first mem- M P MP ber will become — , and the second — , and we have — = — . From IN \^ li Vc^ which we get the proportion INI : N : : P : Q. Let 2 . 4 = 8 . 1. Then 2 : 8 : : 1 : 4. Theorem III. 179. The square of a mean proportional is equal to the product of the other two terms of the proportion. For, from the definition of a mean proportional, we have M : N : : N : P ; hence, by Theorem I., N^" = MP. Let 2 : 4 : : 4 : 8. Then 4^ = 2 . 8 = 16. Theorem IV. 180. W^hen four quantities are in proportion, they will also be in proportion by alternation ; that is, when antecedent is compared with antecedent, and consequent with consequent. For, let M : N : : P : Q; then, by Theorem I., MQ = NP, and GEOMETRICAL PROPORTION. 97 dividing tliis equation, member b}' member, by PQ, we get 1^ = j^ , from whicb M : P : : N : Q. Let 4 : 8 : : 12 : 24. Then, also, 4 : 12 : : 8 : 24. Theoiiem V. 181. When four quantities arc in proportion, they will be in propor- tion when taken inversely/; that is, when the consec^uents take the place of antecedents, and tlic antecedents the place of consequents. Let M : N : : P : Q; then, also, we have MQ = XP, or NP = MQj divide both members by ^MP, and we get ^ = — , from which X : M : : Q : P. Theorem VI. 182. If four quantities are in proportion, they will be in proportion by composition ; that is when the sum of antecedent and consequent is compared with either antecedent or consequent. M P Let M : N : : P : Q; thcn^ also, ^ = — . Add 1 to both membei-s, , , M + X P +Q and reduce to a common denominator, and we have — r^: — = — --r — , from which we get jM + X : N : : P + Q : Q. Let 4 : 12 : : 8 : 24. Then 4 -f 12 : 12 : 8 + 24 : 24. In like manner it may be shown, that when M : N : : P : Q, that we will also have M + X : :M : : P + Q : P- Theorem VII. 183. When four quantities are in proportion, they will also be in proportion by division ; that is, when the difference of antecedent and consequent is compared with either antecedent or consequent. M P Let !iM : X : : P : Q ; then, also, we have — = — . Subtract 1 from X Ki both members, and reduce to a common denominator, we will have M — N P — Q j^ = —T^, from which we get 31 — X : X : : P — Q : Q. Let 12 : 4 : : 24 : 8. Then 12 — 4 : 4 : : 24 — 8 : 8. It may be shown in like manner, that when four quantities are in 7 o 98 GEOMETRICAL PROPORTION. proportion, M : N : : P : Q, tluxt we will also liave M — N : M P— Q:P. Let 12 : 4 : : 24 : 8. Then, also, 12 — 4 : 12 :: 24 — 8 : 24. Theorem VIII. 184. Equi-multiples of any two quantities are proportional to the quantities themselves. For, take the identical proportion, M : N : : IM : N ; then MN = NM. Multiply both members by in, and there results mM . N = mN . M ; and, since mM and ?>iN may be regarded as single terms, we have from Theorem 2d, M : N : : mM : mN. Let 1 and 2 be multiplied by the same number, 3 ; then 1.2:: 1.3:2.3: or 1 . 2 : : 3 : 6. Theorem IX. 185. If equi-multiples of the antecedents of four proportional quan- tities, and also equi-multiples of the consequents be taken, the four resulting quantities will be proportional. For, let M : N : : P : Q ; then MQ = NP. Multiplying both mem- bers by inn, and there results mM . «Q = «N . 5»P. Hence, mM : tcN : : mP : iiQ. Let 1 : 2 : : 4 : 8, and take m = 4, and n — 3. Then 4 . 1 : 3 . 2 : : 4 . 4 : 3 . 8 ; or 4 : 6 : : 16 : 24. It is plain that the above theorem is equally true when m = n. So, that all the terms of a proportion may be multiplied by the same quan- tity without destroying the proportionality of the terms. An infinite number of proportions may then be formed from a single proportion. Theorem X. 186. If there be two sets of four proportional quantities, having two terms, the same in both, the remaining terms will constitute a pro- portion. Let M : N : : P : Q. M : N : : R : S. Then, MQ = NP, or NP = ^^IQ. MS = NR and NR = MS. GEOMETRICAL PROPORTIOX. 99 P Q Dividing these equations member by memocr, we get — - = — -, from which, P : R : : Q : S, or P : Q : : R : S. (Art. 180). And the same property can evidently be shown when any other two terms are equal. Let 1 : 2 : : 4 : 8. 1 : 2 : : 6 : 12. Then 4 : G : : 8 : 12, or 4 : 8 : : (i : 12. Let 1 : 2 : : 4 : 8. 1 : 3 : : 4 : 12. Then 2 : 3 : : 8 : 12, or 2 : 8 : : 8 : 12. It will be seen that the corresponding terms of the proportion must be taken in connection with each other. Theorem XI. 187. Of four proportional quantities, if the two antecedents be aug- mented or diminished by quantities which are proportional to the two consequents, the resulting quantities will be proportional to the con- sequents. For, let M : N : : P : Q. N : Q : : R : S. Then, MQ = NP. (A). NS = QR. Or, QR = NS. (B). Adding and subtracting (B) from (A), there results Q (M =b R) = N (P d= S). Hence, 31 ± R : P ± S : N : Q. Let 1 : 2 : : 4 : 8. And 3 : 6 : : 4 : 8. Then 1 =b 3 : 2 d= 6 : : 4 : 8. It can be shown in like manner, that if the two consequents be aug- mented or diminished by quantities, which are proportional to the antecedents, the resulting quantities Avill be proportional to the ante- cedents. The reciprocal of a ([uantity is unity divided by the quantity ; thus, the reciprocal of A is -r- . A Theorem XII. 188. Two quantities are inversely proportional to their reciprocals. For, take the identical proportion A : B : : A : B ; then AB = B A. 100 GEOMETRICAL PROPORTION. Divide botli members by AB ; then Hence, A : B : : - : -. Thus, 2 : 3 : : -J : ^. j> A Theorem XIII. 189. If there are any number of proportions having the same ratio, any one antecedent will be to its consequent as the sum of all the ante- cedents to the sum of all the consequents. For, let M:N::P:Q. M : N : : A : B. M : N : : C : D. M : N : : E : F. J Tlien, MQ = NP. MB = NA. MD = NC. MF = NE. Adding these equations member by member, we get M (Q + -Tj + D + F) = N (P + A + C + E). Hence, M:N::P + A + C+E:Q + B + D-fF. Let 1:2 1 :2 1 :2 Then, 1 : 2 : : 4 -f 6 + 8 : 8 + 12 + IC), or 1 . 2 : : 18 : 30. It is not necessary that the first antecedent and consequent of each proportion should be M and N; all that is required is, that they have the same ratio ; for, then, we can substitute SI and N for them, and the above demonstration becomes applicable. To show our authority for this substitution, take the two proportions, M : N : : P : Q, and E : S : T : U. II T From the second, we get RU = ST, or — = — ; and, since, by hy- Pi M M T pothesis, TT = ^, there results — = — . Hence, M : N : : T : U. GEOMETRICAL PROPORTION. 101 When we have any number of proportions, like those marlied A, having a common ratio, we can obviously form a continued proportion from them. Thus, M : N : : P : Q : : A : B : : C : D : : E : F. So, also, 1 : 2 : : 4 : 8 : : G : 12 : : 8 : 16. The character : : is written before each new antecedent, and refers it and its consequent back to the first antecedent and consequent. It is obvious that any antecedent may be repeated any number of times, provided, that its consequent is repeated the same number of times ; and, also, that any consequent may be multiplied or divided by HDything whatever, provided, that the same operation be performed on its consequent. Thus, M : N : : P + A + + P: : Q + B -f T) + F, may be written, M : N : : P + A + A + C + mE : Q + B + B + D + m¥, without altering the truth of the proposition. Thus, let 4 : 8 : 6 : 12, and 4 : 8 : : 5 : 10. Then, 4 : 8 : : + 5 : 12 + 10, and 4 : 8 : : 6 + 5 + o : 12 + 10 + 10. And also, 4 : 8 : : 6 X + 5 + 5 : 12 X 9 + Id + 10. Theorem XIV. 190. If four quantities are in proportion, the sum of the first ante- cedent and consequent will be to their difference as the sum of the second antecedent and consequent is to their difference. For, since M : N : : P : Q, it follows that MQ = NP. Add NQ to both members, then Q (M + N) = X (P + Q). Subtract XQ from both members, then X (P — Q) = Ql — X) Q. Multiplying the two equations together, there results (M + X) (P — Q) :=. (P + Q) (M - X). From which we get M + X^ : M — X : : P + Q : P — Q. Let 2 : 6 : : 8 : 24. Then 2 + 6 : 2 — 6 : : 8 + 24 : 8 — 24, or 8 : — 4 : : 32 : — 16. 9* 102 Ci EOMETRICAL PROPORTION Theorem XV. 191. If there are any number of proportional quantities, a single proportion may be formed from them by multiplying together the cor- responding terms of all the proportions. For, let M : N : : P : Q. A : B : : C : D. E : F : : a : H. Then, MQ = NP. AD = BC. EH = FG. Multiplying these equations, member by member, we get (MAE) (QDH) = (NBF) (PCG). Hence, MAE : NBF : : PCG : QDH. Let 1 : 2 : : 4 : 8. 3 : 1 : : 9 : 3. 5 : 10 : : 2 : 4. Then, also, 15 : 20 : : 72 : 9G. Theorem XVI. 192. If four quantities are in proportion, their like powers will be proportional. For, let M : N : : P : Q. Then, MQ = NP. Squaring both members, M'^Q^ = N'P'. Hence, M^ : N^ : : P^ : Ql In like manner it may be shown that M'" : N" : : P™ : Q"". Let 1 : 2 : : 4 : 8. Then, also, 1- : 2^ : : 4" : 8^ or 1 : 4 : : 16 : 04. Likewise, 1= : 2^ : : 4^ : 8^ or 1 : 8 : : 64 : 51 2. And similar proportions can be obtained for the 4th, 5th, &c., powers. GEOMETRICAL PROPORTION. 103 General Remarlcs. 193. All the foregoing theorems have been deduced from the first two ; and it is obvious that an infinite series of proportions might be deduced from them. The test of the truth of any proportion deduced directly or indirectly from the first two theorems, is the product of the extremes being equal to the product of the means. Thus, if M : N : : P : Q; then, also, M + 4 : P + 4 ^ : N : Q. For, by multiplying tlie extremes and means together, we have ]MQ -f 4Q = NP + 4Q, which is a true equation when MQ = NP. In general, no propor- tion is false, however absurd it may seem, when the product of the extremes is equal to the product of the means. Thus, \ : x* : : x~* : 1 is a true proportion, because the pi-oduct of the extremes is equal to unity, and that of the means also equal to unity. The proportion M : N : : P : Q, gives MQ = NP, from which we can get the four equations M P N ~Q' i\r N 1^ ~Q' Q N ~M' Q N P ~M' involving eight distinct ratios. And, since ratio is the quotient arising from dividing one quantity by another of the same kind, we conclude, from the above series of equations, that either the first consequent or the second antecedent must represent quantities of the same kind as the first antecedent; and that either the first antecedent or the second consequent must represent quantities of the same kind as the first con- sequent. We also see that two distinct species of quantities, and but two, can be represented by a proportion. If two terms of a proportion are abstract numbers, the proportion can represent but one kind of quantity. 104 GEOMKTRICAL PROPORTION. 104. In tlie equation MQ = NP, resulting from the proportion 31 : N : : P : Q, if three terms are known, it is plain that the fourth term can be found by solving the equation with reference to it. In the Single Rule of Three of Arithmetic, three terms are given to find the fourth. Thus, suppose the first three terms of a proportion are 3, 4, and G. The fourth term can be found from the proportion 3:4:: 6 : ,r. Hence 3.r = 24, or x = S, and the complete proportion is 3 : 4 : : 6 : 8. So, likewise, let the first three terms be a, b, and c, then the fourth results from the proportion, a : h : : c : x. Hence, .(■ = — . And we see that the fourth term can be found by multiply- ing the second and third together, and dividing their product by the first. If the first, or either of the middle terms is unknown, we have only to represent the unknown term by x, form the equation from the pro- portion, and then solve it with reference to x. Let the last three terms of a proportion be a, h, and c, to find the first. We have, then, x : a : : b : c. Hence, x = — . c So, we see that either extreme can be found by multiplying together the means, and dividing their product by the other extreme. Let the second term be unknown, and the other three be a, b, and c. Then a : x : : b : c. Hence, a; = y. Let the third term be unknown, and the other three be a, b, and c. Then a : b : : x : c. Hence, x .— — . We see that either mean may be found by multiplying the ex- tremes together, and dividing their product by the other mean. EXAMPLES. 1. The first term of a proportion is a^; the third, c^; the fourth, ml What is the second term ? Ans. — ~. c 2. The first term of a proportion is o">+"5 the second, a-° ; the fourth, o""". AVhat is the third term ? Ans. a'". 3. The three last terms of a proportion are a"", £% and (aby^ What is the first term ? Ans. a~^b~\ I ox. 105 4. Given the first three ternit; of a prcportiou, a^ — W^ a -\- b, and a — b, to find the fourth term. A7is. 1. 5. Given three finst terms of a proportion, a', i"", and c"", to find the fourth. \ Ans. b''c'^a~'. PROBLEMS IX GEOMETRICAL PROPORTIOX. 195. 1. Two numbers are to each other as 2 to 3; but, if 8 be added to both, the sums thence arising will be to each other as 4 to 5. What are the numbers ? Ans. 8 and 12. Let X = smaller, then the greater will be %x. For, 2 : 3 : : a- to fourth term |x. Then .r + 8 : -|a; + 8 : : 4 : 5. 2. An author can write 240 pages in ten weeks : how many can he write ii\12J weeks? Ans. 300 pages. 3. Two quantities are to each other as a is to J; but, if c be added to both of them, the resulting sums will be to each other as d to c. What are the quantities ? . (e — J) or , (e — d) be Ans. ~- — and \-^ — . ua — ae bd — ae 4. An author can write a pages in h weeks : suppose that he writes uniformly, how many pages can he write in c weeks ? Ans. X = — pages. What values must be given to the letters in order to make Examples 3 and 4, the same as 1 and 2 ? How can the results in the last two examples be verified ? 5. A father's age is now 3 times that of his son, but in 10 years more the father's age will only be double that of the son. What are their respective ages now ? Ans. Father, 30; son, 10. 6. A father's age is now a times greater than that of his son ; but in b more years the father will only be c times older than his son. What arc the respective ages of father and son ? Ans. Father, ^^ ^ ; son, — . a — c a — c 7. A gcntl,^:;i;ui puts out liis money at a certain rate of interest, and derives $500 income. He puts the same sum out a second year, one per cent, more advantageously, and derives 6600 from it. What are the two rates of interest ? A7is. 5 and 6 per cent. 106 GEOMETRICAL PROPORTION. 8. The governor of a besieged town has provisions enougli to allow each of the garrison 1* pounds of bread per day, for GO days; but, as he learns that succor cannot be expected for 80 days, he finds it neces- sary to diminish the allowance. What must the allowance be ? Ans. IJ lbs. per day. 9. A gentleman, who owns 20 slaves, had laid in a twelve-months' supply for them, when he purchased 10 more. How long will his sup- plies last? A)is. 8 months. 10. A planter, who knows that his negro-man can do a piece of work in 6 days, when the days are 12 hours long, asks how long it will take him when the days are 15 hours long. Ans. 4 days. 11. Three negroes can hoe a field of cotton in 7 days : how long will it take 4 to do the same work? Ans. 51 days. 12. A family of 5 persons use a barrel of flour in C weeks : how long will 2 barrels last 7 persons, using it at the same rate. Ans. 8 1 weeks. 13. Two farmers purchase a piece of land for $1000; whereof the first pays $600, and the second $400, and sell it again for $1200. What proportion of the pi'ofit ought each to receive ? Ans. The first, $120; the second, $80. 14. A bookseller purchased a certain number of books at the rate of 2 for a dollar, and also an equal number at the rate of 3 for a dollar, and sold the whole at the rate of 5 for 3 dollars, and gained $22 by the sale. What was the entire number of books he bought and sold ? A71S. 120. 15. The hour and minute hands of a clock are together at 12 o'clock. When will they be together again ? Ans. 5j^Y minutes past one o'clock. Let X be the space after one passed over by the hour hand before being overtaken by the minute hand. Then, 12 : 1 : : 60 -f- a- : .x. 16. Two pedestrians start from the same point at the same time, to walk around a race-course a mile in circumference. The first walks 11 yards in a minute, and the second 34 yards in 3 minutes. How many times will the first have gone around the track before he is over- taken by the second ? Ans. 33 times. 17. A brick-mason has brick 9 inches long, and 4^} inches wide, with which to build a wall 112^1 feet long. He wishes to place 180 bricks in a row, and to lay some of them with their ends to the front NEGATIVE QUANTITIES. 107 as headers, and some of them lengthwise as stretchei-s. How many headers and how many stretchers must there be in each row? Ans. 60 headers, and 120 stretchers. 18. A brick-mason wislies to place twice as many stretchers as headers in a wall 105 feet long. How many of each kind must he use, supposing the dimensions of the brick the same as in the last problem ? Ans. 112 stretchers, and 56 headers. 19. A Freshman recited 5 times a week in mathematics, and his average for the week was 66. His average for the first three days was to the average for the last two as 7 to 6. What were those averages? A)u. 70 and 60. 20. Three farmers purchase 900 acres of land for S9000 ; of which the first pays $2000, the second 83000, and the third S4000. What share ought each to get, supposing that the land is equally valuable throughout ? Ans. The first, 200 acres ; the second, 300, and the third, 400. 21. A composition of copper and tin, containing 50 cubic inches, was found to weigh 220 7 ounces. Assuming that a cubic inch of copper weighs 4-66 ounces, and that a cubic inch of tin weighs 3-84 ounces, what must have been the relative proportion of tin and copper in the composition? Ans. Tin to the copper as 1 to 2i. NEGATIVE QUxVNTITIES. 19G. Quantities are con.sidered as negative Avhcn opposed, in charac- ter or direction, to other quantities of the same kind that are assumed to be positive. If a ship, sailing at the rate of 10 miles per hour, encounter a head- wind that drives it back at the rate of 8 miles per hour, then its rate of advance will plainly be expressed by 10 — 8 miles per hour. Here the retrograde movement, as opposed to the forward, is considered negative. Suppose the wind to increase in violence until the ship is carried back at the rate of 12 miles per hour. Then its rate of advance will be expressed by 10 — ■ 12, or — 2 miles per hour. The ship will then plainly be carried back at the rate of 2 miles per hour, and we see that the minus sicn has indicated a chamre of direction. 108 NKGATIVE QUANTITIES. If a man be now thirty years old. His age, four years hence, will be expressed by 30 + 4. Four years ago, it would have been ex- pressed by 30 — 4. Here, future time being positive, past time is ne- gative. And we sec that a change of character is again indicated by a change of sign. If a man's money and estate be represented by a, and his debts by h; then a — h will express what he is worth. His debts, as opposed to his property, are considered negative. If h be greater than a, then a — Z> is negative ; and we commonly say that the man is worth less than nothing. We see that there has been a change of sign in the expression for what the man was worth, corresponding to a change in the estimation of that worth. When the worth fell below zero, its ex- pression became negative, because regarded as positive when above zero. If a man agree to labor for two dollars a day, and to forfeit one dollar for every day that he is idle ; and he labor 4 days, and is idle 2 ; then his wages will ^e 4 x 2 + 2 (— 1), or 4 x 2 — 2 ( -f- 1) = 8 — 2 dollars. We see that we have regarded as negative either the for- feiture, as opposed to the gain, or the working days as opposed to the idle. Distance, regarded as positive, when estimated in one direction, must be considered negative when estimated in a contrary direction. A B C Let a distance, AB, estimated towards the right from the point A, be represented by -f- m ; and let a distance, BC, estimated towards the right from the point B, be represented by -f ?i. Then, AC=:AB -f BC = TO + «. Now, suppose the point C be made to fall between A and B. AC B Then, AC = AB — BC = m — n . And we see that the expression for BC has become negative, and that this change of sign corresponds to a change in the direction of BC. It was first estimated towards the right from B; it is now esti- mated towards the left from B. NEGATIVE QUANTITIES. 10?) Now, suppose BC be made greater than AB, tlieii the point C will fall on the left of A. C A B_ And m — n, the expression for AC, will become negative. And we see that the expression for AC, which was positive by hypothesis when the distance was estimated on the right of A, has become negative when the distance is estimated in a contrary direction. 197. The question now arises, as to what interpretation is to be put upon a negative result when it appears in the solution of a problem. Let it be required to find a number, which, when added to 6, will give a result equal to 4.* Then, from tlic conditions, we have the statement cc + 6 = 4. Hence, x ^= — 2. Now, the problem is purely arithmetical, and in arithmetic all numbers are regarded as posi- tive. Therefore, the solution is absurd in an arithmetical sense, but it is true in an algebraic sense. For substituting — 2 for a; in the equa- tion of the problem, we have — 2 -f 6 == 4, a true equation. The value satisfies the equation of the problem. It is, therefore, a true answer to the problem in an algebraic sense. But it is not a true answer to the problem as stated, for we were required to find a number, which, when added to 6, would give a sum ecjual to 4 ; and we have really found a number which, srdjfracfed from 6, gave a difference equal to 4. The negative solution has then satisfied the equation of the problem, but has failed to fulfil the conditions as enunciated. The explanation of this difficulty is simple, when we return to the interpre- tation put upon a negative quantity. We have said that a negative quantity always indicates a change of direction or character. That change must be marked by a corresponding change of condition in the statement of a problem, and then the negative solution will be changed into a positive one. The preceding problem must be changed into this : required to find a number, which, when subtracted from 6, will give a difference equal to 4. Then, G — ./■ = 4, and x = + 2. A negative solution will not, then, satisfy a problem as enunciated, but will be the true answer to another problem in which there has been a change of condition corresponding to the indicated change of charac- ter. But negative solutions can be best explained by the discussion of the " Problem of the Couriers." 10 110 NEGATIVE QUANTITIES. PROBLEM OF THE COURIERS. 198. Two couriers travel on the same road ; the forward courier at the rate of h uiilcs per hour, and the rear courier at the rate of a miles per hour. At 12 o'clock they are separated by a distance of m miles. The problem is, to determine how much time will elapse before they are together, and also the point of meeting. Let the indefinite line A BO represent the road on which they are travelling ; A, the position of the rear courier; B, of the forward courier at 1^ o'clock; 0, the unknown point of meeting, and x the required time of meeting. Then, from the conditions of the problem, ax = AO, hx = BO, and AB =r m ; and since, from the figure, we have, AO — BO = AB — m, we get the statement, ax — hx^=m. Hence, ,r = -. The distance they are apart at 12 o'clock, divided by the difl"erence of their rates of travel, will give the number of hours that must elapse, after 12, before they are together. The time, multiplied by the rate of travel, a, will give the distance, AO, from A to the unknown point of meeting ; or the time, multiplied by the rate h, will give the distance, BO. If, for in- stance, they are separated by a distsince of 48 miles at 12 o'clock, and the forward courier travels at the rate of 4 miles per hour, and the rear courier at the rate of 6 miles per hour, then, m = 48, a = 6, and 48 h = 4. Hence, x = - — ^—7 = 24 hours, ax := Q . 24 = 144 miles b — 4 = AO. ia; = 4 . 24 = 96 miles = BO. Verification, AB + BO == 48 + 96 = 144 = AO. We might make AO the unknown quantity, and call it )/ ; then, BO = y — m. Then, — , the distance the rear courier will have to travel, divided by his rate of travel, will give the time that must elapse before he will overtake the forward courier. So, ' — - — will give the time that the rear courier will be travelling before he is overtaken. And, since these are but difi"erent expressions for the same time, we get the equa- tion ^- = — . Hence, y = j ; and supposing, as before, that li a a — u X E O A T I V E QUANTITIES. Ill G . 4S a= Q, l> = A, and ??i = 48, we have _y = AO = -r- — 7 = 144 miles. Then, — = --- = 24 hours, as before. a b 199. We will confine our discussion mainly to the expression, .r = -, in which x represents the unknown time. Since the value of a fraction increases as its denominator decreases, it is evident tjiat, by making the difference between a and b very small, we can make the time veiy great. This is apparent, too, from the problem ; for, if the rear courier travel but little faster than the forward, he will gain but little upon him. Now, if the denominator be made the smallest possi- ble, the fraction will be the greatest possible. Hence, when the deno- minator is zero, which results from making b = a, the fraction must have the greatest possible value; or, in other words, an infinite value. Therefore, x = — = infinity, synabulizcd by the character 00 . It will then be an infinite time after 12 o'clock before the couriers will be to- gether; that is, they will never be together. This is plain, from tho nature of the problem ; for, if the couriers are separated by a distance of m miles at 12 o'clock, and travel at equal rates after 12, they must always keep at the same distance of m miles apart. The character 00 , then, indicates impossibility. "Whenever it ap- pears, by examining the equation of the problem, we will discover an absurdity arising from the hypothesis that gave the infinite result. In this ease, infinity was the result of the hypothesis, a = b. "When a = h, ax is equal to bx, or AO = BO : a part equal to the whole, which is absurd. TThe character :» , then, indicates two distinct things, viz. : impossi- bility in the fulfilment of the required conditions, and absurdity in the conditions themselves. 200. Now, let b become greater than a, the denominator of the value of X becomes negative, and the whole fraction negative. "What docs this solution mean ? The sign of the time being minus, indicates that it refers to past time, because in the statement we regarded future time as positive. It is plain, also, from the nature of the problem, that the time of the couriers' being together, if together at all, must be past, not future. For the forward courier travelling more rapidly than the rear courier, 112 N i: r. A T 1 V F, Q U A N T I T I E S . ■will never be ovcvtr.ken by him. The solution, then, fliils to apply to the problem as enunciated ; but it is a true solution for another ques- tion, viz. : how long before 12 o'clock were the two couriers together '{ Lot us return to the particular problem : making h = instead of a = C. Then, m = 48, a = -1, h = G, and x = ^^ = — 24. a — h Now, it is obvious, that the forward courier, gaining two miles per hour on the rear courier, must have been with him 24 hours before 12, otherwise the two would not be separated by 48 miles at 12. _, . . . ,. cini Eeturning to the equation in distance, y = ■ -, we see that the distance, AO, also becomes negative under the hypothesis of a<^h. This solution indicates impossibility for a point in advance of B, be- cause, when a <^ h, ax is also <^ bx, or AO <^ BO, which is absurd. But it will be a true solution for a point in rear of A. A B Because, for such a point, the relation ax <^ bx will be true. Resuming the values m = 48, a = 4, and i = G, we get y = A(J =r — 96 miles ; which agrees with the solution above, for the courier A, 24 hours before 12 o'clock, being at 0, must be 96 miles from at 12. When the unknown quantity was time, we have seen that the nega- tive solution indicated impossibility for future time, but gave a true answer to the question in past time. When the unknown quantity was distance, the negative solution indicated impossibility for forward dis- tance, but gave a true answer to the question in distance estimated%iu the opposite direction. In general, a negative solution indicates that the problem, as enunciated, cannot be solved, but gives a true solution when the unknown quantity is changed in character or direction. That being the case, a negative solution may always be changed into a positive solution by changing the character or direction of the quan- tity that produced the negative result. The negative soKition, ;<: = -, may be changed into a positive solution in three ways. First, by changing the direction of the time ; that is, by changing -f x into — X, which is equivalent to asking, how long before 12 o'clock the two couriers were together. Then the equation, ax — bx= m, becomts NEGATIVE QUANTITIES. ll^l hx — aa: = m, or a:; = , a positive result, because h is greater b — a than a. Second, by changing — h into + h, which is equivalent to turning the forward courier around to meet the rear courier. The equation then becomes ./■ — j. The diagram corresponds to this ; for, in this case, the point O Avill be between A and B, and wo will have ax + 6x = AO + BO = AB. A B Third, by changing + a into — a, and — h into + h, which is equi- valent to turning both couriers around, and making the courier B go in pursuit of the courier A, the equation becomes — ax + hx = m, or X = , a positive result. It will be seen that the results in the 6 — a first and third cases are the same, as tliey obviously (night to be. 201. Now, let m = 0, and a > h. Then, x = — - = 0. The a — h quotient of zero by any finite quantity Ls plainly zero; for tlie numera- tor of a fraction indicates the amount to be divided, and the division of nothing must give nothing. Thus, if a man has zero dollars to di- vide among three persons, the share of each will plainly be nothing. In the present instance, the soliition, 0, shows that the couriers will be together at a zero time after 12; in other words, at 12 itself. This ought to be so; for, when m = 0, they are together at 12 ; and, since they travel at unequal rates, ^hey will be together no more. 202. Now, let m = 0, and a = h. Then, x = — . To explain this symbol, it is necessary to notice particularly the conditions. The con- dition, m = 0, places the two couriers together ; the condition, a — h, makes them travel at equal rates. Being together at 12, and travelling at equal rates, they must nlways be together. They will then be to- gether at 1, at 2, 3, 3i, &c. There is, then, no dctenniiuUe time at which they are together. Hence, — is called the si/mhol of Itxhtermi- natlon. It does not indicate that no solution can be found, but, on the contrary, that too many can be determined ; and the indetermination consists in this, that any one of the infinite solutions will answer just 10* II 114 NEGATIVE QUANTITIES. as well as any other. Suppose, for instance, the answer to the ques- tion, who discovered America? was, an inhabitant of Europe some time after the Christian era. The answer would be indeterminate, be- cause equally applicable to countless millions. 0^ U This will be a true equation, when x = 1, 10, 1000, anything what- ever. Hence, we say that x is indeterminate. The diagram, also, shows the same thing. A B For, when m = 0, the points A and B become the same ; and, since a = h, then, AO = BO in all positions of the point 0. Thus, it is shown in three ways that j- is the symbol of indetermi- nation. The symbol also arises from an identical equation ; for, when a = h, and vi = 0, we have ax = ax, an identical equation, in which X may have any value whatever. By recumng to the equation in y, we will observe that the distance, from the common point of starting to the unknown point of meeting, also becomes indeterminate when in = and a = h. This, obviously, ought to be so. 203. There is one remarkable exception to the foregoing statement, that — is the symbol of indetermination. It is in the case of vanishing fractions. A vanishing fraction is one, which becomes — , in conse- quence of the existence of a common factor to the numerator and deno- minator, which has become zero by a particular hypothesis made upon it. 204. Take the expression -— = = — , when m = n, but, bv de- m^ — n'^ ' ' J composing the denominator into its factors, we have . (?n — n) (ni + n) We see that — is caused by the common factor, m — n, becoming by the hypothesis m = n. Divide out this factor, and we have left = r- , when m = n. m -{■ n 2n NEGATIVE QUANTITIES. 115 , „ . ???^ — n^ (m + n) (m — n) , Ajrain, take the traction, = ^^ ■ — - = — -, wnou '^ m — n m — n U m = n. But, divide out m — n, and we have '■ — - — = 2n, when „ (m — 7if (m — n) (m — n) m = n. So, -^ = --^ = -rr, when m = n. But VI — 71 m — 11 , m — '^ r. -I c^ by division, the fraction becomes — - — = U, when m = n. ho. m /M mt «. Q = — , when m = n. But, by divi- {m — nf (m — «) {m — n) 111 sum, we cret = — = x, when m = n. We conclude, that a vanishing fraction has one of three trae values; that it may be either finite, zero, or infinity. How, then, are wc to decide whether the symbol — indicates indetcrmination, or a vanishing fraction ? If the particular hypothesis which gives the symbol, does not make the given equation an identical equation, we may be certain that — points out a vanishing fraction. We will resume the subject of vanishing fractions more at lengt^i hereafter, 205. The foregoing symbols are of the highest importance, and ought to be remembered. Arrangement in tabular furm will assist the meiuory. m that is, a finite quantity, divided by zero, equal to infinity. The sym- bol, oo, indicates impossibility in the fulfilment, and absurdity in the conditions of the problem. that is, zero, divided by any finite quantity, equal to zero. This is a true solution, unless it conflict with the condition of the problem. a; = — A. The negative solution indicates that the problem, as enunciated, can- 116 GENERAL PROBLEMS. not be solveil, but the negative result will be a true answer when tlie unknown quantity is changed in character or direction. the symbol of indetermination, when there is no common factor in the numerator and denominator. By indetermination is meant, that there is an infinite system of values that will satisfy the conditions of the problem. GENERAL PROBLEMS. 206. 1. A person employed a workman for m days, upon condition that the workman should receive n dollars every day that he worked, and should forfeit p dollars every day that he vras idle ; at the end of the time he received c dollars. How many days did lie work, and liuw many was he idle ? Let X = working days, then m — x = idle diiys; and by the con- ditions, wc get )tx — (in — .t) jj = c. Hence, x — -, on — x = idle days = m — — . n + J) n +2) n + p When will the last expression be zero, and when negative, and v/^hat do these solutions indicate ? The negative solution needs some explanation : o, the entire amount received, cannot exceed nm, what was paid for working ni days, unless the workman had not only no forfeiture to pay, but also worked more than the time for which he was employed. The idle days have then changed their character, and become working days beyond the period for which the man's services were engaged. 2. The same problem as before, except the workman, at the end of the time, was in debt c dollars. , 77?;) — c , nm -f- <• Alls. X — —i- , and m — x = . n + m n -f p When will the first solution become zero, and when negative ? 3. The same problem, except that the workman received nothing. mp , nm Ans. X = — - — , and m — ./• = . 11 + 2^ '^ + 1^ The above results are formulas, and may be made applicable to par- ticular problems of the same kind. 4. In a certain college, the maximum mark for a recitation is 100. GENERAL PROBLEMS. 117 A Freshman recited 5 times in mathematics; two days he got 95 for each recitation, and his average for the 5 days was 86. "What did lie average each of the other three days ? Ans. 80. 5. Divide the number a into two such parts that the product of the first by m shall be equal to the quotient arising from dividing the second by n. a - 071711 Ans. = , and ■= . 1 + 117)1 1 + nm What effect has increasing n upon both results? Under what form must the second expression be placed to show that it is directly propor- tional to n and mP When will the two expressions be equal? 6. A father is 32 years old, and his son 8 years. How long will it be until the age of the father is just double that of the son ? • Ans. 16 years. 7. A Either is a years, and his son h years old. How long until the age of the father will be c times as great as that of the son ? . a — ch Ans. :r = —. c — 1 What do those values become when c = 1 ? What, when ch = a? What, when ch^ a'i What do these different solutions indicate ? When c — \, x= cc, the symbol of absurdity, as it ought to be. The hypothesis makes the father ami son equal in age after the lapse (if h years. When ch = a, x = 0. A true solution, since the father is now c times as old as his son. When cb'^ a, x is negative. Fu- ture time being positive, past time is negative. The solution, then, indicates that r- years ago, the father was c times as old as his son. 8. A father is 64, and his son 16. How long will it be until the age of the father is 9 times as great as that of the son. Ans. X = — 10. The solution indicates that 10 j-ears ago the age of the father was 9 times as great as that of the sun. The father was then 54, and the son 6 years old. 9. Bronze cannon (commonly, but improperly, called brass cannon) are composed of 90 parts of copper, and 10 parts of tin : 8j% lbs. of copper lose 1 lb. when immersed in water, and 7j% lbs. of tin lose 1 lb. when immersed in water. The Ordnance Board suspecting that some bronze cannon did not contain copper enough, immersed one of them, 118 GENERAL PROBLEMS. weighing 000 lbs., aud found its loss of weight What was its composition ? Ajis. so parts of copper, and 10 parts of tin. Verify the results. The copper lost 89^;J lbs., aud the tin 13 y lbs. The sum of which is 1033|0 9 lbs. 10. A fox, pursued by a greyhound, is 125 of his own leaps ahead of the greyhound, and makes 6 leaps to the greyhound's 5, but 2 of the greyhound's leaps are equal to 3 leaps of the fox. How many leaps will the fox make before he is caught by the greyhound ? Let X = number of leaps made by the fox; then, since the grey- hound makes but 5 leaps while the fox makes 6, he will make | cf a leap to one leap of the fox. Therefore, he will make |.x leaps' while the fox makes x leaps. But each of the greyhound's leaps are equal to I of a leap of the fox. Hence, the |x leaps of the greyhound are equivalent to | . ^x leaps of the fox ', and since the greyhound has not only to run over the ground passed by the fox in making x leaps, .but ,. also that passed in making 125 leaps, we have the equation of tlie pro- ,,' blem. I . |x = :k + 125. Ans. x = 500 leaps, ;:, Verification. While the fox made 500 leaps, the greyhound made/. 4161 leaps ; and these were equal to | . 4161, or 625 leaps of thq, fox, the entire number of leaps made by the fox. Bcmarks. This problem shows the importance of making the two members ex- press the same thing by referring them to the same unit. The leaps of the greyhound have been expressed in terms of those of the fox. We might have made the unknown quantity represent the number of leaps made by the greyhound, and, in that case, the leaps of the fox must have been expressed in terms of those of the greyhound. 11. A fox, pursued by a greyhound, has a start of a leaps, aud makes h leaps while the greyhound makes c leaps ; but d leaps of the greyhound are equal to e leaps of the fox. How many leaps will tlie fox make before he is overtaken by the greyhound ? , adb ^''s. X = ^^_^^ . When will x = oP When will it be equal to infinity? When will it be negative ? How are these solutions explained ? X will be zero, when a is zero, x will be infinite when ec = Jh ; GENERAL PROBLEMS. 119 that is, when the number of leaps made by the greyhound, multiplied by the value of each leap, is equal to the number of leaps of the fox, multiplied by the value of each of his leaps. In that case, the hound •will evidently never overtake the fox ; and the solution indicates impos- sibility or absurdity, x Tvill be negative when db is greater than ec ; then the fox is running faster than the hound, and the distance re- presented by the x leaps must be estimated in a contrary direction. P' 11 F P Let II be the position of the hound ; F, that of the fox ; and P the point where the fox is overtaken by the hound. Then, when x is ne- gative, we understand either, that the fox pursued the hound and caught up to him at some point, V on the left^ or that, at some time previous to the fox being at F, and the hound at H, they were both together at P', and the fox ruuiiiug faster than his pursuer gained upon him the distance HF — - a. 12. A man, desirous of giving 4 cents apiece to some beggars, found that he had not money enough by 5 cents ; lie therefore gave them 2 cents apiece, and had 15 cents left. IIow many beggars were there, and how much money had he ? Ans. 10 beggars : 35 cents. 13. A man, desirous of giving n cents apiece to some beggars, found that he had not money enough by h cents ; he therefore gave them c cents apiece, and had d cents left. IIow many beggars were there, and how much money had the man ? d + h ^ ^nd -]- rh Ans. bcirirars, and cents. a — c " "" a — f What values must a, I, c, and d have to make this problem the same as the last ? When c ]]> a, the solution is negative; and negative solutions indi- cate a change of direction and character. The beggars become givers : the amount given becomes the amount received : the deficiency be- comes a surplus, and the surplus a deficiency. When c — a, both so- lutions become infinite, and the symbol, cc, here plainly indicates absurdity. When d and L are both zero, the two solutions will be both 0, or §-• Why? 14. A man, desirous of giving 2 cents apiece to some beggars, found that he had not money enough by 15 cents ; he therefore gave them 120 O K N E R A I- 1' R C) B L EMS. 4 cents jipicce, and liad 5 cents left. How many beggars were there, and how much money had tlie man ? Ann. — 10 beggars, and — • S5 cents. How is the solution explained? There were 10 givers, who each gave the man 2 cents, and 15 cents over; or each 4 cents apiece, lack- ing 5 cents in all. 15. A piratical vessel sails at the rate of r miles per hour for a hours, when she sustains some injury, and can only sail / miles per hour. At the moment in which the piratical vessel is disabled, a sloop-of-war starts in pursuit, and sa*ls at the rate of r miles per hour, from the point where the pirate first started. How long before the sloop will overtake the pirate ? Ans. x = > hours. J. — / When will this solution be zero? When negative? When infinite? Under what form must the fraction be placed to show that x is recipro- cally proportional to r ? Make the general solution applicable to a particular example. 16. Divide a number, a, into two parts, which shall be to each other as m is to n. ma ^ va A)is. , and 4- 11 111 + n 17. Divide a number, a, into three parts ; such, that the first shall be to the second as 7i to m, and the second to the third as q to j5. aiiq amq amp Ans. , ) . mp + mq -f nq mp -f mq + nq mp + mq + nq What supposition will make all the parts zero ? What one of them ? What two of them ? 18. Divide a number, a, into four parts; such, that the first shall be to the second as n to m, the second to the third as q to p, and the third to the fourth as r to s. anqr amqr nqr + mqr + mpr + mp)s nqr -f mqr + mpr -f- mps^ ampr omps i\qr + mqr + mpr -f mps nqr + mqr + mpr + mp 19. Milk sells in the City of New York at 4 cents per quart. A milkman mixed some water with 50 gallons of milk, and sold the mix- ture at 3 cents per quart without sustaining any loss by the sale. How much water did he put in the milk? An^. GG§ quarts. GENERAL TROBLEMS. 121 20. Milk sells ia Boston at a cents per quart. A milkman mixed a certain quantity of water with b quarts of milk, and sold the whole at (• cents per quart without losing anything hy the sale. How much water was added to the milk ? ah — he Ans. j: = — quarts. The value of x is zero when c = a. In that case, evidently, no water is added. The value is infinite when c = o. Then the milkman gives away, gratuitously, an infinite quantity of water. "When c ^ h, X is negative. ' Then we understai^l that a certain quantity of water is separated from, not added to the milk, and that the price of the milk is c cents per quart, and of the mixture a cents per quart. 21. How often are the hour and minute hands of a clock together? An>>. J^very Qbj^j minutes. 22. How often are the minute and second hands of a clock to- gether? Ans. Every 13'g minutes. 23. How often are the hour, minute, and second hands of a clock together r Ans. Every 720 minutes. The last problem is solved by means of the least common multiple. 24. There is an island 60 miles in circumference. Three pereons start from the same point to travel around it, travelling at the respect- ive rates of 4, 6, and 1(5 miles per hour. How often will all three be together? Ans. Every 30 hours. This problem is solved by means of the least common multiple. 25. There is an island a miles in circumference, around which three persons start to travel, at the rates of h, c, and d miles per hour. When will they all be together again ? Alls. In a hours, divided by the greatest common divisor of c — h, and d — c. 26. Same problem as 24, except that there are 4 persons travelling, at the respective rates of 3, 6, 12, and 27 miles per hour. Ans. Every 20 hours. 27. A farmer purchases a tract of land for $500, on a credit of 10 months, or S480 cash. What is the rate of interest that makes these sums equivalent. Ans. 5 per cent. Let .T = interest upon $100 fur one month. The statement will be 100 + lO.r : 100 : : 500 : 480. 11 122 GENERAL PROBLEMS. 28. A farmer buys a tract of laud for a dollars, payable iu h moutlis, or for c dollars casli. What is the rate of iutercst i* 100 (a — c) Ans. X = ^-^ -. c/j What suppositiou will make this value zero? What two supposi- tious will make it infiuite ? What supposition will uiake it negative ? and how is the negative solution explained ? 29. Two numbers are to each other as 8 to 3 ; but if 8 be added to both numbers, the first will only be double the second. What are the numbers? Ans. 32 and 12. 30. Two numbers are to each other as a to h ; but if c be added to both of them, the first will only be d times as great as the second. What are the numbers ? ac(d—l) fjcQj — 1) Ans. First, ^ 7-,— : second — ^^ r-~. a — bd a — Ixl What do these solutions become when cZ = 1 ? What, when hd = a? What, when bd'^ al What, when d = 1, and hd = a ? The first hypothesis gives a true solution. The second gives an ab- surd solution, as it ought, since hd can only equal a when the first number, after the addition of c to both numbers, exceeds the second proportionally as much as before. But this is impossible, since the smaller increases most rapidly. See Article 96. The hypothesis, hd, ^ (/, gives a solution impossible for arithmetical quantities, but possible for algebraic. We must either suppose that two numbers are to be found, which result from the subtraction from a number not expressed, or we must change the character of the problem, and make c subtractive. When hd = a, and r? := 1, the two numbers are represented by g, the symbol of indetermination. An infinite, or indeterminate, number of quantities will satisfy the conditions of the problem. The following problems will illustrate the foregoing cases : 31. Two numbers are to each other as 4 to 3 ; but when 5 is added to both, the numbers are equal. That is, tZ = 1 . Ans. Both numbers zero. 32. The same as last, except that, after the addition, the first will be I greater than the second. That is, hd = a. I Ans. Both infinite. GENERAL PROBLEMS. 123 33. The same as 31, except that, after the additions, the first will be twice as great as the second. That is. Id ^ a. Ans. — 10, and — 11. Change the character of the problem, and take 5 from both numbers, and the solutions will be + 10 and + 7. 34. The same as 31, except that the numbers were equal before the addition (5f 5 to both. Ans. Both indeterminate, — . An}- numbers will answer. 35. A fiither divided his estate, worth $1200, among his three sons, so that the share of the first should be to that of the second as 6 to 4 ; and so that the share of the third should be the greatest common divisor of the shares of the first and second. What will be the share of each ? Am. 6600, $400, and $200. 30. A father divides his estate among his three sons, so that the share of the first shall be to that of the second as — is to '-— : and so q s that the share of the third shall be the greatest common divisor of tht- other two. The father's estate is worth a dollars : what is the share of each son ? , miq , . , a second, — ; tmrd. VIS + H J + 1 ' ' ms -\- nq -\- \ ms + nq + 1 When will the shares of the first and second be equal ? What will the share of the third be then ? What will be the efiect upon the 37. The difference of two numbers is a, and the difference of their squares is zero. What are the numbers ? 38. The sum of two numbers is 2a, and the difference of their squares equal to c. What are the numbers ? 4a^ + c , 4«- — r Ans. — , and 4« ' 4« What do these solutions become when c ^ ? What, when c = 4a'? What, when r ^ 4a-? 124 G E N E R A I. 1' R () B L E M S . n such a manner that the track of one cuts the parallel of latitude through the Island of St. Helena 40 miles on the west of that island, and cuts the meridian through the island 40 miles north of it; and that the track of the other cuts the parallel of latitude 80 miles on the east, and the meridian 80 miles on the north ? Where will the two tracks intersect each other ? Ans. 20 miles east, and GO miles north of the island. Call DO, X ; then, PJ) = ./■ + 40 ; and PD = 80 — a: Hence, 40 + .r = 80 - then, ^ = 20 ; and PD = 60. 40. A Yankee mixes a certain number of wooden nutmegs, which cost him I cent apiece, with a quantity of real nutmegs, worth 4 cents apiece, and sells the Avhole assortment for $44; and gains $3-75 by the fraud. How many wooden nutmecs were there ? Ans. 100. 41. A Yankee mixed a certain quantity of wooden nutmegs, which 1th cost him — — part of a cent apiece, with real nutmegs, worth c cents apiece, and sold the whole for a dollars. He gained d cents by the fraud. How many wooden nutmegs were there ? What does this become when f ? = ? What, when h = {)'{ What, when he = 11 What, when he <^1. 42. The sum of three numbers is 200 ; the first is to the second as 5 to 4, and the thii'd is the greatest common divisor of the first two. What are the numbers ? Ans. 100, 80, and 20. 43. The sum of three numbers is a ; the first is to the second as h to 0, and the third is the least common multiple of the first two. What are the numbers ? ah ac nhc Am h -{■ c + he 6 -f (• + he 6 + c + he What single hypothesis will make them all equal ? What will these values become when a — 115, h = 25, and c = I'o. 41. A northern railroad company is assessed $120,000 damages for contusions and broken limbs, caused by a collision of cars. They GENERAL PROBLEMS. 125 pay S5000 for each coutusiou, aud S6000 for each broken limb ; and the entire amount paid for bruises and fractures is the same. How many persons received contusions, and how many had their limbs broken ? Atis. 12 of the former, and 10 of the latter. 45. Same problem as the last, except representing the assessment by a, the price of each contusion by c, and that of each broken limb hjh. a (I A)is. —, and — . 'Sc -u What do these values become when c = y What, when h = ? What, when a = ? What, when h ■=^ c? 4G. The reservoir at Lexington contains 48,000 gallons of water, and supplies the town and Virginia Military Institute. If all the con- ducting pipes were closed, the reservoir would supply the town and the Institute for 57| hours, aud the town alone for 96 hours. How many gallons docs the Institute use per hour. Ans. 333 J gallons. 47. Two pipes will exhaust a cistern containing a quantity of water, represented by q, in a hours, and the first will, alone, exhaust it in h hours. How long will it take the second pipe to empty it, and how much docs it exhaust per hour. Ans. hours, and ^ — gallijus per hour. 6 — a ab What do these solutions become when d = f>, and a~^ h? The negative solution can be illustrated by an example. 48. Two pipes, a and h, will exhaust a cistern in 4 hours; and the pipe a can, alone, exhaust it in 2 hours. In what time can the pipe h empty the cistern ? Ans. In — 2 hours. The solution being negative, may refer to past time, and indicate that, during the two hours before the pipe a began to play, the pipe // had exhausted the cistern. The two hours play of the pipe h, added to the two of the pipe a, give the 4 hours of the problem. Or, we may suppose that the character of the pipe h has been changed, and that it has been a supplying pipe for two hours, and, consequently, the pipe a has been twice the time in exhausting the cistern. 49. A man asks $120 cash for his horse, or §126-30 on a credit of 9 months. Supposing these valuations to be equal, what is the rate of interest ? Ans. 7 per cent. 11* 126 GENERAL PROBLEMS. 50. A man values liis horse at c dollars cash, or h dollars on a credit of a months. What is the rate of interest ? 1200 (i — c) Ans. X = ^: -. ac "What does x become when h = c? What, when a'^ h? What, when or, or c = ? 51. There is an island 32 miles in circumference : two persons start to travel around it, the first at the rate of 11 miles per day, and the second at the rate of 3 miles per day. When will the distance be- tween them be equal, estimated in either direction around the island ? Ans. At the end of the second day. 52. Two persons start to travel around an island a miles in circum- ference, the first at the rate of h miles per day, and the second at the rate of c miles per day. When will the distances between them be equal, estimated in both directions around the island ? Ans. X = - — ; 2 (6 - c) What will this solution become if the travellers move at equal rates ? What, if the second travels the fastest ? What, if the second stops ? 53. Three persons start to travel around an island which is a perfect circle, 120 miles in circumference. The first travels'at the rate of 4 miles per day, the second at the rate of 8 miles per day, and the third at the rate of 6 miles per day. How many days will elapse until the lines joining their respective positions, and the centre of the island, will trisect the circumference, under the supposition that the second, only, has gone past the starting point on a second tour of the island ? Ajis. 20 days. 54. Three persons travel around a circular island, 160 miles in cir- cumference. They start together, and travel at the respective rates of 7, 5, and 2 miles per hour. How long will it be until the third will be between the first and second, and equally distant from them, under the hypothesis that the first and second, only, have passed the starting point ? Avs. 40 hours. 55. Three persons travel around an island a miles in circumference. They travel at the respective rates of h, c, and d miles per hour. How long will it be until the third will be half way between the first and second, and how fiir will the first have travelled ? Ans. Distance, r^-^— — : time, —^r-. — • — 2c/ -f c h — 2rt -j- c GENERAL PROBLEMS. 127 What do these values become when 2d = h -\- c ? "What, when 2d = c? How do you explain these results ? 5G. Divide the number 24 into two such parts that their product be the greatest possible. Let X express the excess of one of the parts over the half of 12, then, Q -{- X, and 6 — x will represent the two parts. And (G + .r ) (6 — a;), 36 — x^ is to be the greatest possible. This result will evi- dently be the greatest possible when a- = 0, or when the parts are equal. 57. Divide the quantity a into two such parts that their product shall be the greatest possible. "What are the parts ? n a Ans. — , and — . 58. Two men have each an end of a pole feet long upon their shoulders, with a burden suspended from it. The share of the load sus- tained by each man is inversely proportional to the distance of the load from his shoulder. The proportion of the weight borne by one man is to that borne by the other as 3 to 2. How far is the burden from the shoulder of the first man ? Ans. T^ feet. 59. Same problem as the last, except the representation of the length of the pole by a, and the proportion by h and c. How far is the bur- den from the shoulder of each man ? ^^ c , (dj Ans. feet, and feet. l> + c b -\- c When will these distances be equal ? "When one double the other ? 00. A general wishing to range his men in a solid square, found that he had too many men by 100. He increased each side of the square by one man, and then found that he had but 39 too many How many men had he ? Ans. 1000. Let J- = side of the square. 61. A general ranged his army in a solid square, and found that he had a more men than would enter the square. He then increased each side of the square by one man, and found that he had h more men than could enter the square. How many men were in each side of the square ? A... X = "-<* + ^l 128 GENERAL PROBLEMS. What does this value become when 6 + 1 = a ? What, when h = — 1 ? How do you expkiin these results ? The last result will confirm a truth hereafter to be demonstrated. 62. Divide the number 20 into three such parts that the half of the first, the one-third of the second, and the one fifth of the third shall be equal. What are the parts? Ans. 4, G, and 10. For, let X represent the equal result after the division of the three parts by 2, 3, and 5; the. parts themselves will plainly be represented by '2x, 3.r, and 5x. Hence, 2a; + 3.7; + bx = 20, or j: = 2. Then, 2a; = 4, Sa- = 6, and 5x = 10. 63. Divide the number a into three such parts that the first divided by h, the second by c, and the third by cl, shall all be equal. What are the parts ? . ah ac , ad ^-- M^TTrf- hTTTd' ^'^'^ h+v^r When will the three parts be equal ? What will be the eifect of making & = ? How do you explain the result? 64. Two laborers are engaged to dig a ditch ; the first can dig it in 5 days, the second in 3i days. How long will it take the two, working together, to dig the ditch ? Ans. 2 days. Verify the result. 65. Two men are employed to dig a ditch ; the first, alone, can dig it in a days, and the second, alone, can dig it in h days. How long will it take the two, laboring together, to perform the work ? , if/, and cd ^ a? What, when h = c? Explain these results, especially the last. C8. The sum of three numbers is 420 : the first, is to the second as G to 7, and the third is just as much greater than the second as the second is greater than the first. What are the numbers ? Ans. 120, 140, and IGO. 69. The sum of three numbers is a : the first is to the second as b to c, and the second exceeds the first as much as the third exceeds the second. What are the parts ? . oc a ^ 2ab — ac ^^"^ HZ • a ^'^^ —36— We observe that, when quantities bear the above relation, the mean is always the third of the whole. What single hypothesis will make all three parts equal ? What is tlie effect upon the three part>s of making c == 2b? Explain this result ? 70. At the Woman's Eights Convention, held at Syracuse, New York, composed of 150 delegates, the old maids, childless-wives, and bedlamites were to each other as the numbers 5, 7, and ?>. How many were there of each class ? Ans. 50, 70, and 30. ELIMINATION BETWEEN SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 207. Simultaneous equations are those which can be satisfied for the same values of the same unknown quantities. Elimination can only be performed upon simultaneous equations. It is a process by which we deduce from two or more equations, containing two or more unknown quantities a single eqiiatiou, containing one unknown quantity. The single equation so found is called the final equation, and since it con- tains but one uiiknown quantity does not lead to arbitrary values. I 130 ELIMINATION BETWEEN SIMULTANEOUS We will first take the simplest case, that of two equations involving two unknown quantities. The elimination of one of these unknown quantities may be effected in four ways. 1. By Comparison. 2. By Addition and Subtraction. 3. By Substitution. ^ 4. By the Greatest Common Divisor. ELIMINATION BY COMPARISON. 208. Take the equations 2y = 2a; + 4, and 3y = 6x — 12. From the first we get y = .x + 2, and from the second, y = 'Zx — 4. Now, since, by hypothesis, the y and x in one equation are equal to the y and .r in the other equation, we can equate the values of y, and make the two x'& represent the same thing. Hence, a; + 2 = 2a; — 4. From which we get a; = 6. Solving the first equation with respect to y X, we get X = y — 2 ; and from the second we get a; = -^ + 2. Now, since the .r's are equal by hypothesis, we have a right to y equate their values. Hence, y — 2 = ^ +2. Assuming that the y's are equal in the two members of this equa- tion, wo have y = S. Hence, the values of x and y are x = 6, and y = 8. It will be seen that these values satisfy both equations. Elimination by comparison between two equations with two unknown quantities consists in solving both equations with reference to one un- known quantity, and equating the values so found. This eliminates the first unknown quantity, and gives a single equation, from which the value of the second can be found. The second unknown quantity being eliminated in like manner, the value of the first can be found. 209. If there had been three unknown quantities, and but two equa- tions, there would have been, after the elimination of y, a single equa- tion with two unknown quantities, and the values of these two un- known quantities would, of course, have been arbitrary. If there had been two equations and but one unknown quantity, the equation could not, in general, be satisfied by the same value for that unknown quan- EQUATIONS OF THE FIRST DEGREE. 131 tity. The equations .r + 2 = a, and x — 2 = a, cannot be satisfied for the same values of x. We see, then, fJiat the nnniher of equations must he preciseli/ equal to the number of unknovni quantities. 210. The equations have been combined under the supposition that they were simui^neous. If they are not so, the hypothesis has been absurd, and the result ought to indicate the absurdity. Take thc' manifestly absurd equations, y = 2./- + 2, and y = 2x — 2. Combining, we get 2x — 2x, or Ox = 4. Hence, x ^ ^ ^ cc. The absurdity of the h}^othcsis is here -oointed out by the symbol of absurdity. 211. If the equations arc the same, or differ only in form, they can, obviously, be satisfied by an indeteriiiinate number of values for one of the unknown quantities, provided that of thc other is deduced after the substitution of the assumed value of the first. Take, y = 2x + 2, and 1/ = 2x + 2. Combining, we get 0^ = 0, or ^ = — . Which indicates that y may have any value whatever. Suppose we assume ^ = 4 ; this value for 1/, substituted in either of the equations, will give x = 1; and the two values, y = 4, and x = 1, will satisfy both equations. Assuming arbitrary values for t/, we get an infinite number of values for x, and the solution becomes wholly indeterminate. ELIMINATION BY ADDITION AND SUBTRACTION. 212. Resume the equations, 2y = 2x + 4, and S>/ = Qx — 12, Multiply the first equation by 3, and thc second in like manner by 2, and we will have 6y = 6x + 12 6y = 12a: — 24 0= Qx — 36. 132 ELIMINATION BETWEEN SIMULTANEOUS If tliGse equations are simultaneous, the 0// in the first equation is equal to the 6y in the second. And, since the x's will also be equal, the result of the subtraction of the equations, member bj member, will be = 6x — 36, or x = 6, the same value as before found. Now, to eliminate x in order to find the value of?/, the first equation must be multiplied by 3, and the second must remain as it is. We will get, Gj/ = 6x + 12, and 3;y = 6x — 12, and by subtraction, Zy = 24, or y = 8. The same value as when we eliminated by comparison. It will be seen that the coefficients of the unknown quantity to be eliminated have bees made equal in the two equations ; and, since they had like signs, the elimination could only be efiectcd by subtraction. Had these coefficients, however, been affected with contrary signs, it would have been necessary to add the equations member by member. Take as an illustration, 2y — x = 3 X — y =: — 1, by addition, x is eliminated. ?/ = 2 To eliminate y, multiply the second equation by 2, and we get 2a; — 2y = — 2, and by addition, x z=.\ The two values are then y =^ 2, and x — 1, and these satisfy both equations. Elimination by Addition and Subtraction consists in making the coefficients of the unknown quantity to be eliminated the same in both equations, and then subti-acting the equations member by member, if these coefficients have like signs, or adding them member by member if they have unlike signs. It will be seen that the method of elimination by addition and sub- traction involves no fractions, whilst the method by comparison, in general, does involve fractions. To eliminate by comparison between the equations 3y = a: + 2, and 2y = a; — 2, X -{- 2 X 2 ■will involve the fractions — ^ — , and — r — . o 2 EQUATIONS OF THE FIRST DEGREE. 133 ELIMINATION BY SUBSTITUTION. 213. This consists in finding the value of one of the unknown quan- tities in one of tlie equations in terms of the other unknown quantity, and substituting the value found in the second equation, so as to deter- mine the valueijf the second unknown quantity in known terms. llesume the equations, and 3y = Gj; — 12. From the first we get .y = x + 2 ; and this value of ?/, substituted in the second equation (since the ^'s and x's are, by hypothesis, the same in both equations), gives 3x + 6 = 6.t — 12. Hence, x = Q. In like manner, x, found from the first equation, x =■ y — 2, substi- tuted in the second, gives 3?/ = 6 (y — 2) — 12, or y = 8. The two values are then x = G, and y = S, as found by the other two methods. This process will also, in general, involve fractions. ELIMINATION BY THE GREATEST COMMON DIVISOR. 214. This process consists in transposing all the terms of the two equations to the first member, and then dividing the polynomials in the first member by each other, as in the method of finding the greatest common divisor, until a remainder is found which contains but one un- known quantity. This remainder, placed equal to zero, constitutes the final equation, and the value of one of the unknown quantities can be deduced from it. The value of the other unknown quantity can be found by a similar process. Resume the equations, 2y = 2x -f 4, and Sj^ = Qx — 12, transposing, 2y — 2x — 4 = 0, and ?>y — Qx + 12 = 0. 12 6y — 12.r 4- 24 2y — 2x — 4 6^ — 6x —12 — Gx — 3G = 3 Quotient = 0, or X = 6. 134 ELIMINATION BETWEEN SIMULTANEOUS To eliminate x, arrange witli reference to x, and we have — Gx + % + 12 I — 2x + 2.y — 4 — 6x + Cy — 12 I ^ 3 Quotient. — 3y + 24 = 0, or ^ = S It only remains to be shown the reason why the remainder is placed equal to zero rather than to 10, or anything else. Let A = represent the first equation after it has been prepared for division. Let B = represent the second equation. Let Q represent the quotient after the division of A by B, then, A |_B_ BQ Q A — QB = 0. Now, since A is zero, and B zero, A — QB is plainly zero. For equations of the first degree there will be but one remainder. But if the equations be of a higher degree th-an the first, there will be two or more remainders. But the same course of reasoning will show that each successive remainder must be zero. For the first remainder having been shown to equal zero, and the divisor B also equal to zero, the second remainder must also be equal to zero. Let B' represent the second remainder, and Q' the quotient resulting from the division of B byB'. Then, B |jy_ Q^B^ Q^ B — Q' B' == 0. Since B =: 0, and B' = 0, plainly B — Q'B' must be zero. The third remainder, the fourth, and so on, can, in like manner, be shown equal to zero. 215. If we combine equations, which are not simultaneous, by either of the first three methods, the absurdity of the hypothesis is shown by cc, the symbol of absurdity in the result. But when we combine such equations by the fourth method, the absurdity appears in the final equation having no unknown quantity. Take the equations ^ = 2x-i-2, and y = 2x + 4. EQUATIONS OF THE FIRST DEGREE. 135 Combining by last process, y — 2x — 1 1 y — 2x — 4 — 2a; — 4 + 2 = 0, which, is absurd. 21G. Though it is usual to say that the absurdity of combining equa- tions which are not simultaneous is shown by the final equation, yet we might retain the trace of one of the unknown quantities, and then we would still have the symbol oo. In the last example, we might retain the trace of y, and the final equation would be 0^ + 2 = ; whence y = a;. Rcmarlis. 217. Of the four methods of elimination, the last is generally used when the degree of the second equation is higher than the first, and the second method (by addition and subtraction) is preferable for simple equations, since it does not involve fractions. Elimination by substitu- tion is generally associated with the other three methods after the value of one unknown quantity i« found ; this value is generally substituted in one of the given equations, and we are thus enabled to deduce that of the other. Take the equations, 2y = 2.r + 4, and Zij = G.c — 12. From the first we have, y = s -f 2, and from the second y = 2x — 4. Equating the two values of y, we get x + 2 = 2x — 4. Hence, x = 6. This value for x, substituted in the first equation, gives 2y = 16, or y =: S. And substituted in the second, gives 3/y = 24, or y = 8. Hence, the given equations are simultaneous. But if the substitution of the value for x in the two equations gave different values for y, we would conclude that the equations were not simultaneous. 218. Examples in elimination heticeen two simple equations o/ the first degree invohiiij two xtnhioicn quantities. 1. Find the values of x and y in the equations, y = ax -{■ h, and y = a'x -\- h'. b — V ^ ah' — a'h Ans. X = ;, and y = r-. 136 ELIMINATION BETWEEN SIMULTANEOUS The hypothesis, a = a', makes both values infinite. The combined equations show that when a = a', a'h must equal ah' ; that is, unequal multiples of the same quantity must be equal, which is absurd. The equations, in fact, represent two straight lines, and the found values of X and y represent their point of meeting. The hypothesis, a = a',^ makes the lines parallel, and their point of meeting is, of course, at an infinite distance. Making a = a' and h = V, x and y both become g, or indetermi- nate. In this case, the two equations become identical, and can be satisfied by any values for x and y, as the symbol g indicates. By this we mean that arbitrary values may be given to either x or y, and these arbitrary values for one of the unknown quantities, taken in connection with the deduced values of the other, will satisfy both equations. When a = a', and b = I', the two lines coincide, and their point of meeting, being any where on the common line, is, of course, inde- terminate. When h = V, we will have x = 0, and y ^=l>. 2. Combine 2^ + 3x = 4, and y-6.x = 7. Ans. X — 3. Combine 2y -f 3:r = 0, y — 6x = 0. Ans. x := 0, and y == 0. When there is no known term or terms in the two equations the values of x and y will always be zero, since these values will satisfy both equations. 4. Combine and 5. Combine and 6. Combine and Am. a; = 8, and y = 30. 2i/ + |— 4^ = 0. 3y + -5- = % Ans. x = 1, and y = 2. f + 1-1 = 0. X + 2y -f 4 = 0. Ans. X = :32|, y = - ■18 J. X ■4 X- +1+ 1 + 1^ = "- 1 + y = 36f . EQUATIONS OF THE FIRST DEGREE. 137 7. Combine 3y — 3x + 6 = 0, and lij — lx^ 14. Ans. y = CO, and x = co, 8. Combine ^^_!^^+2a' = 0. and a'lj "la'x Act! 3 ~ 3 3 ■ .Irts. x= 2, and^ = £. 9. Combine and i + i + 4 = 0. 3 '^ ---—2 = 0. Ans. X = --j^i, and y = |. Regard — , or — , as the quantity to be eliminated. 10. Combine ii 4. ii _4« = 0. y X and =: 2a. y X When one of the unknown quantities is wanting, it may be written with a zero coefficient. 11. Combine y = x + 2, and y = 2,ory = Ox + 2. Ans. X = 0, and y = 2. 12. Combine x = 0, and y = 7x + 4. Ans. X = 0, and ^ = 4. 13. Combine x+y = a and ax + y = b. Ans. x = -, and y = -, a — 1 a — 1 What do these values become when o = 1 ? Why ? What, when h-^a? What;, when a = ? What, when h = a"" ? 14. Combine ■ — + ^ = c. X b and \- y =. a. 12* _ g (5 — 1) _ (c — fQ 5 . «s. a; — ^^_ _—j-, an y _ ]_ _ ^ • 138 ELIMINATION BETWEEN SIMULTANEOUS What do these vahics become when h = 1? What, when c = d? What, wlien c= d, and & = 1 ? What, when & = ? How are the results explained ? 15. Combine — -| = <:•. X y . a' h' , and - — 1 = c X y . oV — a'h ah' — ah ch' — be" ac' ■ — a'c Solve this example by the four methods of elimination. When x is eliminated by the last method, the final equation in y ought to be {ad — a'c) y + a'h — ah' = 0. What do the values of x and y become when a'h = ah' ? Why? What do these values become when hd = cU ? What, when a = a', h = V, and c = d ? 2hc and y = 16. Combine y + |-y + a; = c, and y — x = c. Ans. X = — -T^ ■ 26 + a -^ 26 + a The final equation in x, when y is eliminated, ought to be (26 + a) a; + ac r= 0. What do these values become when c = ? Why ? What do they become when a = ? What, when 6 = 0? 17. Combine -L + y = 2a (1 + a). and y + 2x — 2« = 2«l Ans. X = a, and y = 1o?. Eliminating by substitution, we get y = 2x'^. Hence, the second equation will give 2a;= — 2a^ + 2.r — 2a = 0, or 2 (.r^ — a^) + 2 (x — a) = 0, or 2 (x — a) (x + a + 1) = 0. Dividing out by the factors 2 (a; + a + 1), we have x — a = 0, or x = a. The same re- sult may be obtained by the fourth method of elimination. 18. Combine ^ -f y = 26, and _ + 2?/ = 26 + a. X EQUATIONS OF THE FIRST DEGREE. 139 Combining by fourth method, we have {rj — 2h)x + y I (2y — 2h — a)x + a. Preparing for division by multiplying by (2y-2/.-a) x(y — 2i)(2y — 26 — a) + i/(2y — 26 — a) xly — 2b) (2i/ ~2b — a) + ai/ — 2ba y — 2b. Quotient. 2/ — 2ay — 26y + 26a = 2y (y — a) — 26 (y — a) = 0. Remainder. or, {jj — a) {2y — 26) = 0, the final equation. Divide out by 2y — 26, and we have y — a = 0, ory=:a; and this value for y, substituted for cither of the equations, gives if- = .77 • We might have divided out by y — a, and then we would have had 2y — 26 = 0, ox y = b; and this value for y would liave given x = \. The equations, then, admit of two systems of values, y = a, and ; and y = b, and x = 1. Example 17 gives, also, a 19. Combine and 20. Combine and 26 second system, x = — (a + 1 ), y = 2 (« -f 1/ . ^+- = 2, x y y — x = 0. Ans. X = Q, and y = g. - + J -= 2 X y y — x = 2. Am. x = 2, and y = 4, or .r = — 4, and y = — 2. Combining by substitution, we get cc* + 2x — 8 = 0; or, adding and subtracting unity, x^ + 2x + 1 — 9 = (.f + 1)^ — 3^ = (x + 1 + 3) (X + 1 — 3) = 0. By suppressing the fii-st factor, we get a; = 2 ; and by suppressing the second, we get x = — 4. 21. Combine yx — x = 6, and X — y =z — 2. Ans. X = 2, and ?/ = 4 ; or x = — 3, and y = — 1. In this case, add and subtract J. 140 ELIMINATION BETWEEN SIMULTANEOUS 22. Combine yx — — = 0. y and y -\- X ■= 1. Ans. X = 1, and y = 1 ; or x = 3, and y = — 1. ■ 23. Combine a; + ^ + y .— 2, and x-\-2by = 0. . -46 . 9 .4«s. .T =^ , and y 6 (a + 2)' •^- 1-6 (a + 2)' What do those values become when a = — 2 ? What, when 6=0? 24. Combine x -{- h — a + ^= , 2?/ and 2x -\ ^ = 2a — 26 + 2. a — 6 Ans. X = a — 6; and y ■= a — 6. 25. Combine y = ax -}- 6, and y = 2. 2 - -4ns. X = — — ,andy = 2. "What do these values become when a = ? What, when 6 = 2';' Explain these results. 219. Elimination hdioeen any numher of simidtaneoiis equations. The same principles govern the elimination of any number of simul- taneous equations as have been shown to govern the elimination be- tween two equations with two unknown quantities. No specific rules can be given for elimination, because each equation may contain all the unknown quantities ; or, a part only of them may contain all. It may be even that no equation, or but one, contains all the unknown quan- tities. The main thing to be observed is, to eliminate the same un- known quantity from all the equations that contain it. We will then have one unknown quantity less than before, and one equation less. Continue the process of elimination, until a single equation with a single unknown quantity is obtained. If the number of unknown quantities is greater than the number of equations it will be impossible to obtain a single equation with one unknown quantity, because the number of equations reduced by elimination is always eoual to the number of un- EQUATIONS OF THE FIRST DEGREE. 141 known quantities reduced. Two eliminations reduce the number of unknown quantities and equations by two ; three eliminations by three, &c. It is evident, then, that when the number of unknown quantities exceed the number of equations, the last equation obtained will contain two or more unknown quantities, and will, consequently, be an inde- terminate equation. 220. If the number of equations exceed the number of unknown quantities, it is plain that, before all the equations have been freed from their unknown quantities, we will get a single equation with but one unknown quantity. The value of this unknown quantity then can be determined, and it may not be such as to satisfy all the equations. 221. The number of equations and unknown quantities must not only be equal to each other, but the equations must be different in cha- racter, not inform merely. The equations j/ = 2x -\-2, and 2i/ = 4x -\- 4, differ only in form, and it is impossible to eliminate between them. EXAMPLES. 1. Solve the three equations, 2^ — x-{.z=2, 2i/-\-2x+4z = 8, Bt/ + 13-c + 3z= 19. Ans. 1/ = 1, X =^1, and z=^\. 2. Solve the three equations, y + x + z^^:), y — :c + z=B, y — X — 2 = — 5. Am. y = 2, .r = 3, and 2 = 4. 3. Solve the three equations, y-f-cc-fs = a+ /> + r, y — X -\- z -\-h — c = «, y — 2x — '^z--^a — 2h — 3c. Ans. y z= a, X =hj and z == c, 4. Solve the three equations, x + 2y = 4, x -{- z = a, y — z=l. Am y = 4 — (a + Z-), X r= 2 (a + t) — 4, and 2 = 4 — (a + 26). 112 ELIMINATION BETWEEN SIMULTANEOUS 5. Solve the tlirec equations, X — y z=. a — h, .T + 2 = o, X + y -\-z — c — L Ans. y = h — 2, X — a — 2, and z = S -\- c — (a + Z>). 6. Solve the three equations, y a ^ z c 1 X 2ax + j — 2a + - + y — z = a — c + l. 9/ — 2x + 3z=a + Sc — 2. Ans. y — a, X = 1, and z ~ c. 7. Solve the three equations, Ax +By + Cz + D = 0, A'x + B'y + C'z + D' = 0, A"x + B"y + G"z + D" = 0. , _ (AT'^ — A'V) D + ( A^^C — AC^O ly + (AC^ — AT) D" ^ns. y — ^^,_j^„ _ ^„^,^ ^ _^ ^^„^ _ ^g,,^ ^, _^ ^^^, _ ^,^^ ^,„ _ (B'T^ — BV") D + (BC^ — B'V) D' + (S'G — BCQ D'^ ''^^ ~ (A'B" — A"B') C + (A"B — AB") C + (AB' — A'B) C" _ (A^^B^ — A^B^O D + ( AB^^ — A^^B) J)' + ( A^B — ABQ D^ ^ ~ (A'B" — A"B') C + (A"B — AB") C + (AB' — A'B) C"' What do these become when D, D', and D" are all equal to zero ? Why ? What do they become when either A, A', A", or B, B', B", or C, G', and C" all three are zero ? Why ? What will be . the effect of making D, D', and D" zero, and A = A' = A", B = B' = B", and c .= C = C" ? 8. Solve the equations, X + y - a, X + z = b. y + z = c. a -\- c — b a + b — c b + c — a Ans. y = ~ , X = , z = ^ . What hypothesis will reduce these values to zero ? What will make the first two equal ? What all three equal ? EQUATIONS OF THE FIRST DEGREE. 143 9. Solve the three equations, .X -f- ^ + 2^ — / = 4, x—y — z = 2, 2x-\-2y \z= a, Values indeterminate. Why ? 10. Solve the three equations, X -^^ y — a, X y Ix + cy = 0. Values found from first and second different from those found from second and third. 11. Solve the equations, c c y + X + z~4:(a + b) = —n(a + h), y — X — 2 (a — h) -f a + :: = 2a + h. Arts, y = a — h, x = h — (/, and z ^ a -\- b. 12. Solve the equations, ay + hx = c, y + x = h, y + J—z = a. c — fr ah — r Ans. y = :, X = , and z = b — a. a — b a — b What will be the effect of making h = a? , f^ ^ , . y + X V + z y + t 13. Combine "^—^ + '^~- + '^^-r— = 3. 4 X + z x+J _^ z_+2 ^ 3 5 G 7 X — y + t — z = 2, x + y + l + t — z = 5. Ans. y = 1, X = 2, z = S, and < = 4. 14. Combine xy = 00, .r+^ = 29. A71S. X = 0,y = 10, and z — 20. 144 ELIMINATION BETWEKN SIMULTANEOUS 15. Combine ^ + ^ + 4 f 4" = 3G. o 2 4 o Ans. .-c = 18, y = 20, z == 40, and t = 60. 16. Combine xy = QO, X + z = 29. a; — - = 0. y Equations absurd. Why? 17. Combine a; = 2z + 4, y = 3;s — 6, a; = 2s — 2. Ans. X = ex, 1/ = ac, and z = ao. 18. Combine ' x + ^ + z + t = 4, x + ^ — z — t = 0, I z + t — X — v/ = 0, 7/ + Z — X~f=0, y — z -{- X — t + w = 1. Ans. X = 1, y =1, z =1, t = 1, and to = 1. 19. Combine x + y + z — 4, 2x + 2y + 2a = 5 i/ — z = 0. Ans. X = cc, y = cc, and z = oc. Equations evidently not simultaneous. Tlie combination, tlien, is absurd, and the result shows the absurdity. 20. Combine x + y + z = S, x — i/ — z= 1. 3a; + 3y + 32 = 18. Ans. X = oo, 1/ = cc, and z = cc. The first and third equations plainly conflict Hence, the equations are not simultaneous. EQUATIONS OF THE FIRST DEGREE. 145 The first and second combined give x z=2', and this value substi- tuted in the second equation, gives y + 2; = 1. The values of x and y + z, substituted in the third equation, give 9 =rr 18. But we may have the absurdity shown by its appropriate symbol by combining the first and third equations, and retaining the trace of one of the unknown quantities. Thus, we may have Ox = 3, or a; = 00. And so, likewise, y and z may be shown to be infinite. 21. Combine the equations, — a; + y + :; + ; — t + a = (a — i) (c + (Z + e), x—y — z + t = a — h + (h — a) (c + d — c), x+ ^ -Y-. +-=l(a — h), c d e ' X y z t + -71 r. +— -. a: = 4 + i — a. a — b c (rt — //) (/ (« — b) e (a — h) Am. X = (i — b, y =z{a — by, z = (a — b^d, t = (a — h)e. When will all the values be zero? When three? When the last, only? 22. Combine the equations, 2^-3. ==1, 2y + Sz = r, y + z = S, ylns. ^ ^ 2, and ,~ =: 1. IIow does it happen that true solutions are found for four equations, involving but two unknown quantities ? 23. Combine x+y + z + t-{-u=z 200, 2 + 3 +1 + t "5 + ^=50, x + y- -. = 10, x + z — / = 10, x + t — u = 10. A7is. a- = 20, y-30. Z : = 40, < = 50, and u = = 60. 24. Combine the equations X + y + z + t + 11 -. = 200, X y j + i + i-i 4- f = 50, X +y — z = 10, X+Z~t: = 10, X -{-t — u = 10, 13 K 14G ELIMINATION BETWEEN SIMULTANEOUS (A) y + .- — « = 10, (B) ?/ + < — t( — 20. Ans. Same as for the la.st equation. In tlie last two examples the new equations, marked (A) and (B), were not incompatible with the others, and, therefore, the solutions found, from taking as many equations as unknown quantities, satisfied the additional equations. When, however, the number of equations exceed the nuiubcr of unknown quantities, and the surplus equations are not satisfied by the same values as the other equations, the solu- tions will be contradictory. General Re.marls. 2>22,. If we have m equations involving m unknown quantities, we can find the value of one of these unknown quantities in terms of the others, and this value substituted in all the other equations will give us m — 1, new equations, involving m — 1 unknown quantities. We can now take one of these m — 1 equations, and find the value of a second unknown quantity in terms of the remaining m — 2 unknown quantities, and thus get m —.2 new equations, involving m — 2 un- known quantities. And, by continuing this process, it is plain that, after m — 1 eliminations, we would get a single equation, invohing a single unknown quantity, and, therefore, would be able to find the value of that unknown quantity. But, if we have m equations, involv- ing m -\-h unknown quantities, we can only make m — 1 svibstitutions, and then we will have a single equation, involving h -\- 1 unknown quantities. If Z* = 1, the final equation will contain two unknown quantities, and bo of the form x + i/ = 10, an indeterminate equation, in which the value of x can only be found by attributing arbitrary values to I/. HI) z= 2, the final equation v,'ill contain three unknown quantities, and be of the form of cc -f y + s = 10 ; in which x can only be determined by giving arbitrary values both to y and z. It is plain, then, that when the number of unknown quantities exceed the number of equations, the elimination will lead to indetermination. 223. If, on the contrary, we have m -\- h equations, involving m unknown quantities, m of these equations are suificient to determine the m unknown quantities. The remaining h equations might, or might not, be satisfied by the values found from the m equations. If the I equations are satisfied, they are not independent equations ; if they are not satisfied, they are contradictory equations. Thus, x = 2, and EQUATIONS OF THE FIRST DEGREE. 147 2xz=4: are satisfied by the same value of x ; but the second equation is not an independent equation, since it differs only in form from the first, a; = 2, and a; = 3 are contradictory equations. We conclude, then, in general, that the number of independent equations must be precisely equal to the number of unknown quantities. 224. An artifice sometimes enables us to eliminate between a num- ber of equations more readily than by the usual direct process. As an illustration take the equations, X + y = 5, X -{- z = Q, z +y = 7. Let s = x -{■ 7/ -{- ::. The three equations then become s — z = b s — X = 7 Adding, we get 3s — (x -|- y + ~) = 18. But, since x + 1/ -\- z = s, we have 2s := 18, or s = d. This, substituted in equations marked A, gives 2i= 4, y = 3, and x = 2. Take, as a second illustration, xi/ -\- xz = 27, 7/z-{-y + z = 29, xyz = GO. 27 By factoring and solving the first equation, we get y + 2 = — . 27 This, substituted in the second, gives ys + — =: 29. But from the third equation we find yz = — . Hence, f- — = 29, from which 27 x = ^. Then, y + ^ = — = 9, and xyz = GO, or yz = 20. Elimi- nating, we get z- — dz + 20 == 0, or s« — 10^ + 25 + s — 5 = 0, or (z — by + z — b = 0, or (s — 5) (2 — 5 + 1) = 0. Dividing out the second factor, we have left z — 5 = 0, or z=zb, from which y = 4. Take another illustration, xy + xz = \2 (1), y^-f^=rl08 (2), s — X = 8 (3). 148 ELIMINATION BETWEEN SIBIULTANEOUS Subtracting (1) from (2), we get y {z — j-) -\- z {z — x) == 96, or (z — X) {y + 2:)=r96 (4). Dividing (4) by (3), member by mem- ber, we got y -\- z = 12 (5). Equation (2) may be put under tlie form of z (?/ + z) = 108. Dividing this, member by member, by (5), we will have 2 = 9. Equation (1), put under tlie form of x (y -^ z) =z 12, divided by (5), gives x = 1. The value of z substituted in (5), gives 2/ = 3- Hence, x = 1, ?/ = 3, and z = 9. PROBLEMS PRODUCING SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 225. Many of the problems already given (Arts. 176 and 206) could have been solved as readily, or more readily, with two unknown quan- tities ; and there are many problems wliich can only be solved by the use of two or more unknown quantities. We will solve three of the problems already given, to show the man- ner of using two unknown quantities. 1. The sum of two numbers is a, and their difference I, what are the numbers ? Let X = greater number, and y = smaller. Then, by the condi- tion of the problem, x + y = a, and x — y =zlj. Eliminating y by adding the two equations member by member, we^ get a; = — -|- — . Eliminating x by subtracting the equations member by member, we have ?/ = — -. So, we see that when we know the sum and diffe- 2 2 rence of two quantities, we get the greater by adding the half difference to the half sum, and the less by subtracting the half difference from the half sum. Thus, the sum of two numbers is 20, and their difference 10 ; the greater is, by the formula, 15, and the smaller 5. 2. A fox is 125 of his own leaps ahead of a greyhound, and makes 6 leaps to the greyhound's 5 ; but two leaps of the greyhound are equivalent to 3 leaps of the fox. How many leaps will the fox take before he is overtaken by the greyhound ? Let X = distance passed over by the fox, counted in terms of his own leaps. Let y = distance passed by the greyhound. Then these distances would be proportional to the relative number of leaps taken by the fox and gi'cyhound, if their leaps were equal in length. And EQUATIONS OF THE FIRST DEGREE. 149 we would have x, y : : 6 : 5. But since each leap of the greyhound is equivalent to | leaps of the fox, the 5 leaps of the greyhound are equi- valent to 5 . I = ij^ leaps of the fox. Hence, x : y : : Q : L^, and the distance, y, passed by the greyhound, will be expressed in terms 5x of the leaps of the fox. From the proportion we get y = —, and from the condition of the problem, y — x = 125. Combining, we get X = 500, and y = 625 leaps of the fox. It will be readily seen that this solution has some advantages over that with one unknown quantity. 3. Two couriers start from different points on the same road, and travel in the same direction. The forward courier travels at the rate of b miles per hour, and the rear courier at the rate of a miles per hour. They were separated by a distance of m miles at starting. How long will it be until they come together ? Let X = distance travelled by forward courier, y = distance tra- velled by the rear courier. Then, since the distances travelled must be proportional to the rates of travel, we will have x : y : : h : a, or by = ax. From the conditions of the problem, we have y — xz=m; combining, we get x = r, and y = -,. These di.stances, di- u — (/ u — o vided by the rate of travel, will, of course, give the time elapsed before their junction. Both expressions give for the time, as found, a — b when one unknown quantity was used. This solution, compared with the preceding, shows how much the use of two unknown quantities has shortened the work. 4. A carpenter wishes to saw a piece of timber, 20 feet long, into two parts, so that one of them shall be but two-thirds as loii"- as the other. "Where must he place his saw ? Ans. At 12 feet, or 8 feet from the end. 5. A carpenter wishes to saw a piece of timber 7n feet long, into two such parts, that one shall be the — part of the other. Where must he place the saw ? . . ^ b)}i am Alls. At -— — feet, or feet from the end. b + a b + a What values must b and a have to make the solutions in examples 4 and 5 the same ? When will the parts be equal ? 13* 150 ELIMINATION BETWEEN SIMULTANEOUS 6. A colonel wishes to divide his regiment of 800 men, and 10 com- panies, in such a manner that the two flank companies shall contain each one-third more men than each of the central companies. What must be the number of men in each flank company, and in each cen- tral company ? ^??s. 100 men in each flank company, and 75 in each central com- pany. 7. A colonel wishes to divide his regiment, composed of a men, and b companies, in such a manner that the two flank companies shall each contain — more men than each central company. What must be the composition of the flank and central companies ? Ans. Central companies, each -^ ; flank companies, each-v- 7^ ■ be + Z be + 2 What values must a, b, and c have to make this solution the same as the last ? What do the expressions become when c = ? What do the central companies become in that case ? 8. A man becoming insolvent, leaves $4000 to his two creditors : to one of whom he owes $8000, and to the other $6000. What share ought each to have out of the $4000 ? Ans. The first $2285|, the second |;i714f . 9. A man becoming insolvent, leaves a dollars to his two creditors. To one he owes b dollars, and to the other c dollars. What share ought each to have ? . ab ^ ac A71S. X = , and V == . When will these values becomes equal ? When the first double of the second ? What do they become when he leaves nothing ? What, when he leaves enough to pay his creditors ? 10. A planter hired a negro-man at the rate of $100 per annum, and his clothing. At the end of 8 months the master of the slave took him home, and received $75 in cash, and no clothing. What was the clothing valued at ? Ans. $12 J. Verify this result. 11. A planter hired a negro-man at the rate of a dollars for c months, and was also to give the negro a year's supply of clothing. At the end of b months the negro was taken away, and the planter paid m dollars in cash, and gave no clothing. What was the clothing valued at ? . cm — ab Ans. y = r . EQUATIONS OF THE FIRST DEGREE. 151 When will 2/ = ? When -will it be negative ? Wlieu iufiuite ? The zero solution can be explained most satisfactorily by placing cm = ab under the form — =— . Let the pupil make a problem to explain the negative solution. 12. A planter has 500 acres of cultivated land, which he wishes to plant in such a manner that he may have twice as much cotton as corn, and three times as much corn as small grain. What division of his land must he make ? Ans. 50 acres of small grain; 150 acres of corn; and 300 acres of cotton. 13. A planter has m acres of land and wishes to cultivate it all so that he may have a times as much cotton as corn, and b times as much corn as small grain. How much of each kind must he have ? Ans. z = q , J acres of small grain, y = z^ , — ; — j acres 01 1 -i- b + ab "= ' -^ 1 + b + ab corn, and x = , 5 r acres of cotton. ' 1 +b + ab What suppositions upon a, b, and vi will make this solution the same as the last ? What two suppositions will make the three divisions of land equal ? What will be the effect of increasing b upon the values of z, y, and x? What of increasing or decreasing m? Why does m enter into the three numerators ? 14. In the year 1G92, the people of Massachusetts executed, impri- soned, or privately persecuted 409 persons, of both sexes, and all ages, for the alleged crime of witchcraft. Of these, twice as many were pri- vately persecuted as were imprisoned, and 7j| times as many more were imprisoned than were executed. Ile(][uircd the number of suffer- ers of each kind ? Ans. 19 executed, 150 imprisoned, and 300 privately persecuted. 15. A planter has $2500 to expend in the purchase of 30 head of horses and mules. He wishes his horses all to be equal in value, and his mviles all to be equal in value, but each mule to be one-fourth less valuable than each horse, and the number of mules to be twice as great as the number of horses. What must be the price of each horse, and of each mule ? Ans. Each horse $100, each mule $75. 16. A planter has a dollars to expend in the purchase of b head of horses and umles. He wishes to have c times as manv mules as horses, 152 ELIMINATION BETWEEN SIMULTANEOUS but cacli nnilc to be — times less valuable than each horse. What a must be the priee of each mule, and of eacli horse ? Ans. Each horse , ^ , — , — —- : each mule ~ — , ^ -^. b {d -f- cd — c) ' b (jci + cd — c) What values must be given to a, b, c, and d to make this solution the same as the last ? IIow can the solution be verified ? What will be the effect of making (Z = 1 ? Why ? What is the effect of making c = 1. The expressions for the entire cost of the horses and mules and -3—^ — - — -- . What is the effect of making f = d + cd — c d + cd upon these values ? What of making a = ? How must the expres- sions for the price of eachi liorse and mule be written to show that the former decreases with the increase of d, and that the latter increases with the increase of d ? 17. The sum of two digits is 6, and the second digit is double the first. What is the number made up of these two digits ? Ans. 24. The digits are the individual figures making up a number. In this example 2 is the first digit, and 4 the second. 18. The sum of two digits is a, and the second digit is b times greater than the first. What are the digits, and what is the number ? , a ah ^ 10a -{- ab Ans. X = — -^, 2/ = ^^1 ; number = -y^^^- The number 24 is made up of the digits 2 and 4, the number 42 of those digits in reverse order. To express that, 18 added to the first number would reverse the digits; we represent by x and y the digits in the first number. Then, lOx + y + 18 =zlGy -\- x. 19. A number is made up of two digits, and the first is double the second. If 27 be taken from the number, the digits will be reversed. What is the number? Ans. 63. lOx + y — 27 = IQy + X, and x = 2y. 20. A number is made up of two digits, and the first is a times as great as the second. And if b be taken from the number, the digits will be reversed. What is the number, and what are the digits ? Ans. First digit ^'^^ ; second g^^^- Number, f^^, (10a + I) b EQUATIONS OF THE FIRST DEGREE. 153 What will these results become when a = 1 ? Does the equation of the problem indicate the absurdity '( What will the number become when i = a — 1 ? What, when a <^\'i How is the negative solu- tion explained ? 21. A number is made up of three digits, whose sum is equal to 10. The first digit is double the second, and the second triple the third. What is the number ? Ans. G31. 22. A number is made up of three digits, whose sum is equal to a. The first is h times as great as the second, and the second c times aa great as the third. AVhat are the digits, and what is the number? , a ac acb _ 100oc6 + lOac + a he -\-c-\-\ ■ What will be the effect of increasing c upou the three digits ? What of increasing h ? What upon the digits and number of making 6 = 0? 23. A number is made up of three digits, whose sum is equal to 10. The first digit is double the second; and if 495 be taken from the number, the digits will be reversed. What is the number ? Ans. 631, 24. The sum of the three digits which make up a number is equal to m. The first digit is a times as great as the second, and if h be taken from the number the digits will be reversed. What are the digits, aqd what is the number ? mam — 6 (a -f 1) 99ni + h (99m -\- h) a (2a + 1) 99 ' y~ (2a + 1) 99' (2a + 1) 99' 100a (90??i -f h) + 10 (99m + J) + 99am — b(a + 1) and number = — ^ ^2, + 1) 99 25. A gentleman in Piichmond expressed a willingness to liberate his slave, valued at SlOOO, upon the receipt of that sum from charitable persons. He received contributions from 24 persons; and of these What was the entire amount given by the latter ? A71S. $50 by the former; 6950 by the latter. 26. If 7i be taken from the numerator and denominator of a certain 15i ELIMINATION BETWEEN SIMULTANEOUS fraction, its value will be doubled; but, if 6f be taken from the nume- rator and denominator, its value will be trebled. What is the fraction? Ans. |. Wbat fraction is that from wbich, if a be taken from both its terms, the value will be doubled ; and if b be taken from both its terms, the value will be trebled ? 2a — h ah Ans. 4a — 06' 4a — ob ah 2a — b The first result gives the reduced fraction, the second (which is identical with it) the fraction in which the substitution must be made. We will illustrate by a problem. 27. Find a fraction, such, that if 5 be taken from both its terms, the value of the fraction will be doubled ; but, if 4 be taken from both its terms, the value will be trebled. Ans. f,or^ 9" The subtraction of 5 from the numerator and denominator of the second fraction will give |, and the subtraction of 4, in like manner, will give |. But these results could not be obtained by operating on the first fraction. The reason of the difference is obvious. 28. If A and C can do a piece of work in 3 days, B and C together in 7 days, and A and B together in 3|, in what time can each per- son do the work alone ? Ans. A in 4i days; B in 21 days, and C in lOJ. 29. A and C can do a piece of work in a days, B and C can do the same in b days, and A and B the same in c days. In how many days can each one, alone, do the work ? Ans. A in - — ■ days, B in • --^— days, C in be + a {b — c) -^ ' ac -\- b (^a — c) "^ ' 2^7?;^ be — a (6 — (•) What do these values become when a =zc? What, when b — r? What, when a = b = c? Suppose a {b — f ) ]> be, will c be a co- operator, or a draw-back ? EQUATIONS OF THE FIRST DEGREE. 155 30. If A can do a piece of work in 4i days, B in 21 days, and C in 10^ days, how long will it take them all, working together, to do it ? A71S. X = 21 days. Solved by a single equation, 31. If A can do a piece of work in a days, B in Z» days, C in c days, how long will it take them all, working together, to perform it ? . ale Ans. X = — days. ab + ac + be "^ If & = 0, then, X ^0, an absurd result. But, by going back to the equation of the problem, one of the terms is infinite, aud the ab- surdity appears under its appropriate symbol. 32. A farmer has a piece of land, worth $800, and two negroes. The first negro and land together are worth three times as much as the second negro, and the second negro and land together arc worth just as much as the first negro. What is the worth of the negroes ? Alls. First, 81G00 ; second, $800. 33. A former has a tract of land, worth a dollars, and two slaves. The first slave and the land together are worth b times as much as tho second slave, and the second slave and land together are just equal in value to the first slave. What is the value of the slaves ? Ans. First, — ^^ — dollars; second, " dollars. o — 1 b — 1 What do these values become when b = 1? 34. A has a number of five and three-cent pieces in his pocket ; B wishes to get 24 of them, and gives A one dollar. How many pieces of each kind must he get ? Alls. 14 five-cent pieces, aud 10 three-cent pieces. 35. A has two kinds of pieces of money in his pocket ; the first worth a cents each ; and the second b cents each. B wishes to get m of them, and gives A c cents. How many pieces of each kind must he get ? . am — e , , . , , <" — bm ^ , . , Ans. X = 5- second kind, and y = first kind. a — b -^ a — b Suppose a = b, aud explain the absurdity of the solution. What is the meaning of the solution when am = c? What, when am^<^ c? What, when bm ^ c ? 36. A has two kinds of money. It takes 8 pieces of the first kind, 156 ELIMINATION BETWEEN SIMULTANEOUS and 83i pieces of the second kind to be worth a dollar. 3? offers him a dollar for 27 pieces. How many pieces of each kind must he get? Ans. 2 of the first, and 25 of the second kind. 37. A has two kinds of money. It takes a pieces of the first kind, and b pieces of the second kind to make a dollar. A dollar is offered for c pieces. How many of each kind must be given ? Ans. — r of the first kind, and — y of the second kind. b — a b — a Explain the solution when b = a. When b = c. When a = 0, or c. 38. A certain person has a certain sum of money, which he jjlaced out at a certain interest. A second person has a less sum by $16661, which he puts out at one per cent, more interest than the first got, and receives the same income as the first. A third person has a less capi- tal than the first by $2857^, but invests it two per cent, more advan- tageously, and also receives the same income. What are the three sums at interest, and what the respective rates of interest of each ? Capitals, $10,000, $8333^, and $7142f . Rates of interest, 5 per cent., 6 per cent., and 7 per cent. 39. A gentleman invests a certain capital at a certain rate of inte- rest. A second gentleman has a less capital by a dollars, but, by in- vesting it at one per cent, more advantageously he derives as much income as the first. A third gentleman has a less capital than the first by b dollars, but, by investing it at two per cent, more advantageously, he also receives the same income as the first. Required the three capi- tals, and the three rates of interest. Ans. Capitals, $- Tj^—^i r~)^-r, r"- ^ ^ 2a — b Za — b la — b 2 (b — a) h ^ 2a Rates of mtercst, — r^ ^, ^ r, and =-. ' 2a — b 2a — b La — b Verify these results. Discuss them when b = a. When b = 2a, When b = 0. When a = 0, &c. One hundred parts of gunpowder are composed of the following ma- terials in the following proportions : For -war. For hunting. For mining. Nitre 75 78 65 Charcoal 12 1 12 15 Sulphur 12,^ 10 20 100 100 100 EQUATIONS OF THE FIRST DEGREE. 157 40. At the beginning of the Mexican war, the proprietor of the Du- pont JNIills wished to work up the materials of his powder for hunting and mining, and make war powder out of it. He removed the sulphur by sublimation, and then wished to ascertain what proportion to take of the remaining charcoal and nitre in the two specimens. What pro- portion ought he to have taken ? Ans. The proportion of the hunting materials to that of the mining, as i|g is to 2^5 ; or as 125 is to 30 ; or as 25 to 6. Then, calling 25x the amount of hunting material in the nitre, we "5 X 75 have 25x + ^-^^ = "5, or x = |-|. There ought then to be ' — G X 75 parts of the hunting material, and — -j— parts of the mining material to give 75 parts of nitre in the war mixture. A similar relation can be obtained for the proportion of charcoal. 41. The sura of four numbers is 107. The first, increased by 8, the second, increased by 4, the third, divided by 2, and the fourth, multi- plied by 4, will all give equal results. "What are the numbers? Ah!^. 20, 24, 50, and 7. 42. Tlie sum of 4 numbers is a. The first, increa.sed by b, the second, increased by c, the third, divided by (f, and the fourth, mul- tiplied by/, will all give equal results. AVhat are the numbers? (a _ c) /— (1 -f //) h (c + a)/+ (.//--f l)c — (l+/0?> ^"'- '\2+d)/+l ' (2-1- J)/+l \2/i + (a-c^/ \d 2h+a-c (2 + c/)/+l ' ^(2-|-cO/+r Verify these results by addition. Show that their sum is equal to a. Verify them by adding h to the first, c to the second, multiplying the fourth by/, and dividing the third by J. What single supposition will make tlie first and second equal to each other ? What single supposi- tion will make the third and fourth equal ? What will make the first part zero ? What effect will this hypothesis have upon the other parts ? 43. Four persons owe a certain sum of money: of which the first is to pay one-third, the second one-fourth, the third one-fifth, and the fourth one-sixth. After paying a portion of the money, there is still a deficiency of $36. What portion of it has each to pay ? Ans. The first, Sl2|of ; the second, $dj%\; the third, 67yVT; ^^^ fourth, SG/,\. Let 60.r = the proportion of the first. 14 158 EQUATIONS OF THE FIRST DEGREE. •i4. Four persons owe a debt of a dollars : of wliicli the first is to pay the — tli part; the second, tlic — th part; the third, the — th part; and the fourth, the — th part. What has each to pay? Ans. The first, -^-^ — , ,^ — , — ; r- : the second, ' cd {c -\-h) + be (e + d)' ' aide ^, ^, . , ahce the third, — ^— — — — ; the ed (c + 6) + he (e + (/) ' ' ed (c + 6) + be {e + d)' fourth, (c + 6) + be (e + d) What hypothesis will make the first two results equal ? What the second two ? What all four ? What will be the effect of making either h, c, d, or e equal to zero. How is this explained ? 45. The denominator of one fraction is 4, and of a second fraction, 8 ; and the numerator of the second fraction is 4 times as great as the numer^^or of the first. The two fractions and their greatest common divisor, added together, are equal to 3. What are the numbers ? Ans. i, igi, and |. 46. The denominator of one fraction is h, and that of a second frac- tion is c; the numerator of the second fraction is m times greater than that of the first, and the sum of the two fractions is equal to the least common multiple of their denominators. What are the fractions ? Ans. -, and c + mb c + mb b ~c 47. A 1000 cubic inches of bronze were found to weigh 5100 ounces. A cubic inch of copper weighs b\ ounces, and a cubic inch of tin weighs 4i ounces. What was the proportion of copper and tin in the composition of bronze ? Ans. The copper to the tin as 85 to 15. 48. Some inspectors of cannon weighed in cubic inches of bronze, and found the weight to be to ounces. A cubic inch of copper weighs h ounces, and a cubic inch of tin weighs c ounces. How much copper, and how much tin was in the composition ? to — mc „ ^mb — w . Ans. — ounces of copper, and — ounces oi tin. b — c b — c What will these values become when b z= c ? Going back to the VANISHING FRACTIONS. 159 equation of the problem, -what will b =^c show in regard to tv and mb ? Suppose c = 0, what will the results show ? Suppose ^o =i 0, what will both solutions become ? VANISHING FRACTIONS. The symbol g has been interpreted to sitrnify iudctermiiiation, and this is the true interpretation for solutions of equations of the first degree, when the symbol proceeds from two suppositions, made either upon the values found, or upon the equation of the problem. But the symbol may arise from a single hypothesis, and then it always indicates, not indctermination, but the existence of a common factor. x-a y^ Take the expression, —, which becomes g whcn=j: — y. But, X y by factoring the numerator, we have — = ^-'-A- — -^'2 __ ^ ^ x — y x — y X + y ^= 2y, when x =zy. The true value of g in the present in- stance is 2y, as shown by removing the common factor, x — ?/. Again, . X — y , T-, ■'■ — V take the expression -, = — when x = y. But —, ~ = ^ .(,' — if "^ x^ — y^ - — ^7 — ; = = -r when x := ?/. And the true value of (x — y){x-i-y) x-^y -ly the vanishing fraction is again shown to be finite. But, take -^- —. (•*: — y) = J- when x = y. Factoring, we get -, — ^ . = = — , or Qo when x = y. And the true value of the vanishing fraction in the (x yf present instance is infinity. Take again, ^— = — when x = y. Factoring, we get — — — — ^^^ — z=i x — y = when x = y. So, we see, that when there is a common factor existing between the nu- merator and denominator of a fraction, which factor has become zero^ the fraction may have either of three values, finity, infinity, or zero. To show it more generally, take the expression -~ = — when Q {x — (/)" x = a. There may be three cases, m may be = n, <^ n, or ^ n. 160 VANISHING FRACTIONS. p (x a")™ In the first case, when m==in, the fraction becomes j—^^ r— = Q (x — a)™ P — , a finite quantity. In the second case, divided by (.c — o)"", the P fraction becomes /r- ^ , which is infinite, when x =z a. In the third case, when m ^ n, dividing by (x — o)", the fraction becomes Y (x a)""" — ^^ — = when X = a. And as we have taken a general ex- Q pression, we conclude, in general, that a vanishing fraction is one which assumes the form of g, in consequence of the existence of a common factor, which has become zero by a particular hypothesis, and that the true value of the fraction may be either finite, infinite, or zero. 227. When the common factor is apparent, we have only to strike it out before making our hypothesis, and we get at once the true value of the fraction. But there are many expressions, in which it is diffi- cult to detect the common factor, and it then becomes necessary to know a process by which the common factor may be discovered. We will illustrate the process by a simple example, in which the common X^ ft2 factor is apparent. Take the expression, =: — when x = a. X — a Here, the assumed value of x is a. If, however, we make x = a -{■ h, and substitute this value for x in both terms of the fraction, reduce the result to its lowest form, and then make h = 0, it is evident that we will have done nothing more than attribute to x the value a. Making ,1 , ,., ,. , «' + 2ah + li' — a" 2ali + A' „ the substitution, we get ; = r i=z2a -{-Ji^ 2 - is then 2a, when X =za. As the same process is plainly applicable to all fractions in which the terms are affected with iiumerical exponents, we derive for such fractions the general RULE. Adrihufe to (hat term vpon which the hypothesis is made the value V)hich reduces the fraction to the form of ^ plus an increment h, reduce the result to its lowest form and then make h = 0. VANISHING TRACTIONS. 161 In the above example, x is tlie term upon which the hypothesis is made, and a the value which reduces the fraction to the form of g Hence, by the rule, x =:a + h. 228. — EXAMPLES. 1. Find the value of ■ = — when x = a. X — a Ans. 3al 2. Iind the value of —. = — when x = 1. x' — X — ax -\- a Ans. -. a — 1 o I?- wi, 1 pX^ -\-hx + X + h 6. lind the value oi —. = — when x = — 1. x' — ax -\- X — a Ans. 1 +a 4. Find the value of '^ ^ '. '"" ' ' '^^^ = ^ when x z= m. Ans. X- + nx' — mx — nm- f) x^ -\- ax — mx — am m (1 + 2«) a -\- 111 ;; r- ^ .1 i . x^ — \x^ — xif ■\- h/^ . 5. Find the value of ■ ^ ^ ^ = — when x = y. 6. Find the value of ^ -. —^ := — when x=.\. x^ — x^ — X + 1 Ans. CO. 7. Find the value of the fraction — ^ '- — =: — when x' — ox — ax -\- ab X = h. Ans. X = b -{■ a. 8. Find the value of the fraction '— '—— = — when X — bx — ax -{- ab x=za. Ans. 2a. 14* L 162 YANISIIlNCi FRACTIONS. The two last results arise from the fact of the given fraction being a double vanishing fraction. It can be put under the form of ^ , ^ ^' , , which is a vanishing fraction, when either o: = h, (x — b) (x — a) or a; = n. 229. It is obvious, in all these examples, that x, minus the value of X, which makes the fraction assume the form of ^, is the common fac- tor. If we, then, divide both terms of the fraction by this common factor, and then attribute the appropriate value to x, we will have the true value of the vanishing fraction. But there are many expressions for which this rule fails, as will be seen more fully hereafter. Since it is often difficult, if not impossible, without the aid of the differential calculus, to ascertain the existence of a common factor, if there be one, it becomes important to have a simple test by which we can tell whether £ indicates indctcrmination or a vanishing fraction. Take the fraction tt— ^^ ^, which becomes [{ by the single hypo- Q(.c — a)"' thesis x = a. It is evident that the expression is a vanishing fraction, and that the common factor is some power of x — a. But take the 'P (x a)"' fraction —j r^-, which becomes Q by the double hypothesis x = a, vj (X Uj and x = h. It is plain that the expression cannot be a single vanish- ing fraction like those exhibited in the first six examples ; and if it is a double vanishing fraction of the form exhibited in Examples 7 and 8, it cannot be a true solution to a problem of the first degree, since x cannot have two values. We then conclude, that when g arises from a single hypothesis upon a solution of an equation of the first degree, it indicates the existence of a common factor. But, if it arises from two suppositions, it indicates indetermination. In conformity to this rule, we have interpreted §, in the problem of the couriers, to indicate indetermination, because the symbol proceeded from the double hypo- thesis, VI =: 0, and a :=b. POWERS AND EXTRACTION OF ROOTS. 163 FORMATION OF THE POWERS AND EXTRACTION OF ROOTS. 230. The power of a quantity is the result obtained by multiplying it by itself any number of times. Any quantity is the first power of itself. If a quantity be multiplied by itself once, or enters twice as a factor iu the result, the result is called the second power of the quantity. Thus, 2.2= 2!^, or 4, and a . a =a^, are the second powers of 2 and a. If a quantity be multii^lied by itself twice, three times, four times, &c., or enters into the result as a factor three times, four times, &c., the result is called the third power, the fourth power, &c., of the quan- tity. In general, the number of multiplications of the quantity by itself is one less than tht^quantity which designates the power, and the number of times that the quantity enters as a factor in the result, is precisely equal to that quantity. The quantity which designates the power is called the exponent of the power, and is written a little above and to the right of the given quantity. Thus, 2* = 4 is the second power or square of 2. 2' = 8 is the third power or cube of 2. 2* = 16 is the fourth power of 2. ay is the ?/ power of a, and indicates that a has been multiplied by itself y — 1 times, or that a enters as a factor y times in the expression o>. When no exponent is written, the first power is always understood. Thus, 2 = 2', and a + i = (a + i)'. The quantity to be raised to a power may be expressed numerically, or by letters, and may be entire or fractional, positive or negative. And, since the power in every case is a product, we may define the formation of a power, to consist in finding the product arising from multiplying the quantity, by itself, a number of times one less then that indicated by the exponent of the power. The poxcer differs from an ordinart/ prodxtcf, then, in this essential ■particular, all the factors of the power are equal. 1G4 FORMATION OF THE POWERS AND 231. The root of a quantity is tliat quantity which, multiplied by itself a certain number of times, will produce the given quantity. When a quantity, multiplied by itself once, or taken as a factor twice, gives the given quantity; it is called the square root of the given quantity. Thus, 2 is the square root of 4, because 2.2 = 4; and a is the square root of a^, because a . a ^= a^. Raising quantities to powers is called Involution. Extracting the roots of quantities is called Evolution. Involution deals in equal factors. Evolution finds one of those equal factors. t 232. Involution is a simple process. Evolution is more difiicult, and requires particular explanation. "We will begin with the simplest form of evolution, the extraction of the square root of whole numbers, which is nothing more than evolving one of the equal factors out of the product of two equal factors. It is evident that evolution is the reverse of involution, and that we cannot extract any root without knowing how the powers of that root are formed. To demonstrate the rule, then, for the extraction of the square root of whole numbers, we must first examine and see how the square power of whole numbers is formed. 233. The first ten numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. And their squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. Reciprocally, the numbers of the first line are the square roots of the corresponding numbers in the second line. We see, also, that the square of any number below 10 is expressed by not more than two figures. That is, the square of units cannot give a higher denomina- tion than tens. So, likewise, it may be shown that the square of tens cannot give a higher denomination than thousands, since the square of 99 is 9801. The numbers, 1, 4, 9, 16, &c., and all the other numbers produced by the multiplication of a number by itself, are called pcr/t'c^ squares. We see that there are but nine perfect squares between 1 and 100. The square roots of all numbers lying between 1 and 100 will be found between the consecutive roots of two perfect squares. Thus, the square root of 20 lies between the consecutive roots 4 and 5, being greater than the former, and less than the latter. The square root of 26 lies between the consecutive roots 5 and 6. EXTRACTION OF ROOTS. 165 The square root of all numbers below 10,000 may be regarded as made up of tens and units. Thus, 99, the square root of 9801, is made up of 9 tens and 9 units. The number 32 is made up of 3 tens and 2 units. We have seen that the square of a number containing two figures could not give a higher denomination than thousands ; con- versely, the square root of thousands cannot give a number containing more than two figures ; that is, a number containing tens and units. 284. If, then, we have to extract the square root of a number con- taining more than three figures, and less than five, we know that its root must contain two figures, and, therefore, be made up of tens and units. Before we can deduce a rule for the extraction of the root of thousands, we must know how thousands are derived from squaring tens and units. Let a represent the tens, and b the units, which enter into the square root of thousands. Then, (a + by = a^ + 2ah + V^. Hence, thou- sands arc made up of the square of the tens that enter into its root, plus the square of the units, plus the double product of the tens by the units. Let us square the number 34, made up of 3 tens and 4 units, = 30 4- 4. Then, 30 corresponds to a in the formula, and h correspond.s to 4. Hence, a^ = (30^) = 900, 2rti = 2 . 30 . 4 = 240, b' = (4f = 16, Then, (34)^ = a' -f 2ab + U' = 1156. 235. If we were required to extract the square root of this number, we would have to revei*se the process, and first take out a*, then, 2ah and Z)l The number, 1150, belonging to the denomination of thousands, its root must contain tons and units, and we must first get out the tens by extracting the square root of the square of the tens ; that is, the s/ «^- Now, the square of tens will give 2a -\- b at least three figures, therefore the 60 -f- 4 root of the tens cannot be sought in the two right hand figures, and we, therefore, separate them from the others by a dot, and the given number is then separated into what are a^ + 2ab -f // 11 .56 a b 9 00 = a' 30 -f 4 T 56 = 2ab -f b' 2 40 16 00 Remaifider. 16G FOR-MATION OF THE TOAVERS AND called pcrtoih of two figures each. The square of 30 is 900, and the square of 40 is 1600. The number, 115G, falling between 900 and 1600, its root must contain 3 tens plus a certain number of units. We then extract the greatest square contained in the left hand period, 1100, and set the root, 30, on the right, after the manner of a quotient in division. We have now found a, square it, and subtract a^ or 900, from 1156, and the remainder must be 2oi + 1'^: in the present instance equal to 256. The remainder, 2ah + 1j\ can be written (2a + l)h; and it is evident that, to find h accurately, we must divide by (2a + h). But, as the term h within the parenthesis is unknown, we are com- pelled to use 2a as the aj)proximate divisor. We write 2a, or 60, on the left as a divisor, and divide 256 by it, and set the quotient h, or 4, on the right of the root found, and also on the right of the divisor. Now, multiply the two terms on the left, 2a -f J, or 60 + 4 by Z» or 4, and we evidently form 2a^ + ?/, the parts entering into the remainder, 256. Subtracting the two products thus formed from 256, we find no remainder. Hence, 34 is the exact root of 1156. We have separated the tens from the units, to let the beginner see that he is really taking a^ -}- 2ah + 1/ in succession, out of the given number, 2a + h \ 2 56 = 2a6 + V^ 1156. But we might have indicated 6 4 I 2 56 = 2ah + V their separation by a point above, and written 3 4, instead of 30 + 4, and 6 4 instead of 60 + 4. When the beginner is familiar with the princi- ples, he may omit the dots, -which are intended to guard him against confounding the tens with the units. He must observe, however, that this divisor being the double product of tens, is, itself, tens, and, there- fore, if written out in full, would contain a cj'pher on its right. And since, in dividing by a number whose right hand figure is 0, we point ofi" that figure from the right of the divisor, and also point oif the right hand figure of the dividend, we must be careful to do this in dividing by the double product of the tens. In the present instance, the right ' hand figure, 6, of the remainder 256, must be separated from the other two figures, since, in using 6 as the divisor instead of 60, we have, in fact, pointed off from the right of the divisor. 236. -There is one point of considerable importance that needs some examination. In getting the second figure of the root, we used 2a as tlie approximate divisor of the remainder, 2al> -\- 1/ =z (2a + li) h, u'-\ :- 2ah -Vh' 11 . 56 a -\-h 9 00 = = a^ 8 ■ 4 2 "56"; = 2a6 + 1' 2 56 = = 2ah + V EXTRACTION OF ROOTS. 167 whereas, the true divisor, to find h, is plaiuly 2a + h. Our divisor being too small, the quotient, which is the second figure of the root, can never be too small, but may be too great. It is plain that, when h, in the expression (2a + h), is very small in comparison with '2a, it may be neglected. But b may be so large that the omission of it will give too great a quotient. The square of 35 is 1225 ; the square root of 1225 is then, of course, 35. Now, if we proceed to extract the square root of 1225, the remainder, after taking out a", or 900, will be found to be 325 = 2ah -\- V = 300 + 25. And we sec that the square of the units has added 2 tens to 'lah, the double product of the tens by the units. When, therefore, we point off 5 from the right of 32 5, and divide 32 by 6, it is plain that the dividend, which ought to be 2ah (if the divisor is 2a), is too great by 2 tens. The quo- tient, then, would be too great, if the 2 tens added were divisible by the divisor. Then the second figure of the root would be augmented improperly by the quotient, arising from dividing the 2 tens impro- perly added by G, or 2a. In the present case, however, if we omit the 2 tens, and divide 30 by G, we get the same quotient, 5, as when the 2 tens are retained; their addition has not, then, affected the result. But, square 19, and we get 3G1. In this case, h, in the expression (2a -f 6), is not small in comparison with 2a, and cannot, therefore, be neglected without affecting the re- sult. Now, if we use 2a as the divisor of the remainder, 2ai -f W, to find h, we ought to use lah alone as the dividend; and, therefore, if we use the whole of the expression, 2ah -\- V, our dividend is too great. Dividing 26 by 2, the quotient is 13, which is plainly absurd for the units of the root. But, we see that 180, or lah, ought to have been the dividend corresponding to 20, or 2a, as a divisor. The 26 tens is then too great by 8 tens ; and, since the 8 tens added, give 4 for a quotient when divided by the 2 tens of the divisor; the second figure of the root is too great by 4, and we must write 9 as that figure, and not 13. The 8 additional tens in the 2G tens come from the square of the units, and being divisible by the divisor, 2a, have improperly aug- mented the second figure of the root. Had the 8 tens not been divisible by 2rt, the second figure of the root would not have been increased at all, and the quotient of lah -\- W by 2a would have truly been the "•- + : lah + />2 3G1 a -{■ b 100: = a^ 19 + h 261: = 2ah + b' 2 9 18 0: = 2ah 81: = 1'' 00 Remainder. 168 FORMATION OP THE POWERS AND second figure of the root. In general, whenever the square of the units (i^) incorporate into the remainder, 2ab + Z;'^, tens which are exactly divisible by 2a, the divisor, the second figure of the root will be too great, and must be diminished by the quotient of the incorporated tens by the 2a of the divisor. 237. The foregoing course of reasoning has shown that the second figure of the root may be too great, and the cause of its being too great; but, since the units of the root are unknown, the number of the tens proceeding from their square, that are incorporated with 2ah, cannot be known. We must, then, in practice, form the product of 2a + 5 on the left by the second figure, I, of the root, and compare the result with the remainder. If the product is greater than the remainder, the second figure must be diminished until the product is equal to the re- mainder, or smaller than it. If the given number is an exact square, its root will be exact, and the product will be exactly equal to the re- mainder. When the root is not exact, the product must be made less than the remainder. The preceding principles enable us to deduce for the extraction of the square root of whole numbers, embraced between 100 and 10,000, the following RULE. I. Separate the two rigid hand figures from the other figxires or figure of the given numher, and find the greatest square contained in the left hand period, which may contain hut one figure. II. Set the root of this greatest sq%iare on the right, after the manner of a quotient in division. Sichtract the square of the root thus found from the first period, and annex the second period to the remainder. III. Double the root found and place it on the left for a divisor. Seek how often the divisor is contained in the remainder, exclusive of the right hand fig%ire, and place the quotient on the right of the root already found, separated from it hy a dot above. Place it also on the right of the divisor, separated from it in like manner. IV. Midtiply the divisor thus augmented by the second figure of the root, and suhtract the product from the first remainder. If there is no remainder, the root is exact. Jf the product exceed the first remainder, the second figure of the root must be diminislied until the product is equal to or smaller than the first remainder. EXTRACTIOX OF ROOTS. EXAMPLES. 1. Extract the square root of 225. Ayis. 15. 2. Extract the square root of 7569. Ans. 87. 3. Extract the square root of 2025. Ans. 45. 4. Extract the square root of 841. Ans. 29. 5. Extract the square root of 2500. Ajis. 50. 6. Extract the square root of 7921. Ans. 89. 7. Extract the square root ot 9801. Atis. 99. 8. Extract the square root of 4096. Ans. 64. 9. Extract the square root of 5476. Ans. 74 10. Extract the square root of 7056. Ans.- 84. 238. We have seen that the second figure of the root has frequently to be diminished. We may diminish it too much, and it becomes ne- cessary to know when we have made the second figure too small. The test of this depends upon the principle, that the difference between two consecutive squares is equal to twice the smaller number plus unity. Let a =. smaller number. Then, a + 1 = consecutive number, or the number just above a. And (a + 1)^ = a^ + 2a + 1. Their difference is 2a -j- 1, as enunciated. Now, when there is a remainder, after finding the second figure of the root, and subtracting the product of it by the quantity on the left from the first remainder, it is evident tbat the second remainder ex- presses the excess of the given number, which we may regard as (a 4- l)'^, over the square of the two figures found. If, then, the second remainder be exactly twice the root found plus unity, it is evi- dent that the root found is o, and that the second figure of the root can be increased by unity. To illustrate, suppose 6 to be the square root of 49, (a -{- 1)^ then the remainder being equal to twice the root found 49 I a plus unity, the root can be increased by unity. In 36 = a^ I 6 general, whenever the remainder exceeds twice the 1 ." — 2a -j- 1 root found plus unity, the root can be augmented by unity. If the remainder is exactly equal to twice the root found plus 15 170 FORMATION OP THE POWKRS AND unity, the root, increased by unity, will be the exact root of the given number. 239. This rule is of importance in finding the square root of imper- fect squares. Let it be required to find the square root 1 5G 1 12 of 156. We find a remainder 12, and a root 12. Is 1 12 the greatest root contained in 156 ? Is the root 12 22 I 56 plus a remainder, or 13 plus a remainder? By the 44 principle just demonstrated, the true root of 156 must 12 be 12 plus a remainder, because the second remainder is not double the whole root found, plus unity. 240. This principle also enables us to pass from the square of a number to the square of a consecutive number without raising the second number to the square power. We have only to represent the smaller number by a, then the consecutive number will be a + 1, and its square must exceed a? by 2a + 1. Thus, (100)=^ =r a^ = 10000. a^ (2a + 1) Then, (101)^ == (a + 1)' = 10000 -}- 200 + 1 == 10201. (a + 1)^ a^ (2a + 1) So, also, (102)-^ = (101 + If = 10201 + 202 + 1 = 10404. The following are incommensurable numbers. 1. Find the square root of 1720. J.ns.. 41 + 2. Find the square root of 1445. Ans. 38 + 3. Find the square root of 6411. Ans. 80 + 4. Find the square root of 5555. Ans. 74 + 5. Find the square root of 1755. Ans. 41 + 6. Find the square root of 1960. Ans. 44 + 7. Find the square root of 7777. Ans. 88 + 8. Find the square root of 6666. Ans. 81 + 241. If we square any number, as 12 and 55, containing two figures, and made up then of tens and units, the square will contain two periods, counting from the right, and it is plain that the tens can only be sought in the periods on the left. If we square a number made up of EXTRACTION OF ROOTS. 171 hundreds, tens, and units, the square 'will contain three periods, and the hundreds can only be found in the left hand period, and the tens only in the second period, annexed to what is left of the first periods after the square of the hundreds has been taken from it. In general, the number of periods in the given number always indicate the number of figure places in the root, and each figure of the root has its appro- priate period or periods. The principles that have been demonstrated for the extraction of the square root of numbers between 100 and 10,000 can readily be ex- tended to any numbers whatever. Let a represent the highest denomi- nation in the root, and s all the succeeding denominations in the root. Then the number itself will be expressed by (a -f s)- = o? -\- 2as -\- s^, an analogous expression to (« + by = a- -j- 2ah + I/, and differing only in its more general significance. The a in one formula is not re- stricted to represent tens, as it is in the other, but may represent hun- dreds, thousands, millions, &c. j and the s of the first formula is not restricted to represent units only, but may represent tens and units, hundreds, tens, and units, &c. Let us square 155 by means of the formula. Then, a = 100, and s = 55. a' = (100)- = 10000, 2as = 200'55 =11000, s' =(55/ = 3025, Hence, (155)^ = 24025 We see from the formula that a, it matters not what may be its de- nomination, must first be found; and, that after its square has been subtracted from the given number, the remaiiuler will be 2t(H -f s" = (2a + s)s. The number, 24025, being greater than lOOOO, its root will be greater than 100, and, therefore, a- canuot be found in the two right «" + 2('s + s^ hand periods. We seek it in the 2 40 25 | a + s period on the left, and after placing 1 00 00 I 1 55 it on the right, and subtracting its 2a -f s I 1 40 25 = 2as + 0^ square from the given number, have 255 I 1 11 00 = 2as 14025 for a remainder. We cut off 30 25 = s' the right hand figures, because 2, the 00 00 Remainder, approximate divisor, is really 200; and, after trial, we find 55 to be the right hand figures of the root. 172 FORMATION OF THE POWERS AND The 55 is set on the right of the root ah-eady found, and also on the right of the divisor. The product of 255 by 55 is 14025, and there is, consequently, no second remainder, and the root is exact. 2-42. The approximate divisor is always large for numbers above 10000, and s can only be found by repeated trials. But the above process can be greatly simplified by observing that, since 2a enters into s, representing several denominations, it must enter into each de- nomination separately. Thus, in the foregoing example, 2a, or 2, being a multiplier of 55, is a multiplier of the first 5, regarded as 5 tens, or 50, and of the second 5, regarded as units. We might, then, have found the 5 tens and the 5 units, separately taking care to write the one after the other, so as to make their denomination distinct. In the present example, 2as being equal to 2 . 55, is, of course, equal to 2 (^50 + 5) J s may be regarded as a single term, 55, to be found at once, or it may be regarded as made up of 50 and 5, separate terms, to be found separately. But, if the second and third figure of the root be found separately as independent numbers, they must be sought for in their appropriate periods. It will simplify the work when we proceed in this manner to subtract the square of a from the left hand period, and bring down each term in succession. In this case, since we make two terms of s, let s ^ s' -f s" . Then the root will be a + s" -f s", and the number will be (a + s" + s")^, which, by performing the mul- tiplication indicated, will give us a^ + 2os' -f- s'^ -f- 2as" + 2sV' + s"^ = a" + 2as' + s'^ + 2 {a + s') s" + i>"\ When we subtract o^ from the left hand period, we have, in fact, subtracted 10000 from the given num- ber. After the se- cond period has been annexed to the remainder, the 140 truly represents 14000, and, since the zero, on the right of 140, belongs to the denomination of hundreds, it must be separated from the 14 when we come to seek for the tens of the root, because /, the tens, is sought for in 2as' by using 2a as the approximate divisor. Now, 2as' must be at least thousands, and, therefore, the denomination of hundreds does not contain s'. We write s, when found, on the right of the first term of the root and also on 2a + s' 2 5 (« + s' + s"f 240'25 a -f «' -f s" 1 . . =a' l'5'5 1 40 — 2as' -h s'2 ^ &c. 12 5 = 2as' + s'2 2 (a + 0+5" 30 5 1 52 5 = 2 (a + s') s" + 1 52 5 = 2 (a + s') .s" + Remainder. EXTRACTION OF ROOTS. 173 the right of the divisor. Multiplying the divisor thus augmented by s' in the root, and subtracting the product from the first remainder, we will plainly have left, after annexing the next period, 2 (a -f s').s"+ s"^. And we see that the approximate divisor to find s" is 2 (a + s'). The whole root already found mu.st then be doubled and used as our approxi- mate divisor. The right hand figure of 1525 is cut off, because 2 (a + s')s" gives at least hundreds. After s" is found, we multiply '1 (a -\- s') by it, and have no remainder; the root is then exact. We have used the broken line, in the above example, after each minuend and subtrahend to indicate that there were other numbers to follow them. (a + i/ + a" + a'"/ 2 98'59"84 i a + s' + s" + «'" 1 98 = 2as' + s" + &c. 1 89 rr= 2r,s' + s" 9 2a + 27 („ + /) + ," 9 5 9 =2{a+ sy + s" -f &c. 34 2 2(a + s' + s") + i 3418 684 = 2 (a + s')s" + 2 7 58 4 =zz 2 (a + s' -f /')■•>■'" + «"" + &c. 2 7584== 2 {n + a' + 6-").s"' + s"'\ If wc have a number made up of 4 periods, as 2 98 59 84, we know that its root must contain 4 figure places, which may be represented by a, s', ,s", and s'". And the given number must then be equal to (a 4- s' + s" + s'")^ = a"" + 2as' + s" + 2 (a + s')s"+ s"^ + 2 (a+sf We sec, then, that 2a is the approximate divisor to find .s', the second figure of the root ; (2a + s') the approxim-ate divisor to find s", the third figure of the root; and (2(a-\-sf) + s") the approximate di- visor to find s'", the units of the roots. In other words, we see that the whole root found has to be doubled to find each figure of the root succeeding those already found. It is evident, too, that the right hand figure of each of the successive remainders must be cut off previous to dividing by 2a, 2 (a + s'), &c., because in all these remainders that figure is of too low a denomination to make any part of the product of the figure of the root sought by the approximate divisor. Thus, 8, which belongs to the denomination of tens of thousands, cannot be a part of the product arising from multiplying 2a, or 2000, by s', which is hundreds. This product must give at least hundreds of thousands. 15* 174 FORMATION OF THE POWERS AND 243. It is plain tliat^ from the manner in which tlie square of any number of terms is formed, the foregoing demonstrations for numbers having two, three, and four periods are general ; and we, therefore, have for the extraction of the square root of any number whatever, the general I. SejxtJ'ate the given number into periods of two figures each, begin- ning on the right; the left hand period may contain but one figure. II. Find the greatest square contained in the left hand period, and set its root on the right, after the maimer of a qxiotient in division. Subtract the square of the root found from the left hand period, and to the remainder annex the second period, and use the number tints found as a dividend. III. Doidjle the root found and p>^(-^ce it on the left for a divisor. Seek how often the divisor is contained in the dividend exclusive of the right hand figure, and place the qtiotient on the right of the root already found, and separate the two figures by a point. Set the quo- tient also on the right of the divisor and separate in like manner. IV. Multiply the divisor thus increased by the second figure of the root, subtract the product from the dividend, and to the remainder annex the second period of the given number. Use the remainder and annexed period as a new dividend. V. Double the whole root found for a new divisor, and continue the operation as before until all the periods are brought down. Remarks. 244. If the last remainder is zero, the given number is a perfect square. But, if the remainder is not equal to zero, we have only found the entire part of the root sought, and the given number is incommen- surable. 245. If we take 155, the square root of 24025, we observe that 15 has been derived from the two left hand periods. We might, then, after finding 5, have squared 15, and subtracted its square from 240. So, after finding the last 5 of the root, we have subtracted the square of 155 from the three given periods. In general, when we have found two figures of the root, or three figures, or four figures, &c., we may subtract their square from the two left hand periods, or the three left hand periods, &c. EXTRACTION OF ROOTS. 175 GENERAL EXAMPLES. 1. Extract the square root of 16008001. Ans. 4001. 2. Extract the square root of 4937281. Ans. 2222. 3. Extract the square root of 1111088889. .4ns. 33333. 4. Extract the square root of 1975304G913G. Ans. 444444. 5. Extract the square root of 36000024000004. Ans. 6000002. 6. Extract the square root of 12.59631362889. Ans. 1122333. 7. Extract the square root of 15241383936. Ans. 123456. 8. Extract the square root of 16080910030201. Ans. 4010101. 9. Extract the square root of 123456787654321. Ans. 11111111. 10. Extract the square root of 12345678987654321. Ans. 111111111. 11. Extract the square root of 308641358025. Ans. 555555. OF INCOMMENSURABLE NUMBERS. 246. An incommensurable number is one whose indicated root can- not be exactly extracted. Thus, the \/2, V8, and \/27 are incom- mensurable numbers. Such numbers are also called irrational num- bers, ajid sometimes surds. We have indicated (Art. 240,) that the roots of imperfect square powers were not complete by writing the sign + after the entire parts of those roots. So we may write the -v/ 5 := 2 -f . The number 5 lying between 4 and 9, the square roots of which are 2 and 3, its own root will be greater than 2, and less than 8. May not this root, then, be expressed by some fraction whose value is greater than 2, and less than 3, such as |, or | ? May not the roots of all imperfect square powers be expressed by vulgar fractions in exact parts of unity ? To prove that this cannot be, we will demonstrate a theorem upon which depends the proof of its absurdity. Theorem. 247. Every number^ P, which will exactly divide the product, A X B, of two numbers, and which is prime with respect to one of them, will divide the other. 176 FORMATION OF THE POWERS AND Let A be the number with which P is prime. Let Q be the qua tieut arising from dividing AB by P, then -t— - = Q. We may put this equation under the form A X ^tj = Q. Now, Q is, by hypothesis, an entire number; the second member being a whole number, the first member must also be a whole number, else we would have an irreduci- ble fraction equal to a whole number, which is absurd. The product of A into r— has then to be entire ; now, A itself, is entire and prime T) with respect to P, hence, — must also be entire. For a whole number, multiplied by a fraction, can only give an entire product when the whole number is divisible by the denominator of the fractfon into which it is multiplied. Thus 4, a whole number, multiplied by the fraction #, gives an entire product, because 4 is divisible by 2. But 5 into | does not give an entire product, because 5 is not divisible by 2. Now, the whole number A is, by hypothesis, not divisible by the denomina- B B tor, P, of the expression p- ; the product of A by ^15- cannot, then, pos- sibly be equal to the whole number, Q, unless B is divisible by P. 248. We are now prepared to show that the square root of an im- perfect square, such as 5, cannot be expressed by a fraction. If the root of 5 can be expressed by a fraction, let — be that fraction, a being greater than h, and prime with respect to it. We assume a and b to be prime with respect to each other, because, otherwise, their quotient would be a whole number, and we know that the root of an imperfect power is found between two whole numbers. We have, then, y/5'= — . From which, by squaring both members, there results 5 = -y^- The first member of this equation being a whole number, the second member must be a whole number also. But —^ cannot be a whole number unless a^ is divisible by I ; for, to divide a^ by b^, is to divide it twice by b. Of course, then, if a^ is not divisible by b, it cannot be divisible by b^. But a^ is not divisible by h, for, by the fore- going theorem, a number which exactly divides the product of two factors, and is prime with respect to one of thenj, must divide the EXTRACTION OF ROOTS. 177 otter. Now tte factors of o^ are a and a ; 6 is prime with respect to the first factor, it must then divide the second in order to give an entire quotient. This is plainly impossible, since the second factor is the same as the first; a^ is, then, not divisible by h; still less, then, can a^ be divisible by Ir. The equation 5 = — must, therefore, be absurd ; but that equation has been truly derived from the equation v/? = -r- A correct algebraic operation has led to an absurd result : b but this can only be so when the assumption at the outset, upon which the operation is based, is absurd. The assumption, •>/~o= — was ab- surd, and it has led to an absurd result. The foregoing reasoning has been in no way dependent upon the fact that 5 was the particular imperfect power under consideration, and is, therefore, general. We conclude, then, that the square root of no imperfect power can be expressed by an exact fraction. EXTRACTION OF THE SQUARE ROOT OF FRACTIONS. 249. Since the square of a fraction is formed by squaring the nume- rator and denominator separately, it follows that the square root of a fraction must be taken by extracting the square root of the numerator and denominator separately. Thus, ^^ = 'j, since -5X3= |. ,0, also, \/'jr, = j, because j X - = -^, We may remark that all square roots can be affected with either the positive or negative sign, if these roots are regarded as algebraic quan- tities. Thus, the \/| may be either + |, or — |, because — | X — „ . „ , / a^ 1.1 (I a . a I = |. ho, also, \ / 71 iii'^y be either -f — , or — — ; since — y, multiplied by itself, gives y^. 1. Extract the square root of ^^. 2. Extract the square root of /g. Ans. |, or — |. 3. Extract the square root of i. Ans. ^, or — |. M An^. ^or-J. An?;. ..or — 1. Ans. 4, or — #. Alls. |,or-i. Ans. ior-|. Ans. J^- Sor-V- Ans. 5, or — 5. ITS FORMATION OF THE TOW Ell S AND 4. Extract the square root of -1. 5. Extract the square root of ^V- 6. Extract the square root of ~-f. 7. Extract the square root of ^-^. 8. Extract the square root of f 1. 9. Extract the square root of lY- 10. Extract the square root of ■*^^^. 250. Those examples show that the positive square roots of proper fractious are greater than the fractions themselves. Thus, ^ {- := h. The reason of this is plain : the numerator of a proper fraction being less than the denominator, is not diminished proportionally so much as the denominator by the extraction of the square root. Let a^ and 1/ be two unequal squares, and let a^ ^ b^; then, a ^ h, when a and b are both positive. Extracting the square roots of a^ and b'^, we have n and b for the roots. The extraction of the root has then diminished fi^ a fold, and ?/ only b fold. The greater quantity has, then, been diminished the most. 251. The positive roots of improper fractions aVe less than the frac- tions themselves, because the extraction of the 'root diminishes their numerators more than it diminishes their denominators. The negative roots of all fractions, whether proper or improper, are, of course, algebraically less than the fractions themselves. 252. In the foregoing examples the numerators and denominators were all perfect squares. But, if we have a fraction whose denomi- nator is not a perfect square, we can readily find its exact root to within less than unity, divided by the denominator of the fraction. Let it be required to extract the square root of |. AYe multiply the nume- rator and denominator of the fraction by the denominator, which does not alter the value of the fraction, and we have V I == \ / '? — -?• = " V 5 X 5 v/4f = f +. The root of the numerator lies between 6 and 7. The root of the fraction is greater than ^, and less than |; and we see that I is the true value of the fraction to within less than ^. By this, we mean that when we take £ as the true root of the fraction, we commit an error less than i. We can, of course, get a nearer approximation to the true value of the fraction by multiplying both terms of the fraction by the third, fifth, seventh, or some odd power of the denominator. This will make the denominator a perfect power, and its root can be exactly EXTRACTION OF ROOTS. 179 r 1 m -^ /8 X 5 X 5 X 5 /lOOO 31 , _. found. Thus, v & = \ / t e 5 f = V / ^^^T^ = ^ +• The ' " V 5 X 5 X 5 X 5 V G2o 25 true root is greater than |4, and less than -^'i. Hence, |i diflfers from the true root by a quantity less than ^'^. 253. When the denominator and numerator are both imperfect powers, as in the example just given, we may make the numerator a perfect power by multiplying both terms of the fraction by the first, or some odd power, of the numerator. But, in this case, the degree of ap- proximation is not immediately appai-ent. Thus, \/| = v/k 5 = I approximatively. The true root is greater than |, and less than |. The degree of approximation can only be determined by reducing these fractions to a common denominator. We have, then, ^8^ and |ij, and their diifercnce is ^^tj. Then, ^| differs from the true root by a quantity less than ^%. And ||, or §, also differs from the true root by a quantity less than ^%. It is plain that the degree of approxima- tion can be more readily determined by making the denominator ra- tional than by making the numerator rational. It is even preferable to make the denominator rational when the numerator is already so. though the process of making the denominator a perfect square make the numerator irrational. Thus, to find the approximate root of |, place ^^~ V 5X 5- ^55-5+- 254. If the denominator is already rational, we have only to extract its root for a new denominator, and write over it the approximate root of the numerator for a new numerator. We have, then, for finding the approximate root of any fraction, both terms of which are not rational, the following EULE. Mahe the denominator rational, if not already so, hy nudtiplyhuj loth terms of the fr action hy the first, or some odd poicer of the denomi- nator, according to the degree of approximation required, so that the denominator of the given fraction shall he the square power of the de- nominator of the fraction that marks the degree of approximation. Then extract the root of the denominator for a new denominator, and torite over it the approximate root of the numerator for a neio numera- tor. Affect the nexo fraction with the douhle sign, to indicate that there 180 FOKMATION OF THE TOWERS AND arc txco roofs equal, n-ith contrary air/iis. If the denominator be already rational, and a greater drgree of ajjproximation is required than that indicated hy unity divided hy the root of the denominator, mvltijjly hoth terms of the fraction hy the third, fifth, seventh power, d-c, of the denominator, according to the degree of approximation required, and proceed as hefore. EXAMPLES. 1. Extract the square root of J to witliin less than h of its true value. -^ns. ± *. 2. Extract the square root of I to within less than i of its true value. ^«s. ± |-. . 3. Extract the square root of I to within less than h of its true value. ^"s. ± h. 4. Extract the square root of | to within less than j'g of its true value. Ans. ± ||. 5. Extract the square root of | to within less than 1 of its true value. -Ans. zfc |, nearer 4 than |. 6. Extract the square root of f to within less than ^V of its true value. Ans. ± ^|. 7. Extract the square root of -^^ to within less than ^'^^ of its true value. Ans. ± ^o. 8. Extract the square root of -^^ to within less than ^^^ of its true value. Ans. ±: f |fi. In the first example, the multiplier of both terms of the fraction is 2; in the second, (2)^; in the third, unity; in the fourth, {-^f; in the eighth, (27)^ 255. Since the denominator of the fraction may be raised to as high an even power as we please, it is evident that the degree of approxi- mation can be made as close as we choose to the true value of the fraction. 256. We may, by a similar process, determine approximatively the roots of incommensurable numbers to within less than unity, divided by any whole number. Let it be required to determine the square root of 2 to within less EXTRACTION OF ROOTS. 181 than 1 of its true value. Then, v/2 = \ / " ^ }^^ = V-]| = | +. The true value lies between § and §, and, therefore, | differs from the true value by a quantity less than i. We multiplied and divided the given number by the square of the denominator of the fraction that marked the degree of approximation required. This, of course, did not alter the value of the given number, it simply placed it under the form of a fraction with a rational denominator. The nest step was to ex- tract the root of the numerator to within the nearest unit, and to write the result over the exact root of the denominator. To demonstrate a general rule applicable to any number, and true for any degree of approximation, let a be the number, and — the fraction that marks the degree of approximation. Then, -^"(7= -v / ^ — . Let J* denote the root of ojtMo within less than unity; in other words, let r denote the entire part of the root of an''. We will then have Va = \/-^ > \/^, or -, and < \/^-^, or -— . )• ?•-+- 1 The true root of a, then, lies between the numbers — and , which n n differ from each other by — . Hence, — differs from the tme root by 1 ?• + 1 a quantity less than — . So, also, differs from the true root by 1 r a quantity less than — . We then have a riuht to take either — , or n ^ n r 4- 1 , for the approximate root. That one is taken to which the root lies nearest. , To find the approximate root of any number, a, to within less than — of its true value, we have the following Multiply and divide the given numler hy the square of the denomi- nator of the fraction that marks the degree of ajyproximation. Ex- tract the root of the numerator of the fraction thus formed to within the nearest unit, and set the resvlt over the exact root of the denominator. Give the doidjle sign to the root. 16 182 FORMATION OF THE POWERS AND EXAMPLES. 1. Extract the square root of 2 to within less than ^ of its true value. ^«s- ± I- 2. Extract the square root of 50 to within less than J of its true value. ^i«s. ± 2_8. 3. Extract the square root of 50 to within less than -J^ of its true value. -^'Ihs. ± %^. 4. Extract the square root of 50 to within less than ^^^ of its true value. ^ns. ± fM. It is obvious that, by increasing the denominator of the fraction that marks the degree of approximation, we may make the approximate in- definitely near to the true value of the root of the given number. 257. Apiiroxlviate roots of wliole numhers expressed dedmally. To extract the root of any whole number, a, to within any decimal ^ r >• + ! Hence, -z is the true root to within less than ; that is. (10)"' (10)" — — differs from the true root by a quantity less than . Now, multiplying the given quantity, a, by (lO)^-", is the same as annexing EXTRACTION OF ROOTS. 183 2m cyphers; for, multiplying it by (10)^, annexes 2 cyphers; by (lO)'', annexes 4 cyphers; by (10/, S cyphers, &c. And dividing the ap- proximate root found, r, by (10)°" is plainly the same as cutting off from the right of the root found m places for decimals, for to divide the root by (10)', (10)^, (10)^ is the same as cutting off from the right one, two or three places of decimals. Hence, to approximate to the true root of any given number to within a certain number of decimals, we have this RULE. Annex tioicc as many cyphers to the given ninnher as there are deci- mal places required in the rout, extract the root of the number thus in- creased to toith in the nearest unit, and cut off from the right the re- quired number of decimal places. EXAMPLE.S. 1. Required v/2 to within 1. Ans. 14. 2. Required v/2 to within 01. Ans. 141. 3. Required v/U to within -001. Ans. 1414. 4. Required v/50 to within -01. Ans. 7-07. 5. Required V'SO to within -001. Ans. 7-071. 6. Required v/"o0 to within 0001. Ans. 70710. 7. Required ^9000 to within -1. .l/is. 94-8. 8. Required ^9000 to within -01. Ans. .94-86. 9. Required v/ 900(7 to within -OQl. Ans. 94-869. 10. Required V 145 to within -01. Ans. 12-04. 11. Required v/T45'tp within -001. Ans. 12-041. 12. Required s/ 1000 to within -001. .4ns. 31-622. 1% Required v/lOOO to within -0001 Ans. 31-6227. 14. Required V lOOOOO to within -01. Ans. 316-22. 15. Required yiOOOOO to within .001. Ans. 316-227. Examples 12 and 14 show that, to pass from the root of any number to the root of a number 100- times as great, we have only to remove the decimal point one place further to the right. The converse is evi- dently true also. 184 FORMATION OF THE POWEl MIXED NUMBERS. 258. The approximate root of mixed numbers can now readily be found. Suppose it be required to find the approximate root of 2-5 to witliin -1. If we annexed two cyphers, as before, the result would be 2500 J and then, when we shall have come to point off" into periods, the whole number, 2, will be united with the decimal 5. The root found will be 5, which is plainly absurd. But, 2-5, changed into an equivalent vul- gar fraction, is fg. Hence, by the rule for vulgar fractions, v/2-5 = N^fl = \/y§§ = t| + = 1"5 + • We see, that in the present instance, we hav^ annexed a single cypher, which made the decimal places even, and double the number of places required in the root. We next pointed off from the root the number of decimal places required. If we are required to find the approximate root of 2-5 to within ^i^, yj^^, or •01. Then, V2^ = v/fl - \/ '^\ll^^ = iB§ - 1-58- We have, obviously, added three cyphers to 5, and, therefore, made the number of decimal places even and equal to the number of places re- quired in the root. In pointing off for decimals, we have only pointed off two places, the number required in the root. To demonstrate the rule in a general manner, let a be the entire part of the mixed number, and ^z—;^ the decimal part. Then the given number will be h ah ~ 10"' ■ Hence, \/«' -\/ ab (10)"' V^ b X (10)™ (10)- \/' X h (10)" (10)'"' ■>(ior'»"^ < r+ 1 (10)"' • In wli lich r represents the entire part of the root of a-X h (10)"'. Now, the multiplication of b by (10)"" is the same as annexing m eypliers to b, and whenever m is odd, the number of decimal places will be even, and double the number required in the root. When m is even, the decimals will not be mixed with the whole numbers, as in the mixed number 3-4^ and there need be no cyphers annexed. We have supposed, in the general demonstration, that the number of decimal places required in the root was precisely equal to the num- ber of decimal places in the mixed number. But, if this were not the case, the denominator of the equivalent vulgar fraction has only to be multiplied by such a power of 10 as will make it 10 with an exponent twice as great as the number of places required in the root. The nu- EXTRACTION OF ROOTS. 185 merator, when multiplied by this power of 10, will have its number of decimal places even, and equal to double the number of places required in the root. Annex cyphers until the mauler of decimal places in the mixed number is even, and equal to double the mimber of places required in the root. Extract the root of the result to xcithin the nearest unit, and then point off from the right, for decimals, the number of decimal places required in the root. EXAMPLES. 1. Extract the square root of 4-9 to within -1 Ans. ± 2-2. 2. Extract the square root of 4-9 to within -01. Ans. rb 2-21. 3. Extract the square root of 4-9 to within -001. Ans. =fc 2-213. 4. Extract the square root of 4-25 to within -01. Ans. =t 2-OG. 5. Extract the square root of 425 to within -001. Ans. ±2001. 6. Extract the square root of 9G-1 to within -1. Ans. ± 9-8. 7. Extract the square root of 9-Cl to within -1. Ans. ±3-1 exactly. 8. Extract the square root of 145-755 to within -01. Ans. ± 12-07. 9. Extract the square root of 14575-5 to within -1. Ans. it 120-7. 10. Extract the square root of 101-7 to within -01. Ans. ± 10-08. 11. Extract the square root of 1001-01 to within -1. Ans. ± 31-6. 12. Extract the square root of 10-0101 to within -01. Ans. ± 3-16. 13. Extract the square root of 1728-555 to within -01. Ans. ±41-57. 14. Extract the square root of 172855-5 to within -1. Ans. rb 415-7. 16* 186 FORMATION OF THE POWERS AND 15. Extract tlio square root of 17285550-666 to within -01. Ans. ±4157-58. IG. Extract the square root of 1728555066-6 to within -1. Ans. zfc 41575-8. ROOTS OF NUMBERS ENTIRELY DECIMAL. 259. Let it be required to extract the square root of -4. This deci- mal, changed into an equivalent vulgar fraction, is j%. Hence, v/ -4 — y/-A^ = ■v/j''g% ^ y^o, ^^^ ^^., and < --. It is plain that the multiplication of h by (10)"" makes the number of decimal places even. In pointing off for decimals, as many places must be cut off from the right as there are periods. RULE. Aiinex cyphers to the given decimal until its places are even. Ex- tract the root of the result, as in ivhole numbers, and cut off from the right, for decimals, as man ij places as there are periods in the numher ivhose root was extracted. If it he required to extract the root to within a certain decimal, annex cyphers until the numher of periods is equal to the numher of places required in the root. EXAMPLES. 1. Required V"^, Ans. db -5 exactly. 2. Required %/ -9 to within -1. Ans. =b -9t 3. Required V"-09. Ans. d= -3 exactly. 4. Required v/-009 to within -01. Ans. ± -09. EXTRACTION OF ROOTS. 187 5. Kequired v/-0009. Ans. ±z -03 exactly. 6. Required v/-725. Aus. d= -85 exactly. 7. Required n/-U725 to within -01. --I'^s. ± -27. 8. Required v/ -00725 to within -001. Ans. ± -085. 9. Required V~^. Ans. ± -9 exactly. 10. Required v/-00«l. Ans. rb -09 exactly. 11. Required v/ -00081 to within -001. Ans. ± -028. 12. Requtred V"^000081. Ans. =b -009 exactly. 13. Required V 0000081 to within -0001. Ans. ± -0028. 14. Required v/-0U4. Ans. ± -12 exactly. 15. Required V -04937284. Ans. ± -2222 exactly. 16. Required V -05555555 to within .0001. Ans. ± -2357. 17. Required V -1111088889. Ans. ± -33333 exactly. These examples show that the roots of decimals are greater than the decimals themselves. This ought to be so, for all purely decimal num- bers can be changed into proper fractions. SQUARE ROOT OF FRACTIONS EXPRESSED DECIMALLY. 260. Vulgar fractions may be changed into decimal fractions, and then their roots may be extracted by the last rule. If the given frac- tion be mixed, it must be first reduced to an improper fraction and then changed into an equivalent decimal fraction. EXAMPLES. 1. Required V'J = %/ -6606 = + -81. Ans. d= -81. 2. Required y/~i = V'^- Ans. =t -5 exactly. 3. Required v^^ = v^llllllll. Ans. ± 3333. 4. Required ^~^ = v/-0625. Ans. ± -25 exactly. 5. Required J^j = s/ 012345079012. Ans. =b -111111. 6. Required VT to within -01. Ans. ± -86. 7. Required V2|^5 to within -01. Ans. ± 158. 8. Required v/2l. Ans. ±15 exactly. 9. Required y J~ to within -001. Ans. d= -301. 188 FORMATION OF THE POWERS AND 10. Required Vl2f^ to within -001. Ans. ± 3-523. 11. Required V251 to within -001. Ans. ± 5-011. 12. Required V'S^^ to within -01. Ans. ± 2-25 exactly. The foregoing examples show that a vulgar fraction, which is a perfect square, may or may not have an exact root when changed into a deci- mal fraction. 261. — GENERAL EXAMPLES. » 1. Required V48303584-4856 to within -001. A71S. do 0950-078. 2. Required ^ 25012001-44 to within -1. Ans. ± 5001-2 exactly. 3. Required ^-0289 to within -01. Ans. ± -17 exactly. 4. Required V -000144 to within -001. Ans. ± -012 exactly. 5. Required V^^ to within -001. Ans. ± -242. 6. Required ^/^ to within -001. Ans. ± -125 exactly. 7. Required v/^|^ to within -0001. Ans. ± -0G25 exactly. 8. Required v/^oW *« ^i*^^"" -00001. A71S. ± -03125 exactly. 9. Required V1728 to within -001. Ans. ± 41-569. 10. Required v/1728 to within jl^. Ans. ± 6j?^8_6. 11. Required V16|l| to within -001. Ans. ± 4-103. 12. Required v/^'^ to within -1. Ans. ± -2 exactly. 13. Required s/~^ to within -00001. Ajis. db -03703. 14. Required V^j to within -00001. Ans. ± -01234. EXTRACTION OF THE CUBE ROOT OF NUMBERS. 262. The cuhe or third power of a quantity is the product arising from taking the quantity three times as a factor. Thus, the cube of m is m X m x m = m". The cube root of a quantity is one of the three equal factors into which the quantity can be resolved. The pro- cess of extracting the cube root consists, then, in seeking one of the equal factors which make up the given quantity. When the quantity can be exactly resolved into its three equal factors, it is said to be a EXTRACTION OF ROOTS. 189 perfect cube. But, when one of its factors can only be found approxi- uiatively, it is said to be an incommensurable quantity, or an imperfect cube. Thus, ^S, and ^^27 are perfect cubes; but -i/y, and ) the required 190 FORMATION OF THE POWERS AND number of times, we will get (a + bf = a^ + Sa^b -f Sab^ + i^. l''hat is, a number, whose root is made up of tens and units, is equal to the cube of the tens, plus three times the square of the tens by the units, plus three times the product of the tens by the square of the units, p>his the ctibe of the units. The formula may be written (a + bf = a^ + (Sa^ + Sab + V^h. And we see that the first thing to be done is to extract the cube root of a", then the tens (represented by a) will be known. It is obvious, too, that the true divisor of the remainder, after a^ has been taken out, to find b, or the units, is the coefficient of b, (3a^ + Sab -f V^) ', but. since b is unknown, the last two terms of this coefficient, Sab and //, are unknown. Hence, we are compelled to use Sa^ as an approximate divisor of the first remainder to find b. Now, b, thus found, will, in every case, be too great, because the divisor has been too small ; but it may happen that when b is small, the addition of Sab + b^ to Sa^ would )iot diminish the quotient b by unity, and then there will be no error committed in assuming b to be the true quotient, provided we increase our divisor by Sab -\- b^, and form the parts that enter into the first remainder. We will illustrate by an example. Let it be required to extract the cube root of 1728. We begin by separating the three right hand figures, because the (I + b cube of Srt^ =300 l'72S 12 the tens (Sa -\-b)b = (SO i- 2)2= 64 1000= a^ must be, (Sa' + Sab+P) =364 72S = (Sa' + Sab + b')b ^^ ^^''^^' 728=(Sd^ + Sab + b^)b thousands, and, there- fore, cannot be contained in the right hand period. We next seek the greatest cube in the left hand period, which is really 1000 ; the root is one ten, or 10, we set it on the right, after the manner of a quotient in division. Sa^, or 300, is assumed as an approximate divisor of the remainder after a*, or 1000, has been taken from 1728. The remain- der, 728, being represented by (3(t'^ + Sab + li^)b, the true divisor to find b is, of course, the parenthetical coefficient of b. Having found b, or 2, by means of the approximate divisor, we set it on the right of the ten, separating it by a point to indicate that it is of a different denomi- nation. We next add {Sa + b)b, or 64, to Sa^, or 300, and we have 364 for the true divisor, provided that b has been found correctly. This divisor we will call the supposed true divisor. Finally, we multi- EXTRACTION OF KOOTS. 191 ply (3a^ + Sah + ¥, the supjjosed true dioisor by h, and the product made up (3a^ + 3a6 + i^)i, the parts entering: into the remainder. The product thus formed being exactly equal to the remainder, proved two things : first, that if the true, instead of the approximate, divisor had been used, h would not have been diminished by unity; and, second, that the number 1728 is an exact cube, and that its root is 12. Iq the present example, it is easy to see why b was found correctly by using the approximate instead of the true divisor ; a and h being both small, the omission of ^ah + V^ did not materially aifect the divisor. \V'e have, from the foregoing, a simple test by which to ascertain whether h, found by using the approximate divisor, 3o^, must be di- minished. "Whenever the supposed true divisor will give a less quo- tient than I, we conclude that h was too great, and it must be dimi- nished by 1, 2, &c., until the supposed true divisor will enter into the remainder the same number of times as the approximate. It will never happen when the unit figure of the root is small with respect to the tens, as in 71, 82, 93, &c , that the h, found by using 3a^ as a divisor, must be diminished by unity, or some greater number. But, when the unit figure is great with respect to the tens, as in 18, 19, &c., it may be necessary to diminish h by one or more units. The reason of this is plain. We will illustrate by an example. Kequired the cube root of 5832. "We see that i, found by using 30^^ as a divisor, is IG, and the sup- posed true divisor will then be 1036; but this, 5832 ] instead of entering IG 3a' =r 300 1000 times into the remainder, (30 + 16)16 = 736 4832 ^' 4832, will only enter 4 (1036)16 = 16576 times. Besides, 1036, or the supposed (3a^ + ^ah + b^), when multiplied by b, will give a pro- duct, 16576, greater than the remainder. The unit figure is then too great, and has to be diminished by 8 ; this diminution can only be determined by trial. The process ought to have been 5832 a + b 1000 1'8 3a2 = 300 4832 = (3a2 + 3o6 -f b^)h (30 -f 8) 8 = (3« + h)h = 304 4832 = (3a2 -|- ^ab + V)h 3a2 + 3a5 + h^ =604 192 FORMATION OF THE P AV E R S AND The true divisor by trial was found to be 604. This divisor was repre- sented by (3a^ + Sah + J/), and, when multiplied by b, or 8, gave (Sa^ + Sab + V)b, the remainder. Take,. as another example, G859. 6 859 I a-\-b 3a^ =300 1000 I 19 (3a + &)i = (30 + 9)9 = 351 5 859 = (3a^ + Sab + h^j 3a2 + Sah -^W = 651 5 859 = (3a^ + Sab + W)b By trial, 9 was found to be the second figure of the root, the sup- posed true divisor was then determined to be 651, and this, multiplied by b or 9, gave 5859. The given number was then a perfect cube, and 19 its exact root. In this case, the supposed true divisor is actually the true divisor. 263. The process for extracting the cube root of a number below 10000 may be without any difficulty extended to all numbers whatever. Suppose the number to be 1881365963625, its root may still be re- garded as made up of tens and units, the cube of the tens cannot enter into the last three figures, 625, on the right, and they may, therefore, be separated from the other figures. The greatest cube contained in 1881365963 must have more than one figure in its root, because the number is greater than 1000, which contains two figures in its root. Then the root of 1881365963 may be regarded as made up of tens and units ; and, as the cube of the tens cannot give a less denomination than thousands, the tens cannot be found in the last three figures, 963, which may, therefore, be separated from the other figures. After the separation of 963 the foregoing reasoning may be repeated, and thus dividing the number into periods of three figures until we come, at length, to the place occupied by the cube of the tens of the highest order. The period on the left thus found, may contain but one figure, as in the present example, or it may contain two figures, or even three figures. We will designate the tens of the highest order by a', those of the second order by a", those of the third by a"', &c. In like man- ner, we will designate the units of the highest order by U, those of the second order by b", &c. EXTRACTION OP ROOTS. 193 Sa'ariSOOir 1st Ap. divisor. 1881365*963*625 I l'2'3*4*5 (30+2)2= &i=(Sa'+b')b' 1000 _| 3a'a+3a'6'+6'2=:3B4=True divisor. SSI— (Za'^+3a'b' +1/^)1/+ 64=(3a'+i/)// 728=(3a'^+3a'6'+6'2)6' 4=6'^ 153 365=(.3a"2-j-3a"?y'+fr"a)6"-i- 3a"!»=43200=2d Ap. divisor. 132 ^Q~=(5a"^-{-3a"h"^+b''^)V (360 -f3)3 = 10%9= (Za"+h'')b" 20 498 365=(3a"'a+3a"'V"+i/"3)6"'+ 44289=True divisor. 18213904=(3a"'a+3a"'i/"+^"')6"' 10S9=(3a"+i"j^" 2 285 059 625= (3a""»+3)6"" 9= h""^ 2JS5^59625=(3a"">+3a""&""+6""») V" 3ri"''=4.o38T00=3d Ap. divLsor. (3090+4)4= 14-76 =<3a"'+6"0V" 4553476=True divisor. 14776=(3a"'+6"')'/" 16=6 '"^ (3a''')»=45O>)20S0O=4th .\p. divisor (37026+5)5= 185125 45701 ly25=True divisor. lu this example, we begin by dividing the nuuiber off into periods of three figures each ; the period on the left contains but one figure. We extract the greatest cube contained in this period, and set the root, 1, on the right. We then have found a', or the tens of the highest denomination. We then bring down the next period. Now, since but two periods are under consideration, the a', or 1, found, will be tens with respect to these two periods, and therefore, three times its square, or 3a'^ = 300. This is, then, the first approximate divisor. Dividing 881 by it, we get 2, or the supposed units contained in the root of the two left hand periods, we then add 64, or (Sa' + ?/)6', to the approxi- mate divisor, and, if U be the true units of the root, 364, or 3a'^ -f SaU + h'^, will be the true divisor. Now, since 364 enters into 881, the same number of times that 300 does, b' has been found correctly. We, therefore, multiply 364 by 2, and we form the three parts, (Sa'^ + 'dab' -\- b'^)b', of which the remainder, 881, is composed, except some tens and units which have been incorporated in it from the cube of the tens of a lower denomination. We subtract 728 from 881, and bring down the next period. Now, the next approximate divisor is plainly three times the square of 120, or 43200. But the shortest way of getting three times the square of 120 is to add 64, and the S({uare of b', or 2, to the true divisor, and multiply the sum of the three by 100. The reason of this is apparent. The tens of the next denomina- tion is found by dividing 153365 by 43200 ) the quotient 3 is set on the right of the 2 in the root, separated from it by a dot. We next add 1089, or (3a" + b")b", to 43200, and we have the second true divisor. Multiplying this divisor by 3, and subtracting the product from 153365, and bringing down the next period, we have a new 17 N 194 FORMATION OF THE POWERS AND number, the root of whose unit is to be found. The approximate di- visor to find h'" must be three times the square of 1230 ; and this is most readily found by adding 1089 and the square of h", or 3, to 43200, and multiplying the sum by 100 ; that is, annexing two cyphers to the sum. The approximate divisor thus found, 4538700, enters 4 times in 20498365, and the true divisor, when formed, enters the same number of times ; 4 is then, truly, the fourth figure of the root. Mul- tiply 4538700, the true divisor, by the last figure of the root, subtract the product from 20498365, and bring down the next period. The next approximate divisor is three times the square of 12340 ; this can readily be formed like the preceding. The true divisor is formed when 5, the final unit, has been determined. The true divisor, multiplied by 5, gives a product equal to the last remainder. Hence, the given number is a perfect cube, and 12345 is its exact root. The reason of the above process becomes very plain upon a slight examination. It was established at the outset, that the tens and units contained in the first two periods could be sought, independently of the other periods. Having found found 1 ten and 2 units in the first two periods, it is plain that this number, 12, is the tens in 123, the root found in the first three periods. So, 123 may be regarded as the tens of the first four periods. The root found in these periods is 1234, made up of 123 tens and 4 units. In like manner, 1234 may be re- garded as the tens of the root of the whole five periods. In fact, the root found, 12345, is made up of 1234 tens and 5 units. We then had tens of different denominations, 1 ten, 12 tens, 123 tens, and 1234 tens. To apply the general formula, (a -f lif = a' + 3«^& -f ^aW -f 1/, for the cube of a number made up of tens and units, to determining the tens and units contained in any two consecutive periods, it became necessary to distinguish the tens by dashes to indi- cate the denomination to which they belonged. We will add another example, to show more fully the application of the preceding principles. Required the cube root of 997002999. Sa"* = 24300 = App. divisor. 997'002'999 I 99 9 (3a' -i- h')V = 2511 729 I la" + h")h" 26811 = True divisor. 268 002 2511 241 299 81 26 703 999 2940300: = 2d App. divisor . 26 703 999 26811 2967111 = 2d True divisor. EXTRACTION OF ROOTS. 195 In this example, the cjuotient of the division of 2G8002, by the ap- proximate divisor, 24300, is ] ; but this is manifestly absurd. By trial, 9 is found to be the second figure of the root, because the true divisor, 26811, enters 9 times, neither more nor less, in 268002. We have for the extraction of the cube root of any number above 1000, the foUovring RULE. I. Begin on the right and divide off periods of three jigures each. There will remain on the left a period of one, tico, or three figures. II. Extract the greatest cube contained in the left hand period, and set the root on the right, after the manner of a quotient in division. Subtract the cube of the root from the left hand period, and brivg doivn the next period. III. Talce three times the sqnai-e of the root found, regarded as tens, and set it on the left as an approximate divisor; see how often this enters into the first remainder. The quotient icill be the second figure of the root, or something greater. Add to the ajyproximate divisor the product of three times the first figure of the root, regarded as tens, plu:^ the second figure of the root by the second figure of the root. The sum of this product and the approximate divisor %oiU be the true divisor if it enter into the remainder the same number of times as the ajiproxi- mate divisor. Multiply the true divisor by the second figure of the root, sidttract the product from the first remainder, and bring doicn the next period. IV. Add to the first trtie divisor the same number that was before added to the approximate divisor, plus the square of the second figure of the root, and annex two cyphers to the sum, the residt will be the second approximate divisor. The true divisor is found as before. Continue this process until all the p)eriods are brought doicn ; then, if the last true divisor, midtiplicd by the fined units of the root, gives a product exactly equal to the last remainder, the given number is a per- fect ctdie, and its exact root has been found. 1. Required the cube root of 9261. Ayis. 21. 2. Required the cube root of 85184. Ans. 44. 3. Required the cube root of 8024024008. Ans. 2002. 4. Required the cube root of 1371330631. Ans. 1111. 196 FORMATION OF THE POWERS AND 5. Required the cube root of 10306070(50301. A71S. 10101. 6. Required the cube root of 1307631. Ans. 111. 7. Required the cube root of 1879080904. Ans. 1284. 8. Required the cube root of 95306219005752. Ans. 45678. 9. Required the cube root of 95306219005752000. Ans. 456780. 10. Required the cube root of 468373331006. Ans. 7766. APPROXIMATE ROOTS OF INCOMMENSURABLE NUMBERS. 264. Let a represent any incommensurable number — we have seen that its root cannot be expressed by an exact fraction ; it may, however, be truly determined to within less than any fraction, — , whose numerator mv" , ^ r^ , / (r + ly IS unity. I-tor a = -y, and « ]> — , and <^ 3 ; r denoting the entire part of the root of an^. Then, since a is comprised between y^3 (r 4-iy , , r r 4-1 -^, and ^^ ^— ^, its root will be greater than — , and less than . n n "■ M n r . . 1 Hence, — differs from the true root by a quantity less than — . Now, as — may be made indefinitely small, the approximate root may be found as near to the true root as we please ; the difficulty of extracting the root increasing, however, with the increase of n. Hence, we deduce the following RULE. Midiiphj and divide the nnmher hy the cube 0/ the denominator of the fraction that marks the degree of aj^proximation ; extract the root of the new numerator to within the nearest unit, and divide the result hij the root of the new denominator, lohich will he the denominator of the fraction that determines the degree of approximation. 1. Required the cube root of 4 to within i. Ans. |. Because ^T== -y^^' = \/~ < f, and > |. The true V or V' Ans. 5_3 4 ' Ans. V Ans. V EXTRACTION OF ROOTS. 197 root lies, then, between | and |, but is nearer to | than to |. Hence, I is the true root to within less than i. 2. Required the cube root of 80 to within j^^. Ans. f i. Beea„se4.W=\y^=^y™>.i<.i. " 3. Required the cube root of 90 to within |. A7is. \^. 4. Required the cube root of 712 to within J, Ans. ^^ nearly. 5. Required the cube root of 1820 to within I. Ans. \i. G. Required the cube root of 200 to within A. Am 7. Required the cube root of 2397 to within ]-. 8. Required the cube root of 1531 to within i. 9. Required the cube root of 3575 to within I. 10. Required the cube root of 3375 to within i. Ans. Yf or 15, commensurable. 11. Required the cube root of 10G2208 to within J-. Ans. -S"^. CUBE ROOT OF FRACTIONS. 265. Since the cube, or third power, of a fraction is formed by raising the numerator and denominator separately to the third power, the cube root of a fraction can plainly be found bj- extracting the root of the numerator and denominator separately. The root of the nume- rator written over the root of the denominator will then be the root of the fraction. There are three cases: 1. The numerator and denominator of the given fraction may be both perfect cubes, and then the root of the one written over the root of the other will be the required root. 2. The numerator may be an imperfect cube, and the denominator a perfect cube, then the root of the numerator extracted to within the nearest unit, written over the exact root of the denominator, will be the approximate root to within less than unity, divided by the root of the denominator. 3. Both the num.erator and denominator may be imperfect cubes, or the denominator only may be an imperfect cube, then the denominator must be made a perfect cube by multiplying it by its second power. The numerator must also be multiplied by the same number, otherwise 17* 19S FORMATION OF THE POWERS AND the value of the fraction will bo altered. The root of the new nume- rator, to within the nearest unit written over the exact root of the new denominator, will be the approximate root required. If a nearer degree of approximation be required, both terms of the fraction may be multi- plied by the 5th, 8th, &c. power of the denominator. The reason for making the denominator rational rather than the numerator, is the same as that given in the explanation of the principles involved in ex- tracting the square root of fractions. The rule for the extraction of the cube root of fractions belonging to either of the three foregoing classes, is as follows : RULE. Make the denominator rational, if not already so, hi/ multiplying hoth terms of the fraction hy the square power of the denominator. Extract the root of the new numerator to within the nearest unit, and lorite the root found over the root of the new denominator, which will he the same as the denominator of the given fraction. If a nearer de- gree of approximation he required, hoth terms of the fraction may he multiplied hy the bth, Sth, &c. powers of the denom{?iator. The root (f the neic fraction loill he the root required. EXAMPLES. 1. Required the cube root of tf^. Ans. |. For (I)' = I X f X f = 3^. 2. Required the cube root of ^f to within \. Ans. |. Because (-|)^ = |4, and (f)'' = ^-^if . Hence, the true root lies be- tween I and I ; and |, therefore, difi'ers from the true root by a quan- tity less than -i- 3. Required the cube root of | to within i. Ans. |. — 3 /2~~5^ Because y i = \ / - — — = -y^Q ^ 4, and < 4. The true root V 5 . 5^ o3 ^ o' -- o is nearer 4 than |, and, tliercrnre, | is taken. 4. Required the cube root of | to within I. Ans. |. Because in == \/^ = VW > h a» e- ^"^ < APPROXIMATE ROOT TO WITIIIX A CERTAIN DECIMAL. 267. There are two cases : the given number whose root is to be found may be entire, or it may be mixed ; partly entire and partly decimal. CASE I. Approximate root oftoJiole numbers to ivitJiin a certain decimal. If the decimal fraction that marks the degree of approximation be changed into an equivalent vulgar fraction, the cube of its denominator will be unity, followed by three, six, nine, twelve, or some multiple of three cyphers. In other words, the cube of the denominator will con- tain unity, followed by as many periods of three cyphers each, counting from the right, as there are decimal places in the fraction of approxi- mation. Thus (-1/ = {-.^J = ^J,^ . (-Ol)' = (^1^^ = y.^^^^. Then, to multiply the given number by the cube of the denominator of the decimal changed into a vulgar fraction is nothing more than an- nexing three, six, nine, or some multiple of three cyphers. After a now number has thus been formed, the extraction /of the root is, of 200 FORMATION OF THE POWERS AND course, performed just as when the fraction of approximation was a vulgar fraction. Let it bo required to extract the cube root of 5 to within -01. ^ 3/5.(100)3 V^OOOOOO. ,^^ J/,,, Then, V. = N/-^/ = \/imr > ^^«' '""^ < T^^- RULE. Annex to the given numher three times as many cyphers as there are decimal places required in the roof. Extract the root of the neio num- her thus formed to within the nearest unit, and point off from the right the required numher of decimal places ; tchich amounts to the same thing as dividing the root of the new numher hy the denominator of the fraction of approximation changed into an eQuivalent vulgar fraction. EXAMPLES. 1. Eequired the cube root of 60 to within -1 Ans. 3-9. 2. Required the cube root of 1775 to within -1. Ans. 12-1. 3. Required the cube root of 9 to within -01. Ans. 2-08. 4. Required the cube root of 9 to within -0001. Ans. 2-0801 nearly. 5. Required the cube root of 18G4967 to within -1. Ans. 123-1. CASE II. 268. Approximate roots of mixed numhers to within a certain decimal. Let it be required to extract the cube root of 2o to within -1. Then, y 2^ 3/23 . 10^ 3/2300 ^ , „ ^ ^ , Wc see that when the decimal has been changed into a vulgar frac- tion, and the denominator made rational, the numerator, 2300, is the given number, with the point omitted, and with cyphers enough an- nexed to make the number of places, counting from where the point was, a multiple of three. If it had been required to extract the cube root of 2-3 to within -01,. it would have been necessary to add five cyphers EXTRACTION OF ROOTS. 201 to 3 ; the number of decimal places then would have been made a mul- tiple of 8. Now, in pointing off for decimals, it is plain that we point off as many places for decimals as there are places required in the root. For, if one decimal place be required in the root, the denominator of the root of the equivalent Aiilgar fraction will be 10; if two places be required, it will be 100, &c. RULE. Annex cyphers to the decimal part of the mixed number until there are three places, if one place he required in the root ; six places if two he required in the root, &c. Extract the root of the new number thics formed as a whole number, and point off from the right the required number of decimals. EXAMPLES. 1. Re(|uired the cube root of 2-12 to within -01. Ans. 1-28. 3/212 3/212x100^ 3/2120000^,, For, VTT2 = ^— = V -Too^ = N/ "100^ > • "< or 1-28. 2. Required the cube root of 4-1 to within -1. Ans. 1-6. 3/41 X 10^ 3/4100 ^ , , , / , , For, yCT= V^y = ^___ = y^__ > m, and < jg. 3. Required the cube root of 888 to within -01. Ans. 2-07. 4. Required the cube root of 68-64 to within -1. Ans. 4-1. 5. Required the cube root of 1770-25 to within 1. Ans. 12-1. 6. Required the cube root of 1150-455 to within -1. Ans. 10-4 nearly. 7. Required the cube root of 5011-125 to within -1. Ans. 17-1. In the 4th and 5th examples, one cypher only had to be annexed. In the last two examples, no annexation was required. But if the root in the last two examples is to be determined within -01, then three additional cyphers must be annexed ; if within -001, six addi- tional cyphers, &c. The reason for annexing cyphers until the decimal places can be separated into periods of three figures, is evident, even without ehang- 202 FORMATION OF THE POWERS AND ing the decimal into an equivalent vulgar fraction. For each period of three figures must give one figure in the root , if, then, the decimal places were not made multiples of three, when we come to point ofi" from the right the decimals would be mixed with the whole number. Suppose we were required to find the root of 8-72 to within -1. Now, 8-72 is but little greater than 8, whose root is 2. Hence, the root of 8-72 ought to be but little greater than two; but if we annexed no cypher to 72, we would have but one period, 872, and the root would be 9 approximatively. APPROXIMATE ROOT OF DECIMAL FRACTIONS TO WITHIN A CERTAIN DECIMAL. 269. A decimal fraction changed into an equivalent vulgar fraction will have a rational denominator when its number of decimal places are multiples of three. Thus, -021 = ^§1^, -007681 = toVoUo. -123456789 = j^^WVo- Hence, if cyphers be annexed to the decimal until the number of places are made multiples of three, the new fraction, when changed into an equivalent vulgar fraction, will have a rational denominator. And, since every period of three figures gives one figure in the root, cyphers must be annexed until the number of periods of three figures is exactly equal to the number of places required in the root. RULE. Annex cyphers to the given decimal until it can he divided off into as many periods of three jigures each as there are places required in the root. Extract the root of the oicw decimal thus formed to vnthin the nearest unit, and point off from the right the required number of decimal places. EXAMPLES. 1. Required the cube root of -8 to within -1. Ans. -9. 3/800 For, ^.S=^-,% = s^ j^3- > /„ and < fo. 2. Required the cube root of -08 to within -1. Ans. -4. For, ^■i)>i = ^j^^=V^n^>j'„ and'\'\ Ans. it 12a''&V 4. Extract the square root of 3Ga'^?>'V°(^^l ^^^ _t- Qa^i\^°d'' EXTRACTION OF ROOTS. 207 5. Extract the square root of a'^h'^c-'^^. Ans. ± a^^h'rc-^^. 6. Extract the square root of 256 7. Extract the square root of a^b'c^' Ans. ± a-'h-^c-^. 25 8. Extract the square root of . , > , _, ^ i -. _i 9. Extract the square root of ——-. . , = i ^ 2o Ans. it o~^xyz. 10. Extract the square root of P-^M''-N-"=. Ans. ± ^»3I-^N-^ 11. Extract the square root of , ,., , . , _, ^^ _„7_i„ _i ^ «>i*''r Ans. ± 20a "/^^"c-'. 12. Extract the sciuare root of — j7r?r~ « j /■.-,^\■^-^ mUn ^ 400 .'Ins. it (20) 'a'"i'/v. 13. Extract the square root of o-V>-^*c-'^^ Ans. zt n-7>-'v--^'-. It is plain that a monomial will not be a perfect power when its co- efficient is not a perfect square, and when every exponent is not some multiple of 2. But wlien the exponents of the literal factors are not exactly divisible by the index of the root, the division can be indicated. Tlius, v/4x = =t 2.r-, for, by the rules for multiplication, (db 2.r-) (=h 2x~) = ix. Hence, =t 2a:- is, truly, the square root of -ix. So, also, VTiV = it oH^c^, because (it ah^c^) (=t ahh^) = al^c'. The square roots, then, of all algebraic quantities may be truly ex- pressed whenever their coefficients are perfect squares. SQUARE ROOT OF POLYNOMIALS. 274. A trinomial is the least polynomial th^t is a perfect square. It would be a mistake to suppose that the square root of a^ + Ir is a + h, for (a -f hf = a^ + 2ah + l\ The term, lab, enters into the square of (a -f h), and is not found in a' -\- 1/. Any polynomial, to be a perfect square, must be susceptible of resolution into two equal factors ; and we know that when these factors are multiplied together to reproduce the polynomial, the two extreme terms of the product (if the 208 F O 11 M A T I O N OF T II E P W E R S A N D factors have been arranged with reference to a certain letter) are irredu- cible with the other terms. Hence, the square root of these extreme terms must be terms of the whole root. The extreme terms of s/a^-]-l/ must then be a and h, but the intermediate terras can only be found approximatively. Any binomial, as (a' + s), when squared, will give a trinomial, a'^ + 2a's + s^. Conversely, if we have any trinomial that is a perfect square, its root must be a binomial. Let it be required to extract the square root of ISah + 81o^ + 7/. If this trinomial be arranged with reference to the highest power of one of its letters, as a, and the square root of the first term be taken, we know that we have certainly one term of the root required, Be- cause, from what has been said, we know that the first term of the arranged trinomial is the product of the first terms of the two equal factors into which the trinomial can be resolved. In other words, it is the square of the first term of the required root. The arranged poly- nomial is Sla'^ + ISab 81a- '+ ISah -f h' I 9a + b + h'. We begin by Sla^ = a'^ \lSa + b = 2a's + s extracting the root of 2a's -f s^ = ISab + 6^ the first term, and set 18«6 + b^ this root in the same horizontal line with the polynomial, and on its right separated by a vertical bar. We subtract from the given quantity the square of the first term of the root ; there remain, then, only the two terms, 18a?> + b^. Now, we know that the first term of this remainder, ISab, is the double product of the first term of the root by the unknown second term. If, then, we divide 18aZ» by 2 (9a), or 18a, the quotient, b, must be the second term of the root. The root is then completely known. The result can be verified by squaring the root 9a -\- b ; or, since the remainder ISob + b^ con-esponds to 2a's + s'^, and since 18a corresponds to 2a', if we write 18a below the root, and b, which corresponds to s, on its right, and then multiply 18a + & by Z>, we must evidently form the remain- der, 2a's -f si 275. Since, when a trinomial is a perfect square, its extreme terms must be perfect squares, and its mean term must be the double product of the roots of these terms, we can tell in a moment when a trinomial is a perfect square ; the mean term of the aiTanged trinomial must always be the double product of the roots of the extreme terms. Let us apply EXTEACTION OF ROOTS. 209 tMs simple test to some expressions. 4a^ + 4ma + rr^ is a perfect square, and its root 2a + m ; a + 2 V a6 + 6 is a perfect square, since the test is satisfied, and the root, -^ a + s/h -, .r^ -\-2.x-y~ -\-i/ is a per- 3 3 feet square, and its root, x~ + y~. The root, however, will only be commensurable when the extreme terms of the arranged trinomial are rational. We have, from the fore- going, a simple rule for the extraction of the root of a trinomial that is a perfect square. Extract the root of the extreme terms, and connect their roots together 1)1/ the sign of the mean term. Thus, the square root of m^ — 2mn + n^ is m — n; this is obvious, since {in — iif = m^ — 2mn + n^ ; or it may be seen by going through the steps of the process described. EXAMPLES. 1. Kequircd v/49a^ + Warn -f- m^. Ans. la + m, 2. Required v/49a''^ — lAam -\- m^. Ans. la — vi. 3. Required v/49a^ + 28am + 4ml Ans. In + 2m. 4. Required y/^Qa^ — 28am + 4m''. Ans. la — 2m. 5. Required V4m + IQ^Wn + 16». Ans. 2 n/w + 4 v/TT. 6. Required 74m — IQs/mTi + 16rt7 Ans. 2-/m — 4%/?^ 7. Required -/m^ — 14am + 49al A?is. m — 7a. 8. Required V4hi2 — 28am -f 49a''. Ans. 2m — 7a. Bema7-7:s. Examples 2 and 4, in connection with 7 and 8, show, that when the mean term of the trinomial is negative there will be two distinct roots, according to the arrangement of the terms. The reason of this is plain. It is evident, moreover, that when either of the extreme terms has the negative sign, or when both are negative, the root will be imaginary. 18* o 210 FORMATION OF TUE POWERS AND The rule for the extraction of the square root of a trinomial has an important application in the solution of complete equations of the second degree, and ought, therefore, to be remembered. 270. If the given numher, instead ofheing a trinomial, have a tri- nomial root. Let a + m -\- n represent this root. Then, by representing a + m by p, and the given polynomial by N, we have N = (^ + n)^ = jp2 _|_ 2pn + n^ = a^ + 2am + m^ + 2 (a + m) « + n^. We see that the first three terms is a perfect square, and the root may be found by the rule for the extraction of the root of a trinomial ; then, when we have taken the square of the root, a^ + 2am + ??i^, from the given polynomial, there will remain 2 (a. + m') n -f n"^. The first term of this remainder, 2an, divided by 2a, the double of the first term of the root will give n, the third term of the root. The remainder, 2 (a + m)rt -f n^, may be put under the form (2a + 2m -f 71)%. If, then, the first two terms of the root found be doubled, and the third term added to them, and their sum be multiplied by the third term, the product thus formed will be equal to the remainder. The process for extracting the root of a polynomial which is the square of a trino- mial, is precisely like that for extracting the root of a polynomial which is the square of a binomial. The divisor, to get any term of the root after the first, is twice the first term of the root ; the terms of the root preceding the last term found are doubled, the last term added, and the product of the whole, by the last term, subtracted from the succes- sive remainders. The diifcrcnt steps can be best exhibited by the following example : a" = 4:n' 2m-f3m+5 4n + 3m 2am+7n^-{-2 (^a+m)n-\-nr=127nn + dm^-\-i:nb-\-Qmb-\-h^ 4?i-}-6m-f?» 2am+m''=12«m-f9m^ 2 (a + m)n-irn''=4:7ib+Qmb-\-b'^ (2a + 2m+n)n=4:nb-^Qmb-\-b^ We began by extracting the root of the first term, the root found was set on the right, and its square subtracted from the given polyno- mial. We had, then, taken out a^ of the formula from the given ex- pressions 5 we next doubled the root found, and used this double root (2rt of the formula) as a divisor to find the second term. This, when found (to of the formula), was set on the right of 2??, and connected by EXTRACTION OF ROOTS. 211 its appropriate sign. The double of the first term, and the second term, written directly under the root, were next multiplied by the second term. We thus formed "lam -\- m^ of the formula, which, when subtracted from the first remainder, left -inh + 6mh + l^, correspond- ing to 2 (a + m) 71 + ?i^ of the formula. The first term of this remain- der (2an of the formula), divided by twice the first term (2rt of the formula), obviously, gave the third term (?() of the formula. The first two terms of the root were next doubled, and the third term added to their sum. Having thus formed 2a + 2m + m of the formula, wc multiplied this sum, 4?i + Qm,-\-h, by the third term, h, (or n of the formula). We thus plainly formed the three parts of which the re- mainder was composed; and the product being exactly equal to the last remainder, the polynomial had an exact root, 2« + 3m -|- h. The foregoing reasoning can be readily extended to a polynomial whose root contains four terms. For, let N represent the polynomial, I, the sum of the first three terms of the root, and n the last term of root. Then, '^ = {I -\- nf = F + 2ln -f n^. Suppose the first three terms to be a, h and c; then, N = (a -f i + c)^ -f 2 (a + h + c)n + ii". Now, the first three terms is the square of a trinomial, and the root can, therefore, be extracted precisely as in the foregoing example. After the square of the root has been subtracted from the given polynomial there will remain 2 (a + b + c) n+ii^. The divisor to find n is, ob- viously, still 2a, twice the first term of the root. And, since th( remainder can be placed under the form of | 2 (a -f- S 4- c) -f- » | n, it is plain that, if the last term be added to twice the sum of the first three terms, and the whole sum be multiplied by the last term, we will form the three parts of the remainder. The process for extracting the root of a polynomial which is the square of four terras, is then identical with that for exti-acting the root of a polynomial which is the square of three terms, and that, we have seen, is the same as the process for extracting the root of a polynomial that is the square of a binomial. Now, if the root is composed of five terms, after the square of the first four terms has been subtracted from the given polynomial, the remainder will be found to be the double product of the four terms found by the untnown fifth term, and the divisor to find this term will still be twice the first term of the root. And so the reasoning may be extended to a polynomial whose root is made up of six, seven, or any number of terms. Hence, for extracting the root of any polynomial, we have the followins: 212 FORMATION OF THE POWERS AND RULE. Arrange the polynomial 7vit7i reference to one of its letters. Extract tlie root of the first term, and set the root in the same horizontal line icith the given polynomial, and on its right, separated from it hy a vertical bar. Suhtract the square of the root found from the given polynomial, and hring down the remaining terms for the first remain- der. Write double the first term of the root found immediately be- neath the place of the root, divide the first term of the first remainder by it, and write the quotient, lohich is the second term of the root, on the right of the first term of the root, and also beneath and on the right of double this term. Multiply the double of the first term, and the second term, itself affected with its propter sign, by the second term, and subtract the binomial product from this first remainder, and bring dotcn the re- maining terms for the second remainder. Divide the first term of the second remainder by the double of the first term of the root, and the quotient, affected with its proper sign, will be the third term of the root. Set this third term in the root, and also in another horizontal line on the right of double the sum of the first two terms of the root. Midtiply the three terms in this horizontal line by the third term, and subtract their product from the second remainder. Continue this process until the final remainder is zero, the root will then be exact; or continue until the letter, according to which the polynomial has been arranged, has disappeared from one of the remainders. Tlien, if all the expo- nents of that letter in the given polynomial are positive, we conchide that the polynomial is not a perfect square. Required the square root of 4x'^ + Vlxy + 4a;2 + 16x? -f %^ + 6^2 + 24^? + %lz + 16ZI The polynomial is already arranged. 4z2 4- 12xy+4zz4-162:Z+97/2+6y2;+24?/Z4-8Zz+16Z2+22 4x2 1st Rem. =12zy+42z+16a;Z+V4-Gy3+24yZ+8/z+lGZ2-|-22 ]2zy +V 4X+32/ 42:-f-6y+2 4a;+6y+22+4i 2cl Rem. =:4xz-\-16zl -^Gi/z-\-24yl-^8lz-\-l&-^-\-z^ izz +C.yz 2^ 3d Rem. =16xZ -^24yl-\-8lz-}-16l^ 16x1 -\-24yl-{-Slz-\-lGP EXTRACTION OF ROOTS. 213 277. After the beginner has become familiar with the preceding principles, it will not always be necessary to go through the process of forming the successive remainders. The successive products, distin- guished by parentheses, may be all formed, and their sum taken at once from the given polynomial, as in the following example. TO* — 6mn + Ore* + imp — 2mq — \2np + 6/17 -f 4p* — 4p2 -f 5* (m'-i) + 2(— 6j?m + 9n») + '(4mp — V2np -j- ip"^) -f *(— •Zmg + (mq — ipq + g') 7)1 — 3?i+2p— 7 •2nj— 67i+2;> We have distinguished the successive products by parentheses, affected by exponents written on the left. 278. It is obvious that the successive steps required by the general rule for extracting the square root of any polynomial, amount to nothing more than subtracting the square of the algebraic sum of the terms of the root, as they are found, from the given polynomial. In some cases, then, it may save trouble to subtract the square of the algebraic sum of the first two terms of the root from the given polynomial, then bring down the remaining terms and find the third term. Next, subtract the square of the algebraic sum of the first three terms of the root from the given polynomial, and continue in this way until all the terras of the root are found. 1. Kequircd v/x- + 4^^ — Oxz + 4ay — 12?^ + 92^ Ans. x-V^ — ^z. 2. Required V x" -^-i/ — 'ixY + 4z'' + \z^x' — \zhj\ Ans. -x^ — 7f -f 2z^ 3. Required \ / mi^ -f ]/r — mn + -^ — \xn + '\xm. Ans. in — ^iU -\- Jx. 4. Required \/a^ — 2«./' + x^ + 2am — 'Ian — 'Ixm + 2xn. -f m^ 2m?i + n^ Ans. a — x -\- m — n. 5. Required ^ *® + y** + txhn + 2xy — 'Ix^n + 'li/hn — 'Zijhi + m^ — 2mn + n"^. Ans. x^ -\- y^ -\- m — n. 214 FORMATION OF THE POWERS AND 6. Required W a'- + x^ + 2ax-{-am + a7i + xn + xm+ i7+x + ~9 . m n Ans. a + x + — + — \ Required y/sLc^^ + 9a-?/+-^ + 9x2 + 9xm + ^ + '!^' + i! + 4 '"' _L '" ^ n , y , 2: , m 2"+ 4"- ^ns. dx + ^+- + -. 8. Required W |- + ^ + ^ -f Qmi/ + 6mx + 5^^ + bzx + 36m^ + 202^ + QOmz. . ^ V n A71S. _ + ^ -f 6w + 5z. 9. Required v/^^ + 4x1/ + 4x' + 1 + 4x + 2y. Ans. y + 2x + 1. 10. Required n/^-''+ lOa-y + 4a;'2/ + 12x2/'' + ^V'- Alls X" 4- 2x1/ -f 3y^ 11. Required Vx* -\- 2x^// + 2x^y + 2xY' + x^yt + ^V- Ans. X? -\- xy -{- x?y. 12. Required the square root of ^^ + lOj^y + 4ic^y + 12.r3/3+9y-|- 2X' + ^x^y + 10:rj/2 + 6/ + x'' + 2.CTy + ?/^ Ans. x'' + 2xj/ + 3^/2 + a- + y. Remarlc. The short process, indicated in Art. 277, cannot be followed in the last example. SQUARE ROOT OF A POLYNOMIAL INVOLVING NEGATIVE EXPONENTS. 279. The principles for the extraction of the root of a polynomial containing negative exponents are the same as for the extraction of the square root of a polynomial, all of whose exponents are positive, observ- ing, however, in the arrangement of the polynomial, that that negative exponent is the least algebraically which is the greatest numerically. When, too, the arranged letter has disappeared from any remainder, it EXTRACTION OF ROOTS 215 may be supplied witli a zero exponent, provided, that the exponent of the same letter is negative in some of the terms. Take, as an example, x-"^ + 2,x-^i/ + x^if + 2.xif +2y + y^. Arranging the polynomial with reference to the ascending powers of X, and proceeding as before, we have X-' + 2x-'y + 2xV + xy + 2xf + x^ 2x-' + y 1st Rem. =:2x-'y + 'Ix^y + xY + 2x/ + xhf 2a;-' + 2y + xy 2x-\i/ + xY 2d Rem. = 2x'> + 2:?;y^ + a;^ Ix'y + 2x/ + a;y The first term of the second remainder is 2y ; x, the letter accord- ing to which the polynomial was arranged, docs not enter into this re- mainder until supplied. It, of course, must be introduced with a zero exponent, otherwise the expression 2y would be altered by its introduc- tion. But, since x" = 1, then, 2x°y = 2y. Take, as a second example, x~* 4- 2x~^y -f- ?/- -j- 2 + 2yx' -i- x*. X-* + 2x-'y -f ^2 + 2 + 2yx' + x* ^ 1st Rem. = 2x-2y + / -f 2 + 2yx' -}- x* '^''■rr'y + y' 2d Rem. = 2x° -f 2])x' + x* 2x0 -f 2yx^ + x* X-' + y + ^ 2x- -' + y 2x -2 + 2y + x' The first term of the second remainder is 2 ; x, affected with a zero exponent, was introduced into that terra. 3. Required Vu;"^ -f 2x-'Y^ + y^ + 2x-^ + 2xy-^ -f- x\ Ans. x~^ + y-^ + x. 4. Required the square root of —7--\ ^ — [-"7-- + 1 +'^^—:^ — h-c"- 4 o y o x~'^ i/~^ 5. Required the square root of -j — | ^ ' . i . "' -^ +V + 1 + 0;-' + 2a;-2-f^-|-4:c='-f-4. ,-2 ,,-3 Am. _ -f- ^ + 2 -f xl 216 FORMATION OF THE POWERS AND 6. Required the square root of x^ — 2x^^ -}- x*y^ + 2a;'2/~' — 2x' + 2 — 2x-hj + 2.r-='^-' + a"^ + 3/-I A7is. x" — x^i/ + x-^ +^-'. 7. Required the square root of x~^-{-2?/-\-x^i/'^-{-—'^ — l-a;~^^~'4-l. Ans. X-* H ^ h xy. INCOMMENSURABLE POLYNOMIALS. 280. When the exponent of the letter, according to which the poly- nomial is arranged, is positive in all the terms, we know that the poly- nomial is incommensurable when this letter is not found in any of the successive remainders, or is found affected with a lower exponent than it appears with in the first term of the root. But, if the letter, accord- ing to which the arrangement is made, appear in some of the terms with a negative exponent, we can only tell that the given polynomial is not commensurable by observing that the operation would never end. EXAMPLES 1. Required the square root of a^ -f h ' "^ 2a Sa' ^ 16^ , ^' ^' I' 2. Required the square root of 1 — x^. Ans. 3. Required the square root of a;^ — 1 X X X ^ns. l-^_--__&c. 1 1 1 2x Qx'' Ib.c* 4. Required the square root of 2x^ -f- 2a; + 2. 13 ^ Ans. xV2 -f ^r + = + &c., or V^(x + 1, + -- + &G.) ^/2 4x^2 ^ ~ Sx 5. Required the square root of 2x^ — Gx + 4. 3 1 1 Ans. Xs/2 — __&c., or y/2 (a- — #— &c.") v/2 4x^/2 ^ - 8x ^ 6. Required the square root of to -{- n. Ans. s/m -j =. = -\ ^^ — &c. 2s/m StoVto lGmV»i 7. Required the square root of x~^ -f 2a;~^ -f- 4a~' Ans. x-^ -{■ 1 + 2x -{■ &c. EXTRACTION OF ROOTS. 21' SQUARE ROOT OF POLYNOMIALS CONTAINING TERMS AFFECTED WITH FRACTIONAL EXPONENTS. 281. Since, in multiplication, the exponents of like letters are added, whether they be fractional, or entire, it is evident that the square of m m m Zm 2m m rt"" = a° X a" is a ° . Hence, the square root of a^ is a" . Quanti- ties involving fractional exponents may then be operated upon in the same manner as quantities containing only entire exponents. This will be shown more fully under the head of fractional exponents. Assum- ing a truth which scarcely needs a demonstration, we will give a few examples of polynomials involving fractional exponents. EXAMPLES. 1. Required ^/ x"> + ,/0 — 2xy — 2fz' -f 2x'z^ -f Ans. x" — if'-\- z". T. • ■, \ / '' V X y z x^z y z Required Vj + ^ + -f- + -^- + -2-+V- J i ^ X- y z^ ^"^- Y + T+ 2- Required \/ - + -L + -^ ^ ^ + — h^ 2/^^ . X t/ z ^ns. _ -f- 4^ -f- -. S o 6 4. Required \/x'^ + y'^ + 2x'^y^'° + 2^'0«^ -f 2a; 5 a^ + a*. A71S. x^ -{- 1/^^ -}■ d^. CUBE ROOT OF POLYNOMIALS. 282. The cube of a monomial, as a, is a", for (ay = a x a X a =za^. The cube of a binomial, a + h, is a" + Sa^J -f 3a&' -f b\ For, by actual multiplication, we have (a + ly = (a + li) (a + h) (a + I) = a" + Sa'b + ^a¥ -f V^. Since, then, the cube of a mo- nomial is a monomial, and the cube of a binomial a polynomial of four 19 218 FORMATION OF THE POWERS AND terms, it follows that a polynomial of four terms is the least polynomial which can have an exact cube root. Knowing the third power of a binomial, we have only to reverse the process to obtain the cube root of the power. We see that the first term of the power is the perfect cube of a, and we know, from the man- ner in which a product is formed, that that term, a?, has been derived without reduction from the multiplication of the three factors of which the power is composed. If then we extract the cube root of o?, we know certainly that the root, a, is a term of the required root. Then a will be one of the extreme terms of the root; and, since the extreme terms of the root when cubed give powers that are irreducible with the other terms, it follows that the cube of a or a^ is to be taken from the given polynomial. After this subtraction, the remainder is 3a^6 + 3a6^ + 1/, and it is plain that the second term of the root can be found by dividing the first term of the arranged remainder by 3a^, or by three times the square of the first term of the root. The remainder, 3a^6 + 3aZ>2 .^ ^3 ^^^ ^e p^^ ^^^jgj, tl^g fQ^,j^ Qf ^3^2 ^ 3^ J + 12)&. If then we add three times the square of the first term of the root, three times the first power of this term by the second term of the root, and the square power of the second term of the root, together, and multiply the sum of the three terms by the last term of the root, we will obviously form the three parts of which the remainder is composed. The pro- cess, then, for extracting the cube root, is analogous to that for extract- ing the square root of a polynomial. a? + ^a'h + ^aW + V \ a + h 3a' + 3a6 -f 6^ 1st Eem. = 'da^h + 'iay + W 2d Rem. =0 The root is set in the same horizontal line with the given polyno- mial, the cube of the first term is taken from the given polynomial, and the first term of the arranged remainder is divided by three times the square of the first term of the root, to obtain the second term of the root. Finally, immediately beneath the root is written, three times the square of the first term of the root, plus three times the product of the first and second terms of the root, plus the square of the second term of the root, and the product of the sum of these three terms by the last term of the root is taken from the first remainder. EXTRACTION OF ROOTS. 219 If the root be composed of three terms, a, h and c, we may repre- sent the algebraic sum of a and h by m. Then, let P be the poly- nomial whose root is a + h ■\- c, ytq will have V ^ (a -\- h -{■ cf = (m + cj = m^ -f- ^m^c -\- Smc" + c" = (a + bf + 3(a + bf c + S (a + b) e f c» = CT* -f 3a^i -f ^ab^ + i' + ^ah + Qabc + Z¥c + c^ The first four terms of this polynomial are the same as in the last ex- ample, and therefore their root, a -f i, can be found as in that example. After the cube of the sum of these terms has been taken from the given polynomial, the remainder may be placed under the form of | 3 (a + by -f 3 (a -f 6) c-f c^ I c. Then, having found the third term by dividing the first term, 3aV, of the arranged remainder, by 3a-, it is plain that tha remainder itself can be formed as before, by multiplying the last term found into three times the square of the sum of the other terms of the root, plus three times the fii-st power of the sum of the other terms into the last term, plus the square of the last term. Now, whatever may be the number of terms in the root, we may re- present the algebraic sum of all of them, except the last, by s. Sup- pose n to be the last term. Then P = (s -f ?i)' = s' + Ss^n + Ssn^ + ?i' : in which s represents the algebraic sum of any number of terms. If a is the first term of s, then the first term of the development of 3s^m, will be Sa^n. Hence, the divisor to find n is still three times the square of the first term. After s' has been subtracted from the poly- nomial, the remainder can be put under the form of (3s^ -|- Ssn + n^)«. Hence, whatever the number of terms in the root, the successive re- mainders can be formed just as when there are but two or three terms in the root. The following examples will illustrate the process. 1st R.=3x»+3a;*+9x*y+x=+18x=^+27xV+9-c'i'+27ii/^-|- 27^' 3x5 ^sx* +x'> (3x« + 3x3 _^ a.»)^ (3(x^+^)»+3(xHa;)3y+9y»)3y Rem. = 9x*y + 18x=(/ -f 27xV -|- 9x^y + 27x3/«H- 27 j' 9x«y -I- ISx^y + 27xV + 9x»y + 27xy'' + 27y » Rem. =00 00 00 x'+ Qx'+ 15x*+20x'+lox'+Qx+ 1 x' 1st E. —Qx^i- 15x'-f 20x'+ ldx% Gx +1 Gx'+12x'+ 8:c^ a;^-f2.c-f 1 (3x^-f-6.cH4.r^)2.T (;6(x'+2xy+S(x'+2x)l+l)l = (Sx'+ 12x='-fl2xH3x^-h 6x+ 1)1 2d Rem. = Sx*+ 12x^-f 15x-^+ 6x -j- 1 3.r^+12.T='+15x^-f6x+ l 3d Piem. = T» 0~0 220 FORMATION OP THE POWERS AND RULE. Arrange the given polynomial with reference to the ascending or de- scending powers of one of its letters. Extract the ciile root of the term on the left, and set the root on the same horizontal line toith the given polynomial on the right, and separated hy a vertical line. Subtract the cuhe of this first term of the root from the given polynomial, and hring down the remaining terms for the first remainder. Write three times the square of the first term of the root immediately beneath the place of the root. This will be the divisor to find all the other terms of the root. Divide the first term of the remainder by the divisor, and set the quotient, with its appropriate sign, on the right of the first term of the root, for the second term, of the root. Set, on the right of the divisor, three times the product of the first and second terms of the root, affected with its proper sign, plus the square of the second term of the root. Next, midtiply the algebraic sum of the three terms, thus formed, by the last term of the root, and subtract the product from the first remainder. Bring down the remaining terms for the second re- mainder, and divide the first term by the divisor. The quotient is the third term of the root. TaJce three times the square of the sum of the first two terms, and add to this three times the product of the algebraic sum of the first two terms by the third term, phis the square of the third term. Multiply the algebraic sum of the whole by the third term,, and. subtract the product from the second remainder. Bring doivn the re- maining terms from the third remainder. Find the fourth term of the root as before, and continue the process iintil all the terms of the root are found. Remarks. 1st. When the exponent of the letter, according to which the ar- rangement is made, is positive in all the terms of the given polynomial, we know that the root is not exact, whenever the exponent of the as- sumed letter is less in the first term of any of the successive arranged remainders than it is in the divisor. 2d. It is evident that the steps of the process, according to the rule, amount to nothing more than subtracting the cube of the algebraic sum of the terms of the root, when found, from the given polynomial. It may, therefore, sometimes save trouble to proceed thus. Subtract the cube of the first term from the given polynomial. Find the second term of EXTRACTION OF ROOTS. 221 the root according to the rule. Cube the algebraic sum of the two terms of the root, and subtract the result from the given polynomial. Find the third term, and subtract the cube of the algebraic sum of the first three terms of the root from the given polynomial. Proceed in this manner until we get a zero remainder, or until it becomes evident that the given polynomial is incommensurable. EXAMPLES. 1. Eequired the cube root of x^ -f 3x« + 3x^ + 7x« + 12x* + Qx* + 12x='+12x2+ 8. Aus. x'^+x^^^. 2. Kequircd the cube root of x^ + 6x' + 21a;* + 44a;' + 63^-^ + 54x + 27. Ans. x^ 4. '2x + 3. 3. Eequired the cube root of 8a;« + 48x' + lG8x^ + 352ar' + 504x2 + 432:c + 216. Ans 2x' + 4a; + 6. 4. Required the cube root of tt + x + ~yr +"97"+ — /^ + -^ + 1- , x^ x ^ Ans.- + j + \ r -n . ,,, , , ,. 125a;' 75.?;"' 15a;' a;'' 5. llcquired the cube root of ■ . - -\ -7- 4 1 . ^ 64 ^ 64 ^ 64 ^ 64 Sa;^' la; An,. ^+Y- p T> • J xu 1- i. ^ 125a;' 7bx'' 15.';* a;' 0. Required the cube root of 1 1 1 . 5a;' la; Ans.-~ + -. r, T, . , ,, , ^ ■x\ Sx' 3.r' 7a;« Sx' Sx* 7. Required the cube root of — +-— + —- + —- + __ -| 8 8 8 8 2 4 , 3a;' 3x^ , ^ . ^' ^ , 8. Required the cube root of Scv^ + 12a' J — 12a^m + Qah' — 12amb + Qant" — 3mb^ + Sm'b + 6' — m' Ans. 2a + b^m. 9. Required the cube root of a;' — 9x^^ — Sx^ + 27a;/ + 18a;y + 3x — 27/— 272/^ — 9y — 1. ^«s. a; — 3^ — 1. 10. Required the cube root of x^ + 3x* + 3a;y — Sa;^ + Sa;" + a;' + 6.xy — 6a;'y + 6a;V + SxY — QxY — Zxhj + 3.r/ — 6x/ + 3a;/ + / — 3/ + 3/ — /. Ans. x^ + x + / —y. 19* 222 FORMATION OF THE POWERS AND CUBE ROOT OF INCOMMENSURABLE POLYNOMIALS. 283. — EXAMPLES. 1 . Required the cube root of x^ + 3j;^ + 3.r + 3 Ans. rr + ] 2. Required the cube root of x^ + Qx^ + 12a; + 9. 2 Ans. rr + 1 + g-^+,&c. Ans. a^ + 2 + ^^-^ + ,&c. ox 3. Required the cube root of x^ + o^. Avic nr _L . 3^2 9a-.^' 4. Required the cube root of x^ + c,x^ + 4. a" a" ^ns. ic + 1 , &e. X 5. Required the cube root of x + 3.c^ + 6. 6. Required the cube root of u^ + Zar^ + 3x-i + 2 y/x} Ans. .r~' + 1 + - — ;r- + , &c. 7. Required the cube root of x~^ + &x~'^ + 12.x~' + 9 3^ Ans. X > + 2+o^+,&c. 8. Required the cube root of x~^ + 3a;~^ + 3.t~^ + 2. Ans. x~'^ _j_ 1 --(-- 9. Required the cube root of x'' + 9.r« + 27.r' + 26 ^72S. .T^ -j. 3 10. Required the cube root of x^ + Scc^ + 3a;* _}- a; Ans. x-2 j_ l.f-J_-f &e. ^72s. a-3+3 — — — ,&c. Ans. x^ -\- X — K J &c. EXTRACTION OF ROOTS. 223 CUBE ROOT OF POLYNOMIALS INVOLVING FRACTIONAL EXPONENTS. 284. — EXAMPLES. 3 1 1. Kequired the cube root of x~ + 3^:; + 3x- -f 1. 7 5 2. Required the cube root of x + Sz^ + 3x^ + x 3 9 3. Required the cube root of .t« + 3.c' + x- + 3x- Ans. a;- + 1. 1 3 1 1 Ans. x^^ + x^. I ins. x^ + a:-. 5_1 L3 3 4. Required the cube root of .x'^ + 3x ^ + ox ^ + x'^. Ans. x^ + x^. 5. Required the cube root of .r« + 3x^ + 3a;* + 3ar' + Gx^ + Sx^ + x^ + Bx + 8x2 4. 1. ^„s. 0-2 + x2 + 1. 6. Required the cube root of 8x« + 2-kJ + 24x'' + 48x'^ -f 24x=' + 8x2 4. 24x + 24x2 ^ 8. Aiis. 2x^ + 2x2 + 2. 1 4 ]j? 7. Required the cube root of x2 + 3x^ + 3x •* + x' Ans. x^ + X. 11 17 8. Required the cube root of x^ + Sx*^ + Sx^ -\- x^ . 1 J J Ans. x^ + x'«. 1 fi i_4 4 9. Required the cube root of x^ + 3x ^ + 3x " + x^, 2 4 Ans. x3 -|- x^. 10. Required the cube root of x'^ + 3x^4^ + 3x'^^ + 1 + 3x^ + 3x2 + 6x^4'^ + 3x* + 3x^ + xl Ans. x' + x4 + 1. 11. Required the cube root of x~'^ + 3x~ ^ + 3x~2 -j. 1 _j- S^c » _f- 6x~'4 4- Sx~~ + 3x-' + x~4 + Sx~4. Ans. X-* + x~4 + 1. 224 FORMATION OF THE POWERS AND 12. Required the cube root of x"'^ + 3x~ s + 3x~ § + x~^. Ans. x~^ + *~^- L3 8 13. Required the cube root of cc^ — 3x ^ -f 3x3 — ^. Ans. x^ — x^ . 14. Required the cube root of x^ + 3*^ + Qx^ + ^x'^ + Qx^~ +^x^ + lOx^ + 9«2 + 1x^^ + 6x + 3a;^ + 1. Ans. x^ -^ x -^ x^ + 1. _3 _S 15. Required the cube root of 1 + 3x-' + 6j3~^ + 3a; ~ + 6 ~ + 9.7;~3 + lOx-^ + 9x-^ + 7a;~^ + 6x-^ + 3x~'^' + »-«. Ans. 1 + X-' + x~3 + a;-l IG. Required the cube root of 27xH81x5+162x'' + 81«~ + 162a;^ + 243a;^ + 270a;=' + 243a;H189x"^' + 162x + Slx^ + 27. Ans. 3 (a;'^ + a; + a;^ + 1). CUBE ROOT OF POLYNOMIALS CONTAINING ONE OR MORE TERMS AFFECTED WITH NEGATIVE EXPONENTS. 285. The process is the same as for polynomials, all of whose terms are affected with positive exponents; observing, in the arrangement, that that negative exponent is the least algebraically which is the great- est numerically, and that when the letter, according to which the arrangement has been made, has disappeared from any of the succes- sive remainders, it may be introduced affected with a zero exponent. EXAMPLES. 1. Required the cube root of x~^ + 3 -[- 3x^ -f- x^. Ans. x~^ -f cc*. For, by the rule x-^ + 3 + 3x« + x"^ I x-"- -f a* (3x-* + 3a;2 -f x^'^x" 1st Rem. = 3x° -f 3.^^ + x'^ 2d Rem. = EXTRACTION OF ROOTS. 225 This polynomial miglit have been arranged with reference to the de- scending powers of x. 2. Required the cube root of :r~^ + Sji-^ + 3 + a;'- Alls. cc~^ + X. 3. Required the cube root of x-^ + 3:e-^ + 3x-^ + 3x-2 + 6x-' + 4 + 3^ + 3^2 + x^. Am. x-'' -\- x ■\- 1. 4. Required the cube root of x' + S.r^ + ox + 7 + 12x-' + 6x-^ + 12x-^ + 12x-^ + 8x-«. ^«s. a; + 1 + 2^-^ 5. Required the cube root of I + Gx"' + 21.>:-^ + 44x-^+ 63x-^ + 54x-^ + 27x-«. Am. 1 + 2x-' + Sx"''. 6. Required the cube root of a"^ — Ga-^i + 12a-^Z<' — Sa-«6'. ^Hs. a-' — 2a-^6. 7. Required the cube root of x^ + Gx^ — 40 + GGx-^ — G4x-^. Ans. X -f 2 — 4x~'. 8. Required the cube root of 1 + Gx"' — 40x-^ + OGx-^ — G4x-«. Am. 1 + 2x-' — 4x-2. 9. Required the cube root of 1 — Gx"' + lox-^ — 20x-' + 15x-' + — 6x-5 + x-«. Am. 1 — 2x-' + x-^ QUANTITIES AFFECTED WITH FRACTIONAL EXPONENTS. 28G. Suppose it be required to multiply a" into itself until it is taken three times as a factor. Then, by the rules for multiplication, «" X a" X a = a'-. This result is, evidently, the cube of a". Now, o-jmul- .1 J 3 tiplied by itself once, gives a' X a' = a~ = a ; and ^/ a, multiplied by itself once, gives -J a X \/ a ^ yj a- = a. So, a', and \/a, arc cqui- valent expressions. The expression, a"-, then, which indicates the cube of a", also indicates the cube of sf a. The numerator of the fractional exponent denotes the power to which the quantity has been raised, and the denominator the degree of the root to be extracted. Taking a three times as a factor, we have a X a X a = a^"*"^"*"^ := a. Now, P 226 FORMATION OF THE POWERS AND the ^/ a, raised to the third power, will also plainly be a, because, from the definition of Involution and Evolution, it is evident that the latter undoes what the former has done. The root of any quantity raised to a power indicated by the index of the root, must then be the quantity itself. Hence, ( ^ a)* := a ; then, a= ^ a ; since we have seen that both a , and V a, when cubed, give a. In general, an = ^a". Because aa, taken as a factor n times, gives d^Xa^Xa^ = a" " n = a", and V«"j raised to the n^^ power, is also a™. The fractional index may be regarded as denoting a poioer of a root. The denominator expresses the root, and the numerator the power. The denominator shows into how many equal factors, or roots, the given quantity is divided, and the numerator shows how many of these fac- tors have been taken. The fractional index may be considered to de- note tJie root 0/ a power, as well as the power of a root. Thus, a^ may either denote that the third power of the *J a has been formed, or that the square root of the third power has to be taken. Taking the former view, it indicates an executed operation ; taking the latter view, an unexecuted operation. In general, a" may be read the m^'^ power of the n^^ root of a, or the «."' root of the m}^ power of a. It is more convenient to read the ex- pression thus, a to the m divided by n, power ; or a, raised to a power denoted by the quotient of m, divided by n. So, ai is read, a to the P — power. 9 There are two consequences of notation by means of fractional expo- nents that deserve consideration. 1st. Any multiple of the numerator of a fractional exponent may be taken, provided an equal multiple of the denominator is also taken. Thus, az=a^ is also n=a.^^=a^=a"', &c. So, also, ai = a'l. The reason of this is plain; the increase of the denominator makes the equal roots or factors of the given quantity, a, smaller; but the numerator being increased just as much as is the denominator, the number of these smaller equal roots will be increased enough to make up for their diminution in magnitude. 2d. Since the fractional exponent may be exchanged for any other of equal value, it may be expressed in decimals. Thus, o^ = a"*^ ~ a-^. So, also, «•'» EXTRACTION OF ROOTS. 22 1 = a.037037+_ These decimal indices are called logarithms. The manner of calcu- lating them, and their use, will be shown hereafter. MULTIPLICATION OF QUANTITIES AFFECTED AYITII ENTIRE OR FRACTIONAL EXPONENTS. — MONOMIALS. 287. Quantitities with fractional exponents must plainly be multi- plied in the same way as quantities affected with entire exponents. That is, the exponents of the same letter or letters in the multiplicand and multiplier must be added, and after the common letter or letters must be written tho.ee not common, with their primitive exponents. Let it be required to multiply a^ by o^. Then, a- is to be repeated a^, or a- times. Now, to repeat a-, a- times, we have only to add the exponents of multiplicand and multiplier, for the first exponent 1 3 denotes that a^ has been taken three times to produce the product, a-\ and the second exponent denotes that a^ has been taien four times in 4 1 the product, a^. Hence, a^ must enter seven times in the result of 3 4 .7 the multiplication of a- by a-, and that result must be written a^. To -1.1 i 1 multiply a- by h, is to repeat a-, h times. Hence, a-xh = a^b. RULE. Multiply the coefficients together for the coefficient of the product. Reduce the exponents of the same letters in multiplicand and multiplier to the same denominator, add their numerators, and set their sum over the common denominator. Annex these letters to the new coefficient, and lorite after them all the letters ichich are not common to the multi- plicand and midfiplier. '-. EXAMPLES. 112 3 3 4 1. Multiply a^6°" by a°Z>^c. Ans. n^l>''c. i J 1 7 13 9 „ ,, . . , a^McS ^ aWc+' , a^b 4 c^ 2. Multiply -—^— by ^— . Ans. ^ . 228 FORMATION OF THE POWERS AND 3. Multiply by . Ans. 4. Multiply ^ ^ J by a~x^i/^z^. x-i/z" 5. Multqily a "6 "c p by a"i"'cp . -4/is. aa-^z. Am. 1. s-2r q-3p ^ws. a 2^ 6 3q ^ 1 J^ Ans. — 56x^3/4% '^0. ^ns. 2no+l 3d+1 4p-|-l 60a 2 V^G * . 6. Multiply a ^b "Jc - by a'i^c^ 3 4 4 I. Multiply 2a •'b "c " by — - — . 8. Multiply Sx'^y^z'" by — 7x^bK^. 9. Multiply 12a"'i-cP by 5a'' 6 c*. _^ _1 __L " _L -P 10. Multiply 4a " i =c i bya°6»ci. J-«s. 4. DIVISION OF QUANTITIES AFFECTED WITH FRACTIONAL AND ENTIRE EXPONENTS. — MONOMIALS. 288. Since tbe exponents of tbe like letters are added in multiplica- tion, they must be subtracted in division. For, the object of division is to find a quantity called the quotient, which, when multiplied by the divisor, will give the dividend. Let it be required to divide a^ by a' ; then, we must find a quantity, which, when multiplied by a" will = a — a give a^. This quantity is plainly a'~, for a X a~ = a ^ 3 But the quantity found, a~ , has resulted from the subtraction of the exponent of a"' from 2 = 4, the exponent of a^. So, the result of the I 1 " i_ 1 J._ L ' i division of ci^ by a°^ must be a"' ", since a" '•Xa»=a'". When the denominators of the fractional exponents are diff"erent, the subtrac- tion can only be indicated, not performed, until they are reduced to the same index. If the exponents of the same letter in the divisor exceed that of the dividend, the exponent of that letter in the quotient EXTRACTION OF ROOTS. 229 will be negative. So, also, if there are any letters in the divisor not coiumon to the dividend, they must appear in the quotient, with their primitive exponents taken with a contrary sign ; else, when the quo- tient and divisor are multiplied together, these letters would enter in the product. RULE. Divide the coefficient of the dividend hy that of the divisor, the result icill he the coefficient of the quotient. Write after it all letters common to dividend and divisor, affected with exponents equal to the difference of their exponents in the dividend and divisor ; and, also, cdl the letters common to the dividend only, with their primitive exponents; and all common to the divisor only, icith their primitive exponents taken with a contrary sitjn. 3 1 .1 i } 3 _ 1 1. Divide 4a2c-(?'^ by 2a"c'cZ. Ans. 2a~cd ^^. 11} 1 °^'" '-2« t 2. Divide VZa^b' c d^ by Ga- b"". Ans. 2a »' b'^c d^. 3. Divide 7x"'?/"zp hj x'"y''z~p. Ans. Iz p . _j _j J I 1 _i 4. Divide 72 "rj hjz'y x . Ans. Tar'^-'x . 5. Divide 24a'*i^c^ by 242='a~''i~^c"^. Ans. z'^ahc. h + -} 18 6. Divide l^x" y"" z by — j — j — -. a;~V 2~* Ans. 1. 7. Divide oa-^h ' c" by h^'vh ^d "■ . Ans. 9i ' c^J" . 8. Divide 80m"r'"i^c^ by 5m^j"^;r". Ans. IQm'Hc^x'^. _1 1 _i_ (m+n (l+n) (I+p) 9. Divide 360e -/"^ p by 1 80^7^. Ans. 2e~~^f ^g~'^. 10. Divide4a"c^&'' by a-c^~^e~*. .4ns. ch^d^^e\ 20 230 FORMATION OF THE POWERS AND RAISING TO POWERS QUANTITIES AFFECTED WITH FRAC- TIONAL AND ENTIRE EXPONENTS. — MONOMIALS. 289. Since, to raise a quantity to a power is only to multiply it by itself a certain number of times, the rule for the involution of quantities affected with any exponents whatever is deduced directly from the rule of multiplication. RULE. Raise the coefficient to the required power, and write after it all the literal factors affected with exponents equal to the product of their primitive exponents hy the exponent of the power. Thus, to raise a/ to the third power, is to take a three times as a '' \ i 2" 2+^+^ ^ factor. Then, (a")'' = a Xa''y.a =a " =a-^. The exponent of the result is the product arising from multiplying the exponent of 1 m the quantity by 3, the exponent of the power. So, (Mx" y=(M.yx° , I J_ L 1 because, (Ma;")" = M X M X Ma;"' +'""'''" + &c. If the given quan- tity be a fraction, it is raised to the required power by raising the numerator and denominator, separately, to that power : because, the power of a fraction is nothing more than the product of the fraction multiplied by itself a certain number of times ; and a fraction is mul- tiplied by itself by multiplying the numerators and denominators to- gether separately. EXAMPLES. 1. Find the — ^^ power of a". Ans. a ' . 2. Find the m^^ power of a ' . Ans. a~. 3. Find the f' power of a". Ans. a~^ EXTRACTION OF ROOTS. 231 l_ — m 4. Find the — m'" power of a ' . ^4718. a 5. Find the 4*" power of oaVi^ Ans. 81a^c'-6'«, 6. Find the 3" power of the 4<* power of j-. Ans. 12Sa''b-''. 2a 7. Find the 2'" power of ^. Ans. (2)''a''t~^ 8. Find the —Shi*" power of a^b^ Ans. a-x^Z/-'^" 9. Find the — 5m^^ power of the S"* power of a^P. Ans. a-^-^b-^" 10. Find the — 12«'' power of the — G*" power of ^ + a''b-''. 11. Find the — 12*" power of the — T"" power of j-. Ans. +a'*b-^. 12. Find the — 11"' power of the — 7"" power of — — . A71S. —aPb-"". 13. Find the ?■"> power of the ^n'" power of a". Ans. a°°". 14. Find the — r*"" power of the m^^ power of a~° 15. Find the r*" power of the — m^^ power of a~" 16. Find the — r*" power of the — m*"* power of a° 17. Find the S"" power of . 18. Find the 5'" power of . 19. Find the C" power of 2a^b^. 20. Find the 6"' power of ~2a}b^ 21. Find the 31 power of laH^. An. . a'^^ An. . a""'. An \ a'""^ Ans. ah* C5 • ns. + C5 Ans. Ua'b. Ans. Ua'b. Ans. 8a6^. 232 FORMATION OF THE POWERS AND 1 1 1 22. Find the 3'' power of —2a^h^. A^is. —Sab-. 23. Find the i"" power of Ma "Lp. Ans. (Myall v. EXTRACTION OF ROOTS OF QUANTITIES AFFECTED WITH ENTIRE OR FRACTIONAL EXPONENTS. — MONOMIALS. 290. To extract the m'** root of any quantity, a^, is to find a second quantity which, when raised to the m'^ power, will produce the given quantity, «■'. The extraction of a root is then just the reverse of raising to a power, and, of course, the reverse process must be pursued to find p p the root. The m^^ root of a^ must be am, because the m"* power of am is equal to a^. To raise a quantity, which has a coefficient, to a power, we first raise the coefficient to the indicated power, and write the new coefficient before the literal factors as the coefficient of the power. Hence, to extract the root of a quantity, which has a coefficient, we must first extract the root of the coefficient, and write it before the root of the literal factors. The on'^ power of Ma^ is (M)"'a''°', therefore, the m"" root of (M)"af"" must be Ma''. The ru* power of the fraction - = — . Hence, the wi"" root of -, — = — . That is, the m"" root of numerator and denominator must be taken separately. Extract the root of the coefficient, and write the result be/ore the lite- ral/actors affected with exponents equal to the quotient arising from dividing their primitive exponents hy the index of the root. If the co- efficient of the given monomial is not a perfect power of the degree of the root to he extracted, the operation is impossible. If the exponents of the literal factors are not divisible by the index of the root, the lite- ral factors win appear in the root with fractional exponents. The root of fractions is extracted by extracting the root of the numerator and denom in a tor separately. EXAMPLES. 1. Find the -"^ root of a ^ Ans. a". r m Ans (Z— 2. Find the r"" root of a^. ' '^' EXTRACTION OF ROOTS. 233 3. Find the m'" root of a^ Ans. ar . 1 " 4. Find the '" root of a'. Ans. a-°. 5. Find the — m'" root of a ■• . Ans. a ^ . 6. Find the — j-'" root of a "^ . Ans. a~~^. m I 7. Find the — m'" root of a" r . Ans. a'r . 1 m ^ 8. Find the '" root of a~ ' . Ans. a ' . m 9. Find the 4'" root of Sla'c'W. Ans. 3aVt^ 10. Find the 7'" root of 128a''i-''. Ans. 2ah-\ 11. Find the q"^ root of (2)''a<'i-^. Ans. 2ah-\ 12. Find the — Sm'" root of a-«^t-'^. Ans. aW. 13. Find the r*" root of a"". Ans. a"". 14. Find the 4'" root of l^Jn^. Ans. 2a^^b^. rt"-/)3 a- 6^ 15. Find the 5'" root of —-. Ans. . c^ c 5 5 11 16. Find the 5*^ root of — — ^. Ans. . c" c 17. Find the 3'' root of 8ah^. Ayis. 2ah^. 18. Find the 6'" root of Mci'b. Ans. 2a*6*. 19. Find the 3* root of — 8aZ>i Ans. — 2a*6*. t 21^ j 2 20. Find the t^^ root of (Mya^bp. An-s. Ma°ip. PROMISCUOUS EXAMPLES. -i -i 1 _3 _ 3 1, Raise 2a "b ^c^ to the 3* power. Ans. Sa -b ic. _3 _£ _i _j_ i 2. Extract the 3" root of 8a ^5 p. Ans. 2a 'b pc 20* 234 FOEMATION OF THE POWERS AND 3. Multiply 2a-P6-°'c-'' by ZaPh'^c''. Ans. 6 4. Divide 6 by 3a^h'^c\ Ans. 2a-Pi-"c-". 5. Divide 6 by 2a-P6-"'c-». Aris. Sa^h'^c^ 6. Raise 2a^b^c^ to the 6'" power. Ans. 64a'&c^. 7. Extract tlie 6'" root of Q^a^bS. Ans. 2a^b^c^. 1 1 m— 1 n— 1 8. Multiply (M)mar by (M) ma-. Ans. Ma. m-l E-1 Jl_ 1 9. Divide Ma by (M)"^a"^. Ajis. (M)'nan. 1 1 in— 1 n— 1 10. Divide Ma by (M)^an. Ans. (M)"^a~. 11. Multiply 8a;"V' by ^. Ans. xy. 12. Divide a;y by 8a ^y "*. ^ns. 4 5 3,,4 a;oy 4 5 3 ,,4 13. Divide cry by ^^. ^ns. 8a; ^ ^. o 14. Raise to the third power. Ans. —Vlhx-^y-^z-^a~''b-^. 15. Raise Va^V to the r* power. Ans. (Vya^'b^'. 16. Extract the r'" root of (P)V^6^ ^ws. Va%\ 17. Divide (2)Vi=' by (2ya'"i-". ^ns. 2a' -"'J^'". 18. Multiply 2a''-"i2-m itjy (^2ya"i». Ans. Sa'b'. 19. Multiply 2a^'"&2-m ^^^ (2)V"'J="". ^?js. (2)V"+'">J2('+'">. 20. Extract the s*" root of (W)'^a-'b~'^ . Ans. (W)''a-'i""^. Li 10 20 21. Raise 3a^6'- to the 10*^- power. Ans. (py°a'b'. 10 20 12 22. Extract the 10"^ root of (3)"'a^6^ Ans. Sa^b~'. 23. Multiply (4)'a^^* by (ifah^. Ans. (4)V5. EXTRACTION OF ROOTS. 235 24. Divide (4)V6 by {^fah^. Ans. (4)'a^6*. 25. Divide (4)Vi by (4)'a-6*. Ans. {Afah^. 11 3 3 26. Raise — a"&^ to the 3''* power. Ans. — d^h^. i I 4 27. Raise — a'h^ to the 4*" power. Ans. + a^i^. I 2 3 28. Raise tx^y^z'^ to the c*" power. ^ns. (2yxy^^. 2 2 £ 29. Extract the c*" root of (2)'x/^'. ^ns. Ix^y^z^. 30. Find the r'" root of ^X. Ans. 2a~^. 31. Raise 2a~' to the r"" power. .4hs. (2)'' a"". 32. Raise aVrUo the 10'" power. Ans. a%\ 33. Extract the 10"" root of a'1/. Ans. a'b\ 34. Raise a-^i« to the 2'' power. ^tis. ah''' 35. Extract the square root of aZ*'- 1 vl?«s. a-^i-®. 36. Raise a-'^h-'^ to the 6"" power. Ans. a-°"b-°'\ 37. Extract the 6'" root of a-o'^i-^'l Ans. a'^'b-'^. 38. Raise a'"«6-«» to the 1000'" power. ^Ins. a'b\ 39. Extract the lOOO'" root of a'b^. Ans. a-°°'b-<^. 40. Multiply ah^ by a^^^i-^. A^is. a-'=i« 41. Divide a-"6"^ by a-6". ^?is. a-^b-^. 42. Divide a-^5Z>-« by a-^S-^'. ^ns. a'i-*. 43. Multiply (10)2a-2i-=' by (lO/a- "J- ^ Ans. (lO/a-^S-*. 44. Divide (10)^a-^6'5 by (\^fa=b\ Ans. (\^fa-%-\ 45. Divide (lOya-^i-^ by (lO/a-26-3. Ans. (l^ya-^b-\ 46. Multiply (10)-'a-'6-«by(10)'a--'6-^. Ans. (\Ofa'b-\ 236 FORMATION OP THE POWERS AND 47. Raise 2aZ>V to the yVh power. A71S. (^ly'a-'hh". 48. Extract the yigth root of {^y'a-'h-h^ Ans. 2ah^c\ 49. Raise (2)':r?/V to the /^ih power. Ans. {2y^x-^yh-\ 50. Raise (2)5a'"5" to the /gth power. Ans. 2a-^'"h-^\ It will be seen that decimal powers are smaller, and decimal roots greater, than the quantities themselves. CALCULUS OF RADICALS. 291. Any quantity with a radical sign is called a radical quantity, or simply a radical. Thus, \/a, \^b, Vc^, are radical quantities, or radicals. The coefficient of a radical is the quantity prefixed to the radical sign, and it indicates the number of times, plus one, that the radical has been added to itself. Thus, 2y/a and m^/a, indicate that ^/ a has been added to itself once and vi — 1 times. When no coefficient is written, unity is understood to be the coefficient. Thus, >/ a = ly/a. When the indicated root can be exactly extracted, the radical is said to be commensurable or rational. Thus, ■v/4 and ^8 are rational radicals. When the indicated root cannot be exactly extracted, the radical is said to be incommensurable or irrational. Thus, ■s/2 and ^ 5 are irrational or incommensurable radicals. A root has been defined to be a quantity, which, taken as a factor a certain number of times, will produce the given quantity. An even root is a quantity, which, taken as a factor an eve7i number of times, will produce the given quantity. But no quantity taken an even number of times will produce a negative result. Hence the even root of a negative quantity is impossible. The indicated even roots of negative quantities are called imaginary quantities. Thus, y — 2, ^ — 2, v' — 2, are imaginary quantities. Roots that are not imaginary are called real roots. The term rational is in contradistinction to irrational. The term real is in contradistinction to imaginary. A quantity may be real and not rational ; but no quantity can be rational or irrational and not be real. Thus, V2 is real, but not rational; but v'4 and \/2 are both real. EXTRACTION OF ROOTS. 237 292. It has been shown that all radicals may be changed into paren- thetical expressions ; the numerator of the exponent of the parenthesis denoting the power to which the quantity under the radical sign is raised, and the denominator of the exponent of the parenthesis denoting the index of the radical, or the degree of the root to be ex- tracted. Thus, v/a* may be changed into a-, and -(/a™ may bo changed into a". 293. A simple radical is one, which, when changed into an equiva- lent parenthetical expression, has unity for the numerator of the exponent of the parenthesis. Thus ^a, i/a, &c., are simple radicals. A complex radical is one which has the quantity under the sign raised to some power different from unity. Thus, \/a', ^a", are complex 3 ^ radicals, the equivalent parenthetical expressions («)"-, and (a)", having numerators different from unity. 29-4. Radicals are said to be similar when they have the same index and the same quantity under the sign. Thus, 2yja and 3's/a are similar radicals. The ^ a and ^a are not similar, because, though the quantities under the radical signs are the same, the indices of the radicals are different. The s/ a and \/6 are dis.similar, because the quantities under the signs are different. 295. Similar powers of the same quantity can be added by adding their coefficients. Thus, 2a^ -f 3a^ = 5a^ ; because, since the literal factors are the same, they may be represented by the same letter, x. Then, 2a* -f 3a^ = 2x + 3x = Sx, and replacing x by its value a^, we have the sum of the two quantities equal to ba^. For a like reason, similar powers of the same quantity may be subtracted from each other, by taking the difference between their coefficients and uniting this difference as the coefficient of the common quantity. Thus, 5a' — 2a» = 3a^ 296. In the same way, similar radicals may be added or subtracted, 2s/ a-{- •y/a = 3v/a. For, make V a = a; ; then 2v/a-|- v/« = 2x + X = Zx = Z^ a. So, also, 2s/a — y/a = y/a. 297. The calculus of radicals shows how radicals may be operated upon algebraically — added, subtracted, multiplied, &c., &c. These algebraic operations must be in accordance with certain prin- ciples, which both modify and facilitate them. It would require an extended treatise to embrace all the principles of the calculus of radicals. A few of the most important only are given in this work. 238 FORMATION OF THE TOWERS AND First Principle. 298. Any parenthetical expression, composed of several factors, may be decomposed into as many new parenthetical expressions as there are p p p p factors. Thus, (aZ>c)T = (a)q (5)q (c)q . For by the rules for raising a monomial affected with any exponent, fractional or entire, to a power, each factor has to be separately raised P P £ P to the indicated power. Hence, (cihc)i — (a)q (b)i (c)q . And, in p P L H like manner, (a'"i"c'')q = (a'")q' (6")i (c')'' • 299. This principle has an important application in the case of simple I I I 1 I radicals. For (abed)'" = y abed, is also equal to (a)" (&)n (cf {dy or to ^a . y 6 . ^c ^ d. Hence, ^ abed = )x/12. 18. Reduce 40^^" + SacVc"^ — 9ac\/c to one sum. Ans. 0, 19. Reduce 12iV«"'"'"c'°' + 5m V« to one sum. Ans. (12&ac2 + 5m) Va^ 20. Reduce 3\/aV6^ and — 2abcy/abc to one sum. ^?is. abcy/abc. The whole difficulty in solving these examples consists in finding the incommensurable factor common to all the radicals. Second Principle. 302. The Ji"" root of the quotient of two quantities is equal to the quotient of their w* roots. That is, \ / — = — — V 6 «/6- For, to raise a fraction to any power, we raise the numerator and denominator, separately, to the required power. Hence, to extract any root, as the n'", the root of each term of the fraction must be extracted separately. This principle is used in extracting the roots of fractions. EXTRACTION OF ROOTS. 241 Third Principle. 303. The mn^^ root of any quantity is e(|ual to the m^^ root of thy n^ root of that quantity. That is, '":y~a = k/ ^ya. A quantity, a, may be raised to the sixth power by first squaring a, and then cubing the result. Hence, the sixth root of a® might be extracted by first extracting the square root, and then extracting the cube root of the result. So, a may be raised to the wiJi"" power by first raising it to the ?<'" power, and then raising the result to the m"" power. Hence, the m?*"' root can plainly be extracted by taking first the n"", and then the m"" root, in succession. 304. This principle is of great importance, and extensive applica- tion. It is used to extract high roots in succession, whenever their indices can be decomposed into factors. Thus, V256(W^=:\/y"2o m^iW = ^ l()a^6« = \a^V All even roots, and roots that are multiples of 3, can be extracted in succession. The sixth root is equal to the cube root of the square root. The eighth root is equal to the fourth root of the square root, or it can be found by extracting the sfiuare root three times. But the fifth root, seventh root, eleventh root, &c., cannot be extracted successively. When the principle can be applied, we begin by extracting the low- est root first. The index of the radical must be decomposed into its prime factors, and the root corresponding to the lowest factor ought to be first extracted. It frequently happens that the factors are equal. EXAMPLES. 1. Extract the eighth root of 256a'i^ 2. Extract the sixth root of Tl^a}%^\ 3. Required V 6561 a^^6'2 j^^g g^ejs^ 4. Required ^'25Jbc and ^cb^s/b&= >/96V = s/^V& -Jbc. Hence, sfbc is a common incommensurable factor, and the sum is {2ac -f- 35c) s/ be. 2. Add Ax'^y^s/a^ = y / 3 . 5 ■2^3V6^ and 6a V 240a = V 36 . 3 . 5 . 4V. Hence, y/o.S.a, or \/15a, is a common incommensurable factor, and the sum of the two expressions is (Qah + 24a) \/ 15a. 4. Add together V90 and V'MO. y/90=V3^5.2 and v/810 = n/3\2.5. Hence, v/2.5 is the common incommensurable factor, and the sum is (3 + 9)v/ 10. 5. Add together as/2AQ and Z>n/2400. ^hs. (6a + 206) VG". 6. From a v/300 subtract Z>V"243. Ans. (10a — 9Z>)V37 7. From a V 33750 subtract 6-^/10000. Ans. (Sa — 2h) ^l250" 8. Add the three expressions, hay/ ax, mx^\/Qa'^, na" s/ x. Ans. (ha + 3ma + n^s/ax. 9. Add the three expressions, ha^~ax, mx'^^27a* + na^ ^x. A?is. {ha + 3ma + n)^ax. 'ax + mx* ySla^ + na'^ ^Ic. Ans. (ha + Sma -f- n)^ax. 10. Add the three expressions, haijax -{■ mx'^ y 81a^ + na'^ ij x. Fifth Principle. 307. Any factor may be passed without a parenthetical expression, by multiplying its exponent by the exponent of the parenthesis. p mp p Thus, (a'"hc) q = a 1 (6c) + c)5 Dip p It matters not how many terms there may be within the parenthesis ; they may be all represented by a single letter, after they have been divided by the factor to be passed without. The foregoing demonstra- tion will then be applicable. EXAMPLES. 1. Pass X without the parenthesis (xy^z^)~~^. Ans. x~^(i/^z^~ 2. Pass x~^ without the parenthesis (x~^y^z^y. Ans. X " (i/^z^' Pass X without the parenthesis (:c ah)- Ans. x(ahy 4. Pass 4 without the parenthesis (4a2i')-. Ans. 2(a2?'^)- - £ 5- 1 5. Pass x'' without the parenthesis (a:''?/'"^)P. Ans. x(y'";^y, 6. Pass 8 without the parenthesis (Sahy. Ans. 2 (ahy. 7. Pass IG without the parenthesis (IGahy . Ans. 2(ah) 8. Pass 4 without the . (4« + ?0^- / parenthesis g . Ans. \a .r 9. Pass 5 without the parenthesis (a + 5&)». 0. Pass 2 without the Ans. parenthesis (2a + 4iy. Ans. 125(1 16 (a + 2hy 21* 24G FORMATION OF THE POWERS AND (4 (mx + nx"^) + (m + 2nxf )* 11. Pass 4 without tlie parenthesis Ans 3 (mx + ^ix"^ + 4 (wi + '2-nxy )- 12, Pass the term 4a without the parenthesis (4a +h + cy. Aiif;. la \i-+ ^ / • 13 Pass the term 8a' without the parenthesis (Sa" + i + • Ans. 2a\ o 14. Pass the term bo? without the parenthesis (pci? -{■ h -\- c)" . Any number of terms may be passed without, by representing them by a single letter, and then replacing that letter by its value after the transfer has been made. 15. Pass (a + i) without the parenthesis (a + i + c)". Let a -f h = X. Then, (a + h -\- cf = (x ■\- c)* = x^(l + x-'c)^ = (a+ i)" I 1 + (a + i)-'c 1^ 16. Pass (a + h -\- c) without the parenthesis (a -{- h -\- c -}- dy. .h Ans. (a + i + «)- I 1 + (a + & + cy\l \ ' 17. Pass (a + i 4- c) without the parenthesis (a -\- h + cf. Ans. (a + 6 + cy (Tf or (a + & + cj. 18. Pass a~° without the parenthesis (a"" + ly. Ans. a~p(l + a^hyp . The last example shows that any factor affected with a negative ex- ponent may be made to appear with a positive exponent in the other terms of the parenthesis. This transformation is used in the differential calculus when it is desired to change a negative into a positive exponent. It is evident that the fifth principle is only the more extended appli- cation of the first principle. EXTRACTION OF ROOTS. 247 Sixth Principle. 308. Any factor may be passed within a parenthesis, by multiplying its exponent by the reciprocal of the parenthesis. Because, to pass it out again, we must multiply its new exponent by the exponent of the parenthesis, and when its new exponent has been formed as directed, the factor, after it has been passed out again, will p ^ p. be afifected with its primitive exponent. Thus, a'"(i)'» =(ai'Z>)'i. Because, when av is passed out again, the expression will become mq p 2 p^ OP q(i))"- of its coefficient. Ans. (^ + A-. 7. Clear 64("--^ + -^Y of its coefficient. \4U9ba lb/ 8. Clear 27(2a + |6)- of its coefficient. Ans. (18a + b)^. 248 FORMATION OF THE POWERS AND 9. Clear 32 (2a + Ib)^ of its coefficient. Ans. (8rt + b)^. Ans. (25a + 1256)" --a + -b\^ of its co- efficient. 11. Clear tlie parenthetical expression a~(l + a~')^ of its coefficient. Ans. (a-j- 1)^ It is evident that the sixth principle is only the more extended ap- plication of the fourth principle. Seventh Principle. 309. The denominator of the exponent of a parenthetical expression may be multiplied by any quantity, provided vre raise the quantity within p the parenthesis to a power denoted by the multiplier. Thus, (0)1 = Z pp JL '^ -^ JLJL (a'")""'. For (a)T = an, and (a"')""! = W'l = a ■) . Hence, (a) i = (a'")°"i. If there is more than one term within the parenthesis, their algebraic sum may be represented by a single letter, and the foregoing demon- stration is, therefore, applicable to all kinds of parenthetical expres- sions. 810. The seventh principle has two applications. 1st. It is used to cause complex radicals (or parenthetical expressions with the nume- rators of their exponents different from unity), to be alFected with ex- ponents having a common denominator. Thus, (a + by^ and (a -f i)^, can be changed in accordance with the 3 4 principle into the equivalent expressions ( ( and (a H- J)""", into equivalent parenthetical ex- pressions, the denominators of whose exponents shall be the same. P_ _£_ Ans. ( (a + ^)° ) '"'' and (a + t)""". JL -P. 2. Change (a -f t)™* and (a -f- i)""-, into equivalent expressions. Ans. ( (rt -f hy ) ■»"■' and ( (a + Z^)" ) ■»^°. 311. 2d. But the most important application of the seventh prin- ciple, is in reducing simple radicals to a common index. * ?. — — J -. _ Thus, since a" = o*", or \/ a = ^a^, and a = a", or %/ a = -J/o', it is plain that \/a and ■i^a, can be reduced to a common index. So, ^a and ^a, may be changed into the equivalent radicals '"^^a"' and '"^a". In each instance, the power to which the quantity under the radical is raised is indicated by the quotient arising from dividing the least common multiple of the indices of all the radicals by the index of the radical under consideration. RULE. Form the least common mxdtiple of the indices of all the radicals. Raise the quantity under each radical to a power indicated hy the quo- tient arising from dividing the least common multiple hy the index of the radical under consideration. 250 FORMATION OF THE POWERS AND Reduce v' a, i/ «j and ^/ a, to same index. The least common mul- tiple is 12, and the three quotiente^ ^'i^ ^> V^ ^ ^> ^°^ V' = ". Then the equivalent radicals are -7 a*, J^ a^, and '-^ al EXAMPLES. 1. Ecduce y"a^ Va^ y«^ and ^~a^io same index. ^«s. 'V'^, 7"^, 7"^, and ^"^. 2. Eeduce Va, v/a, ■C/c, '^o, and v'a, to same index. Ans. ^-i/"^5^ ^"^2, ^^^, ^"^, and ^o^.' 3: Reduce ^ a, 'X/a, '-7 a, V a, and ^ a, to same index. ^ns. ""iP/7i"", '""^~a", ""^a"', ""Vo^, and "Va^. 4. Reduce ^a»', and '^a'.' 5. Reduce ^o, ^ a, ^^ a, ^ a, '-^a, Va, and Sya, to same index. Ans. 7^, V~^, 7"< 7^ VTtV7^and ^"^ 6. Reduce '^/a, "yo; ^ct, y/ a, ^a, and '"-7 a, to same index. Ans. "^"^""^ -"7^^ "71:?°, "70°^', "^o^ and '"7"(?^ 7. Reduce -;/«, ^"^ ^^ 4/o7V«7 v' ^7 V^T and 7«7 to same index. ^HS. 2520/-^^ 2520,"^, ^520,^ 2520/-^4^ 2520/-^0^ ^^^O/"^ 2520/^ and 2520/^, It will be seen that simple radicals are changed into complex by the operation of reduction to a common index. Eighth Principle. 312. Any factor of the index of a complex radical may be sup- pressed, provided, (he same factor is suppressed in the exponent of the power to which the quantity under the sign is raised. That is, "7 a" = V a. For, '"7 a'" = (a)"^ = («)" = \^ a- EXTRACTION OF ROOTS. 251 This principle is just the converse of the last, and reverses the results of the last. It is used to simplify complex radicals, and frequently reduces them to simple radicals. EXAMPLES. 1. Simplify Via + hf, V (a + hj, '^ (a + hf and l^(a + h)\ Ans. >J a + 6, V a + 6, y a + h, and ^ a + h. 2. Simplify fi/(a + h)\ 'Via + i)'°, V{ci -f hf, and 8/(a + i/. ^«s. v/(a + hf, (a + ^')', Va + i, and X/(a + 6)'». 3. Simplify v/(a+6)«, V(a+t)', !e /(a+&)'S V (a+^')^ V(«+i)'*- ^ns. (a + />)^ ^(a + Z/), N/(a + 6/, (« + i), (a + i)^ 4. Reduce X/~^, V< ^ < and V'^. Ans. V a, i/c, y/a, and y/a. 5. Reduce i^y"^, 7~^, 7^, V"^ and 7^ Ans. i/'a, i^i, V«, 7"^, and 7 oT 6. Reduce ""Vo^, "'" power of t;/a. Ans. a. 9. Required wm*"* power of "^a. Ans. ^a, or a°. 10. Required m*"" power of '"i/a. Ans. i/a. 11. Required 3* power of lya. Ans. X/O" EXTRACTION OF ROOTS — ' 12. Required 12"' power of ^a. 13. Required 6*" power of v^a. 14. Required S*"* power of l^o. 15. Required pni^^ power of '""^a. TO MAKE SURDS RATIONAL BY MULTIPLICATION. CASE I. Monomial Surih. 315. Suppose the given surd is ^ b ; this is equivalent to i° . Now, it is required to multiply i^ by such a quantity as will make it rational. Since the surd will be rational when the numerator of the fractional exponent is exactly divisible by the denominator j and, since, in multi- plication, we add the exponents of the same literal factors, the multiplier of J " must be h, affected with such an exponent, that, when added to — , the sum of the two exponents will be a whole number. Call x the n unknown exponent of the multiplier, then x -\ = 1, or .r = 1 11 11 n — 1 "-=-' 1 ^jil = . Hence, the multiplier is i " , and we see that h'' X h " = b, & rational product. Had we placed x -\ =2, and found the multiplier under this hypothesis, the product would have been h^. But, when it is required to make the surd rational, and of the first degree, the sum of the primitive exponent, and the unknown exponent, must be placed equal to unity. Let it be required to find a multiplier which will make ?/■* rational. Then, X + i = l, ov x — i. Hence, the multiplier is i/^, and we see 3 1 that ^^ . 1/'^ =y is3L rational product. 22 254 FORMATION OF THE POWERS AND RULE. Place the primitive exponent, plus x, equal to unity ; find the value of X from this equation. The value so found will he the exponent of the midtiplier. The multiplier itself must he the given monomial, ex- clusive of its exponent, raised to a power indicated hy the value ofx. 1. Find a multiplier which will make x - rational. Ans. X " . 2. Find a multiplier that will make y^ rational. Aiis. y". 3. Find a multiplier that will make y ° rational. Ans. y' 4. Find a multiplier that will make y^+'" rational. Ans. y ''-^'" . Corollary. ^ 316. The principles demontrated in Case I. enable us to find the approximate value of fractions whose denominators are monomial surds. RULE. Mxdtiply both terms of the fraction hy such a quantity as loill make the denominator rational. Approximate as near as may he desired to the true value of the monomial surd in the numerator, and then re- duce the fraction to its lowest terms. 5 1. Required the approximate value of to within -01. Ans. 1*58. 6 WW 5(3-16) _ 15-80 _ EXTKACTIOX OF ROOTS. 2c 2. Required tte approximate value of — = to within -001. Aj}s. 1-732. 3. Required the approximate value of —= to within -01. Ans. 2-52. 4. Required'the approximate value of — =:^=r to within -001. VIOOO Ans. 3 162. 6. Required the approximate value of —= to within 01 Ans. 4-65. 6. Required the approximate value of — z=:zr: to within -0001. VIUUO Ans. -189738, nearly. 7. Required the approximate value of — to within -1. ;y4uo A71S. 54-2. 8. Required the appi'oximate value of ___ to within -01. Ans. 2-32, nearly. 9. Required the approximate value of —z= to within -000001. v/320 Ans. 3-577781, nearly. g 10. Required the approximate value of — =iz to within -001. v^2 Ans. 2-519. 15 11. Required the approximate value of — = to within -001. V lo Ans. 3-873, nearly. 30 12. Required the approximate value of to within -001. Ans. 3-873, nearly. 13. Find the approximate value of —= to within -01. Ans. 5 04, nearly. 256 FOR 31 ATI ON OF THE TOWERS AND 14. Find the approximate value of — ^= to within -01. Ans. -89. 18 15. Find tlie approximate value of — = ^18' Ans. 6-87. CASE II. 317. To find a multiplier that will make rational an expression, con- sisting of a monomial surd, connected with rational terms, or consisting of two monomial surds. Let it be required to make rational ^/p ■}- y/q hj multiplication. From the principle demonstrated in Case 1., it is plain that y/jj can only be made rational by multiplying it by v^^j, and V q can only be made rational by multiplying it by Vq- But, unless v/p and <>/ q, in the multiplier, are connected by the sign minus, there will be two terms in the product remaining irrational and unreduced. Hence, the mul- tiplier must have the minus sign between its terms. Thus, \/p + Vq Vp — Vq p-\- s/pq — Vpq- -9 p — q If the given expression is-Jp — Vq, the multiplier must be, for a like reason, Vp -}- V q- If the given expression contain but one monomial surd, and is of the form J) -f s/g-, it may be written y/p^ + Vq, and reduced, as before, by multiplying by Vp^ — V q- So, p — Vq may be written Vp'' — \/q, and may be made rational by multiplying it by Vj^" + y/q. It matters not how many terms may be under the radical, or how many rational terms may be outside of the radical, the foregoing proces- ses will still be applicable ; because the sum of the quantities under the sign may be represented by a single letter, and the sum of the rational terms outside of the radical may be represented by a single letter. Thus, a -\- b + -v/m — n = p + Vq = V p^ + V q. Multiply now by Vp^ — Vq, and replace p and q by their values. The result will be (a + Vf — (w — ii). EXTRACTION OF ROOTS. 257 Let it be required to make y/ p + v^g rational by multiplication. The operation is as follows : p + Vfq + Vp'f- + ? __ _ — Vfq — Vp(l p -j- (J = Product. The multiplier, to make ^j^ rational, must be ^jr ; and the multi- plier, to make ^q rational, must be ^ qK But, after multiplication ^y Vl^ 4- ^^2^ there remained two uncancelled surds, "/ q rational. The operation is as follows : 4/p + ^q ■■ Given surd. ^p' - + -s/ q = Second multiplier, ji — q=^ Second product. Any two monomial surds, whose common index is some power of 2, may be reduced in the same manner. And, since each multiplication 1258 FORMATION OF THE POWERS AND <;ive.s a product containing two surds, witli a common index one-lialf as great as the common index previous to multiplication, it is evident that the number of multiplications will he indicated by the exponent of the power of 2 in the primitive index, common to the surds given to be reduced. Thus, %/j> -}- \/<2 can be reduced by three multiplications, since 8 = 2'. And lyp + v/ + y/q. + >/« = Given expression, ^/p — v/ J + x/« = First multiplier, — y/pi — ^/(^H — q^ p — q -\- 11 -\- 'l^pn = First product, p — q -\- n — 2^pn = Second multiplier, (P — ? + «)* — -^Pf- = Second product. Let it be required to make rational ^jj + v/'i — \/»- The process is as follows : v^p + v/? — v/'i = Given expression, \/i> + n/? + \/^i ^^ First multiplier, p + \^pq — s/pn + Vpq -\- q ~ >/q>^ + s/pn + \/9» — n p -{■ q — " + ~\/pq = First product, p -h q — n — ~\/pq = Second multiplier, (P + q — '0^ — -^Pq = Second product. Take as a third example ^'p — y/q + ^lii + v/«. ^P — s^T + Vm + y/n y/p + >/q — ^m + ^n p — \/pq + -^pm + s/pn n + -^pn — \/ nq -\- ^/ am —q + ^pq + \/nq + ^qm — m — s/jjni — y/mn + \/qm, (jj + % — q — m) + 2-^pn + ^^ qm = First product, (j3 + « — g — m) — I'sf pn + 2 y/qm = Second multiplier, P* + 4P%/^TO — 4j9tt + 45'7?i = Second product reduced, and in which P represents the. rational term. 264 FORMATION OF THE POWERS AND We may represent P^ — ij^n + 4(pn by M^, and then we will have M** -f AY^qm = Second product, M^ — 4Pv^"^ = Third multiplier, M^ — IQF'qm = Third product. CoroUari/. 320. The principles of Case III. enable us to approximate to the /alue of a fraction containing three or more monomial surds. EXAMPLES. 1. Reduce — =: ^r^ to an equivalent fraction having a rational denominator. Ans. 5+^/5 — v'30. For _ ^^_ _ 10(^5 — VQ^+ 1) ^ ^ V5 + V6 + 1 ' (V5 + V0"+ 1) (v/T— ^6"+ 1) 10( s/5 — V'G + 1) 10 v'"5( V 5"— ^/ 0"+ 1) 2^5" 2(5) VSO + v/6. The approximate value of the result can be found, if desired. 2 2. Reduce —z= — ^rr. to an equivalent fraction with a ra- V6+v/8+v/14 tional denominator. A^is. (\/6 — \/8-f V14(6 — V 84) ' =48 ■ 20 3. Reduce iz= :z= = to an equivalent fraction with a 1+ V5+ v6+ v/10 rational denominator. Ans. (1 — V5 — V6"+ n/TO~) (v/TO + v/"30) __ _. Q 4. Reduce — == = = — — =. to an equivalent fraction with ^1 + ^/2 + v/3 + ^4 ^ ^ a rational denominator. Ans. (3 — ^2 — ^3) (4 + 2^/6). EXTRACTION OF ROOTS. 265 2 5. Reduce — = = z= r=:to an equivalent fraction with ^3 + ^5 + ^6 + ^/8 a rational denominator. Ans (v/3"— v/5"— v/6 + ^/8j (^24 + ^30) — 6 2 6. Reduce — = -= = = to an equivalent fraction. v/3 -f v/5 + >/6 + v/10_ _ _ Ans. v/3 — v^S"— v/(r+ v/TO. EXTRACTION OF THE SQUARE ROOT OF A MONOMIAL SURD CONNECTED AVITU A RATIONAL TERM, OR OF TWO MONO- MIAL SURDS. 321. Before we proceed to extract the root, it will be necessary to demonstrate three principles upon which the extraction depends. First Principle. The square root of a quantity cannot consist of the sum of two parts, one of which is rational and the other irrational. For, if possible, suppose \/ a = x + y/y. Then, by squaring both members, we get a = x^ + 2xi/y + y- From which, ^/y = ^ . That is an irrational quantity equal to a rational one, which is absurd. But the absurdity has not resulted from an error in the analysis, and must, therefore, be in the condition. Second Principle. When the first member of any equation contains a monomial surd, connected with rational terms, and the second member is made up in the same manner, the rational quantities in the first member are equal to the rational quantities in the second, and the irrational quantities in the two members are respectively equal also. Let a 4- \/b = X + i/i/, then a = x, and ■/ — 20. ^ns. V^ + v/ — 4, Verify the result by squaring it. 9. Required the square root of 1 + V — 288. Ans. 3 + ■/■ Verify the result by squaring it. 10. Required the sum of \/l+s/—2W, and \/l— v/^2BB. Ans. 6. We see that the square root of an expression containing an imagi- nary quantity, cannot have all its terms real. But the sum of the roots of two expressions, involving imaginary quantities, may be real and even rational. The formulas (A) and (B^t are generally applied to expres- sions for which y/ a" — b is rational. 11. Reduce — = to an equivalent fraction with a rational V9 -I- V*i _ denominator. Ans. 10 (y/o — v/2). For ^^ = 10 ( vw - vw - Vim+i/Wf) ^ = 2 (v/27 — ^Y— ^/T8'+v/T2)=2(3v/'3 — 2 ^2"— 3 ^2+ 2^/8) = 2 (5v/3"_ 5^/2) = 10 (v/3~— ^2). Or, the reduction may be performed thus, -^— — - = — ^ (^3-fv^2)(^3_^2 3-2 ^^ ^ ^ 4 12. Reduce — r= = to an equivalent fraction with a rational V5-ft/4 ' _ _ _ denominator. Ans. 4 ( (2 + v/o)(n/5)— 2^/2 — y/KJ). 23* 270 FORMATION OP THE POWERS AND 4 _ ( yy — 4/(4)^ — V(5/4 + V5(4)^) _ _V5+Vi~_ _ 5-4 _ 4(^5 4/5 — v/8 — v/l<^ + 2^5) = 4((2+^5)(^5)_2^2 — n/10). 5 13. Reduce — = ^^ to an equivalent fraction with a rational y4 + i/2 _ _ denominator. Ans. | (^4 -(- (3/2 — 1)2). p^^ _5 ^ ^ (y (4)^ + V(2)^ _ 3/(4) (2) y4 + 3/2_ (y4 + y2) (^(4)^ + V(2)^ - VC4) (2) ) = I (2^2 + 3/4 _ 2) = 1(3/4 + (^2 _ 1)2). 14. Reduce — t=z= — to an equivalent fraction witli a rational y27+V8 denominator. Ans. ^. For _^ = 6 (^(27)^+ V(8r-y(27) (8)) ' V27 + y8 35 The result is rational because the given expression was really rational, though under an irrational form. 15. Reduce :=^ to an equivalent fraction with a rational de- n -f- V'wi nominator. . m (n — \/m) Ans. 3 . IMAGINARY QUANTITIES. 323. Any expression whatever, made up of monomial surds and rational terms, or monomial surds only, may be rendered rational by repeated multiplications. The few examples given will show the man- ner in which these multiplications must be made. No general rule can be given in regard to them. There is a particular class of monomial surds, which, when rendered rational, present some differences from other surds in regard to their algebraic signs. These are imaginary surds. An imaginary quantity has been defined to be the even root of a negative quantity, because no quantity, taken as a factor an even num- ber of times, can give a negative result. Thus, y/ — a, %/ — o, %/ — o, ^;/ — a, are imaginary quantities. EXTRACTION OF ROOTS. 271 If tlie indicated root be of the 2ni"' degree, then ^^ — a may be written ^^a ^^ — 1 = 6\/ — 1. In which, h represents the 27H*'' root of a, whether that be rational or irrational. And we see that all imaginary quantities may be decomposed into two factors, one of which is an indicated even root of minus unity, and the other of which is real, and sometimes rational. This decomposition must always be first effected previous to opera- ting upon imaginary quantities. To square v/ — «; "we first write the expression v/o v/ — 1 ; these two factors will both drop the radical when squared. Hence, (^Z — a)^ = W^ V-^^= («) (- !)=—«. So, also, (y—of = (v/a v/— 1)' = (ds/a) (—1 y/—l) = Qx/ — a- (^— «)^ = (v/<( ^/— l)' = Or) (+ 1) = + a' (v/— af=(ya ^— iy=(a' ^a) (+^/— l) = aV- (v/— a/=(v/a v/- !/=(«') (- 1) = — «'• The table shows that tlie even powers of imaginary monomials are always real, and that their signs are alternately plus and minus. 324. A table of products will show the modifications of their alge- braic signs. ( + x/— ")( + n/— »):^+(v/av/— l)(v/av/— 1) = (^a) (+v/— «) (— y/ — g)^ (^^^ZTT) (_^^^^Q = (y-^ C-v/«)(n/-1)(n/— !) = +«. C+yZ—g-) (+v/— ^) = i^a ^— 1) {^b y/— 1) = (^a) (x/i) (>/— 1) (v/=l) = -x/^ (— n/— «) (— %/— 6)=(— v/a y/— 1) (— v/6 y/— 1)=(— ^a) (-V/^ (y/=^ (x/^^) = — x/"^ 272 ro UMAX ION or the powers and (-fV-<0 (— v/— /^) = (v/a V— 1) i—s/h V— 1) = (v/a) (— v/TT) (v/^^1) (v/— 1)=+^"^. We see that like signs produce minus, and unlike signs produce plus, wlien imaginary monomials are multiplied together. 325. "We will now form a table of quotients. ^■v- — a _ v/av/— 1 _ v/6v/— 1 + v/« + v/- + v/6 + v/— a 1" -v/^x/— 1 v/^ — v/— a -v/-6 = - — v/T v/^1 - -v/i v/a — V- -a —v/a v/— 1 -v/TT +>/"= =^' v/6 v/— 1 v/6 And we see that, in division, like signs produce plus, and unlike signs produce minus. It would seem that, since the rule for the signs in division is different from that in multiplication, the product of the quotient by the divisor might not give the dividend. But any of the preceding quotients can be readily verified. Take^^=±^ V—h +v/i _ + -y/g _ _ 4- v/a Multiply the quotient ~j by the divisor v/ — h. Then, — — -= + v/^ X v/ — b = ^ y/b v/ — 1 = -f \/a >/ — 1 = \/ — a, the dividend. 326. Imaginary quantities can be operated upon just as real quan- tities, provided that care be given to attribute the proper algebraic signs to the results. Let it be required to divide 4v/ — i* by 2by/ — 1. The division cannot be performed until 4v/ — i^ is transformed into the equivalent expression, 4&v/ — 1? tlic quotient will then be 2. Let it be required to multiply 4v/ — b^ by 2&v/ — 1- Then, 4v/^T'' X 2b^'^l = Uy^l X 2b^'^^=—Sb^ EXTRACTION OF ROOTS. 273 Now let it be required to divide — 86^ by 2.h^ — 1. Then, —86^ SiV^^^v/"^^^^ ^ _ 4^ xTZT: 1h^—\ 26v/— 1 EXAMPLES. 1. Multiply + x/ — a by — ^ — h. Ans. + y/ah. 2. Divide ^ab by x/ — a. Ans. — ^ — 6. y/ab _ s/a y/h _ y/ZT _ — v/6^n/— Iv/— 1 _ — n/6'n/— 1 = — V^^>. The minus sign is placed before the square root of h, in the expres- sion, — v/6 v/ — 1 v^ — 1, in order to make the result positive. Since ^ — 1 y/ — 1 = —1. 3. Divide v'"*^ by — V — ^- ^"S- + v^ — <*• 4. Multiply ai^ — c" by •/ — a^bc. Ans. — a^bc ^bc. 5. Divide — a^bcy/bc by y/ — a-bc. Ans. -f- abc y/ — 1. 6. Divide — a%Cy/bc by (fi^/ — ^^- '4?is. + « s/ — be. 7. Multiply — V— abc by — abc y/— abc. Ans. — a?b^/ — 1. Ans. a" + '^aj'h JZI\ — QaW — ^aV V— 1 + h\ 19. Subtract 2 2a^ from 10vA=^. Ans. 8v/^=^. 20. Add together 8x7"=^ and 22/Zr2. Ans. lOs/— 2. 21. Muhiply V— a + -J—h by v — «— %/— 6. Ans. h — a. 22. Divide h — « by v' — ^ + 7 — a. Ans. — ■/ — h + V — a. 23. Multiply J- + 2 s/^=r^ by J — 2 ^—a. Ans. \ + 4a. 24. Divide I + 4a by i + 2v7Zr^. ^^s. i — 2x/'^^.. 25. Multiply 2 t/^Ta by 3 v/^T^. ^ns. 6V^^ V— ^ 26. Divide 6v/— a ^ — a by 2^^^^. ^hs. 3s/-^^. 27. Divide 6v/— a V^=^ by 3 v/^^f~ ^hs. 2y^^. 28. Multiply V^^^+ t/^^6'by y^=^— y"^^=^. J.J2S. V — a — v/ — i. 29. Multiply 1 + 6v/'^=T+ cV^irTby l_Z)V^^^nr+ cV^lT ^}»S. 1 + ^2 _ g2 ^ 2cn/— 1. 80. Multiply 1 + 6v''=T + c^/— "l by 1 + Z/^I^ — Cy/^^TT" EXTRACTION OF ROOTS. 31. Multiply 5v/— 1 + V— 3 by 3v/— 1 — V— ^7- Ans. —15 — 3v/3'+ 15^/3~+ 9. 32. Required the second power of 5v/ — 1 + ^Z — 3. Ans. — 25 — lOv/o — 3. 33. Required the square root of — 25 — 10^/3"— 3. Ans. 5v/IIT— 3. 34. Divide 6 4- v/ — 4 by 2 + y/ — 9, and make the denominator of the quotient rational. ^ 18 — 14*/ — 1 Ans. -^3 . For ^+ ^- j = (6+ v/-4) (2 - y/ - 9) 2 + v/— 9 (2+s/— 9) (2 — ^—9) 12 + 2^— 4 — 6^— 9— y/ 3(j y/— 1 y/— 1 _ ___ 4 + 9 12 + 4v/— 1 — 18y/"=nr+ 6 _ 1« — 14v/^^ 13 ~ 13 ■ 35. Divide 2 + V — 2 by 2 — v^ — 2, and make the denominator of the quotient rational. , (2 + y/— 2)^ 2 + 4y/ir2 ^ns. -^^ -^ or G (5 30. Divide 1 -|- y/ — 1 by 1 — ■/ — 1, and make the denominator of the quotient rational. Ans. y/ — 1. 37. Required the quotient of Gy/ — 4 divided by 2y/ — ~^. Ans. 2. Rcmarhs. An examination of the foregoing results will show some properties of imaginary quantities. 1. Real, and even rational results, may be obtained by the ordinary algebraic operations upon imaginary quantities. 2. Two monomial imaginaries raised to an even power, will always give a rational result. 3. Two monomial imaginaries multiplied together, or divided, the one by the other, will always give a real result. 4. A monomial imaginary connected by the sign, plus or minus, with one or more rational terms, and then raised to any power, will give at least one imaginary term in the result. 5. Any expression involving one or more monomial imaginaries, may be rendered real by one or more multiplications. 276 EQUATIONS OF THE SECOND DEGREE. EQUATIONS OF THE SECOND DEGREE. 827. An equation, involving a single unknown quantity, is said to be of the second degree, wlicn the highest exponent of that unknown quantity in any one term is 2. Thus, x^ -\- x = m, and x^ -{- a = h, are equations of the second degree with one unknown quantity. An equation, involving two unknown quantities, is said to be of the second degree when the highest sum of the exponents of the unknown quantities in any term is equal to 2. Thus, xt/ = ?h, a;^ -f ?/ = ??, and y'^ -{- x=p are equations of the second degree with two unknown quantities. For they may be written, x'j/' = in, x^if -f ^ ^= n, and, y^x^ -f a- = w, and, thei'efore, come within the definition. We will begin with equations of the second degree, with a single unknown quantity. These are divided into two classes, complete and incomplete. A complete equation of the second degree, with a single unknown quantity, is one which contains the second and first powers of the unknown quantity. Thus, a-^ + a = 0, clx^ + x = m, and ax'^ -f hx = c, are complete equations of the second degree, with one unknown quantity. There may or may not be other terms, besides those involving the unknown quantity. The first and second powers of the unknown quantity may or may not have coefficients different from unity. An incomplete equation of the second degree is one containing the second power only of the unknown quantity, and it may or may not involve terms in which the unknown quantity does not enter. The unknown quantity may or may not be affected with a coefiicient diffe- rent from unity. Thus, ax^ = x^, x^ -\- m == n, ~ — h c = — q are incomplete equations of the second degree. Equations of the second degree, with one unknown quantity, are frequently called quadratic equations, because the unknown quantity is raised to the second degree, or squared. INCOMPLETE EQUATIONS. 328. We will first examine incomplete equations. hx^ The general form of such equations is ax^ \- d =^e. By transposition and reduction, we get (ca — h')x'^ = ce — cd, or c'^ = EQUATIONS OF THE SECOND DEGREE. 2i( ~ -^ = q, by substitutinsr for the known terms in the second ca — 6 member a single term, equal to them in value. Hence, ever}- incom- plete equation may be reduced to two terms, and the equation be placed under the form of x^ = q. Owing to this circumstance, incom- plete equations are sometimes placed in the class of hinomial equations, or equations involving but two terms. There is no difficulty in solving the equation x^ = q. The square root of the first member will give us x ; but if we extract the square root of the first member, we must extract that of the second also, or the equality will be destroyed. Extracting, then, the root of both members, we have dti x = zt: ^/ q. We have prefixed the double sign to both members, because this equation, when squared, must give back the original equation, x^ = q, from which it was derived, and either + X or — X, when squared, will give + x'^, and either + V q, or — ^/q will, when squared, give q. The value of the unknown quantity, or the root of the equation, as it is generally called, has been defined to be that which, substituted for the unknown quantity, will satisfy the equation, that is, make the two members equal to each other. We sec that an incomplete equa- tion of the second degree has two values, numerically ccjual, but aff"ectcd with different signs. Either value or root will satisfy the equation, for the result of the substitution of either V*/, or — y/q for X, in the given equation, will he q z=q. We have prefixed the double sign to both members, but it is usual to prefix it only to the root. In that cape, however, we must understand that the sign of x is not neces- sarily positive, it being afi'ected with the positive sign only, when it corresponds to the positive root. It becomes, then, necessary to have some notation to distinguish the values. This we will do by dashes. The X that corresponds to the positive root we will write x' ; and the X that corresponds to the negative root we will write x". The fir.st is read x prime, and the second, x second. It will be shown that a com- plete equation has two roots, generally unequal in value. We will also distinguish these roots or values by writing them x', and x". Some equations of high degrees have three, four, &c., values. These will be written x', x", x", x^'', &c., and read x prime, x second, x third, x fourth, and so on. To solve an incomplete equation, it must first be put under the form of x'^ = q. That is, all the known terms must be transferred to the second member, and the coefficient of x"^ must be made pins unity, if 24 278 EQUATIONS OF THE SECOND DEGREE. not already so, by dividing both members by the coefficient of x-. After the equation has been put under the form of x" = q, extract the square root of both members. EXAMPLES. 1. Solve the equation, 2x^ + 1=4. Transposing, and dividing by the coefficient of x"^, we get x^ = ^. Hence, x = ± vT. Then, x' =+y/J, and x" = —Vj. The solu- tions may be left thus, or we may extract the indicated root approxi- matively. 2. Solve the equation, 2x^ + 4 = 1. Reducing, we get x^=±^ — ^. Then, x' = -\- y/ — ^, and x" These roots are imaginary. How are we to interpret them ? An imaginary quantity indicates an impossible operation. Ought not the equation which produces it involve an impossibility ? In this instance, 2x'^, an essentially positive quantity, is added to 4, and their sum is required to be less than 4. The condition of the problem is, then, absurd, or impossible, and the result is impossible, as it ought to be. It will be shown more rigorously, hereafter, that an imaginary solution always indicates absurdity in the conditions of the problem. The ima-/ ginary values, though they fail to satisfy the conditions of the problem, yet will satisfy the equation of the problem ; as they manifestly ought to do, since they have been truly derived from it. Substituting either + v/ — ?>, or — v^ — -?, for x, in the equation, 2.x^ + 4 = l,we get — H + 4 = i, or 1 = 1. ~ 329. We observe, then, this remarkable analogy between imaginary values in equations of the second degree, and negative values in equa- tions of the first degree. The values, in both cases, will satisfy the equation of the problem, but will not satisfy the required conditions. There is, however, this difi"erence : to convert a negative solution into a positive one, numerically equal to it, we have only to impose a single condition; but, to convert an imaginary solution into a real solution, precisely equal to it numerically, we have generally to impose two conditions. Thus, the equation which gives the imaginary solutions, expressed in words would read thus : required to find a number, twice the square of which, augmented by 4, will give a sum equal to 1. EQUATIONS OF THE SECOND DEGREE. 279 The values are cc' = + ■«/ — 3, and a;" = — \/ — |. Now, to get real values numerically equal, and not to make any change upon the arithmetical values of the known quantities in the equation, it must be changed into 2aj^ — 4 = — 1. This equation, expressed in words, would read thus : required to find a number, twice the square of which, diminished by 4, will give a difference equal to — 1. The values of the new equation are cc' = + \/ -|- |, and x" = — V+l- These values are real, and differ from those in the first equation only in the signs of the quantities under the radicals. To get these new values, we made no change, arithmetically, upon the numbers in the first equation, but we made two changes of sign, or, in other words, we imposed two conditions upon the equation of the problem. 3. Solve the equation, x^ + h = a. Ans. x'=z-^\/a — bj x" z=z — s/ <-'' — h. Now, let a =4:, and 6 = 6. Then, a;' = -f 1/ — 2, and x" = — y/ — 2. To get the real values, x' = + >/ 2, and x" = — %/2, we must make h and a interchange values, b must be 4, and a, G ; that is, two conditions must be imposed. 4. Solve the equation, ^ — 1 + x" — 4 + 5 = 3x=' — 7. Ans. x' = +2, x" = — 2. 5. Solve the equation, - - _ 1 -|- .^2 _ 9 _,_ 5 = 3_,.2 _ 22. Ans. cc' =: +3, x" = — 3. x^ 6. Solve the equation, — — 1 + .x^ _ IC + 5 = Sx^ — 43. Ans. a;' = + 4, x" = — 4. 7. Solve the equation, 'jx^ —^ + 4x' — 20 + IOO.t;^' — 500 = o 995 — 199x1 Ans. x' = + v'S^ x" = —^'b. 8. Solve the equation, ^+ ^ _|._ 3+ 7^2— 42 + 999x2== 5994. Ans. x' = + ^6, x" = — v^ey The preceding examples are simple, and the solutions can be readily obtained. But there are many of a more complicated character, and of course more difficult to solve. No general rules can be given to aid the student 3 he must exercise his own ingenuity. It is well, however, 280 EQUATIONS OF THE SECOND DEGREE. to make the clearing from fractions the firet step in every reduction, and then, if there is a single radical in the equation, it ought to be placed in one member by itself. 1. Solve the equation, ^ 3 = -r '44 Clearing of fractions, we get ^/x^ + 144 + 12 = x^. Placing the radical by itself, we have y/x"^ + 144 = x^ — 12. And squaring both members, there results x^ + 144 ^= x'^ — 24x'^ + 144. Hence, a" = ^bx", or x^ = 25. Then, x'= + 5 and x" = — 5. x 10 2. Solve the equation, + ^x' + 44 22- Then, 22a; = lOx + 10^/x^ + 44, or 12x = IQ ^ x^ + 44. Squaring both members, we get 144^^^ = lOOx^ + 4400, or 44a;'' 4400. Hence, x' = 10 and cc" = — 10. ^7'2~Iir77v 3. Solve the equation, — '■ — ^ + Z> = x^ Ans. x = — — — , X ■ = — — . c c These results can be verified by substituting either the value of x' or x" for X in the given equation. Then we will have \ / ^ \- V(? A + 2lc^ A , . , , , 1 + Sic'^ ,,, = cl -^ 61, or, by squaring both members, ^ 1- 6-r Hence the equation is satisfied. 4. Solve the equation, ;==r = — . X + ^x^ -\-a a + & „ -6 ^??S. X' := -v/a — 26 v/^f EQUATIONS OF THE SECOND DEGREE. 281 Verify these results. What do they become when a =2b? "Why ? What, when a = ? Why ? What, when 26 > a ? Why ? 5. Solve the equation, ^p^ + x^ + x = mjc. Av -' — ' ^ -" — ^ ins. X = Verify these results. What do they become when m = 2 ? ^^hy ? What, when wi = ? Why ? What, when m = 1 ? Why ? What, when p = ? Why ? When there are two similar radicals, it is best to unite them in the same member. p o , ,, ,. h-\-hy/\ — 3? a 0. bolve the equation, = — , i + mv/l — a;^ » An. x'=+ ^-'{-^'-l')+-^anHI>-m) nh — am ' ^a? jm^ — b') + 2an/j (6 — m) ni — am For, clearing of fractions, we get nh + nh^l — x^ = ab + am -v/1 — x'^, or, («i — am) ^/l — ic^ = 6 (a — n), or, (hJ — amf (l — x^) = U'(a—ny. Developing, we get n~b'' — 2nba7n + a-m^ — x'' (nb — ajuf = b^a^ —lanb^ + bhi^. Hence, a== {in^—b^) + 2a?j6 (6— «i) = x^{nb — amf. Hence, .• = + ^^llMH^IIl^mHS, ." = _ {lib — ani) ' V a"- (w^ — Z.^) + 2a?i6 (6 — m) 7i6 — am 7. Solve the equation, n-^p^ -{■ t? ■\- ax= xy/p^ + a;* + an. Ans. x' = + -v/a'^ — p^ and a;" = — V «^ — P^ Verify these results. AYhat do they become when a^ = p2 ? Why ? What, when / > a^ ? Why ? 330. There is another value which does not appear, it is xz=n. For, by the second principle, Article 321, the rational terms in the two members must be equal. Equating them, we get ax = an, or a; = n. This ought to be so, for the given equation can be put under the form of (n — a;) y/p^ + x'' — a (n — x) = 0, an equation which can plainly be satisfied when x = n. The given equation, previous to the 24* 282 EQUATIONS OF THE SECOND DEGREE. division by the common factor, n — oc, was really a cubic equation, and contained three values. We are sometimes enabled to detect a value, as in the above example, by equating the rational factors. When the root of this equation will also satisfy the equation formed by equating the radicals, it is, of course, a value in the given equation. 8. Solve the equation, p-,/x^ — ct^ + a = n^/ x^ — a^ + x. ( (P — '0' + 1) Ans. x=^a, ox x^=.a \^ — -. — ~. 1 — (i> — ») 331. There arc some expressions in a fractional form, which must be changed into equivalent fractions with rational denominators. n o , ^1 X- ni + X + y/ 2mx + x'^ 9. Solve the equation, - m + X — v^ 2mx + x^ . , (1 + ^li)' „ (1 - JTi)' Ans. X ^m :=r: , X = m ^— . 2(2+ Vw) 2(2— v/n) For, by multiplying numerator and denominator of the fraction in , „ , , T (m -\- X -\- -J 2mx ■\- x^^"^ the first member by the numerator, we get ^ =^ n. Clear the equation of fractions, and extract the square root of both members. Then, m + cc + s/'^mx + a;^r= d= m-s/ n, or in (1 dz Vn) — x=i — ^2mx + x^. Squaring both members, we get m"^ (1 rt \/ rif — 2mxy\- ± V 7() + a;'^ =r 2mx + x^, and, by transposition, m^ (1 =h s/nf m(l ± ^nf 2mx (2 rb \/n). Hence, x =z 2(2 ± Vn) 10. Solve the e<:|uation, — ^^^==z z==^^= = 1. ^x^ — a'—^x'^ — a'' Making the denominator rational as before, we have (v/.-+aHV^--a-)- ^^^^^^,^^.^,^-^^^,^^,_^,^^^. Then, 2^x' — a' = 2a^ — 2x\ or ^x' — a' = x- — al Squaring again, we get x* — a'* = x* — 2a;V + a*, or 2a;V == 2a\ or x^ = a^ Hence, x' z= -}- a, and x" = — a. It will be seen that these values satisfy the equation. In this ex- ample, the second step was not to extract the square root of both mem- bers, as in the last example, because the double product of the radicals EQUATIONS OF TUE SECOND DEGREE. 283 gave a simple result. When the radicals are of such a form that their double product will not give a simple result, it is best to make the second step the extraction of the square root of both members. 11. Solve the equation, — . ==- = 1. Ans. x' =^ + h, x" 12. Solve the equation, = — -». ^x^ + 8m^— ^x^ — m^ Ans. x' = + jn, x" — Sometimes the first step is squaring both members. ,001 XT- X- n/" + ^c + v/a — X Im 13. Solve the equation, = = \ / — • y/'lm V n , '"^ ^a -9 II ^'l -: n Ans. X =^ -\ y/ Ian — ??r, .7: = y^/'lan — m^ 14. Solve the equation, ^/a + x + ^/a — x = y/la. Ans. x' =z + a, x" = — a. - _ „ , , . v/fl + X + \/'i — X X 15. Solve the equation, - — ■ -^ = — , Ans. x = + 2>/am — m^, x" = — 2V am — m'. What do these values become when m = a? What, when w ^ a ? 16. Solve the equation, ~ =- = -. ^ .T + 1 2x — 4 Ans. x' = + y/b, x" = — >/5. 2.7; 4- 4 X 2 17. Solve the equation, ~~ p- =^ , . ^ ' X + 2 2x — 4 Ans. x' = -\- 2, x" = — 2. What does the second member become when the first of these values is substituted ? How do you explain the result. 18. Solve the equation, " = . X — o X — a Ans. Both values infinite. How are these results explained ? 284 EQUATIONS OF THE SECOND DEGREE. nx + a X + h 19. Solve the equation, — b nx — a What do these values become when « = 1 ? "Why ? What, vyhcn 7/ > a\ 2x + 1 X + 2 20. Solve the equation, = ^- * X — 2 Zx — 1 Ans. x'=z + V^^^, x" = — \/-^ir Why are these results imaginary ? 21. Solve the equation, ax -\ =^ —„ a. X x^ — x 22. Solve the equation, 2x + Ans. 2 8a;. 23. Solve the equation, ox -\ = Ans. x' = + 2, x" = — 2. 27ic — 3 „ Ans. x' = + S, x" = — 3. PROPERTIES OF INCOMPLETE EQUATIONS. 332. 1st. Every incomplete equation of the second degree has two values, and but two, and these values are equal with contrary signs. For, the general form of the equation is x^ =:zq or x^ — q = 0. The first member may be regarded as the difference of two squares, and can, therefore, be placed under the form of (x — \/ q) (x + n/ q}- Hence, the equation, x^ = q, may be written (x — %/§') (^c + \/q)^^- Now the product of two factors being equal to zero, the equation can be satisfied by placing either factor equal to zero. Therefore, x — y/q=^Q, and x -{• y/q = 0. From which, we get x = + y/q, and ar = — -Jq, or, distinguishing the values by dashes, a;' = + >/ q, x" z=^ — ^q. Since there are but two factors, the equation can be satisfied in but two ways. Hence there are but two values, and Ave sec that these are equal with contrary signs. By solving directly the equation, x^ = q, we would obtain the same results. But the process we have adopted is to be used hereafter for EQUATIONS OF THE SECOND DEGREE. 285 complete equations, and it is well to know that the properties of com- plete equations are also those of incomplete equations. 2d. Every incomplete equation of the second degree can be decom- posed into two binomial factors of the first degree with respect to .r ; the first factor being the algebraic sum of x and the first value with its sign changed, and the second factor being the arithmetical sum of a- and the second value with its sign changed. The factors have already been obtained, and are (x — V q), an J (x + %/(/); and we see that these correspond to the enunciation of the second property. The product of these factors is zero. Hence, when wc know the two values of an incomplete equation, we can always form the incomplete equation itself which gave those values. We have only to change the signs of the values, and connect them with x. We will then have the binomial factors, and, placing their product equal to zero, we will have the equation required. Thus, form the equation that gives for x the two values, -f 2 and — 2. The two factors are (x — 2) and (x + 2), and the equation is (x — 2) (a; -f 2) = 0, or x^ — 4 = 0. The result can be verified by solving the equation, x'^ — 4 = 0. Or, since in the equation, (x — 2) (x + 2) = 0, we have the product of two factors equal to zero, the equation can be satisfied by placing either factor equal to zero. Hence, x — 2 = and .r + 2 =: 0. And these equations, when solved, give the values, + 2 and — 2. EXAMPLES. 1. Find the equation that gives for x the two values, -f a^, and — a^. Ans. x^ — a' = 0. 2. Find the equation whose values arc + -/oi, and — y/ab. Ans. x^ — ai = 0. 3. Find the equation whose values are -f o — I, and -f h — a. Ans. x' + 2ah — 1/ — a^=0. 4. Form the equation whose values are -f y/ m — a, and — ^ m — a. Ans. jp- -f a — m = 0. 5. Form the equation whose values are + ^f m — a — s/ in^ — a^ and — \/ in — a -{- ^m^ — a^. Ans. x^ — (m — a) — (m^ — a^) -f 2^w— a ^vv' — a'= 0. Verify this result by solving the equation. 286 EQUATIONS OP THE SECOND DEGREE. 6. SolvG tho equation whose values are a — ^Z d^ — m?, and — a + ^a? — mK Ans. x^ — 2a?+m?+2a v/m^ — a" z= , Verify this result by solving the equation. 7. Form the equation whose values are -f- — , and _ ^iiv" + a" 8. Form the equation whose values are — ■ , and Ans. (m^ — ?i^) cc'^ = an^. 9. Form the equation whose values are w? — n^, and — rti^ -j- n^. Ans. x^ = m* — 2m^n^ + %*. What do the values become when m = ?i ? "What, when m = ? BINOMIAL EQUATIONS. 333. Any binomial equation of the n^^ degree can be solved as the preceding equations. We have only to extract the h*"" root of both members. Let a;^ = 27; then, x =\/W = 3. Let x"" = q; then, x == ^~q. If n be an even number, the radical must have the double sign, and two values, at least, of the equation will be known. If n be an odd number, only one value will be known. It will be demonstrated here- after, that every equation of the ?i*'' degree has n values. The method of finding the other values will then be explained. Let a;" = a^ ; then, x = ±yd^ = ±\/ ^/^(^ = ±n/± a. Then , x' = + ^ + a, x" = —^-\-a, x'" ^ + ^"-^a, xJ'" = — v/— a. In this example the four values of x have been determined, but it is not often the case that they all can be found. Let x^ = «%• then, x = ± V*' = zfc^Y v^'t? = ± V± a, and four of the six values are known. EQUATIONS OF THE SECOND DEGREE. 287 Let a;'2 = a%- then, x = ±'y^ = ±\J l/cc" = ±i/a, and only two of the twelve values are known. EXAMPLES. 1. Solve the equation, ^x* — 6az + oa^ = z — 2a. Ans. x' — + ^z + a, x" = — ^z '\- a, x'" = + ^117(7+^, x"" v/— (z + a). 2. Solve the equation, x* = 256. Ans. x' = + -i, x" = — 4. 3. Solve the equation, ^a^ + x^ + y/a} — x^ = ni^2. Ans. a/ = + i/a* — {m? — a^f, x" = —i/a* — (m''—a^f. 4. Solve the equation, a;'" = 32. Ans. x' = + ^% x" = — ^2^ 5. Solve the equation, -^^-^ _: ^ =2^^-- s/m Ans. x' = + j9 Vm (2 — m), x" = — j^l/m (2 — m). Qj Solve the equation, x* = 1. Ans. x'=-\-l, x" = — l, x'" ± + v/^=T, ./■"" = — vZ-'^nr Verify the preceding results by substituting them in the given equations. GENERAL PROBLEMS IN BINOMIAL EQUATIONS. 334. 1. The sum of two numbers is «, and the ratio of their squares m : what are the numbers ? Ans. X = r=r, X = :^ • 1 + v/m 1 — -v/"^ For, let X = one number ; then, a — x = the other, and, by the x^ conditions, -, = in. Now, extract the square root of both mem- (a — xy ' ^ x' ziz y/m. Hence, • , r= -\-^m, and 11 ^ a — X, — ^/n. From the first equation, u^ ^= ■\- a-^m — x^ y/m, a — X or 288 EQUATIONS OF THE SECOND DEGREE. — , + a »/«i. ,, , _ x'(l -^ .y^n) =a yin, or x' zr= r=. l^rom the second equa- 1 H ^m tion, we get x" r= — a y/ m + x"^m, or x" =i z;^. Now, if 1 — y/m m = \, tlic second value is infinite. How is this result to be inter- preted ? By going back to the equation of the problem, we see that x^ it becomes, under this hypothesis, -^ = 1 ; an equation which can only be true when a z=. 0, or when x ^= — . The first supposition contradicts the enunciation, the second is the value found for x'. The second value can only exist when there is a contradiction to the state- ment : it ought, then, to appear under the symbol of absurdity. We may explain the result otherwise; thus: when m = 1, the equation in x" becomes x" — x" =z — a, or • j, = + —^, or = -| ^,. This equation is plainly absurd for any finite value of x". The value, x" = — --, satisfies the equation of the problem, as it manifestly ought to do. We have — = 1, or — — = 1, IF (a\ ' "' (Oa — aY or, -^[ = ^^^^^, or O'a' = 0' (Oa — af, or = 0. 2. The sum of two numbers is a, and the ratio of their fourth powers is m. What are the numbers ? , a ij m „ a yj m 1 + V?u' 1 — Xjm 3. Two numbers are to each other as m to «, m being greater than n, and the ratio of their squares is equal to (.^ divided by the square of the greater. What are the numbers ? ^. an „ , , ai^ Ans. First, ± , Second. ± — 5-. m ' vv 4. Two numbers are to each others as m to n, and the sum of their fourth powers is equal to a". What are the numbers ? Ans. First, ± :, Second, ± -- EQUATIONS OF THE SECOND DEGREE. 289 5. Two numbers are to each other as ni to n, and the diflforeuco of their fourth powers is O*. What are the numbers ? A71S. iirst, it , Second, y m* — n* %/ «t* — ,!•• What do these values become when m ~ n ? Why ? What when m=0? Why? 6. The cube root of twice the square of a number is 2. What is the number ? Ans. Either -|- 2, or — 2. 7. The cube root of a times the square of a number is h. Wliat is the number? Ans. X '-Wl--\/l What do these values become when a = ? What when h =0? 8. The square of a number multiplied by the first power of the same number is equal to 64. What is the number ? Ans. 4. 9. A man put out a certain sum of money at G per cent, interest. The product of the interest upon the money for G months by the interest for 4 months was 600 dollars. What was the sum at interest ? , A71S. 81000. 10. A man put out a certain sum of money at 6 per cent, interest. The product of the interests upon it for 3, 6 and 9 months was 820 i. What was the sum ? Ans. 6100. 11. The successive quotients, of a quantity divided first by a, and then by h, will, when multiplied together, give a product equal to mn. What is the quantity ? Ans. a;' = + ^ahmn, x" = — y/nhmn. COMPLETE EQUATIONS OF THE SECOND DEGREE. 335. The most general form of such eciuations is, 1 — n=-r; b m clearing of fractions, we get, — amx^ + hex — Imn = hmr. Dividing by the coefficient of x^, — am, we have x' — 1 = : and am a a by transposition, x^ == ^ \ Now, the second member am a 25 T 290 EQUATIONS or THE SECOND DEGREE. may be plainly represented by a single letter, — q, and tlie coefficient of the second may be represented by — p. Hence, the given equation as- sumes the form of x^ — px = — q; in which the highest term of the unknown quantity has a coefficient plus unity. Had the coefficient, — - of a;^ in the original equation been affected with the positive sign, it would have reduced to the form of x^ + 2^-^ — 5- ^^^ ^^^ coefficient of x^ been positive, and that of x negative, the equation would have assumed the form of x^ — px ^ q. Had the last conditions been ful- filled, and the prevailing sign in the second member at the same time negative, the equation would have assumed the form of x^ — px = — q. And, since every change that may be made upon the signs of the coefficients of x^ and x, and upon the signs of the known terms, will eventually lead us to one of the preceding forms, we conclude that every complete equation of the second degree may be placed under one of the following forms : x^ + px = q, First form, a;2 — jyx ^ -\-q, Second form, x^ + pjx = — q, Third form, x^ — px ^= — q, Fourth form. When a complete equation is given to be solved, it must first be placed under one of these forms. To do this, clear it of fractions, if it con- tain any, and then make the coefficient of the first term plus unity, if not already so, by dividing by this coefficient. The resulting equation will then be under one of the required forms. The form that the resulting equation assumes will, of course, depend upon the prevailing signs in the given equation. x^ Ix Reduce the equation, — ^ 8 = ^ — ■ -4j to one of the four forms. Clearing of fractions, we have cc^ — 2a;+12=— 6, or .r^ — 2x= — 18. Hence, the equation has assumed the fourth form. Reduce the equation, — + o = — f to an equivalent equation, which will appear under one of the four forms. Multiplying both members by 20, the least common multiple of the denominators, we will get 5x^ — lOx -f 00 = — 24. Now, divide by -|- 5, to make the coefficient of x^ plus unity. Then, x^ — 2x + 12 24 ^__or-4f EQUATIONS OB' THE SECOND DEGREE. 291 We see in this particular case, that we might have obtained the sam-e re- sult by multiplying the original equation by 4. This -would have made the coefficient of the first term unity, which is the main point to be attended to ; then, by transposition, the equation would have become x^ — 2x = — 16|. And we see tha-t the equivalent equation is of the fourth form. It frequently happens that it is impossible to make all the terms entire, and the coefficient of the first term at the same time plus unity. But, since this coefficient must always be plus unity, we derive the following rale for reducing any equation to one of the proposed form. RULE. Multiplij both members of the equation by the reciprocal of the co- efficient of the first term, and then transjxjse all the known terms to the second member. The multiplier in the equation, | o;^ — 2.r =: 4 is, + |. The multi- plier in the equation, — | cc^ — 2x = 4 is, — -^ . The reason of the rale is obvious, and needs no explanation. EXAMPLES. 1. Keduce the equation, | x^ — J a; + G = 4, to one of the four forms. Ans. X? — -= — 3. 2. Keduce the equation, — | a;^ — \x + 6 = -j- 4, to one of the four forms. , x Ans. x^+- = + ^. 3. Reduce the equation, ^'^ x"^ + lx=zl, to one of the four forms. Ans. x^ -\- ^x = 25. 5 Ans. x^ — bx = 25. 5. Reduce the equation, — f .r'^ — J j' -f fi = + 4, to one of the four forms. X Ans. x^ + '- = o. 336. If the first member of the equation put under one of the four forms is a perfect square, the solution can be as readily effected as in the case of an incomplete equation. For, we will only have to extract 292 EQUATIONS OF THE SECOND DEGREE. tlic square root of botli mciubcrs, and then transfer the known term or terms in tlie first member to the second member, and then the solution will be complete. Suppose that the equation is a;^ + 2aa; + a^ =&. Extract the root of both members, then a; + a = =b ^b. Hence, x' = — a -\- -s/h, and a;" = — a — -Jh. We see that the equation has two values, and that these are not numerically equal, as in the case of incomplete equations. Now if, by any artifice, we can make the first member a complete .square, it is plain that there will be no difficulty in the solution of a compl-ete equation of the second degree. Let us then assume the equa- tion, x^ -\- "px = q, and examine what modification the equation must undergo, in order that its first member may be made a perfect square. The square of a binomial is composed of the square of the first term, plus the double product of the first by the second, plus the square of the second term. The square of a binomial is, therefore, a trinomial. .ir^ + jjx must then be made a trinomial by the addition of some quan- tity before the first member \Vill become a perfect square. What is the quantity to be added ? Take the expression, (x + a)^ = x^-\- 2ax -f- a'. We see that the third term of the trinomial is the square of the second term of the binomial in the first member ; and we sec that 2a,r, the second term of the trinomial, when divided by 2x, twice the first term of the binomial, gives a quotient, a, which is the second term of the binomial. This quotient, squared, is the third term of the trinomial. Now, suppose we only knew the first two terms of the trinomial, x^ and 2ax, and wished to ascertain what was the binomial, which, when squared, would give these for the first two terms of its square. We would know that the first term of the binomial must be x ; and, since 2ax, the second term of the trinomial is twice this first term by the second terra, it is plain that the second term can be found by dividing by 2a,', twice the first term. Having found a, the second term of the binomial, we have only to square it, and the third term of the trinomial will be known. Apply these principles to the expression, a;^ -f px, regarded as the first two terms of a trinomial that is a complete square. It is plain that X is the first term of the binomial, the first two terms of whose square are x^ + ^«. The second term of the binomial can be found by dividing j^x by 2x, twice the first term : the quotient, ~, is the second term of the binomial, and its square, ^, is the required tliird EQUATIONS OF THE SECOND DEGREE. 293 term of the trimonial. If then ^ be added to x^ -\- V^, the first member will be a perfect square. But, if we add ^ to the first mem- ber, we must also add it to the second, eke the equality of the two members will be destroyed. Hence, the equation, x^ -\- px =^ q, can be changed into the equivalent equation, x- + px + x ~ 2 "^ jT" in which the first member is a perfect square. Now, extract the square root of both members, X + ^ = ± \/y + q. Hence, x' = — ^ + ^/^ + q, and Either of these values when substituted for x will satisfy the equation. The first, when substituted, will give ( — ^"'"V/'T'^!?^"^ ^ ("~f + \/f + '^) = *?' ""' ^-^ \/f' +1+J + 1- 'h^'sji + ' Py t + 2 + T "^ '^ ~ 2 ~^'V J + 'i = !?> or !? = q. Hence, the second value also satisfies the equation. We have taken an equation under the first form, but it is obvious that the principles deduced are applicable to equations under either of the three other forms. Hence, for the solution of a complete equation of the second degree, we have this general RULE. Reduce the equation to one of the four forms, hy multiplying hath members hy the reciprocal of the coefficient of the highest power of the unknown quantity, and by transposing all the knoicn terms to the second 25* 291 EQUATIONS OF THE SECOND DEGREE. ■memher. Next, add the square of half the coefficient of tlie fir&t poiccr of the unknmon quantity to loth memhers, and then extract the square root of hoth memhers. Finallij, separate the unhnown quantity from the knoion quantities hy leaving it alone in the first memher. A common error with beginners is, to complete the square without observing whether the coefl&cient of the first term is plus unity. But it will be seen that the rule requires the first step to be the making of this coefiacient plus unity, if not already so. EXAMPLES. 1. Solve the equation, — x^ + 16a; = 28. Ans. x' = + 2, x" = + 14, 2. Solve the equation, x^ — 4:X = — 4. Ans. x' = + 2, x" = + 2. 3. Solve the equation, x^ + 4a; = — 4. Ans. x' = — 2, x" = — 2. It is generally best to reduce the terms under the radical to a common denominator. Take the equation, — ax^ — cx= — m. This, when solved, gives X = — -— d=\/ 1- T— 0- The two terms under the radical can be 2« V a 4a'' reduced to a common denominator by multiplying the numerator and denominator of the first by 4a, If m had been divisible by a, then a would not appear in the denominator of the first term, and the multi- plier then would have been 4a^. The single term, into which all the known quantities in the second member have been collected after the (Coefficient of x^ has been made plus unity, is called the ahsolute term. In the preceding equation, — is the absolute term. Then, to reduce the terms under the radical to the same denominator, we multiply numerator and denominator of the absolute term by either 4 tim.es the coefficient of the second power of x, or 4 times the square of this coefficient. This rule is only applicable when the reduced term in the second member was entire, previous to making the coefficient of x^ plus unity. EQUATIONS OF THE SECOND DEGREE. 295 4. Solve the equation, 2x^ — Sx = 5. Ans. x' = + 2J-, x" = — 1. For, solving, we get, x = | ±^^ + J>^ = ^dtz v/|g + /g = | ± 1= -\-2h,ov-l. 5. Solve the equation, 2x^ — Sx = 2. Ans. x' = + 2, x" = — A- For, solving, we get, x = | ± N/T+7g = | ± v/ -^"^ = |± I = +2, or— *. ,. CI 1 -1 ,• -^^ ^'^ "^ ^'* o. bolve the equation, — , H = -— + — . wr c ?>!'' c ^i??s. x = + o, X = — a . c 337. These few examples are given to show how complete equations can be solved. But the solutions can be better understood when some of the properties of these equations arc known. First Projjcrf^. Every complete equation of the second degree has two values, and but two, for the unknown quantity. For, resume the equation, x^ + px = q. Complete the square of the first member, and we have x' -\- px + -^ = q + ^. Now, the 4 4 first member being a square, the second member may be represented by a square also. Hence, x'^ -}- j^x -{-—-= ni^, or (x + j^Y — "^^ = ^• Now, the first member being the diflerenee of two squares, may be re- solved by the principle of the sum and difi"erence into two factors. Then the equation, (x + pY — m^ = 0, will be changed into the equiva- lent equation, (x -i- p -{- m) (x -{- p — m) = 0. And since, when the product of two factors is equal to zero, the equation can be satisfied by placing either factor equal to zero, we have, x + p -\- m = 0, and x-\-p — m = 0. From which we get X = — p — m, iindx = — p + 7n ; or, distinguishing the values by dashes, x' = — p — m, and x" =z — p + m. Now, if we replaced m by its value, we would have the solu- -UC) EQUATIONS OF THE SECONIJ DEGREE. tions previously obtained. We sec that there are two values nume- rically unequal, and, since the equation, (x + p -\- m) (cc -f 7) — m) = 0, contains but two factors, it can be satisfied in two ways, and only in two ways. Hence, there are two, and only two, values for the unknown quantity. Sfcond Pmpcrty. 338. The first member of every equation of the second degree can be decomposed into two factors of the first degree with respect to x^ the first factor being the algebraic sum of x and the first value of x with its sign changed ) and the second factor being the algebraic sum of x and the second value with its sign changed. The second member of the equation after this decomposition will be zero. The factors have already been obtained, and are (a^ + p -f 7)1) and [x, -\r p — ■?«)• By comparing these factors with x' and x" , we see that Ar p -\- in is equal to — x' , and that + p — «i is equal to — a;", and these factors then can be changed into the equivalent, {x — cp') and {x — a;") ; and the equation, {x ■\- p -\- m) (x. + j) — m) = 0, can be changed into (x ■ — x') {x — x") = 0. 339. This important property enables us to find the equation whose values are known. We have only to change the sign of the first value, and prefix -f a? to it, and the first factor will be known; and then to change the sign of the second value, and prefix + x to it, and the second factor will be known. The product of these factors placed equal to zero is the equation required. Find the equation, whose values are -f 2 and — 3. The first factor must be (x — 2) ; the second factor must be a; + 3. Hence, (x — 2) (x+3) = 0, is the equation required. Expanding the first member, we have x^ -{- X — 6 = 0, or a;^ -f a; = 6. Completing the square, and solving, we get a- = — J db ^Q + I = — i rt v/-^' = — J d= | = -f I, or + 2, or — | = — 3. Hence, the process is verified. But the verification might have been made more readily, thus : the equation (x — 2) (.r 4- 3) = 0, can be satisfied by placing either factor equal to zero. Hence, x — 2 = 0, and a; + 3 = 0. These equations give the preceding values, + 2 and — 3. A few examples will make the beginner more familiar with the second property. EQUATIONS OF THE SECOND DEGREE. 291 EXAMPLES. 1. Form the equation whose values are both — 2. Ans. (x + 2) Qc + 2) =r 0, or a-^ 4. 4.^ a. 4 = 0. 2. Form the equation whose values are (f + ^h, and a — ^/i. Ans. (x — a — ^h) (x — a+ ^V) = 0, or x^ — 2«a- = h — a^. 3. Form the equation whose values are — 2m + y/n + Ani^, and - 2w? — y/7i + 4ml Ans. x^ + Amx = n. 4. Form the equation whose values are a — n+ s/m — 2an + n'^, and a — n — ^m — 2an + ?i^. Ans. x^ — 2 (a — «) ./; = m — a^. Verify these results by solving the equations. 340. The second property of complete equations sometimes enables us to solve an equation very readily. For, whenever the factors of an equa- tion are apparent, we need not solve the eciuation itself, but only those factors separately placed equal to zero. Thus, take the equation 01? + Ix = ax. By transposition we get, x^ ■\- hx — ax = 0. We see that a; is a common factor to the first member, and the equation may be written x (x-\- h — a) = 0. And since, when the product of two factors is equal to zero, the equation can be satisfied by placing either equal to zero, we have a; = 0, and x-\- h — a = 0. Hence, the two values are 0, and a — h. The general equation, x"^ -{■ hx — ax =z 0, slioios, furthermore, that when the unknown quantity enters into all the terms of the first member of an equation, ichose second member is zero, one value of the iinknown quantit}/ must nlwai/s be zero. Take, as a further illustration of the use of the second property, the equation, ax^ — bx^ -[-ax — bx = bx — ax. This equation can be written (a — b) x^ +2 (a — b) x = 0, ot x { (a — b) x + 2 (a —b) ) = 0, or X (a — b) (x + 2) = 0, or x(a — b) = 0, and x + 2 = 0. Hence, x' = 0, and x" = — 2. Again, x^ — ax -\- x — a = 0, may be written x (x — a) -f ■'" — ('' = 0, or (x — rt) (x -f- 1) = 0. Hence, x' = a, and x" = — 1. We have seen that the first two properties of complete equations are also properties of incomplete equations. The remaining properties are also, as we shall see, common to both classes of equations. 298 EQUATIONS OF THE SECOND DEGREE. Tliird Properti/. 341. The algebraic sum of the values of every complete equation of the second degree is equal to the coefficient of the first power of the unknown quantity, with its sign changed. For the equation, si:^ + px = q, when solved, gave the two values, ^' = -f + \/^ + <1 ^' = -f-\/^+- Adding these equations, member by member, we get x' + x" — — p, as enunciated. The third property of complete equations is also a property of incom- plete equations. For, an incomplete equation of the form, x^ = q, may be written, x^ ~\-0x = q. The two values of this equation, solved either as a complete or as an incomplete equation, are, cc' = -f ^q^ and x" = — ^q. The sum of these values, x' + x", is plainly zero, which is the coefficient of the first power of x. An incomplete equa- tion then may be regarded as a complete equation, whose two values are numerically equal, but affected with contrary signs. Fourth Pfop)erty. 342. The product of the values in every complete equation of the form, x^ + P^ = (J^ is equal to the second member or absolute term with its sign changed. For these values are cc' = — "o" + \/ T"^ S' andx" = —^ — S^/^+q. and their product, x'x" = -{- ^ U- -\- q\ = I- This property belongs also to incomplete equations, for the two values of the equation, x^ = q, are x' = + y/q and x" = — .^q, and their product, x'x" =■ — q. EQUATIONS OF THE SECOND DEGREE. 299 "We will see, hereafter, that the third and fourth properties belong to equations of every degi-ee, and that the first and second, with some modifications, are also properties of all equations. EXAMPLES. 1. What is the product, and what the sum, of the values in the equation, x^ + 4x = — 4. Aiis. x'x" = + 4, x' + x" = — 4. 2. What the product, and sum, in the equation, x^ + x = 1. Ans. x'x" = — 1, x' + x" = — 1. 3. What the product, and sum, in the equation, x^ — px = — tj. Ans. x'x" = -\- q, x' + x" = + p. Fifth Property. 343. The value of x, in every complete equation of the second degree, is half the coeflScient of the first power of x, plus or minus the square root of the square of half this coefficient increased by the absolute term. For the equation, x^ + px = tp when solved, gives x = — ^ ± \ / -J + q^ which agrees with the statement. It is obvious that this is also a property of incomplete equations, since, for such equations, ;? = 0. We have shown the foregoing properties by operating upon an equa- tion of the first form, but, since the demonstrations have in no way been dependent upon the signs of p and q, it is obvious that equations under the three other forms enjoy the same properties. The fifth property enables us to solve an equation directly, without completing the square in the first member j but it is well to require beginners to complete the square, until familiar with the principles. 300 EQUAXrONS OF TUE SECOND DEGREE, GENERAL EXAMPLES. 1. Solve the equation, ax^ + ex = ma. Ans. c / f2 „ c / C" 2« V 4a^ 2rt V 4a2 or, .r = -^ — , x" = ^—T . 2a la 2. Solve the equation, Ix^ — Ux = + 280. Ans. a:' = 1 + v^ll^ ■^■" = 1 — n/'IT- 3. Solve the equation, nx^ — mx = nm. ^ , 111 + y/m^ + 4n^m ,, m y/m^ + 4«^?w ^9(S. X= ■\- -r , if == . 2n 2n 4. Solve the equation, ax^ — hx = mx A7 5. Solve the equation, 4.t^ — 5.r = 7a- J.ns. 0^' = 0, a; = . Ans. x' = 0, a" = 3. 6. Solve the equation, nx^ — mx = 6x. Ans. x' = 0, .c' = — ■ . How might the answers to 5 and 6 be deduced directly from the answer to 4. 7. Solve the equation, — x^ — px =^ + §'• 8. Solve the equation, x^ -\- ^x — # = 0. Ans. ^' ~ + 3j x" = — 3. 9. Solve the equation, 4.^' + lOx — 6 = 0. Ans. a:' = + ^, x" = — 3. 10. Solve the equation, x^ — | a: — # = 0. ' Ans. x' = + 3, x" = — A. EQUATIONS OF THE SECOND DEGREE. 301 11. Solve the equation, x^ + x -\- -:^\ = 0. Ans. ; 12. Solve the equation, x^ — x + -f^ = 0. Ans. x' = + |, x" = + |. 13. Solve the equation, 16:/;^ — 16x + 3 = 0. Ans. x' = + I, x" = + f 14. Solve the equation, x^ — ax + ah = + hx. Ans. x' = + a, x" = + h. 15. Solve the equation, x^ + ax + ab = — bx. Alls, x' = — a, x" = — b. 16. Solve the equation, mx' -\ m = -——. i , ^ n ^ Ans. X = -J-, X = . a ^_ _, , , . „ vibx max li. Solve the equation, mx' vi = ; — . ' « b 4 , ^ ,i ^ Ans. X = + —, X = ^. a b __ „ , . . , rt.r mx px 18. Solve the equation, x^ -. = — ex — ^—. b n q Ans. x' = 0, x" = ---{ c — — . b n q -„ r, , , . , , ax mx px 19. Solve the equation, « + -y- + — = ex + ~. '■ b H q A inn pa m Ans. .r' = 0, x" = c + =^ . q b n 20. Solve the equation, x^ + mx + ex + \/n x-{- m\r = 0. Ans. x' = 0, cc" = — m — c — y/n — «t^. 21. Solve the equation, x^ — mx — ex — ^n x — m^x = 0. Ans. x' = 0, x" = m + c + ^n + m^. These equations show that, when the sign of the coefficient of x has been changed, and the other terms left unaltered, the values will be numerically the same, but affected with contrary signs. They show 26 802 EQUATIONS OF THE SECOND DEGREE. also tliat, when all the terms after x^ are negative, the second member being zero, the values will be both positive ; and when the terms are •all positive, the second member being zero, the values will be both negative. 22. Solve the equation, mx^ + px = — q. Ans. x' = — P+ ^p' — Amq ^„ ^ — p—^p^ — 4mq "What will these values become when p^ = Amq ? What, when / < ^mq ? 23. Solve the equation, 4:?^ ^ 24a- = — 36. Ans. x' = — S, x" = — 3. 24. Solve the equation, 4x^ — 24:r = — 36. Ans. a-' = + 3, x" = + 3. 25. Solve the equation, 4x^ — 24.r = —40. Ans. .x' = + 3 + ^—1, x"= + 3 — ^— 1. The following important consequences flow fi-om the last four ex- amples. Whenever the square of the coefficient of x is not greater than four times the product of the coefficient of x"^ into the absolute term, the values will be imaginary, if the sign of the absolute term is negative. And, whenever the square of the coefficient of x is equal to this nega- tive product, the two values of x will become identical. 344. Some equations may be treated either as complete or in- complete. x^ 1. Solve the equation, -, = n. (a — xj Ans. X = ^, X = . I + ^n \ — s/n For, expanding the denominator and clearing of fractions, we have, a-2 — n (a^ — 2ax + x^), or (1 — n) x^ + 2anx = na^, or x^ + 2an na^ ^^ , an . / ncv^ 'a^ :- = -z . Hence, :r = — —- h \ / ^ ■ H = 1— H 1 — n \ — n V l—n^(l — ',if — an 4- ^na? (1 — n) + ahi^ — an -f a^n _ a^/n (1 — ,yn) 1 — n 1 — n 1 — n ' Now, multiply numerator and denominator by 1 -f ^/n, and we have, EQUATIONS OF THE SECOND DEGREE. 303 a^n(l — n)^ = ''^" . In the same way, x" cau be shown (1 — ?0(l + \/w) 1+ n/" equal to =. These values for x' and x" are identical with ^ 1— v/n those before obtained, when the equation, ^ = », was treated as ^ (rt — x)2 an incomplete equation. x^ 2. Solve the equation, :, = m^, as a complete equation. (rt + xy , , am ,, — am Ans. X = ;. — ■ , x = X Solve the equation, :=- = q, as a complete equation (x — ^pf Ans. x' = 1 -f m' 1 4- m a complete equation. s/pq ,, s/pq ^q — 1 y/q + l 345. In the foregoing examples, the unknown quantity has been freed from radicals. Whenever it is connected with radicals, it must be freed from them by the preceding rules., and then the process will be the same as in the examples already given. 1. Solve the equation, y/mx^+ nx = y/p n 4- v/?i^ + 4m» „ n — ^n^ + 4mp ^"^- •'^ = ^ ' '■ = 2^1 • 2. Solve the equation, V^V + nx = p, as an equation of the second degree. . , _ j^ n — P ~ m + ?i' m — n 3. Solve the equation, ^mx^ + nx = p, as an equation of the second degree. , , P tr V Ans. x — — == ,x = =r . •J m -{- n y/m " 4. Solve the equation, ^If + ^ "^ "^ =Sh '^r+x Vl+.r A7}s. a:' = + 8, x"=—^. 5. Solve the equation, ( —=. = m. « + '-« ^/a + x Aus. X ^^ ■ . X ^^^ ^r What will these values become when m = 2? What, when m <^ 9? oO-i EQUATIONS OF THE SECOND DEGREE. 6. Solve the equation, -5-— h ■ — " = -• Ans. x' = — 2, a;" ^ , , • n/1 + .c 1 + j- 7. Solve the equation, -r; — ;— — 1 = 1 J.JIS. a;' l-\-x ^1 + — 3+v/— 3 „ — 3 — v^^ITs a; + a + c 8. Solve the equation, = = 1. ■,/2cx -\- 4ac Ans. x' = — a + ^/2ac — c^, x" = — a — v/2ac — c\ "What do these values become when c^ = 2ac ? What, ■when c'^2aP ^ -|_ 2 + 4 9. Solve the equation, — =z 1. s/Sx +32 Ans. x' 10. Solve the equation + 2 + 5 v/10^ + 40 ^/!s. x' = — 2 + ^/^^^, x" v/— 5. 11. Solve the equation, -J x^ -\- hx =^ y/ ax -{■ y/bx. Ans. x! = 0, a;" = a + 2 s/ ah. Verify all the preceding results. DISCUSSION OF COMPLETE EQUATIONS OF THE SECOND DEGREE. 346. The four forms of these equations are 3? — px = q, X? + px = — 2', 0? — px z= — q, and the corresponding values. First form. EQUATIONS OF THE SECOND DEGREE. 305 ='=f+\/^+^' ="=f-\/^+. — f^\/?- --f-\/^-^> '=f-\/?- f-\/^-^. Second form. Third form. Fourth form. Now, we observe that the first and second forms diflfcr only in the sign of X, and that x' in the first form is the same as x" in the second, taken with a contrary sign ; x" in the firet form is the same as x' in the second, taken with a contrary sign. A similar remark may be made in regard to the values of the third and fourth fonus. We conclude, then, that to change the signs of the values, without altering them nu- merically, we have only to change the sign of the coofilacient of the first power of X. Thus, x^ + .x = 2, when solved, gives x' = 1, x" = — 2; and x^ — .X = 2, gives x' = 2, x" = — 1 . In like manner, x^ — 5x = 4, when solved, gives the two values, x' = + 4, .x" = + 2 ; and x^ + 5.r = — 4, when solved, gives If we proceed to extract the scjuarc root of V+g', in the two values of the first form, we will get jr-, plus a series of other terms. Hence, the two values will become x = + 4- + other terms ; and x" P P — ^ ^ other terms. Hence, the first value is positive and the second negative, and the second is numerically greater than the first, because the radical is added to the quantity without the radical in the second value, and subtracted from it in the first value. 26* u 306 EQUATIONS OF THE SECOND DEGREE. By extracting tlie square root of "^ + f? approximatively in the vaJucs of tlic second form, we would see that the first value was positive and the second negative, and that the second was numerically greater than the first. We have anticipated this, from what has been said before, as to the interchange of position and the change of sign between the values of the first and second form. In the third form both values are negative, because — ^ is ^ \ If / -J — q. The second value is numerically greater than the first. In the fourth form both values are positive, and the first is numeri- cally greater than the second. These results might have been antici- pated from our knowledge of those in the third form. 347. The discussion of the signs of the values in the four forms may be made otherwise, thus : we know, from the fourth property of equations of the second degree, that th.e product of the values in the first form must be equal to — q. Hence, the values must have con- trary signs. And we know, from the third property, that the sum must be equal to — p. Hence, the negative value is the greater. For, when we add a negative and positive quantity together, and their sum is negative, we know that the negative quantity is greater than the positive. For the second form, the product of the values is equal to — q. Hence, the values have difi'erent signs. Their product is equal to 4- p. Hence, the positive value is greater than the negative. For the third form, the product of the values is equal to -|- q. Hence, the values are both positive, or both negative. But their sum is equal to — p. Hence, they are both negative. "We cannot decide from this, however, which is the greater, the first or second value. For the fourth form, the product of the values is equal to -\- q. Hence, the values must have like signs, and must be both positive, or both negative. But their sum is positive, being equal to -j- p ; hence, they are both positive. We cannot decide from this, however, which is the greater, the first or the second value. EQUATIONS OF THE SECOND DEGREE. 30' IRRATIONAL, IMAGINARY, AND EQUAL VALUES. 348. The values, in all tlie forms, will be irrational whenever the root of the radical cannot be extracted exactly. In that case, the ap- proximate value of the radical can be determined to within a vulgar or decimal fraction, and then the approximate values of the unknown quantity will be known. There can be no imaginary values in the first two forms, because the quantities under the radical are affected with positive signs. "When- ever, then, the absolute term of an equation is positive, we know, cer- tainly, that there are no imaginary values in the equation. The third and fourth forms contain imaginary values, only when q is f? . ^ ^. For, in that case only, is the prevailing sign under the radical negative. Whenever, then, an equation is put under the third or fourth form (the unknown quantity being freed from radicals, and the coeflS- cient of x^ being made plus unity), we can readily tell whether there are imaginary values. "We have only to square half the coefficient of the first power of x, and compare the result with the absolute term. The equations, x^ -|- lOx = — 30, and a? — lOx = — 30, both contain imaginary values, because (5)* <^ 30. Imaginary values, being always similar in form, are said to enter the equation in pairs. Since the two values of each of the four forms differ only by the radical, it is obvious that the disappearance of the radical will cause these values to become identical. But, the radical cannot be made to vanish in the first and second forms, because the quantities under the sign are positive. It will vanish, however, in the third and fourth forms, whenever ^=zq. The values of the third form will then both become — i^, and of the fourth form + ^. The equality of the values, in the third and fourth forms, differs from the equality of the values of incomplete equations. In the latter, it is equality only in a numerical sense; in the third and fourth forms, absolute equality. If we substitute ^ for q in the third and fourth forms, the equations will become x^ -}- px = — -^, and x^ — px = — ^, or, by transposi- tion , x^ + px + ^ = 0, and x^ — px + ^ = 308 EQUATIONS OF THE SECOND DEGREE. The first member of one equation is the square of (x + ^), and of the other (x — ^). Hence, when there are equal values in the third and fourth forms, the first members of those equations will be perfect squares, provided the second members are zero. The following are illustrations : x^ + Gx = — 9, and cc^ — 6x = — 9; x^ + Sx = —lQ,andx^ — Sx = — lQ. SUPPOSITIONS MADE UPON THE CONSTANTS. 349. Known terms are frequently called constants, and unknown terms, variables. In the equation, x^ -\- px = q, p and q are the con- stants, and x the variable. Let us make the constant q — o, in the four forms. By going back to the solved equations, we see that the values will become ^" _ p p 2"" 2 ~ 2h First form. ^■ = f + f=., Second form ^' = -f + f = o, P, Third form. Fourth form We see that the first and third forms have the same values, and that the second and fourth forms have the same values. This ought to be so, for the hypothesis, g- = 0, makes the first and third equations iden- tical, and also makes the second and fourth equations identical. EQUATIONS OF THE SECOND DEGREE. 300 We have gotten the foregoing series of values by making (j' = iu the solved equations. Ought we not to get the same results by opera- ting upon the given equations themselves ? When 3 = 0, the first and third forms become x^ -{- px = 0, ov x (x -{- p) :^ ; an equation which can be satisfied by placing either factor equal to zero. Hence, X = 0, and X + p = ; or, a;' = 0, and x" = — p ; the same as before obtained. The second and fourth forms become, when q = ^, x^ — 2)x = 0, or X (x — p) = 0. Hence, x" = 0, and x' =z + p, as before. When 2> = 0, we have this series of values : c' = -\-\/q, x" =z — v/gr, in the first and third forms. x' = 4-v/ — 'j> •^■" = — \/ — 'J, in the second and fourth forms. These results can be obtained directly from the equations ; for the first and second become x^ = q, and the third and fourth, x^ = — q. These two equations will plainly give the four preceding values. When p = 0, and q = 0, the system of values reduces to x' = + 0, x" = + 0, in each of the forms. This ought to be so ; for, in that case, the four equations reduce to the same, x^ = 0. The solutions of the four equations give formulas which can be ap- plied to particular examples. Thus, the values of the first form are Let it be -'=-f + \/f + q, and a^" = -J^-syj^ + required, now, to apply these results as formulas to find the values of the equation, x^ + x = 2. Then, jj = 1, and q = 2. Hence, x' z= — h + v/i- + 2= —^.^3^1. And x" = —i — ^i-^2 = — k — #= — 2. We have used as formulas the values x' and x", belonging to the first form, because the equation, x^ + x = 2,\s of that form. And, of course, we must always apply as formulas the values found in the form to which the given equation belongs. EXPLANATION OF LAIAGINARY VALUES. 350. We have seen that the values of the third and fourth forms be- come imaginary when q was made greater than ^. It remains to ex- plain the cause of the imaginary results, and to ascertain what they mean. In the third form, the product of the values is equal to + q, and their sum equal to — p. The values, then, are both negative; and 310 EQUATIONS or THE SECOND DEGREE. we liave previously seen that tliey were unequal wlicu ^ was unequal to q. If tlicy were equal, caeli must be — ^. Let x represent the execss of the greater over — ^. Then the value which is numerically the greater will be represented by — ( {y + x\, and the smaller nu- merically will be represented by — ("^ x\. Their product will be ^^ x^. Now, it is evident that this product will be the greatest possible, when x = 0. This condition makes the two values equal, and makes their greatest product ^ . But q represents their product, and it, therefore, can never be greater than -^. Hence, when we make 4 -7- ^ In"- ? Why ? 812 GENERAL PROBLEMS INVOLVING 2. Required to divide the number 10 into two sucli parts that their product shall be equal to 25. Ans. x' = 5, x" = 5. Is the disappearance of the radical always connected with maximum values ? 3. Required to divide the number 10 into two such parts that their product shall be equal to 2G. Ans. x' = b -\- >/— 1, x" = 5— V — 1. Why are these values imaginary ? Verify them by substitution in the equation of the problem. 4. The sum of two numbers is a, and the sum of their squares is h, what are the numbers ? , a + ^-Ih — a^ „ a — ^2b — a^ A7IS. X z=: -: , X = ^ . When do these values become imaginary ? when equal ? What is the least value that the sum of the squares can have ? 5. The sum of two numbers is 10, and the sum of their squares is 52, what are the numbers ? Aris. x' = G, x" = 4. G. The sum of two numbers is 10, and the sum of their squares is 50, what are the numbers ? Ans. x' r= 5, cc" := 5. 7. The sum of two numbers is 10, and the sum of their squares is 40, what are the numbers ? Ans. x' = 5 + J^^, if" r= 5 — -J — 5. How may the values, in the last three examples, be deduced directly from the general values in example 4 ? 8. A Yankee pedlar bought a certain number of clocks, which he sold again for m dollars. His gain per cent, on his investment was expressed by the number of dollars in it. How much did he pay for the clocks ? Ans. a;' = — 50 + v/(25 + m)100, a;" = — 50 — V(25 + m)100. 9. Same problem as the last, except that the pedlar sold his clocks for 375 dollars, instead of m dollars. Ans. x' = 150, x" = — 250. What is the meaning of the negative value? EQUATIONS. OF THE SECOND DEGREE. 313 10. A number of partners in business owe a debt of m dollars, but, in consequence of the failure of one of their number, each of the solvent partners has to pay n dollars more than his proper proportion of the debt. How many partners were there. A -' — V n „ _ V n 11. Same problem as last, except that the debt was 728 dollars, and that the increased portion of the solvent partners was 60 j dollars. Ans. x' = 4, x" z= — 3. The negative value is readily explained. An examination of the equation of the problem, after — x has taken the place of + x, wDl show that there have been two changes of condition, and the correspond- ing enunciation will be : " Several partners in trade owed a debt of 728 dollars, and, by the accession of another partner to share the debt with them, their individual liability was diminished by GOf ; required the number of partners." These changes in the enunciation give -f- 3 for the number of partners, and the result can be verified. For, the share of each in the debt is, 242 f dollars before a new partner was added to the firm, and but 182 dollars afterwards. And, in general, for problems involving equations of the second degree, there must be two changes in the enunciation to convert a negative into a positive solution. 12. The difference between the cube and the square of a number is equal to twice the number. Required the number. Ans. x' = -\- 2, x" ^ — 1. 13. Required to .divide a quantity, a, into (wo such parts, that the greater part shall be a mean proportional between the whole quantity and the smaller part. What are the two parts ? Am. Greater part, either , or —^ — ^— ; and lesser part, either |a — la ^5, or -S« + la \/b. Both values will satisfy the enunciation, in one sense ; for, a(%a — Because, by performing the indicated operations, we have for the first 27 314 GENERAL PROBLEMS INVOLVING ((2 ^2 g^2 value, 2a- — .UV^ = ^ — -5- v/5 + — r= -Ja" — ^"V"^J and, for tlie second value, §a« + Ws/^ = ^ -\- W^b + -j- = 2"' + iaVST The second value, however, does not satisfy the conditions of the problem, understood in a literal sense, for the greater part, — v/^j is less than the corresponding smaller part, %a + -^\/~^, and this corresponding part is greater than the whole quantity, a. The explanation of the negative solution is simple, when we return to the equation of the problem, x^ = a (a — x). When x is negative, this equation becomes, x^ = a(a -{• x), and the corresponding enunciation ought to be required to find a number, which shall be a mean between the whole quantity, a, and the sum of a and this number. The given quantity, a, might have been represented by a straight line, -f J-, and the problem then would have been to find a part, BC, which should be a mean between the whole, BA, and the part, AC, left after BC was taken from it. Now, when the ex- pression for BC became negative in the solution of the problem, it in- dicated that BC must be laid oif on the left of the point B, because this distance was laid off on the right of B, when its expression was positive. The diagram becomes -, and AC is greater than AB. This agrees with the second value of AC, 3 a — a + 7^ v/^ which is greater than a or AB. The negative solu- tion here indicates then a change of direction, and the equation being of the second degree, there are two corresponding changes of condition. The first is expressed by the unknown distance being sought upon the prolongation of AB, instead of upon AB itself; the second is expressed by the unknown distance being a mean between AB, and A'B j'lus BC, instead of between AB and AB minus BC. The explanation of a negative solution need never be difiicult; we have only to change + x into — a; in the equation of the problem, and^ then examine and see what the resulting equation means. Some of the following problems will involve one of the four methods of elimination. Whenever the equations between which the elimina- EQUATIONS OF TUE SECOND DEGREE-: olo tion is to be effected is of a higher degree than the first, the method of elimination by the greatest common divisor ought to be employed. 14. Find two numbers whose sum and product are both equal to a. . , a + ^o? — 4a ,, a — ^/a^ — 4« Ans. X = :; , X = ,;^ , , a — v/a^ — 4a ^/a- — 4a + a y = 2 — '^ "" — 2 • When will the two numbers be equal ? "When imaginary ? 15. Find two numbers whose sum and product shall be equal to 4. Ans. 2 and 2. 16. Find two numbers whose sum and product shall be equal to 2. Ans. x' = 1 + y/—l, x" =1 — >/^^^, y' = l — y/'-^, xf' = 1 + ^/^=^. 17. Find two numbers whose sum and product shall be both equal to 5. , 5 + v/5" ,, 5 — ^o~ , 5 — v^o" ,, 5 + Jb Ans. x' = ^ > *■ = 2 ' ^ ^ 2 ' ^ ^ 2 — " Do the values in problems 14, 16 and 17, indicate that there are two distinct sets of numbers, or that the second set is the same as the first, differing only in the position of the numbers. 18. Two capitalists, A and B, invested different sums in trade. A invested half as much as B, and kept his money in trade 6 months, and gained one-twentieth of his investment. B kept double the amount that A had, in trade for 9 months, and gained §100 more than A. Supposing that their respective gains were proportional to their re- spective capitals and the periods of investment, what were their capitals ? Ans. A's 01000, B's 82000. Why was the second value of x rejected ? What was the gain per cent of A and B? 19. The difference between two numbers is a, and the difference be- tween their cubes is h ; what are the numbers ? Ans •" = 2-+ \/-T2^ -^ =Y-VT2^' ^=-2- + \/-T2^'^ ==-Y-\/-l2^- When will x" and y" be real, but negative ? 316 GENERAL PROBLEMS INVOLVING AVhat do the negative values satisfy ? When will the two values of X be equal ? How will the two values of y be in that case ? When will the solutions be indeterminate? How many conditions must be imposed ? 20. The difference of two numbers is 2, and the difference of their cubes 152 ; what are the numbers ? Ans. First number, + 6, or — 4 ; second number, + 4, or — 6. How are the negative values explained ? 21. The difference of two numbers is 4, and the difference of their cubes 16; what are the numbers? Ans. First number, + 2 ; second number, — 2. 22. The diffcrenee of two numbers is 4, and the difference of their cubes 15 ; what are the numbers ? -^W^ 23. The difference of two numbers is zero, and the difference of their cubes zero ; what are the numbers ? A71S. x' and x", both = -—, and i/' and ^", also both = -— . 24. The year in which the translation began of what is called King James' Bible, is expressed by four digits. The product of the first, second and fourth, is 42 ; the fourth is one greater than the second, and the sum and difference of the first and third are both equal to one. Required the year. Ans. 1607. 25. The year in which Decatur published his official letter from New London, stating that the traitors of New England burned blue lights on both points of the harbour to give notice to the British of his at- tempt to go to sea, is expressed by four digits. The sum of the first and fourth is equal to half the second ; the first and third are equal to each other; the sum of the first and second is equal to three times the fourth, and the product of the first and second is equal to 8. Ee- quired the year. Ans. 1813. 26. The year in which the Governors of Massachusetts and Connec- ticut sent treasonable messages to their respective Legislatures, is ex- pressed by four digits. The square root of the sum of the first and EQUATIONS OF THE SECOND DEGREE. 317 second is equal to 3 ; the square root of the product of the second and fourth is equal to 4 ; the first is equal to the third, and is one-half of the fourth. Required the year. Ans. 1812. 27. A gentleman puts out a certain capital at an interest of 5 per cent. ; the product of the interest for 12 mouths by the interest for one month, is just one fourth of the principal. What is the capital ? Ans. a/ = 0, x" = $1200. 28. Some of the New England States were fully, and some partially, represented in the Hartford Convention, which, in the year 1814, gave aid and comfort to the British during the progress of the war. If 4 be added to the number of States fully and partially represented, and the square root of the sum be taken, the result will be the number of States fully represented; but if 11 be added to the sum of the States fully and partially represented, and the square root of the sum be taken, the result will be equal to the square root of 8 times the number of States partially represented. Required the number of States fully and partially represented. Ans. Three fully represented ; two partially represented. 29. The sum of two numbers is a, the sum of their squares h, and the sum of their cubes c. What arc the numbers ? A71S. x' ^+ .s/.+^^- -/>) ,. 2 .J\/«= .,^ -ab) I V" .2 , 4(C— «^) andf = h i.\A= c -ab) I Have we two distinct sets of values, or but one set, with an inter- change of position ? When will the values of x and y be equal ? When imaginary ? When negative ? 30. The sum of two numbers is 7, the sum of their squares 25, and the sum of their cubes 91. What are the numbers ? Ans. x' = 4, a;" = 3 ; i/' = 3, //" = 4. 31. The sum of two numbers is 6, the sum of their squares is 18, and the sum of their cubes 54. What are the numbers ? Ans. Both 3. 32. The sum of two numbers is 5, the sum of their squares 20, and the sum of their cubes 50. What are the numbers ? Ans. x! = % +K/— 15, x" = 4 — \ ^— 15 ; y = 4 — * n/— 15, 27* 318 GENERAL PROBLEMS INVOLVING 33. Two travellers started at the same time from two cities, C and W, and travelled toward each other. They found, on meeting, that the traveller from C had travelled 150 miles more than the other traveller, and that by continuing at the same rate, he could reach W in five days; whereas, it would take the traveller from W twenty days from the time of meeting to reach C. Required the distance between W and C, the rate of travel per day of the two travellers, and the time that had elapsed before their meeting. Ans. Distance between W and C, 450 miles; the rates, 30 and 15 miles per day; the time elapsed before meeting, 10 days. 34. The sum of three numbers is 15, the sum of their squares 93, and the third is half the sum of the first and second. What are the numbers ? Ans. 8, 2 and 5. 35. A man bought a tract of land for 811 per acre, and sold it again at a less price, his loss per cent, on the sale being expressed by the price per acre which he received. What did he sell the land for ? Ans. 10 dollars per acre. 36. In the year 1G37, all the Pequod Indians that survived the slaughter on the Mystic River were either banished from Connecticut, or sold into slavery. The square root of twice the number of survivors is equal to jLth that number. What was the number ? Ans. 200. 37. A. Southern Planter bought m acres of cultivated, and as many of uncultivated, land. He got b more acres of uncultivated than of cultivated land per dollar, and the whole cost of the cultivated exceeded that of the uncultivated by c dollars. How much did he pay per acre for each kind of land ? 1 Ans. x' = -— ^ , 1 . / i (4m + be) . 1 -i> a /b (4m -f be ) for cultivated land ; y = ^ _j. . / b (im + be) , y" = '!__ y^ ./ b (4m + be ), for uncultivated land. 38. A Southern Planter purchased 100 acres of cultivated, and 100 acres of uncultivated land, the former costing him $500 more than the EQUATIONS or THE SECOND DEGREE. 319 latter ; the smaller cost of the latter resulted from his getting for every dollar y'^th of an acre more of the uncultivated land than of the cultivated. What was the cost per acre of the cultivated and uncultivated land. Ans. Former, $10 per acre; latter, $5 per acre. Verification. 100 acres at $10 per acre will cost SIOOO, and 100 acres at $5 will cost $500. A dollar will buy ^^ih of an acre of the culti- vated, and ith of an acre of the uncultivated land ; and the difference between ith and y^th is jgth. So, a dollar will buy -jjjih of an acre more of the uncultivated than of the cultivated land. .39. In the year 1853, a number of persons in New England and New York, were sent to lunatic asylums in consequence of the Spiritual Eapping delusion. If 14 be added to the number of those who became insane, and the square root of the sum be taken, the root will be less than the number by 42. Required the number of victims. A71S. 50. Why is the second value of x rejected ? 40. Two farmers, A and B, invest each a certain amount in the Central Railroad of North Carolina. After a time, A sells his stock for $150, and gains as much per cent, on his outlay as B invested. B also sells his stock' and gets S122- more than he gave for it, but his gain per cent, on his outlay is only half as great as that of A. Required the amount invested by each. Ans. A, $100; B, $50. Why is the negative solution rejected? Verification. 50 per cent, on $100 is $50. And since A sold his stock for $50 more than he gave for it, he gained 50 per cent, on the $100 of outlay. B gained 25 per cent on his outlay, and 25 per cent, upon $50, is $12^. 41. Two travellers set out at the same time, the one from A, and the otlier from C, and travel towards -, ■ each other at uniform rates. After meeting at B, the traveller from A is a days in reaching C, and the traveller from C, c days in reaching A. How long was it after the time of starting until they met at B, and how long was each traveller in performing the distance A C. Time of meeting, dc y^ac. Traveller from A, a ± x/ac; traveller from C, c ± y/ac. 320 a EN Ell A L niOBLEMS INVOLVING What is the moaniug of the negative sign in the values of the time ? When only can the time occupied by the traveller from A be negative ? How then will both expressions for the time occupied by the other traveller be affected, with the positive or negative sign ? 42. Same problem as the last, except that the traveller from A is 9 days between the points B and C, and the traveller from C, 25 days between the points B and A. Ans. Time of meeting, ± 15 days ; time occupied by traveller from A, 24, or — 6 days ; time occupied by the other traveller, 40, or — 10 days. 43. In a certain bank there are $438 worth of 5 and 3 cent pieces, the number of the latter is exactly the square of the number of the former. Required the number of pieces of each kind. Ans. 120 five cent pieces ; 14,400 three cent pieces. 44. Required to find two numbers, such that their sum, their pro- duct, and the difference of their squares, may all be equal to each other. Ans. x' = % + ^J, x" = ^-^~^', y' = ^ + ^J, y = ^ — ^/"|: 45. In the year 1706 the French made a descent upon Charleston; but " South Carolina," says Bancroft, " gloriously defended her territory, and, with very little loss, repelled the invaders." A certain number of the French were killed and wounded, and 100 were taken prisoners. The number of killed and wounded was to the number of uninjured, including the prisoners, as 1 to 3. And the square of the number that escaped in safety from the expedition, was to the square of the number killed and wounded, as 6i to 1. Required the number of invaders, and the number of killed and wounded. Ans. 800 invaders, and 200 killed and wounded. * Verification. If 200 were killed and wounded, then 600 were un- injured, and 200 : 600 : : 1 : 3. And, since 100 were taken prisoners, 500 escaped without harm from the expedition, and (500)^ : (200)^ : : 6J :1. Why is the value connected with the negative sign of the radical rejected? 46. In the year 1842 South Carolina converted the citadel at Charleston, and the magazine at Columbia, into military academies, which were to be supported by the sum of money appropriated annually EQUATIONS or THE SECOND DEGREE. 321 previous to this time to a guard of soldiers. The interest upon this sum for 21 months, amounted to $19G0, and the square of the interest for 6 months exceeded the square of 100'" part of the sum by $288,000. Required the sum appropriated annually to the military academies, and the rate of interest. Ans. 816,000, sum ; 7 per cent, interest. 47. A man in Cincinnati purchased 10,000 pounds of bad pork, at 1 cent per pound, and paid so much per pound to put it through a chemical process, by which it would appear sound, and then sold it at an advanced price, clearing 8450 by the fraud. The price at which he sold the pork per pound, multiplied by the cost per pound of the chemical process, was 3 cents. Required the price at which he sold it, and the cost of the chemical process. A71S. He sold it at G cents per pound, and the cost of the process was i cent per pound. 48. The fore wheel of a wagon makes 12 more revolutions than the hind wheel, in going 240 yards; but if the circumference of each wheel be increased one yard, the fore wheel will make only 8 more revolutions than the hind wheel, in the same space. Required the circumference of each. Ans. Circumference of fore wheel, 4 yards ; circumference of hind wheel, 5 yards. 49. In the year 1853 there were a certain number of Woman's Rights conventions held in the State of New York. If 6 be added to the number and the square root of the sum be taken, the result will bo exactly equal to the number. Required the number. Ans. 3. How does the negative solution arise ? Why is it neglected ? 50. A planter purchased a number of slaves for $36,000. If he had purchased 20 more for the same sum, the average cost would have been $150 less. Required the number of slaves, and their average price. Ans. 60 slaves ; average price, $600. 51. A planter purchased a number of slaves for m dollars. If he had received n more for the same sum, their average price would have been c dollars less. Required the number of slaves and their average price. V 822 GENERAL PROBLEMS INVOLVING n /4:nm + en' ,, n / inm -\- cn^ 2 ' V 4c ' 2 V 4c ' x' and a;" exjiressing the uimiber of slaves. Then, —, and -^ will ex- press the average price. Now, when n is zero, the two values of x ought both to be infinite, indicating an absurdity. But the expressions will not point out an absurdity unless reduced to their lowest terms by extracting the indicated roots. Then the two values of x may be written — — dz m — -| -^, plus any other terms containing the higher powers en ( of X in the denominators.) Now, make n = 0, and the two values of x both become infinite. An expression can never be correctly interpreted unless it is reduced to its lowest terms, for, as in the present instance, there may be a common factor in all its terms, and a particular hypothesis made upon that common factor may lead to absurd results. What do the values become when c = ? what when n :i= and c = ? 52. The field of battle at Buena Vista is 6 J miles from Saltillo. Two Indiana volunteers ran away from the field of battle at the same time ; one ran half a mile per hour faster than the other, and reached Saltillo 5 minutes and 54^*^ seconds sooner than the other. Kequired their respective rates of travel. Ans. 6, and 5^ miles per hour. 53. The New York shilling is 12| cents. A merchant bought a quantity of cloth for $60. The number of shillings which he paid per yard was to the number of yards he bought as 1 to -i-ff^. llcquired the number of yards and the price per yard. Ans. 48 yards ; 10 shillings per yard. 54. A grocer sold 1000 pounds of coffee and 1500 pounds of rice for $240 ; but he sold 500 pounds more of rice for $80 than he sold of coffee for $60. Required the price per pound of the cofiee and rice. Ans. Cofi"ee, 12 cents per pound ; rice, 8 cents per pound. 55. A, B, and C, entered into partnership, and gained as much as their joint fund, wanting $200. A's gain was $240. He invested EQUATIONS OP THE SECOND DEGREE. 323 $100 more than B ; and the joint investment of B and C was S700. Required the gain per cent., and the amount invested by each partner. Ans. 80 per cent. A's capital, S800 ; B's, 6200 ; C's 3500. 56. There is a vessel containing 25 gallons of -wine ; a certain quan- tity is drawn out, and its place supplied by water. As much is drawn out of the adulterated liquor as was first drawn out of the pure wine, and there is now but 16 gallons of pure wine left in the vessel. Re- quired the quantity of pure wine drawn out each time. Ans. First draught, 5 gallons ; second draught, 4 gallons. 57. Same problem as last, except that the vessel contains m gallons of wine, and that, after four draughts, — gallons of pure wine are left. . Ans. Inrst draught, m — \/m; second, = — ; third, ; _ ^/m m . _^, m — y/m fourth, ■ These results can readily be verified ; for, when the four draughts are taken from m, there will remain — ^m — m ^m — m ^m — m ,. , . , \/m -I = — H 1 ^^— — which is equal to v/w + i — vm -\ — — 1 -f =^ or ^=^ -f — , or _ y/in ^m^ ^m v^m -|- 1 — v^wi 1 , or — . m m The equation to be solved was really one of the fourth degree ; but, owing to its peculiar form, it was readily reduced to an equation of the second degree. Only one of the four values of the unknown quantity have been used. 58. Same problem as last, except that the vessel contains 64 gallons of wine, and that, after four draughts, the gLth of a gallon is left. Ans. First draught, 56 gallons ; second, 7 ; third, || ; fourth, ^f^. Verify these results by adding them together, and subtracting their sum from 64. 324 GKNEIIAL PROBLEMS INVOLVING. 59. Two cotton merchants rent a house for a certain sum, with the understanding that each shall pay in proportion to the number of bales of cotton he puts in the house. A puts in 500 bales-, and B as many bales as makes his proportion of pay amount to $200. B afterwards puts in 500 more bales, and then his proportion amounts to $225. Re- quired the sum paid for house rent, and the number of bales first put in by B. Ans. House rent, $300. B first put in 1000 bales. 60. A gentleman has two sums of money at interest, amounting to $1150. The larger sum, being put out stii per gent, less advanta- geously than the smaller, brings only the same amount of yearly inte- rest. At the end of ten years, the smaller sum, added to its simple interest for the whole period, is to the larger sum, added also to its simple interest, as 11 is to 11 2 . Required the two sums, and the per cent, on each. Ans. $550 at 10 per cent., and $600 at 9i per cent. 61. A gentleman bought a rectangular piece of laud, giving $10 for every yard in its perimeter. If the same quantity of ground had been in a square shape, it would have cost him $20 less. And, if he had bought a square piece of land, of the same perimeter as the rectan- gle, it would have contained 6|^ square yards more. Required the sides of the rectangle. Ans. 4 and 9 yards. G2. A and B, together, invested $800 in a speculation. A's money was employed 3 months, and B's 5 months. When they came to set- tle, A's capital and profits amounted to $451 ; B's amounted to $375. Required the capital of each, aud their gain per cent, Ans. A's, $440 ; B's, $360. 10 per cent. gain. 63. The seventh page of a treatise on Analytical Geometry has 11 more lines than the twentieth page of a treatise on Optics ; but the seventh page of the Analytical Greometry has 4 letters less in each line than there are lines in the twentieth page of the Optics, whilst the twentieth page of the Optics has as many letters in each line as there are lines in the seventh page of the Optics. The total number of let- ters on both pages is 3542. Required the number of lines and letters in each page. Ans. Analytical Geometry, 46 lines, and 42 letters in each line; Optics, 35 lines, and 46 letters in each line. EQUATIONS OF THE SECOND DEGREE. 325 64. Required to divide the quantity, a, into two such parts, that the sum of the quotients, arising from dividing each part by the other, shall be equal to m. . T^. ^ .. ^ , <^ /"«^ — ^ f (i /'" — 2 Ans. First part, — - + --\ / , or \ / — : ^ ' 2 ^ 2 V "i + 2' 2 2 V w( + 2 ' , a a I )ti — 2 a a /m — 2 second part, — — -r-\ / -, or — - + — -\ / . ^ '2 2 V «^ + 2' 2 ' 2 V 7n + 2 Are there four independent values? When only will these values be real ? When will one always be negative ? When will the two parts be equal ? 65. Required to divide 10 into two such parts, that the sum of the quotients, arising from dividing each part by the other, shall be equal to 2Jffi. Ans. 7 and 3, or 3 and 7. 66. Required to divide the number 100 into two such parts, that the sum of the quotients, arising from dividing each part by the other, shall be equal to 2. ^Ih.s. 50 and 50. 67. Required to divide 15 into two such parts, that the sum of the quotients, arising from dividing each part by the other, shall be equal to9f .4h.s-. 13 i, and li. 68. Four numbers are in a continued proportion, each number being an exact number of times greater than that which precedes. The dif- ference between the means is 8, and the difference between the ex- tremes 28. Required the sum of the means, and the terms of the proportion. Ans. Sum of the means, 24. The numbers are 4, 8, 16 and 32. In this example, let the unknown quantity be the sum of the means. 69. Same example as preceding, except making the difference of the means, a, and the difference of the extremes, h. Ans. Sum of the means + o \ /^ -r--. greater mean a + V 6 — 6a ^ / h -\-a -, / h -\- a h a K/ ^ ; smaller, a W a; greater extreme -^ -f Ma^ + b^(b — 3a) „ b , /4a' -f b' V riTs^, ^; smaller-- +1\/-^^ 28 \ 326 EQUATIONS OF THE SECOND DEGREE. The negative values liave been rejected. When will these values become infinite? When imaginary ? 70. There are two numbers, such that the square of the first added to their product is equal to m, and the square of the second added to their product is equal to n. What are the numbers ? , „. m ^ , n A71S. First, =b — _ ; second, ^m -\- n ^m -\- n 71. Two numbers are to each other as m to n, and the square of the first added to the product of its first power by m, is equal to the square of the second, added to the product arising from multiplying its first power by n. What are the numbers? Ans. First, 0, or — m; second, 0, or — n. 72. The sum of the squares of two numbers diminished by twice their product and by twice the first number, is equal to unity; and the sum of their squares added to the first power of the second number, is equal to twice the product of the numbers. What are the numbers ? Ans. First, 0, or — | ; second, — 1, or — -^. TRINOMIAL EQUATIONS. 353. A trinomial equation is one of the form, x^" + x" = q, involving but one unknown quantity and three terms, and having the unknown quantity in one term affected with an exponent double of that with which it is affected in the other. A trinomial equation then contains two terms, in which the unknown quantity enters, and an absolute term. Let it be required to solve the equation, x'^ + cc^ =: 20. Let x^ = y. Then the equation becomes y^ -\- y = 20. The last equation gives y = + 4 and y = — 5. But, x^ = y. Hence, x" = 4, o r — 5. Then, x' = + 2, x" = — 2, x'" = + V 5^ x"" = — ^/ — 5. Either of the four values will satisfy the given equation, a-'' -\- x^ = 20. The substitution of the last value gives (— x/— 5/ + (— ^^^bf = 20, or, 25 — 5 = 20, a true equation. Solve the equation, x^ — x^ = 702. Make x^ — y. Then the equa- tion becomes y"^ — ?/ = 702. From which, ?/=-{- 27, or, — 26. But x^ = y. Hence, x = 4/27 = 3, and x = ^ — 26. There are really four more values for x, since, as will be shown here- after, the number of values is exactly equal to the degree of the equa- EQUATIONS or THE SECOND DEGREE. 327 tion. But the method of determining the other values belongs pro- perly to the general theory of equations. The two values found can be readily verified. The examples given are of a simple character. The equations were already of the proposed form. But it frequently happens that an arti- fice must be employed to put the equation under the form of x^" -|- a;" = q. Take, as an example, x^ -f ^x^ 4- 9 = 21. Add 9 to both membe rs, and w e have x^ -\-^ + ^x^ + 9 = 30. Make a:^ + 9 = /, then, v^a;^ -f 9 = y, and the equation becomes y^ ■\- y = 30. From which we get, ?/ -^ -f 5, or — 6. Then, a;^ + 9 = 25, or, j;^ -f 9 = 36. The values of x are -f 4, — 4, + v/27, and — v/27. The negative sign of the radical, ^az -(- 9, must be taken in connection with the last two values. This is indicated by the equation, y/x^ -f- 9 = y = — 6. Again, take ^^^^' + 1 = '^-^^ + 01- This may be written, ^:i±^' — (^!±J^' = 90. XX a-^4- 9 Let =r ?/. The equation then becomes if — ?/ = 90. Hence, 2/ = -f- 10, or — 9. Then, ^-^ =:. 10, and "^^ = — 9. These 9 .9 equations gives the four values, -f 9, -f 1, and — - -\- 2 ■ ii)<;- the root, we get = rb — -. We will, for convenience, call m — x ^i .r' that value of x which is connected with the positive sign in the second member, and we will call od' that which is connected with the negative sign. Hence, -. = — ^, and r, = — — ^. And m — x ^i m — x" ^5 1 • xi 1- , , m^/a , „ my/ a solving these equations, we get x: = _ — —^ and x'' = — =:--^^ — -^. Now, since there are two distinct values for rr, we conclude that in general there will be two points of equal illumination. By this we do not mean that there will be two points of equal brilliancy, but that there will be two points where the intensities of the two lights will be equal to each other. By subtracting cc' and x" in succession from m, we get , mh . ,, m^h ^ -' - and — -" y/a + y/h ^a — y/h Hence, we have the system of values: x' = — ^= :=, distance from A to first point of equal illumination. v/a + v/6 , m^h T « -r. m — x = — = m, distance from B to same point. ^/a+ ^h x" = —^ -=, distance from A to second point of equal illumina- m — x" = := -^_, distance from B to same point. Now, these expressions have been deduced upon the supposition that the point, I, was between A and B, and, consequently, we have assumed that the distance A I is positive, when estimated on the right of A, and the distance B I positive, when estimated on the left of B. Hence, if either x' or ic" becomes negative in consequence of any imposed EQUATIONS OF THE SECOND DEGREE. 331 condition, the corresponding point of equal illumination will be found on the left of A. And, in like manner, if either m — x' or m — x" becomes negative, the point will be found on the right of B. We will begin the discussion by supposing a = b. Then, x' = mVa ms/a m AB • x • i, u- — = =. = = = -C-, or A i = — -— . Hence, the point is halt Va-\-Vb 2v/a ^ ^ way between the lights, as it obviously ought to be. Now, m — x', which expresses the distance from B to I, ought to give the same point. , , „ , «i-«/6 niy/b m ^^ ... It does so, for m — x = -rm =r- = = = -r-- >v e will next Va+ ^b 2V6 ^ examine whether there can be a second equally illuminated point, wheu the intensities of the two lights are the same. Wc have x" = , „ vi\/b ^, , . ,. , = cc, and m — x = jr— = — oo. ihcse values indicate that the second point is at an infinite distance on the right of B ; or, in other words, that there is no such point at all. This ought to be so, for, when the two lights are of equal intensity, the one cannot throw its beams with the same power to a greater distance than the other, and this must be the case if there be a second point. The second values (x" and m — x") then denote impossibility. By going back to the equation of the problem, wc see that it will become, when a = b, x^ = (jn — xy, an equation which can only be true when m = 0, or x = — . But wi = is contrary to the hypothesis ; hence, x = -— is the only true value. The assumption then, that there was a second point (a being equal to b), was absurd, and led to oo , the appropriate symbol of absurdity. "We will nest suppose a > i. Then, x' = —= n becomes "> — , because, if the denominator ^/a + v/i 2 m were 2 ^/a, the value of the fraction would be "i^, and since the deno- VI minator is less than 2^ a, the value is greater than tt- Hence, the point, I, is nearer B than A. In this case, A IB r SoZ EQUATIONS OF THE SECOND DEGREE. m — x' <^ ^, for, were the denominator of the fraction — ~— ^ — __ m exactly 'Z\/b, the value of the fraction would be ^^J t)ut since the deno- minator is greater than 2, ^h, the value is less than — . This result corresponds to the former, and places I nearer B than A ; that is, nearer the light of feebler intensity, as it ought to do. Wc see that x" is greater than m, for were the denominator of x", the -s/a, its value would be equal to m, but as y/ a — y/ b is less than V a, x" is greater than m. Hence, the second point of equal illumination is beyond B. T 1 . ,f m y/ b ^ . . , In this case, m — x = = =^ becomes negative, since the ^a—^/b ^ numerator is negative and denominator positive, and, therefore, the second point must be on the right of B. The results then agree with each other, and agree with the fact. For, after passing the point I, the intensity of the 1st light becomes feebler than that of the 2d. At the point B, there is the greatest diflference between their intensities. Beyond B, both lights diminish in brilliancy, but the 2d more rapidly than the 1st, because of its greater feebleness; and we at length reach a point, V, where the illumination is equal. Next, suppose h^ a. Then, x! r= — ^;= ^ would have three values and three quotients, which is ab- surd. So, if — = any fraction, there would be more than one quotient. m . We are now prepared to assume the form of development of any al- gebraic fraction, arranged according to the powers of a certain letter, and having that letter in the first term of the numerator affected with an exponent equal to, or greater than that of this letter in the first or last term of the denominator. The exponents of the aiTanged letter are also assumed to be positive and entire, both in the numerator and denominator. Let us then place — ; — = A -f- B.c + Gx^ + J)x^ + Ex* + &c., in a -j- X which A, B, C, D, &c., are independent of x, and in which all the ex- ponents of x are positive and entire. A is the representative of that term which does not contain x, or contains x affected with a zero expo- nent. There is generally such a term in a development, and there must, of course, be a representative of that term. Clearing of frac- tions and arranging according to the ascending powers of x, we have a = Aa + a B I i- -f rt C I x2 -f « D I x^ -f a E I x*-{-&c. (31). +A I -f B I +C\ +cl Now, since the coefficients are supposed to be independent of x, their values will not be affected by making x — 0. If then we can deter- mine these values when x = 0, they will be true when x = 1, 2, 3 ; anything whatever. Making x = 0, we have a = Aa. Hence, A = 1. 29 w 338 UNDETERMINED COEFFICIENTS. Now,, since a = Ka, tlaese two terms cancel each other, and (M) becomes +a| +b| +c| +d| Dividing both members of this equation by x, which we have a right to do, we get = a'Q + aG \x + aJ) \x^ -\- ar& \x^ + kc. (N). +A +B I +C I +D I Making, again, a; = 0, wc have left, = aB -f A. Hence, B = — A 1 — = . Now, since «B + A has been found to be zero, they may a a be stricken out of (N), and that equation will become +b| +c| +d| And again, dividing out by x, we have 0= aQ + a D I x^ + a E I .7j2 + &c. (P). +B +C| +d| Making cc = in (P), we have left, = a C + B. Hence, C = B 1 = H 2- Omitting the term, oC + B = 0, in equation (P), and dividing again by x, we have = a D + a E j a; + &c. (2). Making a: = 0, we have left, = aD + C. Hence, J) = = a 3 . Dividing (2) by x, and again making a; = 0, we have aE + D = 0. Hence, E = = -| — j. Now, if we substitute for A, B, a a* ' ' 0, &c., their found values, in the original equation, = A + Ba- ft + x + Cx^ + Da-3 + Ea;^ + &c., we will have — "—^ = 1 _ i^ + ^ _ a -\- x a a^ 7? X^ X^ -, -{ — J r + &e. : the same result that we obtained bv division rr a* a" -^ By transposing a to the second member in equation (M), we have = A a I x° + a B I a; + a C I a;" + rr D I a''' + &c. — « +a| +b| 4-C UNDETERMINED COEFFICIENTS. ooi> And, since we have found A« — a =zO, a A 4- A = 0, aC + B = 0, aD+ C = 0, &c., we conclude, that, if ice have an equation wliose first member is zero, and xohose second member contains all its terms ar- ranged according to the ascending powers of a certain letter, the expo- nents of this letter being all positive and entire, and its coefficients in- dependent of it, these coefficients will be sep)arately equal to zero. This is the first enunciation of the principle of undetermined co- efficients, and ought to be remembered. The second enunciation, (which is an immediate consequence of the first), is as follows : if we have an equation of the form, a-\-hx+ cx^ -|- dx^ + ex* + &c., =a' -\- b'x + c'x^ + ^'^^ + '^^'^ + &c., which is satis- fied for any value of x, then the coefficients of the like powers of x in the two members will be respectively equal to each other. For, by transposition, we have r= (a' — a) x° -\- (Jj — b) x -\- (c' — c) x^ ■\- (d' — d') x^ -f (e' — c) x* -\- &c. And, since these coeffi- cients are independent of x, we must have, by what has just been shown, a' — a = 0, b' — b = 0, d — c=0, d' — d = 0, ^ — e = 0, &e. Hence, a' =^ a, b^=b, cf = c, d = d, e' =^ e, as enunciated. The preceding equation, a + ix -f cx^ -f dx' -f ex* -f &c. = a' -f b'x -f c'x^ -\- &c., and all other equations to which the second enuncia- tion is appliablc, belong to a class of equations called identical equa- tions, the two members of which diffi^r only in form. It is, in general, most convenient to develop expressions in accord- ance with the second enunciation. 1 — X Required the development of . Let J-=^ = A -f B.r + Gx' -{- J)x' -{- Ex^ -f &e. Clearing of fractions, we have 1 — x = A -f (B + A) a; -f- (C + B)x'-hCG + D) a;=» -f (E + D) X* -f &c. By placing a- = 0, we have A = 1, and we might proceed as we did with the fraction, . ' a -f- X continually dividing by x, and making x := in the resulting equation. But, by the second enunciation, we have at once A = 1,B + A = — 1, C + B = 0, C + D = 0, and E -f D r= 0. From which, A := 1, Br= — 2, C = +2, D= — 2, E=— 2. And, substituting these 1 ^ values of A, B, C, D, E, &c. in the equation, , 1= A -f Bx + Cx^ 1 -\-x I X + &c., we get y-— =. 1 _ 2x -f 2x^ — 2x^ -f 2x* — &c. 340 UNDETERMINED COEFFICIENTS. This is tlic same result that we would obtain by actually performing the indicated division. We have assumed the form of development to be A + Bx + Cx^ + &c., in which the exponents are all positive and entire. Now, if we have an expression whose development must necessarily contain negative or fractional exponents, it would be absurd to place it equal to A + B + a; + C.T^ + &c., and the result will make manifest the absurdity by the symbol, cc . Suppose it be required to develop -. It is plain that the first term of the development is x"' ; if, then, we attempt to develop the expression by the method of undetermined coefficients, we commit an absurdity, and that absurdity ought to be made manifest in the result by the appropriate symbol, co . Let, then, — —, = A + Bx + Cx^ + J)x^ + Ex" + &c. X — x/ From this we get l=Ax + (B—A)x^-{-(G — B)x^-].(D — C)x*+&c. Now, in accordance with the first enunciation, make x = 0, and we get 1=0, an absurd result. But, if we first divide both members by X, and then make x = 0, we will have i = go = A, and the absurdity is pointed out by its appropriate symbol. The above expression can be developed by changing its form. De- compose 5 into its factors, — X j; , then develop bv X — a; X 1 — X ^ 1 — X the method of undetermined coefficients, and multiply the development by — . By developing, we find = 1 + x +x^ -}• x^ -\-x*+ &c. ; 111 , „ consequently, --^ = — X -, =r x"' + x° + x + x^ + ^^ -f .x" + &c. The series is evidently arranged according to the ascending powers of x. For most expressions, a slight inspection will show whe- ther their development will contain negative or fractional exponents. and, consequently, whether the assumed form is right. Thus, 1 — x~- cannot be placed equal to A + Bx + Cx^, &c., because its development must contain negative exponents ; cannot be placed equal to 1 +x^ the assumed form, because its development must contain fractional ex- ponents These expressions, and others like them, can, however, be UNDETERMINED C OE FT IC I E N T S . 341 espauded into a series by changing their form. Place x~^ = z, then, 1 1 1— x-^ 1 — z • Let 1 — z = A + Bz + Cz^ + D2' + Ez' + &c. We will find -— — = 1 + z + z' + z" + z' + kc Now, replace z by its value, and we have ^ = 1 -f x~^ + x~'' + x-^ + X-' + &c. i 11 In like manner, place x'~ = z. Then, = ■ = A + B~ 1 + x-^ ^+' -\- C-s^ -f Ds^ + E2'* 4- &c. We will find by the method of undetermined 1 1 +^ 1 J|, 3 value, we have j- = 1 — x'~ + x — x'- + x'^ + &c. 1 + a-'^ FAILING CASES. 356. Any fraction, all of whose exponents are positive and entire iu the letter according to which the series is to be arranged, can be de- veloped, provided the denominator contains a constant. Again, any fraction, all of whose exponents arc positive and entire in the letter according to which the series is to be aiTanged, can be developed ; pro- vided, that there is one terra, at least, in the denominator, the expo- nent of whose arranged letter is equal to or less than the exponent of the same letter in, at least, one term of the numerator. A fraction that fulfils either of the foregoing conditions can always be developed, and the development will be the same as the quotient. The reason of this is plain ', the method of undetermined coefiicients gives a develop- ment for fractions, which is the same that would be obtained by actual division, beginning with the lowest power of the arranged letter. Now, a fraction that fulfils either of the foregoing conditions, will, when ex- panded by division, necessarily give positive and entire exponents in the arranged letter, when the division has been begun, with the nume- rator and denominator arranged according to the lowest power of this letter. Division would give us two quotients for zr, according as we X -\- L made x, or 1, the first term of the divisor; but the method of undeter- 29* 342 UNDETERMINED COEFFICIENTS. mined coefficients gives but one development, whicli is the same as that obtained by arranging the numerator and denominator with refe- rence to the lowest power of x. Therefore, the formula, which is ap- plicable only to positive and entire exponents, must fail, when the least exponent of x in the numerator is less than the least exponent of x in .x^ 4- c^ the denominator. Thus, '—^ cannot be developed, because o^, which involves the lowest power of x in the numerator, gives a quotient affected with a negative exponent when divided by x, the correspond- ing term of the denominator. GENERAL EXAMPLES. 1. Develop \ -{■ x. An$. 1 -f x. 2. Develop = ^ into a series. Am. 1 — X -{■ X? — x^ -f cc® — x} -{■ x^ — &c. ] ,5^ 3. Develop ^j -r- into a series. Am. \—x — ^x^ — 16,x^ — 64cc'' — 256a;5 — 1024a;« — &c. 1 ^2 4. Develop into a series. . ., ^ 1 -f- a; Ans. 1 — X. x-'- 5. Develop = ^ into a series. ^1 — X ' Ans. x-^ + X-' + ^~' + ^"' + ^~'° + &c. X 6. Develop '■ — - into a series. l—xT' 1 ~ 4 5 Am. x'^ -\-x^ -f a; + o;^ -f x^ + a-^ -f &c. 1 + a; 7. Develop :j ^ into a series. Am. 1 -f 3x + 6a;2 + Vlx^ -f 24.x^ + 48x5 -f &c. Am. 1 — X + ic" — x^ -f X* — x^ + x'-' — x'^ + &c. 8. Develop ^ _^ ^ ^ ^2 ^ ^ i'^to a series UNDETERMINED COEFFICIENTS. 343 9. Develop z, s ^ -, into a series. Ans. 1 — X -i- x' — x^-\-x'° — x''+x'' — x'^ + x'° — x"' -^ &e. X^ -rS 10. Develop -. Am. a + a -\- x^ 11. Develop — — r- into a series. X + 1 Ans. x^ — x^ ■\- X* — x' + x^ — x' + &c. x^ -\- x^ . 12. Develop '- ^ into a series. x" 4.r.' 4.7:^ 4x' , 2o 12o b2o 13. Develop ' — into a series arranged according to the powers X y of y. Ans. a;"-' + x'-'^y + x"'"'/ + x'-^y^ + . . . . y». 14. Develop -^— — . . x^ -\- X Ans. X. _ ^ , .r^ + 3x2 + 3x + 1 . ^ 15. Develop = into a series. Ans. x^ + 2x + 1. X'' 16. Develop -5 into a series x^ + X Ans. X* — x' + x'' — x^ + X® — x' + &c. xV — into a series. Ans. X + x' — x* + X' — X® + x^ — x^ + &c. x^ + x^ + X' . — 5 into a series. x^ + X Ans. 1 + cc'' — x^ + 2x'' — 2x^ + 2x« — 2x' + 2x^ — 2x' + &c. velop — ; — :— - — -. into a series. X + x^ + x^ + x" Ans. l — x" + x^ — x' + 2x« — 3x" + 2x + x^ _ 5x'". , x^* + 3x2 -\ velop ~ x^ -+ The process fails. Why X* -{. x^ 4- x^ 17. Develop ' ^ ^ — into a series X 4- x^ 4- x'' + X' 18. Develop j— into a series OA -n 1 ■'?^' + 3x2 + 3x + 1 . 20. Develop into a series. x'= + X Ans. . .>44 UNDETERMINED COEFFICIENTS. 1 _|_ 7.x -f 01? . 21. Develop ■ — ^ — into a series. ^ 1 -{- X Aus. 1 + 6a: — 5x^ + 5x^ — 5x* + 5.r-' _ bx^ + &c. r^ a^ 22. Develop '- -. X — a A71S. x -\- a. a'^ a^ . 23. Develop ^ into a series. SC-* -f x~^ + 2 . 24. Develop ^ 1^ into a series. Ans. 2 — x-' + 2x~' — 2x-^ + 2x-' — 2a;-"' + &c. Make x~^ = z. Develop, and replace z by x~'^. Remarhs. 357. It will be observed that, in most of the preceding examples, any term of the series after a certain number from the left, can be formed from the term immediately preceding, or from two or more preceding terms, according to some fixed law. Thus, in Example 2, every alter- nate term of the series is formed from that which precedes it, by multi- plying by — .T. The series in 5 and 6 are formed in the same way, the multipliers, or all the terms, being x~'^ and x\. In 7, every term after the second is formed from that which precedes it, by multiplying by 2x. In 8, we may regard the constant multiplier as — cc, and omit every alternate set of two terms. In 9, the multiplier may be regarded as — x, and, when three terms occur togethei*, these are omitted. In 11 and 16, the multiplier is — x. In 12, the multiplier, for all terms after the second, is — \x. In 13, the multiplier is x~'^y. In 17, the con- stant multiplier, after the second term, is — x. In 18, the multiplier, after the fourth term, is — x. In 19, there is no law of formation. When the terms of a development are formed from those which pre- cede according to some fixed law, the development is called a recurring series. Multiplication is not the only mode of forming the succeeding terms; sometimes they are formed by addition or subtraction, and even by a combination of two of these methods. Thus, ; ^ —^ expanded, becomes 1 + a; + 2^^ 4- 4x^ + Ix^ -\- 13a;^ -|- 24x« + &c. UNDETERMINED COEFFICIENTS. o45 The literal parts are formed by multiplying by x, the first two terms have the same coefficient, and each succeeding coefficient is equal to the sum of 1 4- 2a; the three which precede it, &c. Again, ^ ^^— — ^ = 1 + 3a; + 4x* + Ix" + lla=^ + 18a;5 + 29:r« + 47x^ + &c., in which the coeffi- cients after the second are equal to the sum of the two preceding coefficients. PARTICULAR CASES. 358. A fraction involving irrational monomials may be treated as in the following example. Required, the development of -^ ^. Let x = z^. Then, ai = 2^, x'l = z"^, z ^ xl. Then, the fraction becomes ~, 3, and, by the method of undetermined coefficients, .j . = A + Bz + Csr^ 4- Ds' 1 — 2r + Ez^ + Fs^ + &c. From which we get A = 0, B =r 0, C = 0, D = 1, E = — 1, F == 0, G = 1, H = — 1, &c. Hence, ^^=4 z=^ — z'^ -\- z^ — z' +-^ — 2'° 4- &c, and, by replacing z by its value, we have \ ^xl — | -\- x — x]. + a-| — a-J + &c. i — x^ Since all examples of the same kind may be treated in the same way, we derive the RULE. Place the variable equal to a new variable, affected with an expo- nent equal to the least common multiple of the denominators of the exponents of the old variables. Develop the neto fraction, and replace the neio variable by its vahie in terms of the old. 1, Expand -y ^ into a series. Ans. x\ — xf 4- ^g — ^1 4- &c. 2. Expand ' * ^^ . Ans. x\ — a;l + x | — x% + &c. 1 4- a; b b 4 346 UNDETERMINED C E F FI CI E N T S . 859. Fractions involving parentheses, differing only in their expo- nents, may be treated as the following. Required, the development of Let 1 + X = z^. Then, ^^ , ^, ^-, , ^ = -, which, when (l+aOi — (1 +:r)- X _ ^' — 1 developed, gives z + z^ + z'^ + z* -\- &c. Now, replace z by its value, andwehave ^^_^ ^^,^^^_^^^ = (1 + ^)J + (1 + .)> -f (1 + cc)i + (1 + ^y + &c. RULE. Place the common parenthesis equal to a Jietv variable, raised to a power deviated hy the least common multiple of the denominators of the p>arentlieses, and proceed as hefore. EXAMPLES. 1. Expand -^ ^- — ^-—^ ^J^to a series. {l — x)}^ — {l — xy\ Ans. (1 — x)i + (1 — x)\ + (1 — xf + (1 — X) ] +&C. 2- Expand ^_^^_^, into a series. Ans. (1 — a;)f — (1 — x)+(l — x)--4 — (1— a-)-|+(l— a;)-2 — &c. m Expressions of the form, ( — j— 7-) , iiiay be expanded by placing a — bx = 2°, deducing the value of a -f bx, and proceeding as before. Take, (l^^\^. Place, 1 — x = z'. Then, x = 1 — z^, and '\1 + x/ ' , XT /l—X\i Z^ Z^ l + x=2-z\ Hence, (^-^) = 43347^^76 = ^=4^+4- Develop this fraction and replace z by its value (1 — cc) J. 360. It is obvious that there are an infinite number of irrational expressions that may be developed by first making them rational in terms of a new variable, expanding the new expression, and replacing the new variable by its value iu terms of the old. So, there are an infinite number of expressions, involving negative exponents, that may DERIVATION. 347 be expanded by substituting, for the old variable, a new variable affected with a positive exponent, and proceeding as before. Thus, let it be required to expand -^j ^. Let x~'^ = z. Then, x~* = z^. Hence, jj-^ + x-' — x-' + x-"' — x-''+&c. There are many expressions, however, involving negative and frac- tional exponents, that cannot be developed by the method of undeter- mined coefficients. DERIVATION. 361. When one quantity depends upon another for its value, it is said to be a function of the quantity upon which it depends. The dependent quantity is the function, and that upon which it depends is called the variable. Thus, in the equation, y = 2x, y is the function and X the variable, because, by changing the value of x, y may be made to have any system of values. Thus, when a; = 0, *, 1, 2, 3, &c j y=0, 1,2,4, 0, &c., it is evident that, in every single equation involving two unknown quantities, there is a function and variable. Either of the unknown quantities may be assumed at pleasure as the function, because both unknown quantities are mutually dependent upon each other. But it is usual to regard that unknown quantity as the function, with refer- ence to which the equation has been solved. If the equation, y = 2a", is solved with respect to a;, we have a; = -^. Then, x is the function a and y the variable, and, by assuming arbitrary values for y, x may be made to have an infinite system of values. The most general form of a solved equation of the first degree with two unknown quantities is, y = ax -f b. In this, a and b are called constants, because their values undergo no change. It is plain that y depends for its value upon x, a, and b, but, as its value only changes with x (since x is the only variable), 7j is called a function of x only. If we have a single equation involving three unknown quantities, any one of these unknown quantities is a function of the other two. If we have y = ox, ov y = ax + b, and x — viz, or x = viz -f v. 348 DERIVATION. we call y an explicit fuuetion of x, and an implicit function of z, because y directly depends upon x for its value, and indirectly upon z. It is plain that we may have explicit functions of any number of variables, and, also, implicit functions of any number of variables. We will, however, confine ourselves to explicit functions of single variables. If we take the equation, y = ax + h, and attribute to x arbitrary values, 0, 1, 2, 3, 4, &c., and call the corresponding values of y, y, y', y", y"', &c., we will have y = h, y' = a + h, y" =2a -\- b, y" = 3a + h. Then, y' — y = a, y" — ?/' = «, y'" — y" = a, &c. That is, the difierence between any two consecutive states of the function is constant. This constant increase of the state of the function is called the differential or derivative of the function. It is evident that, when the function and variable are both linear, that is, both of the first degree, the difference between two consecutive states of the function will always be constant, and then will be truly the derivative of the function. Thus, take the equation, y =: 2x -\- 2, and let x have the constant increment J, beginning at ; that is, let a; = 0, + i, J + ^, 1 + i^j I2 + J; then the difference between two consecutive values of y will always be constant. For, we have y = 2, y' =zl + 2 = 3, y" = 2 + 2 = 4, y'" = 3 + 2 = 5, &e., and, consequently, y' — y = l=y"—y'= y'" — y", &c. But, when the variable is of the second degree, and the function of the first, the difference between two consecutive states of the function, corresponding to constant increments of the variable, will not be con- stant. Thus, take the equation, y = x^, and let x have the constant increment, 1, beginning at zero. Then, y = 0, y' =zl, y" = 4, y'" = 9, &c., and y' — ^=1? y" — y' =^^j 1/'" — y = 5. The student of geometry will understand that there ought to be no constant increment to the function, corresponding to a constant increment to the variable in the equation, y =^ 0?. For, y expresses the surface of a square of which X is the side, and the surface of course increases more rapidly than the side. If the function is of the first degree, and the variable of the third, it will be seen that the difference between two consecutive states of the fuuetion varies still more widely from a constant. The function and variable must then both be of the first degree, in order that the difference between two consecutive states of the function may be con- stant. But, to return to the illustration of the square, it is plain that, if the constant increment to the side of the square was indefinitely small, the difference between two consecutive states of the function, which represents the increment of the fuuetion, would be so nearly con- DERIVATION. 349 stant that it might be regarded as constant. Suppose y = a:^ and that x=^\ foot. Then the surface of the square is one square foot. Now, suppose the side of the square to receive the constant increment, 1 2 1 j-^ part of a foot. Then, y == 1, y = 1 + ^-^-^ + ^j^^^, 4 4 2 1 y" = ^^ 10000 + (TOOOO/' *^''- ^^^°' •^' "^ "^ 10000 + (10000?' 2 2 ?/" — 7/ = 1 -„. And we see that the diflPerence between ■^ ^ 10000 ^ (10000)'' the second and first states differs only from the difference between the third and second, by jq^oo- Upon this principle the derivative or increment of a function is found. It is the difference between two consecutive states of the function, when these states are indefinitely near to each other, and it expresses the increment of the function. The word increment is used in its algebraic sense. When any state of the function is less than the preceding, the derivative is truly a decrement. To find the derivative of the function, y = a-^, let h represent the infinitely small constant increment to x. Then, y = x^, y'= (x + h)^ = x"^ + 2x7* + h^, and y' — y =^ 2xh -f h^. Now, since h is infinitely small by hypothesis, h^ will be an infinitely small quantity of the second order, and may therefore be neglected. Thus, -z — rrp— is very small, ' •' ° '1 million •' but -rz, 777^ — r% is much smaller. Now, by making h indefinitely small, (1 million/ 7 .; o J ? we have taken the states indefinitely near to each other, and, conse- quently, y' — y is the derivative of y = x?. Hence, the derivative of 0^ is 2x7t. It is usual to represent h the increment of the function by dx, read differential of x. The derivative of the square of any variable function is then twice the first power of the variable into the differential of the variable. The differential coefficient of a function is the differential of the function divided by the increment of the variable, and is generally expressed by '^— — '- . In the preceding example the differential coefficient of x^ is 2.r ; and, in general, knowing the differential of the function, we get the dift'erential coefficient by dividing by the increment or differential of the variable ; and, conversely, we get the differential of the function from the differential coefficient, by mul- tiplying by the differential of the variable. It will be seen that, when 30 350 DERIVATION. wc make h = 0, or indefinitely small, that the differential coefficient assumes the form of ^, for then i/' = y. To find the differential coefficient, we have the following RULE. Give to the variahle a varialle increment, h, and find the neiv state of the function. Take the difference between the new and old states of the function, and reject the terms involving the higher powers of the in- crement, as heing infinitely small quantities of the second, third, &c., orders. Next change h into dx, dy, or dz, &c., according as the va- riahle is X, J, or z, &c. We then have the differential of the function. Divide the differential of the function hy the differential of the va- riahle, and we have the differential coefficient. Required the differential coefficient of the function, y = x^. By the rule, we have y' = (x + hy = x^ + 3x-/i + oh^x + 7/,l Hence, //' — y = Sx^h + Sh^x -}- A" = SxVi, or Sx^dx, when h is infinitely small. Therefore, ^~"~ '^ = 3.x-l h Newton regarded all algebraic expressions as the representatives of lines, surfaces, or solids; and supposed lines, whether straight or curved, to be generated by the flowing of points according to fixed laws, surfaces to be generated by the flowing of lines, and solids to be generated by the flowing of surfaces. Thus, a point, moving or flowing according to the law that it shall always be in the same plane, and at the same dis- tance from a point, will generate the circumference of a circle; and a straight line, flowing with the two conditions of being constantly parallel to, and at the same distance from a fixed line, will generate the surface of a cylinder. A straight line flowing in the same direction with the condition that, in all its positions it shall continue parallel to itself in its first position, will generate a plane. One line revolving around another, to which it is perpendicular, _L, will generate the surface of a circle. A flowing square will generate a cube; a flowing semicircle, a sphere. Newton regarded the increment of the function as expressing the uniform rate of increase of the function, and called it fiuxion. The thing generated he c^W^di fluent. It is evident that the fluxion does not express the uniform rate of increase, except when the states are taken indefinitely near to each other. Suppose a cannon-ball to leave the mouth of a piece with a velocity of 2000 feet per second, and that this velocity is reduced to 1200 feet per second at the end of the DERIVATION. 351 third second. It is plain, that the diiFerence between the spaces passed over in any two consecutive instants of time will not be equal to the difference between the distances passed over in any other two consecu- tive instants, unless those instants are inappreciably small. But, for the millionth part of a second, the velocity might be regarded as con- stant. If the instant was then taken thus small, the difference between the spaces in two consecutive instants would be constant. The differential of the function has, in accordance with the New- tonian theory, been defined to be the uniform rate of increase or decrease of the function. The differential of a constant then, must be zero, since a constant admits of neither increase or decrease. Theorem I. 362. The differential of the algebraic sum of any number of func- tions of the same variable is equal to the algebraic sum of their diffe- rentials. Let ?/=-±iudczvzt:w±z; «, v, lo, and z, being functions of the same variable, x. Give to the variable a variable increment, h, and call the new states of the function «', v', xo', and z. Then y = «' =h v' zh w' ± z', and y' — 1/ = u' — m =fc (v' — v) ± (ic' — ic) ± (/-^ z) as enunciated, since u' — « is the differential of m, v' — v, the diffe- rential of V. A shorter notation is dy = du rt dv rfc: dw ± dz, and is read differential of y, equal to the differential of ?<, plus or minus the differential of r, &c. Let y = ax^ + ex. Then, dy = 2axdx -f- cdx. Theorem II. 363. The differential of the product of a constant by a variable, is equal to the product of the constant by the differential of the variable. Let y = Ax. Then, dy = Adx. For, y' = A (x -{- h), and y' — y = Ah, or dy = Adx. Let u = my. Then, du = mdy. Theorem III. 364. The differential coeflacient of the power of a quantity is equal to the exponent of the power into the quantity affected with an expo- nent less by unity than the primitive exponent. 352 DERIVATION. Lety = a;". Then, ^^^- = «x"-'. For, give to x a variable increment, h ; then, y' = (x ■{■ hy, and y — ?/ (a? + A) " — a;" (x + 7i) ° — x" «" — x" h ~ h ~ (x + A) — X ~ a — a; ' "^ -P ' iuij X + 7t by a. But, ■ = «°~' + «°~^j:; + (/""'x^ + a^'^x^ + a — X &c., on to n terms. Now, when h is made indefinitely small, x is equal to a, the first member of the equation in y then becomes equal to the diflferential coefficient, and the second member becomes a"~' -f a"~' + «"~' + &c. = ««""', as enunciated. Hence, we have --- =■ ??«""', and, consequently, dy = na^~hlx. The differential, as well as the differential coefficient of a powei-, is then known. Theorem IV. 365. The differential of the product of two functions of the same variable is equal to the sum of the products which arise from multiply- ing each function by the differential of the other function. Let y =1 uv. Then dy = udv •-(- vdu, in which u and v arc both functions of the same variable, x; the new state of the function is n' = u + du, and the new state of the function v' = v + dv. Hence, y' = u'v' = (u -\- du) (v + dv) ^uv -{• udv + vdu + diidv. But, since du and dv are indefinitely small, their product, dudv, is an in- definitely small quantity of the second order, and may therefore be neglected. Hence, y' = uv + udv + vdu, and y' — y, or dy = udv + vdu, as enunciated. Corollary. 366. 1st. The differential of the product of any number of functions of the same variable is equal to the sum of the products which arise from multiplying the differential of each function by the product of the other functions. Let y = swuz, in which s, w, %i, and z are functions of the same variable, z. Then, dy = icuzds + suzdw -f swzdu -f sicudz. To show this, we will first get the differential of the product of three variables. Let y = szv. Make sz = u. Then y = uv, and, from what has just been shown, dy = udv + vdu = szdv -f vd (sz) = szdv -f- V (sdz -\- zds) = szdv -f vsdz -\- vzds, as enunciated. DERIVATION. 853 Now, knowing the differential of the product of three variable func- tions, we can readily pass to that of four. For, let y = swuz. Then, Ji/ z= svmJz + zd (^Rwu) = sicuch -f z (sic(7v -f sudw + icuds) = swudz + zswdu -\- zsudw + zicuds. And, it is plain that the same process can be extended to the product of any number of functions. 2d. Since the differential of the product of Av is Adv -f vdA = Adv, A being a constant, we have Theorem II. demonstrated in another way. Theorem V. 367. The differential of a fraction is the denominator into the diffe- rential of the numerator, minus the numerator into the differential of the denominator, divided by the square of the denominator. Let y = — , s and v bciuu; both functions of .'-. Then, di/ = V vds — sdv For, since y = — , we have no = s ; and yv being equal to s, the iu- V crement of yv must be equal to the increment of s. That is, d (yv) = ds ydr ds sdv vds — sdv ds, or ydv + vdy = as, or i/y = ^ — ' = ~ = , as enunciated. In the expression, "^ — , we substituted for y its value, — , V V and then reduced the two fractions to a common denominator. CoroUary. 368. 1st. When the denominator is constant, dv is zero, and dy =3 vds — sdo vds ds , . ,.„ • 1 p 1 = —-T- = — , equal to the dmerential ot the numerator v^ v^ V divided by the constant denominator. "We might arrive at this result s Is in another way, for, when v = a constant, — may be written — , and, 1 Is since — is constant, the differential of — will, by Theorem II., be V ' V Ids ds ' di 30 , or dv 354 DERIVATION. 369. 2(1. When the numerator is constant, ds = 0, and dy = vds — sdr - , ado . , , becomes c/y = — —2}^ negative result. This ought to be so, for, when the numerator is constant and the denominator variable, any increment to the variable, x, upon which v depends, will decrease 1/ ; and, consequently, cTy, which expresses the algebraic increment of 7/, is truly a decrement, and ought, therefore, to have the negative sign. In the case supposed, y is a decreasing function of the variable, x, and we see that the differential of a decreasing function is negative. We are now prepared to find the differential of any function affected with any exponent, whether positive or negative, fractional or entire. Theorem VI. 370. The differential of a quantity affected with any exponent, is the product arising from multiplying the exponent of the power into the quantity affected with an exponent algebraically less by unity than the primitive exponent, into the differential of the variable. We will first suppose the quantity to be affected with a negative exponent. Let y = r~° = —^. Then, by Corollary 2d, of the last Theorem, dy = ^^^ = -^ = — nv~''~\ as enunciated. Next, suppose the exponent to be a positive fraction, ■>/ z=v\ Then, 7/ r= \/v'', or 1/' = v'. And, taking the increments of both members, we have, by Theorem III., since r and s are positive and entire, , , -. rv^~^dv r if~^dv r v^~'dv .«?/-' dy = rv'-h/i', or di/ = j— = — X — —7- = — X — r r^~^dv r v~^dv r — r -i_i V ' v~ enunciated. Finally, suppose y= v~ ^ ■ we will find by proceeding exactly in the :?ame way as when the exponent is a positive fraction, dy ^= - x s V » dv, as enunciated. DERIVATION. 355' Corollary. 370. 1st. The differential of any parenthetical expression is equal to the exponent of the parenthesis into the parenthesis, raised to a power algebraically less by unity than at first, into the differential of the quantity within the parenthesis. This is evident, since we may represent the quantity within the parenthesis by a single variable, v. The parenthetical expression will then assume the form of v", and may be differential according to the rale. Let y = (a + hx^y, place a -\- hx^ = /•, then y = z", and dy = nv^-klv = n (a + hx^y'-\l (a + hx^) = n (a + bx^)"^^ 2l)xdx = 2bn (a + hx^y-^xdx. 372. 2d The differential of a radical expression is equal to the diffe- rential of the quantity under the radical, divided by the index of the radical into the radical raised to a power algebraically less by unity than the primitive index. This is merely a particular case of the preceding, for a radical is nothing more than a parenthetical expression affected with a fractional exponent, the numerator of the fraction being unity. Let -yU be the radical; this is equal to v~, and its differential then must be — r^"' dv = — v~^ dv = — j^ = — ■, as enunciated. n n nv— „ ;/ i^n-i' "When n = 2, the radical is of the second degree, and the expression becomes — =, that is, the differential of a radical of the second decree is equal to the differential of the quantity under the radical sign divided by twice the radical. EXAMPLES. 1. Required the derivative oi y = l/a -\- x^. . ^ 2xdx Ans. dy ■ Si/(a + xy 2. Required the derivative of y = v'" + bx'- . -, 2bxdx Ans. dy 2^ a + bx^ 3. Required the derivative of "^ ax + ix°. (a + 9iZ»x°~') dx Ans. - m'^(ax -f ftx")™-' 356 BINOMIAL FORMULA. GENERAL EXAMPLES. 1. Required the derivative of « + ix — cx^ + mx^. Alls, (h — 2cx + omx^) (h\ 2. Required the derivative of — . ^ cx^hilx — (a -f Jjx) 2cxdx Ans. ■ — 3. Required the derivative of cc" (a + Ixy. Ans. m(a + bxyx.'^~^ dx + nx"' (a -f ia;)""' hdx. BINOMIAL FORMULA. 373. By actual muhiplicatiou we can readily find (a + xf = a'' 4- 2ax + o:% (a + xf = a^ + Sa\v + Sax'' + x% (a + xy = a* + 4a''a; + QaV + Ao'x^ + x\ We observe a simple law in" regard to the exponents in these three developments. The exponent of a in the first term is the power of the binomial, and this exponent goes on decreasing by unity unto the last term, in which it is zero. For, x^, x^, and x* may be written, a^x^, a".*', and a°x*. The exponent of x is zero in the first term, and goes on increasing to the last term, in which it is equal to the power of the binomial. With regard to the coefficients, we observe that the coeffi- cients of the extreme terms are unity, that the coefficient of the second term is the degree of the binomial, and that the coefficient of the third term can be found by multiplying the coefficient of the second term by the exponent of a in that term, and dividing by the number of terras which precede the required term. Thus, in the development of (a + xy, to find this coefficient, multiply 4, the coefficient of the second term, by 3, the exponent of o in that term, and divide by 2, the number of terms which precede the required term, the quotient 6 is the co- efficient of the third term. So, to find the coefficient of the fourth term of (a + xy, multiply G, the coefficient of the third term, by 2, the ex- ponent of a in that term, and divide the product, 12, by 3, the number BINOMIAL FORMULA. 357 of terms whicli precede the required term, the quotient, 4, is the co- efficient of the fourth term. This remarkable law in regard to the co- efficients, holds true even for the second and last terms. We observe, furthermore, that the sum of the exponents of a and x in every term is equal to the exponents of the binomial and that the coefficients, at equal distances from the extremes, are equal. Newton showed that the law in regard to the exponents and coefficients was general for any binomial of the form (a + x)"", and that we would have (a + x)'" = a" + ma'"-^x -i ^^ ^-^ \- m ( m — 1) (m — 2) a'^^x^ 2T3 + ^ ■ DEMONSTRATION. 374. To demonstrate this useful and remarkable Theorem, let us assume (a -f x)"'=A -f A'x -f A'V + A''^ + A'"x"' ; in which m is taken positive and entire, and A', A", &c., independent of x. Then, by the principle of undetermined coefficients, we have a right to make a; = in both members of the assumed identical equation. Making a: ^ 0, we have A = a". It was shown that, when two functions of the same variable were ecjual, that their diffiDreutials were equal. Hence, taking the derivations of both members of the assumed equation, we have ?»(a + j)'"~hlx = (zV + 2A"a; -f 3A'V -f &c.)'/^, or, dividing by (Jx, m(a + j-)"-' = A' -f 2A"a; -f ^A"'x^ + &c. (N). Now, make x = 0, and we get, ma"""' = A'. Again, taking the derivatives of both members of (N), and dividing by dx, we get, m (m — 1) (a + a-)""^ = 2A" + 2.3. A"'x + &c. (P). Making x = 0,we have A"= "^ ("^ — 1) """' ^ Again, taking the derivation of (P), dividing by dx, and making a: ^ 0, there results A = — ^ o o. • -In hke manner, we can find A"" = ^»0»'— 1)C» ' — -) ('« — 3) ^"-^ 2.3.4 And A" will plainly be equal to m (m — 1) (m — 2) .... (m — (m — 1) ) a"-" _ 1.2.3.... {m — 2) (m — 1) m ' and, since the factors of the denominator are the same as those of the numerator, only written in reverse order, we have A" = 1. 358 BINOMIAL FORxMULA. Now, replace A, A', A", &c., by tlielr values iu the assumed identical equation, and we have (a + x)- = d'" + ma'"-^ x + m {m — 1) ff™""^ X- m (m — 1) (m — 2) n"'~^ x^ 1.2 ' 1.2.8 m (to — 1 ) («i — 2) . . . {m — n + 2) 0"--"+' x<^-' ■ 1 . 2 . 3 . 4 . . . (n — 1) "^ ^ • The term, m (m — 1) (:m — 2) . . . (»t — n + 2) an>-n+' x"-' 1 . 2 . 3 . . , (ri _ 1) ' is called the n*"", or general term. An inspection of the formula will show that the first factor of the numerator of the coefficient of a iu any term is m, the next factor (m — 1), and so, on decreasing by unity, unto the last factor, which is m, diminished by two less than the place of the term. Thus, the last factor of the numerator of the coefiicient of the 4tli term is (m — 2). So, the last factor of the numerator of the coefficient of the n^^ term must be (m — n + 2). The denominator of the coefficient of a, in any term, is always the continued product of the natural numbers, from 1 up to one less than the place of the term. Hence, the denominator of this coefficient in the n^^ term must be 1 . 2 . 3 . . . . (n — 1). In regard to the exponents, we see that a, in any term, is always affected with the exponent of the binomial, diminished by the number of the term less one, and the exponent of x, in any term, is always one less than the place of the term. Hence, for the ?i*'' term, we have «"'-''+' a;"~'. The binomial formula is usually written iu the ascending powers of «, instead of x. To develop (x + «)" in the powers of a, it will be necessary to regard a as the variable, and x as the constant, to make a = in the successive steps of the operation. As we would obviously obtain the same result, with the exception of an interchange of x and a, we may at once write , VI (m — 1) .t'"~^ a^ (^x 4- <^'T = •^'" + '^^^ « H ■ — I — 2 — ~~~ "^ TO (to — 1) (1^1 — 2) X™-' a' TO (?» — 1) . . . (m — n + 2) cc°'-°+' a"- ' 1.2.3 "^■■■" 1 .2.3. . . (n— 1) We will repeat the laws in regard to the exponents and coefficients. 1. The first letter of the binomial appears as the first term of the BINOMIAL FORMULA. 359 development, affected with an exponent equal to tliat of the binomial, and the exponents of this letter in the successive terms decrease by unity unto the last term, where its exponent is zero. The exponent of the second letter of the binomial is zero in the first teroLi of the develop- ment, and one greater in each of the successive terms, and, in the last terra is m, the exponent of the binomial. 2. The sum of the exponents of x and a, in every term of the development, is equal to the exponent of the binomial. A simple in- spection of the formula will show this. 3. The coefficient of the first term of the development is unity; that of the second term is equal to the exponent of the binomial; that of the third term is formed by multiplying the exponent of the first letter of the second term by the coefficient of that letter, and dividing by one less than the number denoting the place of this required term ; that of the Ji*'' term is formed from the (/i — l)"* term, by multiplying together the coefficient and exponent of the first letter of that term, and divi- ding this product by {n — 1), that is, by one less than the number of the required term. 4. There will always be «z + 1 terms in the development. For, we get one term, a™, or of, without differentiating, and, since the exponent of m is diminished by unity for each differentiation, it is plain that m derivatives can be taken of (a + a:)", and, since we get a term of the development by each derivation, we must have in all (m -f 1) terms. 5. The coefficients at equal distances from the extremes are equal. For the developments of (a -\- xj" and (x + a)" must be identical, differing only in the order of the terms; the first term of the develop- ment of (a -j- a:)" being the last term of that of (x -\- «)"", the second from the left of the development of (a + x)" being the second from the right of that of {r, -f a)" , &c. Hence, the terms of the develop- ment of (a -f a;)", taken in reverse order, will constitute the direct development of (x -f- «)"• I*^ is plain, then, that the second term from the left must have the same coefficient as the second term from the right, the one being formed from a" in the same manner as the other from a™, and so, in like manner, all the other coefficients at equal dis- tances from the extremes must be equal. 6. The sum of the coefficients of the development of {x -f a)*" is equal to the m^^ power of 2. For, if we make a- = 1 and o := 1 in the equation, (x + a)™ = x"" -\- ^/i.r"""' a -\ ^^ z — -^ + 360 BINOMIAL FORMULA. m (m — ) (m — 2) x""'" a' ^ j v, -n j — ^^ i~o — ^^ .... a , the second nieinber will reduce to the sura of tlic coefficients, and the first member will become 2"", and we will have (1 + i)" = 2- = 1 + m + m (m - 1) + ^^'^On—l)(^m—2) + &c. We make x and a unity, in order to reduce the terms in the second member to their coefficients; and, it is plain, that if x and a had another value than unity, the second member would not be the sum of the coefficients. In accordance with the demonstration, the sum of the coefficients in the development (x + of, ought to be 4 ; for, in this case, m = 2, and (2)"" = (2)^ = 4. And, in the development of (x + a)^, m = 3, and (2)"" = (2)^ = 8, which agrees with the fact. So, likewise, in the development of (x -\- ay, m = 4, and (2)" = (2)'' = 16, which also agrees with the fact. FORMATION OF POWERS BY THE RULE. 375. Let it be required to find the development of (x -\- of by the rule. The first term is x^, the second term has 5 for a coefficient, and the exponent of x is one less in this term than in the preceding ; a also enters to the zero power in the first term ; and, since the expo- nents of a go on increasing by unity, a must enter to the first power in the second term. Hence, the second term is 5x'*a. The coefficient of x in the third term is formed by multiplying 5, the coefficient of x in the second term, by 4, its exponent, and dividing the product by 2, 5x4 the number of terms preceding the required term. Hence, —^ — =10, is the coefficient of the third term, and, from the law of the exponents, X must enter to the third power, and a to the second power in the third term. Hence, that term is lOx^a^ For the coefficient of x in 10 3 the fourth term, we have — ^ — = 10, and the fourth term must be 10 . 2 lOx'^a". For the coefficient of the fifth term we have — - — — 5, and that term itself must be 5xa''. For the coefficient of the sixth term 5 (1) we have — = — = 1, and that term itself must be x^a\ or a'. For the 5 coefficient of the seventh term we have — ^ = 0. 'Hence, there is no BINOMIAL FORMULA. 3G1 seventl> term. Then, the development of {x + a)* is x^ + ox*a + lOx'a^ + lOx^a^ + 5xa* + o', and we see that the sum of the coeffi- cients is equal to (2)^ = 32, and that the sum of the exponents of x and a in each term is 5. The preceding development might have been obtained directly from , ^ , , . , m(m — l).c'"~V ,, the general formula, (^x+ a) = a;" -f mx^ 'a -\ — h &c., by making m = 5. But the above process is generally shorter when m is positive and entire. DEVELOPiMENT OF [x — a)'". 376. The development of (x — a)" may be obtained in the same way as that of (x + a)". But it may be gotten at once from the gene- ral formula by changing + a into — a ; the terms involving the odd powers of a will all be negative, and we will have (j: — a)° = af — mx'"-ki + ^ ^ . ^^ ^ ^ ^' g + ± a". The last term will be positive when m is an even number, and nega- tive when m is an odd number. It will be seen that all the laws in regard to the development of (x 4- a)" hold good in regard to the development of (x — a)", except that the alternate terms must be affected with the negative sign, and that the sum of the coefficients is zero. A few examples will illustrate the use of the two formulas for the development of (.x -f «)■" and (x — a)". EXAMPLES. 1. Required the sixth power of (.x + a). Ans. x^ + Gx'a -f Idx'a'' + 20.x='a=' + 15x^a' -f Gxa' + a^. For, m = G,m — l = 5, (vi — 2) = 4, «i — 8 = 3, m — 4t = 2, „ , ^ , m (in — l)^"-^ wi — = 1. Hence, {x + a)"" = x'" + mx'"-^ -\ ^ -y-- -f- .... a™ becomes, in this case, {x + ay=x^-\-Qx^ -i '■ — a* = x^ + Qx' -f 15xW -f a«. 2. Required the development of (x + ay. Ans. cc" + 7x^a + 2lx'a^+S6xW+S5xhi* + 21x^a^+ Ixa^ -f a?. 31 362 BINOMIAL FORMULA. o. Develop (.c + ay. Ans. .x^ + 8x-^a + 28.rV+5GxV+70.^V+56*V-f28.i-aH8xa'+a'. 4. Develop (x — af. Ans. x^ — Q,x^a + 15.xV — •ZQxht' + 15a;V — Qxa^ + a\ 5. Develop (,r — o)'. Ans. x' — nx^a + 21.'r5a2 — 35a;V+35:r'a^ — 21a;2rt + ^Ixa^ — a\ 6. Required the development of {x — a)". Ans. x^ — ^x^a + 36xV — 84xV + 126xV — 126a'V + 84xV — 36a;V + 9xa« + a». 377. The binomial formula has been deduced upon the hypothesis that the coefficients and exponents of x and a were unity, but it can be applied to binomials in which these conditions are not fulfilled, by substituting for the letters within the parenthesis other letters whose coefficients and exponents are unity. Thus, to get the fifth power of z^ + Zy", let z^ = X, and 3y/^ = a ; then, {z^- + ^iff = (« + aj = x' -\-bx'^a + lOcc'a^ + lOx^a? + bxa'^ + a^, which, by substituting for x and a their values, will be found equal z^° + Ibz^f -\- OO^y + 270^y + ^{)bzhf -f 243^'°. In like manner, we may find the development of (nx^ -\- ra^)'" = m (m — 1) r2«"'-=£cP'"--''a2^ jr.rP" + 7?ir?i'"-'.r'""-P«i + — ^i ^-—^ 1- ■ • ■ r'"a'^"\ (A) In this general formula, make x = 1, and a = 1 • the second mem- ber will reduce to the sum of the coefficients of x, and the first member will become (n -f i-)"". Hence, the sum of the coefficients of any bino- mial develojnnent is equal to the m** power of the sum of the coefficients within the 2}cirenthesis. In accordance with this general formula, the sum of the coefficients of the development of (2^ + oy^y ought to be equal to (4)^ = 1024; and this agrees with the development above, for 1 + 15 + 90 + 270 + 405 + 243 = 1024. 378. Formula (A) shows, moreover, that if the exponents of x and a are equal, that is, p — q, the sum of the exponents of x and a in each term of the development will p?n. Whenever, then, the exponents ofs. and a arc equal in any binomial, the sum of the exp)onents of these letters in each term of the development will be equal to the common ex- ponent xoithin the 'parenthesis into the exponent of the poxoer to xohich the binomial is to be raised. BINOMIAL FORMULA. 863 The two important laws just deduced from formula (A) are general for all binomials. The second and sixth laws, observed to govern the development of (x + a)", are but particular cases of the foregoing. The binomial formula can be extended to polynomials, by representing the polynomial by the algebraic sum of two letters. Thus, to obtain the square of cc' + 4:X^ — 8x — 8, represent x'^ -f 4:X? by x, and — Sx — 8, by a. Then, (x» + Ax"" — Sx — Sf = (x + of = x" + 2,ax 4- a^ = (by replacing x and a by their values), x® -f 8x* — 80x' + l'28.c + 64. In like manner, to obtain the cube of x^ -\- 2x — 4, let x" = X, and 2x — 4 = a; then, (x' + 2x — 4)' = (x + af= x" 4- 3x^a + 3a^x + a', which, by replacing x and a by their values, becomes x« + 6x^ — 40x' + 96x — 64. GENERAL EXAMPLES. 1. Find the coefficient of the twelfth term of the development of (.X + a'y, by means of the formula for the n^^ term. m(m. — \)(m — 1)(m — Z) (m — i) (m — 5) (to — 6) (m — 7) (m — 8) (»i — 9) (m — 10 ^^' 1X2X3X4X6X6X7X8X9X10X11 2, Find the twelfth term of the development of (x + a)^. 50 X 49 X 48 X 47 X 46 X 45 x 44 x 43 x 42 x 41 x 40a3V' Ans. 1x2x3x4x5x6x7x8x9x10x11 3. Find the development of (x + a)^°. Ans x^o + 20x'»a + lOOx'^a^ + 1140x'V + 4845x'V + 15504x%^ +38760x'V+77520x'V + 125970x'V + 167960x"a«+ 184756xV + 167960xV' + 125970.rVt'2 + 77520x^a" + 38760x«a'* + 15504x^a'^ + 4845x''a'« + 1140x='a'^ + 190x^a'« + 20xa'« + a^". 4. Find the development of (x^ + 2ax — Aa?y. Ans. x^ + Qax^ — 40o'x^ + 96alr — 64a«. 5. Find the 4th power of Sa^c — 2hd. Ans. 81aV — 216aVifZ-f- 2lQa\%hP — maWiP + lU'd\ 6. Find the cube of x^ — 2x + 1 Ans. x« — 6x5 ^ i^.jA _ 20a;3 + 15x- — 6x + 1. 7. Required the coefficient of the sixth term of the development of (x + ay. Ans 7(7-1) (7-2) (7-3) (7-4) 7.x6.5x4x3 _ 1.2.3.4.5 ^1.2.3.4.5 364 BINOMIAL FORMULA. DEMONSTRATION OF THE BINOMIAL FORMULA FOR ANY EXPONENT. 379. 1. Let m be supposed negative and entire, then (x + a)"" = — , which, hy ae- {x + a)'" .x" + ?7ij"~'a -f m(m — l)x^~hi'^ + tually performing the division, will be found equal to x~"' — mx~"^^a — m( — m — l'),^-'""^ m( — m — 1) ( — m — 2')x-'^-^a'^ rr2 1.2.3 . . . ± Px~^'"«"', &c. Hence, we have the same law of formation as when the exponent is positive and entire. We might demonstrate this truth more rigorously by assuming (x + a)""" = A + A'a + A"a^ + A."'a^, &.C., and proceeding just as we did when m was positive and entire, regard- ing a as the variable. But it is not necessary to repeat the operation. We have a right to assume the exponents of a to be positive and entire in all the terms of the development; because, if we expand any frac- tion whose numerator is a constant, and the first term of whose deno- minator is a constant, all the exponents of the variable will be positive and entire in the development. 2. When m is positive and fractionah Let m = — , and assume q (a +x)^ = A + A'x + A"x' + M"x^ A-^r-'CP). Make ^ = 0, p_ and we have A = as.. Taking the derivatives of both members of (P), and dividing by dx, we have —(a + x')'i = A' -f 2A"a; -f 3A"V -V &c. (2). Making a; = 0, we get A' = ^a~'- From (2), there results U^ — l\ {a -f 3;)Z~'= 2 A" -f 2 . ^A"'x + &c. (R). q\q J Making x = 0, we get A" = — (— — l)-^^ ^^^^^ (^)' ^^^^^'^ ^^" suits ^{^ — l\{^ — 2\ (a + .t)^~' = 2 . 3 A'" -f &c. From which, 2 .3 BINOMIAL FORMULA. 305 Continuing the process and replacing the constants, A, A', &c., by 2 j: « 1_ 1 their values, equation (P) becomes (a -f .r) i = a i + —a i x -\- il(^_l)„^V+ii(ii-l)(^-2U-V ±PaT-»-':c-V q^q ' q\q / \q / rr2 1.2.3 DEVELOPMENT OF BINOMIALS AFFECTED WITH NEGATIVE AND FRACTIONAL EXPONENTS. 380. When m, the exponent of the parenthesis, (x -\- dy°, is posi- tive and entire, it is evident that successive differentiation will even- tually reduce that exponent to zero, and then the next differential will be zero. And, since the constants, A, A', &c., of the assumed develop- ment, have been determined by the successive differentiation of (x + ay, it is plain that the development will terminate. But, when the exponent of the parenthesis is negative or fractional, successive differentiation can never reduce it to zero, and, therefore, an infinite number of constants can be determined in the equation, (j: -\- a)~'", or, p. {x + a)^. . = A+ A'x + A"x' + A'"x' + &c. The series will then never terminate. EXAMPLES. 1. Develop :j = 0- — ^)~' ^°*-*' ^ series. Ans. 1 -^ X + x^ + x'^ + x* -\- x^ -{■ &c. Make m = — 1, 1 = a", and — x = a in the formula (x + a)~° = 1 . 2 2. Develop = into a series. ^ 1 + a- Ans. 1 — X + X- — x^ + x* — x^ -\-&c. 1 _j_ 7^ -L a;2 3. Develop — :, ^ = (1 -\- 7 x -\- x^ (1 -{- a;)-' into a series. I -\- X Ans. 1 + 6x — 5x^ + 5x^ — 5x' -f 5x' — &c. 4. Develop into a series. a -{- X A 1 X , X' X^ X* ^ , a a^ o? a* a* 31* 366 BINOMIAL FORMULA. 6. Develop .y/l + X into a scries. Ans. =t a; X- . x^ /-, , X X- X" p V 6. Develop ^x + 1 into a series. Ans. ±( /—. — ____^^ zzz — &c.y 7. Develop ^/2 = ^1 + 1 into a series. Ms. ±(l+i_i +Jg_-5^ + &c.). 8. Develop ^/5 = v/4 + 1 into a series. Ans. -t(2 + i — J^+rh — &c.). 9. Develop (^Ji* + ?i'')'' into a series. ^?is. ± (m + I m-V — gS^m-'ji^ + _|gm-"w'^— &c.). 10. Develop (a + x)* into a series. 11. Develop (m^ + rr") into a series. 12. Develop (1 + cc^)^ into a series. , 01? x^ 5x® ., ^'''- i + T-y+sT-'-^'^- 13. Develop -■ into a series. ^/7l2 + x" ^ns. ± -(1 - ^, + g-, - le^^e + 728^3 - &c. I 14. Develop 4/a + a; into a series. ,_ 1 X 2 cc'' 6 :e^ 21 a;" ■. 15. Develop 4/1 + ^= into a series. , , la; 2x^ 6a;' 21x* , . ^'''- ^+-5-25+125-625 +*'• BINOMIAL FORMULA. 367 16. Develop v/1 — ^ ^^^^ ^ series. 17. Develop 1/2 into a series. Ans. 1 + I- ^+ -,&^- ^-, + &c. CONSEQUENCES OF THE BINOMIAL FORMULA — SQUARE OF ANY POLYNOMIAL. 381. We have seen that (a-{-by=a'' + 2ab + b% that is, the square of a binomial, is equal to the sum of the squares of its two terms, plus the double product of those terms. To find the square of a tri- nomial ; a + b -\- c, let a + b = s. Then, (a + i + c)^ = (s + c)* = s2 + 2sc + c'^^a + by +2c(a + b) + (? = ct" + i^ + c^ + 2ca -(- 2cb + 2ab; that is, the square of a trinomial is equal to the sum of the squares of its three terms, plus the double product of these terms taken two and two. To find the square of a polynomial of four terms, a -{-b + c + (7, let a + t + c = s' Then, (ii ^ b -\- c ■\- df = (s' + dj = s'2 + 2.i'il+ rp = a' + b' + c"" + (P + 2r/a + 2db + 2dc + 2ca + 2cb -\- 2(ib ; that is, equal to the sum of the squares of all its terms, plus the double product of these terms taken two and two. Now, it is evi- dent that the same law will hold good for the square of a polynomial composed of five terms. For this square is composed of the square of the first four terms plus the square of the fifth term, plus the double product of the fifth term by the sum of the first four terms. And, since the square of the first four terms will give the sum of the squares of all these terms, plus the double product of these terms, taken two and two, it is evident that, when the double product of the fifth term by the sum of the other four, is added to the other double products, we will have the double product of all the terms, taken two and two ; and it is plain, moreover, that when the square of the fifth term is added to the sum of the squares of the other four terms, that we will have the sum of the squares of all the terms. The law is then true for five terms, and it is plain that the same reasoning can be extended to sis, seven, and any number of terms. Hence, we conclude, that the square of any polynomial is equal to the sum of the squares of all the terms, plus the double product of all the terms, taken two and tico. 368 BINOMIAL FORMULA. 1. Kequired the square oi a -\- b + c -\- d -i- e. Ans. a' + h"^ + c' + cl? + e^ -f 2ah + 2ac ^- 2ad + 2«e + 2hc + Ihd + 2he + 2cd + 2ce + 2dc. 2, Required the square oi a -{- h -\- c -{• d -{- e -\- f. Ans. a^ + Z.2 + c^ + d' -{-c' +f^ + 2ah + 2ac + 2ad + 2ae + 2rr/+ 2hc + 2M + 2&e + 2//+ 2cf7 + 2ce+ 2c/ + 2c?e+ 2d/+2cf. CUBE OF ANY POLYNOMIAL. 382. The cube of (a + &), is a? + Sa^i + 3a6^ + h\ To fiud the cube of a + i 4- c, let a + hz= s. Then, (a + 6 + cf =z (s + c)' = s^ + 3s2c + 3sc^ + c« = a^ + ?y^ + c' + ^a?h + Sa^c + 36^c + 3c^a + 3c^6 + Qahc ; that is, the cube of a trinomial is com- posed of the sum of the cubes of its three terms, plus the sum of the products arising from multiplying three times the square of each term into the first power of the other terms, plus six times the product of all the terms, taken three and three. By pursuing precisely the same course of reasoning that we have already employed, it can easily be shown that the law is general for any polynomial. EXAMPLES. 1. Required the cube of « + & + c + d. Ans. a^ + P + c" + d^ + 3a^6 + ^ah -f 2,ahl + U^a + Wc + Wd + 3c^a + 3c^6 + 3A? + M^a + M^h + M^c + Qahc + Qahd + f\nr/7 _L (\Ti/^i7 Qacd + Qbcd, BINOMIAL FORMULA. bOU EXTRACTION OF THE Xth ROOT OF WHOLE NUMBERS AND POLYNOMIALS. 383. The most important consequence of the binomial formula is, that we are enabled by it to extract high roots of whole numbers and polynomials. We will begin with the simplest case, the extraction of the «" root of whole numbers. EXTRACTION OF THE Xih ROOT OF WHOLE NUMBERS. Let a represent the tens, and b the units of the required root. Then, the given number will be (a + />)", and from the binomial formula wo have (a + iy = a- + na--^b+ ^ ^ ^ + ^ tV^S^ + '-^'''- That is, a number, whose n^^ root is composed of tens and units, is made up of the ??<■' power of the tens, plus n times the (it — 1)"" power of the tens into the first power of the units, plus, &c. It is evident that the Jt" power of the tens will give, at least, ?* + 1 figures, therefore, the »*■» root of the tens cannot be sought in the 7i right hand figures. These must, therefore, be cut off from the right. We next seek the greatest n^^ root contained in the left hand period, and, when we have found it, we will have a of the formula. Raise the root found to the Ji*' power, and subtract this power from the left hand period. The remainder will correspond to na^~^h + — -^— ^~^ + kc, of the formula. The appi-oximate divisor to find h (the unit of xi x^ • n I mi X T • • „ 1 " (" — V)n'^^h the root) is na""'. The true divisor is ;ia°~' + -^= — = — -^ + n (n _ 1) (n — 2)«"-^i= , , , , , • , ' ^' , 1~9 — 5 1 ^^ ' "'^^> ^^ '■' ^^ unknown, we can only use the approximate divisor. The n — 1 figures, on the right of the remainder, must be separated from the other figures, because n times the (n — 1)"" power of the tens will give, at least, n figures. Dividing the period on the left by the approximate divisor, we get the units of the root, or, gene- rally, a number greater than the units, because our divisor is too small. It is plain that the number of ternis rejected when using the approximate divisor, depends upon the number of units in n, and that the value of each term depends mainly upon a. When, therefore, n and a are both large, or when one only is large, the approxiinate divisor is very consider- ably too small, and the quotient, therefore, will be too great. Raise the Y 370 N BINOMIAL FORMULA. two figures of the root found to the «*'■ power, and compare the result with the given number; if it be greater than this number, the last figure of the root must be reduced by one, or more. Let us illustrate by an example. Required the fifth root of 33554432. 885 54432 I 32 243 I «a»-' = 5 X 81 = 405 92 "54432 We began by cutting off the five right hand figures. The next step was to find 3, the greatest fifth root contained in 335, the left hand period. Next, the approximate divisor, 405, was formed, and this was found to enter twice in the left hand period of the remainder, after subtracting the n^^ power of the tens from the given number. Upon trial, 32 is found to be the true root, for, (32)* = 33554432. In this case, the unit figure of the root not being large in comparison with the tens, the approximate did not differ so materially from the true as to give a quotient figure too great by unity, or some other number. But, suppose it were required to extract the fourth root of 230625. Then, 39*0625 1 25 16 jia"-' =4 X {2f = 32, 230 625 The approximate divisor gives 7 as a quotient ; but, upon trial, it has (o be diminished by 2; and 25, not 27, is the true root. It will be seen that the units of the root can only be ascertained by trial. If the units of the root be large in comparison with the tens, the quotient obtained by dividing the left hand period of the remainder by the approximate divisor, will often differ considerably from the true units of the root. The following example will illustrate this : KilMUTATIONS AND COMBINATIONS. APPROXIMATE Nth ROOT OF NUMBERS ENTIRELY DECIMAL. RULE. 387. Annex ciphers, if necessary, until there arc as many p>ei'iods of n figures each as there are places required in the rout. Extract the n"* root of the 11 ew number thus formed, and point off the required numher of decimal places. EXAMPLES. 1. Eequired ^.00:^45 to within .1. Ans. .3. 2. Required ^.0028629161 to within .01. Ans. .31. 3. Required V-0004084201 to within .01. Ans. .21. 4. Required ;/. 51676701935 to within .01. Ans. .91. 5. Required y. 96059690 to within .01. Ans. .99. 6. Required ^.1245732577 to within .01. Ans. .66. ^ Required ^ .464404086785 to within .01. Ans. 8. Required V/.0000000000285311070612 to within .01. Ans. .11. 9. Required 1^.00048828126 to within .1. Ans. .5. 10. Required '.y. 600000000000000000000285311670612 to with- in .001. Ans. .011. 11. Required y. 2706784158 to within .01. Ans. .77. PERMUTATIONS AND COMBINATIONS. 388. Permutations are the results obtained by writing n letters lu sets of 1, 2, 3, ... . or n letters, so that each set shall difter from all the other sets in the order in which the letters are taken. Thus, the permutations of the three letters, a, h, and c, taken singly, are a, h, c ; and in sets of two letters each, ah, ac, ha, he, ca, cb ; and in sets of three letters each, abc, ach, hac, hca, cha, cab. PERMUTATIONS AND COMBINATIONS. 875 It will be seen that the sets differ only in the manner in which the letters are written, the letters in all the sets being the same. 389. Let it he required to determine the number of permutations of m letters, taken n in a set. Suppose the letters to be a, h, c, d, &c., and let us first permute them singly, we will evidently have m permutations of the m letters taken one in a set. Now, let us reserve a, there will remain m — 1, letters b, c, d, &c., which, permuted singly, will give (m — 1) permutations, b, r, d, e, f, &c. : write a before each of these letters, and we will have in — 1 permutations, ah, ac, ad, &c., of two letters, with a as the first letter of each set. Next, let us resei-ve b out of the letters, a, b, c, d, &e., and then permute a, c, d, &c., singly. We will again have (m — 1) permutations of the letters, taken in sets of one letter each. Writing b before each of these sets, we will have (m — 1) permutations of n». letters, taken in sets of two, with b as the first letter. Reserving c in like manner, we can again form (m — 1) permutations of m letters, taken two in a set, with c as the first letter of each set. Reserving all the m letters in succession, we can evidently form m (jm — 1) permu- tations of m letters, taken 2 in a set. And, of these, a will be the first letter of (m — 1) sets, b the first letter of (nt — 1) sets, c the first letter of (in — 1) sets, and so on. It is plain, moreover, that a will not only be the first letter of (»i — 1) sets, but that it will also be the last letter of (m — 1) sets, and that it therefore has been made to occupy all possible positions. As the same remark may be made of b and all the other letters, it is obvious that m (ni — 1) truly expresses the number of permutations that can be made of m letters, taken two in a set. That is, the number of permutations of m letters, taken two in a set, is equal to the total number of letters into the total number of letters, less one. In like manner, we may find the number of permutations of m letters, taken three in a set. For, if we omit one of the letters, as a, there will be m — 1 letters left, and these, permuted in sets of two letters, will give(w — 1) (m — 2) permutations. For, we have just seen that the number of permutations, taken in sets of two, was expressed by the number of letters into the number of letters, less one. Writing the reserved letter, a, before each of these (in — 1) (m — 2) sets, we will form (m — 1) (in — 2) permutations of m letters, taken three in a set, with a as the first letter of each set. Reservino; in succession each of 370 PERMUTATIONS AND COMBINATIONS. the tn letters, wc can plainly form m {in — 1) (in — 2) permutations of m letters, taken three in a set. By the same course of reasoning it can be readily shown that the number of peirmutations of m letters, taken four in a set, will be ex- pressed by m (m — 1) {in — 2) (m — 3) ; and of m letters, taken five in a set, by m (m — 1) {m — 2) (w — 3) (m — 4). We observe, in all these expressions, that the first factor is the number of letters, and that the last factor is the number of letters, diminished by the number in a set less one. Moreover, each factor after the first is one less than the preceding factor. Hence, the number of permu- tations of m letters, taken n together, will be expressed by m (in — 1) {m — 2) (m — 3) (in — n + 1), or, denoting by A the num- ber of permutations of m letters, taken n in a set, wc have A = in (in — 1) (in — 2) (m — n + 1). To apply this formula, it is best to determine in the first place the last factor, we then know where to stop. Since the first factor is always the number of letters, and the law respecting the mean factors is known, the formula can then be readily applied. 1. Required the number of permutations of 10 letters, taken 7 and 7. Ans. 10 (10 — 1) (10 — 2) (10 — 3) (10 — 4) (10 — 5) (10 — 6), Orl0x9x8x7x6x5x4 = 604800. For, 11 = 7, and in = 10, then in — n -j- 1 = 10 — 7 + 1 = 10 — 6, the last factor is then 4, and the first 10, the intermediate factors can easily be formed. 2. Required the number of permutations of 5 letters, taken 4 and 4. Ans. 5 (5 — 1) (5 — 2) (5 — 3) = 5 X 4 X 3x2 = 120. 3. Required the number of permutations of 12 letters, taken 10 and 10, Ans. 12 (12 — 1) (12 — 2) (12 — 3) (12 — 4) (12—5) (12 — 6) (12 — 7) (12 — 8) (12 — 9). 4. Required the number of permutations of 12 letters, taken 11 and 11. Ans. 12x11x10x9x8x7x6x5x4x3x2. 5. Required the number of permutations of 12 letters, taken 12 in a set. Ans. 12 (12 — 1) (12 — 2) (12 — 3) (12 — 4) (12 — 5) (12 — 6) (•12 — 7X(12 — 8) (12 — 9) (12 — 10) (12 — 11) ; or, reversing the factors, 1x2x3x4 x5x6x7x8x9x 10 X 11 X 12. PERMUTATIONS AND COMBINATION; Remarks. 390. If, in the general formula, A = m(m — 1) (m — 2) . . . . (m — n -\- 1), we make m = n, and call B the corresponding value of A, we will have B = n (?i — 1) (n — 2) .... 4 x 3 x 2 X 1 ; or, reversing the order of the factors, B = lx2x3x4 {n — 2) (n — V)n. Hence, the number of permutations of n letters, taken all together, is equal to the product of all the natural numbers from 1 to n, inclusive. Example 5 is an illustration of this. It is plain that, by increasing n (jti remaining constant), we will in- crease the number of permutations, because we will have more factors in the product that expresses the number of permutations. And, when n is made equal to m, this product is the greatest possible. Thus, the result is greater in example 4 than in 3, and greater in 5 than in 4. When n = m + 1, the whole formula reduces to zero. This plainly ought to be so, since it is impossible to permute m letters in sets of m+ 1 letters. Zero, here, Ss in many other places, is the symbol of impossibility. • When n = 1, the last factor, m — n + 1, is equal to m. Hence, the first and last factors are the same, and we have A = ^- = m ; that is, the number of permutations of m letters, taken one in a set, is equal to m. COMBINATIONS. 391. Combinations are the results obtained by writing any number of letters, as m, in sets of 1 and 1, 2 and 2, 3 and 3, .... w and n, m and m, so that each set shall differ from all the other sets by at least one letter. In permutations, the sets differ in the order in which the letters are written ; in combinations, the sets differ in the letters them- selves. Thus, the letters, abc, taken all together, give but one combi- nation, abc; but will give six permutations, abc, acb, bar, bca, cab, cba. If the same letters are taken two and two, they will give but three combinations ab, ac, be, while each combination is susceptible of two permutations ; and there 32* 378 PERMUTATIONS AND COMBINATIONS, are, therefore, six permutations of the three letters, taken two aud two, thus, ab, .„ . } cc, ^ ^ .^^ . I he. 1 "^^} ) ah will give f , ac will give > and he will irive , , ca, '^ J ch. 392. Let it he o-equircd to determine the total immhcr of comhtna- nations of m letters, tahen n in a set. It is evideut, from what has been shown, that if we knew the total number of combinations of m letters, taken n in a set, that we could get the number of permutations of m letters taken n in a set, by mul- tiplying the number of combinations by the number of permutations obtained by permuting each combination ; or, in other words, by multi- plying the number of combinations of m letters, taken n in a set, by the number of permutations of n letters, taken all together. Conversely, when the number of permutations of m letters, taken n in a set, and the number of permutations of n letters, taken all together, are known, we can, by dividing the former by the latter, determine the number of combinations of m letters, taken n in a set. 393. To illustrate more fully by an example. Let us* combine the four letters, a, h, c and d, in sets of three, we will have the four combinations, ahcy ahd, acd, hcd. If, now, we permute each of these sots, taking all the letters in each set, we will have the total number of permutations of four letters, taken three in a set, that is, the total number of permutations of vi letters, taken n in a set. "Writing the results in tabular form, we have . ahc ahd acd hcd r ahe ahd acd hcd ach had adc hdn Permutations of each combination, taken 3 hac Ida cad eld and 3, or all together. hca dah cda cdh cah dba dac dbc cha adh dca dch We see that each of the four combinations, in sets of 3, gives six permutations, taken 3 and 3. There will, therefore, be 4 X 6 = 24 permutations of four letters, taken three in a set. Now, as a corres- ponding table could be formed for any number of letters, it is plain that PERMUTATIONS AND COMBINATIONS. 379 the total number of permutations of m letters, taken n in a set, is equal to the number of combinations of m letters, taken n in a set, multi- plied by the number of permutations of n letters, taken all together. Hence, if X = the number of combinations of m letters, taken n in a set ; Y = the number of permutations of n letters, taken all together ; Z = the total number of permutations of m letters, taken n in a set, we shall have Z = X . Y. Hence, X ^ y. But Z and Y are already known. For Z = m (m — 1) (m — 2) {m — ?i + 1), Formula (A) ; and Yr= 1 . 2 . 3 . 4 . 5 n, Formula (B). Hence, we have X== m(m — 1) . . . . (m — n -f 1) ..^ ... i j xi i. xi — ^^ — - — r^^ . Irom which, we conclude that the 1.2.3.4....» number of combinations of m letters, taken n in a set, is equal to the number of permutations of m letters, taken n in a set, divided by the number of permutations of n letters, taken « in a set. In the application of this formula, it will be of service to remember that the first factor of the numerator is the number of letters, that each successive factor is one less than the preceding, and that the last factor is the number of letters diminished by the number in a set, less one. It is well, then, to determine first the extreme factors of the numerator; the mean factors can then be readily supplied. In regard to the denominator, the last factor is the number in a set, and the factors counted to the left go on diminishing by unit}'^ unto the first factor, which is always unity. 1. Required the number of combinations of 6 letters, taken 4 in a set. 6(6 — 1) (0 — 2) (6 — 3) G.x5x4x3 ^^ 1 . 2 . o . 4 1.2.3.4 For m = 6 and n = 4. Hence, the first factor of the numerator is 6, and the last m — n ■{■ 1 = 6 — 4 + 1 = 6 — 3. The two mean factors of the numerator are then readily formed, the extremes being known. 380 PKRMUTATIONS AND COMBINATIONS. 2. Required the number of combinations of 12 letters, taken 7 and 7. 12(12 — 1) (12 — 2) (12—3) (12—4) (12 — 5) (12 — 6) ^'''- 1.2.3.4.5.6.7 12x11x10x9x8x7x6 12x11x10x9x8 1x2x3x4x5x6x7 1x2x3x4x5 X 9 x 8 = 792. r^ll 3. Required the number of combinations of 12 letters, taken 5 and 5, L)(12 — 2)(12— 3)(12— 4) _ 12xllxl0x 1x2x3x4x5 ~" 12x10 12(12 — 1) (12 — 2) (12— 3) (12— 4)_12xllxl0x9x8 = 792. 4. Required the number of combinations of 12 letters, taken 9 and 9. 12(12—1) (12—2) (12—3) fl2— 4) (12—5) (12—6) (12—7) (12— 8) _ '■ 1.2x3x4x5X6x7x«Xy ~" 1X2X3X4X5X6X7X8X9 5. Required the number of combinations of 12 letters, taken 3 and 3 12 (12-1) (12 -2) _ 12 X 11 X 10 _ Ans. ^2.3 - 6 - •'^^- 6. Required the number of combinations of 10 letters, taken 9 and 9. Ans. 10. 7. Required the number of combinations of 10 letters, taken 1 and 1. Ans. 10. 8. Required the number of combinations of 25 letters, taken 21 and 21. Ans. 12650. 9. Required the number of combinations of 25 letters, taken 4 and 4. Ans. 12650. 10. Required the number of combinations of 50 letters, taken 47 and 47. Ans. 19600, 11. Required the number of combinations of 50 letters, taken 3 and 3. Ans. 19600. Scholium. 394. The last ten examples show that the same number of letters, combined differently, may produce the same number of combinations, that is, the number of combinations of m letters, taken p in a set, may be equal to the number of combinations of ??i letters, taken g' in a set. We propose to show that this will always be so when m =p -^q. Thus, PERMUTATIONS AND COMBINATIONS. 381 in example 10, vi = 50 and p = 47 ; in example 11, m = 50 and q = o. The results have been the same in those examples, because m = p + q. To show this, suppose p^ q. Then the number of combinations of m letters, taken p m sl set, will be expressed thus : ^ _ m{m — 1) .... (ot — g4l) (??i — q) (m — q — 1) (711— p + 1 ) 1 . 2Ts (p-iy^ ■ And for the number of combinations of m letters, taken g' in a set, we will have the formula, _ «i (m. — I) (m — 2) (m — 9 + 1) ^ - 1.2.3 (q-l)q — Hence, D = C (^- 9) (^ -9- 1) ■ • • • (^ -^ + 1 ) (? + 1) (? + 2) p Now, it is plain that D will be equal to C, when (m — q) (m — q — 1) &c. =p(p — 1) . . . . (q + 2) (5' + 1). Or, since tlie foctors iu both members go on decreasing by unity, and since the number of them is also equal, the last equation will be true when m — q =:p, or in = q -\- p>, fis enunciated. This rule is of importance whenever _?J ^ -^^ • For, then, we have only to take the difference between /) and m, and the number of combinations of m letters, taken m — p and m — p, will be the same as the number of combinations of m letters, taken p and p. Thus, in example 10, instead of taking the number of combinations of 50 letters, taken 47 in a set, we may take the number of combinations of 50 letters, taken 50 — 47, or 3 in a set. In like manner, the number of combinations of 100 letters, taken 90 in a set, would be the same as the number of combinations of 100 letters, taken 10 in a set. 395. If, in the formula, D = C (^-g) («^-g-l) • • • -(^^-p + D (9 + 1) (9 + 2) . . . . p ' we make q = v — 1, we will have D = C : for, in that case, J) will have only one more factor than C. Hence, the number of covibinations of m letters, taken p and p, is equal to the number of combinations of m letters, taJcen p — 1 and p — 1, multiplied by the factor When p = 2, 3, 4, &c., or the sets are made up of 2, 3, 4, &c., letters, the factor ~ will become —^ — , 382 PERMUTATIONS AND C Jl B I N ATI O N S . ■ffi 2 m 3 — ^T — , — J — . Suppose, for example, that we have 8 letters, or m =z 8, then the successive multipliers will be |, |, |, |, |, '^. And, since the number of combinations of 8 letters, taken 1 and 1, is 8, we will have the number of combinations, taken 2 and 2, 3 and 3, 4 and 4, &c., expressed 8 X | = 28, 28 x | = 56, 56 X | = 70, 70 x | =56, 56 X i = 28, 28 x' f = 8. This remarkable law has had many important applications, one of the principal of which is, the determining of the coefficients in the binomial formula. It is plain from what has been shown that, if we write in succession the number of combinations of m letters, taken 1 and 1, 2 and 2, . . . . n and n, we will have a series of numbers increasing to the middle term, and then repeated in retrograde order. Thus, 6 letters combined, 1 and 1, 2 and 2, 3 and 3, 4 and 4, 5 and 5, give the series 6, 15, 20, 15, 6. In like manner, 7 letters give the scries, 7, 21, 35, 35, 21, 7. The law of formation is precisely that which we have observed in the binomial development; and, in fact, in this development, the coefficients, after the second, are the combinations of m letters, taken 2 and 2, 3 and 3, &c., . . . . n and n. 396. Let us now seek the greatest term of the series formed by com- bining m letters, 1 and 1, 2 and 2, .... n and n. The factors, which m — 1 TO — 2 m — 91 4- 1 determine the successive terms, are — - — , — ^^ — , .... . &c. Now, since these numerators go on decreasing, and the denomi- nators increasing, it is evident that the successive products will go on increasing until w — n -\-l = n, and after that will decrease and be repeated in reverse order. Suppose m an odd number, then placing m — m -f- 1 = n, we get n r= ^ (m 4- 1). Now, if we include unity, the last term of the series, there will be m terms in all, and in case of m being odd, the n*"" term will obviously be even, and, of course, have an even number of terms preceding and succeeding it. Thus, in the series, 7, 21, 35, 35, 21, 7, 1 ; « = —-, — = 4, will represent the 4th term, and we see that it has three terms before and three after it. It is plain, moreover, that tlie 7i^^ term, being formed from the (ji — l)*"" term by , . , . - , ,171 — n -{- 1 ^ . , , multiplying the latter by =1, is, also, equal to the (n — ] )<" term. Hence, when m is an odd number, there will be two equal central PERMUTATIONS AND COMBINATIONS. 383 terms, the series will go on increasing unto the first of these, and decreasing from the second unto the last term, unity. Thus, the com- binations of 5 letters, taken 1 and 1, 2 and 2, 3 and 3, 4 and 4, 5 and 5, give the series 5, 10, 10, 5, 1. 397. When m is even, the factors increase until n = J m. We can- not, in this case, have n z=z — - — , for that would make n fractional, a manifest absurdity. No factor, in this case, is unity, and, of course, there are no equal central terms. The terms go on increasing until n = ^m, and are then reproduced in reverse order. It is obvious that whenever n '^Im, the factor will become a proper fraction, and, of course, the term of the scries formed by multiplying the pre- ceding term by this factor, will be less than the preceding. Ten letters, taken 1 and 1, 2 and 2, &c., give the scries, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1. 398. If we add together the values of C and ]), as found in Art. 394, and at the same time make q = 2) — 1, wc will have C + I) = C + C(m-p-f l )^ C(m + l) ^ m-f 1 ^ ^ X - = / "^ + h p p 1 p \ 1 / (m \ tm — 1 V {jn — p + 3) {in — p -f 2) T/ V "3 / p — \ p ■ The second member of this equation plainly denotes the number of combinations of m + 1 letters, taken p andpj whilst the first member denotes the sum of the combinations of m letters, taken ^^ andp, and p — 1 and p — 1. This important relation enables us to get the number of combinations of m -}- 1 letters from that of m letters. Thus, make m = 4, p = 3, and p — 1 = 2, then 4 letters, taken 2 and 2, give 6 combinations, and taken 3 and 3, give 4, and we find that 5, or m -f 1 letters, taken p, or 3 in a set, give a number of combinations equal to 4 + 6, or 10. 399. Upon this principle, the following table (p. 384) has been con- structed. The first vertical column is made up of ones ; the second column contains the natural numbers from 1 to 20, and expresses the number of combinations of these numbers, taken 1 and 1; the other vertical columns express the combinations of the same numbers, taken 3 and 3, 4 and 4, 5 and 5, &c. These vertical columns result from the hori- zontal, which are constructed according to the principle demonstrated in Article 398, observing that every horizontal column must close with unity. Thus, the numbers in the second horizontal column are 1, 2 384 P E R I\I U T A T I O N S AND COMBINATIONS. and 1, the numbers in the next column are, 1 + 2 = 8, 2 + 1 = 3 and 1, and express the number of combinations of three letters, taken 1 and 1, 2 and 2, and 3 and 3. Now, prefix unity, and we have the third horizontal row made up of 1, 3, 3 and 1, and the fourth row will be made up of 1, 3 + 1 ^ 4, 3 + 3 = 6, 3 + 1 = 4 and 1, the last four numbers expressing the number of combinations of four letters, taken 1 dnd 1, 2 and 2, 3 and 3, and 4 and 4. All the other horizontal columns are formed in the same manner. r- ARITHMETICAL TRIANGLE OF PASCAL. 1 2 3 4 5 1 1 3 6 10 4 10 1 5 1 7 8 9 10 15 21 28 36 45 20 35 56 84 120 15 35 70 126 210 6 21 56 126 252 1 28 84 210 1 8 36 120 1 9 45 1 10 1 n 55 165 330 462 462 330 165 65 11 12 66 220 495 792 924 792 495 220 66 13 78 286 715 1287 1716 1716 1287 715 286 U 91 364 1001 2002 3003 3432 3003 2002 1001 - 15 105 455 1365 3003 5005 6435 6435 5005 3003 16 120 560 1820 4368 8008 11440 12870 11440 8008 17 136 680 2380 6188 12376 19448 24310 24310 19448 Il 18 153 816 3000 8568 18564 31824 43758 48620 43768 19 171 969 3876 11628 27132 60388 75582 92378 92378 20 190 1140 4845 15504 38760 77520 125970 167960 184756 1 and 1 2 and 2 3 and 3 4 and 4 5 and 5 6 and 6 7 and 7 8 and 8 9 and 9 10 and 10 400. We can now readily find the entire sum of the combinations formed by any number of letters, taken in every possible way. Let the numbers in the ^"' row be expressed hj 1, a h . . . m, m . . . b, a, 1; t being supposed an odd number. Then their sum will be 2 (o + 6 -}- m + 1). The numbers in the next row will be expressed by 1 + a, a + i . . . . + . + m, + m + m . . . i + a, a + 1 and 1, and their sum will be 4 (a + h . . . + m + 1) + 1. Hence, the sum of the numbers in the (t + 1)'" column, will be double of the preceding, and one more. When t is an even number, the same law can be shown to be true But the sum of the numbers in the second horizontal column, is 2^ — 1 Hence, in the third column, it will be 2 (2' — 1) + 1 = 2* — 1, and in the fourth, 2 (2='— 1) + 1 = 2^ — 1. And it is plain that, in the f" column it will be 2* — 1. It will be seen that the ones in the first PERMUTATIONS AND COMBINATIONS. 3S5 vertical column are omitted. Hence, 2* — 1, espresses the sum of the combinations of t letters, taken 1 and 1, 2 and 2 .... t and t. Thus, to apply the formula, let it be required to determine the number of combinations of three letters, taken 1 and 1, 2 and 2, 3 and 3. Then t =3, and 2* — 1 = 2=' — 1 = 8 — 1 = 7. This agrees with the fact, for the three letters, a, h, c, give us the seven combinations, u, h, c ; ah, ac, be, and ahr. So, likewise, the four letters, a, b, c, J, by the formula, give a number of combinations equal to 2^ — 1 =16 — 1 = 15; and we, in fact, have a, b, c, d ; ab, ac, ad, be, bd, cd ; ahc, abd, acd, bed ; and abed. 401. We will now deduce an analogous formula to the above, for the sum of the permutations of n letters, taken 1 and 1, 2 and 2, 3 and 3, n and ?i, when each letter is combined with itself, so as to be taken t(j the second, third, fourth, and . . . n"" powers. PERMUTATIONS IN WIIICII THE LETTERS ARE REPEATED. Two letters, a and b, permuted in this way, give the four permuta- tions, aa, ab, ba, bb. Hence, the number of permutations of two let- ters, taken 2 and 2, when each letter is associated with itself, is equal to 2^. Three letters, a, b, c, give the nine permutations, aa, ab, ae, ba, ea, bb, be, cb, ce. Hence, the number of permutations of three letters, taken 2 and 2, is equal to 3^ Four lettci-s, a, b, c, d, give, when taken ' 2 and 2, the sixteen permutations, aa, ab, ac, ad, ba, ca, da, bb, he, bd, ch, dh, cc, cd, de, dd. Hence, the number of permutations of four letters, taken 2 and 2, is equal to 4^. And, in general, it is plain that the number of permutations of n letters, taken 2 and 2, is equal to n^. We will next show that the number of permutations of n letters, taken 3 and 3, is equal to n^. The three letters, a, h, and c, when permuted in this way, give the twenty-seven permutations, aaa, aah, aae, aba, baa, aca, eaa, ahc, cba ; hbh, bbc, bba, bcb, ebb, bah, abb, bac, hca ; cec, ccb, eca, cb.-, bee, eac, aec, cab, aeh. Hence, the number of permutations of three letters, taken 3 and 3, is equal to 3^ And, in general, of n letters, taken 3 and 3, the num- ber is 11^. In like manner, the number of permutations of n letters, taken 4 and 4, is expressed by W*. Hence, for the entire sum of the permutations, taken 1 and 1, 2 and 2, 3 and 3 . . . . n and n, we have 33 z 38G PERJIUTATIONS AND COMBINATIONS. , («" 1)?!, , , , . . n -\- ir -}- ir -(- n° = ^^ ^ — , the whole, constituting a geo- metrical series, whose common ratio is n. Thus, to apply the formula, suppose it be required to determine the number of permutations of five n ■,(«•' — 1> (5^ — 1)5 15620 „,,^, ^ letters, then n = 5, and ^ ^ — = ^^ -. = — -. — = o'JUo, and ' ' 11 — 1 4 4 /'■^^n \)7l if n = 24, the number of letters in the alphabet, we have ~ = n — 1 (24^^^^^ = 1301724288887252999425128493402200. Thus, let it be required to determine the entire number of changes that can be made with the three vowels, a, e, and /. The formula ^ r^— , becomes -^^ — ^ = — - . 3 := 39, and we nnd that we have 11 — 1 ' 2 2 the thirty-nine permutations. a, €, t, taken 1 and 1 ; ae, at, ea, ia, ei, ie, aa, ee, ii, taken 2 and 2, aei, eai, iea, eia, iae, axe, taken 3 and 3, in the usual way; aaa, aae, aai, eaa, iaa, aea, aia, eee, eea, cei, aee, iee, eae, eie, tii, iia, He, aii, eii, iai, ici, taken 3 and 3, when the letters are combined with them- selves. PARTIAL PERMUTATION. ' 402. Sometimes the nature of the problem is such, that some of the quantities cannot be permuted with each other. Thus, let it be re- quired to determine how many words of two letters each can be formed out of the letters a, e, c, d, admitting that the consonants associated together will not form a word. The formula for the number of per- mutations of m letters, taken n and n, gives us 4 (4 — 1) = 12. But we have to reject the number of permutations of two letters, c and (/, taken 2 and 2, or 2 (2 — 1) = 2. Hence, 12 — 2 = 10, the number of permutations that can be formed in the required manner. They arc, ac, ac, ad, ea, ca, da, ec, ed, ce, de. RULE. Find the entire nvmber of ^'""'"t^'^'totions of n letters, taken n and n, nnd from this subtract the number of permutations of the p especial letters, taken n and n. GENERAL EXAMPLES IN PERMUTATIONS, ETC. 387 EXAMPLES. 1. How many words of two letters each can be formed out of the 4 first letters of the alphabet, assuming that consonants alone will not form a word ? A71S. 4 (4 — 1) — 3 (3 — 1) = 6 ; words, ab, ac, ad, ha, ca, da. 2. How many numbers composed of 3 digits each can be formed from the number 24371, in such a way that none of the numbers con- tains only odd digits ? Ans. 5 (5 — 1) (5 — 2) — 3 (3 — 1) (3 — 2) = 54. Numbers, 243, 234, 342, 324, 432, 423 ; 247, 274, 472, 427, 742, 724 ; 241, 214, 412, 421, 142, 124; 237, 273, 372, 327, 732, 723; 231, 213, 312, 321, 132, 123; 271, 217, 721, 712, 172, 127; 437, 473, 374, 347, 734, 743 ; 431, 413, 341, 314, 134, 143 ; 471, 417, 741, 714, 174, 147' 3. How many words of letters each can be formed out of G vowels and 6 consonants, assuming that the consonants by themselves cannot form a word ? Ans. 12 (12 — 1) (12 — 2) (12 — 3) (12 — 4) (12 — 5) — 6 (6 — 1) (6 — 2) (6 — 3) (6 — 4) (6 — 5) = 6G45G0. 4. How many words of six letters each can be formed from the 20 consonants and G vowels of the alphabet, assuming that the consonants alone cannot form a word ? Ans. 137858400. GENERAL EXAMPLES IN PERMUTATIONS AND COMBI- NATIONS. 1. Three travellers on a journey reach three roads: in how many different positions, with respect to each other, may they continue travelling, provided, that no two of them pursue the same road ? Ans. G. A may take the first road, B the second, and C the third; or, A the first, C the second, B the third, &c. 2. How many days can 3 persons be placed in a different position at dinner ? Ans. G days. 3. How many days can 8 persons be placed in a different position at dinner? Ans. 40320 days. 888 GENERAL EXAMPLES IN 4. In how many different positions can a stage driver place the 4 horses of his team in harness ? Aiis. 24. 5. How many changes may be made in the words of the sentence : " Lazy boys make worthless, vicious men ?" Ans. 720. G. How many changes may be struck with 10 keys of a piano ? Ans. 3028800. 7. How many words of five letters each can be made out of the first 24 letters of the alphabet, assuming that no letter is repeated more than once in the same word ? Ans. 5100480. 8. How many words of three letters each can be made out of the 24 first letters of the alphabet, with the same proviso as the preceding ? Ans. 12144. 9. How many numbers of two digits each can be made out of the numbers 1243 (provided, that none of the digits of 1243 appear twice in the same number), and what are they ? Ans. 12. Numbers, 12, 43, 14, 13, 21, 34, 41, 31, 24, 23, 42, 32. 10. How many numbers of five digits each can be formed out of the number 187G543, provided, that no digit of the given number appear twice in the required results ? Ans. 2520. 11. How many changes can be made in every file of 2 men in a squad of 20 men, so that the files shall differ by at least one man ? Ans. 190. 12. How many changes can be made in the position of the first 12 letters of the alphabet ? Ans. 479001600. 13. How many numbers of 1, 2, and 3 digits can be formed out of the number 476, provided, that no digit is repeated in the isame number ? Ans. 15. Numbers, 4, 7, 6: 47, 46, 74, 64, 67, 76; 476, 467, 746, 764, 647, 674 14. How many nuiubers of 1, 2, and 3 digits can be formed out of the number 476, when each digit is repeated 1, 2, and 3 times in the respective numbers ? Ans. 39. Numbers, 4, 7, 6, 44, 47, 46, 74, 64, 67, 76, 77, GO, 476, 467, 746, 764, 647, 674 ; 444, 447, 446, 744, 644, 474, 464 ; 777, 774, 776, 477, 677, 747, 767; 666, 664, 667, 466, 766, 646, 676. PERMUTATIONS AND COMBINATIONS. 380 15. How many numbers of 1, 2, 3, 4, and 5 digits can be formed out of the number 56789, provided, that no digit occurs more than once in the connection with exactly the same digits ? Ans. 31. Numbers, 5, 6, 7, 8, 9 ; 56, 57, 58, 59, 67, 68, 60, 78, 79,89; 567, 568, 569, 578, 579, 589,678, 679, 689,789; 5678, 5679, 5689, 5789, 6789 ; 56789. 16. Three travellers, A, B, C, on their journey come to three roads, and may either travel all together on one of the three, or singly, on the three roads, or two on the same road, and the third on a different road. How many selections may they make of their routes ? Ans. 27. Three, when all together; six, when they go separately; six, when A and B go together, and C goes by himself; six, when A and C are together, and B by himself; six, when B and C are together, and A by himself. 17. IIow many numbers composed of three digits each can be made out of the number 123456780, provided, that all the numbers thus formed shall differ by at least one digit. Ans. 84. 18. How many numbers of 4 digits each can be formed out of the same number, 128456780, so as to fulfil the above condition ? Ans. 126. 10. How many numbers composed of single digits, of 2 digits, of 3 digits, of 4 digits, and of 5 digits, can be formed out of the number 12345, in such a way that each number shall differ from all the other numbers by at least one digit. A71S. 31. Numbers, 1, 2, 3, 4, 5 ; 12, 13, 14, 15, 23, 24, 25, 34, 35, 45; 123, 124, 125, 134, 135, 145, 234, 235, 245, 345; 1234, 1235, 1345, 1452, 2.345 ; 12345 20. How many numbers composed of 1, 2, 3, 4, and 5 digits can be formed out of the number 12345, in such a way that the same digit may be repeated once, twice, &c., in the same number, and the several numbers necessarily differing only in the position of the digits. Ans. 3005. Numbers, 1, 2, 3, 4, 5; 12, 13, 14, 15, 21, 31, 41, 51, 23, 24, 25, 32, 42, 52, 34, 35, 43, 53, 45, 54, 11, 22, 33, 44, 55; 111,112, 113, 114, 115, 211, 311,411, 511, 121, 131, 141, 151, &c. 33* 390 LOGARITHMS. 21. A gentleman being asked what he would take for a valuable horse, replied, that his price was a cent for every change that he could make in the position of the 32 nails in the horse's shoes. What was his price? A71S. 2'631308S69336'93530l'672180121600600 dollars. LOGARITHMS. 403. Tlie logarithm of a quantity is the exponent of the power to which it is necessary to raise an invariable quantity, called the base, to produce the quantity. Let a be the invariable quantity or base, x its logarithm, and i/ the quantity given ; then, a^ = y. In this equation, x is the logarithm of 1/. It is usual to write logarithm, log., or simply, 1. Hence, X = log. y, or 1. 1/. We will begin the discussion by supposing a ^ 1, and X positive, and, though the general principles of logarithms are true for quantities as well as for numbers, yet, as iu practice the logarithms of algebraic quantities are seldom used, we will first show the relations between numbers and their logarithms. Resuming the equation, a* = y, we see that when i/ = 1, x = 0, for, a" =: 1. Hence, log. 1 = 0, and, since this will be so, whatever may be the value of a, we see that the log. 1 is always zero. Every value given to x, above zero, will cause 1/ to increase more and more ; when a; = 1, ,?/ = a. When x has values attributed to it greater than 1, y will exceed a more and more, until x = CO gives also y = oo . Suppose, for instance, a = 8, then x = and x = 1 give y == 1 and y = S. Moreover, x = 1, 2, 3, 4, &c., gives 1/ = 8, 64, 512, 4096, &c. Finally, a; = oo gives y = go , for, 8^" = CO . It is plain that, when x has an intermediate value between and 1, that y will have a value between 1 and 8, and that by making x pass through all possible positive values, entire and fractional, between and 1, and 1 and oo , y will be made to pass through all possible positive values between 1 and 8, and 8 and cc . The contrary of what we have just seen will be the case when X is negative. For the equation, a~'' =y, will become — = y; and it is plain that, as x increases y will decrease, and that when x = cc , y = 0. To illustrate this, let a = 8, and let cr r=: 0, 1, 2, 3, 4, &c., LOGARITHMS. 391 then, y = 1, |, g'j, ^jj, 4 g^g, &c. And it is evident that, by giving to X all possible positive values, entire and fractional, between U and oo , y may be made to pass through all possible positive values between 1 and 0. Suppose now, a <^ 1. It will still be true that all possible values may be formed from the powers of one number, but the order of the numbers will be reversed. For values of ar, between the limits and — CO , we will get all possible positive numbers between 1 and + 00 ; and for all values of x, between and + oc , we will get all possible positive numbers between 1 and 0. The remarkable fact that all positive numbers might be regarded as the powers of one invariable number, led to the invention of logarithmic tables by Napier, for the purpose of abridging complicated numerical calculations. The invariable number is called tlie base of the si/sfem of logarithms, and may evidently be any number whatever, except unity. Since all the powers of unity are unity, it is plain that this number cannot be taken as a base. The base of the common system of loga- rithms is 10. The powers of this number are more readily formed than the powers of any other number whatever, and this was the main reason for its selection. We have (10/ = 1, Hence, log. 1 =0, (10)' = 10, '' log. 10 =.1, (10)^=100, " log. 100 =2, (lOy = 1000, " log. 1000 = 3, &c. &c. We see that, as the number changes (the base being the same), the logarithm also changes. It is plain, moreover, that any change in the base, the number being unaltered, will produce a change in the loga- rithm. Thus, if the base is 12, since 12' = 12, we have log. 12 := 1. The logarithm of 10 in this system will then not be 1, as in the system whose base is 10, but will be some fractional number less than 1. Hence, we see that logarithms depend upon the number and upon the i base. That part of the logarithm in any system which depends upon the base is called the modulus of the system. A table of loijarithms is a table exhibiting the logarithms of all numbers between certain limits, as and 100, or and 10000, calculated to a particular base. The general properties of logarithms are independent of any particular j base. A few of these general properties will now be demonstrated. 392 LOGARITHMS. First Property. 404. The logarithm of the product of any number of factors taken in the same system, is equal to the sum of the logarithms of those factors. For, let a represent the base of the system, y, y', ?/', y'", &c., several numbers, and x, x', x", x'", &c., their corresponding logarithms. Then, cv^ = y, o"' =: y' , a"" = y", a""" = y"', &c. Multipl3'ing these equa- tions together, member by member, we have 0^+^'+""+^'"+ ^'- z=i y ij y" y'" , &c. Then, since the exponent of the base is (from the definition) the logarithm of the second member, we have x + x' + x" + a;"' + &c. = log. yy'y"^", &c. But the first equations give x = log. y, x' = log. y', x' = log. y", &c. Hence, log. y -\- log. y' + log. y" + log :!/"' + &c. = log yy'y"y"' &c., as enunciated. Then, since the logarithm of the product of any number of factors is equal to the sum of the logarithms of these factors, it follows that to multiply by means of loga- rithms, we have only to add together the logarithms of the factors, and to find the number corresponding to this sum. This number will be the product required. Thus, let it be required to multiply 100 by 10, by means of logarithms. The log of 100 is 2, in a system whose base is 10 ; the log of 10 is 1 ; the sum of these logarithms, 3, is the loga- rithm of the product. Look for the number corresponding to 3 as a logarithm, and you will have 1000 as the product required. To exhibit the whole in an equation, we have, log. 100 x 10 = log. 100 -f log. 10 = 2-1-1 = 3; the number corresponding to which is 1000. We have, then, for multiplication by means of logarithms, this RULE. Add toycther the logarithms of the several factors. This sum will he equal to the logarithm of the product. Looh in the table of logarithms for the number corresponding to this logarithm, and you loill have the product required. 1. Required the equivalent expression to log. a b c d. Ans. Log. a + log. b -\- log. c -f- log. d. 2. Required the equivalent expression to log. (San. Ans. Log. 6 -f- log. a -f- log. n. LOGARITHMS. 393 3. Required the equivalent expression to log. (a + by c" d^. Ans. Log. (a + i)" + ^og. c" + log. d^. Second Projperty. 405. The logarithm of the quotient arising from dividing one quan- tity by another is equal to the logarithm of the dividend minus the logarithm of the divisor. For, let x equal the quotient of two quan- 11 ^''^ 4 1-1 titles, m and n, then x = — , or nx = m. And, since, when two n quantities are equal their logarithms must be equal, we have log. nx = log. m, or log. n + log. x = log. w. Hence, log. x = log. m — log. n, as enunciated. From this we derive for division, by means of loga- rithms, the following RULE. From the logarithm of (he dividend subtract the logarithm of the divisor; the remainder will be the logarithm of the quotient. The number in the table corresponding to this logarithm will be the quotient required. Thus, to find the quotient of 1000 by 10, we have log. 1000 — log. 10 = 3 — 1 = 2, and the number corresponding to 2 as a logarithm is 100. 1. Find the equivalent expression to log. - — . bn Ans. Log. am — log. bn, or log. a + log. m — log. b — log. n. 2. Find the equivalent expression to log. ^ — ~. Ans. Log. X -f- log. ij — log. (b -\- n). Find the equivalent expression to log "^ bn Ans. Log. (x -f _y) — log. b — log. n. 4. Find the equivalent expression to log. j^ bn Ans. Log. X -f log. y — log. b — log. n. 394 LOGARITHMS. 5. Find the equivalent expression to log. (y^ )• Ans. Log. (.'r?/)P — log. (in)p. G. Find the equivalent expression to log. (- V Ans. Log. (« + i) — log. c. Third Property, 406. The logarithm of the power of a number is equal to the expo- nent of the power into the logarithm of the number. For, take the equation, as=y, and raise both members to the m*" power, we will have a"^ = 3/° (A). In equation (A), m is any number whatever, positive or negative, entire or fractional. But, since the exponent of the base is the logarithm of the second member, equation (A) gives mx = log. ?/™ ; and the first equation gives x = log. y. Hence, m log. y = log. ?/■", or log y"^ = m log. y, as enunciated. It follows, then, that to raise a number to a power by means of logarithms, we have only to apply this RULE. Multiply the logaritlim of the numher, as found in the tahic, hy the exponent of the poioer to which it is to be raised. This product xciJl he the logarithm of the poioer, a7id the number found in the table corres- ponding to this logaritlim loill he the required poioer. Thus, to raise 10 to the second power by means of logarithms, we multiply the logarithm of 10, which is 1, by 2, the exponent of the power. The product, 2, is the logarithm of the power, and the number corresponding to this logarithm is 100. So that we have log. (10)^ = 2 log. 10 = 2 ; the number corresponding to which is 100. 1. Find the equivalent expression to log. (a + bj^. Ans. m log. (a H- b). 2. Find the equivalent expression to log. c" (a + t)'". Ans. n log. c + m log. (a + b). The 1st. property is here used in connection with the 3d. LOGARITHMS. 3. Find the equivalent expression to log. a" {mry. Alls, n (log. a + log. r + log. m). 4. Find the equivalent expression to log. a» (-A . Ans. 11 (log. a + log. h — log. c). The 2d. and 3d. properties both employed. 5. Find the equivalent expression to log. ^^ — j^ — . Ans. m log. (a -\- I) — 2 log. h. 6. Find the equivalent expression to log. — ^-g. Ans. Log. (a — x) — 2 log. (a -j- x). (a -f xY 7. Find the equivalent expression to log. — ^ ^. Ans. 2 log. (a + x) — log. (a — x). Fourth Properti/. 407. The logarithm of the root of a number is equal to the logarithm of the number divided by the index of the root. For, take the equation, a^ = y, and extract the m"" root of both X 3j members. Then, Va" = '2 — n=r2 = i — ^ + J — J + i — i + &c., Z'4 = Z'3 + i- J3 + hV- 3k + T2V7r-&c., /'5 = Z'4 + 1 — 3^. + ih — TTR4 + i,~ho — &c- The first series enables us to calculate the Napierian logarithm of 2. The second series enables us to calculate the Napierian logarithm of 3 when that of 2 is known ; and the formula, evidently, only can be used when the logarithm of the next lowest number is known. Second Transformation. 421. The preceding formula does not give a sufficiently converging series for small numbers. A better series can bo obtained in the fol- LOGARITHMS. 403 lowing manner. In the equation, ?'(! + x) ^=x — tt -^ o r + x^ x^ — ^4- &c., we get, by changing + .r into — x, /'(I — a-) = — x 3,2 ^3 ^4 j.a ^.6 — — -^ -. ^ — -; &c. Subtracting the second series 2 3 4 5 ° from the first, we get l'{l -f x) — ?'(1 — .r) = I' L "*" ^ \ = 2 (x + ^ + ? + T + '^ + n + S + *°-')(*=)- p'-l^ = i + -, in which z is a whole number. Then, x = ~ =-, and replacing x in (E) by this value, we get ^'(1 + -), or I' (I + z) — l' z=2 (- - verges more rapidly than (D), and will answer even for small numbers. Let z = l,2, o, 4, 5, 6, &c., we will have re - « = 2 (i + -1^ + ^^ + ^-2_ + _2_ + fe. f6 — ?5 = 2(T'r + 8(ir)+5Tn)»+f(Tl)'+9(TT7+ll(nv' + &c.) 1. For small numbers it is necessary to take a good many terms of this series, but when the numbers are large one or two terms will answer. Thus, let a- = 1000, then nOOl—nOOO =2(5 J^-j- + o-oiTr-.^., + FrF^jK(yf\5 + ^cV We see that the fi;-st term of this series will only add about jq^^o *° *^^ logarithm of 1000, and the second term will add less than godoFoo^oo- And, since logarithms are seldom 404 LOGARITHMS. carried further than the Tth place of decimals, all the terms after the first may be neglected. 2. In constructing a table it is only necessary to calculate the loga- rithms of prime numbers, since the logarithm of any number made up of factors is equal to the sum of the logarithms of its factors. Thus, no = n + I'b, Z'12 = Z'3 + Z'4 = Z'3 + 2Z'2, &c. 3. Knowing the Napierian logarithm of 10, we can fiud the modulus of the common system. For, log.' 10 : log. 10 : : 1 : M. Hence, M = -^rrcT ' ^^^ ^*-*S- 1^ "= ^^ ^"^^ *'^® calculated value of log.' 10 is 2.302585093. Therefore, M = ^ oaolo^ac,^ = 0-434294482. If we now multiply the Napierian logarithms found from the series by this modulus, we will have a table of common logarithms. ADA^ANTAGES OF THE COMMON SYSTEM. 422. 1. A fixed law for the characteristic. The characteristic of a logarithm is its entire part. Thus, 2 is the characteristic in the logarithm 2-302585093. Whatever may be the base of the system, the logarithms of the powers of that base will contain no decimals, and, therefore, be charac- teristics only. Thus, let 7 be the base, then the log. 7 — 1, log. 49 = 2, &c. In the common system, with the base 10, we have log. 10 = 1, log. 100 = 2, log. 1000 = 3, &c., and we see that the characteristic is always one less than the number of figure places in the number. This is true, whether the number be an exact power of the base or not ; thus, 512, lying between 100, whose logarithm is 2, and 1000, whose logarithm is 3, has for its logarithm 2 and a certain decimal. Now, 10 is the only number in the whole range of numbers, which, taken as a base, will give logarithms whose characteristics are formed according to a fixed law. In a table of common logarithms, the characteristics are not written, since they are always one less than the mimher of figures in the given numbers. 2. A tahle of logarithms for decimals need not be constructed. Since, Log. -1 = log.* -J^ = log. 1 — log. 10 = — 1. Log. -01 = log. jJ = ^og- 1 — 'og. 100 = — 2. Log. -001 = log. j-^oT) = log- 1 — log. 1000 = — 3. LOGARITHMS. 405 And, since, in general, log. := log. 1 — m log. 10 = — m, it is plain tliat the characteristic of a decimal is negative, and one greater than the number of ciphers between the decimal point and the first significant figure. This is true, also, when the decimal is not the re- ciprocal of some power of the base. Thus, log. -08 = log. -f^ = log. 8 — log. 10 = a decimal — unity. Hence, the log. -08 must have a negative characteristic equal to minus unity. Moreover, if we have a decimal, such as -00278 = jxhhj1)0' ^^ ^^'^ ^^^'^ 'og- '00278 = log. 278 — log. 100000. And, since log. 278 is 2 whole number, plus a cer- tain decimal; and, since the log. 100000 is 5, the log -00278 = — 3, whole number plus a certain decimal. We see, from this example, that the logarithm of a decimal is obtained by regarding the decimal as a whole number, and prefixing a negative characteristic, one greater than the num- ber of ciphers, between the decimal point and the first significant figure. The foregoing reasoning can be extended to any decimal whatever, be- cause, change the decimal into an equivalent vulgar fraction, it will be seen that the number of figure places in the denominator exceed those in the numerator by one more than the number of ciphers between the de- cimal point and the first significant figure of the given decimal. Now, if the base were any other number whatever than 10, it is plain that a table of logarithms for whole numbers would not be applicable to decimals. 3. The logarithm of a mixed decimal, such as 42-733, can he found from a table of logarithms constructed to the base 10, b^ regarding the mixed number as a whole number, and prefixing to the decimal p«r< (f the logarithm found in the table a characteristic one less than the number of figure places in the entire p)art of the mixed number. For, log. 42-733 = log. *f^^\^ = log. 42733 — log. 1000 = 4 + a certain decimal — 3 = 1 + a decimal. The foregoing rule for finding the logarithm of a mixed number can evidently be applied to any mixed number whatever. 4. The logarithms of numbers which difi"er only in the number of annexed ciphers, will differ only in their characteristics. Thus, the logarithms of 125, 1250, 12500, and 125000, difi'er only in their characteristics. For, log. 1250 = log. 10 -f log. 125 = 1 -f- log. 125 ; log. 12500 = log. 100 -f log. 125 = 2 -|-log. 125 ; log. 125000 = log. 1000 + log. 125 = 3 -f log. 125. This property of common logarithms is very important, since it saves the trouble (as will be seen more fully hereafter) of constructing a table for numbers whose figure places exceed four. 406 T. O (J A U 1 T II M S . APPLICATION OF COMMON LOGAPJTIIMS. 423. The base of the coiniuon system being 10, the characteristic or entire part of the logarithms need not be written iu the tables, since, as we have seen, it is always one less than the number of figure places iu the given number. The decimal parts of the logarithms are then only written in the tables, and are generally carried as far as six places. The first vertical column on the left in the tables, contains the given numbers ; the last vertical column, marked D, contains the tabular dif- ference corresponding to two given logarithms, and is constructed by a method hereafter to be explained. The intermediate vertical columns, marked at top, 0, 1, 2, 3, &c., contain the decimal parts of logarithms. The following table exhibits the logarithms and numbers between 100 and 130. When the given number contains four figure places, the 000000 •4321 8P«0 01 2837 7033 "5306 9384 033424 7426 041393 5323 9218 053078 6905 0GO698 '4458 SI 86 071882 5547 079181 082785 0360 9905 093422 6910 100371 3804 7210 110590 113043 ■3S0 4230 8046 1S29 5580 9298 2985 6640 •206 3861 7426 •963 4471 7951 1403 7071 •706 4451 8094 1707 5291 S845 5806 9335 2777 6191 9579 2940 6276 first three will be found in the first vertical column on the left, and the fourth must be sought at the top of the page, among the numbers be- tween and 9, inclusive. Thus, 121 may be found in the first vertical column on the left, and the last figure of 1215 will be found in the column marked 5 at top. The logarithm of 121 is written just oppo- site to it iu the column marked 0; 1210 will obviously have the same LOGARITHMS. 407 decimal part of its logaritlim, but a characteristic greater by unity ; the logarithm of 1215, on the other hand, must not only exceed the loga- rithm of 121 in characteristic, but also in decimal part, and its decimal part is then taken from the column marked 5. TO FIND THE LOGARITHM OF A GIVEN NUMBER BELOW 10,000. 424. When the number is less than 100, the decimal part of the logaritlim is found in the table just opposite the given number. The characteristic will be determined by the number of figure places in the number. For numbers above 100 and below 10,000, we have the following RULE. Loohfor the first three figures of the given number in the column marked N, and if it contains hut three, the decimal part of its loga- rithm icill he found Just ojijmsite these three figures in the column marked 0. But if there are only four figures in this column, opposite the given numher, the two left hand figures above, in the same column, where six figures aretcritten, must he prefixed to the four figures found. Thus, the decimal part of the logarithm of 110, is 041398, and the entire logarithm, 20i1SdS. The decimal 2mrt of the logarithm of 112 is Oid-218, the first tu-o figures being found opposite 110. The entire logarithm is then 2-049218. The logarithm of 117 is 2-06816, and of 120, 2079181. 425. When the given number contains four figures, look for the first three in the column marked N. The last four of the decimal part of its logarithm will be found just opposite, in the column marked at top by the fourth figure. To these four figures must be prefixed two deci- mals in the column marked zero, either found opposite the first three figures, or above, where six figures occur. Thus, to get the logarithm of 1213, we first find 121 in the first column, and then look just oppo- site, in the column marked 3, where we find 3801, and prefix to it 08, taken from the column marked 0. Hence, the logarithm of 1213 is 8-083861. The logarithm of 1283 is 3-108227, the two left hand figures of its decimal part being found above, where six figures occur. When, however, we pass over a decimal with a point prefixed, the first two figures are to be found below, and not above. Thus, the loga- rithm of 1234 is 3-091315. In like manner, the logarithm of 1231 is 3-090258. The place of the point is always supplied by zero. If 408 LOGARITHMS. the left hand figures were not taken below, the logarithm of 1231, 1232, &c., would be less than the logarithm of 1230, 1229, &c. EXAMPLES. 1. Required the logarithm of 129. Ans. 2-110590. 2. Required the logarithm of 1290. Ajis. 3410590. 3. Required the logarithm of 1297. A^is. 3-112940. 4. Required the logarithm of 1285. Ans. 3-108903. 5. Required the logarithm of 1233. A71S. 3-090963. 6. Required the logarithm of 1236. Ans. 3-092018. 7. Required the logarithm of 119. Ans. 2-075547. 8. Required the logarithm of 1191. A71S. 3-075912. TO FIND THE LOGAPJTIOI OF A NUMBER ABOVE 10,000. 426. Cut off on the right all the figures, except the first four of the highest denomination. Find the decimal part of these four left hand figures, as before, and prefix a characteristic, one less than the number of figure places in the given number previous to cutting off. jNIultiply the tabular difference by the figures cut off on the right, and from this product cut off as many places for decimals as there are figure places in the multiplier. Add the result to the logarithm before found. Thus, to find the logarithm of 129451, we cut off 51, and find the decimal part of 1294 to be 111934. To this we prefix a characteristic, 5, and we have 5-111934, which is truly the logarithm of 129400 (Art. 422). Next, we multiply the tabular difference, 335, by 51, and cut off 85 from the product, 17085 ; we would then have 170 to add to 5-111934 ; but, as -85 is greater than |, we increase 170 by 1, because 171 is nearer to 170-85 than is 170. Adding then 171, we have 5-112105 for the logarithm of 129451. In general, whenever the decimal cut off' in the product, which results from multiplying the tabular difference by the right hand figures of the given number, exceeds ^, the last figure of the entire part of the product must be increased by 1. The explanation of the process is simple. The tabular diffe- rence expresses the difference between the logarithms of consecutive LOGARITHMS. 409 numbers. Thus, the logarithm of 1293 is 3-1599 ; and of 1294 is 3-1934, and their difference, 335, is written in the column marked D. This column, it is plain, will moreover express the difference between the logarithms of numbers which differ by 10, 100, 1000, &c. "We have seen that the logarithm of 129400 was 5-11934, but we were required to find the logarithm of 129451, and it remains to be seen how much the logarithm of 129400 is to be increased. We employ the principle, that the difference between any two numbers is to the difference of their logarithms, as the difference between any two other numbers is to their difference of logarithm. Hence, 129400 — 129300 : 335 : : 129451 — 129400 : x, or 100 : 335 : : 51 : z, in which x represents the augment to the logarithm of 129400, to give the logarithm of EXAMPLES. 1. Required the logarithm of 1309958. Ans. 6-117259. In this example three places arc cut off for decimals, because we took the difference between 1309000 and 1308000. The first term of the proportion was then 1000. 2. Required the logarithm of 13080000. Ans. 7-116608. 3. Required the logarithm of 13087654. Aiis. 7-116862. TO FIND A NUMBER CORRESPONDING TO A GIVEN LOGARITHM. 427. When the decimal part of the logarithm can be found in the tables, the first three places of the required number will be found on the left, immediately opposite this logarithm, and in the column marked N. The fourth figure, which is to be annexed to the three already found, is to be taken from the number of the column in which the decimal part of the logarithm was written. Then point off, from the left of the number thus obtained, one more place for the entire part, than is indi- cated by the number of units in the characteristic. Thus, the number corresponding to the logarithm 2-107888 is 128-2, and to 1-107888 is 12-82. So, the number corresponding to the logarithm 3-088136 is 1225, and contains no decimal part. But, when the decimal part cannot be exactly found, take the next less logarithm from the tables, and subtract this from the given loga- 35 410 LOGARITHMS. ritlim. Annex two or more ciphers to this difference, and divide by tlic tabular difference. Annex the quotient to the number correspond- ing to the next less logarithm, and you will have the required number. Thus, let it be required to find the number corresponding tS the lo- garithm 4-1132800. The next less logarithm is 4-1132750, and the corresponding number 12980 ; annexing 4 ciphers to 50, the difference of logarithms, and dividing by 335, the tabular difference, we have a -7. 446. There are ten important principles, belonging to the subject of inequalities, which require to be demonstrated. 1. If the same quantity be added to, or subtracted from, the two mem- bers of an inequality, the resulting inequality will subsist in the same sense. 432 INEQUALITIES. For, let A ^ B, and m the difference between A and B, then, A = B + m. Now, add or subtract the same quantity, c, from both members, and there results Adbc = B + m=bc. Hence, of course, A ± c ^ B ± c, and the inequality subsists in the same sense. This principle enables us to transpose terms from one member of the inequality to the other. Take, x -\- 2 ^ a ; add — 2 to both members, and there results x^ a — 2. So, transposition is effected in inequalities, as in equations, by a change of sign. 2. If the two members of an inequality are multiplied by a positive quantity, the resulting inequality will subsist in the same sense. For, let A ^ B, and A = B + m. Multiply both members by c, and we get Ac = Be -f mc. Hence, Ac ^ Be. This principle serves to free the fractions, if any, of their denominators. Take -^ -r-'^l/. Clearing of fractions, there results 2x — a ^ A1j\ 3. If both members of an inequality are divided by a positive quan- tity, the resulting inequality will subsist in the same sense. For, let A ]> B, and m the difference between A and B. Then A = B + m. Dividing both members by c, there results, — = 1 . Hence, — ■>■ — . The last principle serves to clear the unknown quantity of its coeffi- cient when positive. Take, 2a; ^ a + 4&^, then, x ^ — . Li The three foregoing principles serve to solve inequalities, when the coefficient of the unknown quantity can be made positive. 4. If two inequalities, subsisting in the same sense, are added together, member by member, the resulting inequality will subsist in the same sense. For, let A > B, and C > D. Let m be the difference between A and B, and n the difference between C and D. Then, A = B + m, and C = I) -f n. Adding the two equations, member by member, there results A + C = B + D + m + «. Hence, A + C > B + D. 5. If an inequality be multiplied by a negative quantity, the result- ing inequality will subsist in a contrary sense. For, let A ^ B, or A = B + m. Multiply both members by — c, and there results — Ac = — Be — w, or m — Ac = — Be. Hence, INEQU..LITIES. 433 Ac being numerically greater than Be, we have, algebraically, — Ac <^ — Be. The foregoing principles enab'e us to eliminate between inequalities as between equalities, when we have two inequalities involving two un- known quantities whose signs are known, and that of one of them in one inequality contrary to that of the same unknown quantity in the other inequality. Take x + y > 12, Adding member by member we get x^ 10, and this substituted in the first equation, gives y > 2. Let x = 10 + m. Then, from the 2d equation i/ <^2 + m. The value of i/ is then fixed between the limits of 2 and 2 + m. 6. If two inequalities, subsisting in the same sen.-: •, are subtracted member by member, the resulting inequality may subsist in the same or a contrary sense. For, let A > B, and C > D. Then, A = B + w, and C = D + ». Subtracting member by member, there results A — C = B + m — (D + n); or. A — C = B — D + m — «. Now, if m ^ «, A — C will be equal to B — D increased by a posi- tive quantity, and, of course, will be greater than it. But, if 71 "^ m, A — C will not be equal to B — D, until it has been diminished by the diflference between n and m. In that case, A — C < B — D. Take, 8 > 4, m = 4, 6 > 3, n = 3. 8— 6>1. The resulting equality subsists in the same sense, since m > n. But, take 8 > 6, m = 2, 6 > 1, n = 5. 8 — 6 < 5. The resulting inequality subsists in a contrary sense, since n > m. 8. If both members of an inequality are essentially positive, they may be squared without altering the sense of the inequality. For, let A > B, or A = B -f m, then A^ = B^ + 2Bm + m^. Hence, A^ > B^ 37 2c 434 INEQUALITIES. 9. If the two members of an inequality have contrary signs, the re- sulting inequality, after squaring, may silhsist in the same or a contrary For, let A > — B, or A = — B + m. Then, A' = B^ — 2Bm + m^ If m^ > 2Bm, then, A^^B^; for, A'' is equal to B^, increased by the difference between m^ and 2Bm. But, if ni^ <^ 2B?k, then, A^ <^ B'^ by the difference between 2Bm and m^. Take, 3 > — 2 ; then, m =5, m^ = 25, and 2Bm = 20. Squaring, we have, 9 > 4, since m^ > 2Bm. But take, 3 > — 6 ; then m = 9, m^ = 81, and 2Bm = 108. Squaring, we get, 9 35 — 20 — 2m; or y > 15 — 2m. Suppose m = J, then. y < 151, and 7/ > 14. 4. A person desiring to give some money to 11 beggars, found that, to give them 3 cents apiece, would require more than twice as much money as he had about him, and that, to give them 2 cents apiece, would require more than the difference between 37 cents and the num- ber of cents about him. Required the number of cents he had. A71S. X <[ 16 J, and a; ^ 15; hence, x = 16. The nature of the problem makes the solution exact in this case, but it is very seldom so. 5. Find the negative limits of x in the inequalities, cc-f 3< — 5, a; — 3 > — 7. Ans. x<^ — 8, and x^ — 4. 6. Find the limit of the value of x in the inequality, ax+b -^ c -yd^fx. d—l — c Ans. Inienor limit, r— «+/ 436 GENERAL THEORY OF E(^UATIONS, GENEKAL THEORY OF EQUATIONS. 447. The general theory of equations has for its object, the investi- gation of properties common to equations of every degree and of every form. We will confine ourselves mainly to the examination of equations in- volving but one unknown quantity. Tke most general form of an equation of the m*'' degree with one unknown quantity, is »"• -f Px-"-' + Qx'"-^ + Rx"-'' -f Tie -f U = 0. The coefficient of the first term is plus unity, and the other coeffi- cients, positive or negative, entire or fractional, rational or irrational. A value has been defined to be that which, substituted for the un- known quantity, will make the two members equal to each other. Since, in the general equation of the m*'' degree, the second member is zero, a value substituted for x, must reduce the first member to zero also. Hence, in our discussion, we may define a value to be that which, sub- stituted for the unknown quantity in the equation, will reduce the first member to zero. GENEKAL PROPERTIES OF EQUATIONS. First Property. 448. If any quantity, a, be a value of x in the equation, a;" + Pa;""' + . . . Taj -f U = 0, the first member of this equation will be exactly divisible by a; — a. Fffr, let Q' be the quotient resulting from the division of the first member by a:; — a, and let R' be the remainder, if any, after division. We shall then have the identical equation, x"" -f Pa;""' -f Qx'"~^+ .... -f- Ta; -f U = Q' (a; — a) -f R'. But, for x = a, the first member is, by hypothesis, equal to zero, and the second member reduces to R'. Hence, we have = + R', and, therefore, R' = 0, and the division is exact. Second Property. 449. If the first member of an equation, of the form of a:" + Px"^' + Qx"-^ -t- . . . . Tx -f- U = 0, be divisible by x — a, then a will be a value in this e(juation. GENERAL THEORY OF EQUATIONS. 437 For, calling Q' the quotient, we will have x^ + Px""' -j- Qx""^ -f- . . . . Ta; + U = Q(x — a). Hence, then, Q'(.r — a) = 0. But, when the product of two factors is equal to zero, the equation can he satisfied by placing either factor equal to zero. We have then a right to place X — a = 0, from which there results x = a. Therefore, a is a value in the equation, Q'(aj — «) = 0, and, consequently, in the given equation. Corollary. 1st. It follows that, in order to ascertain whether any polynomial is divisible by x — a, we have only to substitute a for x, wherever x occurs, and see whether the polynomial reduces to zero. 2d. Hence, also, if a polynomial is divisible by x — a, a will be a value in the equation formed by placing the polynomial equal to zero. 3d. We may also diminish the degree of an equation, when we know one or more of its values, by dividing its first member successively by the binomial factors corresponding to these values. The division by each binomial factor will reduce the degree of the equation by unity. 4th. Numerical equations can frequently be solved by means of the first two properties. Literal equations can also be solved in the same way, but more rarely : we have only to ascertain what number or quan- tity will satisfy the equation ; then, by dividing out the binomial factor corresponding to this value, we will have a now equation of a degree lower by unity. A second value may be found frdm this equation, and the factor corresponding to it divided out. Thus, we may continue tho process until the given equation is reduced to one of the second degree, which can be solved by known rules. EXAMPLES. 1. Solve the equation, x^ — Gx-^ + 11^ — 6 = 0. We see that a: = 1 will satisfy this equation. Hence, x — 1 is a divisor. Dividing by x — 1, we get a quotient, x^ — 5x -f 6 = 0, by solving which we get a; = 2 and 3. Hence, the three values are 1, 2, and 3. 2. Solve the equation, x* — bx^ -f 4 = 0. We find +1 to be a value, and, dividing by a; — 1, we get a new equation, x^ -f x^ — 4x — 4 =, which is satisfied for x = — 1. Dividing by a; + 1, we get a new equation, x^ — 4 = 0, which gives 37* 438 GENERAL THEORY OF EQUATIONS. the two values, x= -\- 2, and — 2. The four values are then, + 1, — 1, H- 2, and — 2. 3. Solve the equation, x^ + Sx* — 5x' + 4cc + 12 — ISa;^ = 0. Ans. + 1, — 1, + 2, — 2, and — 3. 4. Solve the equation, x* — a^x^ — l^x^ + a^h'^ = 0. A71S. + a, — a, -f 5, — h. 5. Solve the equation, of — a^x'^ — I'^x^ + a^x^ + I'^x^ — x'^ + a^¥x — aW = 0. Ans. + a, — a, + b, — b, and + 1. 6. Solve the equation, sc^ + x^ — 4x — 4 = 0. Ans. + 2, —2, —1. Third Property. 449. Every equation, with one unknown quantity, has as many values for this unknown quantity as is denoted by the degree of the equation, and has no more. Let us assume the equation, a:;" + Pa;"""' + Qx""^ + . . . Ta; + U = 0, which is of the m*'' degree. It is to be shown that it has m values and no more. We will also assume that every equation has at least one value. Let a be one value, then x — a is a divisor. Let jpm-i _j_ p'-^m-z _f. Q'a;"-' + T'ic + U', represent the quotient of the division of the first member by cc — a. Then the given equation will assume the form, {x — a) (x"-' + P'x"""^ -f Q'x"""^ + &c.) = . (A), in which, the coefficients, P', Q', are different from the original coefficients, P, Q, &c. Now, since, in equation (A), we have the product of two factors equal to zero, we have a right to place either equal to zero, let us place a;-""' + Fx"-^' + Q'cc"-^ + &c. = (B). Equation (B) will also have one value ; suppose it negative and equal to — h. Then, by the first property, x — ( — Z)) = x ^- & will be a divisor. The first member of (B) can then be decomposled into two factors, one of which will be a; + &, and the other the quotient arising from the division of the first member by a; + &• Hence, (B) becomes (a; + h) (a;'"-^ + F'a;'"-^ + g'tt""-" + &c.) = (C). And (A) can be put under the form of {x — a) (x + I) (x'"-^ + T"x^-^ + Q"x"'-^ + &c) =0. It is plain that, by putting the second factor of (C) equal to zero, we will get another value, and, consequently, another divisor. By GENERAL THEORY OF EQUATIONS. 439 continuing this process, the degree of a; in the successive quotients will be diminished by unity each time, and, after m — 1 divisions, we will obtain a quotient of the first degree in x, from which, of course, one value of X will be found. "We supposed the first value positive, and the second negative, let us attribute the double sign, ± , to the remain- ing values, c, d, e, &c., since they may be either positive or negative. The first member of the given equation can then be decomposed into m binomial factors of the first degree in x, and we will have, a;" + Px"-' + Qx'"-' + +Tx + V= (X — a) (x + b) (xzpc-) (x^zd) &c. = 0. But, since each factor corresponds to a value, and since there are m divisors, or factors, there must be ?n. values. Scholium. 1st. Had we known that the first member of an equation of the wi*"" degree could be decomposed into m binomial factors of the first degree with respect to x, we could readily have shown that the equation must contain ??i values. For, it can be satisfied in m ways, by placing each of its 7)1 factors equal to zero, and each factor, so placed, will give a value. Moreover, since the equation can be satisfied only in m ways, it contains no more than m values. 2d, It does not follow that all the values must be difi"erent. Any number of them, even all of them, may be equal. Thus, the equation, a? — 2a; + 1 = 0, contains two values, each equal to + 1. The equa- tion, (x — a)"" = 0, contains m values, each equal to -j- a The equa- tion, (x — a)'°(x-{-by=0, contains m values, each equal to + a, and n values, each equal to — b. And so for other equations. 3d. An equation of the third degree, such as (x — a) (x — b) (x — c) = 0, will contain three divisors of the first degree, {x — a), (x — b), and (a; — c) ; three of the second degree, (a; — a) (x — &), (x — a) (x — c), and (x — h) (x — c) ; and one of the third, (x — a) (x — b) (x — c). These divisors are evidently equal to the number of combi- nations which can be formed by combining 3 letters in sets of 1 and 1, 2 and 2, 3 and 3. So, likewise, if we take the m factors of an equation and multiply them two and two, three and three, &c., we shall evidently obtain as many divisors of the second degree as we can form combina- tions of VI letters, taken two and two ; and as many divisors of the third degree as there are combinations of ??i letters taken three and three. 440 GENERAL THEORY OF EQUATIONS. &c. The given equation will then have m divisors of the first degree, m(m — 1) m(m — 1) (m — 2) in X, ■ — — - — divisors oi the second degree, — ^= ^ — ^-^ of the third degree, &c. Fourth Property. 451. All the coefficients, after the first, of an equation of the m*'' degree, are functions of the values. Suppose the general equation of the ?n*'' degree, a;" + Pa:;"'-' -f Qaj"'-^ + .,..+ Tec + U = 0, contains the m values, ± a, d= 6, ± c, ± c7, &c. The equation can then be put under the form of (x rp a) (x =F 6) (x =F c) (x =fi d), &c., to m factors = 0. By actually per- forming the indicated multiplication, we will get. =Fa x»-' X ah =F& Xac qpc Xad z^d t&c &c. Xhc &c. ' zp aJc qr ahd zp ac(^ =p &c. x""-^ =p ahcd, &c. = 0. The upper row of signs belonging to the positive value, and the lower to the negative. We observe the following relation between the coefficients and the values : 1. The coefficient of the second term is the sum of all the values of the unknown quantity, with their signs changed. 2. The coefficient of the third term is the sum of the products of all the values, taken two and two, with their respective signs. 3. The coefficient of the fourth term is the sum of the products of all the values, taken three and three, with their signs changed. 4. The last term is the continued product of all the values of the unknown quantity, with their respective signs if the degree of the equation be even, or with their signs changed if it be odd. We have supposed in the preceding demonstration that all the values were positive or all negative. But the law of formation for the coeffi- cients is evidently the same when some of the values are positive and some nejrative. GENERAL THEORY OF EQUATIONS. 441 Thus, let tte values be + a, + h, and — c. Then, (x — a) (x — h) (x4-c)=0, orx^ — a \x^-\-a'b — 5 — ac + c\ — he X + ahc = 0. And we see that coefficients are formed in accordance with the above 452. The preceding properties show several important things in relation to the composition of an equation. 1. If the coefficient of the second term of an equation be zero, and, consequently, the second term be wanting, the sum of the positive values must be equal to the sum of the negative values. 2. If the signs of the terms of an equation be all positive, the values must be all negative. For, an equation cannot have all its terms posi- tive unless its binomial factors are of the form, (x + a) (x + h) {x + c), &c. ; which factors, placed equal to zero, will give the negative values, — a, — h, — c, &c. 3. If the signs of the terms of an equation be alternately + and — , the values of the unknown quantity will be all positive. For, in this case, the first member of the equation must be made up of the factors, (x — a) (x — U) (x — c), &c., corresponding to positive values. 4. Since the last term, irrespective of its sign, is the continued pro- duct of the values, it follows that, when the last term is zero, one value, at least, must be zero. 5. It follows, also, from the composition of the last term, that, in seeking for a value, we need only seek among the divisors of the last term, since every value must be a divisor of that term. 6. If we know one value, the coefficient of the second term will give the sum of all the rest, and the last term, divided by this value, with or without its sign changed, will give the product of all the rest. Corollary. 453. 1st. The last two consequences enable us to solve equations with facility when all the values except two are known. Take, as an example, the equation, x^ — Ix'^ — |x + | = 0. The values must be sought among the divisors of the last term : + 1 is a divisor of the last term, and may, therefore, be a value ; upon 442 GENERAL THEORY OP EQUATIONS. trial, we find that it will satisfy the equation, and is, therefore, a value. Calling the other two values x and y, we will have, — (1 -\- x -}- 1/) = — I, the coefficient of the second term. Hence, x + y = — ^. We xy have, also (sixth consequence), -f = — i- Combining these two equations, we get x ( — i — a;) = — ^, or cc^ + |x = + |. And the values of x and ^ are found to be + ^, and — 1. The equation has, of course, but three values, since it is an equation of the third degree, (Art. 450). 2d. When all the values are known except two, we can tell whether these values are real or imaginary without actually finding them. Thus, take the equation, x* — 6x^ — IQx + 21 = 0. The exact divisors of the last term are +1, +3, and +7, — 1, — 3, and — 7 ; the values must be sought among these numbers. We find, upon trial, that + 1 and + 3 are values. The second term of the equation being wanting, the sum of the other two values must 21 be — 4, and their product :j — - = 7. But 4 is the greatest product which can be given by two numbers whose sum is 4. Hence, the other two values of the equation are imaginary. And, in fact, by pur- suing the preceding process, we find them to be — 2 -\-y/ — 3, and _ 2— ^ITS- EXAMPLES. 1. Solve the equation, x^ — 2ax^ — da'x + ISa^ = 0. Ans. 4- 2a -f 3a, and — 3a. 2. Solve the equation, x" — 4x^ + x — 4 = 0. Ans. + 4, + v/ — 1, and — ^ — 1. 3. Solve the equation, x* — 2x' — S = 0. Ans. + 2, — 2, + ^/■=^, —^':^^. 454. 3d. If the values are known, the equations which give those values can be formed in two ways, either by multiplying together the binomial factors corresponding to those values (Art. 450), or by form- ing the coefficients, (Art. 451). GENERAL THEORY OF EQUATIONS. 443 EXAMPLES. 1. Form tte equation whose values are + 1, — 1 and — 2. Ans. x'' + 2x^ — x — 2 = 0. For the factors, multiplied together, and placed equal to zero, give (x — 1) (x + 1) (a; + 2) = (A) ov x'' + 2x'' — x — 2 =^ 0. We see that the three factors of (A), placed separately equal to zero, will give the values, + 1, — 1, and — 2, or the problem may be solved by Art. 450. Coefficient of 2d term = — (+1 — 1 — 2)= + 2 3d term = (+1) (-l)+(+l) (-2)+(-l) (-2)=-l « 4thterm=— (4- 1) (— 1) (— 2) = — 2. 2. Form, by both methods, the equation whose values are + 4, + ^^^, and — v/— 1. Tins. ar' — 4x^+x — 4 = 0. 3. Form, by both methods, the equations whose values are + 2, — 2, + ^^-2, — ^^^. Ans. x* — 2x^ — S = 0. 4. Form, by both methods, the equation whose values are — 1, — 2, — 3, and — 4. Ans. x* + lOx'' + 35x2 + 50x + 24 = 0. Fifth Property. 455. Every equation may be transformed into another, in which the values of the unknown quantity shall be equal to those of the proposed equation, increased or diminished by a certain quantity. Let the given equation be ' x" + Px»-' + Qx"' + + Tx + U = 0, and let it be proposed to transform it into another in y, so that y = x± a. The principle of transformation is, of course, the same when y = x + o, or = x — a ; we will then confine our discussion to the case, when y =■ x — a. From this equation there results, x — y + <^- Substituting this value in the proposed equation, it becomes {jy + a)"* + P (^ + a)-' + Q(y + ay-^ + T (// + a) + U = 0. Developing the different powers of y + a by the binomial fonnula, and arranging the development according to the descending powers of y, we have the transformed equation, 44 GENERAL THEORY OF EQUATIONS. 2/° + mo + P „_, in(m — 1) a^ ^ 1.2 + «!, (m — 1) Pa + Q y"'-^ . . . . + a" + Pa'°- + Qan-= 4-Ta + u = 0. If the values of y in this equation can be found, the corresponding values of x in the given equation will be equal to these values in- creased by a. Corollary. 456. The preceding transformation enables us to reduce the number of terms in an equation. For, since the quantity, a, is entirely arbi- trary, such a value may be given to it as will reduce the coefficient of any term in the transformed equation to zero, and consequently make the term itself disappear. Suppose we wish to free the transformed equation of its second term; we have only to place the coefficient of that term equal to zero, and find the value of the quantity, a, that makes it zero. Of course, then, this value attributed to a, will cause the dis- appearance of the second term of the equation. Placing ma -f P = 0, P P we get a = . Then, x i= y . Hence, for transforming one equation into another, in which the second term is wanting, the following RULE. Substitute for the unknown quantity a new unknown quantity, con- nected with the quotient arising from dividing the coeffi/^ient of the second term of the given equation, with its sign changed hy the degree of the equation. EXAMPLES. 1. Transform the equation, x^ — ^x = 5, into another, in which the second term shall be wanting. Ans. x = y — ( — |) = y + 2. The transformed equation will then be {y -\-Tf — 4 (y -f 2) = 5, or y — 4 = 5. Hence, y'^ = 9, and y = =1= 3 ; and, cc = y -j- 2 = 5, or — 1 ; the same result that we would obtain by solving the given equation. GENERAL THEORY OF EQUATIONS. 445 2. Transform the equation, x^ — Zx^ = — 2, into another, in •which the second term shall be wanting. Ans. y^ — 3y = 0. The transformed equation can be readily solved, one value of y being zero. Hence, one value of the given equation is 1. The transformed equation, in this case and in many others, is simpler than the given equation, and, therefore, more readily solved. The chief object of the transformation is to simplify the form of the equation. Scholium. 457. 1. The third term of the transformed equation may be made to disappear by giving to a such a value as will satisfy the equation, — ~ ~ — + (m — 1) Pa + Q = 0. And, since this equation is of the second degree in a, there are two values for a, and, con- sequently, the third term can be made to disappear in two ways. In like manner, the fourth term can be made to disappear in three ways, the fifth term in four ways, &c. j and the last, or {in + l)**" term, in m ways. By recurrence to the derivations of these coefficients from the values, we see that the above results are true. The last term, for in- stance, being made up of the product of the m values, can be made to disappear by placing either of the m values equal to zero. 458. 2. It may happen that the value of a, which makes the second term disappear, will also cause the disappearance of the third or some other term at the same time. Let us examine under what circumstan- ces the second and third terms will disappear together. By placing the coefficient of the second term, Pa + m, equal to zero, we get a = , and, by placing that of the third term, ^ — ^;~— -f (m — 1) P^ /P 2Q P /P^ Pa + Q, also equal to zero, we get « ^ d= \ / — m \ m m(vi — 1) Now, it is plain that the values of a in this equation will be identical P with the last, that is, both , when the radical disappears. Placing 2mQ the radical equal to zero, we get P^ ^ — -. "Whenever, then, the m — 1 square of the coefficient of the second term is equal to twice the quo- tient arising from dividing the product of the desree of the equation 38 446 GENERAL THEORY OF EQUATIONS. into tte coefficient of the third term by the degree of the equation, less one, the second and third terms will disappear together. The same truth may be demonstrated more elegantly by the principle of the greatest common divisor. Dividing the coefficient of the third term by that of the second, we get, after two divisions, a remainder, mQ + ^^ ^ • And, it is evi- dent that, when this remainder is placed equal to zero, we will have made manifest the circumstances under which there will be a common factor between the two coefficients. Placing the remainder equal to zero, we get P'^ = zr, as before EXAMPLES. 1. Make the second and third terms disappear from the equation, x^ — Bx^ + Sx — 28 = 0, and find one value of x. Ans. Transformed equation, ^^ — 27 = 0, and a; = 4. 2. Find two values in the equation, x* -f 4a;' -f 6x^ -\-4x — 15 = 0. Ans. Transformed equation, i/* — 16 = 0, then, y = ± 2, and x = + 1, or — 3. Sixth Property. 459. Every equation having the coefficient of the first term plus unity, and the other coefficients entire, will have entire numbers only for its rational values. Let the proposed equation be ^m + p^m-1 _|_ Qa.m-2 + Tx + U = 0. In which, P, Q, &c., are whole numbers. If X can have a fractional value in this equation, let this value, re- duced to its lowest terms, be — . Substituting this value for x in the given equation, it becomes Multiplying by h^~\ and transposing, we get ^ = — Pa"-' — Qa^-^i — TaS^-^ — U6"-». GENERAL THEORY OF EQUATIONS. 447 But tlie second member is an entire quantity, since all its terms are entire. Hence, the first member must be entire. But, since a, by hypothesis, is prime with respect to &, a" must also be prime with re- spect to b. The supposition of the proposed equation containing a ra- tional fractional value, has then resulted in the absurdity of making an entire quantity equal to an irreducible fraction. "We conclude, there- fore, that this supposition is wrong, and that the rational values are all entire. The demonstration is restricted to rational values, because the assumed value, -^, is rational in its form. CoroEari/. 460. Articles 452, 456 and 459, are used in solving numerical and literal equations by changing their forms. Let the proposed equation be y* — -ii/^ — 8?/ + 32 = 0. "We know that all the rational values are entire (Art. 459), and that they must be found among the divisors of the last term (Art. 452). But as there are so many divisors of the last term, it will be more convenient to employ (Art. 456) to transform the equation into another, whose last term admits of fewer divisors. Make re =y — (—j-\ = i/-\-l. The transformed equation in x, is x* — 6.7=^ — IGx + 21 := 0, and the divisors of the last term, are -}- 1, + 3, + 7, + 21, and — 1, — 3, — 7, — 21. On trial of these divisors, wc find that -f 1 and + 3, will satisfy the transformed equation, and, consequently, are values. Hence, cc = 2/ + 1 = 2 and 4, in the given equation. Dividing the first member of the given equation by (x — 2) (Art. 449), it will be reduced to a cubic equation, and again dividing by (x — 4), it will be reduced to a quadratic, which can be solved. Seventh Projierii/. 461. Imaginary values enter equations hi/ pairs. We are to show that if a-\-hy/ — 1 is a value in the equation, x" -f Pa;"-' + Qx'"-^ + Tx + U = 0, a — h^'^^ will also be a value. The truth of the proposition is an evident consequence of Art. 451, for U, the last term, is the product of all the values, and it has been assumed real. But it can only be real, when the imaginary values (if 448 GENERAL THEORY OF EQUATIONS, any), enter by pairs, and of such a form as to give a real product when multiplied together. Thus, if we have two imaginary values, m + v/ — n and ^ — 4, we must also have two other values, m — ^ — n and — v/ — 4, otherwise, U will not be real. Or, we may demonstrate the property otherwise, thus : by substituting the assumed value, a + h^ — 1, in the given equation, we will have, (a -f l-y — 1)" + P (a + i^/— I)"-' + T (a + hs/^^) + U = 0. When these terms are expanded, the odd powers of t v' — 1 will be imaginary, and the even, real. Representing by A all the real terms involving a or h, and by B^-^ 1 aU the imaginary terms, we will have A + B v' — 1 = 0. Now, since a + h^ — 1 is, by hypothesis, a value, the last equation must be satisfied, but this can only be so when A = 0, and Bv/ — 1 = 0, since imaginary terms cannot be cancelled by real ones. If « — h^ — 1 be substituted in the equation, the expanded results will be precisely the same, except that the odd powers of hy/ — 1 will be negative. Hence, we will have A — Bv^ — 1 = 0. But in order that a -f h^ — 1 should be a value, we have seen that A = and B = 0. Hence, the equation, A — B v' — 1 =; 0, is satisfied, and that being so, a — h-y — 1 must be a value. The absurdity of the hypothesis of a single imaginary value may be illustrated by an example. Let us assume that the three values of an equation of the third degree are + a, +1, and + v/ — 1. The equation must then be (x — a)Xx — 1) (x — vA=n:) = 0, or (x — a) (x" — (1 — v/=l) 35 + >/ — 1) = 0. Hence, a? — a = 0, andcc'' — (l + \/ — V)x-\-y/ — 1 = 0. The second equation, when solved, will give x = ~ \/-^..-r+ i.2^-i-i ,L±v^i^^/. !^/-l 2 And we se e that we do not get back again the two values, + 1, and +V-1. Remarks. 462. 1. The product of imaginary values is always positive, and, therefore, the absolute term of an equation, whose values are all imao-i- nary, must be positive. 2. If the second term of an equation containing only imao-inary values, is wanting, these values will all be of the form, ±v'— ^• 3. Every equation of an odd degree has at least one .real value forr, GENERAL THEORY OP EQUATIONS. 449 and the sign of this value will be contrary to that of the last term of the equation. 4. Every equation of an even degree, whose last term is negative, has at least two real values for x ; one positive, and one negative. EXAMPLES. 1. One value of cc is 4 + ^ — 10, what must a second value be? Anz. 4— ^=IOr 2. Two of the values of an equation of the fourth degree are + a, and — hj and the last term is — m. How are the remaining values ? Ans. Real. Eighth Property. 463. If the real values of an equation, taken in the order of their magnitudes, be n, h, c, d and e ; a being ^6, h'^ c, &c. ; then, if a series of quantities, a', h', c', d', &c. ; a' taken greater than a, V be- tween a and h, c' between h and c ; be substituted for x in the pro- posed equation, the results will be alternately positive and negative. For, since a, &, c, d, &c., are assumed to be values, the proposed equation can be put under the form of (x — a) (x — h) (x — c) (x — d) &c. = 0. Substituting, for x, the proposed series of quantities, a', h', d, d', &c., we get the following results. (of — a) (a' — b) (a' — c) (a' — d) &c, = -{- result, since all the factors are positive. (6' — a) (b' — I) (V — r) (U — d) kc, = — result, since first factor only is negative. (c' — a) (d — h) (d — c) (c' — d) &,c., = + result, since first two factors only are negative. (df — a) ((Z' — 5) (d' — c) (d' — d) &c., = — result, since first three factors only are negative Remarks. 1. We see that, between the quantities, a' and h', which give the first plus result and the first minus result, there is one value, a. And be- tween the quantities, a' and d', which give the first plus result and the second minus result, there are three values, a, h and c. And, as the 38* 2d 450 GENERAL THEORY OF EQUATIONS. same is evidently true for any number of odd values, we conclude that, if two quantities be successively substituted for x in an equation, and give results, with different signs, between these quantities, there will be one, three, five, or some odd numbers of real value. 2. If any quantity, m, and every quantity greater than m, give re- sults all positive, then m is greater than the greatest value of x in the equation. 3. If the results obtained by substituting two quantities have the same sign, then, between these quantities there are two, four, six, or some even number of values, or no value at all. Thus, between a' and c', which give + results, there are two values, a and h. Scliolium. 464. The preceding property may be demonstrated differently, by employing a principle of frequent application in all branches of mathe- matics, viz., that a quantity changes its sign in passing through zero and infinity. Let the quantity be x = a ; suppose x ^ a, the expres- sion is positive ; it is zero when x = a, and negative when x<^a. Hence, the quantity x — «, changes its sign in passing through zero. + M Again, take , This quantity is positive when x'^a ; it is in- finite when X = a ; and negative when x <^a. Now, in the results of Art. 463, between the first + result and the first — result, the first member of the proposed equation must have passed through zero, and consequently through a value. In like manner, between the first — result, and the second + result, the first member must again have passed through zero, and consequently through a second value. Then two values must have been passed through between the first two + re- sults, &c. By continuing the same course of reasoning, we might readily show that three values must be passed through between the first positive and second negative result; five values between the first posi- tive and third negative result; two values between the first two negative results; four between the first and third negative results, &c., &c., Limit of the Values. — Ninth Property. 465. If we substitute for the unknown quantity the greatest coeffi- cient plus unity, the first term of the equation will be greater than the sum of all the other terms, provided we always affect the greatest co- efficient with the positive sign. GENERAL THEORY OF EQUATIONS. -151 Let us resume the equation, x^ + Prr-"-' + Qx"-^ + Ta; + U = 0. It is plain that the most unfavourable case will be when all the co- efficients have the same sign, -f or — . We will assume all the coeffi- cients after the first to be negative ; the given equation will then be of the form, x'^ — Tx^-' — Px"-^ — ¥x'^-^— P:c»-^ — P = 0, or a;" _ P(a;"-' — x""'' — x"'"^ — 1) = 0. Disregarding the sign of P for a moment, we wish to ascertain what value X must have, in order that cc" ^ P(x°~' -f- cc"^^ + a;"~* . . . -f x + 1). Inverting the order of the terms within the parenthesis, we will have a geometrical progression, whereof, 1 is the first term, x°^' the last term, and x the ratio of the progression. The formula, S' = — , will then give the value of the quantity within the parenthesis, ^ a;" — 1 _ , ^ ^ /:,•■»— K ^ Px" P (7-m |v rX ^_^. ....... ,..^..... ^.^^_^^,... ^x — 1 x — r Px" This inequality will evidently be true, when x" = -. Dividing both members of this equation by a;", and clearing of fractions, we find a; = P + 1, as enunciated. It is to be observed that, in this demonstration, we assume a; ^ 1. EXAMPLES, 1. What number substituted for x in the equation, x^ — 7x* -f Gx'^ -f 6x^ -f X = 0, will make the first term greater than the sum of all the other terms ? A71S. 8. 2. What in the equation, x*— lOa;^ + 12x + 13 = ? Ans. 14. 3. What quantity in the equation, x^ — 2aa;^ -f- 6ax — 12a = 0. Ans. 1 4- 12a. JScJiolium. In seeking for the positive values of an equation, we need not go beyond the greatest coefficient with a positive sign prefixed to it, plus unity. Thus, in example first, there is no positive value greater than 8, because 8, and all numbers greater than 8, will continually make the first member positive (Art. 463). 452 GENERAL THEORY OF EQUATIONS, Second Limit. — Tenth Property. 466. If we substitute for x in the equation, x" + Pa:"-' + Qx"-^ . . . , Ta; + U = 0, unity, increased by that root of the greatest nega- tive coefficient which is indicated by the number of terms preceding the first negative term, we will have a superior limit of the positive values. Suppose Nx"""" to be the first negative term, and suppose W to be the greatest negative coefficient. The most unfavourable case obviously is that in which we suppose all the coefficients after N, negative, and equal to W. It is plain, moreover, that any value of x which makes a." ^ W(x'"-° + x"-"-' + «+!), will, of course, make the first member of the given equation positive. We are to find the value which fulfils this condition. Resuming the inequality, x" ^ W(a;"-° + a"-"-' + ic""-""^ .... ^ TTT /*"-"+' — 1\ ^ W£c"-'+' -j- X + 1), we have, also, x" > W ( z — ), or x" > -— ' "^ \ X — 1/ -^x — 1 W .... 'Wx'"~°+' -. And this inequality will be true, when x" = 1 W From which, x — 1 = —^^ , or (x — l)x"~' = W. Now, if (x — 1) (x — 1)°-' = W, then {x — l)x"-' will be greater than W. If, how- ever, we place it equal to W, we will have (a; — 1) (x — 1)"-' = W, or (x — 1)° = W. Hence, x = 1 + y W, which agrees vsdth the enunciation. It will be seen that in two respects we have taken an unfavorable case. The second limit t en may considerably exceed the greatest positive value, but it is smaller than the first limit, and, therefore, the most used in practice. EXAMPLES. 1. Required superior positive limit of the values in the equation, x« -f 2x* — Sx" + 7x=^ — 17x -F 5 = 0. Am. 1 -^JW. 2. Required superior positive limit of the values in the equation, a;< + 4x» -J- 12x2— 5x — 16 = 0. Am. 1 + yle: GENERAL THEORY OP EQUATIONS. 453 Eleventh Property. 467. The equation, x"" + Pa;'"-'+ Qa;'"-^ + + Ta; + U = 0, can be transformed into another in y, or some other variable, such, that the values of the new unknown quantity shall be some multiple of those of the old unknown quantity. For, let y = nx, then x = —. Substituting for x, wherever it occurs 'in the given equation, its value in terras of y, we have -- + P ^^ri + Q^V....T^^+ U = 0, or y- + 7iFy-i + n'Qy-' w^-'Ty + n"!! = 0. And we sec that the transformation is effected by changing x into y, multiplying the coefficient of the second term by the multiple n, that of the third term by n^, &c. For instance, let it be required to transform the equation, x* + 2x' + 4x^ + 8ar -f- 1 = 0, into another, in which the values of the new unknown quantity shall be twice as great as those of the old. Then y ^ 2x, and we have y* + 4/+16/+64y + 16 = 0. RemarJcs. The principal use of this transformation is in clearing an equation of fractional coefficients, and at the same time making the new equa- tion preserve the proposed form ; that is, the coefficient of the first term Pa;""' is still to be plus unity. Let us take the equation, as™ -\ 1- Qx"-' . Kx"-' Tx U ^ . , . , — ■ H ^ 0, in which we assume mn to be n mn n n the least common multiple of the denominators. Then y = mnx. The transformed equation will then be ^-^^ — I h^— — ;-! ~ 5 T» m-3 Tl TT m^b^Ry'"-^ .... m"-'w"'-'^Ty + m^Ti^-'U = 0. And we see that the transformation is effected by changing x into y, and multiplying the second term by the first power of the least common multiple, the third term by the second power of that multiple, &c. Thus, the transformed equation of x^ + |- — | + 1 = 0, is y' + 4/ — 36y + 1728 = 0. 454 GENERAL THEORY OE EQUATIONS. 468. The two rules just given are, of course, only applicable when the coefficient of the highest power of x is plus unity. When that is not the case, we may either first make this coefficient plus unity, and apply one of the preceding rules, or we may at once make y = nx, n in this case representing the product of the least common multiple of the denominators by the coefficient of the highest power of x. EXAMPLES. 1. Transform the equation, x^ — 4x' + 2x -f 2 = 0, into another, in which the values shall be twice as great as in the given equation. Ans. y — Stf + 8y + 16 = 0. 2. Transform the equation, x* — ox'^ + 2a;^ + a? + 1 = 0, into an- other, whose values are treble those of x. A71S. / — 9if -f 18/ + 27y + 81 = 0. 3. Transform the equation, x'^ — "9- + -vr + k + 1 = 0, into an other, containing only entire coefficients. Ans. y — 15/ + 300/ + 5400?/ + 810000 = 0. 4. Transform the equation, 2x^ — "5"+ p + 1 = 0, into another, o O whose coefficients shall all be entire. Ans. 2/ — 2/ 4- Gy 4- 216 = 0. 5. Transform the equation, 2x^ — "o" + F "^ 1 = 0, into another, o o whose coefficients shall all be entire, and the coefficient of the first term, plus unity. Ans. / — 2/ + 12y + 864 = 0. In this, make y = 12a:;. 469. When the coefficient of the first term is some whole number different from unity, and the other coefficients are entire, make n, in the equation y = nx, equal to the coefficient of the first term. Thus, to transform the equation, Qx^ — 19^;^+ 19x — 6 = 0, make y = Qx, the transformed equation will be / — 19/ + 114y — 216 = 0. Twelfth Properti/. — Process of Divisors. 470, In every equation in which the coefficient of the first term is unity, and all the other coefficients are entire, a value will divide the last term, the sum arising from adding the quotient so obtained to the GENERAL TS'EORY OF EQUATIONS. 455 coefficient of the second term from the right, the sum arising from adding this second quotient to the coefficient of the third term from the right, and will thus continue to be an exact divisor of all the successive sums so formed, until the coefficient of the second term from the left has been added to the previous quotient, when the last quotient will be minus unity. Let us resume the equation, x^ + Px"-' + Qx"'-^ + Ta; + U = 0, in which all the coefficients are entire. Suppose a to be a value, the equation will then become, after transposition and division by «, a"~' + Pa""-^ + Qa°~^ T = . Now, since the first member is made up of entire terms, the second member must be entire also. Hence, U is divisible by a, as it ought to be, since U is the product of all the values. Transposing T to the second member, we have a"—' + P(j°'-2 ^ Qa'o-s + &c. = — T = U'; U' representing the alge- braic sum of T and — . a Dividing both members again by a, we get the equation, a""" + Pa"-' + Qa"-^ + &c. = . a The first member being entire, the second member must be entire also. Hence, the second condition to be fulfilled by a value is, that it must be an exact divisor of the sum formed by adding the quotient of the last term by the value to the coefficient of the first power of x. This was also to have been anticipated, because the coefficient of the first power of x is made up of the values, taken m — 1 and m -^ 1 ; one of these combinations will not contain the value a, and will have a sign contrary to the first quotient, and will be cancelled by it. The re- maining terms making up this coefficient all contain a, hence, a will be a divisor. By transposing the terms in succession, and continuing the . . . P' division, we will find, after m transpositions and divisions, — = — 1 ; P' representing the sum of the coefficients of x""-' added to the pre- ceding quotient. Hence, the last condition to be fulfilled by a value is, that it must be an exact divisor of the sum formed by adding the coefficient of x"-' to the preceding quotient. 456 GENERAL THEORY OF EQUATIONS. We, therefore, have the following rule for finding the rational values of an equation. RULE. I. Transform the equation, if necessary, into another, in lohich all the coefjicients shall he entire; that of the first term heing plus unity. All the rational values will then he entire (Art. 459). II. Find the superior positive and superior negative limits of the values. III. Write down in succession, in the same horizontal line, all the entire divisors of the last term hetween zero and these limits. IV. Divide the last term hy each of these divisors, and write the re- spective quotients hcneath the corresjyonding divisors. V. Add the coefiicient of the first power of x to each of these quo- tients, and tvrite each sum thus formed heneath the corresponding quotient. VI. Divide these sums hy the corresponding divisors in the column of divisors, and write the new quotients under the corresponding sums, rejecting those divisors lohich give fractional quotients, and crossing out the vertical column in which they occur. VII. Proceed in this way until the coefficient of x""' has heen added, to the preceding quotient, then those divisors that give minus unity for quotients, are values, and they are the only rational values. EXAMPLES. a;3 _ 2a;2 — 5x + 6 = 0. The superior positive limit is 1 +^5 = 6. The superior negative limit is found by changing + x into — x; the superior positive limit of the transformed equation will then be the superior negative limit of the given equation. Changing + x into — x, we get — x^ — 2x^ -f- 5a; + 6 = 0. But the coefficient of the first term must always be plus unity, hence, by multiplying the equation by minus unity we have, x^ + 2x^ — bx — 6 = 0. _ Then — (1 + v'e) = superior negative limit. Assume — 4 to be this limit, the square root of 6 being between 2 and 3, we take 3 to be the root, since it is better to have the hmit too great than too small. Hence, we have these divisors below the limits, +1, + 2, + 3 ; — 1, -2,-3. GENERAL THEORY OF EQUATIONS. Positire. Nesative. Divisors, + 1, + 2, + 3, - 1,-2, Quotients, + 6, + 3, + 2, - 6,-3, Sums, +1,-2,-3, -11,-8, Quotients, +1,-1,-1, + 11, + 4, Sums, —1,-3,-3, + 9, + 2, Quotients, — 1, x , — 1- - 9,-1. 457 Positive. Divisors, + 1, 2, +4, + 8, Quotients, — 8, -4,-2, — 1, Sums, — 8, -4,-2, — 1, Quotients, — 8, — 2, X, X, Sums, — 10, — 4, X, X, Quotients, — 10, -2, X, X, Sums, — 10, — 2, X, X , Quotients, — 10, — 1, X, X, Hence, we have + 1 , + 3, and — 2, for the three values of the given equation. Let us take, as a second example, the equation, x* — 2x^ — 8 = 0. Superior positive limit, = + 9 ; superior negative limit, = — 9. The equation may be written x* ± Ox'* — 2x^ ±0x — 8 = 0. NegatiTe. — 1,-2,-4,-8, + 8, + 4, + 2, + 1, + 8, +4, +2, +1, -8,-2, X, X, — 10,-4, X, X, + 10, + 2, X , X , + 10, +2, X, X, — 10,-1, X, X, Hence, + 2 and — 2 are values in the given equation. And by dividing the first member of the given equation by the factors (x — 2) and (x + 2) corresponding to these values, we obtain the quotient, x^ + 2. By placing this quotient equal to zero, the remaining two values of the equation will be found to be + ^Z — 2, and — ^ — 2. 471. The process of divisors is applicable to literal, as well as nu- merical equations, when all the coefficients are entire. A literal equa- tion, too, will best show why it is that the successive sums are divisible by the values. Take, as an example, x^ — (a +6 + c)x'^ + (o6 + ac+ hc^x — ahc = 0. Since the terms are alternately plus and minus, the values must all be positive. 39 458 GENERAL THEORY OF EQUATIONS. Divisors, + a, -\- b, -{- c. Quotients, — he, — oc, — ah. Sums, ah -j- ac, ah + he, ac + he. Quotients, h -^ e, a -{- c, a -\- b. Sums, — a, — h, — c. Quotients, — 1, — 1, — 1. Hence, the three values are -\- a, + h, -\- c. We will take, as a second example in literal equations, a;3 4- (a — 6 -f 2')x^ + {2(a — h') — ah)x — 2al = 0. Positive. Divisors, + o, + &, + 2, Quotients, — 2h, — 2a, — ah, Sums, (2a— ha — 4&), — h(a + 2), 2(a — h — ah), Quotients, X, — (a-\-2,)(a — h — ah}, Sums, X,—h, 2(a+l)—h(a. + l), Quotients, X , — 1, x. Negative. - «,— ^,— 2, + 2h, + 2a, + ah, a(2 — &),4a — &(2 + a),2(a — S), + (h-2),X, + a, X , + 2, — 1, X,— 1. (a -6), Hence, the three values are + h, — a and — 2. The process of divisors being of such high practical importance, we will give another example, in which one of the sums is zero. Take the equation, x'' — x" —ix + 4 = 0. Superior positive limit, = 5 ; superior negative limit, = — 3. Tositive. Divisors, + 1, + 2, +4, Quotients, + 4, -f 2, + 1, Sums, 0, — 2, — 3, Quotients, 0, — 1, X , Sums, — 1, — 2, X, Quotients, — 1, — 1, X , Negative. -1,-2, -4,-2, -8,-6, + 8, + 3, + 7, + 2, -7,-1. Hence, the three values are + 1, + 2, and — 2. 472. The process of divisors enables us to find all the rational values of any equation. We have only to make all the coefl&cients entire, if not already so, that of the highest power of x being made plus unity. All the rational values of y in the transformed equation will be entire ; find these values, and then those of x will be known from the relation between x and y. Take 6x' — Idx^ + 19x — G = 0. Making ^ = Qx, we have the GENERAL THEORY OF EQUATIONS. 459 transformed equation in y, ^ — 19j/^ + 114y — 216 = 0. By pi*o- ceeding as before, we find the three values of y to be + 9, + 6, and + 4. Hence, the corresponding values of a; are | + 1 and + |. 473. In this, and in many examples, the last term has a great num- ber of exact divisors, and the process is therefore tedious. It is often convenient to transform an equation, whose last term is too large, into another whose values shall be greater or smaller by a constant quantity. Making y = z + 2, in the preceding equation, the transformed equa- tion in z '\^ ^ — 132r^ + 50z — 56 = ; and the three values of z are 7, 4, and 2. Hence, those of y are 9, 6, and 4, and those of x, -f |, -}- 1, and -f I, as before. 474. If we solve the equation, x^ — 2x'' — 3a;^ -f 8x — 4 = 0, by the process of divisors, we will get the three values, -f 2, — 2, and — 1. And, by dividing the first member of the equation by the factors corresponding to these values, we will have left, x -f- 1 = 0. Hence, the value — 1 enters twice in the given equation. We see, from this example, that while the process of divisors gives the rational values, it does not show whether any or all of them are repeated once or more. This remark is evidently applicable to all equations whatever. Some test is then necessary, by which we can ascertain the equal values. EQUAL VALUES. 475. Let the equation be (x — a)" {x — hy {x — cj (x—d) {x — e) &c. = (A), in which there are m values equal to a, to values equal to h, j^ values equal to c, and all the other values unequal. Calling D the differential coefficient of this equation, we will have (Art. 366), D = m (x — a)"-' (x — bf (x — c)p (x — d) (x — e) &c. + n (x — 6)"-' (x — a)" (x — cf (x — d) (x — e) &c. -\-p{x — c)p-' (x — a)" (x — 6)„ (x— c?)(x — c)&c. + (x— a)"'(x — J)"(x — c)P (x — e) &c. -f (x — a)" (x — 5)° (x — c)p (x — d) &c., plus other terms (B). It is plain that the greatest common divisor, D', between (A) and (B), will be D' = (x — a)"-' (x — &)"-' (x — c)P-'. We see that this divisor contains (m — 1) values equal to «, (n — 1) values equal to h, and (p — 1) values equal to c. Hence, it is plain that the number of equal values in each set of the greatest common 4G0 GENERAL THEORY OF EQUATIONS. divisor is one less than in the giv equation. Therefore, to ascertain whether there are equal values, we have this RULE. Find the differential coefficient of the first memher of the given equa- tion [or, as it is generally called, the first derived polynomial), next get the greatest common divisor hetiveen the first member of the pro^wsed equation and this differential coefficient, place this common divisor eqtial to zero, and find its values. Each independent value so found loill he repeated once oftener in the given equation than in the greatest common divisor ; if the latter, for instance, contain two values equal to a, and four values equal to b, the former loill contain three values equal to a, and five equal to b. If the greatest common divisor he of too high a degree to solve, we may find the second differential coefficient, or second derived polynomial of the given equation, then the greatest common divisor between this polynomial and the first memher of the given equa- tion. Each independent value of this common divisor will he re- peated tioice oftener in the given equation. EXAMPLES, 1. Does the equation, x* — 2x^ + ^x^ — x-\-h = 0, contain equal values ? First derived polynomial, 4x' — 6a:^ + 3a; — 1. Greatest common- divisor, x — 1, which, placed equal to zero, gives cc = l. Hence, th|| given equation contains two values equal to 1. If we divide that equatipn by (x — Vf, w e will hav e left , x^ + l = Q. Hence, the other two values are + v/ — I, and — s/ — J- 2. Does the equation, x^ — x^ — 14a;'' — 2Qx^ — 19a? — 5 = 0, con- tain equal values ? First derived polynomial, 5x^ — Ax^ — 420;^ — 52x — 19. Greatest common divisor, (x + 1)'. Hence, the given equation eon- tains four values equal to — 1, and, by dividing the equation by (x -f- 1)^, we find the other value to be + 5. 3. Required the equal values in the equation, x* — 2x^ -f 1 = 0. Ans. Two values = + 1 and two values = — 1. GENERAL THEORY OF EQUATIONS. 461 a;" 4. Required all the values of the equation, x^ — — — 2x^ + a;^ + x — J = 0. Ans. + 1, + 1, + J, — 1 and — 1. 5. Required all the values of the equation, x* ■\- x^ — 6x^ — 4x 4-8 = 0. Ans. + 1, + 2, — 2, — 2. After dividing out the equal factors, (x + 2) (x + 2), we had a quadratic left, which was solved by known rules. 6. Find the values of the equation, x^ — .r" — 2x^ + 2x^ + x — 1 = 0. J[ns. + 1, + 1, + 1, — 1, and — 1. The first greatest common divisor is x^ — x^ — a: + 1. As this, when placed equal to zero, is too high to solve, we find the greatest common divisor between the second differential coefficient of the given equation, and the first member of that equation, or, what amounts to the same thing, the greatest common divisor between x^ — x^ — x-\- 1, and its derived polynomial, Zx^ — 2a; — 1. This common divisor is (a; — 1)^ Hence, the given equation contains three values equal to plus one, and, by dividing out by the factors, (a; — 1) (x — 1) (a; — 1), or, x^ — Zx^ + 3a; — 1, we have left, a;^ + 2a; + 1 = 0. Hence, the other two values arc, — 1 and — 1. 7. Solve the equation, a;^ + a;^ — 5a;' — a;^ + 8a; — 4=0. Ans. a: = + 1, + 1, + 1, — 2, — 2. 8. Find the five values of the equation, a;^ + 2a;^ — 16a; — 32 = 0. Ans. +2,-2,-2, +v/^=4; — v/^iT 9. Find the values of the equation, x^ -\- x^ — 2a* — 16a:^ — 16x + 32 = 0. ^7w. + 1, + 2, —2, — 2, +^"=^47—^:34: 10. Find the values of the equation, x" — (2a + h)x^-\- (0^+ 2ah)z — en = 0. Ans. + a, + or, + J. 11. Solve the equation, x^ — 2 (a + i)x=' + (a^ + 4ai + l'^)x^—.2 (a'h + aW)x + aW = 0. Ans. + a, -\- a, + b, + b 39* 462 GENERAL tii:kory of equations. DERIVED POLYNOMIALS. 476. We have liad occasion to use the relation, x = a-\- i/,m trans- forming the general equation in x into another in i/, such, that the second term was wanting. But the transformation in terms of y and a is of more general application. Sometimes a is an undetermined con- stant, whose value is afterwards determined in such a manner as to make the equation fulfil a given condition. Sometimes a is a deter- minate number or quantity, which expresses the difference between the values of the given equation and another which we wish to form. Suppose, for instance, we wished to transform the equation, x^ — x'^ + 4a; 4- 5 = 0, into another in ?/, such, that each value of x should exceed by 2 each corresponding value of i/. Then, a must be made equal to 2 in the relation, x = a + i/, and we must substitute 2 + y for x wherever it occurs. Hence, the equation in 7/ would be (2 -f- i/y — (2 -f 7/y + 4(2 -|- y) -f 5 = 0, or, expanding and reducing, ^ -f- 5^^ -f 12y -f 17 = 0. In this case, the transformation is easily effected by actually substituting the value of x in terms of y, and developing the several binomials in the new equation. But, when the given equa- tion is of a high degree, this process is tedious and impracticable. Suppose, for example, it were required to transform the equation, a;'^ — 40a;" + x}°—2x^ + 8a;^ -f- 5x' — 4a;« + 9x' + 12.r^ — x^-{-Sx + 15 = 0, into another in y, such, that the values of x should be less by unity than the corresponding values of y. Then, a = — 1 in the re- lation, X = a + ^, and we must substitute for x, wherever it occurs, (y — 1). It is plain that the development of the several binomials in the new equation will be exceedingly tedious. The method of derived polynomials enaUes us to effect the required transformation in y without suhstiiuiion and development. Let VIS resume the equation, ^m ^ pa;n.-l _|_ Qa;m-2 -f To! -f U = 0, and let us suppose, as before, x = a-{- 1/. The new equation in y will then be (a + 2/T + P(« + ^/r-' + Q(« + y)"-' T(a-f y)+ U = 0. Let us assume (a -f 2/r + P(a +y)-' + Qia + ijT-' T(«. -f 2/) + U = A GENERAL THEORY OF EQUATIONS. 463 Making y=0, we get a" + Pa"-' + Qa"-' + Ta + U = A. Differentiating tlie given equation, and dividing by dy, we get K« +3/)"-' + P(»^ — 1) (« + VT-'' + Q("^ — 2) (a + yY^ -I- T = B + 2Cy 4- 3D/ + 4E/ (R). Now, make y = 0, and denote by A' what B becomes in tbat case, we get ma-"-' 4- (wi— 1) Pa'"-^ + (m— 2) Qa""-^ + T = B = A'. Differentiating (R), we get m(m — 1) (a + ^)"-2 + (m — 1) (m — 2) P(a + y)"-^ 4. (m — 2) (m — 3) Q(a + y)°'-» + &c. = 2C + 2 . 3Dy + 3 . 4E/ + &c. (S). Making y = 0, and representing by A" what 2C becomes, we get m(m — 1) a—2 + (m — 1) (m — 2) Pa""^ + (w — 2) (m — 3) Qa"-* + &c. = A" = 2C. Then C == — ^ ^ (»^— 1) ft""" + (^?^ — 1) (m — 2) Pg-"-^ + &c. 2 2 Differentiating (S), we get m(m — 1) (m _ 2) (a + y)"-^+ (m — 1) (m — 2) (m— 8) P(a+^)°'--» + (m — 2) (m — 3) (m — 4) Q(a + y)""^ + &c. = 2 . 3 . D + 2 . 3 . 4Ey + &c. Making 3/ = 0, and preserving a similar notation, we have _ A^' _ ?7i(m— 1) (m— 2) a"-^+(ni— 1) (m — 2) (m — 3) Pa^-^ 2.3 273 («i — 2) (m — 3) (m — 4) Qa-' H 2~-g h &c. By proceeding in the same way, all the other coefficients could be determined. But the law of formation is already apparent, A is what the given equation becomes when a (or the difference between x and^) is substituted for x ; A' is formed from A, by multiplying the coeffi- cient of a in every term by its exponent, and then diminishing that A" exponent by unity ; -^ is formed in precisely the same way, except A'" that the result is divided by 2 ; -5- is formed in like manner from A", o except that the result is divided by 3. A"" -J- is formed according to the same law from A'", the result being divided by 4. 464 GENERAL THEORY OF EQUATIONS. + -Si + + + + 5=5 + =!^ rG =§ + ^ ^ + r + o CO Is, B • 6 to cid 1 + S "f s s 1 e 1 a §^ ^ ,.—^ 1 T §^ § v^/ y — ^ ^~-- CO CO 1 1 g vi w (M /--N 1 (M 1 1 -S 1 ^ Y k + I "^ £ CO ^ /~~. * ^^ CO IM CO 1 ^ i (M /-^ 1 (M 1 1 SJ SS ^^ 1 ^-^ 1 p-l g 1 1 + s 1 v_/ E + 7 X E e IM v^ 1 ^ §^ 1 S: v.^ 1— 1 .'^ 1 7 1 g i ^ — • S- s s 0^ ^- ^ CO Ul % x'^ — 1, X2 = 6x, X3 = 6, X4, X5, &c. = 0, and equation (D) becomes ox^ — 1 -\- Zxy + if = ^ ; or, Zx^ + Zxy -\- if — 1 = ; and, preparing the given equation for division by multiplying by y, we have GENERAL THEORY OF EQUATIONS. 469 3X' — 3a; — 18 | 3z'+3g y+y'— 1 3x^+3x^!/+xy^—x a>-]/=Quotient. —3x^y—(y''+2ix—18 I Sx-'+Zxy+y^-l -Sx^y-Sxy'^y'+y 2(jf»-l) 2 {y''—\)x + y^ — y—l'i \ 6(y'^—l')x^+(5iy^—l)xg+2{y'*—l)^\ 2d Quotient = 3x. e(y'^—l)x^+3y^x—Sxy—Ux 3{y''—l)xy+biv-\-2{y^—l) = 2d Remainder. or, 3(y^—y+lS)x+2(y^r)'^ Multiplying by 2(y«-l) 2(1/^1) Hy' - y + 18) x(y» - 1) + 4(y» - ir I 2(y'^l)x+y°-y-I8 ^ jy'—y + 1 8)ar(ya— 1 )+3[ y'— y + 18)( y'-y— 1 8) | 3(y=—y+18)=:3d Quotient. 4(3/' — l)'-3(y'-y + 18)(y»-y — 18)=0 or, developing 4(/ — 3^/* + fi/ — 1) — ^f + St/* + 5ii/^ + St/* — 3/ — 54j^ — 54y + 54y'+ 3(18)^ = ; or, reducing / — 6y^ + V 4- 968 = 0, wliicli is the equation of differences sought. Now, make y^ = z, and we have for the equation of the square of the differences a' — 6^^ + 9z + 968 = 0. In the above development, the alternate terms were struck out. This was to have been anticipated from what had been said. But we may show, in another manner, that the equation of differences can contain only even powers of y. For, let a, b, c, d, &c., be the numerical differences of the values of the given equation, then, + «, — a, -f i, — h, &c., will be values in the equations of differences; and, since the factors corresponding to these values, permuted in pairs, give us (t/ + a) (3/ — a), (j/ + l) (j/ — J), &c., the first member of the equation of differences will be, (jj^ — a^ (^'—-^^) (/-cO,&c. Now, making ^ = z, we will have (z — a") (~ — l") (z — c^) &c. = for the equation of the square of the differences. The degTce of the last equation will only be half as great as that of the equation of dif- ferences. EXAMPLES. 1. Given, x^ — 9 = 0, to find the equation of differences. Ans. / — 86 = 0. 2. Given, x' + 9x 4- 4 = 0, to find the equation of differences. Ans. / + 54/ + 81/ + 3348 = 0. After the first division the remainder will be, 2(//^ ^ 9) a; + y + 9?/ + 12. Multiply the last divisor twice by 2(/ + 9). Then, after two divisions, you will have 4(/ + 9)' — 3(/ +9y — 12) (2/^ + ^2/ + 12), which will reduce to the expression above. 40 470 GENERAL THEORY OP EQUATIONS. 3. Given, x^+ ax + J = 0, to find the equation of differences. Ans. if + Qai/ -\- Da^ if + 4a^ + 27b"- = 0. First remainder is, 2(y^ +a'^)x + ?/ + oy + oh. Use 2(y^ + aF) twice as a multiplier. 4. Given, x^ — a; = 0, to find tte equation of differences. Ans. y^ — 61/ + dy' — 4 = 0. First remainder is, 2(y'^ — 1) x + y^ — y. The coefiicient of x is used twice as a multiplier. It will be seen that + 1 and — 1 are values in the equation of dif- ferences. Dividing by y^ — 1, and solving the resulting equation, y'* — by'^ + 4 = 0, by the rules for binomial equations, we get the four values, -f 2, — 2, + 1 and — 1. The six values, then, in the equation of dif- ferences, are, + 2, — 2, and + 1 and — 1, repeated twice. These ought to be the values, since those in the given equation are, -f- 1 and — 1, the differences between which are those given above. IRRATIONAL VALUES. 480. An equation, freed from the factors corresponding to the ra- tional values, will contain only irrational or imaginary values, or both irrational and imaginary values. We can best explain the process of finding the irrational values by an example. Let us take the equation, x^ — 2x'^ — 2 = 0, to find one positive rational value. The superior positive limit is, 1 + 1/2 = 3. Substi- tuting the natural numbers from up to 3, we find that and 1 give negative results when substituted for x in the equation, and that 2 gives a positive result. Hence, a value of the equation lies between 1 and 2, and 1 is the entire part of the irrational value. Now, make x = y + liu the given equation, then the new equation in y will have values less by unity than those in x. Hence, y will con- tain the decimal part of the value of x. This transformation can be most readily effected by the formula of Article 477. We have (1)5 — 2(1)^ — 2 = —3 = A. 5(1)* — 6(1)2 = — 1 = A'. 20(1)3 — 12(1)' = 8 ^j^n^ 60(1)2 _ 12 = 48 == A'". 120(1)» = 120 = A-. 120 = 120 = A^ GENERAL THEORY OF EQUATIONS. 471 Then, A 4- AV + 1-^ + ^i-^ + 1.2 ' 1.2.3 ' 1.2.3.4 ' 1.2.3.4.5 + &c. = becomes, — S — y + Ay'' + Sf + 5y* + y^ = ', or, changing the order of terms, y^ -f- 5y* + Sy^ + 4?/* — y — 3 = 0. If, now, we make z = lOy, then y = — . The first figure, then, of the value of z, in the transformed equation, will be tenths in the value of y ; and, consequently, tenths in the value of x. The transformed equation (Article 467) is, ^ + 502" + 800;^^ + 4000;^^ _ lOOOOj/ _ 300000 = 0. S. p. Limit = 1 + V300000 = 1 + 23 = 24. The limit is here too great for any practical use. Substituting, be- tween and 10, we find a change of sign in the results corresponding to the substitution of 5 and 6. For 5, we have, (5)H50(5)"+800(5)'+4000(5/— 10000(5) — 300000=— 115625. And 6 gives, (6)H50 (6)"+ 800 (6)='+4000 (6)^-10000(6)— 300000 = + 29376. Hence, 5 is the entire part of the value of z, and this corresponds to five-tenths in y and x. Tliereforc, 1 . 5 is the approximate value of a;, to within tenths. To get a nearer approximation, let us transform the equation in z into another in w, so that the values of to shall be less by 5 than those of z. Then iv will contain the remaining part of the deci- mal value of z. The equation in w is, lo' -f 75^" -f 2050w' + 24750ii;=' -f 118125io — 115625 = 0. Making t = 10?p, or lo = — , the transformed equation in t (Art. 467), will be, i*-f750<"+205000i=' -1-24750000/2+1181250000^— 11562500000 = 0. On trial, we find that 8 and 9 give results with contrary signs ; hence, 8 is the entire part of the value of t, and corresponds to -8 in w, and to .08 in y and x. We have, then, for the approximate value of x, 1-58. The process would be, in all respects, the same, were a negative value to be found, except that we would find the negative limits. Let us take, as a second example, x^ — lOx' -{- 6x + 1 = 0. We find that and 1 give results with contrary signs, and so do 3 472 GENERAL THEORY OF EQUATIONS. and 4. Let us find the decimal part of the value, whose entire part ia 3. Mate x = y + S. The equation in y will be, rf + m/ + SOy' + ISOy" + Uly -8=0, and that in z, z^ + I50z* + 80002» -f ISOOOOz^ + UlOOOOz — 800000 = 0. Substituting and 1, we find a change of sign. Hence, is the tenths of the given equation. Making z = + «;, we have, to' + IbOw* + 8000!«=' + 180000i«^ + 1410000«; — 800000 = 0. The equation in t, is, /'+1500<''+800000f^4-180000000«^+ 14100000000^— 80000000000=0. We find that 5 and 6, when substituted, give a change of sign. Now, make f = s+ 5, the equation in s will be, .s5_^175s^4-830250s^+192226280s2+15960753125s— 4899059375=0. By changing this into an equation in r, making r = 10s, we will find a change between 3 and 4. Hence, 3 053 is the approximat-e value of a; to within thousand tlis. 481. In this process we have proceeded upon the hypothesis, that but one real root lay between two successive integers. To ascertain whether this is the case, we have only to transform the given equation in X into another in y, so that the values of y shall be the squares of the difi"erences of those of x. Next, find the inferior limit of the posi- tive values of y. Suppose D^ to be this limit ; then, since D^ is less than the least value in the equation of the square of the differences, v^D^, or D, will be less than the least difference between the values in the given equation. Now, if D be ^ 1, it is plain that no real root will be comprised between two successive integers, and the process described above can be pursued. A similar course of reasoning can be applied to D'^, the inferior limit of the negative values in the equa- tion of the squares of the differences. But, if D <^1, then two or more real roots may be comprised be- tween two consecutive integers. In this case, we have only to substi- tute a series of numbers, whose common difference shall be = or <^ D. Then, those numbers, which give results with contrary signs, will have but one real value between them. Another method of frequent appli- GENERAL THEORY OF EQUATIONS. 473 cation, when D is a proper fraction, -, is to transform the equation in x into another in y, by making x = -. Then, the differences of the values of y will be greater than unity, and only one real root will lie between the successive integers in the transformed equation. For, let x' and x" be consecutive values of x, then x' = -, and x" = — . Hence, r r (x' — x") r = y' — y". Then the differences between the consecutive values of 2/ is r times greater than between those of x, and, as r is the denominator of D, y' — y" must be greater than unity. EXAMPLES. 1. Find one irrational value in the equation, x^ — 8.1'+ Ix^ — 56 = 0. Ans. 2-828. 2. Find one irrational value in the equation, a;* + 3ic' — 4x^ — 15a; — 5 = 0. Ans. 2-236. 3. Find one irrational value in the equation, x^-\- 2x' — 2ar^ — 4 = 0. Ans. 1-264. 4. Find one irrational value in the equation, .x' — 1x +7 = 0. Ans. —3-048. 5. Find one irrational value in the equation, x* -\- x^ — 12:c^ — 17a; — 85 = 0. Ans. 4-123. 6. Find one irrational value in the equation, x^ — 2 = 0. Ans. 1-414. 7. Find, by the process of irrational values, one value in the equa- tion, x^ — 4 = 0. Ans. x^2, the decimal part in all the transformed equations being zero. 8. Find one irrational value in the equation, x* + x^ — 25x^ — 26a; — 26 = 0. Ans. 5-099. 482. If we apply the foregoing process to an example of the form, x* + 14x^ — 49 = 0, the consecutive numbers will give no change of sign. One value in the equation of differences will be found to be zero, and also one in the equation of the square of the differences. 40* 474 GENERAL THEORY OF EQUATIONS. D'^ and D are tliea botli zero. "When this is the case, we may infer the presence of equal values. Ou trial, as in Art. 475, we find x^ — 7 the greatest common divisor between the first member of the given equa- tion and its derived polynomial. We see, by this example, that the preceding method is only applicable to equations which have been freed from their equal values. While the foregoing method afi'ords a complete solution to the pro- blem of finding the irrational values of numerical equations, yet it is of diificult application to equations of high degrees, and, in all equa- tions, whether of low or high degrees, the difl&culty increases with the number of decimal places sought. NEWTON'S METHOD OF APPROXIMATION. 483. This is known as the method of successive substitutions, and consists in substituting, in the given equation, the natural numbers be- tween the limits, until results with contrary signs are obtained. Let a be the least of two consecutive numbers which give results with con- trary signs. Then a is the first figure, or entire part of the value sought. Substitute a -{- y for x in the given equation, y being a small fraction, whose second and higher powers may be neglected. Hence, y may be found in the transformed equation, and a + y will constitute the first approximation to the value of x. Let a -1- y be denoted by h, and make x = h -{■ y', y' being a small fraction, whose higher powers may be neglected. The transformed equation will give the value of y, and this, with 6, will constitute the second approximation to x. Calling h + y', c, and making x = e + y" , we can get a third approxi- mation to the value of x, and may thus carry the approximation to as many places of decimals as we please. Let us take, as an illustration, the equation, a;* — 2 =0. Substi- tuting between the limits, we find that 1 and 2 give different signs. Then 1 is the entire part of the value. Make a; = 1 -f y, and reject y^, we will have 1 -j- 2^/ — 2 = 0, or y = J = -5. Hence, for first approximation, x = 1-5. Now, make x = 1-5 +y, and we will have, •25 after rejecting y'\ 2-25 + 3/ — 2 = 0, or, / = — -^ = — -0833. Then, for the second approximation, we get cc = 1 -5 — -0833 = 1-4167. Place X = 1-4167 + y", we get 2-8834/' = — -00303889, or, y" = — -00107. Then, for a third approximation, x = 1-4167 — -00107 = 1-41563. GENERAL THEORY OF EQUATIONS. 475 484. The accuracy of the process evidently depends upon the un- known quantity introduced being a small fraction. "Whenever, then, the substitution of the natural numbers make y an improper fraction, more minute substitutions must be made, unless we intend to carry the approximation to several places of decimals. Thus, take the example, x'' + x — 8 =; 0, we find that 1 and 2 give results with contrary signs. Then, 1 is the entire part of the value sought. Making x = y + 1, we have, 3/^ + 3y + 3^^ + 1 -|- y + 1 — 8 = 0, or, rejecting the higher powers of y, 4^/ — 6 = 0. Hence, y ■= &=. 1*5, an improper fraction. Then, for the first approximation, we have, x ^ 2-5, a result obviously too great. Next, place x = 2-50 + ?/, we have, after rejecting higher powers of y, (2-50)» + 3 (2-50)2y + 2-50 + y — 8 = 0. Then, y' = — -512, and, for the second approximation, x = 1-988. Making now x ^ 1-988 + y" , we will have, after rejecting the higher powers of if' ■> (1-988)'' + 3 (l-988)y' 4- 1-988 + y"— 8 = 0, or, 7-856862272 + ll-856432y'+ 1-988 -f y" — 8 = 0, or, 11-856432/' = _ 1-844862272. Hence, y" = — -144 nearly, and, for a third approximation, x = 1-844, which difi'ers from the true value by less than one hundredth. We see, from this example, that the error was considerable when y was an improper fraction, but became reduced by carrying the approximation farther. A closer approximation could have been obtained, without carrying the operation so far, by making minute substitutions. A change of sign would have been found between the results corresponding to 1-8 and 1-9. LAGRANGE'S METHOD. 485. This differs from Newton's, in that the unknown expression added to complete each successive value is fractional in form, and in that the transformation is made into an equation involving this un- known quantity. Thus, let a be the entire part or number next below the value, we make x = a H — , and get a transformed equation in y, such, that the values of y must be greater than unity. Let h be the entire part of the value of y, then, for the first approximation, we have a: = a 4- - = — . Next, place y = & A — . The transformed equation iu ^ must have its values greater than unity. Let c be the 476 GENERAL THEORY OP EQUATIONS. entire part of tlie value of y' . Then, approximatively, y = h ■{ — = , and, for a second approximation to x, we have, x ^ a -^^ - = a c . . 1 + -. To find a third approximate value for x, let y = c -| — j^ CO -\- I y 1 and we may thus continue to approximate to the value of x until the result is as accurate as desired. To apply these principles, let us take the equation, x^ — x — 5 = 0, We see that 1 and 2 give contrary signs, hence, cc = 1 + -. Substi- tuting this value of cc, we have 5^^ — 2f/^ — 3y — 1 = 0. The entire part of the value of y is 1, hence, for first approximation to x, we have X — \ ■\ — = 1 + 1 = 2, which is plainly too great. Next, making y = 1 + -, we get y'^ — %y'^ — 13/ — 5 = 0. In this, 9 and 10 give contrary signs. Then, y = 1 + -^ = Lo, and a; — 1 + y^^ = 12^ fbr the second approximation. Now, make ?/' = 9 -| — ^, and the transformed equation in y" will be, 4iy"' — 86^"^ — 19y' — 1 =: ; a change of sign between the re- sults given by 2 and 3. Hence, y' = 9| = i^^, y = 1 + -2^ = |^, and a; = 1 -f ^a = |fl, for the third approximate value of x. We find on trial the third approximate value a little too great, and the second a little too small. Hence, the true value lies between 4f ^^^^ io> ^^^ either of these numbers diflPers from the true value by less than ^ jq. 486. Let us take the simple example, x^ — 2 = 0. Place x=^\ •\ — , y then, y — 1y — 1 = 0. We find that 2 and 8 give contrary signs. Hence, for first approximation, x = 1 + i = |. Now, make ?/ = 2 -f- -» and the equation in y will become y'"^ — 2y — 1 = 0; and, again, 2 is the entire part of the value. Hence, y = 2 + i == 55 and x=^\ -\- \ = I, for a second approximation. On trial, we find | too great, and | too small, and these differ from each other by a tenth j therefore, the true value differs from either by less than a tenth. For a third approximate value we would find, x = |-|, which is a little too great, but is within g'jj of the true value. The fourth approximate value, ||, is too small, but differs from the true value by less than g-J-, • By continuing thus the process, we would find the approximations alternately too great and GENERAL THEORY OF EQUATIONS. 477 too small, and thus can tell at every step how near we have come to the true values. We see that Lagrange's method enables us to determine the proximity to the true value at every stage of the work, and this is the chief advantage claimed for it over the process of Newton. EXAMPLES. 1. x^ — 7x + 7 = 0. Ans. X = f I, after three approximations. 2. X* — a- — 6 = 0. Ans. .r = |, after third approximation. 1st approximation, 2 ; 2d approximation, ^ ; 3d approximation, |. GENERAL SOLUTION OF NUMERICAL EQUATIONS, 487. When we have a numerical equation of any degree to solve, we must first find its rational values by the process of divisors, if it is under the proposed form. But, if it is not, we must transform it into another equation in y, so that the coefficient of the first term shall be plus unity, and the other coefficients entire. Then, knowing the relation be- tween 1/ and X, we can determine the values of x when those of y have been found. Next we must ascertain whether any of the values are repeated (Art. 475). Having thus found all the rational values, we next divide by the factors corresponding to them. Then, by means of the equation of the squares of the difi'crenccs, we can ascertain what series of numbers to substitute in the reduced equation (Art. 481). The final step is to find, approximatively, the irrational values by either of the three processes explained. Now, if the number of rational and ir- rational values be subtracted from the degree of the given equation, the difference will be the number of imaginary values. GENERAL EXAMPLES. 1. Find the three values of the equation, 2x' ~ f- 106a; — 20 = 0. A71S. a; = 1, 5, and 10. 2. Solve the equation, x^ — Ux'' + 7x^ + 28a; — 28 = 0. Ans. x = ±2, 1-356, 1-692, and —3-044. 17a^ 3. Solve the equation, x'' -\ — |a; — i = 0. Ans. X = ^, — I, and — 3. 478 Sturm's theorem. 4. Solve tte equation, x" — lO^c^ — 12^2 + 92x + 280 = 0. A71S. X = 10, and 4-302. 5. Solve the equation, 2x^ — 2x* — + -^ -{■ x — 1 = 0. A71S. a; = 1, ^, — J, and =h ,y2. 6. Solve the equation, x'' — 4x^ — 4x^ + Sdx* — 60x' — 28a:^ + 224a; — 224 = 0. 7. Solve the equation, x* — 2x^ — ^\x^ + 22a; + 280 = 0. Ans. X = 4, and 5-134. STUEM'S THEOREM. 488. When we have, by means of the process of divisors and the method of equal values, detected all the rational values, and then, by the aid of the equation of differences, discovered all the irrational values, we can determine the number of imaginary values, by subtract- ing the number of rational and irrational values from the degree of the equation. When this difference is zero, there are, of course, no imagi- nary values. But this method is circuitous, and is, moreover, rendered tedious by the employment of the equation of differences. Sturm's Theorem de- termines directly the number of imaginary values, and dispenses tcith the equation of differences. Let X = x™ -f Pa;-"-' -f Q^c--^ + +Ta: + U = ie aji equa- tion cleared of its equal values. Take the derivative of X, and call it X' ; then, X' = mx^-'-\- (m — l)Px"»-2-f (m — 2)Qx'°-^ -f- -f T. Now, divide X by X', and continue the division until we get a remaiur der in a; of a lower degree than X'. Call this remainder, with its sign changed, X". Divide X', in like manner, by X", and continue the division until the remainder is a degree lower than the divisor. Change, in the same way, the signs of all the terms of this remainder, and call the resulting expression X'". Pursue the same process until we get a remainder independent of x, which must be so eventually, since the given equation, by hypothesis, contains no equal values. This will be the (m — 1)*'' remainder, and all the expressions, including X and X', Sturm's theorem. 479 will make up (m + 1) functions. Hence, designate by X^+'j tlie last remainder, with its sign changed. We will then have the following expressions, Q, Q', &c., designating quotients : X =X'Q -X" (A), X' =X"Q' —X'" (B), X" = X"'Q" — X- (C), X'" = X'vQ"' — X- (D), X- = X-Q-— X- (E), X»-' = X-'Q"->— X»+' (W). In preparing the dividends for division, we may, as in the process for finding the greatest common divisor, multiply by any positive con- stant. In like manner, we may suppress any positive constant common to all the terms of any of the successive remainders. 489. The theorem we have to demonstrate may be enunciated as follows : If we substitute any number, tn, for x in the above series of func- tions, X, X', X", &c., and count the variations of signs in the results, and then substitute any other number, as n, for x in the same series, and count again the variations of signs in the results, the difference between the number of variations in the tico cases icill exjjress exactly the number of real values between the numbers m a7id n. If, then, m be taken as the superior limit of the positive values, and n as the superior limit of the negative values, the difibrencc between the number of vari- ations will express the total number of real values. When this difiFe- rence is zero, there are no real values ; when it is equal to the degree of the equation, all the values are real. 490. The demonstration of the theorem depends upon the four fol- lowing principles : 1. M) tico consecutive finctions can vanish for the same value of x. For, if two functions, as X" and X'", could disappear together, equa- tion (C) would give = — X'^. Hence, X'^ = 0, and X"' and X*'' being zero, equation (D) would give X^ = 0. And so all the func- tions in succession would become zero, until we would finally have X^^' ^ 0. But this would indicate a common divisor, and, conse- quently, equal values, which is contrary to the hypothesis with which we set out. 480 Sturm's theorem. 2, When such a value is given to x as to make any function other than'K equal to zero, the functions adjacent to the one disajij^earing will he affected with contrary/ signs. For example, let X'^ = 0. Then, equation (D) will give X'"=— X^; and, since the own sign of X'^ may be either plus or minus, this equa- tion may be written X'" = — (± X^). Now, if X'^ be affected with the positive sign, the second member will be negative; and, since the signs of the two members of an equation must ^ways be the same, X"' will be negative, and, therefore, of a contrary sign to X'^. But, if the own sign of X^ be negative, then the second member will be positive. Hence, X'" will be positive, and, therefore, of a contrary sign to X^. 3. The disappearance of any function other than 'Kfor a piarticidar value ofx, will neither increase nor decrease the variations of signs in the results of the sej'ies, X, X', X", d;c. Suppose that one of the intermediate functions, as X'"', becomes zero, for the value x = b. Then, by the first principle, neither X'" nor X'^ can be zero for x = I ; and, by the second principle, they must have contrary signs. There is, then, one variation of sign between X'" and X"^ when x = b. We are to show that this variation was not caused by the disappearance of X'''. For, however little x ■=h may differ from a value of X'" or X''^, h may be taken smaller than this difference. And, moreover, since a quantity can only change its sign in passing through zero or infinity, it is plain that, between x = b and x = b — h, neither X'" nor X^ can undergo any change of sign. They will still be affected with contrary signs, and, for cc = & — h, we will have either + X'", ± X'', — X^, or else, — X'", ± X'^, + X\ We affect X'' with the ambiguous sign, since we do not know whether its sign is positive or negative. Now, read the double sign either way, and take either of the preceding expressions, and you will have one variation. Hence, before x arrives at the value b, which causes the disappearance of X"', the three successive functions give one variation. Now, when x passes beyond b, and becomes b + h (h being less than the difference between b and a value of either X'", or X"), X'" will change its sign, but X'" and X" will not. Hence, forx:=h-\- h, we will have either + X'", =f X'", — X^ or — X'", =p X'", + X'. And, reading the ambiguous sign either way, in either of the above expres- sions, we will have one variation. Hence, both before and after x reached the value b, which made X''' = 0, we find a variation of sign among the three functions. Then x:=b, or the vanishing of the inter- mediate function, X'\ has not caused that variation. Sturm's theorem. 481 4. Every passage ofS. through zero causes a loss of one variation xchen x is ascending towards the superior positive limit, or again of one variation throughout the series, X, X', &c., when x is descending toioards the superior negative limit. For, let a; = a be a value in the equation, X = 0, and take u lesp than the difference between a and a value of X'. Then, X will have a different sign, when a + u is substituted for x, from what it would have were a — u substituted for x. Hence, in the passage of x from a — u to a + M, X will undergo a change of sign, but X' will not. since, by hypothesis, no value of X' is passed over. Neither will any intermediate function, X", X'", &c., undergo a change of sign unless it pass through zero, and, by the third principle, such passage will not affect the total number of variations. Therefore, in the passage of x from a — « to a + u, either a permanence, existing between X and X', is changed into a variation, or else a variation is changed into a permanence. We will now examine the order in which this change takes place. Designate by X, what X becomes when x = a — u. Next, develop all the terms of X by Art. 456; we will have Xi= A — A'm + A"«^ A"'m' -j — -y — ^j — -^ — r, + &c> (R)) in which A denotes what X becomes when xz=a; A' is the derivative of A, A" the derivative of A', &c. But, since a is a value of .r, A must be equal to zero. Then, equation (R) may be written, X,= — u {M — i-^ + ,^" — &c.) Now. u may be taken so small that the first term will be greater than the algebraic sum of all the other terms within the brackets. Hence, the sign of the quantities within the parenthesis will depend upon that of A'. If, therefore, the own sign of A' is positive, the second member will be negative. Then, X, must be negative, and, therefore, of a con- trary sign to A'. But, if the own sign of A' be negative, the second member will be positive. Then, X, will be positive, and X, and A' will be affected with contrary signs. And, since X, and A' represent what X and X' become when a — tt is substituted for x, we see that there is a variation of signs between these functions below a value, x = u, which satisfies the equation, X = 0. Now, suppose x = a -\- u, and, therefore, above a value in the equation, X = 0. Develop as before, designating by Xa what X becomes when a -f u takes the place A"?f A"'u^ of X. Then, X2 = u (A' + ~ — -^ + ,^ ,^ + &c.), in which u is 41 *2p " 482 Sturm's theorem. so small that the sign of A' controls that of the parenthesis. We see, that, whether the own sign of A' be positive or negative, Xg must be affected with the same sign. Hence, when x = a -{- ^l, and, therefore, above a value, there is a permanence of sign between X and X'. Therefore, in the passage of x towards the superior positive limit, a variation is lost, and changed into a permanence between X and X', whenever X becomes equal to zero ; and, as the other functions neither gain nor lose variations, one variation only is lost throughout the entire series whenever X = 0. If we had begun with a + m, and passed down to a — u, it is evi- dent that one variation would have been gained, every time that x reached a value in the equation, X = 0. So, then, whenever x, increasing or decreasing by insensible degrees, reaches a value which will satisfy the equation, X = 0, one variation is lost or gained in the signs of the series, X, X', X", &c. Hence, if we take any number, as m, and substitute it for x in all the functions, and count the number of variations in the signs of the results, and then take any other number, as n, and treat it in like manner, the difference between the number of variations in the two cases will express the number of real values lying in the given equation between m and 7i. Moreover, it is plain that if «i = + oo, and n — — go, this difference will express the total number of real values in the equation. This would be equally true if we used the superior positive limit and the superior negative limit. But, there is this advantage in the employ- ment of -I- oo, and — oo ; the substitution need only be made in the leading term of the successive functions, for then the sign of this term would control the signs of all the other terms in the same function. 491. Having determined the number of real values, the next step is to determine the initial figure of each one of theSe values. To do this, we substitute the natural numbers, 0, 1, 2, 3, &c., until we get the same number of variations of signs in the entire series as was given by + 00. The number that gives the same variations, as + oo, is the superior positive limit. Substitute, in like manner, 0, — 1, — 2, — 3, &c., until we get the same number of variations as given by — oo. We then have the superior negative limit. The least of the numbers, whether positive or negative, between which a variation is lost or gained, is the initial figure, or entire part of a real value. Should there be two or more variations lost or gained between consecu- Sturm's theorem. 483 numbers, then there are as many values as there are changes of sign, which have the same initial figure. We will illustrate by a simple example, -■^'i^ Take equation, x^ + 3*'' — 1 = =*a:."~ • Then its derivative, X' = 3x2 ^ g^ Now, multiply all the terms of X by 9, and divide by X'. After two divisions, we will get a remainder of a lower degree than X' ; this is — 2x — 9. Change its sign, and call it X". Then, X" = 2a; + 9. Multiply X' by 4, and, after two divisions, by X", we will have a re- mainder, + 207. Hence, X'" = — 207, and we will have X = x'-f a;2 — 1=0 X' =3^2 + 2a; X" = 2a; -f 9 X" = — 207. Making a; = + oo, in the leading terms of these functions, the order of the signs will be + + -1 , one variation. And for 03 = — oo, it will be 1 , two variations. Hence, 2 — 1 = 1, real value. To find the initial figure of this value, make a; = 0, 1, 2, 3, &c., and write the signs of the results beneath their respective functions. We will have X, X', X", X'", then, a; = 0, — 1, 0, -f 9, — 207, two variations. X = 1, ''^• 1, +5, +11, — 207, one variation. Hence, 1 is the superior limit of the value sought, and is its initial figure. The decimal part of the value can be found by the process for the irrational values. The transformed equation in y,\s, y^ -\- y^ — 1 =: ; and that in z, z' + \^z^ — 1000 = 0. And, since 7 and 8 give results with con- trary signs, 7 is the tenths of the required value. Then, a; = 0-7 is an approximate value of x. The approximation may be carried as far as desired by the method of irrational values. 484 Sturm's theorem. We need not substitute 0, — 1, — 2, &c., in the equation, since there is but one real value, and that has been shown to be positive. 492. Whenever only one variation is lost between two consecutive numbers, Sturm's theorem affords no advantage over the process for finding the irrational values, except that of detecting directly the num- ber of imaginary values. But, when there are two or more real values comprised between two consecutive numbers, we are enabled by means of the theorem to dispense with the equation of differences, which otherwise must be employed to find the decimal part of those values. We will now explain the chief advantage of the theorem. Let us take the equation, x" — loi? — 1x^ -f 4 = 0. We will have the following series of functions : X = cc^ — 2x' — 2a;=' + 4 = 0, X'^Sx" — 6x2 — 4x, X" = 4x' + 6x2 — 20, X'" = — 42x2 — 168x + 300, X'' = — 2880x + 3840, X' = — , a constant. + oo gives 4- + + H , one variation, — cx) gives 1 1- H ) four variations. Hence, there are three real values. Proceeding as before, we have X, X', X", X'", X''', X\ When, X = " x = 1 " x= 2 + - + + + + + + + + - three variations, three" variations, one variation. Since + 2 gives the same number of variations as -f oc, it is the superior positive limit. Moreover, as there are two variations lost be- tween 1 and 2, there are two real values between them. Had we sub- stituted the natural numbers, 0, 1, 2, 3, &c., in the given equation, we would have found no change of sign, because an even number of values lay between consecutive numbers. To detect these values, without the aid of Sturm's theorem, we must either have recourse to the equa- tion of differences, or to minute and tedious substitutions. Sturm's theorew. 485 So, then, it is evident that whenever an equation contains two, four, six, or any number of even values between consecutive numbers, Sturm's theorem enables us to detect these values in the shortest possible manner. 493. But, in addition to this, the theorem gives us the means of finding the decimal part of the irrational values in a shorter and better manner than by the usual process, as we will now show. Since 1 is the entire part of both the irrational values, the first step, according to the usual process for finding the irrational values, is to transform the equation in x into another in y, so that the values of i/ shall be less by unity than those of x. The transformed equation in y is / + 5/ + 8/ + 2y ^ - 5y + 1 = 0, and that in z, is z^ + 502" + 800z' + 20002^ — 500002 + 100000 = 0. We find, on trial, that 0, 1, and 2 give positive results ; 3 and 4 give negative results; 5, and all numbers above 5, give, again, positive results. Hence, 2 and 4 are the tenths of the sought values; and 1-2, and 1'4 are those values, approximatively. Now, if, in the transformed equation in z, two values had lain be- tween consecutive numbers, we must have had recourse either to mihute substitutions, or to the equation of differences. If, for instance, the second value had been 1-26 instead of 1-4, the tenths would have been the same for both values, and the substitution of the natural num- bers in 2 would have given no change of sign. To obviate a difficulty that might occur in more than one of the transformed equations, we proceed thus. We take all the functions, X, X', X", &c., and transform them into others in y, so that the values of y shall differ from those of x, by the initial figure, which, in this case, is unity. We will then have the Y = / + 5y' + 8/ + 2/ — 5y + 1, Y' = by' + 20y« + 24/ + 4y — 5, Y" = 4/-M8/-f24y_10, Y'".==- 42/ -252^ + 90, Y'^ ^ — 2880^ + 960, Y" = — , a constant. 41* 486 s»urm's theorem. After having found Y, as indicated, we may get its derivative, Y', and then divide Y by Y', and proceed as we did when getting the functions, X, X', X", &c. But, in genei-al, the better method is to transform the functions in x into others in y, so tliat the values of y shall be less than those of x, by the initial figure of the sought value. msformed functions in z are : = z5 ^ 50z^ + 800z» + 2000z^ — 50000z + 100000, ' = 5^" + 2002^ + 24002^ + 4000z — 50000, The transformed functions in z are : Z = Z'_.„ . , , Z" = \z^ + ISOz^ + 2400z_ 10000, Z'" = _42z2 — 2520z + 9000, Z" = — 2880z + 9600, Z' = — , a constant. And we have the following results : z, Z', Z", Z'" Z", Z\ When z = + — — + + — , three variations, " z = l + — — + + a u " z = 2 + — — + + cc a " z = S — — — + + — , two variations, « z = 4 — — + — — ___ U li " z = 5 + + + — — — , one variation. We need go no further in the substitution, since 5 gives the same number of variations as + oo. We see that there is a variation lost between 2 and 3, and another between 4 and 5. Hence, 2 and 4 are the tenths in the required values, as we before found. Now, if there had been two, or any number of values having the same initial decimal figure, 3, for example, there would have been as many variations lost or gained as there were values having the same initial decimal. 494. We see, then, the two great advantages of Sturm's theorem in finding the irrational values over the other three processes described : 1st. When an equation comprises an even number of values between two consecutive numbers, it enables us to detect these values without minute substitutions, or the employment of the equation of diiferences. 2d. When two or more values have the same initial decimal figure, it enables us to tell the exact number of those values. If we add to these two advantages the one first mentioned, that of detecting directly the Sturm's theorem. 487 number of real, and, consequently, the number of imaginary values, we can see how important the theorem is. To get the hundredths, the functions in z may be transformed into others in s, so that the values of s shall differ from those of z by the initial decimal figures ; in this case, 2 and 4. Then, again, transform the functions in s into others in lo, so that the values of w shall be ten times greater than those of s. We will then have a series of functions, W, W, W", W", &c., in which we may substitute the natural num- bers, 0, 1, 2, &c., until we get as many variations as -j- oo gives in the series. The least of the consecutive numbers, between which a loss of variation occurs, will be hundredths in the sought value. We, of course, will have two series of functions in to, the one corresponding to 2, as the initial figure of decimals, and the other to 4. 495. Since there were three real values in the equation, x^ — 2x' — 2x^ + 4 = 0, and we have found but two positive values, the other must be negative. To determine this negative value, let us resume the functions, X = x5 — 2x' — 2x^ + 4 = 0, X' = 5a;'' — Qx^ — 4:X, X" = 4x» + 6x'' — 20, X'" = — 42a;2 — 168x + 300, X'' = — 2880x + 3840, X" := — , a constant. 2, — 3, &c., in the foregoing series, we will X", x^ + — , three variations, + — , three variations, + — , four variations. And, since — 2 gives the same number of variations as — oc, it is the superior negative limit. And, since a gain of variation occurs be- tween — 1 and — 2, the entire part of the negative value is — 1. In this case, we know that there is but one negative value ; we may then proceed at once to find the decimal part of this value by the usual process for irrational values, without forming the functions, Y, Y', Y"; &c. Substituting 0, — 1 _2,_3,& have X, X', X", X' When, x= + - + " x = —l + + — + " x=-2 - + — + 488 Sturm's theorem. . The transformed equation in ^ is 2^' — 5/ + 8/ — 6^^ + 3y + 3 = 0, and that in z. z5_ 502^ + SOO^^' — 6000/ + 30000z + 300000 = 0. A change occurs between 4 and 5. Hence, 4 is the tenths of the sought value, and we have x — — 1-4 for the approximate value. Now, diminish the values in the equation in z by 4, and the trans- formed equation in s will be 8^ — 208" + 1760s=' — 21040s2 + 130480s + 21536 = 0, and that in w will be, Mj5— 200wHl76000w;='— 21040000«;2-^1304800000w+2153600000=0. A change of sign occurs between the results after the substitution of — 1 and — 2. Hence, 1 is the hundredths of the negative value. And we have, for a second approximation, x = — 141. Hemarh. 496. There is one point of considerable importance in the demon- oration, which deserves to be attended to. It has been shown that a variation is lost or gained between X and X', every time that X be- comes equal to zero, or that x passes a value. It might be asked, then, Why not confine the substitutions to X and X' ? It is to be observed, that a change of variation only takes place be- tween X and X' when very minute substitutions are made. For, in the demonstration of the fourth principle, we supposed ii to be indefi- nitely small. Now, if we substitute a number, p, which will make the signs of X and X' contrary, and again substitute another, jj', there being two values of X, and none of X', between p andp', then X and X' will still be affected with contrary signs. So, that, between p and p', there will be no change of variation, though there are two real values. But, if one value of X' lies between p and p', there will be one variation lost, and but one. For, whilst X changes its sign twice, X' will change its sign once. Take X = ex'' — 5u; ^- 1 ; then, X' = 12x — 5. stuem's theorem. 489 There is a variation between X and X' when x = 0, and this is changed into a permanence when x = 1. There are two values of X between and 1, and one of X'. Again, if there were three, five, or any odd number of values of X passed over, and none of X', there would be one variation lost or gained, and but one. But, if, at the same time, an odd number of values of X' were passed over, there would be no change of variation. It is plain, then, that the loss and gain of variation between X and X' will only correspond to the number of real values, when the substitu- tions are so minute that no value of either X or X' is contained be- tween them. Scholium. 497. Whenever any function is constantly positive for all values of X, we need not form any other functions, but only count the number of variations given by + oo, and — oo, in the constantly positive func- tion, and in the functions which precede it. For, if we formed the succeeding functions, they would give tlie same number of variations for -(- CO that they would for — oo. Hence, the difference between the number of variations of these functions must always be zero. To show this, we take for granted that a function, which always remains positive for all values of x, must be of an even degree, since it will contain only imaginary values. The next succeeding function may have its leading term either positive or negative, but the nest must be negative, since any value that reduces the function, consecutive with the positive one to zero, must cause the adjacent functions to be affected with contrary signs. All the functions of an odd degree may be either positive or negative, but those of an even degree must be alternately positive and negative. Let x^ + mx^ + &c., be the constantly positive function. We will then have the series, + x^ + mx^ + &c. = X', ± n^K^ + &c. = x^n — px'^ — &c. = X^+^ dr rx'^ + &c. = X^, + sx^ + &c. = x^, ±tx -\- &c. = x^^ — A, a constant, = x-n 490 Sturm's theorem. Now, suppose the ambiguous sign to be plus througliout, we will have for -f oo, these results, + H h + H J 3 variations, and for — oo, these results, -\ 1 , 3 variations. Next, suppose the ambiguous sign to be minus throughout, then, — 00 will give + -\ H + H J 3 variations, and + oo will give H 1 , 3 variations. We will find like results when some of the ambiguous signs are taken as positive, and the rest as negative. GENERAL EXAMPLES. 1. Given, x^ — x^ — 7 = 0, to find x. X = x'' — x' — 7 = 0, X' = Sx'' — 2x, X" = 2x + 63, X'" = — , a constant. + oo gives + + H , 1 variation, — oo gives (- ^ , 2 variations. Hence, one real value, and x = 2-310. 2. Find one value of x in the equation, x'^ — 2x' -f x^ — 5 = 0. Ans. X = 2-076. X = x* — 2x^-j-x^ — 5 = 0, . X' = 4x^ — Qx^ + 2x, X" = x' — x + 20, X'" = 2a; — 1, X'^ = — A, a constant. 4- 00 gives + + + H , 1 variation, — oo gives -\ 1 , 3 variations. Hence, two real values. DESCARTES RULE. « 3. Given, a;'' — a; — 7 = 0, to find x. Ans. x = 2-086. X = a;=' — a; — 7 = 0, X' = 3x^—1 X" = 2x + 21, X'" = — A. + 00 gives + + -I , 1 variation, — ex gives 1 , 2 variations. Hence, one real value. 4. Given, x^ — x^ + 7 = 0, to find x. Ans. a: =: — 1-63. X = a:^ — a;* + 7 = 0, X' = 3x'' — 2a:, X" = 2a; — 63, X'" = — A. + 00 gives + 4- H J 1 variation, — oo gives 1 ,2 variations. Hence, one real value. DESCARTES' RULE. 498. ^4.?!. equation cannot have a cp-eater number of negative values than there are permanences of sign from + to +, or* from — to — ; nor can it have a greater number of positive values than there are variations of sign from + to — , or from — to -\- . When the adjacent terms of an equation are affected with the same sign, a permanence is said to exist between them ; and when they are affected with contrary signs, there is a variation between them. In applying Descartes' rule to an equation, every sign is read twice, except the first and last. Thus, in the equation, x* — 4x^ -f 12a;^ + X — 5 = 0, there is a variation between the first and second terms, a variation between the second and third terms, a permanence between the third and fourth terms, and a variation between the fourth and fifth terms. In all, there are three variations and one permanence. 492 DESCARTES' RULE. When a term is missing from an equation, it must be supplied witli tlie coefficient, ± 0. Then, in reading the signs, we must first count the variations and permanences, regarding the ambiguous sign as posi- tive, and then count again, regarding the ambiguous sign as minus. Thus, taking the equation, x^ — 4 = 0, we must write it x^ ±0x — 4 = 0. Counting the upper sign, we have a permanence between the first and second terms, and a variation between the second and third terms. Counting the lower sign, we have a variation between the first and second terms, and a permanence between the second and third terms. In whatever way, then, we read the ambiguous sign, there will be one permanence, and one variation. If there are any number of missing terms, they must be supplied in like manner with positive or negative zero coefficients. 499. To demonstrate the rule of Descartes, it is necessary to show that the multiplication of any equation by a factor, corresponding to a negative value, will introduce into the new equation at least one more permanence than existed in the old ; and that the multiplication by a factor, corresponding to a positive value, will introduce at least one more variation. 1st, Let us take the equation, x"' — Px"-' + Qx'^-^ + Rx™"" — Sa;""^ . . . . + Tx — U ^ 0, and multiply by a factor, x + a, corres- ponding to a negative value. The resulting equation will be a;m+l_p|.^n, + Qbm-> + R|x'»-^— S|x»-=' -fTlx"— Ulx = -fal — Pal +Qa\ +'Ra\ — Sal +Tal— Ua ^■' Now, it is plain that, if we suppose the coefficients in the upper column are, throughout, greater than the corresponding coefficients in the lower column, there will be the same number of permanences in the new as in the old equation, until we get to the last two coefficients (the coefficients of x and x°), we will then have one more permanence than in the given equation. And, if we suppose the coefficients of the lower column to be greater throughout, there will be a new permanence between the first and second terms, and the same succession of signs in the remaining terms. Moreover, it is evident, that if the lower column sometimes prevailed, and sometimes did not, there might be more than one permanence introduced. For instance, if a were greater than P, Pa less than Q, Ra greater than S, T greater than Sa, Ta greater than U, there would be but one variation in the new equation. DESCARTES' RULE. 493 It is even possible to change all the variations into permanences, by multiplication by a factor corresponding to a negative value. As an illustration, take the equation, x^ — 2a;* + \2x* — 8x^ + 36x- — x •\- 15 = 0, and multiply by cc + 4. The new equation will be, x-' — 2.\x^ + 11\x^— Slx^ + 36|a;3— 1 j x^ + 15 Ix = + 41 —si + 48 I — 32 I + 144 I — 4 | + 60' or, x' + 2a;« + ^x' + 40^" + 4x=' + 143x'= + llx + 60 = 0. By this example we see that an equation, containing only variations, is changed into another containing only permanences, by multiplying the former by a factor corresponding to a negative value. Seven per- manences have been introduced where none existed before. And, by recurrence to equation (M), we see that it is impossible to read the signs in any order, without having one more permanence than in the given equation. Now, the given equation may have been one of the first degree, m being equal to one, and P, Q, K, S, and T being equal to zero. Then, if the sign of U were positive, there would be one negative value, and conversely. The new equation (after multiplica- tion by x + ci) would be of the second degree, and would contain, at least, one more permanence than the old. And, by multiplying this new equation by another factor corresponding to a negative value, we would introduce, at least, one more permanence. And so, by contiuu ing the process, it could be shown that, whatever might be the degree of the equation, the number of permanences must always be equal to, or exceed the number of negative values. 500. 2d. By a similar course of reasoning, we could show that the the multiplication by a factor corresponding to a positive value would introduce, at least, one variation ; or, in other words, that the number of positive values can never exceed the number of variations. Take, as an illustration, the equation, a;6 _ 4x5 + i2x' + 8.x3 + SO.r^ — x -f 15 = 0, and, multiply it by the factor, x — 2, the resulting equation will be, a;' — 6x« -f 20x5 — lOx'' + 14x' — 61x2 -f 17x — 30 = 0, and has gained three variations. 501. It is plain that the preceding reasoning has been on the sup- position that a was a real value, otherwise we could not, in equation 42 494 DESCARTES' RULE. (M), have instituted any comparisons betweeij a and — P, — Pa, and 4- Q, + R«> and — S, &c. In case, then, that there are imaginary values in an equation, the rule of Descartes only points out limits, be- yond which the positive and negative values cannot go. But, when the equation contains only real values, the number of positive values will be exactly equal to the number of variations, and the number of negative values exactly equal to the number of permanences. To show this, let m = degree of the equation, ?i = number of real negative values, p = number of real positive values, h = number of variations, b' = number of permanences. A slight inspection will show that, in case of real values, b + b' = 7n ; and we know that n + p = m. Hence, b -\- b' = n -\- p. But, since n cannot exceed b', and p cannot exceed b, we must have n = b'. For, if ra <^ 6', then necessarily p^ b, which cannot be. In like manner, we could show that we must havej9 = 5. Corollari/. 502. In case of there being one or more missing terms in an equa- tion, the rule of Descartes will enable us to detect imaginary values. We have only to supply the missing terms with plus or minus zero coefficients, count the variations and permanences when the upper sign is taken, and then again, when the lower sign is taken. If there be any discrepancy in the results, there will be imaginary values. Thus, take cc^ + 4 = ; supplying the missing term, we have x^ ± Ox -^ 4: = 0. The upper sign taken in connection with the other two, gives two permanences, whilst the lower gives two variations. These dis- crepant results indicate imaginary values. Take x*-\-Sx^^4: = 0, then, x* db Ox' + Sx^ ± Ox — 4 = 0. In one case, we have one variation and three permanences ; in the other, three variations and one permanence. The difference in the two readings again indicates imaginary values. GENERAL EXAMPLES. 1. What are the values in the equation, x^ — 2x^ — 7x -|- 1 = ? 2. What are the values in the equation, x" — 7x -f 1 = ? 3. What are the values in the equation, x'' — 6x* -f 7x^ — Sx* -|- 9x«4- 10x2 — 11x4- 12 = 0? ELIMINATION BETWEEN TWO EQUATIONS. 495 4. Wliat are the values in the equation^ x" — 1 = 0? 5. What are the values in the equation, x^ — 1 = 0? 6. "What are the values in the equation, x^ + 1 = ? 7. What are the values in the equation, x^ + 4a;^ + 6x^ + 4a; + 1 = 0? ELIMINATION BETWEEN TWO EQUATIONS OF ANY DEGREE. 503. When one quantity depends upon another for its value, it is said to be a function of the quantity upon which it depends. Thus, in the equation, y = 'Zx — 4, ^ is a function of x, because every change in the value of x will produce a corresponding change in the value ofy. The mathematical symbol to designate a function may be F, or/, or the Greek letter, $. Thus, to indicate that i/ is a function of x, we may employ the notation, y =. F(:c), or y =/(x'), ox y ^ 4'(a:)- The second member may contain constants, as well as the variable, x. Thus, 3/ is a function of x in the foregoing equation, y = 2.x — 4, the con- stants being 2 and — 4. 504. The most general form of an equation of the m*" degree be- tween two variables, x and y, is, cc" -f Bx"-' + Cx---^ + Dx"-^ + Ex"-^ + U = ; in which B, C, D, &c., are functions of ?/. B is supposed to be of the first degree in y, and of the form, a -\-hy. C is of the second degree in y, and of the form, c -\- cly -\- ey^. D is of the third degree in y, and of the form, f -\- gy ■\- hy^ + 1y^- E is of the fourth degree in y, &c. &c. U is of the m}^ degree in y, and of the form, u -f my + +!/°7 and it does not contain x. 505. The equation is said to be complete when x enters into all the terms but the last, y into all the terms but the first, and when, also, the sum of the exponents of x and y in each term is equal to m. 496 ELIMINATION BETWEEN TWO EQUATIONS. 506. Eliminatioa between equations of a degree higher than the first is usually effected by means of the greatest common divisor. This method of elimination has already been explained (Art. 214), but we propose to demonstrate the process more rigorously, in two ways, 1. Let A = 0, and B = 0, be the proposed equations containing both X and y. Now, if we knew beforehand a value m of y, that was common to the two equations, A = 0, and B = 0, and substituted this value in them, the new equations. A' = 0, and B' = 0, would contain only x and constants. Now, if n be a value of a;, in the equation. A' = 0, its first member must be divisible by x — n, and it is plain that the given equations would not be simultaneous unless n would also satisfy the equation, B' = (Art. 207). Hence, x — n must also be a divi- sor of the equation, B' = 0. We see, then, that the hypothesis of a common value in y results in the condition of a common divisor in x. Conversely, if we can force the given equations to have a common divisor in x, they must have a common value in y. It is upon this principle that we seek for a common divisor between the first members of the equations, A = 0, and B = 0, and continue the process until we get a remainder freed from x. It is plain that, if we place this remainder equal to zero, and substitute the value of y, found from it in the last divisor, it will be an exact divisor, and will be the one sought. From the foregoing reasoning it is evident, that if it be absurd to place the remainder, freed from cr, eqvxal to zero, the given equations are not simultaneous. 507. 2d. Let the successive quotients, in the process of dividing A by B, B by the remainder, &c., be designated by Q, Q', &c., and the successive remainders by R, R', &c. Then we will have. A = BQ 4- R, (M) B = RQ' + R', (N) • R = R'Q" + R", (0) &c. &c. Now, since, by hypothesis, A = 0, and B = 0, equation (M) will give R = 0. And, since B = 0, and R = 0, equation (N) will give R' := 0. From this we see that we have a right to equate, with zero, that remainder which is freed from cc, and contains only y. 508. The above series of equations show, moreover, that if the re- mainder, which is freed from a-, and the preceding divisor be placed OF ANY DEGREE. 497 equal to zero, the system of values so found will satisfy tlie given equations. For, if E." be that remainder, and R' the preceding divisor, when E,", and R' are equated with zero, equation (0) shows that R also = 0. And R and R' being equal to zero, from (N) we get, B = 0. And, since B = 0, and R = 0, equation (M) shows that we will also have A=:0. Hence, the values of ?/, found by placing the last remainder equal to zero, may be substituted in the preceding divisor, in order to deduce the corresponding values of x. The importance of this remark consists in the fact, that the preceding divisor is of a lower degree than either of the original equations, and, therefore, more readily solved. 509. The reasoning in Art. 507 proceeds upon the supposition that the successive quotients, Q, Q', Q", are all finite, and then, of course, the successive products, BQ, RQ', R'Q", &c., will all be zero, when B = 0, R = 0, R' = 0, &c. But, if any of these quotients be frac- tional in form, it may happen that the value of y, found from placing the last remainder equal to zero, will reduce the denominator of the preceding divisor to zero also. In that case, we would have the product of zero by infinity, which is indeterminate.* Suppose, for 4 example, R = y — 1, Q = —^ . Then, A = BQ -f R becomes A = ^(-A—\ + y — 1 : or, when R = 0, A = B^ = B oc ; or V — y (since A and B are zero), ^ oo, which may, or may not be, a true equation. 510. To avoid the fractional form of quotient, it may sometimes be necessary to multiply the dividend by some function of y, though this multiplication may possibly introduce some foreign values into the equation, as will be shown more fully hereafter. 511. We will now illustrate the foregoing principles by a few examples. * Let 00 :iz A, dividing both members by 0, we get oo == — - =: oo, a true equation. And this will evidently be true when A = 1, 5, 10, 20, or anything whatever. 42* 2a 49S ELIMINATION BETWEEN TWO EQUATIONS Let a;2 + J/'' — 8 = 0, and 2a; — 3y + 2 = 0. 4 4x2+4/— 32 \2x — ^y + 2 4ic^+6.ry+4.x 2a: + (By — 2) = Quotient. 1st Kemainder = 2(3?/— 2)a; + 4/— 32 2(3y— 2)a:— 9/+12^— 4 2d Remainder = 13^2—12^—28=0. From which we get y = 2, and y" = — i| ; and these, when sub- stituted in either of the given equations, give x' = 2, and x" = — 1|, Let x^ — / — 7 = 0, and a; — y — 1 = = 0. Combining, we will get a;^ + (^ + \)x + (/ + y) + (^ + 1) for a quotient, and /+ t/ — 2 for a remainder. The equation, formed by placing the remainder, freed from x, equal to zero, is called i\i.e final equation. In this example, the final equa- tion, y^ + y — 2 = 0, gives %f = 1, y" = — 2. And these values for y, when substituted, give x' — 2, and x" = — 1. Let x" — Syx^ + 3/a; — 5a;* -f lO^a; + 6a; — / — 5y^ — Qy = 0, and a;' — 6yx^ + Sy^x — x — 4/ -|- y = 0. Then the first quotient is -|- 1, and first remainder (2y — 5)x^ — 5y^x -\-7x + IQyx + By^ — 5/ — 7y. Preparing the last divisor for division by multiplying by (2y — 5)^, we have x^—Syx^-^SiJ'x—x—iy^+T/ (2^-5)" (2y-5)^x'-20y^x^+32y''x+im>/^x+lOy^a^-12byx'^-imy^x+20yx-25x-16y^+80y*-96y^-20y<^+25]/, and this, when divided by (2y — 5)x^ — by^x + 7x + lOyx + 3/ — 5/, gives as a quotient, (2y — 5)x — 5/ -f 15y — 7, and as a re- mainder y*x — 10/a; + B5y^x — bOyx + 24a; — y^ + lOy* — 35^' + 50/ — 24y. And, by factoring this remainder, we get (/ — 10/ -f- 35/ — bOy + 24)x — 3/(/ — 10/ + 35/ — 50y + 24) ; or, (a; —y) (/ — lOy^ + 35^2 — 50?/ + 24). Rejecting the factor, x — y, as leading to arbitrary values, we have the final equation, y* — lOy^ -\- 35/ — 50y + 24 = 0. This equation, when solved by the process of divisors, gives y' = 1, / = 2, /" = 3, and y'"^ = 4. And these, when substituted, give x' = 3, x" = 5, x'" = 5, a;'^ = 7. 512. Two things are suggested by this example. 1st. May not the multiplication by the factor, (2y — by, involving One of the unknown quantities, have introduced foreign values ; that is, values which did not enter the given equations ? 2d. x — y being a common factor to OP ANY DEGREE. 499 the remainder, is found to be also common to both the given equations. How are such factors to be treated ? We will examine these subjects separately. 513. 1st. In regard to foreign values, we have this simple test. Place the multiplier equal to zero, find the value for y, and, from either of the given equations, the corresponding values of .r ; if these values be the same as some of those found from the final equation for y, with the corresponding values of x, then the system of common values in both X and y must be rejected. In the example, placing {2.y — 5)* ^ 0, we get i/ = I, a value difi'eient from those before found. Hence, the multiplier has not introduced a foreign value. But, if the multi- plier, placed equal to zero, had given us, for example, y = \, with the corresponding x = 3, then this system of values must be rejected. 514. How are factors to be treated which are common to both of the first members of the given equations ? There may be three cases, but all lead to arbitrary values. 1st. The common factor may be a func- tion of X only, fix). 2d. It may be a function of y only, f{y). 3d. It may be a function of both x and y, f(x, y). When the common factor is /(x) only, x will have determinate values, and y indeterminate. For, by placing f(x) = 0, we will get true values for x ; but, when f(x) = 0, both equations will be satisfied, whatever values y may have. Take the equations, (a -\- hx') (x?y + 2fx — ay) = 0, and (a -f hx) (Axhf — 2yx + my"-) = 0. Placing a + Ja; ^ 0, we get x = , and both equations will be satisfied when a + hx = 0, whatever may be the values of y. Hence, /(a;)=0 gives x determinate, and y indeterminate. In like manner, a common factor, f(y), would give y determinate, and x indeterminate. In the the third case, the common factor, /(x, y) = 0, will satisfy both equa- tions ; but, since we have a single equation, f(x, y) = 0, containing two unknown quantities, the values of both x and y must be indeter- minate. Thus, take the equations, (2x — 4ty) (a? -}- ly — 5) = 0, and (2x — 4y) (xy — 1x^ + bx^ — 7/) = 0. It is plain that both equations will be satisfied when 2x — 4y = ; but the equation, 2x — 4y = 0, will give indeterminate values for both x and y. Hence, we conclude that, when the given equations contain a com- mon factor, it must be divided out. For, a common factor, /(x), would give determinate values for x, but indeterminate for y; a common factor, f(y), would give determinate values for y, and indeterminate 500 ELIMINATION BETWEEN TWO EQUATIONS for x; and a common factor, /(a;, ^), would give x and y, both inde- terminate. 515. Take the equations, (x — 1) (x^ — 2xy -fa; — 2) = 0, and (x^ — XT/ — 2) (x — 1) = 0. Suppressing x — 1, we have x^ — 2xy + X — 2 = 0, and x^ — xi/ — 2 = 0, from which we get y = 1, and x = 2. Take the equations, (2y — 6) (xy + 5a-?/ — y -f 1) = 0, and (2i/ — 6) (xy -f- 4y — 4) = 0. Suppressing 2y — 6, and combining the resulting equations, we get IQy^ — 53j/ -f 37 = for the final equation. From which, ?/' = 1, y = ^l, x' = 0, x" = — ||. Take the equations, (2x — 7y) (xi/ — y -f 5a;^ — 5a;) = 0, and (2a; — 7y) (xi/ -f- 7a; — 7 — y) = 0. Removing the common foctor, we get, after combination, 5a;^ — 12 -f 7 = for the final equation. From which, x' = |, x" = 1, and, by substitution, y' = — 7, and 516. When the first members of the given equations can be resolved into factors of the first degree, or of a low degree, the elimination will be greatly facilitated. Take the equations, (x — 1) (?/x — 3) (a;^ — 2xy') — 0, and (^x — 2a;) (x^ — 2^) = 0. These equations can, obviously, be satisfied when X — 1=01 when a; — 1 = 01 when ya; — 3 = 01 and yx — 2a; = 01^ ''and x^— 2y = o|*^ -^ and i/x — 2x = o\^ '' when 7/x — 3=0 I when x^ — 2xy=0 I when x^ — 2xy=0 j and x'—2y=0\^Knd x^— 2y=0 1^ \nd i/x— 2x=o\^ ^' From (A) we get the system of values, ic = l, and y = 2. " X = 1, and y = i. " X = ^, and y = 2. " a; = VF, and x =: ^=z. • ^6 " X = 1, and y = g. « x' = 0, x" = 4, and / = From (B) From (C) From (D) From (E) From (F) y" = 2. An artifice will sometimes enable us to decompose the first members of the given equations into their respective factors. OP ANY DEGREE. 501 Take the equation, cc^ — lyx, + 6x 4- 2/* — % + 5 ^ 0, (A), and X* + 2^a; + / + 6a; + 6y + 5 = 0, (B). From (A), we get x^ — lyx + / + 6x — 6j^ + 5 = 0, or (x — ^)^ + 6(x — ^) + 9 — 4 = ; or (since tte first three terms constitute a perfect square), (x — y+3y _4 = (x— y + 3/ — (2y = (a; — y + 5)(x— y + l)(Art. 50). From (B), we get (x + y^ + 6(x + ^) + 9 — 4 = 0, or (x + y + 3)* — 4=(x+y + 3)2 — (2)^ = (x+3/ + 5) (x+y + 1) (Art. 50). Hence, we have (x — y + 5) (x — y + 1) = 0, (A'), and (x + y + 5) (a; + y + 1) = 0, (B'). And (A') and (B') will evidently be satisfied, when X — ^y-}-5 = 0| when x — y + 5 = 0| when x — y-\-\=.^\ and x+y + 5=0l'^ -^ and x+y + l=Or^ and x^y^b=^^' and when x — y + 1 = and X + y + 1 = (Q). From (M) we get the system of values, x = — 5, and y = 0. From (N) " " « x = — 3, and 3/ = + 2. From (P) " « « X = — 3, and y = — 2. From (Q) " " « x = — 1, and y = 0. Take the equations, x* — 3yx — 3x + 2/ + 7y — 4 = 0, (A), and x^ — 4x + x^^ — 4y = 0, (B). From (A) we get (by making 2/ =l -| ^, and adding and sub- tracting I), (-l)"-^(-D+l-(l-l+?)=o or, (-^ - 1 - I) - (f - |-)'-:(^-y-4) (a;-2y+l) (Art. 50). From (B), we get x(x — 4) + y (x — 4) = (x + y) (a; — 4). Hence, we have (x — y — 4) (x — 2y + 1) ^ 0. and (x + 2/) (a; — 4) = 0. From which we get the equations, a;_y_4=0| a;— y — 4 = 1 cc — 2y+l=0 and x+y = Ol^^^' x — 4 = 0r^' x + 2/ = F^' x-2y + l = 0| X — 4 = 0P^ 502 ELIMINATION BETWEEN TWO EQUATIONS From (R), we get the system of values, a; = 2, and y = — 2. From (S), " " " x=i 4, and i/ = 0. From (T), " " « x = — {, and ?/ = + \. From (U), =z 4, and y = ^. Take the equations, x^ + y^ — 2yx — 4^ + 4a3 = 0, (A), and a-y — y^ — 6a;y + 5 + 4y = 0, (B). From (A), we get (x — yf + 4 (x — y) = 0, or, {x—y)(x — y+^) = (), (A'). And from (B), we get (by adding and subtracting 4), xY — ^xy + 9 +4y— / — 4 = 0, or, (xy-%r-(^y-2J = ^, or, {xy^y-h^ {xy-y-V,=.^, (B'), (A') and (B') give the system of equations. X — y = and ccy + y — 5 = X — y + 4 = 0" ^y — y — 1 = m^ x—y = Q ^y—y—l=0 (H), x — 2/4-4 = 01 xy-{-y — b=(i\ (I), (K). From (G), we get the values. ..^_, + 41,.^^=:zW2i, y' — l + v/21 1 — ^/21 From (H), we get the values, — -- 2 J' l+v/5 x/5 2/" = 2 _^ 1 — ^/5 From (K), we get the values. 5+^/29 ,,_ — 5 — -v/29 ^ - 2^3; , 3 — ^29 y =+ r, — . 2 _' 3 +^/29 OF ANY DEGREE. 503 517. One of the equations only may be capable of decomposition into factors. The equations may be of the form, A = 0, and BD = 0. We will then have two systems of values j one resulting from A = and B = 0, the other from A = and D = 0. Take the equations, xy — Ixy — 20x = 0, (A), and {xy — 5a; + 3) (x — 2) = 0, (BD). We get the system of equations, xY — Ixy — 20x = 0, (A), and cry — 5x + 3 = 0, (B). From which, x' = 3, and x" = |, y = 4, and y" := — |. (A) and (D) give xhf — Ixy — 20x = 0, and re — 2 = 0. , . , , ^ , 7-^^/209 ,, i — ^lm From which we get x =z 1, y = , y = -r . 518. If any of the successive remainders be capable of decompo- sition into factors, which are functions of x or y, these factors may be placed, separately, equal to zero, and the deduced values of x or y sub- stituted in the preceding divisor. For, let B, be one of the dividends, R' the divisor, Q the quotient, and/(x) X f{y) the remainder. Then, |, = Q + -^/^l or R = R'Q + f{x) x /{y). And this equation will be satisfied when R' =: and /(x) = 0, or when R' = &nd/(y) = 0. Take the equations, yx'^ +y^x^ — x^ — y^x -^ yx -{- ]/ — 1=0, (R). and x^ — y + 1=0, (R'). Dividing R by R', we will geflfe remainder, (x^ + 1) (y^ — 1) Placing a;* -f- 1 = 0, we get x = ± -y — 1, and this, substituted in (R), gives 2/ = 0. Placing y^ — 1 = 0, we get y = =t 1, and these values, when substituted, give x' = 0, and x" = db y/ — 2. Hence, we have the system of values, a/=+ v/— 1| x"=—^— 1 1 x'" =z I a;'^ = y/—2\ x^ z= — ^A^ 2/ = 0| y" = o\y"' = l\ y- = — 1| r = — 1 All of which will satisfy the given equations. 504 ELIMINATION BETWEEN TWO EQUATIONS 519. When it is necessary to multiply (A) by either a function of * or 2/ to make it divisible by (B), we can tell whether the multiplier has introduced foreign values, by combining it, placed equal to zero, with B = 0. If any of the values thus found are the same as those resulting from the combination of the given equations, we must reject these common values from the solutions of the given equations. For, Jet A = 0, and B = ; and suppose that (A) will not be divi- sible by B until it has been multiplied by /(a;). Then we will have A(/x) = 0, and B ^ 0, which can be satisfied when A=0, and B =0; or, when f(x) = and B = 0. Now, it is plain that, if the combination of the new equation, A.f(w) = 0, with B = 0, gives, among its system of values, the same as given by /(a;) = 0, and B = 0, we must reject the common values as having been introduced by the multiplication of /(x). Take the equations, x^y — 2xy -{- x^ = 0, (A), and x^y — 2x^ -\- x = 0, (B). Multiplying (A) by x^, to prepare for division, we get for the final equation, a;^(x' ^• 2x^ — 5x + 2) :^ 0. From which we get, x = 0, , ^ „ — 3+VT7 ,„ —3—^17 „, -. X = 1, X = , X = ^ . The corresponding 1 . / V 6v/T7-26 .„ _(26 + 6^T7) values of y are, v = ti, v = — =» ¥ = — ^^ =-^ "' ^ 38 — 10^17 38 + 10^17 (M). Now, placing /(x) = 0, and B = ; or, x" = 0, and cc' — 2x^ + x = 0, we get x=zO, and y = § ; and, since these values are the same as the first of those, marked (M), we must reject them from (M), and leave but three values for x, and three for y. 520. If A be exactly divisible by B, the values of x and y will be indeterminate. For, then, ^ =Q,; and, since A = 0, and B = 0, we will have B I] = Q, or = OQ . (P). Now, it k plain that Q may be /(cc), or f{y),oxf(x,ij). But, equation (P) will be satisfied, whatever may be the form of Q, and whatever may be the values of x or y. Take the equations, a;« _ ^x^y _ 2^'' 4- Zy^x" + 6xV — 2/'x — Qy^x + 2f = 0, and x' — 2xy + y' = 0. We get as a quotient, x^ — xy — 2x + 2y, and a remainder zero, and any value whatever of x, with the corresponding or deduced value of y, will satisfy both equations. OF ANY DEGREE. 505 Remark. The case exhibited in 520, differs only from that in 514 in this respect, there is a common factor to the two equations in both instances, but 520 does not manifest that factor. The indeterminate nature of the given equations, when the final equation in y is zero in both members, may also be shown by retaining the trace of y. For, when we place the zero remainder equal to zero, we have Oy = 0, or 2/ ^ %■ 521. When the final equation reduces to a constant, the value of y will be infinite, and the given equations will be contradictory. For, then we will have 0^/ = A, constant ; or ^ = 00, the symbol of absurdity. Hence, the combination of the given equations has led to an absurd- ity, and, therefore, these equations must be contradictory. Take the equations, y^ — 0? — Sx* + 9 — Zx = 0, and y — x — 1=0. Combining, we will have, for the final equation, 10 = 0; or, retaining the trace of y, 10 -f Oy = 0, or y = — 'gO = 00, The given equations are not simultaneous. GENERAL EXAMPLES. |/_2xy — 4x+ a;2 = 0, I y + a; — 4 = 0. Am. a/ =4, a;" = 1 ; y = 0, y = 3. \y'' — 1xy — ^x — x^ = 0, I y2_2xy — 5x — 2a;« + 2 = 0. _ Am. iK = 1, or — 2, y = 1 ± v/5, or — 2 ±x/— 4. \yx- — Zfx^ + 3y V — Ay*x^ + bifx — 2y^ = Q I a;'' — 3yx + 2/ = 0. Ans. X and y indeterminate. ^S — 2yx' + x'-\-x^ — 2yx+f = 0, \ x" — 2yx + 1 = 0. Ans. x = =i= 1; or dr^ — 3, 2/ = =t L 506 5. ELIMINATION BETWEEN TWO EQUATIONS 9. 10. 11. 12. x^ — 2i/x^ ^x" + x" — 2yx + 1=0, a;2 _ 22/^ + 1 = 0. Ans. x = ^, y=%. y"" — 2xy -{■ x^ — 2y — 1 z= 0, y^ — 2xt/ + x'^ + x = 0. Tins. x = — 1, or — ^ ; 1/ = 0, OT — |. 2/2 _ 2x1/ + 2x^ — 2y + 4: = 0, 2/" — 4xy + 7a^ — 22/2 — 5 = 0. y^ — ^xy + 5x' + 2cc + 1 = 0, y — 2a; = 0. Ans. Values imaginary. cc = — l,y = — 2. 3^2 + CC2/ — 2a;2 + 3x — 1 = 0, y'^ — X^= 0. Ans. 35 = J, or 1, or I, and y = f , or — 1, or — J. (ya;-l)(x-2)(y_4) = 0, (x + 2)(2/x-a;).(a> — 3) = 0. J.ns. Values of x. Values of y. x = —2 y = -J x^\ y= + l a; = 3 y=+i a; = y incompatible a; = 2 2, = 1 a; incompatible y incompatible a; = — 2 y = 4 a; = y = 4 a; = 3 y = 4 / — 2ar2/ + x'' + 2y=0, y2_ 2xy + cc^ — 2y — 1 = 0. }, a; = — }±v'^. 2^V — 2/x* — S^x — 2yx^ + 8yx + 16y + 2a;=' — 16 = 0, yX +^ — 03— 1 = 0. "or Final equation, y^ + y — 2 = 0. OF ANY DEGREE. 507 13. y-^' + 7 = 0, (A) /_x^ + 3 = 0. (B) Ans. a; = 2, and y = 1. After the first division, multiply (B) by (cc^ — 3)^, we will get for the final equation, Qx" — Ux^ — 21a^ + 7Q = 0. The only rational value in this equation is a; = 2. 14. y- 15 = 0, 7 = 0. Ans. x = l, and y = 2. Final equation, (x* + 15)=" — {x" + 7)* = 0. Required the values and the final equation belonging to 15. I ^6_a;s_2101 = 0, 1 2/' — X* — 369 = 0. THE END. - THT RETURN TO: CIRCULATION DEPARTMENT 198 Main Stacks LOAN PERIOD 1 Home Use 2 3 4 5 6 ALL BOOKS MAY BE RECALLED AFTER 7 DAYS. Renewals and Recharges may be made 4 days prior to the due date. Books may be renewed by calling 642-3405. DUE AS STAMPED BELOW. APR 1 6 2000 FORM NO. DD6 UNIVERSITY OF CALIFORNIA, BERKELEY 50M Berkeley, California 94720-6000 UNIVERSITY OF CAUFORNIA LIBRARY .J.*«JfQ. 'm^.y U.C.BERKELEY LIBRARIES CD^bD^3^bb