GIFT OF ROBINSON'S MATHEMATICAL SERIES. NEW UNIVERSITY ALGEBRA: THEORETICAL AND PRACTICAL TREATISE, MANY NEW AND ORIGINAL METHODS AND APPUCATl COLLEGES AND HIGH SCHOOLS. IPO BY HORATIO N. ROBINSON, LL. D., LATH PROFESSOR Or MATHEMATICS IS THE UNITED STATES NAVT, AXT> A FULL COCRSB OF MATIiEiLATlCS. NEW YORK: IYISON, PHINNEY & CO., 48 AND 50 WALKER ST. CHICAGO : S. C. GRIGGS & CO., 39 AND 41 LAKE ST. 1804. ROBINSON'S SERIES OF MATHEMATICS, The most COMPLETE, most PRACTICAL, and most SCIENTIFIC SE- RIES of MATHEMATICAL TEXT BOOKS ever issued in this country. (IN TWENTY-TWO VOLUMES.) 1. Robinson's Progressive Table-Book, ....................... 2. Robinson's Progressive Primary Arithmetic, .............. 3. Robinson's Progressive Intellectual Arithmetic, .......... 4. Robinson's Rudiments ol Written Arithmetic, ........... 5. Uobiiison's Progressive Practical Arithmetic, ............. 6. Robinson's Key to Practical Arithmetic, .................... 7. Robinson'* Progressive Higher Arithmetic, ................ 8. Robinson's Key to Higher Arithmetic, ...................... 9. Robinson's New Elementary Algebra, ........................ 10. Robinson's Key to Elementary Algebra, .................... 11. Robinson'* University Algebra, ........................... 12. Robiiisou'v Key to University Algebra,.... .................. 13. Robinson's New University Algebra, ....... . ................. 14. Robinson's Key to New University Algebra, ............... 15. Robinaon's New Geometry and Trigonometry, ............ 16. itobi'ison's Surveying and Navigation, ....................... 17. Robinson's Analytical Geometry and Conic Sections,... 18. Robinson's Din", and litteg. Calculus, (in preparation) ........ 19. u^biiison's Elementary Astronomy, ....................... 20. 'lobinson's University Astronomy, .......................... 21. Wobinson's Mathematical Operations, ....... .............. *<> Kobinson's Key to Geometry and Trigonometry, Conic Sections and Analytical Geometry, .......................... Entered, according to Act of Congress, in the year 1862, by DANIEL W. FISH, A. M., in the Clerk's Office of the District Court of the United States for the Northern District of New York. TRCAIR, SMITH A MILES, STEREOTYPERS, STRACUSR. N. T. PREFACE. In tlic preparation of the New University Algebra, care has been taken to preserve every feature of the original work, on which rested, in any degree, its claims to superiority. The aim has been to make that which was good, decidedly better. Hence the changes that have been made, con- sist, for the most part, in more apt arrangement, in large additions of orig- inal matter, and in presenting the whole in more attractive form. The treatise, as now submitted to the public, is, indeed, far more com- plete than the former, not only in the range of topics, but also in gen- eral discussions and practical applications. In many parts the methods of investigation are essentially different, the object being, in some in- stances, to secure simplicity in logical arrangement, and in others, to es- tablish principles and rules by more general and rigorous demonstrations. The articles on Inequalities, Differential Method of Series, and Interpo- Intior, which, in the old treatise, appear as an appendix, have been elabo- rated, and made to take their appropriate place in the body of the work. The section on Radical Quantities is quite full, embracing the more important properties of Imaginary Quantities and Quadratic Surds, be- sides a complete logical development of the Theory of Exponents. As, in the author's New Elementary Algebra, the Binomial Theorem / has been fully investigated with reference to integral exponents, it has been deemed unnecessary to repeat here the particular demonstration. Accordingly, the whole subject is deferred till the section on Series is reached, where a general demonstration of this theorem is given in a con- cise way, and a full variety of applications added. The whole subject, as presented in this connection, with the accompanying illustrations, can not fail to interest the lovers of Algebra. The General Theory of Equations is treated in two sections, the one embracing the general properties of equations, and the other the solution IV PllEFACE. of numerical equations of all degrees. The whole subject is here pre- sented, however, in a condensed form, the student being conducted, in a manner direct as possible, from the theoretical to the practical. The section on the Properties of Equations, it is proper to say, owes its im- proved character to the able hand of Prof. I. F. QUINIJY, of the University of Rochester, whose services in perfecting other books of this Series de- serves especial mention. The effort which has been made in this treatise, to combine the lest practical with the highest theoretical character, is specially commended to the notice of the true educator. Great care has been taken everywhere to set forth in distinct form the principles of the science, their exact log- ical relations being noted by proper references ; while due prominence has been given to those numerous precepts and expedients which are so nec- essary to the constitution of an expert Algebraist. The design throughout has been, not to conceal, but fully to reveal the difficulties of the science, and to encourage the learner, not to avoid, but to grapple with, and overcome them ; since, to the student of Mathemat- ics, labor rightly directed, is discipline, and discipline, after all, is the true end of education. It is but just to state, that J. C. PORTER, A. M., has had the constant care and supervision of the present work, having also rendered important assistance in the preparation of some other works of the Series, a fact which, considering his long and distinguished success as a teacher of Mathematics, and his acknowledged ability as a mathematical scholar, ought to afford a sufficient guarantee* for the utmost accuracy and class- room fitness on every page. Thus distinguished for fullness of matter ; for scientific arrangement; for ample discussion ^nd rigid demonstration ; for clear statement and close definition; for rules brief and of easy application; for examples numerous, apt and strictly practical; for the nicest adaptation to the purposes of teaching , for the finest mechanical execution ; for whatever, in short, care, skill, science and taste can accomplish ; the New University Algebra is submitted to the public. July, 18G2. CONTENTS. SECTION I. DEFINITIONS AND NOTATION. PAGE, General Definitions 9 Symbols of Quantity 10 (symbols of Operation 10 {symbols of Kelution 12 Composition of Algebraic Quantities T 14 Axioms 15 Exercises in Algebraic Notation 16 Computation of Numerical Values - 17 Signification of Plus and Minus Signs 18 ENTIRE QUANTITIES. Addition 20 Subtraction 25 Multiplication 31 Formulas and General Principles 36 Division 39 Exact Division 43 General Relations in Division 44 Reciprocals, Zero Powers, and Negative Exponents 45 Divisibility of a b m 46 Greatest Common Divisor 52 Least Common Multiple f 60 FRACTIONS. Definitions and General Principles 64 Reduction 66 Addition , 74 Subtraction 76 Multiplication v 77 Division 79 Reduction of Complex Forms .-. : 81 (v) VI CONTENTS. SECTION II. SIMPLE EQUATIONS. Definitions 83 Transformation of Equations. 85 Reduction of Simple Equations 89 Problems 94 Two Unknown Quantities 103 Elimination 104 Three or more Unknow r n Quantities 112 Problems 118 General Solution of Problems 124 Discussion of Problems 130 Nothing and Infinity 135 Interpretation of Anomalous Forms 136 Problem of the Couriers 138 Inequalities 145 SECTION III. POWERS AND ROOTS. Involution 151 Powers of Monomials 151 Powers of Fractions 154 Discussion of Negative Indices 155 Powers of Polynomials 157 Polynomial Squares 158 Evolution ^ ICO Roots of Monomials 161 Square Root of Polynomials 164 Square Root of Numbers 166 Cube Root of Polynomials 172 Cube Root of Numbers 176 SECTION IV. RADICAL QUANTITIES. Reduction of Radicals 182 Addition of Radicals 187 Subtraction of Radicals 189 Multiplication of Radicals IflO Division of Radicals 191 Powers and Roots of Radicals 193 General Theory of Exponents 197 Imaginary Quantities : . . .201 CONTENTS. Vll Properties of Quadratic Surds 204 Square Root of Binomial Surds 206 Rationalization 208 Radical Equations 212 SECTION V.; QUADRATIC EQUATIONS. Pure Quadratics 216 Affected Quadratics 218 Second Method of completing the Square 221 Treatment of Special Cases * 224 Equations in the Quadratic Form 228 Examples of Equations Solved like Quadratics 232 Promiscuous Examples in Quadratics 234 Simultaneous Equations containing Quadratics 236 Examples of Simultaneous Equations 243 Theory of Quadratics 247 Discussion of the Four Forms 250 Discussion of Problems 252 Interpretation of Imaginary Results 253 Problem of the Lights 254 Problems producing Quadratic Equations 258 SECTION VI. \ PROPORTION, PERMUTATIONS AND COMBINATIONS. Proportion 265 Propositions in Proportion 267 Problems in Proportion 274 Permutations and Combinations 278 Examples of Permutations and Combinations 283 SECTION VII. OF SERIES. Arithmetical Progression 285 Application of the Formulas 287 The Ten Cases 290 Problems 291 ^Geometrical Progression 293 Application of the Formulas 294 Problems 298 Identical Equations 301 Decomposition of Rational Fractions 306 VI 11 CONTENTS. The Residual Formula 808 Binomial Theorem 310 Application of the Binomial Formula 313 Method of Substitution 317 French's Theorem 318 Development of Surd Roots into Scries 320 Expansion of Fractions into Series 323 Method of Indeterminate Coefficients 325 Reversion of Series 328 Summation of Infinite Series 331 Recurring Series % 332 Differential Method 336 Interpolation 340 Logarithms 343 Properties of Logarithms 343 Rules for Computation 345 The Common System 34G Computation of Logarithms 347 Use of Tables 353 -Exponential Equations 857 SECTION VIII, PROPERTIES OP EQUATIONS. General Properties 359 - Commensurable Roots 370 __ Derived Polynomials 374 __ Composition of Derived Polynomials 375 - Equal Roots 370 _ Transformations 380 Detached Coefficients 388 ^Synthetic Division 392 vSurd and Imaginary Roots 398 Rule of Des Canes 400 ^. Cardan's Rule for Cubics .401 SECTION IX. SOLUTION OF NUMERICAL EQUATIONS OP HIGHER DEGREES. _- Limits of Roots : 405 Limiting Equation 408 Sturm's Theorem 410 . Homer's Method of Approximation 416 A TREATISE ON ALGEBRA. SECTION I. DEFINITIONS AND NOTATION. 1. Quantity is anything that can be increased, diminished, or measured ; as distance, space, weight, motion, time. A quantity is measured by finding how many times it contains a certain other quantity of the same kind, regarded as a standard. The conventional standard thus used is called the wilt of measure. . Mathematics is the scieuce which treats of the properties and relations of quantities. It employs a peculiar language, con- sisting of symbols, to express the values of quantities, and the operations to which these values are subjected. The symbols are of three kinds, as follows : 1st. Symbols of Quantity, consisting of figures or numerals used in arithmetical computations, letters and other characters used in general analysis, and graphic representations or drawings used in geometrical investigations. lid. Symbols of Operation, consisting of the signs or characters employed to indicate those mathematical processes by which quanti- ties are made to undergo changes of value, such as addition, sub- 'traction, multiplication and division. 3d. Symbols of Relation, consisting ot the signs used in com- paring quantities with respect to their relative magnitudes, and certain abbreviations employed in the process of reasoning. 3. Algebra is that branch of mathematics iu which quantities are represented by letters, and the operations arid relations are indicated by signs. The object of algebraic notation is to abridge and generalize the analysis of mathematical problems. Algebra is therefore a species of universal arithmetic. (9) 10 ALGEBRAIC .QUANTITIES SYMBOLS OF QUANTITY. 4. An Algebraic Quantity is a quantity expressed in algebraic language. There are two kinds of algebraic quantities known and unknown. o. Known Quantities are those whose values are given ; when these are not expressed l>y figures they are represented by the leading letters of the alphabet, as a, b, c, d. . Unknown Quantities are those whose values are to be deter- mined ; they are represented by the final letters of the alphabet, as u, x, y, z. f . The small italic letters just given are the more common sym- bols of quantity. In addition to these, capital letters are sometimes employed, as A, B, 6', D, X, Y, Z, etc. Quantities which have like relations to a series of quantities in any investigation, are sometimes represented by a single letter repeated with different accents, as a, a', a", a'" a"" , read, a, a prime, a second, a third, etc.; or by a letter repeated with different subscript figures, as a, a,, 3 , a 3 , a 4 , etc., read, a, a sub one, a sub two, a sub three, etc. In certain investigations it is convenient to represent quantities by the initial letters of their names. Thus, S or s may represent sum ; D or called plus. It indicates that the quantity written after it is to be added to the other quantity or quantities in the expression. Thus, in a-{-b, the sign indicates that the quantity b is to be added to the quantity a; and the expression is read, a plus b. 9. The Sign of Subtraction is a short horizontal line, , called minus. It indicates that the quantity written after it is to bo subtracted irom the other quantity or quantities in the expression. DEFINITIONS AND NOTATION. 11 Thus, in a b, or 6-j-a,the minus sign indicates that the quantity I is to be subtracted from the quantity a ; and the expression is read, a minus b, or minus b plus <(. The sign may be written between two quantities to indicate that their arithmetical difference is to be takejp, when it is not known which is the greater. 10. The Double Sign, , is written before a quantity to indi- cate that it is to be both added and subtracted ; it serves to unite in a single expression two combinations of the same quantities. Thus, ab is equivalent to a-^-b and a b, and is read a plus 01 minus b. 11. The Sign of Multiplication is the oblique cross, x. It in- dicates that the quantity before it is to be multiplied by the quantity after it. Thus, in ax>, the sign indicates that a is to be multiplied by b. Instead of the sign, x, a point is sometimes used to denote multiplication ; as 3 >'x m y, which signifies the same as o x x x,y. The multiplication of quantities which are represented by letters, is generally indicated by writing the factors one after another with- out any intervening sign. Thus, 3abc signifies the same as 3 xa x b x c, or 3-a-b-c. It is evident that this notation cannot be employed when the several factors are represented by figures. We cannot represent 3 times 4 by simply writing the factors together, thus, 3 4 ', for the product thus indicated could not be distinguished from the number 34. NOTES. 1. The result of any multiplication is called a product, and the quantities multiplied are called factors. 2. When the quantities to be multiplied are represented by letters, they are called literal factors ; when they are represented by figures, they are called numerical factors. 12. The Sign of Division is a short horizontal line with a point above and one below, -^. It indicates that the quantity before it is to be divided by the quantity after it. Thus, in a--b, the sign indicates that a is to be divided by b. Division is also expressed by writing the dividend above, and the divisor below, a short horizontal line ; as 1. b 13. The Sign of Involution is a number written above and to the right of a quantity, to indicate how many times the quantity is 12. ALGEBRAIC QUANTITIES. to be taken as a factor. Thus, in a 6 , the number 5 indicates that a is to be taken 5 times as a factor; and the expression is equivalent to aaaaa. A factor repeated to form a product is called a root } the product itself is called a power ; and the figure which indicates how many times the root or factor is taken, is called the exponent of the pow- er. Thus, in the indicated product a 6 , a is the root, a 5 is the power, called the 5th power of a, and 5 is the exponent of this power. When no exponent is written over a quantity, the exponent 1 may always be understood. NOTE. For the sake of brevity, the exponent of the power may be called the exponent of the letter or quantity over which it is placed. Thus in 5 , 5 may be called the exponent of a. 14. The Sign of Evolution, or Radical Sign, is the character \/ . It indicates that some root of the quantity after it is to be ex- tracted. The name or index of the required root is the number written above the radical sign. Thus, \/a denotes the cube root of a] Va denotes the 4th root of a; and so on. When no index is written over the sign, the index 2 is understood ; thus, v/ denotes the square root of a. Fractional Exponents are also used as the sign of evolution, the denominator being the index of the required root. Thus, in a a , the denominator, 3, indicates that the cube root of a. is required, and the expression is equivalent to V- Fractional exponents are used to denote both involution and evolution in the same expression, the numerator indicating the power to which the quantity is to be raised, and the denominator the required root of this power. Thus, the expression a* signifies the 4th root of the 3d power of a, and is equivalent to V s . SYMBOLS OF RELATION. 15. The Sign of Equality is two short horizontal lines, . It indicates that the two quantities between which it is placed are equal. Thus, in a=u--c, the sign, ==, indicates that a is equal to b plus <;. An expression of equality between two quantities is called an equal ' . It indicates that the quantities between which it is written are unequal, the opening being always turned toward the greater. When the opening is toward the left, it is read greater than ; wher the point or vertex is toward the left, it is read less than. T^us, a>b signifies that a is greater than b ; x+yx*i/*. 31. A Monomial is an algebraic quantity consisting of only ono term ; as 3x, or 7xy. DEFINITIONS AND NOTATION. 15 . A Polynomial is an algebraic quantity consisting of more than one term; as x-\-y, or 4a 2 3x-\-m. 33. A Binomial is a polynomial of two terms; as a-{-b, or 3* z. 34. A Residual is a binomial, the two terms of which are con- nected by the minus sign ; as a b, or 4.r 3y. 3>. A Trinomial is a polynomial of three terms; as x+y+z, or la W+d. G. The Degree of a term is the number of its literal factors. Since the exponents show how many times the different letters are taken as factors, the degree of a term is always found by adding the exponents of all the letters. Thus, x and 5y are terms of the first degree ; a 2 and 4a& are terms of the second degree ; x 3 , 3j 2 iy, o.ry 2 , and xyz are terms of the third degree. 37. A Homogeneous Quantity is one whose terms arc all of the same degree ; as x 3 bx*y-\-%xyz. 38. A Function of a quantity is any expression containing that quantity. Thus ax* is a function of x' f 3y-|-% 4 is a func- tion of y. AXIOMS. 3O. An Axiom is a self-evident truth. The following axioms underlie the principles of all algebraic operations : 1. If the same quantity or equal quantities be added to equal quantities, the sums will be equal. 2. If the same quantity or equal quantities be subtracted from equal quantities, the remainders will be equal. ;>. If equal quantities be multiplied by the same, or equal quanti- ties, the products will be equal. '4. If equal quantities be divided by the same, or equal quantities, the quotients will be equal. 5. If a quantity be both increased and diminished by another, its value will not be changed. B. If a quantity be both multiplied and divided by another, its value will not be changed. 7. Quantities which are respectively equal to the same quantity, are equal to each other. 16 ALGEBRAIC QUANTITIES. 8. Like powers of equal quantities are equal. 9. Like roots of equal quantities are equal. 10. The whole of a quantity is greater than any of its parts. 11. The whole of a quantity is equal to the sum of all its parts. EXERCISES IN ALGEBRAIC NOTATION. 4O. In the examples which follow, it is required of the pupil simply to express given relations in algebraic language. 1 . Give the algebraic expression for the square of a increased by by 4 times b. Ans. a a -j-4&. 2. Give the algebraic expression for 7 times the product of x and y, diminished by 5 times the cube of z. 8. Indicate the quotient of 12 times the square of a minus 5 times the cube of b, divided by the sum of a and c. 4. If d represent a person's daily wages, what will represent his wages for 6 days ? Am. Qd. 5. An army drawn up in rectangular form, has b men in rank, and a men in file ; of how many men is the army composed ? 6. If a man labor m days in a week at c dollars per day, what will his earnings amount to in 7 weeks ? 7. The length of a prism is a, the breadth c, and the altitude a c; required the solid contents. Ans. ac(a c). 8. A has 4m dollars, B has m times as many dollars as A, and C has 3 times as many dollars as B wanting d dollars; how many dollars has C ? 9. A dealer sells b sheep and c calves, at an average price of m dollars per head ; how much does he receive for all ? 10. A man has 3 square lots measuring m rods on a side ; how many acres in the 3 lots ? . 3m a ' 160* 1 1 . From a rectangular piece of land whose length was a rods and whose width was b rods, there were sold c acres ; how many acres remained unsold ? 12. A ship laden with a barrels of flour, valued at m dollars per DEFINITIONS AND .NOTATION. 17 barrel, met with a disaster by which I barrels were lost, and the remainder damaged to the amount of d dollars per barrel \ what was the worth of the remainder ? Ans. (a U)(in d). 13. A man having c acres of land worth dollars per acre, divi- ded its value equally between m sons and one daughter) how many dollars did each receive 1 14. A company of n persons began business with a joint capital of c dollar*. The first year they gained b dollars, the second year they lost d dollars, the third year they doubled the capital with which they began that year, and then dissolved partnership, sharing equally their accumulated capital ; what was each man's share ? COMPUTATION OF NUMERICAL VALUES. 41. The Numerical Value of an algebraic quantity is the num- ber obtained by assigning numerical values to all the letters, and performing the operations indicated. 1. Whit is the numerical value of (a 2 bc)a, when a=30, 6=25, and c=2S ? OPERATION. (a 3 bc)a= (30x30 25x28) x 30 =200x30 = 6000, Ans. Find the numerical values of the following expressions, in which 0=12; =10; c=8; m=G' } w=5; d=2. 2. a' be. Ans. 64. 3. (a+ld)m. Ans. 192. 4. ani-f-c 2 md*. Ans. 40. 5. (a-f 6)m (c+. Ans. 82. G. [4a 2 -(36 a 2c)]J. Ans. 584. 7. (a 2 i) (&' a). Ans. 11792. 4 a c Ans. 8. 3 flm _-(6+2c) a 1 d* _ -^ ' _ -. __ Ans. 24. 26+n 2* , 18 ALGEBRAIC QUANTITIES. Find the numerical values of the following expressions, in which a = S ; b=G ; c=4 ; d=2 ; m=3 ; n=l. ______ 11 ' ' _Z __ . Ans. 6. ' a-j- b -fc 12. (6a s vi 4mV) (m'-7w). ^Lns. 336. 13 (5a+26X ^ 11. 15. |2c(aV m 8 ) 2(56'+4m) -j-cjd 1 . -4ns. 3424. 1 /am 9 t? 4 1 - m 17. - - --h-i-r \ c ' w a 18. - [ a -|-2cxw^ c?]i 2 (ai-f-m') * 19. 20 fa b \ / a 5\ >!! 21. -TT-f itl-Hhl 7 -- TT)- -4m. 1. \a-f-6 ' a &/ \a b a--lJ -\-l 6a 3 -22a-f-18 __ ' -4ns. a _6a 2 -fll 6 . 1 T !U. SIGNIFICATION OF THE PLUS AND MINUS SIGNS. . The signs, -\- and , have heen denned as symbols of operation, the former indicating addition, and the latter subtraction. Now when we meet with detached or single terms affected with the plus or minus signs, as for instance in the examples of addition, subtraction, multiplication or division, we are to consider the positive DEFINITIONS AND NOTATION. 19 terms as quantities taken adtlitwely, and the negative terms as quantities taken suhtr actively ; and this is the only signification that need be attached to these signs, at present. It is obvious that positive and negative terms, as just explained, are in an important sense opposites. And *it will be shown in a future section (182 ^ that quantities sustaining to each other various other relations of opposition or contrariety, are distinguished by the plus and minus signs. 4:3. In order to establish general rules for algebraic operations, it will be necessary in this place to recognize the following principles, consequent upon the peculiar manner of considering quantities in Algebra : 1. A quantity may be considered as having two kinds of value, an absolute or numerical value determined simply by the number of units it contains, and an algebraic value depending on the sign. 2. Two quantities having the same absolute value, but affected with unlike signs, are not algebraically equal. Thus, 5a is not equal to 5, for the former expression signifies that a is taken additively five times, and the latter signifies that a is taken subtract- ively five times. 3. Two quantities having the same absolute value, but affected with unlike signs, are together equal to zero. Thus if a denote tho absolute value of any quantity, then -fa a=0 -f2a 2a=0 =0,etc. 20 ENTIRE QUANTITIES. ADDITION. 44. Addition, in Algebra, is the process of uniting two or more quantities into one equivalent expression called their sum. 4>. The term, addition, has a more general meaning in Algebra than in Arithmetic, because the quantities to be added may be either positive or negative. 4G. The Arithmetical Sum of two or more quantities is the sum of their absolute values, and has reference simply to the num- ber of units in the quantities added. 47. The Algebraic Sum of two or more quantities is a quantity, which, taken with reference to its sign, is equivalent to the given quantities, each taken with reference to its particular sign. 48. To deduce a rule for addition, which will conform to the nature of positive and negative quantities, let us consider the fol- lowing examples : 1. Add 4, 3, and 5. Since in these quantities a is taken additwely, 4, 3, and 5 times, or 12 times, the algebraic sum required must be -|-12a; or simply, 12a. That is, 4a-f3a-f-5a=12a 2. Add 4a, 3a, and 5. Since in these qantities a is taken subtr actively , 4, 3, and 5 times, or 12 times, the algebraic sum required must be 12a. That is, 4 tt 3 a _ 5 a 12a Hence, The algebraic sum of two or more similar terms having like signs, is the sum of their absolute values taken with their common sign. 3. Add la and 3 a. From Ax. 11 we have 7o=4, labd*, abd*, 5ad a , ISaid*, and 7 aid*. Ans. oW. 12. Add 2x^ 2a 2 , 3a*+2xy, a'-f x,y, 4a 2 3xy, and 2xy2a*. Ans. 4rt*-|~4ay. 13. Add 8aV 3./y, 5ax 5^y, 9.ry 5rjKc, 2a a x a -{-xy, and Ans. lOaV a ADDITION. 14. Add a'2ac+cd+b, ?>*Sac3cd2b, 2a f + ac 6b, and a 2 4ac-j-2cd 36. -4ns. 7 a 2 Sac 5cd+2Z>. 15. Add 2aV 3mx-f-4m a <:?, 3m a d-f5aV 5mx, 6mx 4w 3 d _ 3aV, and 2mx 3V -Sm'e?. Ans. aV. 16. Add 2bx 12, 3z 26*, 5* 1 3 1 /ay3 1 /x+12, and x a +3. 4ns. 9z a -f 3. 17. Add 106' 36x a , 2ft V6 1 , 10 26x a , />V 20, and 36x a + 10. 18. Add 96c' 18ac a , 155c 3 H-ac, 9ac 9 246c s , and 9ac a 2. Ans. ac 2. 19. Add Gm'+Som+l, 6aw^ 2m a -f-4, 2m 1 8aw-f-7, and 3m a . 9m 2 20. Add 5x 4 3x-f4x 2.r-f 10, 7x 4 +2x f +2x f 4- 5x+2, and Ans. 12r 4 -f6x a -f 12. 21. Add 3*y 5xV xV ^- , and aty 1 ay > -2x 22. Add 5a-j-3l/^ a ^i4-4, 7a T/V 1 5, 3a 5l/m a 1 8, and 2a-f 2l/m 2 ^14-2. 4ns. 17a T/m 2 1 7. 23. What is the sum of 3aV 2c a a^+ a^c^, 2a a c2-|-3c a a^ 24. What is the sum of 9a(o 6) 4?7iV m c, 7ml/ 6a(a 6), and 12 mVm^c 8a(a &) ? m c 25. What is the sum of a-\-b -\-c-\-d-\-m, a-\-b-{-c-\-d m, CT-|- 6-|-c d m, a-|~^ c ^ m > an( ^ a ^ c ^ Wl ? 4ns. fta-|-36-|-f e? 3m. 5O. The Unit of Addition is the letter or quantity whose co- efficients are added, in the operation of finding the sum of two or more quantities. Thus, in the example, the letter x is the unit of addition. Also, in the example, the quantity, V^cr+c, is the unit of addition 24 ENTIEE QUANTITIES. 51. When dissimilar terms have a common literal part, this may be taken as the unit of addition. The sum of the terms will then be expressed by inclosing the sum of the coefficients in a parenthesis, and prefixing it to the common unit. EXAMPLES FOR PRACTICE. (2.) (3.) \laxy* (5a (2c (12a+2m).ry 1 (la+c (5.) ( ' Si) O 9 1) (3a +3/Q (m a - (a ^/,-c) (x 6. Add ax, 2cx, and 4c?x. Ans. (a+ 2 7. Add ' y-j-^? 3flry-j-2ca;, and , and (a+fyx+Zcdxy. Ans. 9. Add rar-j-Ty, Tax oy, and -4ns. (8a Z 10. Add (Z> d)i/x, and (c-f 2a i)|/a;. Ans. (c-fa^/ie. 11. Add (a + 26) m Cv/m, (2a 6c) 771 3a^m, (5c 4a) 7?i >/w, and (2a 3i)??i-f-4a %/"-. -4?w. (a ^ c) ( 12. Add ax-\-y-\-z, x-\-ay-\-z, and x-\-y-\-az. Ans. (a+2) ( SUBTRACTION. 25 SUBTRACTION. * . Subtraction, in Algebra, is the process of finding the differ- ence between two quantities. 53. It is evident that 5 units of any kind or quality subtracted from 8 units of the same kind or quality, must leave 3 units of the same kind or quality. That is, Also, 8a (. 5a)= 3a But these remainders are the same as we shall obtain by changing the signs of the subtrahends and then adding the results, algebra- ically, to the minuends. Thus, -f8a (+5a) 8 a ( 5a)= Hence, in Algebra, Subtracting any quantity consists in adding the same quantity with its sign changed. 54. This principle may be established in a more general manner as follows : Let it be required to subtract the quantity I c from a. OPERATION. We first subtract b from a, indicating the Minuend, a operation, and obtain for a result, a b. Subtrahend, b c But the true subtrahend is not 5, but bc; - and, as we have subtracted a quantity too Difference, a b-\-c great by c, the remainder thus obtained must be too small by c ; we therefore add c to the first result, and obtain the true remainder, a b-\-c. But this result is the same as would be obtained by adding b-\-c to a. 55. It follows from the principle enunciated above, that any quantity is subtracted from nothing or zero, by simply changing its sign or signs. Thus, (+a) = -a ( ) -]- a (a b)=-a+b 3 26 ENTIRE QUANTITIES. >. From these principles and illustrations we deduce the fol- lowing RULE. I. Write the subtrahend underneath the minuend, placing the similar terms together in the same column. II. Conceive the signs of the subtrahend to be changed, unite the similar terms as in addition, and bring down all the remaining terms with their proper signs. EXAMPLES FOR PRACTICE. (1.) (2.) (3.) (4.) 6x a y 4mc a ba*bc Zx'y'z (5.) (6.) (7.)_ ax 2y 6a (8.) (9.) 5m ^ a -f c c a 3a*x+c*dcd* -\-7md 2 10. From 2x a 3x-f y a subtract a x* 4x. 11. From 7a 5c-f-2 subtract a-fc-f 2. Ans. Sa 6c. 12. From 8x a 3xy+2/+c subtract x 3 6^+3.y a 2c. -4ws. 7o: a +3'^ /-f-3c. 13. From a-f-6 subtract a 6. Ans. 2b. 14. From ^x-\-iy subtract \x \y. Ans. y. 15. From a-j-6-j-c subtract a b c. Ans. 16. From 3a-6 2x-|-7 take 8 36+ a-f 4x. 17. From 6y 9 2y 5 take 8/ Ans. 14/-f 3y 17. SUBTRACTION". 27 18. From 3p+q+r 3s take q 8r-f2s 8. ^Tis. 3p-j-9r 5s+8. 19. From I3a*2ax-\-9x* take 5a 3 Tax x\ Ans. 8a a -f-5aa;-j-10a; 8 . 20. From a 4 Sx'-j-So; 9 7x+12 take zV-4x s -f 2x a 6x-fl5. Jbis. x 3 -|-3x a x 3. 21. From a 6 3a 4 c+5aV 2aV-}-4ac 4 c 5 take a 5 -4a 4 c+ 2aV 5aV-f3ac 4 c 5 . ^Lrw. a 4 c4-3a 3 c a -|-3aV-j-ac 4 . 22. From 2x 4 +28x 3 -|-134ar l 252z-f-144 take 2x 4 -|-21x 3 + 67 x 2 _63x4-84. ^Lns. 7z 3 +67z a 189a;+60. 23. Fromx 6 --5x10x-lOx--5 4 take x 6 5x 4 24. From the sum of 6x a y llax 8 and 8x 2 ^+3ax 3 , take 4x> . Ans. 10x 2 ^ 4ax 3 a. 25. From the sum of 8cx-\-cx take x-\-ax-\-bx. Ans. (c l)x. 7. From (a-f 26-f-c) V~xy take (25 c)Vxy. Ans. (a+2c)l/xy. 8. From (3a 2m)x s -f(5a+2m)x a +(4a m)x take (am)x* 28 ENTIRE QUANTITIES. 9. From l+2az-f'3aV:f 4aV+5aV take z*+ 2az 4 +3aV-f 4oV. Ans. USE OF THE PARENTHESIS. SS. The term, parenthesis, will be employed hereafter as a gen- eral name to designate the various signs of aggregation employed in algebraic operations. The following rules respecting the use of the parenthesis should be thoroughly considered by the learner, if he would acquire facility in algebraic transformations. 59. From the definition of the signs of aggregation, (IT), we understand that if the plus sign occurs before a parenthesis, all the terms inclosed are to be added, which does not require that the signs of the terms be changed ; but if the minus sign occurs before a parenthesis, all the terms inclosed are to be subtracted, which re- quires that the signs of all the terms be changed. Hence, 1. A parenthesis preceded by the plus sign may be removed, and the inclosed terms written with their proper signs. Thus, a b -\- (c d-\- e) = a b -j- c d-\- e 2. Conversely : Any number of terms, with their proper signs, may be inclosed by a parenthesis, and the plus sign written before tlie whole. Thus, a b-\-c d-\-e=a-^-( 5-j-c d-\-e) 3. A parenthesis preceded by the minus sign may be removed, provided the signs of all the inclosed terms be changed. Thus, a (b c-}-d e)=a b-\-c d-\-e 4- Conversely : Any number of terms may be inclosed by a paren- thesis, preceded by the minus sign, provided the signs of all the given terms be changed. Thus, a b -j- c d-\- e = a b -j- c (d e) GO. When two or more parentheses are used in the same express- ion, they may be removed successively by the above rules. Thus, a | b c (d e) j =a j b c d-\-e j = a b-\-c-\-d e Or, in a different order, a | b c (d e) l=a 6-f c-\-(d e)=a b-\-c-\-d e SUBTRACTION. 29 EXAMPLES FOR PRACTICE. Gl. Remove the parentheses from the following express-ions, and reduce the results : *: 1. 3a-|-(2& 2 or d+m). Ans. 2a+2b*d+m. 2. 4 x _y_(3x 7y+5)+2x. Ans. 4x 9 -f Qy x 5. 3. a -t-2c (4c 3a-]-2m 9 ). ^ras. 4a 2c 2m 9 - 4. 4x 3 2x 2 HX (2^'+5x 7) 6x+l]. Ans. 3z 8 +lLc 8. 5. a-[-2m j c-f-x [a 7/1 (c 2x)] J . ^HS. 2a-f m 2c+x. 6. 3z 9 4x am / x 8 a [Sam (2x-f 2am)-|-2a; a ] 5am } . Ans. 4x a 5x-|-5a77i. 7. 3a |2m a -j-[5c 9a (3a+m 2 )]+6a (m 2 -|-5c)}. . 9a. 8. x 2 |5mc a [a: 9 (3c 3mc 8 )+3c (x 9 2mc 3 c)] }. 9. m 9 m 1 | m 9 2m 2 [m 9 3m 3 (m 9 1m 4)] } . Ans. 2m-f 2. 10. 52 s 32 2 +4z 1 [2z 8 (3z 9 2z-fl) z*+z~\. Ans. z*-\-z. 11 4 c _2c 9 -f-c4-l (3c 8 c 9 c 7) (c 8 4c 2 +2c+8). Ans. 3c 9 ., 12. 3a 2 t 4:cd (Bed 2a 9 6)'~ [a 8 + c (5crf+ 3a 2 6) + (3a r +2ca')+a 3 ]. Ans. 8a 13. -- /4a 9 m3mV 7m 9 a^ 9a 9 m TI i 2 ] 5a 9 m J 12a 9 m Ans. SmV+Gn 5a 9 m 3ai 9 . . In Algebra, addition does not necessarily imply augmenta- tion, nor does subtraction always imply diminution, in an arithmet- ical sense. We have seen that one quantity is added to another by annexing it with its proper sign ; but a quantity is subtracted from another by annexing it with its sign changed. Hence, 3* 30 ENTIRE QUANTITIES. 1st Adding a positive quantity has the same effect as subtracting a negative quantity ; and adding a negative quantity has the same effect as subtracting a positive quantity. 2d. If to any given quantity a positive quantity be added, the result will be greater than the given quantity; but if a negative quan- tity be added, the result will be less than the given quantity. 3d. If from any given quantity a positive quantity be subtracted, the result will be less than the given quantity ; but if a negative quantity be subtracted, the result will be greater than the given quantity. 63. Let a denote any negative quantity. Add b to this quantity, and subtract -{-& from it ; and we have a+(-.b)=-a-b But according to the last two propositions, the result, a 6, should be less than the given quantity, a. That is a b < a Now, the quantity, a 5, contains a greater number of units than a. These cases, however, are not exceptions to the laws enunciated above ; for in an algebraic sense, the less of two nega- tive quantities is that one which contains the greater number of units. (SeelOT). 64. If a represent the greater of the two numbers, and b the less, then a+b is their sum and a b their difference ; and the sum and difference may be combined in two ways, as follows : 1st; To a+b 2d; From a+b Add a b Subtract a b 2a 2b Hence, 1. If the difference of two numbers be added to their sum, the result will be twice the greater number. 2. If the difference of two numbers be subtracted from their mm, the result will be twice the less number. MULTIPLICATION. 31 MULTIPLICATION. ~ *' Go. Multiplication, in Algebra, is the process of taking one quantity as many times as there are units in another. G6. In order to establish general rules for multiplication, we must first consider the simple case of multiplying one monomial by another; and we will investigate, first, The law of coefficients; second, The law of exponents ; third, The law of signs. 1st. The law of coefficients. Let it be required to multiply 5a by 36. Since it is immaterial in what order the factors are taken, we may proceed thus : 5X3=15; aX& al; and 15X& 15ai. Or 5aX3k=15a&. Hence, The coefficient of the product is equal to the product of the coeffi- cients of the multiplicand and multiplier. 2d. The law of exponents. Let it be required to multiply a*b* by cfb*. Since a*b* aaaa bub, and a?b*=aaa bb, we have a t b 3 Xa 3 b y =aaaabbbaaabb=a' T b*. Hence, The exponent of any letter in the product is equal to the sum of the exponents of this letter in the multiplicand and multiplier. 3d. The law of signs. In Arithmetic, multiplication is restricted to the simple process of repeating a number ; and the only idea attached to a multiplier is, that it shows how many times the multiplicand is to be taken. In Algebra, however, a multiplier may be affected by either the plus or the minus sign ; and it is necessary to consider how the sign of the multiplier modifies its signification. For this purpose, suppose it were required to multiply any quantity, as a, by c d. Now it is evident that a taken c minus d times, is the same as a taken c times, diminished by a taken d times ; or aX(c d)=ac ad. In the first term of this result, a is taken c times additively, or a-f-a+a+a etc., to c repetitions; and this is the product of a by -f c. In the second term, a is taken d times 32 ENTIRE QUANTITIES. subtractively, or a a a etc., to d repetitions ; and this is the product of a by d. Hence we conclude that the signs, + an ^ , when prefixed to a multiplier, must be interpreted as follows : The plus sign before a multiplier shows that the multiplicand is to be successively added ; and the minus sign before a multiplier shows that the multiplicand is to be successively subtracted. To exhibit the law which governs the sign of a product, ac- cording to this principle, we present the four cases which involve all the variations of signs. It will be observed that according to the above interpretation, the multiplicand is to be repeated with its proper sign when the multiplier is positive, but with its sign changed when the multiplier is negative. We shall therefore have the fol- lowing results : 2. -f-X( &)= o> a a etc. = ab. 3. aX(+&) = a a a etc.= ab. 4. X( &)=+a-fa+a-fetc. = -fa&. Comparing the first result with the fourth, and the second with the third, we observe that When the two factors have like signs, the product is positive ; and when the two factors have unlike signs, the product is negative. T. This law applied in the case of three or more negative factors gives the following results : (-a)X(- &) =+ab (_a)X(-&)X( c) =(+** )X(-O= o*c ( ) X (&) X ( c) X ( <0 =( abc ) X ( *)= +alcd (a) X (6) X ( c) X ( d) X ( ) = (-|-a&crf) X (e)=abcde Hence the general truth : The product of an even number of negative factors is positive ; and the product of an odd number of negative factors is negative. CASE I. 68. When both factors are monomials. From the principles already established we derive the following RULE. I. Multiply the coefficients of the two terms together for the coefficient of the product. MULTIPLICAT II. Write all the letters of both terms for each an exponent equal to the sum of its exponents in the two terms. III. If the signs of the two terms are alike, prefix the plus sign to the product ; if unlike, prefix the minus sign. EXAMPLES FOR PRACTICE. (1.) (2.) (3.) (4.) lx*y a*cm? 5c 4 m* xy* Gac'd 3c*d? 5. Multiply 17a 3 &V by lac. 6. Multiply UaWc by 10a 6 6V. 7. Multiply U7ab*c*x by 2a 8 6V. 8. Multiply 7x*yz* by xyz. Ans. 2 9. Multiply 12c ) a by (zx). Ans. _4m a (rr ^) 8 . 22. Multiply (a c)" +1 by (a c)"- 1 . ^ws. (a c) 2 *. 34 ENTIRE QUANTITIES. CASE II. 69. When one or both of the factors are polynomials. 1. Multiply x y-\-z by a-\-b c. OPERATION. X y +Z Product by a, ax ay-\-az Product by 6, bx ky-{-bz Product by c, CX-\-Cy CZ Entire Product, ax ay-}-az-\-bx by-\-bz cx-\-cy cz Hence the following general RULE. Multiply all the terms of the multiplicand by each term of the multiplier, and add the partial products. EXAMPLES FOR PRACTICE. a.) (2.) (3.) (4.) (5.) a Ixy lOy 6. Multiply 3a 3 x 2 ifz-\-z* by 2a^ a . 7. Multiply x 4 3x 3 -f 2x 2 5x-f3 by 3x 2 . Ans. 3a; 6 9x 6 +6x 4 8. Multiply V 3a 2 c 3 +a a c ac a -f-o c+1 by or MULTIPLICATION. 35 9. Multiply 2ax 3x by 2x-j-4#. Ans. 4ax*-\-8axy 6x a 12xy. 10. Multiply 3a 2 2al 6 a by 2a 46. 11. Multiply x 2 xy+y* by x-f-y. ^ns. x 8 -]-?/ 8 . 12. Multiply a 2 3ac-f-c 2 by a c. .4ns. a 3 4a 2 c-f4ac 2 c 8 . 13. Multiply 2x 9 3x-|-2 by x 8. 4ns. 2x 8 19x 2 -f26x -16. 14. Multiply a s +2a 2 i+2a6 2 +& 8 by a 3 2a 2 6+2a& 2 1\ Ans. a 6 & fl . 15. Multiply a w +6 W by a n -{-&". 16. Multiply 4x 3 +8x 2 +16x--f-32 by 3x 6. Ans. 12x 4 192. 17. Multiply a s -{-a 2 &-f-a& 2 -f 6 3 by ab. Ans. a 4 5 4 . NOTE. The product of two or more polynomials may be indicated, by inclosing each in a parenthesis, and writing them one after another, with or without the sign, x , between the parentheses. Such an expression is said to be expanded, when the indicated multiplication has been actually performed. 18. Expand (a-j-m) (a-f d). Ans. cf+am+ad+dm. 19. Expand (a-j-2m 1) (a-fl). Ans. a a +2am+2m 1. 20. Expand (z 9 + 4s*-f-52 24) (z* 4^-fll). Ans. z B +151z 264. 21. Expand (a 8 4a f +lla 24) (o"+4a-f-5). Ans. a 5 41a 120. 22. Expand (m 3) (m 1) (m-f-1) (w-f 3). Ans. m 4 10m 2 -f 9. 23. Expand (x 8 2a; a -f-3a; 4) (4x s +3x 2 -j-2x+l). 24. Expand (^+2^+^-4^ 11) (y 2y+3). J.WS. y+lOy 33. 25. Expand (c 2 c-fl) (c a -f c + 1) (c 4 c 2 4-l). JTZS. c 8 -f c 4 -f-l. 26. Expand (x 6 5x 4 +13x 3 x 2 x-f 2) (x 2 2x 2). Ans. x 7 7^ 6 -f-21x 5 17x 4 25x 3 -f6x 3 2x- -4. 27. Expand (16x 4 8x s -f4x 2 2x-fl) (2x-fl). Ans. 32^-f-l 36 ENTIRE QUANTITIES. FORMULAS AND GENERAL PRINCIPLES. TO. A Formula is the algebraic expression of a general truth or principle. The following formulas are useful, as furnishing rules for obtain- ing the products of certain binomial factors. If a and b represent any two quantities whatever, then a-\-b=. their sum, and a &:= their difference; and we have, after performing the indicated operations, the results which folhow: Or, expressing the result in words, The square of the sum of two quantities is equal to the square of the first, plus twice the product of the first -and secondj plus the square of the second. II. (<* &) f =(a &) (a fc) = a a 2a&-f-Z> 9 Or, in words, The square of the difference of two quantities is equal to the square of the first, minus twice the product of the first and second, plus the square of the second. III. (a+ft) (a &)=a f V Or, in words, The product of the sum and difference of two quantities is equal to the difference of their squares. By the aid of these formulas we are enabled to write the square of any binomial, or the product of the sum and difference of any two quantities, without formal multiplication. EXAMPLES FOR PRACTICE. 1. What is the square of 3a-f-2a&? The square of the first term is 9a a , twice the product of the two terms is 12a a &, and the square of the second term is 4a 2 a ; hence , by the first formula, (3a+2at) a =9a a -f 12a 2 &H-4a 9 5 9 , Arts. MULTIPLICATION. 37 2. What is the square of 2:c a 5 ? The square of the first term is 4x 4 , twice the product of the two terms is 20x a , and the square of the second term is 25 ; hence by the second formula, (2z a 5) a =4r 4 20x a +25, An*. 3. What is the product of 5x-{-j/ a and 5x y* 't The square of 5x is 25x a , and the square of y* is # 4 ; hence by the third formula, (5x+# a ) (5x-y)=25z a --y, 4ns. 4. What is the square of c+m? Am. c a -j-2cm-J-m a . 5. What is the square of x y ? Ans. x* 2xy-|- < y a . 6. What is the product of x-{-y and x y ? Ans'. x 3 y*. 7. What is the square of 3z a -f-4y ? Ans. 9x 4 -f24xV-f ]% a . 8. What is the square of 5c 8 2cd ? 4ns. 25c 6 2 9. What is the product of 4^ 9 +3^ and 42 a Zyz ? 10. What is the square of 3a*z-|-2ay ? 11. What is the square of x+l ? Ans. x* -\-2x-\- 1. 12. What is the square of 2^ a 1 ? 4ns. 4z 4 4z a -f 1. 13. What is the product of m-j-1 and m 1 ? 4?is. m a 1. 14. What is the square of z a 30 ? Ans. z* 60z a -|-900. 15. What is the product of 3a a 6-|-cF and 3a a 6

x-f7 is arranged according to the descending powers of x. 73. A polynomial is said to be arranged according to the ascend- ing powers of any letter, when the terms are so placed that the exponents of this letter increase from left to right throughout the terms that contain it. Thus, the polynomial d ax-\-cx* bx* is arranged according to the ascending powers of x. 74. A term or quantity is said to be independent of any letter, when it does not contain that letter. 70. The product of two polynomials has certain special proper- ties, which may be stated as follows : 1. If both polynomials are arranged according to the descending powers of the same letter, then the first term obtained in the partial products will contain a higher power of this letter than any of the other terms ; and as this term can not be reduced with any of the others, it will form the first term of the entire product. 2. If both polynomials are arranged according to the ascending powers of the same letter, then the last term obtained in the partial products will contain a higher power of this letter than any of the other terms ; and as this term can not be reduced with any of the others, it will form the last term of the entire product. 3. If both polynomials are homogeneous, then the product will h*. iomogeneous ; and the degree of any term will be expressed by the sum of the indices denoting the degrees of its two factors. DIVISION. 39 DIVISION. t; 76. Division, in Algebra, is the process of finding tow many times one quantity, called the divisor, is contained in another quantity, called the dividend; the result of division is called the quotient. It follows, therefore, that the quotient must be a quantity which multiplied by the divisor, will produce the dividend. Thus, revers- ing the process of multiplication, we have, abc-^-a=bc 1 because bc^a-=abc 7 7. It was shown in the multiplication of monomials, (GO), that the coefficient of the product is found by multiplying together the coefficients of the factors ; and that the exponent of any letter in the product is found by adding together the exponents of this letter in the factors. Hence, in division, 1. The coefficient of the quotient must be found by dividing the coefficient of the dividend by that of the divisor ; and 2. The exponent of any letter in the quotient must be found by subtracting the exponent of this letter in the divisor from its exponent in the dividend. Thus, It was shown in multiplication, (66), that when two factors have like signs, their product is positive ; and that when two factors have unlike signs, their product is negative. In division, therefore, when the dividend is positive, the quotient must have the same sign as the divisor ; and when the dividend is negative, the quotient must have the sign unlike that of the divisor. And there will be four cases, with results as follows : 2. +ab+(a)=b 3. a&-K-i-a)= & 4 . ab -4- ( a) = -f- b Hence, 3. If the dividend and divisor have like signs, the quotient will be positive ; but if the dividend and divisor have unlike signs, the quotient witt be negative. 40 ENTIRE QUANTITIES. ' CASE I. 78. When the divisor is a monomial. From the principles already given we have the following RULE. To divide one monomial by another ; I. Divide the coefficient of the dividend by the coefficient f tin divisor, for a new coefficient. II. To this result annex the letters of the dividend, with the expo- nent of each diminished by the exponent of the same letter in the divisor, suppressing all letters whose exponents become zero. III. If the signs of terms are alike, prefix the plus sign to the quotient j if they are unlike, prefix the minus sign. To divide a polynomial by a monomial ; Divide each term of the dividend separately, and connect the quo- tients ly their proper signs. NOTE. It may happen that the dividend will not exactly contain the di- visor ; in this case the division may be indicated, by writing the dividend above a horizontal line, and the divisor below, in the form of a fraction. The result thus obtained may be simplified, by suppressing all the factors common to the two terms; thus, But as this process is essentially a case of reduction of fractions, we shall omit such examples till the subject of fractions is reached. EXAMPLES FOR PRACTICE. 1. Divide IQab by 4. Ans. 4fc. 2. Divide 21aVd by 7ac 9 . Ans. 3a'J. 3. Divide 42afys 4 by 6zV. Ans. 6x*yz. 4. Divide 2a by a*. Ans. 2a*. 5. Divide a 7 by a 9 . Ans. a. 6. Divide 16^ by kx. Ans. 4.c 2 . 7. Divide Ibaxy 3 by Say. Ans. 5xy*. 8. Divide 117a 5 6V by 3a 6 &c a . Ans. 39i 3 c. 9. Divide 63a 8 & 4 c) 2 3m(a 6) by 3(a 6). Ans. am 9 Im* m. 21. Divide 7a(3m 2a) (3m 2a) a by (3m 2a). -4ns. 9a 3m. CASE II. 79. When the divisor is a polynomial. Suppose both dividend and divisor to be arranged according to the descending powers of some letter. Then it follows, from (75, 1), that the first term of the dividend must be the product of the first term of the divisor by the first term of the quotient similarly ar- ranged. We can therefore obtain this term of the quotient, by simply dividing the first term of the dividend by the first term of the divisor, thus arranged. The operation may then be continued in the manner of long division in Arithmetic; each remainder being treated as a new dividend, and arranged as the first. 1. Divide 6a 4 +a 3 Z> 20aV+17a& 8 4Z> 4 by 2a' 3a&+ V. OPERATION. 9/. 3a'-|-5a& 4i a , Quotient 8a a -8a a 4* 42 ENTIRE QUANTITIES. Hence we have the following RULE. I. Arrange both dividend and divisor according to tht descending powers of one of the letters. II. Divide the first term of the dividend by the first term of the divisor, and write the result in the quotient. III. Multiply the wlwle divisor by the quotient thus found, and subtract the product from the dividend. IV. Arrange the remainder for a new dividend, with which pro- ceed as before, till the first term of the divisor is no longer contained, in the first term of the remainder. V. Write the final remainder, if there be any, over the divisor in the form of a fraction, and the entire result will be the quotient sought. EXAMPLES FOR PRACTICE. 1. Divide a*+3a*x-\-3ax*+x 9 by a-f-ar. Ans. a* -\-2ax-\-x*. 2. Divide a 3 4a 2 c-f 4ac 9 c s by a c. Ans. a 9 3ac-fc*. 3. Divide a' 6a a -j-12a 8 by a 3 4a-{-4. Ans. a 2. 4. Divide 3z a 2x 4 -f #' x 8 2x 15 by x* 5 4x. Ans. x 2 2x-f-3. 5. Divide 25z 6 z 4 2x f 8*' by 5z 4x a . Ans. 5x s +4x 2 -f-3x-{-2. 6. Divide 6a 4 -f9a 2 15a by 3a 2 3a. Ans. 2a 2 -f-2a-f 5. 7. Divide x' y* by x s +2xV+2x/-f T ^ 8 . Ans. x* 2x*y-\-2xy* y* 3. Divide ax 3 (a 2 -f-5)x 2 -f6 9 by ax b. Ans. x>axb 9. Divide a 4 -f46 4 by a' 2aZ>-f 2*. Ans. a 2 -f 2ab+2b* 10. Divide x 6 z 4 +x s x 2 -f-2x 1 by a 2 -f x 1. 11. Divide l+3x by 1 5x. Ans. l+8z+40x 2 +200x 8 +etc. 12. Divide 1 x x* by l+x-{-z 2 . J.jw. 1 2x-f2a; s 2x 4 -\-2x* 13. Divide x' 2x-f 1 by x 9 DIVISION. 43 14. Divide a'-f '-f c 8 3ac by a+b+c. Ans. a 2 -f-& a -r*c 2 6c ac db. 15. Divide 2x^ 5xy~ by a 4 4x 3 y+xy 3.r/. 16. Divide a 6 -fc 6 -fa 4 -f- c 4 a 4 c ac 4 2aV b^ a'+c 3 a 2 c ac 2 . 17. Divide 4x 6 5x*+8z 4 lOa; 8 8z 2 5x- 4 by 4x s -f 3x 2 -f-2^ +1. Ans. x 9 2x s -f 3z 4. 18. Divide x* a 6 by x a. J.TIS. cc 4 -f-x*a-j-x a a -j-xa 8 -j-a 4 . 19. Divide a'-j-o; 3 by a a. a x 20. Divide x m xy^x^y+y by x y. Ans. x^y^. 21. Divide a e+m a c l" a m b*-\-b n+d by a m b n . Ans. a c l d . 22. Divide x* n 2x**y* 2xy* n +y** by cc n +y n . EXACT DIVISION. 80. Division is said to be exact when the quotient contains no fractional part ; the quotient in this case is said to be entire. 81. It follows from the rule of division, (78), that the exact division of one monomial by another will be impossible under the following conditions : 1. When the coefficient of the divisor is not exactly contained in the coefficient of the dividend. 2. When a literal factor has a greater exponent in the divisor than in the dividend. 3. When a literal factor of the divisor is not found in the divi- dend. 8SJ. It is also evident, from (79), that the exact division of one polynomial by another will be impossible, 1. When the first term of the divisor arranged with reference to any one of its letters, is not exactly contained in the first term of the dividend arranged with reference to the same letter. 2. When a remainder occurs, having no term which will exactly contain the first term of the divisor. 44 ENTIRE QUANTITIES. GENERAL RELATIONS IN DIVISION. 8.3. The algebraic value of a quotient depends upon the compar ative values and relative signs of the dividend and divisor. Now if either the dividend or the divisor be changed with respect to its value or sign, the quotient will undergo a change, according to a certain law. As these mutual relations are frequently concerned in algebraic investigations, we present them in this place, considering first the law of change with respect to absolute value ; and second, the law of change with respect to algebraic signs. 1st. Change of value. 84. In any case of exact division, the quotient is composed of those factors of the dividend which are not included among the factors of the divisor. It is evident, therefore, that if we introduce a new factor into the dividend, the divisor remaining the same, we shall introduce the same factor into the quotient ; and if we exclude a factor from the dividend, the divisor remaining the same, we shall exclude this factor from the quotient. Again, if we introduce a factor into the divisor, we shall exclude it from the quotient ; and if we exclude a factor from the divisor, we shall introduce it into the quotient, the dividend remaining the same in both cases. Hence we have the following general principles : I. Multiplying the dividend multiplies the quotient, and dividing the dividend divides the quotient. II. Multiplying the divisor divides the quotient, and dividing the divisor multiplies the quotient. III. Multiplying or dividing both dividend and divisor by the same quantity does not change the quotient. 2d. Change of signs. 85. To show in what manner the sign of the quotient is affected by changing the sign of dividend or divisor, we observe that two signs can have only three relations, as follows : DIVISION. 45 Now if one of the signs, only, in any of these couplets, be changed, the relation of the signs in that couplet will be changed, either from like to unlike, or from unlike to like ; but if loth of the signs in any couplet be changed, their relation will not be altered. Hence, I. Changing the sign of either dividend or divisor, changes the sign of the quotient. II. Changing the signs of both dividend and divisor, does not alter the sign of the quotient. RECIPROCALS, ZERO POWERS, AND NEGATIVE EXPONENTS 86. The Reciprocal of a quantity is the quotient obtained by di- viding unity by that quantity. Thus, _ is the reciprocal of x\ x is the reciprocal of a c. 87. In dividing any power of a quantity by any other power of the same quantity, we subtract the exponent of the divisor from the exponent of the dividend, to obtain the exponent of the quo- tient. Thus, And in general, we have a"~~ a If in this expression n=m, the exponent of the quotient will be 0; and if n>m, the exponent of the quotient will be negathe. Thus, a 3 a 3 a* -=a' ; -=a'- i =a- 1 ; -=a t - 4 =ar* l etc. a a a 88. It has been found useful for certain purposes in Algebra, to employ the notation, a, a" 1 , a~ 2 , a~ 8 , etc. We will therefore proceed to interpret the meaning of zero and negative exponents, in general. Let a represent any quantity, and m the exponent of any power whatever. Then by the rule of division, =a =' 46 ENTIRE QUANTITIES. But the quotient obtained by dividing any quantity by itself must be equal to 1. That is _-& DIVISION. 47 nent, m, is an odd number, the mth remainder will be b m (a m ~ l m ~ m ~}= b m (a Z>)= b m (l 1)=0 and the division will therefore be exact. But if m be even, the mth remainder will be and the division will not be exact. Hence, The sum of the same powers of two quantities is divisible by the siim of the quantities, if the exponent is odd, but not otherwise. 2. Divide a m +b m by ab. Commencing the division, we have b m a b 1st rein. _|_ a -'&+ b 1 * =-f&(a'*- 1 -f 6 1 "- 1 --a"*" 1 ^ a"^~*b 2drem. If this operation be continued, it is evident that whether m be odd or even, the mth remainder will be and the division can not, therefore, be exact. Hence, The sum of the same powers of two quantities is never divisible by the difference of the quantities. 3. Divicje a m b m by a-{-b. Commencing the division, we have a+b 1st rem. a M -'& l m = b(a 1 *' 1 -\-b m ~ 1 ') i^b 2dre m . , _f- a -V b m ==-.b*(a'-' > b m -*) If this operation be continued, then it is evident that when m is odd, the mth remainder will be 7/(a r - m -f-& m - nl ) 6 m (a+6 ) = & m (l-f-l)= 2b m and the division can not be exact. But if m be even, the mth re- mainder will be _|_j( tt _j-) = _|_&( a _&) = _|_j*(l__l) = and the division in this case will be exact. Hence, 48 ENTIRE QUANTITIES. The difference of the same powers of two quantities is divisible by the sum of the quantities, if the exponent is even, but not otherwise. 4. Divide a m b m by a b. Commencing the division, we have 6 a m < ab 1st rem. 2drem. _[_*-&_ fr _}_&' ( a *-_&*-') If this operation be continued, then it is evident that whether m be odd or even, the rath remainder will be -f l m (cr- I-} = -j-b TO (a Z>) = -f 6 m (l 1) =0 and the division will therefore be exact. Hence, The difference of the same powers of two quantities is always divisible by the difference of the quantities. OO. If we continue the division in the 1st, 3d, and 4th of the preceding problems, then in the cases of exact division, the form of the quotients will be as follows : a b (1) b- 1 (2) *- i +&"~ 1 (3) 91. By giving particular values to m in (1), (2) and (3), we btain the following results, which maybe useful for reference: a-f 6 (2) DIVISION. ab =a*+a1>+l* In like manner we may obtain (3) ? 1 . , =x 1 x+l ft "1 =T i " !h=i : a:* 1 a; 1 (5) (6) FACTORING. The Factors of a quantity arc those quantities which, being multiplied together, will produce the given quantity. OS. A Prime Factor is one which can not be producedjby the multiplication of two or more factors; it is therefore divisible only by itself and unity. 5 D 50 ENTIB.E QUANTITIES. . An algebraic expression may be factored by inspection, by trial, or by its law of formation. To express the prime factors of a monomial, we have only to factor the coefficient, and repeat each letter as many times as there are units in its exponent. Thus, 15a 8 x*y =3 X 5 X aaaxxy @5. The following remarks will aid in factoring polynomials : Is*. If all the terms of a polynomial have a common factor, the quantity may be factored by writing the other factors of each term within a parenthesis, and the common factor without. Thus, 2V 6aV-f4a 2 .T 10a'=2a a (x 8 3x a +2x 5a) 2d. If two of the terms of a trinomial are perfect squares, and the other term is twice the product of the square roots of the squares, the trinomial will be the square of the sum or difference of these roots, (7O, I and II), and may be factored accordingly. Thus, in the trinomial, 4a 4 20o 2 &-{-256 2 , the two terms, 4a 4 and 25& a , are the squares of 2a 2 and 5& respectively, and the other term, 20a 2 6 is equal to 2x2a a X5&; and we have 4a 4 20a 2 i-h256 2 ^(2a a 5&)(2a 8 56) 3d. If a binomial consists of two squares connected by the minus sign, it must be equal to the product of the sum and difference of the square roots of the terms, (7O ? III). Thus, 9x 2 y=(3x+y) (3.r y) 4th. Quantities in the form of a w & w may be factored by refer- ence to the principles and formulas relating to these quantities. Thus, a s +Z, 8 =(a-f&) (a 2 a&+ Z> 2 ) NOTE. It may happen that when there is no factor common to all the terms, a portion of the polynomial'may be factored. EXAMPLES FOR PRACTICE. 1. Factor a'6-f a 2 6 2 -f-a 2 6c. Ans. a'i(a 2. Factor 3xy~ 3xy+3;ey 6^y. Ans. 3. Factor 5a 8 6c 2 15a a 6V 5a a icV. Ans. DIVISION. 51 4. Factor a'-j-c'tf-f cmx. Ans. a'+c(c+m)x. 5. Factor x* x'y+xy'y*. /^Vy^-v/ 6. Factor a 4 & a -j-2a 8 6'-}-a a & 4 . Ans. ' a'l*(a-\-b) (a+fi). 7. Arrange (x 9 x')a-\-(x*-\-x)(5bc)q according to the pow- ers of x. Ans. (a-f-3& c)aj f ^a 36+c)ic #. 8. Factor a'm 9awi 8 . Ans. am(a* 3m) (a a -{-3m). 9. Factor 8a 8 x 8 . 4. (4a"+2cKC+x s ) (2a ). 10. Factor y'-}-243. ^TIS. (y Sy'+O.y 1 27y+8l) (y+3). 11. Find the factors of x' y*. Ans. (a^+ay+y) (* ay+y") (a;+y) (x y). 12. Find the factors of a 8 a&*-f-2a&c ac*. J.TIS. a(a-|-& c) (a &+c). SUBSTITUTION. 00. Substitution, in Algebra, is the process of putting ono quantity for another, in any given expression. 1. Substitute y 1 for x, in x'-\-x* 5z 3. OPERATION. 5x = 5(y 1) = = --3 =/ 2^4^+2, Ans. ITence, for substitution we have the following RULE. Perform the same operations upon the substituted quantify as the expression requires to be performed upon the quantity for which the substitution is made EXAMPLES FOR PRACTICE. 1. Substitute a b for a in a a -fa&-[-3*. Ans. a? a5-f b*. 2. Substitute a-f 2 for a in a* 2a-fl. Ans. cc s -(-2x-f 1. 8. Substitute a-f 3 for y in y 4 2y 8 +y"~6. . .T 4 -f lOa'- 52 ENTIRE QUANTITIES. 4. Substitute s+r for x, in x a -j-az-|-&, and arrange the result ac- cording to the descending powers of r. Ans. r a -f(2s+a>-f s a -j-as-}-. 5. What will a 4 -f a 8 &+a 7 6 a +a&'+& 4 become, when I a ? Ans. 5a*. 6. What will x*-f ox a -f a*x-{-a* become, when m-f-1 is put for x and m 1 for a ? Ans. 4m(ra a +l). 7. What will x 4 -f-y 4 become, when a-}-b is put for x and a b fory? ^ns. 2(a 4 +6a 2 &'+6 4 ). 8. What is the value of (x+0+b+c)*+ (z a 6 c) 6 , when 9. In x 8 7a+6 substitute y 2 for a. -4ns. y' 6y'-|-5y. 10. In x 6 2* 4 -f 3z 7x'+8a; 3 substitute y+1 for a;. 11. If a 6=z, b c=y, and c a = as, prove that 2(o 6)* THE GKEATEST COMMON DIVISOR 97. A Common Divisor of two or more quantities is a quantity which will exactly divide each of them. 98. The Greatest Common Divisor of two or more quantities is the greatest quantity that will exactly divide each of them } it is composed of all the common prime factors of the quantities. The term, greatest, in this connection, is used in a qualified sense, and has reference to the degree of a quantity, or of its leading term, not to its algebraic or its arithmetical value. Thus, if x 3 and iC 3 _|_4x-j-2 are 'the prime factors common to two or more quantities, then according to the above definition, (x a +4o;-j-2)(a; 3)=# 8 -{- x a 10x 6, is the greatest common divisor. But this product is not necessarily greater in value than one of the prime factors. For, if a; 4, then we have -f 4x-f2=34, and x 8 -j-o; a lOz 6=34. 99. Several quantities are said to be prime to each other when they have no common factor. DIVISION. 53 CASE I. 1OO. "When the given quantities can be factored by inspection. It is evident from (81. 2) that no factor of the greatest common divisor can have an exponent greater than the least with which it enters the given quantities. Hence the following obvious EULE I. Find ~by inspection, or otherwise, all the different prime factors that are common to the given quantities, and affect each with the least exponent which it has in any of the quantities. II. Multiply together the factors thus obtained, and the product will be the greatest common divisor required. EXAMPLES FOR PRACTICE. 1. Find the greatest common divisor of a* 2a*x*-}-ax' > and a 4 2a a x-\-a*x*. Factoring, we have a 6 2aV+a x 4 =a (a 4 2aV+ z 4 )==a(a x)\a+ x) 9 a 4 2a 8 x -f-aV=a a (a a 2ax -f x a )=a a (a xf The lowest powers of the common factors are a and (a x) 9 ; and we have a(a x) 2 = a 8 2a a x + ax a the greatest common divisor required. 2. Find the greatest common divisor of 2a*bc*, 6a&V, and IQa 3 bc\ Ans. 2abc\ 3. Find the greatest common divisor of 5rE 2 ,y V, 6x 8 ^ 2 , and I2x*yz*. Ans. x*yz*. 4. Find the greatest common divisor of x* y 3 and x a 2xy-\-y* Ans. x y. 5. What is the greatest common divisor of a 2 m b*m and 2ac 2 m 2c*bm ? Ans. m(a &). 6. What is the greatest common divisor of a a x 8 3a a x a +a 2 a: and az 3 ? Ans. a(x 2 3z+l). 7. What is the greatest common divisor of 16x a 1, x 4x a , and ? Ans. 54 ENTIRE QUANTITIES. CASE II. 101. "When the given quantities can not be factored by inspection. 102. The greatest common divisor is found in this case by a process of decomposing the quantities by division. But in order to deduce a rule for the method, it will be necessary first to establish certain principles relating to exact division. 103. First, suppose A to be a quantity which is exactly divisible by another quantity, D, and let q represent the quotient. Then, A D =q If we now multiply the dividend by m, we shall have, from (84: I), ^=gm D " in which qm is entire. Thus we have shown that if D divides A, it will also divide Am. Hence, l.I/a quantity will exactly divide one of two quantities, it will divide their product. Again, let A and B represent any two quantities, and S their sum. Now suppose both A and B are exactly divisible by D, and let -=2, and yy=2'- We shall have And dividing each term by Of in which _ must be entire, because its equal, q-{-q r , is entire. Hence, 2. If a quantity will exactly divide each of two quantities, it will divide their sum. Finally, let d represent the difference of A and B, and suppose A and B to be divisible by D, q and q e being the quotients, as before. We shall have A B=d And dividing every term by D, in which _ is entire, because q q' is entire. Hence, DIVISION. 55 3. If a quantity icill exactly divide each of two quantities, it will divide their difference. 1O4. We may now show, by the aid of these principles, wln-t relation the greatest common divisor of two quantities bears to the parts of these quantities when decomposed by Division. Suppose two polynomials to be arranged according to the powers of the same letter, and let A represent the greater and B the less. Then let us divide the greater by the less, the last divisor by the last remainder, and so on, till nothing remains. If we represent the several quotients by q, q', q", etc.; and the remainders by R, R', R"j etc., the successive operations will appear as follows : (1.) (2.) (3.) By** *) *' R R? To investigate the mutual relations of A, B, R y and R'j we ob- serve that in division the product of the divisor and quotient, plus the remainder, if any, is always equal to the dividend. Hence, from the operations above, we have the three following conditions : R'q" =R Rq' Bq Now from the first equation it is evident that R' divides R with- out remainder; it will therefore divide Rq', (1O3 ? 1). And since R r divides both Rq' and itself, it must divide their sum, Rq'-\-R', or B, (1O3, 2) ; consequently, it will divide Bq, (1O3, 1). Finally, since it divides both Bq and R, it must divide their sum, jBq-\-JR, or A, (1O3, 2). Hence, I. The last divisor, R', is a common divisor of R, B } and A, or of all the dividends. Again, the dividend minus the product of the divisor and quo- tient, is always equal to the remainder. Therefore, from the first and second operations above, we have ABq =R 56 ENTIRE QUANTITIES. Now any expression which will divide B, will divide By, (1O3, 1) ; hence, any expression which will divide both A and B, will also di- vide A Bq, or R, ( 1O8, 3). Whence it follows that the greatest common divisor of A and B will divide R, and is therefore a common divisor of B and R. For like reasons, referring to the second equa- tion, the greatest common divisor of B and R will also divide R', and is therefore a common divisor of R and R'. But the greatest common divisor of R and R' is R f itself. Consequently, R' is the greatest common divisor of R and B } and also of B and A. Hence II. The last divisor, R f , is the greatest common divisor of the given quantities, and also of the dividend and divisor in each subse- quent operation. 1. What is the greatest common divisor of 12x 8 2x* 7-r 3 FIRST OPERATION. 12z s 2x 2 Ix 3 12x*2x*4x Go; 2 2x X 1 1st Rem. SECOND OPERATION. ' 2x 1 X 1 Ans. x 1. The process here employed for finding the greatest common divi- sor of two polynomials, is subject to two modifications, which we will now investigate in their order. 1st. Suppressing monomial factors. It is evident that any monomial factor common to the given poly- nomials, may be suppressed in both, and set aside as one factor of their greatest common divisor. We may then apply the process of division to the resulting polynomials, and obtain the remaining fac- tor or factors of the greatest common divisor required. Again, if either polynomial contains a factor which is not common to both, this factor can form no part of the greatest common divisor DIVISION. 57 required, and may therefore be suppressed. And since the greatest common divisor of the given polynomials is the same as that of the dividend and divisor in each operation following the first, (II), it is evident that we may suppress the monomial factors in every remainder that occurs. And it should be observed, that if all the monomial factors of the given quantities have been previously sup- pressed, no monomial factor of any one of the remainders can belong to the greatest common divisor sought, or be common to any two suc- cessive remainders, (II). This modification of the process will be illustrated by the example which follows : 2. What is the greatest common divisor of 12x 6 -f-22x 3 -f-6x and 6x B 15x s 36x? The first polynomial contains the monomial factor 2x, and the second contains the monomial factor 3x. We therefore suppress these factors, setting aside x, which is common, as one factor of the greatest common divisor sought. We then apply the process of di- vision to the resulting polynomials, as follows : FIRST OPERATION. 6x 4 -fllx 9 -L 3 6x 4 15x' 36 2x 4 5x a 12 26x a +39 Suppressing the factor 13 in this remainder, we have 2x a -f-3 for the next divisor. SECOND .OPERATION. a 12 2x a +3 2x 4 -f3x a 8x a 12 8x a 12 Taking the last divisor, and the common factor, x, which was set aside at the beginning, we have (2x a -f-3)Xa=2x'-|-3x. Ans. 2d. Introducing monomial factors. It may happen at any stage of the process, that after suppressing every monomial factor of the divisor, its first term will not be exactly contained in the first term of the dividend. In such cases, the dividend may be multiplied by such a factor as will render its first term divisible by the first term of the divisor. No factor thus 58 ENTIRE QUANTITIES. introduced can be common to the dividend and divisor, since by hypothesis all the monomial factors of the divisor have previously been suppressed. Consequently, if the process of division be con- tinued under this modification, the last divisor must be the greatest common divisor sought. This point will be illustrated by the fol- lowing example : 3. What is the greatest common divisor of 2x* 12x 3 -f-17x 2 -}- Qx 9 and 4z 8 18x 2 -fl9z 3 ? "We first multiply the greater polynomial by 2, to render its first term divisible by the first term of the other polynomial. FIRST OPERATION. 4 24x 8 -f34z a +12:r 18 3 ,-1 2z 3 -f- 5z 2 -t- 5x 6 4-10x 12 New prepared dividend. a 19a- 3 In the above operation, we suppress the facto.r 3 in the first remainder, and multiply the result by 2, to render the first term divisible by the first term of the divisor. We thus obtain 4x 3 -f- 10x 2 4-10x 12 for the second dividend. As the two partial quotients, x and 1, have no connection, they are separated by a comma. Multiplying the last divisor by 2 for a new dividend, we proceed as follows : SECOND OPERATION. 8x a +29x 15 s 36x 2 + 38x- 6 7x 2 + 23x 6 56x 2 -}-184x 48 New prepared dividend. -105 I9x+ 57 Dividing this remainder by 19, We have x 3 for the next divisor. DIVISION. 59 THIRD OPERATION. X 3 15 15 Thus we find that the greatest common divisor is x 3. Had we suppressed -f-19 instead of 19, in the final remainder of the second operation, we should have obtained .z-f-3, or 3 x for the greatest common divisor. It should be remembered, however, that the term greatest, in this connection, has reference to exponents v and coefficients, and not to the algebraic value ; (O8). Consequent- ly either x 3, or 3 x may be considered the greatest common divisor of the given polynomials. And it is immaterial what sign is given to any monomial factor which we may suppress or introduce at any stage of the work. 1O*>. From these principles and illustrations we deduce the fol- lowing general RULE. I. Arrange the two polynomials with reference to the same letter ; then suppress all the monomial factors of each t and if any factor suppressed is common to the two polynomials, set it aside as one factor of the common divisor sought. II. Divide the greater of the resulting polynomials by the less, and continue the division till tlie first term of the remainder is of a lower degree than the first term of the divisor ; observing to suppress the monomial factors in every remainder, and to introduce into any dividend, if necessary, such a factor as will render its first term exactly divisible by the first term of the divisor. III. Take the final remainder in the first operation as a new divisor, and the former divisor as a new dividend, and proceed as before ; and thus continue till the division is exact. The last divisor, multiplied by the common factor, if any, set aside at the beginning, will be the greatest common divisor required. TV. If more than two polynomials are given, find the greatest common divisor of the first and second, and then the greatest com- mon divisor of this result and the third polynomial, and so on. The last will be the greatest common divisor required. 60 ENTIRE QUANTITIES. EXAMPLES FOR PRACTICE. Find the greatest common divisor, 1. Of x 4 2x 3 4* 2 -f llx 6 and x 3 8x 9 -fl7x 10. Ans. x* 3x42. 2. Of 6x 3 -|-x 2 44x+21 and 6x 3 26x'-f 46x 42. 4n*. 3x 7. 8. Of x 8 6ax 2 -f-10a a x 3a 3 and 3ax a 14a a x -J-15a. J.jis. x 3a. 4. Of x 4 8x s -f 14x 2 -fl6x 40 and x 9 8x a -j-19x 14. 5. Of a 3 -j-5a 2 +5a+l and a s -fl. Ans. a-fl. 6. Of 2a 4 5a 3 i 3a^ 3 -f7ai 8 + 3i 4 and4a 3 2a a 5 4a& a 3 ^4?is. 2a 3i. 7. Of 3x 8 4xV+3^ 2 2?/ 3 and 4x 2 7^+3.y a . Ans. x y. 8. Of 4x 6 2x 4 -f-4x 3 27x 9 +4o; 7 and 5. Ans. 9. Of a*c 4a 3 cm-f 3acm 3 and aV 6a a c a m-|-5c 2 m 9 . ^4s. c(a a m). 10. Of x 4 4x s 16x a +7x+24 and 2x 3 15x a -h9x-j-40. Ans. x a 5x 8. 11. Of 15x 9 4-71o; 4 -f 60x a 56 and 3x 6 17x 4 20z a +84. Ans. 3x 2 -j-7. 12. Of 3a 4 4-14a 2 m a 5wi 4 , 6a 4 14a 3 m a -f4i 4 , and 3a 4 22a 2 m -f-7m*. 13. Of 2aV 2 ^u?w. ax by. 14. Of 9a 4 -f 12a 8 +10a 3 +4a-f-l and 3a 4 -f-8a 3 -}-14a 2 +8a-f-3. Ans. I LEAST COMMON MULTIPLE. 1O6. A Multiple of any quantity is another quantity exactly divisible by the given quantity. It follows from this definition that if one quantity is a multiple of another, the multiple must be equal to the product of the other DIVISION. 61 quantity by some entire factor. Thus, if A is a multiple of JB, then A=JBm ) in which m is entire. 107. A Common Multiple of two or more quantities is one which is exactly divisible by each of them. 108. The Least Common Multiple of two or more quantities is the least quantity which is exactly divisible by each of them. CASE I. 109. "When the quantities can be factored by inspec- tion. From the principles of exact division, we may make the follow- ing inferences : 1. A multiple of any quantity must contain all the factors of that quantity. 2. A common multiple of two or more quantities must contain all the factors of each of the quantities. 3 The least common multiple of two or more quantities must contain all the factors of each of the quantities, and no other factors. Hence the following RULE. I. Find by inspection all the different prime factors that enter into the given quantities, and affect each, with an exponent equal to the greatest which it has in any of the quantities. II. Multiply together the factors thus obtained, and the product will be the least common multiple required. EXAMPLES FOR PRACTICE. 1. What is the least common multiple of *-}-&, a?d b*d, and Factoring, we have a 3 -f ab a(a-\-b) a?db*d = k and x=. 3. Then 4a 2x 8^-6 2 The real sign of this fraction therefore is minus, though its apparent sign is plus. 132. Each term in the numerator and denominator of a fraction has its own particular sign, distinct from the real or apparent sign 6* E 66 FRACTIONS. of the fraction. Now the essential sign of any entire quantity is changed, by changing the signs of all its terms. Hence, I. Changing all the signs of either numerator or denominator , changes the real sign of ike fraction; (8f ? I). II. Changing all the signs of both numerator and denominator, does not alter the real sign of the fraction; (8o ? II). III. Clianging the apparent sign of the fraction, changes the real sign. REDUCTION. 123. The Reduction of a fraction is the operation of changing its form without altering its value. CASE I. 124. To reduce a fraction to its lowest terms. A fraction is in its lowest terms, when the numerator and denom- inator are prime to each other. And since it does not alter the value of a fraction to suppress the same factor in both numerator and denominator, (HOj III), we have the following RULE. I. Resolve the numerator and denominator into their prime factors, and cancel all those factors which are common Or, II. Divide both numerator and denominator by their greatest common divisor. EXAMPLES FOR PRACTICE. 1. Reduce to its lowest terms _- * ' a'-j-a 3 " a'(a 2 -|-l) a 2. Reduce ^7"^"^ ^ to its lowest terms. 4a* 2a* 3a-f 1 KEDT7CTION. 67 The greatest common divisor of the numerator and denominator, as found by (1O5), is a 1 ; hence, (3a a 2a l)-4-(a l)=3a+l (4 tt _2a a 3o+l) -T- (a 1)= 4a a -f 2a 1 And we have for the reduced fraction, 3a+l a j Ans. Reduce each of the following fractions to its lowest terms : 3. ^. Ans. 4 X *~~ l . Ans. * 1 *y+y y a* al* A a* ab 5. . jSLns. x 6 &V A 6. r j- Ans. 2 X * I6x 6 7 ^Ins.- _. ' 3*' 24x 9 3 Ans - 4 X 4 x _f =* v< s i o 2; 10 / a i ; a-j-o 10. a 8 x a a-fx 11. ^^ x'-^-.+l^ ^ns.^ 1 .. " a;_ x '_x 2 -fa; x (x-fy) s -x-y 3 (3x a 1) (2x a 1) x a (5x > 7) . 1 U-: 1 1 .; L+ Ans. XO 9 t-\9 l / O N I OS FRACTIONS. CASE H. 125. To reduce a fraction to an entire or mixed quan- tity. The division indicated by a fraction may be at least partially performed, when there is any term in the numerator whose literal part is exactly divisible by some term in the denominator. Hence, RULE I. Divide the numerator by the denominator as far as pos- sible, and the quotient will be the entire quantity. II. Write the remainder over the denominator, annex the fraction thus formed to the entire part, with its proper sign, and the whole result will be the mixed quantity. EXAMPLES FOR PRACTICE. Reduce the following fractions to entire or mixed quantities : x Ans. a-f T-. bx Ans. a-\ a Ans. 5a+l+ . y Ans. 2( Ans. x 4a 3-f-^ ox - Sx' 7x'+7z+30 <-H~ 3x-10 An >- 3 * 5 REDUCTION. 69 10. ^ ' *y x_6.r 4 -f 10*' 3 CASE m. To reduce a mixed quantity to the form of a fraction. This case is the converse of the last, and may be explained by it. Hence the following RULE. Multiply the entire part by the denominator of the frac- tion ; add the numerator if the sign of the fraction be plus, but subtract it if the sign be minus, and write the result over the denomi- nator. EXAMPLES FOR PRACTICE. Reduce the following mixed quantities to fractions : 2. 2b-^=^ Am. c b o An,. 6. 3o 9 a+3 a+3 ^ . Ans. . xy xy (x 2)' a 2 TO FRACTIONS. 9. a > aJ>l >-L. An,. - o6 2y 4 -4-2y 1+y 1 A "I I Q,. 1 Q -.9 I f>j. 8 I * y !____. ^i?lS. CASE IV. . To transfer a factor from the denominator to the numerator, or the reverse. Let us take any fraction, as - , and multiply both numerator by* and denominator by ^~ n , observing that any factor having zero for its exponent is equal to unity, (88 ? 1), and may therefore be omitted. We shall have ax If we multiply both numerator and denominator of the same frao tion by x~ m j we shall have ax m ax m ~ m ax* a In like manner we may transfer any factor having a negative ex ponent. For example, let us take the fraction, ax , and multiply b both numerator and denominator by x' ; we shall have b ~ bx bx'~bx' By the same principle also, any fraction may be reduced to the form of an entire quantity ; thus, In all operations of this kind, the intermediate steps may be omitted, and the results obtained by the following REDUCTION. *-. 71 RULE. I. To transfer any factor from the denominator to the nu- merator, or the reverse : Change the sign of its exponent. II. To reduce any fraction to the form of an entire quantity : Transfer all the factors of the denominator to the numerator, ob- serving to change the signs of the exponents of the /actors transferred. EXAMPLES FOR PRACTICE. In each of the following fractions, transfer the unknown factors, or factors containing unknown quantities, to the numerator. ax axi/~* * c 3a 9 . 3a 3 *-' Ans. 1. 5ni 3 _I_ AI *-*-'-. axyz a c 4A . Ans. > 5cm(a ar)~* 5c 4m In each of the following fractions, transfer the known factors to the denominator, and the unknown factors to the numerator. '&cV o - , Ans. (x a) l (a ^)" (a &)* HAns. - 1 ^ 7 i ft A r i o ft 72 FRACTIONS. In the following fractions, transfer the factors having negative exponents. 12. 5i Ans. *?L 5m*- 13. M*'-l) ' 14 * '* Reduce each of the following fractions to the form of an entire quantity. 15. !1 An*. 5a 16. L^L. Ans. am* 17. *L. 18. CASE V. 128. To reduce one or more fractions to a common denominator. We have seen (124) that a fraction may be reduced to lower terms by division. Conversely, a fraction must be reduced to higher terms by multiplication, and each of the higher denominators it may have, must be some multiple of its lowest denominator. Hence, 1. A common denominator to which two or more fractions may be reduced, must be a common multiple of their lowest denomina- tors ; and 2. The hast common denominator of two or more fractions, must be the least common multiple of their denominators. 1. Reduce _f_ and .to their least common denominator. a*b al* REDUCTION. 73 "We find by inspection that the least common multiple of the given denominators is a*b a . And If, therefore, we multiply both numerator a^d denominator of the first fraction by I 9 , and of the second by a, we shall reduce the two fractions to their least common denominator, cfl?. Thus, cXb*=b*c, new numerator of first fraction ; : new numerator of second fraction. TT c d b*c ad A Hence _ , ___ . _ , Ans. a*b aV a*b* a*b* From these principles and illustrations we deduce the following RULE. I. find the least common multiple of all the denominators, for tlie least common denominator. II. Divide this common denominator by each of the given denom- inators, and multiply each numerator by the corresponding quotient. The products will be the neic numerators. NOTE. Mixed numbers should first be reduced to fractions, and all fractions to their lowest terms. EXAMPLES FOR PRACTICE. In each of the following examples, reduce the fractions and mixed quantities to their least common denominator : 2a 36 4ac Bbx 1. and ~ Ans. , -= x 2c 2cx 2cx 4ac Ans. , 2c 5 3& 4. ^1, and b' acx H- a a c (bx -\-b} (x -j- a) icy ' &c.r -j- a6c bcx ~\- abc bcx -j- abc T C JL y ay 1 ay' y ay' y FRACTIONS. 5 x 1 4x' 4* 5 x 5 ADDITION. 12O. We have seen (115) that a fraction is equal to the re- ciprocal of its denominator multiplied by the numerator. Hence, if two or more fractions have a common denominator, they will have a common fractional unit, which may be made the unit of addition. Thus, a b 1 1 1 Or thus, a b a-\-b - + - = ac-'-f bcr l =(a+V)cr l = - The intermediate steps may be omitted ; hence the following RULE. I. Reduce the fractions to their least common denomi- nator. II. Add the numerators, and -write the result over the common denominator. NOTES. 1. If there are mixed quantities, we may add the entire and fractional parts separately. 2. Any fractional result should be reduced to its lowest terms. ADDITION. 75 EXAMPLES FOR PRACTICE. 3* 2x * LAdd-g, _ and _. An*. _ a ,a+b ac+ab+b* 2. Add T and - Am. - - be vc a 1 . a*+a? 3. Add -z and f Ant. a+x . 4. Add and a-b"^a+b ' a a -6 a i 5 5. Add 2cH -and * 2 . 2x 3 M . 5x* 4x 9 6. Add 5x+-^ and 7. Add JL, J-,and-^-. An*. tf. a-f-c a c a-j-c a c 8. Add 9 * Add (b c) (c a)' ( c a) (a )' &nd (a 4) (i c)' Ans. 0. Aril a '~ b &'+ , 1-f a6 ' ' ai . 0. be ac , 11. Add , - - - -, - - T-T-, - r,and 12. Add (a b) (a c) (6 c)(i a)' (c a)(c , X 3 05 2 031 P x a - J.71S. -^ 6 ; 8 2- a 3x 76 FRACTIONS. SUBTRACTION. 13O. If two fractions have a common denominator, they will have the same fractional unit ; and the one may be subtracted from the other, by taking the difference of the numerators. Thus, c c c c c Or thus, a b Hence the following RULE. I. Reduce the fractions to their hast common denominator II. Subtract the numerator of the subtrahend from the numerator of the minuend, and write the result over the common denominator. EXAMPLES FOR PRACTICE. 3x x 2z 13* i From - subtract -. -Ans. -^-. 7 y v)o 7 X 2z 1 17^4-2 2. From subtract - Arts. -! . 23 o From subtract ; . Ans. j-. Ua 10 . . ' , 3a 5 4. From 3a -\ -- take 2a -) -- - . lO 7 a 4- b , a 6 5. 1< rom - - take - - _ An*, -r - _ l a _|_ 5 a* b* 6. From x + *~ y take %JL. Ant. x -J?- * * '* -, . x , . x a 7. From 3x -J take ic -- . be x 9 + x 5 a;* 4- x 1 1 -llx + 12 4x+ 8 9. From MULTIPLICATION. 77 3a-f-6 , 2(a 5) J.NS. . A T, 10. From . 7a6(a 6) 2(a' 4') MULTIPLICATION. 131. Any fraction may be multiplied by an entire quantity in two ways : 1st. By multiplying its numerator; or 2d. By dividing its denominator ; (119, I and IT). 139. A general rule for the multiplication of fractions is fur- nished by the following example : 1. Multiply ~ by 1. 6 a OPERATION. 6 d By observing the result, we find that the new numerator is the product of the given numerators, and the new denominator is the product of the given denominators. Hence the following RULE. I. Reduce entire and mixed quantities to fractional forms. II. Multiply the numerators together for a new numerator, and the denominators for a new denominator, canceling all factors com- mon to the numerator and denominator of the indicated product. EXAMPLES FOR PRACTICE. _ ,, ... , a , b a 1. Multiply - by Ans. b x x 78 FRACTIONS. _, . . . - , 2a 3. Multiply - by _. Ans. 4. Multiply ^L^r'by-^-- A*. 2y a-fx 5. Multiply L_ < nr ,.. , o , a 6. Mulbply _by ___ Ans. 1. Multiply =by 2" ^~- 8(-+). 8. Multiply a+ X - by a--. II I* 9. Multiply * X '-? X by Ans. * ax ~*> a t * J 1J. "V *"+*> ,, and =1*. rj a a -j-2uc-fc* a' 2ac-j-c a ac 2 ^ ax Ar ii- i "-' (;rt - c u . Multiply I T A -r by & c (c & a)(6 c a) DIVISION. . Any fraction may be divided by an entire quantity in two ways : 1st. By dividing its numerator ; or 2d. By multiplying its denominator; (119, I and II). We may, however, derive a general rule for the division of frac- tions, from the following example : 1. Divide by 6 d OPERATION. 4=*^.-rf4g=g 6 d cd~ l Ic By inspecting this result, we find that the new numerator may be obtained, by multiplying the numerator of the dividend by the de- nominator of the divisor ; and the new denominator may be obtained, by multiplying the denominator of the dividend by the numerator of the divisor. Hence the RULE. I. Reduce entire and mixed quantities to fractional forms. II. Invert the terms of the divisor, and proceed as in multiplica- tion. EXAMPLES FOR PRACTICE. . _. .. 5x , I . xc 5cx 1. Divide by Ans. Xr = r a o ao by -1-,. c * a-f b 80 FRACTIONS. . 8. Divide - by -r - ; - An*- -7-^ * * 2c a x a x 2x a 7 , a* (2x a 7) (x+a) 4. Divide ; by nrns ; ' Ans - x+a J x a -j-2ax-f-a a a* 5. Divide 3 X '~ l \ 78 by ^ ^?is. x a +Z> a . x* 2bx-4-l* J x b 2ax+ x* , x 6. Divide -H^- by - 70x 15 Ans. -TT- - lO.r 4 8. Divide ~~ . by Jr -4.w*. - 6x 7 x 1 18av-21 9. Divide y by Ans. - - . ^- -j '+* , 2ax-f2ax a 7 10. Divide -g^- by - ^Lns. ^ 11. Divide -5-^ 12. Divide -^-^ by t- wo TIX ma mx 13. Divide - - 7 - by a-j-6 *m x a 1* -n- -j 3(x 7 1) /X+l\ (XI\ 15. Divide -7-7 - by I -^r 1 1 1 7 ] - Ans. 3a. 2(a-f-i) J \ 2a / \a-\-b/ IP iv M 10^-f3a'+3^ a 16. Divide , _. ., a a 1 , a 1 1 17. Dmde - 3 -f- by -,--+- ^, 18. Divido a 4ns. 1. COMPLEX FORMS. 81 REDUCTION OF COMPLEX FORMS. 134. A fraction is said to be simple, when both numerator and denominator are entire ; otherwise it is said to* be complex. 1 Ho. To reduce a complex to a simple fraction, we may regard the quantity above the line as a dividend, and the quantity below it as a divisor, and proceed according to the last rule. A more convenient method may be derived from the following observations : 1. If a fraction be multiplied by its own denominator, the product will be the numerator. 2. If a fraction be multiplied by any multiple of its denomina- tor, the product must be entire. Hence to simplify a complex fraction, we have the following RULE. Multiply both numerator and denominator by the least common multiple of the denominators of the fractional parts. EXAMPLES FOR PRACTICE. 1. Simplify a-j-a? Multiplying both numerator and denominator by (a a;) or by its equal a* x*, we have a _i a x a*4-ax a*4-x* ax-\-x* ' ' .. - . __! , Ans. a a? x a a*-\-ax ax x 9 a x 2. Simplify __i. Ans . 82 FRACTIONS. ac 3. Simplify _JL'. ^ s _ i'c + ao 3 4. Simplify m n n a 6 + + 0+1 ( Xl * x-\-cd = ax cx-{-a*. Ans: ex ax x = a 3 5ai* 4crf. In the following, transpose the unknown terms to the first mem- ber, and the known to the second, by (II). 5. ax-\-bc = a*+c*x. Ans. c*x ax = bc a 3 . 6. 4c^ a*x 3cm = ax m*. Ans. a*x-{-ax = 4ceZ* 3cm-j-w 8 . 7. ax 7-|-5ceZ = bc-\-a*cx 4m 9 . Ans. a* ex ax = 5cd 7 &c-f-4m a . 8. a 9 c*x Sdx = c*d>x W. Ans. c'cPx-}- 3. A root of an equation is said to be verified, if the two members of the equation prove to be equal after the root has been substituted for the unknown quantity. 154:. A simple equation may be reduced, by transforming it in such a manner that the unknown quantity shall stand alone, and con- stitute one member of the equation } the other member will then ex- press the value of the unknown quantity, or the root of the equation. Let it be required to find the value of x in the equation Clearing of fractions, we have 20x 8 3z-f-2l = 48+10x. (2) By transposition, we obtain 20x 3x lOx = 48+ 8 21. (3) Uniting similar terms, 7x = 35. (4) And dividing both members by 7, we have x = 5. (5) To verify this value of x, substitute it for x in equation (1) ; wo shall have 252 5 7 25 3 4 "6" Reducing each term to its simplest form, we obtain 7f+J = 4+4l; whence, by addition, we have 8* = 8t, and the value of x is therefore verified. loo. It should here be observed that an equation of the first degree, containing but one unknown quantity, can not have more than one root. For, whatever the equation may be, suppose it to be 8* 90 SIMPLE EQUATIONS. cleared of fractions, and the unknown terms transposed to the first member, and the known terms to the second. Then if we represent the algebraic sum of the coefficients of x by a, and the second member by b, the equation will take this general form : ax = b. .(1) Now, if possible, suppose that this equation has two roots, r and r f . Then since every root must satisfy the equation, (14O), we shall have, by substituting r and r' successively in (1), ar = b, (2) ar'= b. (3) whence, by Ax. 7, we shall have ar = ar'-, (4) or, by transposing and factoring, a(r r') = 0. (5) But equation (5) is impossible, since, by supposition, r r' is not zero, and a is not zero. Hence, An equation of the first degree can not have more than one root. lt>O. From these principles and illustrations we derive the fol lowing RULE. I. If necessary, clear the equation of fractions, and per- form all the operations indicated. II. Transpose the unknown terms to the first member and the knoicn terms to the second, and reduce each member to its simplest form, factoring, when necessary, with reference to the unknown quantity. III. Divide both members by the coefficient of the unknown quan- tity, and the second member will be the value required, or the root of the equation. The three principal steps in the reduction of a simple equation, containing but one unknown quantity, may be briefly stated as follows : 1st. Clearing of fractions. 2d. Transposing and- uniting terms. 3d. Dividing by the coefficient of x. REDUCTION. 91 PRACTICAL SUGGESTIONS. There are certain cases in which the preceding rule may be mod- ified, with advantage, by special artifices. 1 When the equation contains similar terms, or fractions having a common denominator, these should be united as far as possible before clearing of fractions. Thus, Given, 5C = 1 00 transposing and uniting terms, 2x -j -- = 44 ; clearing of fractions, 14x -f & 7 = 308 ; 15x = 315 ; x= 21. 2. When the equation contains fractions whose numerators or denominators are polynomial, we may clear the equation of its sim- pler denominators first, uniting the entire quantities at each step, if possible. Thus, multiplying by 9, 6x-f 7 -f ~ = 6x-fl2 ; transposing and uniting, &x | JL clearing of fractions, 2lx 39 = lOa-f-5 ; whence, by division, x = 4. 3 When the equation contains but a single numerical term, we may simply indicate the multiplication of this term, in the clearing of fractions, until the final step in the reduction is reached. Thus, Given ^ + f_ + |L =8 8 ; multiplying by 84, 2lx + 12x+ Ix -f- 4x = 88 X 84 ; 44x = 88 X 84 ; dividing by 44, x - 2X84; x = 168. 92 SIMPLE EQUATIONS. EXAMPLES FOR PRACTICE. Find the value of x in each of the following equations. 1. 7x 16 = 3x 4. Ans. x = 3. 2. Zx + 9 = 5*+ 1. Ans. x = . 3. 4x-\-7 = x-}-2l 3-f-x. Ans. a = 5J. 4. 5a;-f 16 = a + 52. Ans. a = 9. 5. 5aa; c = I Sax. Ans. x = "*" c 8a 6. ax4- 6 = 9a; 4- c. 4ns. *= " 9 7. 9. 8 10. + - = -. ^In.. = 20. 7x-16 17x 3 3 2 + 5 ~ 8 2 12 ^._i_?-4_^ ^_i_??~18 1Q 17x 12 5^ .16 10* 3 6* 13. . 3~ 4 6 2 . a = 16 3^ 11 5x-5 . 97-7^ Ans x-Q 14. 21 + jj-- g- + 2 ^ a ~ 9 15 z JL . A**. = ~21 4x 11 ~3 9x+20_4 -12 ^ aj = 16 ' "36 5x-4 + 4 20, 36 5x+20 to 86 "' "25 + 25 ^ 9z 16 5 T 25 ONE UNKNOWN QUANTITY. 93 3x x 1 20x4-13 M K 18. -=6* Z A*. x = 5. 19 . *_=? + * =20 -?+l ^..9. 20. f+- 1 + f+- 2 = 16-?+ 3 . 21. 2z : + 15 = JT 3f- "- 23 - ^-^V = 7 - ^ a5 = 5 - 24. iirr_(i_^_r) = 7*. ^,. x = J- y / v */ r^ A. x | ^- ^* t/^~^*X j 25 -T-+-l-= f -^ +3 ' ^.^ = 1. a; = 60. ^-U? = 130. An*. aj = 120. -Aiw. a; = 120. a; = 84. x = 1260. a c a-j-c a 32. 43)x 2a = 3a5 ftb*x. Ans. x = ^y a J_Lac+&c = 0. ulws. x = VV-T 94 SIMPLE EQUATIONS. 34. a'O-l)+am(a; 2) = m*. Ans. x = ^^- b fix b 35. ax+cx+x = l-\ - Ans.x=-- c a-f-c ~ , ~ c x a 36. = = Ans. x = 6 d b o 7 , _ __ 1 ' a 1 ^ i 1 a+1 i-f-1 2(a a 2+& a 38. c-f-1 c 1 ' (c I) 1 c 1 39. ''- + *j- -f - = a b c 41. 1.25* 6.125+.25* = .625*. Ans. x = 7. 42. 3.164* 4.266 = .24*+.08z. Ans. x = 1.5. 2.4x .12 4.6* 3.6 .64* .048 28 ' 4 = 7 ' X = PROBLEMS PRODUCING EQUATIONS WHICH CONTAIN ONE UNKNOWN QUANTITY . A Problem, in Algebra, is a question requiring the values of ODG or more unknown quantities from given conditions. 158. The Solution of a problem is the process of finding the values of the unknown quantities. lt>9. Every problem in Algebra contains a statement of the rela- tions between certain known and certain unknown quantities. When these relations are such as to furnish one or more equalities, the process of solution consists in expressing these equalities algebraic- ally, and in reducing the equations thus obtained. ONE UNKNOWN QUANTITY. 95 I GO. There are two classes of problems which may be solved by the use of a single equation. 1st. Questions referring to a single unknown quantity. 2d. Questions referring to two or more unknown quantities, so related that when one is known, the others may be determined directly by the given conditions. The following are examples of the first class : 1. What number is that the sum of whose third and fourth parts is 21? Let x represent the number ; then by the conditions clearing of fractions, 4x-j-3x = 21X^2, 7x = 21X12, whence x = 36, Ans. 2. A and B have each the same annual income. A's yearly ex- penses are $800 and B's $1000, and A saves as much in 5 years as B saves in 7 years ; how much is the annual income ? Let x = the income ; then x 800 = A's annual savings ; and x 1000= B's " " Now by the conditions of the problem, we have 5(z 800) = 7(z 1000); whence 5x 4000 = 7s 7000, x = 1500, Ans. The following are examples of the second class : 3. Three men form a copartnership with a joint capital of $7200. A put in a certain sum, B put in three times as much as A, and C put iu as much as both A and B ; how much did each nian furnish ? Let x = A'? share ; then 3* = B's " and 4x = C's " By the conditions, Sx = $7200 whence, x = $900, A's share, Bx = $2700, B's share, 4x = $3600, C's share. 96 SIMPLE EQUATIONS. 4. There are two numbers whose difference is 6 ; and if J of the less be added to ^ of the greater, the sum will be equal to $ of the greater diminished by of the less. Required the two numbers. Let x = the less ; then x-{-G = the greater. By the conditions of the problem, x x-\-Q ce-j-6 x 3"^ ~~5~ = ~~3 5> clearing of fractions, 5x+Sx+l8 = 5x+30 3x; whence, 6x = 12 ; x = 2, the less, ^ oj-j-6 = 8, the greater. These examples illustrate the three essential steps in the solution of any problem requiring but one equation; and we may derive from them the following GENERAL RULE. I. Represent one of the unknown quantities by some letter or symbol, and then from the given relations find an algebraic expres- sion for each of the other unknown quantities, if any, involved in the question. II. Form an equation from some condition, expressed or implied, by indicating the operations necessary to verify the value of the unknown quantity represented by the symbol. III. Reduce the equation thus derived. The three steps in this process may be named as follows : 1st. The notation. 2d. The equation. 3d. The reduction, REMARKS. 1 By the first two steps, the conditions of the problem are translated from common into algebraic language. This is called the Ktatement of the question. 2. The chief difficulty in the solution of a problem is generally experienced in obtaining the statement. This arises in part from ONE UNKNOWN QUANTITY. 97 the fact that among the problems which may be proposed, there exists an indefinite variety of conditions ; and the operator is left very much to his ingenuity, both in adopting suitable notation for any problem, and in deriving the equation. 3- Algebraic problems present two kinds of conditions, Explic- it and Implicit. An explicit condition is one which is distinctly and formally expressed in the language of the problem. An implicit condition is one which is not directly expressed, but only implied, or left to be inferred from other conditions. 4. In any problem there are always as many conditions as there are quantities to be determined. And if we represent one of the unknown quantities by an arbitrary symbol, and then proceed to derive expressions for the other unknown quantities, if any, each from a separate condition, there will always remain a final condition, cither explicit or implicit, from which to derive the equation. PROBLEMS FOE SOLUTION. 1. What number is that from which if 6 be subtracted, and the remainder multiplied by 11, the product will be 121 ? Ans. 17. 2. A man holds a lease for 20 years, and J of the time past is equal to ^ of the time to come ; how much of the time has passed ? Ans. 12 years. 8. What number is that which being increased by 4, ^ and j of itself is equal to 250 ? Ans. 120. 4. Divide 77 into two such parts that if one part be divided by 7 and the other by 3, the sum of the quotients shall be 15. Ans. 56 and 21. 5. The sum of two numbers is 75, and their difference is equal to ^ of the greater ; what are the numbers ? Ans. 45 and 30. 6. After paying away | and i of my money, I had $66 left ; how much had I at first ? An*. $120. ^4 7. After paying away of my money, and \ of what remaned, and losing i of what was then left, I still had $24 ; how much had I at first ? Ans. $60 9 98 SIMPLE EQUATIONS. 8. What number is that from which if 5 be subtracted, f of the remainder will be 40 ? Ans. 65. 9. A men sold a horse and a chaise for $200, and 4 of the price of the horse was equal to ^ of the price of the chaise. What was the price of each? Ans. Chaise, $120; Horse, CSO. 10. Divide 48 into two such parts, that if the less be divided b/ 4, aiid the greater by 6, the sum of the quotients will be 9. Ans. 12 and 36. 11. An estate was divided among 4 children in the following manner : The first received $200 more than | of the whole, tho second $340 more than 1 of the whole, the third $300 more than of the whole, and the fourth $400 more than | of the whc-e ; what was the value of the estate ? Ans. $4800. 12. What number is that from which if 91 be subtracted, ^ of the remainder will be equal to -Jg of the number ? Ans. 130. 13. Four men take stock in a railroad company, amounting in the aggregate to $73,500. A takes a certain sum, B takes three times as much as A, C takes three times as much as A and B togeth- er, and D takes one third as much as B and 0. How much of the stock does A take ? Ans. $3500. 14. Divide 105 into two parts which shall be to each other as 3 to 4. Since the parts are to each other as 3 to 4, let 3x = the less part ; then 4x = the greater ; whence 7x = 105, x = 15 ; 3x = 45, the less ; 4;e m 60, the greater. 15. A and B shared between themselves a bequest of $2000, in the proportion of 7 to 9. How much did each receive ? Ans. A, $875; B, $1125. 16. A farmer made a mixture of rye, oats, and peas, using 3 bush els of rye as often as 4 of oats and 5 of peas The whole amount of grain used was 72 bushels; how many bushels were there of each kind? Ans. Rye, 18 bushels; Oats, 24 bushels; Peas, 30 bushels. ONE UNKNOWN QUANTITY. 99 17. From two casks of equal size were drawn quantities which are in the proportion of 6 to 7 ; and it appears that if 16 gallons less had been drawn from that which now contains the less, only one half as much would have been drawn from it as from the other. How many gallons were drawn from each ? ,4ns. 24 and 28. 18. It is required to divide the number 204 into two such parts, that of the less being taken from the greater, the remainder will be equal to | of the greater subtracted from 4 times the less. Ans. 154 and 50. 19. A man bought a horse and chaise for 341 dollars. Now if | of the price of the horse be subtracted from twice the price of the chaise, the remainder will be the same as if | of the price of the chaise be subtracted from 3 times the price of the horse. Required the price of each. Ans. Horse, $152 ; Chaise, $189. 20. A certain sum of money was put at simple interest ; in 8 months it amounted to $1488, and in 15 months it amounted to $1530. What was the sum at interest? It is often convenient, in the solution of a problem, to avoid the multiplication of large numerals. This may be done by represent- ing a given number by a letter, as follows : Put a = 1488 ; then a-f-42 = 1530. Let x = the sum at interest ; then a x = interest for 8 months ; and a+42 x = 15 " Equating two expressions for the monthly interest, a x a-j-42 x f ~8~ ~T5~ whence 15o 15;c = 8a+336 8x, lx=la 336, and x = a 48 = $1440, Ans. 21. A prize having been captured by a privateer, the sum of $7560 was awarded to the officers, and the residue was divided equally among the crew, consisting of 27 men. If the officers had received $9560, and the crew had consisted of 25 men, each private would have received the same sum for his share ; what was the value of the prize? AUK. $34560. ' 100 SIMPLE EQUATIONS. 22. A .merchan t allows $1000 per annum for the expenses of his family, and annually increases . that part of his capital which is not so expended by a third of it ; at the end of three years his original stock is doubled. What had he at first ? Ans $14,800. 23. A man having a lease for 99 years, was asked how much of it had already expired ; he answered that f of the time past was equal to | of the time to come. Required the time past and the time to come. Ans. Time past, 54 years ; time to come, 45 years. 24. In the composition of a quantity of gunpowder, the niter was 10 pounds more than | of the whole, the sulphur was 4^ pounds less than of the whole, and the charcoal was 2 pounds less than 4 of the niter. What was the amount of the gunpowder ? Ans. 69 pounds. 25. Divide $183 between two men, so that ^ of what the first re- ceives shall be equal to T 3 of what the second receives. Ans. 1st, $63 ; 2d, $120. '26. Divide the number 68 into two such parts, that the differ- ence between the greater and 84 shall be equal to 3 times the difference between the less and 40. Ans. Greater, 42 Less, 26. 27. Four places are situated in the order of the letters A, B, C, D. The distance from A to D is 34 miles ; the distance from A to B is to the distance from C to I) as 2 to 3 ; and ^ of the distance from A to B, added to one half the distance from C to D, is three times the distance from B to C. What are the respective distances? Ans. 12, 4, and 18 miles. 28. A man driving a flock of sheep to market, was met by a party of soldiers, who plundered him of j of his flock and 6 more. Afterward he was met by another company, who took ^ of what he then had and 1 more ; after that he had but 2 left. How many had he at first ? Ans. 45. 29. A boy engaged to carry 100 glass vessels to a certain place, and to receive 3 cents for every one he delivered, and to forfeit 9 cents for every one he broke. On settlement, he received 240 cents; how many vessels did he break ? -4ns. 5. BO. A person's entire indebtedness to A, B, and C, was $:?70. His ONE UEKXOWK indebtedness to B was twice as much as to A, and his indebtedness to C was twice as much as to A an,d 13. How much did he owe each ? Ant. A, $30 ; B, $60 ; C, Si 80. 81. A company of 4 laborers received $315. B received li times as much as A, C received 1^ times as rfiuch as A and B, and D received 1| times as much as A, B, and C. What did each laborer receive ? Ans. A, $24 ; B, $36 ; C, $80 ; D, $175. NOTE. Let Gx represent A's share, and 9z B's share. 32. A gamester, after losing i of his money, won 4 shillings ; he then lost \ of what he had, and afterward won 3 shillings ; he then lost ^ of what he had, and found that he had only 20 shillings re* maining. How much had he at first? . Ants. 30 shillings. 33. A gentleman spends | of his yearly income for the support of his family, and | of the remainder in improvements on his premises, and lays by $70 a year. What is his income ? Ans. $630. 34. Divide the number 60 into two such parts, that the product of the two parts may be equal to 3 times the square of the less part. Ans. 15 and 45. 35. My horse and saddle are together worth 90 dollars, and my horse is worth 8 times the value of my saddle. What is the value of each ? Ans. Saddle, $10 ; Horse, $80. 36. Divide 8462 between two persons, so that for every dime which one receives, the other may receive a dollar. Ans. $42 and $420. 37. The rent of an estate is 8 per cent, greater this year than last. This year it is 1890 dollars ; what was it last year ? AM. $1750. 38. The sum of two numbers is 840, and their difference is equal to ^ of the greater. What are the numbers ? Ans. 504 and 336. 39. A person, after spending 100 dollars more than A of his income, had remaining 35 dollars more than of it. Required his income. Ans. $450. 40. Divide $1520 among A, B, and C, so thatB shall have $100 more than A, and C $270 more than B. Ana. A, $350 ; B, $450 ; C, $720. 9* 102 SIMPLE EQUATIONS. 41. A and B htive the same income. A contracts an annual debt amounting to of it ; B lives, upon J of it ; at the end of two years B lends to A enough to pay off his debts, and has 82 dollars to spare. What is the income of each ? Am. $280. 42. A sets out from a certain place, and travels at the rate of 7 miles in 5 hours ; and 8 hours afterward B sets out from the same place in pursuit, at the rate of 5 miles in 3 hours. How long and how far must B travel before he overtakes A ? Ans. 42 hours ; 70 miles. 43. A can perform a certain piece of work in 8 days, and B can do the same in 12 days ; in how many days can both, working together, do it? ARK. 4J. 44. A person has just 6 hours at his disposal ; how far may he ride in a coach which travels 8 miles an hour, that he may return home in time, walking back at the rate of 4 miles an hour ? Ans. 16 miles. 45. A can dig a trench in one half the time required by B, B can dig it in two thirds of the time required by C, and all together can dig it in 6 days ; find the time that each alone would require. Ans. A, 11 days; B, 22 days; C, 33 days. 46. A and B start from opposite points and travel toward each other, A at the rate of 3 miles an hour, and B at the rate of 4 miles an hour. At the same time, C sets out with A and travels at the rate of 5 miles an hour. After meeting B he turns back and travels until he meets A ; he then finds that the whole time elapsed since starting is 10 hours. How far apart were A and B at the beginning ? Ans. 72 miles. 47. Two farmers owning a flock of sheep, agree to divide them. A takes 72 sheep ; B takes 92 sheep, and pays A $35. Required the value of the flock. Ans. $574. 48. A crew which can row at the rate of 12 miles an hour in still water, finds that it takes 7 hours to come up a river a certain distance, and 5 hours to go down again. At what rate does the river flow ? Ans. 2 miles per hour. TWO UNKNOWN QUA SIMPLE EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. 161. TVe have seen that every equation containing one unknown quantity can be satisfied with one value, and only one value, of the unknown quantity, (155). But if we consider a single equation containing two unknown quantities, we shall find that for every value which we please to give to one of the unknown quantities. we can determine a corresponding value of the other unknown quantity, such that the set of values will satisfy the equation. Thus, let = 17. (1) Put x = 1, and substitute this value in the given equation ; we have 24-3y = 17, >w& Now the set of values, x 1, y 5, will satisfy the equation ; for, by substitution, we have 2+15 = 17. In the same manner, we may obtain the following sets of values, each one of which will satisfy equation (1) : 1. x=l, y = 5. 2. a = 2, y = 4f 3. * = 3, y = 3. 4. a = 4, y = 3. It is evident that there is no limit to the number of sets of val- ues that may be obtained. The equation, and also the quantities, in such cases, are said to be indeterminate. Hence, 169. An Indeterminate Equation is one which is satisfied by an infinite number of values of the unknown quantities. Every sinyle equation containing two unknown quantities, is indeterminate. 163. If we take two equations with two unknown quantities, as = 31, (1) 2ij = 19, (2) it is evident that we may obtain as many sets of values as we please, 104 SIMPLE EQUATIONS. winch will satisfy each equation, considered separately. Thus, pro- ceeding as before, we find that the set, x = 5, y = 41, will satisfy the first equation ; and a different set, x = 4, y = 3i, will satisfy the second equation. Now suppose we are required to satisfy Loth equations with the same set of values for x and y. Multiplying (1) by 3, and (2) by 2, we have 6-r-f 15.y = 93, (3) 6x+ 4y = 38. (4) Subtracting (4) from (3), we have lly = 55, (5) whence y 5. (6) Substituting this value of y in (1), we have 2z+25 = 31, (7) whence, x = 3. (8) Thus we have a single set of values, x = 3, y = 5, which will satisfy both equations. For, let these values be substi- tuted in the given equations j we shall have 6-4-25 = 31, 9+10 = 19. Equations thus related are said to be simultaneous. Hence, 104. Simultaneous Equations are those which must be satisfied by the same values of the unknown quantities which enter them. When two or more simultaneous equations are given, the values of the unknown quantities are determined by a process called ELIMINATION. 1O5. Elimination is the process of combining equations in such a manner as to cause one or more of the unknown quantities con- tained in them to disappear. There are four principal methods of elimination ; 1st, By addition and subtraction ; 2d, By comparison ; 3d, By substitution ; 4th, By indeterminate multipliers. TWO UNKNOWN QUANTITIES. 105 CASE I. 166. Elimination by addition and subtraction. I. Given 3z-f 2y = 23, and 4x 3y = 8, to find the values of x d^r. OPERATION. 23; (1) 8. (2) Multiplying equation (1) by 4, and equation (2) by 3, we have I2x+8y = 92, (3) 12* 9y = 24; (4) subtracting (4) from (3), 17y = 68; (5) whence, y = 4. (6) , Thus, we have eliminated x, and found the value of ^. Again, multiplying equation (1) by 3, and equation (2) by 2, we have 9*+6,y = 69, (7) 8** 6y = 16 ; (8) adding (7) to (8), I7x = 85 ; whence, x = 5. We have thus eliminated y, and found the value of x. Hence, RULE. I Multiply or divide the equations by such numbers or quantities that the coefficients of the quantity to be eliminated shall be made equal in the two equations. II. If these coefficient* have like signs, subtract one of the prepared equations from the other, member from member; if they have unlike signs, add the equations, member to member. NOTES. 1. In preparing the given equations by multiplication, find the least common multiple of the coefficients of the letter to be eliminated, and divide this multiple by each coefficient ; the quotients will be the least multiplies that can be used. If the coefficients are prime to each other, it is evident that each equation must be multiplied by the coefficient in the other equation. 2. It is generally convenient to clear the equations of fractions, before applying the rule. This is not necessary, however. For if any letter has fractional coefficients in the two equations, the fractions may be reduced to a common denominator ; it will then be necessary to render the nume- rators equal by multiplication or division, according to the rule. 106 SIMPLE EQUATIONS, CASE II. 167. Elimination by comparison. 1. Given ox-\-5y = 42, and 2x+y = 14, to find the values of x and y. OPERATION. From (1), by transposition, etc., (2), therefore, by Ax. 7, clear! ng of fractions, 84 lOy = 42 3y ; whence, ly = 42, y 6. Substituting the value of y in (3), x = 4. Hence, we have the following RULE. I. Find the value of the same unknown quantity, in terms of the other, from each of the given equations. II. Form an equation by placing these two values equal to each other. CASE in. , 168. Elimination by substitution. 1. Given Zx+2y = 16, and bxZy = 14, to find the values 3f :c and y. OPERATION. Zx+2y = 16. (1) 5x 3y = 14. (2) From (1), we obtain y =. . /g\ 2 ' TWO UNKNOWN QUANTITIES. 107 Substituting this value of y in equation (2) ; , 48-9* 5* __ = 14 ; clearing of fractions, 10x 48-f9x = 28, whence, x = 4. ' < Substituting the value of cc in (3) } ^ = 2. Hence the following RULE. I. Find the value of one of the unknown quantities, in terms of the other, from either of the given equations. II. Substitute this value for the same unknown quantity in the other equation. CASE IV. 169. Elimination by indeterminate multipliers. 1. Given 2x-f 3y 23, and 5cc-j-2# = 30, to find the values of x and y. If we multiply the first equation by a quantity, m, which is as yet undetermined, we have 2mx+3my = 23m, (1) 5*+2y = 30. (2) Subtracting (2) from (1), and factoring with reference to x and y, we have (2m 5)x-f (3m 2)y = 23m 30. (3) It is evident that equation (3) is true, whatever be the value of m. We may therefore assume m to be of such a value that the co- efficient of one of the unknown quantities shall become zero ; this will eliminate that unknown quantity, by causing the term contain- ing it to disappear. Thus, assume 2m 5 = 0, (4) whence, m = f . (5) But if 2m 5 =-0, the first term of (3) is 0, and that equation becomes (3m 2)# = 23m 30 ; (6) 23m 30 108 SIMPLE EQUATIONS. If we now substitute in (7) the value of m, as given in (5) r wo shall have . 23x^30 _ 11560 __ 55 : 3x| 2 : 15-4 -TI = In like manner we may eliminate x from (3). To accomplish this, assume 3m 2 = 0, (9) whence, m = |. (10) But if 3m 2 = 0, equation (3) becomes (2m 5)a; = 23m 30 ; (11) 23m 30 whence, x = (12) 2m o Substituting in (12) the value of m, as expressed in (10), we have 23XI-30 _ 46-90 _ -44 ~2xf^ 1=15 =n- This method of elimination is due to the French writer, Bezoufr. It is called the method of indeterminate multipliers, because the multipliers which we employ are at first undetermined. Strictly speaking, the multipliers thus used are not indeterminate ; for, in order to effect the elimination of the unknown quantities, they must have certain definite values, which values are always determined in the course of the operation. Recurring to the operation above, we notice that m has two values, thus : * = *, (1) m = |. (2) The first value of m is the one by which x was eliminated ; tho second, the one by which y was eliminated. Resume the given equations, 2x+3y = 23, (1) 5ar+2y = 30. (2) It will be readily seen that if the first equation be multiplied by the first value of m, the coefficients of x will be alike in the two equations ; and if the first equation be multiplied by the second value of m, the coefficients of y will be alike in the two equations. It is obvious, therefore, that the method of indeterminate multipliers is but a modification of the method f addition and subtraction. TWO UNKNOWN QUANTITIES. 109 Recurring ngnin to the above operation, it is evident that if the sum instead of the difference, of equations (1) and (2) had been taken, the elimination might have been effected with equal facility. Hence, RULE. I. Multiply one of the given equations by the indefinite factor m, and then take the sum or difference of this result and the other equation, fcrc for iny with reference to the unknown quantities. II. Put the coefficient of one of the unknown quantities in this last equation equal to zero, and determine the value of m; then substitute this value in the equation Containing m, and the result will be an equation of but one unknown quantity. 17O. In the reduction of simultaneous equations containing two unknown quantities, sometimes one of the preceding methods of elimination is preferable, and sometimes another, according to the special relations of the coefficients. EXAMPLES FOR PRACTICE. Ans. x = 6 ; y = 4. 2. Given \ 6x + 2y = 19 I to find* and y. | 7 x _6y= 9 f Ans. x = 3 ; y = 2. 3. Given j 3 *+^ = I to' find x and y. ( *4-4y = 38 j Ans. x = 10 ; y = 7. 4. Given \ 5x-3y = 36 ) ^ d I 2ic+9y = 96 f Ans. x = 12 ; y = 8. 6 - GiTen { sES^S } *-** ^Lws. a; = 20 ; y = 2. 6. Given { &*-4y= 40 ) fid d I a; 5,y = 97 J = 28 ; y = 25. iven j ^+J^= 1 to find and y. ( Qx I2 = I J 7. Given Qx I2y 110 SIMPLE EQUATIONS. 8. Given j 7x+7j/ = 30 ) to ^ x and y> ( 3x-f-4y = 17 ) Ans. x = 4 ; y = 4J. 9. Given I ? Z+ ^ = 1 to find* and y. ( 5x 6y = 55 j Ana. x = 5 ; y = 5. 10. Given J ^"^ * [ to find x and y. ( lOx-fty = 43 ) ^Liu. x = 2i;y = 5J. 11. Given j ^~^= ^ [ to find, and y. ( 2x 3y = 450 ) Ans. x = 300 ; y = 350. f x v *\ V = 20 i A 2 4 r 12. Given to find x andy Ans. x=l6: y = 24. TWO UNKNOWN QUANTITIES. Ill 17. Given 18. Given 19. Given 21. Given .4ns. cc = 3 ; y = 7. 4a>fl7 68 20. Given <^ "^ > to find a; and y. 5y-f27 54 J.TIS. a; = 13 ; y = 3. 23. Given find x and y. 733 ' to ^ n( ^ x Ans. x = 21 ; y = 20. 151-16. 7 112 24. SIMPLE EQUATIONS. . Given I a f + ^ = d jt \ to find x and y. ( a'a-f&'y == d' ) 25. Given I'd Id' a'dad' to find x and y. cz-f ay = ac az-f-cy = -^r 26. Given 4 i > to find x and ^ a c Ans. x = --;y = - SBIPLE EQUATIONS CONTAINING MORE THAN TWO UNKNOWN QUANTITIES. 171. If we have three or more simultaneous equations, they may reduced by successive eliminations, as follows : C 2x+ 4y+ 4^ = 18; (i) n < I Multiplying (1) by 3, (2) by 2, subtracting (5) from 4, multiplying (1) by 5, ' (3) by 2, subtracting (8) from (7), multiplying (6) by 4, (9) by 3, subtracting (11) from (10), yhence, substituting value of a in (9), ". values of y and z in (1). 3^_j_ 3y_|_ 2z = 17 ; 5*+ %+ 5^ = 32. (2) (3) (4) (5) (6) (T (9) (10) (11) (12) 6z+12y-fl2^ = 54, 6z+ 6y-f 4^ = 34; 6y+ 8z = 20; 10a;+20y-f 20z = 90, 10z4-12y+102 = 64 ; 8y+10^ = 26 ; 24y-j-32^ = 80, 24y-f 30^ = 78 5 2^ = 2; -i . 9), y-2- TWO UNKNOWN QUANTITIES. 113 INFERENCES. 1. If we have n equations, and proceed as above to combine one of them with each of the others, eliminating the same letter by each combination, we shall have n 1 derived^equations containing all of the unknown quantities except the one eliminated. 2. If then we combine one of these derived equations with each of the others, eliminating another letter, we shall have n 2 derived equations, containing all of the unknown quantities except the two eliminated quantities. 3. Since each succeeding group of derived equations consists of one less equation than the preceding group, it follows that if this process of successive elimination be continued, the (n l)th group will consist of a single equation ; and this will contain all of the unknown quantities except the n 1 eliminated quantities. Hence, 4. If the number of original equations equals the number of un- known quantities, the final equation will contain but a single un- known quantity, the value of which may be found. By substituting this value in one of the equations of the preceding group, the value of a second unknown quantity may be determined; and so on. 5. But if the number of original equations is less than the number of unknown quantities, the final equation will contain more than ne unknown quantity, and will be indeterminate, (162); and consequently, the given equations will be indeterminate. 6. In the solution of two or more simultaneous equations of the first degree, by successive eliminations, the value of each letter is determined finally by a simple equation containing only that letter. And since every such equation can have only one root, (155), it follows that any group of simultaneous equations can be satisfied by only one set of values of the unknown quantities. 172. From the foregoing inferences we derive the following RULE. I. Combine one of the given equations with each of tlie others, eliminating the same unknown quantity by each combination; then combine one of the new equations with each of the others, elim- inating a second unknown quantity, and thus continue till a final equation is obtained, containing out one unknown quantity. 10* H 114 SIMPLE EQUATIONS. II. Reduce this final equation, and find the value of the unknown quantity which it involves ; substitute this value in an equation con- taining two unknown quantities, and thus find the value of a second; substitute these values in an equation containing three unknown quan- tities, and find the value of a third} and so on, till the values of all are found. PRACTICAL SUGGESTIONS. This rule may be modified in certain cases, as follows : 1. Instead of combining the first equation with each of the others, we may pursue any order of combination, or adopt any one of the four methods of elimination, which seems best suited to tho mutual relations of the coefficients. The following example will il- lustrate the precept just given : ( x+ y+ z = 9 ; (1) Given *+ty-f-3z = 16 ; (2) Subtracting (1) from (2), y+2z = 1 ; (4) (2) from (3), ?/+ z = 5 ; (5) ' (5) from (4), z = 2 ; substituting value of z in (5), y = 8 ; " values of y and z in (1), .JL = 4. 2. If two or more of the equations, taken together, contain less than all the unknown quantities, it is generally most convenient to employ these equations first, in the process of elimination. Thus, s 2ux+ fy+Zz = 19 ; (1) \ 3w-f5x 4# = 23 ; (2) Given ^ 4u+ 3* = 32 ; (3) ^ 2u+5a; __ =30. (4) Multiplying (4) by 2, 4u+10x = 60 ; bringing down (3), 4w+ 3x = 32 ; by subtraction, 7-r = 28, x= 4; substituting in (3), u = 5, (2), y = 3, (1), z = 2. TWO UNKNOWN QUANTITIES. 115 3. When the coefficients sustain to each other relations of equal- ity or symmetry, it is often convenient to employ an auxiliary quan- tity, as in the two examples which follow : \ u+v+x+z = 15 ; 1. Given < u+v+y+z = 16 ; ( v+x+y+z = 18.' Since in each equation one letter is wanting, let ( 2 ) (3) (4) (5) then s-z = 14, * y = 15, x = 16, s v = 17, = 18. L - }y = 5, B uf <. # = 4, tion, ) v = 3, (M=2. Hence, by substitu- ting the value of <^ ; x = 4, s in each equatk By addition, 5s s = 80, f 2. Given < Assume equation (1) becomes (2) " (3) Multiplying (4) by 8, (5) by 4, " (6) by 3, Adding (7), (8) and (9), = 450; = 490. x + y+ z = s-, 4s Bx = 300, 7s 6y = 450, 9s 8z = 490. 32s 24* = 2400, 28s 24y - 1800, 27s 24s = 1470. 87s 24s = 5670, 63s = 5670, s= 90. Substituting value of s in (4), x = 20, (5), # = 30, " " " (G), 2 = 40. (1) (2) (3) (4) (5) (C) (7) (8) (9) 116 SIMPLE EQUATIONS. EXAMPLES FOB PRACTICE. Required the values of the unknown quantities in the following equations : Ans. -( y = 6x-|-7y f 2x+4y3z = 22 ^ I 6x4- ly z = 63 J r 3x497,48* = 41 1 5. < 5x44y 2z = 20 V I lLc47# 6* = 37 J /- x4^4^ = 31 ^ j. J x+y-, = 25 I I xyz = 9 J f x4^4^ = 26 ^| t. -| xy = 4 V I x z = 6 j / xyz= 6. j >. J 3.T/ x~^ = 12 V ( 7^_ ?/ _a; = 24 ) 11x47^ Qz x4y4^ = 31^ (x=20, f y= 8, x y- x4^4^ = 26 }> Ans. \ z = 6. x y z = 6 ) / ic = 39, u4.ns. -(^ == 21? -2/ x = z* / ( 2 = 12. NOTE. In the last example, assume x+y+z = 5, and add this equation to each of the given equations. Then determine s as hi (3, 2). 6. 7. 8 4x43y43 = 8 Ans. TWO UNKNOWN QUANTITIES. 117 / n+v+x+y+2* = 52 \ I U + V + X + Z + 2y =T 50 I / u+v+y+z+2x = 48 \ } u +x+y+z+2v = 46 I ( t4.*+y+*+2 = 44 J 9. / u + v + y +z+2x = 48 > Ant. y 2^ = _ 04-8* = 35 10. 3+ <=13 Ans. cy+3t u = v+ y z = l - 11. < 8*+3y-6* = l V r x+ y z = l \ . < 8.r+3^ 6^ = 1 V ( 3z 4x = l J Ans. { x=^ a , ') y = /T^ z = T '- a. X = 7 TT 118 SIMPLE EQUATIONS. ( lx By = a I I x = 4a, 17. I 5# llx = a > Ans. \ y a ~*~6 ~~6 a ax _ c , = - . x 7* &c ac ac ai ex + y + 02 = 2a 20. acx y-f~ ac * = a*-f- c * a a a?-|-a y-f- ,,.i ... i 1 * = a 3 > PROBLEMS PRODUCING EQUATIONS CONTAINING TWO OR MORE UNKNOWN QUANTITIES. 173. Two or more equations are said to be independent, when they are not derived one from the other, and can not be reduced to the same form ; as 3*+ y = 17, 2x+8y = 28. Equations derived from the same problem are independent, when they express different conditions of that problem. TWO OR MORE UNKNOWN QUANTITIES. 119 We have seen tbat a group of equations will be determin- ate, when the number of equations is equal to the number of un- known quantities, but not otherwise, (172, 4 and 5). Hence, A problem icill be capable of solution only iclien its conditions fur- nixli as many independent equations as there age unknown symbols employed in the notation. 1. Find two numbers, such that twice the first plus three times ihe second is equal to 105 ; and three times the first plus twice the second is 95. Ans. First, 15; Second, 25. 2. Find three numbers, such that the first with ^ of the sum of the second and third shall be 120; the second with J of the sum of the third and first shall be 90; and the sum of the three shall be 190. Ans. 50, 65, 75. 3. A sum of money was divided among three persons, A, B, and C, as follows : the share of A exceeded | of the shares of B and C by $120 ; the share of B exceeded f of the shares of A and C by $120; and the share of C exceeded | of the shares of A and B by $120. What was each person's share? Ans. A's, $600; B's, $480; C's, $360. 4. A and B, working together, can earn $40 in 6 days ; A and C can earn $54 in 9 days; and B and C can earn $80 in 15 days. How much can each person alone earn in one day ? Ans. A,-$3f; B, S3; C, $2i. 5. A man has 4 sons. The sum of the ages of the first, second and third is 18 years; the sum of the ages of the first, second and fourth is 1 6 years ; the sum of the ages of the first, third and fourth is 14 years; the sum of the ages of the second, third and fourth is 12 years. What are their respective ages? Ans. 8, 6, 4, and 2 years. 6. Three persons engaged in throwing dice, on certain conditions. In the first game A forfeited to B and C, respectively, as many shil- lings as each of them had ; in the second game B forfeited to A and C, respectively: as many shillings as each of them then had; in the third game C forfeited to A and B, respectively, as many shillings as each of them then had; they had then 16 shillings apiece. How many shillings had each at first? Ans. A, 26; B ; 14; C, 8. 120 SIMPLE EQUATIONS. 7. A gentleman left a sum of money to be divided among four si rvants. so that the share of the first should be | of the sum of the shares of the other three, the share of the second J of the sum of the other three, and the share of the third -} of the sum of the other three. On making the division, the fourth had 14 dollars less than the first. Required the sum divided, and the several shares. Ans. Sum divided, $120; shares, $40, $30, $24 and $26. 8. A person has two horses and two saddles, the saddles being worth $15 and $10, respectively. Now the value of the better liorse with the better saddle is | of the value of the other horse and saddle ; but the value of the better horse with the poorer saddle is || of the value of the other horse and saddle. What are the values of the two horses ? Ans. $65 and $50. 9. A vintner, in mixing sherry and brandy, finds that if he takes 2 parts of sherry to 1 of brandy, the mixture will be worth 78 shillings per dozen ; but if he takes 7 parts of sherry to 2 of brandy, the mixture will be worth 79 shillings per dozen. What are the sherry and brandy worth per dozen ? Ans. Sherry, 81 shillings ; Brandy, 72 shillings. 10. Two persons, A and B, can perform a piece of work in 16 days. They work together for four days, when A is called off, and B is left to finish it, which he does in 36 days. In what time would each do it separately ? Ans. A, in 24 days ; B, in 48 days. 11. What fraction is that, whose numerator being doubled, and denominator increased by 7, the value becomes f ; but the denomi- nator being doubled, and the numerator increased by 2, the value becomes f ? Ans. 4 . 12. Two men were wishing to purchase a house together, valued at 240 dollars. Says A to B, " If you will lend me ~ of your money I can purchase the house alone f but says B to A, " If you will lend me | of yours, I can purchase the house alone." How much money had each? Am. A, $160; B, $120. 13. A pleasure party, having chartered a boat for a certain sum found, on settling, that if ' their number had been 4 more, they would have had a shilling apiece less to pay ; but if their number TWO OB MORE UNKNOV'X QUANTITIES. 121 had been 3 less, they would have had a shilling apiece more to pay. What was their number, and what had each to pay ? Ans. 24 persons; each paid 7 shillings. 14. A certain number consists of two places of figures, units and tens; the number is equal to 4 times the sum of 4ts digits, and if 27 be added to the number, the order of the digits will be inverted. What is the number ? NOTE 1. Let x represent the digit in the place of tens, and y the digit iu place of units ; then \Qx+y will express the number. Ans. 8G. 15. A number is expressed by three figures whoso sum is 11 ; the figure in the place of units is double that in the place of hun- dreds ; and if 297 be added to the number, the result will bo expressed by the same figures with their order reversed. What is the number ? Ans. 326. 1(3. Divide the number 90 into three parts, such that twice the first part increased by 40, three times the second part increased by 20, and four times the third part increased by 10, may all be equal to one another. Am. First part, 35 ; Second, 30 ; Third, 25. 17. A person placed $100,000 out at interest, a part of it at 5 per cent., and the rest at 4 per cent.; the yearly interest received on the whole was S4G40. Required the two parts of the principal. Ans. $64,000 and $36,000. 18. A person put out a certain sum of money at interest at a certain rate. Another person put out $10,000 more than the first, at a rate per cent, greater by 1, and received an income greater by $800. A third person put out $15,000 more than the first, at a rate per cent, greater by 2, and received an income greater by $1 ,500. Required the three principals, and the respective rates of interest. NOTE 2. To avoid the inconvenience of large numbers in the operation, put a = 5000; then 2a - 10COO, 3 = 15000, ^- = 1500, and |^ = 800. In the final result, the value of a may be restored. Ans j Principals, $30,000. $40,000, $45,000. ( Hates, 4, 5, 6, per cent. 11 122 SIMPLE EQUATIONS. 19. If B's age be subtracted from A's, the difference will be C's nge ; if 5 times B's age and twice C's age be added together, and from their sum A's age be subtracted, the remainder will be 147; and the sum of the three ages is 96. Required the ages of A, B, and C, respectively. Ans. A's, 48 ; B's, 33 ; C's, 15. 20. Find what each of three persons, A, B, and C, is worth, knowing, 1st, that what A is worth added to 3 times what B and C are worth, is equal to 4700 dollars; 2d, that what B is worth added to 4 times what A and C are worth, is equal to 5800 dollars ; 3d, that what C is worth added to 5 times what A and B are worth, is equal to 6300 dollars. Ans. A, $500 ; B, $600 ; C, $800. 21. A grocer sold 50 pounds of tea at an advance of 10 per cent, on the cost, and 30 pounds of coffee at an advance of 20 per cent, on the cost, and received for the whole $27.40, gaining $2.90. What was the cost per pound of the tea and coffee ? Ans. Tea, $.40; Coffee, $.15. 22. Five persons, A, B, C, D, E, play at cards ; after A has won one half of B's money, B one third of C's, C one fourth of D's, D one sixth of E's, they have each $30. How much had each to begin with? Ans. A, $11 ; B, $38 ; C, $33 ; D, $32 ; E, $36. 23. Three brothers desired to make a purchase, requiring $2000 of each. The first wanted, in addition to his own money, 4 of the money of the second ; the second wanted, in addition to his own, | of the money of the third ; and the third wanted, in addition to his own, I of the money of the first. How much money had each ? Ans. 1st, $1280; 2d, $1440; 3d, $1680. 9 24. A courier was sent from A to B, a distance of 147 miles; after 28 hours had elapsed, a second courier was sent from the same place, who overtook the first just as he entered B. Now the time required by the first to travel 17 miles, added to the time re- quired by the second to travel 56 miles, is 13| hours. How many miles did each travel per hour? Ans. 1st, 3 miles; ?d, 7 miles. 25. Find two numbers, such that if J of the greater be added to J of the less, the sum shall be 13 ; and if 4 of the less be subtract- ed from | of the greater, the remainder will be nothing. Am. 18 and 12. TWO OR MORE U1 T KNOWN QUANTITIES. 123 26. Find three numbers of such magnitudes, that the first added to \ of the sum of the other two, the second added to ^ of the sum of the other two, and the third added to ] of the sum of the other two, may each be equal to 51. Ans. 15, 33, and 39. 27. Said A to B and C, " If each of you wifl give me 4 sheep, I shall have 4 more than both of you will have left." Said B to A and C, " If each of you will give me 4 sheep, I shall have twice as many as both of you will have left." Said C to A and B, " If each of you will give me 4 sheep, I shall have three times as many as both of you will have left." How many sheep had each ? Ans. A, 6 ; B, 8 ; C, 10. 28. "What fraction is that, to the numerator of which if 1 be added, the fraction will be ^ ; but if to the denominator 1 be added, the fraction will be | ? Ans. T 4 -. 29. What fraction is that, to the numerator of which if 2 be added, the fraction will be | j but if to the denominator 2 be added, the fraction will be -g ? Ans. ^. 30. Four persons, A, B, C, D, were engaged together in mowing for 4 successive days. The first day A worked 1 hour, B 3 hours, C 2 hours, and D 2 hours, and all together mowed 1 acre ; the second day A worked 3 hours, B 2 hours, 4 hours, and D 11 hours, and all together mowed 2 acres ; the third day A worked 5 hours, B 4 hours, C 12 hours, and D 5 hours, and all together mowed 3 acres ; the fourth day A worked 9 hours, B 7 hours, C 6 hours, and D 8 hours, and all together mowed 4 acres. How many hours would each alone require to mow 1 acre ? Ans. A, 5 hours ; B, 6 hours ; C, 12 hours ; D, 15 hours. 31. If A give B $5 of his money, B will have twice as much money as A has left ; and if B give A $5, A will have thrice as much as B has left. How much has each f Ans. A, $13; B, $11. 32. A corn factor mixes wheat flour, which cost him 10 shillings per bushel, with barley flour, which cost 4 shillings per bushel, in such proportion as to gain 43| per cent, by selling the mixture at II shillings per bushel. Required the proportion. Ans. The proportion is 14 bu?hels of wheat flour to 9 of barley. 124 SIMPLE EQUATIONS. 33. There is a number consisting of two digits, which number divided by 5 gives a certain quotient and a remainder of 1, and the same number divided by 8 gives another quotient and a remainder of 1. Now the quotient obtained by dividing by 5 is twice the val- ue of the digit in the tens' place, and the quotient obtained by divi- ding by 8 is equal to 5 times the digit in the units' place. What is the number ? Ans. 41. 34. The four classes in a certain college are to compete for four prizes, amounting in the aggregate to $119, and the prize money is to be raised by contribution, on the following conditions, namely : that the members of the class whose candidate obtains the 1st prize shall each pay one dollar, and the class whose candidate obtains the 2d prize shall pay the remainder. Now it is found that if a senior gets the 1st prize and a junior the 2d, each junior will pay ^ of a dollar ; if a junior gets the 1st prize and a sophomore the 2d, each sophomore will pay \ of a dollar ; if a sophomore gets the 1st prize and a freshman the 2d, each freshman will pay ^ of a dollar ; and if a freshman gets the 1st prize and a senior the 2d, each senior will pay | of a dollar. Of how many members does each class consist ? A j Freshman, 104 ; Sophomore, 9o ; ( Junior, 88 ; Senior, 75. 35. Find four numbers, such that if 3 times the first be added to the second, 4 times the second be added to the third, 5 times the third be added to the fourth, and 6 times the fourth be added to the first, each sum shall be 359. Ans. 95, 74, 63, 44. GENERAL SOLUTION OF PROBLEMS. fi 7o. In the preceding problems, the given quantities have been expressed by numbers, and it has been required simply to determine the values of the unknown quantities from the numerical relations thus expressed. Tf, however, the given quantities in any problem be represented by letters, the solution will give rise to a formula, showing not only the value of the unknown quantity, but indicating the precise ope- GENERAL PROBLEMS. 123 rations to be performed in order to obtain this value. This is called a general solution of the problem. 176. When any particular problem has been proposed, we may, by simply varying the numbers, form other problems of the same kind or class ; and the solutions of all the problems of the class will require exactly the same operations. Hence, 177. The General Solution of a problem is the process of obtaining a formula which shall express, in known terms, the values of the unknown quantities in the given problem, or in any problem of its class. 178. An Arbitrary Quantity is one to which any value may be assigned at pleasure, in a general formula or equation. 179. For illustration, let the following questions be proposed : 1 What number is that whose third part exceeds its fourth part by 6? Instead of confining our attention to the particular numbers here given, we may first investigate the problem under a general form, as follows : What number is that whose mth part exceeds its nth part Ly a ? Let x represent the number ; then by the conditions, (1) clearing of fractions, nx mx = amn, (2) amn whence, x = (3) n m Equation (3) is the formula which indicates the operations to be performed in solving all questions of this class. If in this formula we put m = 3, n = 4, and a = 6, we shall 43 the number required by the particular question as at first proposed. 2. What number is that whose fifth part exceeds its seventh part by 12 ? To obtain the number by the formula, let m = 5, n = 7, and a = 12; then 12x5x7 x = - = = 210, Arts. 7 o 126 SIMPLE EQUATIONS. EXAMPLES FOR PRACTICE. 1. Divide the number n into two such parts that the increased by a shall be equal to the less increased by b. n-f-' a n-\~a b Ans. Greater, : Less, - 5 4 u 2. In the last example, what will be the two parts if n = 84, a = 16, and b = 58 ? Ans. 63 and 21. 3. The sum of three numbers is s ; the second exceeds the first by a, and the third exceeds the second by b. Required the numbers. s 2a b s-f-a b a-fa+26 4. My indebtedness to three persons, A, B, and C, amounts to a dollars ; and I owe B n times the sum which I owe A, and C m times the sum which I owe A. What is my indebtedness to A ? Ans. -iii* 5. In the last example, what is the sum due to A when a = $786, n = 2, and m = 3 ? Ans. $131. 6. A person engaged to work a days on these conditions : For each day he worked he was to receive b cents, and for each day he was idle ^e was to forfeit c cents ; at the end of a days he received d cents. How many days was he idle ? ab d , b+c 7. My horse and saddle are together worth a dollars, and my horse is worth n times the price of my saddle. What is the value f Cach? . A. Saddle, -^L j Horse, ~ 8. The rent of an estate is n per cent, greater this year than it was last. This year it is a dollars ; what was it last year ? lOOa Ans - d 9. A person after spending a dollars more than J of his income, *iad remaining b dollars more than ^ of it. Required his income. Ans. doUars . GENERAL PROBLEMS. 127 10. A person after spending a dollars more than ^th of his in- come, had remaining b dollars more than -^th of it. Required his ncome - dollars . mn in n 11. If A can perform a certain piece of worfcf in a days, and B can do the same in b days, and C the same in c days, in how many days can all together perform the work ? abc Ans. j. 7-7 days. ab-\-ac-\-bc 12. In the last example, what will be the time required, when a = 6, b = 8, and c = 12 ? Ans. 2f days. 13. If from a times a certain number c be subtracted, the remain- der will be equal to b times the number increased by d. llequired the number. A c-\-d Anx. ! __ a b 14. A farmer would mix oats worth a cents a bushel with peas worth b cents a bushel, to form a mixture of c bushels worth d cents a bushel. How many bushels of each kind must he take ? Ans . Oats, 1^; Peu, fc9. a b a b 15. There were a boys in one party, and b boys in another party, and each party had the same number of nuts. Each boy in the first party snatched m nuts from the second party, and ate them ; then each boy in the second party snatched m nuts from the first party, and ate them. Each party then divided the nuts remaining to it equally among its members, when the boys in the two parties found that they had the same number of nuts apiece ; how many nuts had each party at first ? Ans. r/t(a-j-/). 16. Find four numbers, such that if a times the first be added to-the second, b times the second be added to the third, c times the third be added to the fourth, and d times the fourth be added to the first, each sum shall be m. iit(b( d cd-\-d 1) m(acd ad-\-a 1" 1st. r~7~~i > 2d ' nr~i ' i aueu 1 liocd I m(abd ab-{-b 1) m(abc bc-\-c 1} 3d, - = : 4th, ^-7 r abed 1 abed 1 128 SIMPLE EQUATIONS. 17. A sent n pupils regularly to a certain school during a term of a days, and B sent m pupils regularly to another school for a term of I days. The two schools had the same number of pupils in attendance, and raised the same amount of money by rate-bill. There were c days' absence allowed for at the school to which A sent, and d days' absence at the school to which B sent ; and A and.B found that they had equal sums to pay. What was the number of pupils attending each school ? bcm arfn Ans. r-. 7 18. Divide the number m into four parts, such that the second shall be a times the first, the third a times the second, and tho fourth a times the third. in An*. 1st part, 19. The sum of two numbers is s, and their difference is d. Required the numbers. s-\-d s d Ans. Greater, - ; Less, = J ^ 20. There are three numbers, such that the sum of the first and second is a, the sum of the first and third is I, and the sum of the second and third is c. What are the numbers ? a-\-b c a-4-r b , l-\-c a Ans. 1st, -^2 ; 2d, -^ ', 3d ; -3L -- 21. There is a number consisting of two digits ; the number is equiil to a times the sum of its digits ; and if c be added to the number', the order of the digits will be reversed. Required tho two digits. t .(10_a) V Digit in units' place, -- -; Ans. * c ( a n Digit in tens' place, 22. Find what each of three persons, A, B, and C, is worth, knowing, 1st, that what A is worth added to I times what B and C. arc worth is equal to p ', 2d, that what B is worth added to m times what A and C are worth is equal to q ; 3d, that what C is worth added to n times what A and B are worth is equal to r. We give here a solution of this example, partly to illustrate tho method of simplifying algebraic formulas by the use of auxiliary quantities. GENERAL PROBLEMS. 129 Let x = A's money, y == B's money, z = C's money. = p> (!) Then, by the conditions, 4 y-j-mx-j-mz = ^, (2) ( z-\-nx -\-iiy = r. (3) Assume x+y-\-z = *r (4) Multiplying (4) by Z, wi, and n, sue- / x = - - j (5) cessively, and subtracting (1) from the 1 first product, (2) from the second, and / y = y j (6) (3) from the third, and reducing the J respective remainders, we have f z = -=> (7) N Adding (5), (6), and (7) 3 we obtain ^-f-^T' (8) i 1 ' n 1 S ~ (ll + m-1 + nl) "~ (ll + m 1 + n 1 j (9) Now the parenthetical expressions in equation (0) are known quantities. Hence, to simplify the results, I m .n -T + ; ~T + T' (10) . 1 771 1 n L Put < \ m ft ** (11) m _l ' n _ 1 Equation (?) then becomes s=as 6, (12) b whence s = ol Substituting value of * in (5), x = ^^ =-^- - ^- > (I \.)(a L) a 1). " (7), = 130 SIMPLE EQUATIONS. DISCUSSION OF PROBLEMS INVOLVING SIMPLE EQUATIONS 180. The Discussion of a problem consists in attributing certain values and relations to the arbitrary quantities which enter the equa- tion, and in interpreting the results. 181. When a problem has been solved in a general manner, we may proceed to make an unlimited number of suppositions upon the arbitrary quantities involved in the formulas, and thus obtain a va- riety of results. But our experience of algebraic equations would lead us to expect that the problem might not be rational, or possible, under every hypothesis. Now the principal object in the discussion of a problem is to examine the peculiar or anomalous forms which present themselves, and ascertain whether the problem is rational or absurd, or how it is to be understood, under the suppositions which lead to these peculiarities. We shall commence with the INTERPRETATION OF NEGATIVE RESULTS. 1. What number must be added to a that the sum may be b ? Let x represent the required number. Then by the conditions of the question, a-fz = 1} (1) x = I a . (2) This is a general solution, a and b being arbitrary quantities. First, suppose a = 20 and b = 28; then by the formula, x = 2820 = 8, a result which satisfies the conditions ; for, we perceive that 8 is the number which must be added to 20, or a, to make 28, or b. Second^ suppose a = 20 and b = 12 ; then by the formula, x = 12 20 = 8, a negative result. In order to ascertain the meaning of the minus sign in this case, let us enunciate the question according to the supposition that gave this result ; thu-, What number must be added to 20, that the sum may be 12 ? DISCUSSION OF PROBLEMS. 131 Now as 20 is greater than 1 2, no number can be added to 20, arithmetically, to make 12. The problem is therefore impossible under the second hypothesis, if understood in an arithmetical sense. We shall find, however, that if we change the words added to, and sum, to their opposites, the result will be a rational question, of which 8, the absolute value of x, is the answer.* Thus, What number must be subtracted from 20, that the difference may be 12 ? Ans. 8. We observe, moreover, that the negative result, 8, will satisfy the equation of the problem, under the second hypothesis. Thus, 20+(-8J=12; or, 208 = 12. That is, 12 is really the algebraic sum of 20 and 8. 2. A man dying left two sons, the elder of whom was a years of age, and the younger b years of age. In how many years after the death of the father was the elder son twice as old as the younger son? Let x represent the number of years ; then by the conditions, a+x = 2(6+*) ; (1) x = a 26. (2) Since a and b are arbitrary quantities, suppose a = 30 and b = 12. Then by the formula, x = 3024 = 6. This result will satisfy the conditions arithmetically ; for, if the elder son was 30, and the younger son 12 years old, at the death of the father, then in 6 years the age of the elder was 30-f-6 = 36 years, and the age of the younger was 12-J-6 = 18 years. Again, suppose a = 30 and b = 18. Then by the formula, x = 3036 = 6. To interpret the negative result in this case, we observe that the problem under the second hypothesis is impossible, if understood iu the exact sense of the enunciation. For, when the elder son was 30 and the younger sen 18 years old, the younger son was already more than one half as old as the elder ; and as their ages are equally increased by any lapse of time, it is evident that the elder 132 SU1PLK EQUATIONS. son could never become twice as old as the younger son, after tho death of the father. Let us therefore modify the general problem as follows : A man dying left two sons, the elder of which was a years of age, and the younger b years of age. How many years before the death of the father was the elder son twice as old as the younger? If we let x represent the number of years, then the solution will be as follows : a x = 2(b x); (1) x = 21 a. (2) Now suppose, as before, that a = 30 and I = 18. Then by tho new formula, x = 8630 = G, a result which will satisfy the modified conditions; for, six years before the death of the father, the age of the elder was 30 6 = 24, and the age of the younger was 18 6 = 12. From the foregoing discussions we draw the following inferences : 1. When the solution of a problem by a simple equation gives a negative result, the minus sign indicates that the problem is impossible, if understood in the exact sense of the enunciation. 2. The impossibility thus indicated consists in adding a quantity when it should be subtracted ; or in treating a quantity as reckoned or applied in a certain direction, when it should be reckoned or ap- plied in an opposite direction. 3. In all such cases, an analogous problem may lie formed, in- volving no impossibility, by changing the terms of the absurd condi- tion to their opposite*; and the answer to the new question icill be found by simply changing the sign of the negative result already obtained. 18S. The foregoing discussions give a more extensive significa- tion to the plus and minus signs, and lead to a more general view o positive and negative quantities, than was presented in a former sec- tion. Let us recur to the problem of the two sons. In the solution of this problem, wo employ the signs, -f- and , in the statement, mere- NEGATIVE RESULTS. 133 ly to indicate addition and subtraction. But in the result, these signs have a very different use ; they enable us to distinguish the circumstances or conditions of the quantities which they affect. Thus, under the first hypothesis, the period of time represented by a- oc- curred after the death of the father, and in the result is found to be affected by the plus sign; but under the second hypothesis, the pe-" riod represented by x occurred before the death of the father, and in the result is found to be affected by the minus sign. Thus we perceive that plus and minus, in Algebra, are not symbols of operation merely, but also symbols of relation, serving to dis- tinguish quantities in opposite conditions or circumstances. It should be observed, however, that this enlarged use of the plus and minus signs is not entirely conventional or arbitrary, but is necessarily involved in the nfore extended signification given to the terms addition and subtraction, in Algebra. Indeed we shall never meet with a negative result in the solution of problems, so long as the language conforms, in the exact arithmetical sense, to the facts of the case. EXAMPLES FOR PRACTICE. 1. What number is that whose fourth part exceeds its third part by 12 ? AM. 144. The question is impossible, if understood in an arithmetical sense. Let the pupil modify the enunciation, and solve the new problem. 2. A man when he was married was 30 years old, and his wife 15. How many years must elapse before his age will be three times the age of his wife ? Ans. 7^ years. That is, their ages bore the specified relation 7^ years before, not after, their marriage. 3. The sum of two numbers is s, and their difference d, what are the numbers ? s d s d Ans. Greater, -j- > Less, ^- How shall the result be interpreted when s = 120 and d 160 ? 4. Two men, A and B, commenced trade at the same time, A liaving 3 times as much money as B. When A had gained $400 12 134 SIMPLE EQUATIONS. and B $150, A had twice as much money as B ; how much did each have at first? Ans. A was in debt $300, and B $100. 5. A man worked 7 days, and had his son with him 3 days, and received for wages 22 shillings, and the board of his son and him- self while at work. He afterward worked 5 days, and had his son with him one day, and received 18 shillings. What were his daily wages, and what the daily wages of his son ? Ans. The father received 4 shillings per day, and paid 2 shillings for his son's board. 6. A man worked for a person 10 days, having his wife with him 8 days, and his son 6 days, and he received $10.30 as compen- sation for all three ; at another time he wrought 12 days, his wife 10 days, and sou 4 days, and he received $13.20 ; at another time he wrought 15 days, his wife 10 days, and his son 12 days, at the same rates as before, and he received $13.85. What were the daily wages of each ? Ans. He received $.75 for himself, $.50 for his wife, and paid $.20 for his son's board. 7. A man wrought 10 days for his neighbor, his wife 4 days, and eon 3 days, and received $11.50 ; at another time he served 9 days, his wife 8 days, and his son 6 days, at the same rates as before, and receivecT $12.00 ; a third time he served 7 days, his wife 6 days, and his son 4 days, at the same rates as before, and he received $9.00. What were the daily wages of each ? Ans. Husband's wages," $1.00; Wife's, ; Son's, $.50. 8. What fraction is that which becomes | when 1 is added to its numerator, and ^ when 1 is added to its denominator ? Ans. In an arithmetical sense, there is no such fraction. The algebraic expression, zjf, will give the required results. How shall the enunciation be modified, to form an analogous question involving no absurdity ? 9. Four merchants, A, B, C, D, find by their balance sheets that if they unite in a firm, receiving the assets and assuming the liabil- ities of each, they will have a joint net capital of $5780. If A, B, and C unite on the same conditions, their joint capital will be $7950 ; if B, C, and D unite, their joint capital will be $2220; and NOTHING AND INFINITY. if C, D, and A unite, their joint capital will be $7320. Required the net capital or the net insolvency of each. 10. Two men were traveling on the same road towards Boston, A at the rate of a miles per hour, and B at the rate of b miles per hour. At 6 o'clock A was at a point m miles f'rjpm Boston, and at 10 o'clock B was at a point n miles from Boston. Find the time when A passed B upon the road. A m n 4& Ans. - _ - hours after 6 o clock. a b 11. What time of day will be indicated by the preceding formu- la, if m = 36, n = 28, a = 5, and b = 3 ? Ans. 4 o'clock. 12. There are two numbers whose difference is a; and if 3 times the greater be added to 5 times the less, the sum will be b. What are the numbers ? b--5a b 3a Ans. Greater. 5 ; Less, 5- o o How shall this result be interpreted if a = 24 and b = 48 ? NOTHING AND INFINITY. 183. The limits between which all absolute values are comprised, are nothing and infinity ; and the symbols by which these limits are denoted, are and oo. 184:. In certain algebraic investigations it is convenient to em- ploy these symbols in connection with each other and the ordinary symbols of quantity. They may thus sustain the relations of divi- sor, dividend, quotient, or factor. Such relations, however, can not really exist except between symbols of quantity. Hence, in Algebra, does not always signify merely absence of value ; nor does oo represent infinity, in the highest sense of the word. The more complete definition of these symbols may be given as follows : 185. The symbol 0, called nothing, or zero, may be used to denote the absence of value, or to represent a quantity less than any assignable value. 186. The symbol GO, called infinity, is used to represent a quan- tity greater than any assignable value. 136 SIMPLE EQUATIONS. INTERPRETATION OF THE FORMS TT> - , ~ AND - GO A 1 187. In order to understand the signification of the expressions, A A U o' - !' and o' we may consider the symbols and oo as resulting from an arbitrary or varying quantity, made to diminish until it becomes indefinitely small, or to increase until it becomes indefinitely great. , a 188. Let 7 represent a fraction, a and b being arbitrary quan- tities. And let it be remembered that the value of a fraction depends simply upon the relative values of the numerator and denominator. 1 If the denominator b is made to diminish, becoming less and less continually, while the numerator a remains unchanged, the value of the fraction must increase, becoming greater and greater continually, (110, II); and thus when the denominator b becomes less than any assignable quantity, or 0, the value of the fraction must become greater than any assignable quantity, or OO. Hence, we conclude that ^ = oo. That is, A finite quantity divided by zero is an expression for infinity. 2. If the denominator b is made to increase, becoming greatei and greater continually, while the numerator a remains unchanged, the value of the fraction must diminish, becoming less and less con- tinually, (110. II) ; and when the denominator b becomes greater than any assignable quantity, or oo, the value of the fraction must become less than any assignable quantity, or 0. Hence, That is, oo ~ A finite quantity divided by infinity is an expression for zero or nothing. 3. --If the numerator a is made to diminish, becoming less and .ess continually, while the denominator b remains unchanged, the ANOMALOUS FORMS, 181 value of the fraction must diminish continually, (119, 1) j and when a becomes less than any assignable quantity, or 0, the value of the fraction also must become 0. Hence, - = 0- That is b Zero divided by a finite quantity is an expression for nothing or zero. 4. If both a and b are made to diminish simultaneously, but in such a manner as to preserve their relative value, then the value of the fraction will remain unchanged, however small the terms become, (119, III) \ and when both a and b become less than any assignable quantity, or 0, we shall have the expression ^- reprcscnt- a ing the value of j-. And since this value may be any quantity whatever, we conclude that ^- represents an indeterminate quantity. That is, Zero divided by zero is a symbol of indetcrmination. NOTE. If it should be difficult for any one to conceive how both terms of a fraction may, b}' being diminished, become nothing at the same time, and yet preserve the same relative value to the last, it may be useful to consider the following illustrations : c Take the fraction -5- , in which d represents the diameter of a circle, and c the circumference. Now the diameter and circumference of a cir- cle have the same ratio to each other, whatever the dimensions of the circle. Hence, if the circle be made to diminish until it shall become a point, or vanish, both terms of the fraction, ^ , will diminish, and become at the same instant, the value of tJie fraction remaining the same t.'irougJi- out, and reducing to the form, Q-> at the instant the circle vanishes. Now the ratio of the diameter to the circumference of a circle is known to be .1416 ; hence, hi the present case, we shall have O = 8.1416. Again, let s represent the side of a square and d the diagonal. Then we have the well known ratio d_ If the square is supposed to diminish by insensible degrees, both d and 9 will vanish at the same instant, and we shall have finally, 0_ 12* 138 SIMPLE EQUATIONS. PROBLEM OF THE COURIERS. 18O. The anomalous forms which have been explained in the last article will now be viewed in connection with a general problem, involving certain relations of motion, time and distance. The dis- cussion will also confirm our interpretation of negative results. PROBLEM. Two couriers, A and B, were traveling along the same road and in the same direction, namely, from C' toward C ; the former going at the rate of a miles per hour, and the latter at the rate of b miles per hour. At 12 o'clock, A was at a certain point P, and B was d miles in advance of A, in the direction of C. It is required to find when and where the couriers were together. This problem is entirely general, and we do not know from the enunciation whether the couriers were together after, or before 12 o'clock ; nor whether the place of meeting was to the right, or to the left of P. But in order to effect a statement of the problem, we will suppose the required time to be after 12 o'clock. Then we must regard time after 12 o'clock as positive, and time before 12 o'clock as negative; also, distance reckoned from P toward C as positive, and distance reckoned from P toward C' as negative. Ac- cordingly, Let t =. the number of hours after 12 o'clock j x =. the distance from P to the point of meeting. And since A traveled at the rate of a miles per hour, and B at the rate of I miles per hour, we have x =. at = distance traveled by A after 12 o'clock ; lit = " " " B t( " " But since A and B were d miles apart, at 12 o'clock, we nave at It = d, t =-*-., a) a b x = ^L (2) a b "We may now discuss this problem with reference to the time , and the distance x, which are the two unknown elements PROBLEM OF THE COURIERS. 139 I. Suppose a>6. Under this hypothesis the values of both t and x will be positive, because the common denominator, a b, is positive. Now since t is positive, we conclude that the two couriers came together after 12 o'clock j and as oc is positive, we infer that the point of meeting is somewhere to the right of P. These conclusions agree with each other, and are consistent with the conditions of the problem. For, the supposition that a is greater than b implies that A was traveling faster than B. A would therefore gain upon B, and overtake him sometime after 12 o'clock, and at a point situated in the direction of C. II. Suppose a<6. Then in equations (1) and (2) the denominator, a &, is negative, and consequently both t and x will be negative. This implies that t and x must be taken in a sense contrary to to that in which they were employed under the hypothesis, (I), where they were positive ] that is, the time when the couriers were together was before 12 o'clock, and the place of meeting was sit- uated to the left of P. This interpretation, also, agrees with the conditions of the prob- lem, under the present hypothesis. For, if a is less than b } then B was traveling faster than A ; and as B was in advance of A at 12 o'clock, he must have passed A before that time, somewhere to the left of P, in the direction of C'. III. Suppose a = b. Under this hypothesis we shall have a b = 0, and d ad t 3= jT- = oo> and also x = =r- = oo. Now, according to these results, t, the time to elapse before the couriers are together, is greater than any assignable quantity, or infinity ; therefore they can never be together. And likewise x, the distance from P of the supposed point of meeting, is greater than any assignable quantity, or infinity j hence there can be no such point, however distant from P. This interpretation is in accordance with the conditions of the problem, under the present hypothesis. For, at 12 o'clock the two 140 SIMPLE EQUATIONS. couriers were d miles apart ; and if a = 6, they were traveling at equal rates, neither approaching nor separating. Hence, they could always continue in motion, and remove to any distance from P, without meeting. IV. Suppose d = 0, and a>6 or a6 or a<&, the couriers were travel- ing at different rates, and must be either approaching or receding from each other at all times except at the moment of passing; hence, they could be together only at a single point. Y. Suppose d 0, and a = b. We shall then have Here the values of both t and x are represented by the symbol of indetermination, which signifies that the time and the distance may be anything whatever ; and we infer that the couriers must be together at all times, and at any distance from P. And this conclusion is evidently confirmed by the conditions of the problem. For, if d = 0, the couriers were together at 12 o'clock ; and if a = l>, they were traveling at equal rates, and would never separate. 1OO. To the foregoing interpretations, there is an apparent exception in the case of the expression jr. For, a fraction which is not indeterminate will reduce to this form, if its terms contain a common factor that becomes zero under the hypothesis. Thus, in the solution of a problem, suppose DISCUSSIONS. 141 If we put a = b ) which implies that a b = 0, then and the value of x appears to be indeterminate. Let us, however, cancel the common factor, a ft, from both terms of the fraction in equation (1), we shall obtain m-Q If in this reduced equation, we make a = i, as before, we shall have a determinate' value for x. Thus, _3tt Hence the following practical direction : In the discussion of a problem, a fractional result should be reduced to its lowest terms before maJcing the hypothesis. 1O1. We are sometimes liable to an error in the reduction of an equation, in consequence of a false assumption respecting the valuo of an expression reducible to the form of indetermination. 1. Let us take the equation, x+2 ~ x2 Deducing second member, -- = 6, $) X-\-+j clearing of fractions, 6x-|-7 = 6z-f-12, (3) transposing and factoring, (6 6)x = 5, (4) 55 ,~ whence, x = ^- - Q = g-, or, by (188, 1), x= co. This result is erroneous. To obtain the true root of equation (1), multiply both members by (ai+2) (x 2) ; we shall obtain 6x a 5,r 14 = 6x a 24 5 whence. 5x = 10, or, x = 2. Now we observe that if this true value of x be substituted in the second member of equation (1), it will reduce to the form ^r > 142 SIMPLE EQUATIONS. and the mistake in our first solution was made in assuming that 6x 12 ^- = 6, a conclusion which would be correct in all cases except when x = 2. 2. If we make two assumptions that are inconsistent, respecting the values of quantities reducible to the form of in determination, the result will be an algebraic absurdity. Thus, take the identical equation, 8+20 = 8+20. (1) By transposition, 88 = 2020, (2) 8-8 2020 dividing by 4 4, T~\ ~ ~Z~f~' 5(44) factoring, ____ = J, suppressing common factor, and 2 = 5. (5) Equation (5) is absurd. But this equation is not correctly derived from (3) or (4). In equation (3), both numerators and both denomi- nators are zero. Hence (3) may be written, (L __0. ~~ 0' a result which involves no absurdity, and certainly gives no author- ity for saying that 2 is equal to 5. 1O3. To afford the pupil further exercise in the interpretation of anomalous forms, we give the following EXAMPLES. 1 . A cistern has four pipes communicating with it. If all be opened together, and left running for 1 5 hours, the cistern will be filled ; but if the first run only 5 hours, the second 8 hours, the third 7 hours, and the fourth 3 hours, the cisteru will be but one half full ; if the first run 3 hours, the second 4 hours, the third 3 hours, and the fourth 1 hour, only -J of the cistern will be filled ; and if the first run 4 hours, the second 2 hours, the third 3 hours, and the fourth 2 hours, only ] of the cistern will be filled. In what time would the cistern be filled by each pipe alone ? . An Inequality is an expression signifying that one quan- tity is greater, or less, than another ; as a>&, and c 5, because 3 ( 5) = -f-2, a positive result. 13 K 146 INEQUALITIES. 198. Two inequalities are said to subsist in the same sense, when the first member is the greater in both, or the less in both. Thus a > d and c > d, or u < z and x < y, are inequalities which .sub- sist in the same sense. But the inequalities, m > n and p < q, subsist in a contrary sense. PROPERTIES OF INEQUALITIES. 199. Inequalities are frequently employed in mathematical in- vestigations j and to facilitate their use, it is necessary to establish the following properties : I. An inequality will continue in the same sense, if the same quan- tity be added to, or subtracted from, each member. For, suppose a > b. Then according to (196), a b is positive. Hence, (ac}(bc) is positive, and consequently ac > bc. It follows obviously from the principle just established, 1. That a term may be transposed from one member of an ine- quality to another, by changing its sign. 2. That if an equation be added to an inequality, member to member, or subtracted from it in like manner, the result will be an inequality subsisting in the same sense. II. If an inequality be subtracted from an equation, member from member, the sign, of inequality -will be reversed. For, suppose x=y, and a > 5; then we shall have /)o a.\ /y fj *. a negative quantity, (196) ; hence, x a <^ y b. III. If the signs of all the terms of an inequality be changed, the sign of inequality will be reversed. GENERAL PRINCIPLES. 147 For to change the signs of all the terms is equivalent to subtract- ing each member from 0=0. IV. If twu or more inequalities subsisting '.n the same sen.se, be added, member to member, the resulting inequality will subsist in the same sense as the given inequalities. For if a > b, a' >V, a" > b", , then from (196), ab, a'b', a"b", are all positive ; and the sum of these quantities, o_6_|_a' 7/+o" b", or (a-f a'-f-a") (&-f &'+&"), is therefore positive. Hence, a_j_ a '_|_ a " > 6+64-6". It is evident that if one inequality be subtracted from another established in the same sense, the result will not always be an inequality subsisting in the same sense. Thus, it is evident that we may have a > b and a 1 > b 1 , in which a a' may be greater than b b' , less than b b', or equal to 66'. V. If one inequality be subtracted from another established in a contrary sense, the result will be an inequality established in the same sense as the minuend. For, if a > b (1) and . a'<6', ( 2 ) then a 6 is positive and a' b r is negative ; therefore, a 6 (a' 6'), or its equal (a a') (6 b') must be positive, and wo shall have a a ' > 66', an inequality subsisting in the same sense as (1). If (1) be subtracted from (2), member from member, it can be shown, in like manner, that a '~ a < b' 6. VI. An inequality will still subsist in the same sense, if both members be multiplied or divided by the same positive quantity. 148 INEQUALITIES. For suppose m to be essentially positive, and a>6. Then since a b is positive, we shall have both m(a 6) and (a 6) positive. Therefore, ma >> nib and "> m wi VII. T/" &o& members of an inequality be multiplied or divided by the so.me negative quantity, the sign of inequality will be reversed. For, to multiply or divide by a negative quantity will change the signs of all the terms, and consequently reverse the sign of inequal- ity, (III). VIII. // two inequalities subsisting in the same sense be multiplied together, member by member, the sign of inequality remains the same when morf than two of the members are positive, but is reversed when more than two of the members are negative. That is, Multiply a > b a > 6 a > b a^> b By a'>b' a' > b' a' > V a' > V Products, aa' > bb' aa f < bb' aa' < bb' aa' > W The first two results are evident from the fact that when the two members of an inequality are both positive, the greater member has the greatest numerical value ; but when the two members are both negative, the greater mem 1 ,v has the least numerical value. The other two results are evident from the fact that any positive quantity is greater than any negative quantity. It will be found thai if two of the four members arc positive and two negative, iLe result will be indefinite. REDUCTION OP INEQUALITIES. 2OO. The Reduction of an inequality consists in transforming it in such a manner that one member shall be the unknown quantity standing alone, and the other member a known expression. The inequality will then denote one lim'U of the unknown quantity. REDUCTION. 149 2O1. The principles just established may now be applied in the reduction of inequalities of the first degree. Thus, let it be required to find the limit of .t in the inequality, x 2x 3x 9 2 "T^T+r Multiplying both sides by 20, transposing and collecting terms, 3z>45; dividing by 3, x 15. EXAMPLES FOB TBACTICE. 2x 2x 2x 2. ___>__ 2 . 6x 5 11 Ix 3x x 1 20*-fl3 4. -j --- ^- < 6x -- j ^Lns. re > 5. l+d 5. ax 6 > cx-\-d. Am. x > -- a e a: a x 6. = m ?i 2O2. If there be given an inequality and an equation, contain- ing two unknown quantities, the limit of each unknown quantity may be found, by a process of elimination. 1. Given 2x-\-5y > 16 and 2x-\-y = 12, to find the limits of x and y. 13* 150 INEQUALITIES. If we subtract the equation from the inequality, the result will be an inequality subsisting in the same sense, (19O ? I, 2), and x will be eliminated. Thus, From 2x+5y > 16, (1) subtract 2x+y = 12 ; (2) If we substitute 1 for y in the equation, the first member will be made less than the second ; and we shall have 2x+ 1 < 12, whence, x < 5^. The limit of x may be found in a different manner, as follows: From equation (2), y = 12 2x. Substituting this value of y in (1), we have 2x-f60 10* > 16, whence, Sx ]> 44, or, x < 5^. Thus we may eliminate between equalities and inequalities, either by addition and subtraction, or by substitution. Let it be remem- bered, however, that when an inequality is subtracted from an equa- tion, the sign of inequality will be reversed; (1O0 5 II). EXAMPLES FOR PRACTICE. 1. Given 2x-\- 4y > 30 and 3x+ 2y = 31, to find the limits of a and y. Ans. x < 8 ; y > 3j. ' 2. Given 4x 3y < 15 and 8x+2y = 46, to find the limits of x and y. Ans. x < 5| ; y > 2. 3. Given 7* Wy < 59 and 4x-\- 5y = 68, to find the limits of x andy. Ans. x < 13 ; y > 31. 4. Given 5z-|-8y > 121 and 7x-{-4y = 168, to find the limits of x and y. Ans'. x < 20 ; y > 7. x 4 ?/ 10 , 3# 24 a y 5. Given -g ^-g > 1 and ^ 1- -^ = 13, to find the limits of x and y. Ans. x < 22f ; y < 17f . POWERS OF MONOMIALS. 151 SECTION III. POWERS AND ROOTS. INVOLUTION. . A Power of a quantity is the product obtained by taking the quantity some number of times as a factor ; the quantity is then said to be raised, or involved. 2O4. Involution is the process of raising a quantity to any given power. 2O>. Involution is always indicated by an exponent, which expresses the name of the power, and shows how many times the quantity is taken as a factor. .Thus, let a represent any quantity whatever ; then, The first power of a is a = a 1 j " second " " aa = a* ; " third " " aaa = a*', " fourth " " aaaa = a 4 ; " nth " " aaa...=a n . 2O6. The Square of a quantity is its second power ; and The Cube of a quantity is its third power. 2O 7. A Perfect Power is a quantity that can be exactly pro- duced by taking some other quantity a certain number of times as a factor. Thus, x 9 Ixy+y* is a perfect power, because it is equal to (xy) (xy). POWERS OF MONOMIALS. 2O8. A simple factor may be raised to any power by giving it an exponent which expresses the name or degree of the required power. And if a quantity consists of two or more factors, it is evident that as often as the quantity is repeated, each factor will be repeated. Thus, a = aby ab = aaXbb = a*b\ 152 INVOLUTION. And in general, if abc ..... k represent the product of any number of factors, and n any exponent, we shall have (ale ..... A:)* 1 rr= a n i*c ..... k*. That is, The nth power of the product of two or more factors is equal to the product of the nth powers of those factors. 2O9. If it be required to involve a quantity which is already a power, the exponent of the quantity will be taken as many times as there are units in the exponent of the required power. Thus, ()* = a m Xa m = 0+ = a* m ; (a" 1 ) 3 = a m X"*X" w = a" 1 *" 1 *"* = a 3 "*. And in general, a n raised to the nth power will bo (a m ) n = a mn . That ig, If the mth power of a quantity be raised to the nth power, (lie result will be a power of the quantity expressed by the product of m and n. ^1O. "With respect to signs, it is obvious that if a positive quan- tity be involved to any power whatever, the result will be positive. But if a negative quantity be involved, the successive powers will be alternately positive and negative ; for, it has been shown that the product of an even number of negative factors is positive, and the product of an odd number of negative factors is negative, (G7). To deduce this law of signs in an experimental way, let it bo required to involve a to successive powers. By the principles of multiplication, we shall have, (-a)' =(+a)X (-) = -<*; (_a)> = (-f-a')X(-a) - And in general, the plus sign in the second member being used when n is even, and the minus sign when n is odd. Hence, 1. All powers of a positive quantity are positive. 2. The odd powers of a negative quantity arc negative, but the even powers are positive. POWERS OF MONOMIALS. 153 211. From the foregoing principles relating to the involution of a monomial, we derive the following RULE. I. Raise the numeral coefficients to the required power. II. Multiply the exponent of each letter by the exponent of the required power. III. When the quantity involved is negative, give the odd powers the, minus sign. EXAMPLES TOR PRACTICE. 1. Raise x* to the 4th power. Ans. cc". 2. Raise y 1 to the 3d power. Ans. y". 3. Raise x n to the 6th power. Ans. x* n . 4. Raise x m to the wth power. Ans. a" 1 * 1 . 5. Raise ax* to the 3d power. Ans. a 8 x 8 . 6. Raise ab*x* to the 2d power. Ans. a*b*x*. 7. Raise 5 a x to the 3d power. Ans. 125aV. 8. Raise 8a a & 3 to the 2d power. Ans. 64a 4 & 8 . 9. Raise 4a to the 4th power. Ans. 256a 4 . 10. Raise 4a to the 3d power. Ans. 64a 8 . 11. Required the 7th power of a a x $ . Ans. a l *x* 1 . 12. Required the 4th power of 3ccP. Ans. Find the values of the following indicated powers : 13. (6a& 3 ) 8 . Ans. 15. (a m by. Ans. a**b in . 16. ( a*") 3 . Ans. a 3 ". 18. ( 3a c ^) J . t Ans. 20. ( abc) m . Ans. a m b m c m . 154 INVOLUTION. 212. If it be required to raise a m to the mth power, we shall have (a m ) m = a mxw = a m \ an expression which denotes that power of a whose index is m*. If we put m = 3, then a m = a". Expressions like the above may frequently occur in algebraic operations. EXAMPLES. Find the value of each of the following expressions : 1. (x"y)*. Ans. x mn y n *. 2. (a-y)- Ans. x m *y mn . 3. (x^y. Ans. x m *. 4. \jx j . Ans. x 5. (-c" v ~ 1 y) TO+1 . Ans. x m ~ l y m+l . 6. (ab*c n d n ) n . Ans. o n i n c n c? 1 . POWERS OF FRACTIONS. 213. If a fraction be raised to any power by multiplication, both numerator and denominator will be raised to the same power. a 1 . Required the 3d power of a \* a a a Hence, to raise a fraction to any power, we have the following RULE. Raise both numerator and denominator to the required power. EXAMPLES FOR PRACTICE. 1. Involve =-= to the 2d power. Ans. ^-.* 6tr 6V a a "* 2. Involve =~^ to the 3d power. Ans. NEGATIVE INDICES, 3. Involve = to the 5th power. Ans. 4. Involve -- to the 4th power. W 5 5. Raise =- to the 6th power. 2x m G. Raise to the 5th power. 7. Raise -- to the nth power. ryz 8. Find ( - 9. Find 6)' 155 1024g"& B 16807x> ' Ans. 77- Ans. >. Ans. 15625 a n l> . Ant. Ans. -TV DISCUSSION OF NEGATIVE INDICES. 314. It has been shown in previous articles that a m a w X n = m+n ' =""*> and (a m ) n = a"" , where m and n are positive whole numbers. It remains to be shown that the above relations hold true when one or both of the expo- inents are negative. And in this investigation it is sufficient to re- \member that a quantity with a negative exponent is equal to the re- jciprocal of the same quantity with a positive exponent ; (88, 2). I. To prove that a m xa n = a m+n universally, m and n being integers. 1. Suppose one of the exponents to be negative ; or let n = n'. 2- Suppose both exponents are negative ; or let m = m' and n = n'. 1 Then ~ a 1 *' X a"' ~ 156 INVOLUTION. II. To prove that = a m - n universally, m and n being integers. 1. Suppose the exponent of the numerator to be negative ; or a m a-"' _ a n a" '+ = 2. Suppose the exponent of the denominator to be negative ; or let n = n'. Then - = = a w X B ' = a"* 4 *' = a"-*. 3. Suppose both exponents are negative ; or let m m' and n = n r . III. To prove that (a n ) n = a mn universally, m and n being integers. 1. Suppose n to be negative ; or let Then (a-) = (-)-' = = - 7 = a 2. Suppose m to be negative ; or let / 1 \ A 1 OT = c*-o- = 5=) = 5== = -"= -. Then 3. Suppose both m and n to be negative ; or let m = m r and n = ji'. Then (a")" = (a- w/ )- n/ = (~\ = (^-)* = a m/n ' = a Hence, in all algebraic operations, the same rules will apply to negative exponents as to positive. That is, if two powers of the same quantity be given, then the exponent of their product will be equal to the algebraic sum of the given exponents, and the exponent of their quotient will be equal to the algebraic difference of the giv- on r* YYWn rn fa * en exponents POWERS OF POLYNOMIALS. 157 EXAMPLES. 5. Find the value of each of the following expressions: 1. (a- 7 //) 3 . Ans. a~*l\ 2. (i~V)- a . Ans. Vc- 4 . 8. 2x H "~ i . Ans. ar m . 4. (4a w &-*) a . 5. (_c a tZ-*m 4 )\ C. (Sa-'ay- 1 )- 4 . -4ns. 7. Co'V")" -4** a~"'"- 8. (ar" 1 ') 1 *" 3 . 4s. a;-*" 1 . 9. (4a'*- 2 )' X (a-V). ^n. 16a. 10. (a a "6--) a X(a- t "i" m )"*. ^*. a-i"- 4 ". POWERS OF POLYNOMIALS. S^IG. A polynomial may be raised to any power by actual mul- tiplication Thus, if the quantity be multiplied by itself, the prod- uct will oe the second power; if the second power be multiplied by the quantity, the product will be the third power ; and so on. Hence the following RULE. Multiply the quantity by itself in continued multipli- cation, till it has been taken as many times as a factor as there are units in the exponent of the required power. NOTE. It may be well to observe that in involution we may often reach the same result by different processes. Thus, we have a 6 =a 5 xa = EXAMPLES FOB PRACTICE. Expand the following expressions : Ant. 2. (5z-y). Ans. 125*' 3. (l-f-2x $')'. Ans. l-J-4z 2z 12z'-|-9z 4 . 14 158 INVOLUTION. 4. (3a+2b+cf. Am. 27a'+54a'6-f 27a a c+36a&'+36a6c+8& 8 +9ac-f 12i f c+ 66c-fc. 5. (+&). ^*s. a'-f 7a e &+21a>& a +35a 4 &'-|-35a 8 & 4 -f 21a'i*4- 7a&'-f& T . 6. (x-y}\ Am. x 9 8zV-f-28*y 56xy + 70xy 56*y +28u:y 7. (aV-'-fa-V) 1 . 4ns. a 4 c~ 4 -f 2-j-a~ 4 c*. 8. (a'-f 1-fa-') 8 . 4ns. a'-f 3a 4 +6a a +7H-6a- 3 4-3a- 4 4-a- $ . 9. POLYNOMIAL SQUARES. . We have seen that the square of any binomial may be written without the labor of formal multiplication, (7O). Thus, if x and y represent the terms of any binomial, then This formula for a binomial square furnishes a simple rule for writing out the square of any polynomial, in the same direct manner. To deduce the method, let it be required to square the polynomial, Put x = a and y = l-{-c-\-d-\-e-{- ..... Then the square of x-f y will be equal to the square of the given polynomial ; or And the three parts of the required square will be of = a', (1) 2xy = 2a6-f2ac+2a^-f2ae+ ____ , (2) Now y represents a polynomial ; and to obtain its square, we must proceed as at first. Thus, put x' = b and y' c -\-d-\-e-\- Then the square of x'-fy' will be equal to the square of l-\- c- e-f- And we have POLYNOMIAL SQUARES. 159 x 13 = V, (3) 2x'y' = 2bc+2bd+2be+ . . . . , (4) y" = (<:+ +25c+< 2. Find the square of a-\-b+c-\-d. Ans. a a -f2a6+2ac-f2ac?-f-i a -|-2 3. Find the square of a-\-l>-\- c+d-\-e. Ans. a 2 + 2al + 2ac +2 4. Square x y+z. Ans. x* 2xy-}- 2xz-\-y* 2yz-\-z*. 5. Find the square of a 26-{-3a5 c. Ans. a a 4a6-f6a 9 6 2ac-f46 a 12a& a +46c-f9a a 6 a 6a5c+c*. 6. Find the square of 1 a-j-a 8 a 1 . Ans. l_2a-|-3a' 4a"+3a 4 2a B -}-a i . 7. Find the square of 3ax-}-2a 9 4x s 5. Ans. 9aV+ 12a'x 24ax 8 30ax+4a 4 16a s x 20a'-f IQx* -f40x a +25. 8. Find the square of 1 2x y*-\-xy x*. Ans. I4x2y*-\- 2xy-\- 2x*+4:xy*4x*y + 4x'+ < y 4 2xy' 2x 3 y + 3;c 2 y 2 + x\ 918. In a future section we shall give a formula, called the Binomial Formula, by means of which any power of a binomial may be obtained without the labor of multiplication. 160 EVOLUTION EVOLUTION. 219. A Root of any quantity is one of the equal factors -which, multiplied together, will produce the given quantity. 220. The name or degree of a root corresponds to the number of equal factors into which the quantity is supposed to be divided. Thus, The square root of a is one of the two equal factors whose product is a. The cube root of a is one of the three equal factors whose product is a. The fourth root of a is one of the four equal factors whose product is a ; and so on. 221. Evolution is the process of extracting any root of a given quantity ; it is the converse of involution. 222. There are two methods of indicating evolution : 1st. By the radical sign, j/. When this method is employed, the name or degree of the root is denoted by a figure or letter written above the radical, called the index of the root. Thus \fa denotes the cube root of a; and \/a denotes the fourth root of a. When no index is written, 2 is un- derstood. Thus i/x denotes the square root of x, and signifies the same as \fx. 2d. By fractional exponents. To explain the origin of this method of indicating roots, we observe that a quantity is raised to any power, by multiplying its exponent by the exponent of the required power. Conversely, any root of a quantity may be obtained, by dividing the exponent of the quantity by the index of the required root. Thus, the cube root of a, or a 1 , is written a , and the cube root of a a will be a . Hence, a fractional exponent may be analyzed as follows : 1. The numerator denotes the power of the quantity , whose root is to be extracted. 2. The denominator sJiows what root of that power is to be extracted: BOOTS OP MONOMIALS. 101 333. The two methods of indicating roots may be illustrated by equivalent expressions, as follows : I/a, or a , denotes the square root of a ; Va, or a*, " " cubo " < " a; Va, or a", " " nth " " a. And if a m represent any power of a, then m I/a, or a~, denotes the square root of a m ; Va m , ora^, " " cube " a"; m Va M ,ora% wth " " a w . 334. A Surd is a root which cannot be exactly obtained ; as , % Va 9 or I/a 3 2a6. A surd is called an irrational quantity, while a root which can be exactly obtained is called a rational quantity. A root will bo rational when the given quantity is a perfect power corresponding in degree to the required root ; otherwise it will be a surd. The root of a number which is an imperfect power, may always be obtained approximately. Thus, j/6 is a surd ; but we have ^6=2.44, nearly; for (2.44) 9 =5.9336. 33t>. An Imaginary root is one which is known to be impossi- ble on account of the sign of the given quantity. Thus, the square root of a a , or V a a , is impossible, since no quantity raised t n the second power will produce a 9 . A root which is not imaginary is said to be real. ROOTS OP MONOMIALS. 33O. It has already been shown thnt the root of a simple algebraic quantity may be expressed by dividing the exponent of the quantity by the index of the required root (333). And it is evident that if the exponent of the quantity will not exactly con- tain the index of the required root, the result must be a surd. 14* L 162 EVOLUTION. . We have seen that a quantity composed of several fac- tors, is raised to any power by involving each factor separately to the required power ; (2O8). Conversely, we shall obtain the root of a quantity by extracting the root of each factor separately. Thus, if abc ..../; represent the product of any number of factors, then ____ k = Va V& Vc ____ V& ; or, with fractional exponents, -L .1 JL JL JL (abc. . . .&) = a n b n c . . . .&*. That is, The nth root of the product of two or more factors is equal to the product of the nth roots of the factors. 228. There are certain properties of roots which depend upon the law of signs in involution : 1. Every odd root of a quantity is real, and has the same sign as the quantity itself. For, any positive quantity raised to an odd power is positive ; and any negative quantity raised to an odd power is negative; (21O). 2. Every even root of a positive quantity is real, and may be either positive or negative. For, either a positive or negative quantity raised to an even pow- er is positive ; (21O). 3. Every even root of a negative quantity is imaginary. For, no quantity, whether positive or negative, raised to an even power, will give a negative result. 59. From the principles now established, we have the follow- ing rule for extracting the roots of monomials : RULE. I. Extract the required root of the numeral coefficients for a new coefficient. II. Divide the exponent of each literal factor by the index of the required root. III. Prefix the double sign, , to all even roots } and the minus sign to the odd roots of a negative quantity. NOTES. 1. When the required root of any factor is a surd, it may be indicated either by a fractional exponent, or by the radical sign. 2. The root of a fraction may be obtained by taking the root of the nu merator and denominator separately. ROOTS OP MONOMIALS. 163 EXAMPLES FOR PRACTICE. 1. What is the square root of 49a a :e 4 ? .4ns. 7a;c'. 2. What is the square root of 25c 10 6 a ? Ans. 5c 5 6. 3. What is the square root of 144aVa:y? Ans. 12acVy. 4. What is the cube root of 125a 9 ? -A*. 5a. 5. What is the cube root of 64x" ? Ans. 4a; a . 6. What is the cube root of 216ay ? Ans. Say*. 7. What is the cube root of 729aV 2 ? Ans. 9aV. 8. What is the 4th root of 256aV ? Ans. 4ax*. 9. Find the 4th root of 16a. Ans. 2a*, or 2Va. 10. Find the cube root of 27a*^. Ans. Sa 3 ./*, or3V a *x. 11. Find the*5th root of 82x'y- J.n*. 2x'i/*,or2x* Vy 4 . 12. Find the nth root of a* n l m . Ans. a'b n . 13. Find the square root of 81a- 4 6*. Ans. 9a~ a i 3 . X^. j. iuvx vnv; vt*^ iv -^ v * ---- ^- ~ . ----- ~ . 15. Find the 5th root of 243cr 6 &- 10 . Ans. 3a~ J &- a . 16. Find the mth root of cry*- Ans. a n y m . 17. Find the nth root of afVV 4 . ^Lws. ^" 9 2 ni . ^ 4a'x 4 2x 18. Required the square root of -^-5- - -4n. -g- fl 19. Required the cube root of g , $ Ans. - 4 200a T 5a 8 20. Required the square root o*~9' 21. Required the nth root of f^!T. Ans. . a a _ 22. Required the nth root of-r ,4ns. a n b n c n . 23. Find the square root of (a x)*/* ^4ns. (a x}y*. 24. Find the cube root of (x I) 3 (x+l) . -4s. (** 1) (^+1 25. Find the square root of x*y* (x y)*. Ans. (x*y 9 xy z \ 164 EVOLUTION. SQUARE ROOT OF POLYNOMIALS. 23O. To deduce a rule for the extraction of the square root of a polynomial, let us first observe how the square of any binomial, aa a-f-&, is formed. We have And the last two terms may be written as follows : Let us now consider how the process of involution may be re- versed, and the root, a-j-&, derived from the square. Extracting the square root of a', OPERATION. we obtain a, the first term of the root. Taking a a from the whole expression, a*-[-2ab-\-l*\a-{-b we have 2aZ-j-& 2 , or (2a-f &)&, for a a 9 remainder. Dividing the first term 2a-f-6 2ab-\- u* of this remainder by 2a, as a partial 2aZ-f-& a divisor, we obtain 6, which we place in the root, and also at the right of the 2a to complete the divisor, 2a-|-Z>. Multiplying the complete divisor by &, and subtracting the product from the dividend, we have no remainder, and the work is finished. By the same process continued, we may extract the root of any quantity that is a perfect square. To establish the rule in a general manner, let represent any polynomial. By a previous article, the square of this polynomial consists of the square of each term, together with twice the product of each term Li/ all the terms which follow it ; and the square may be written as follows : And it is evident that if the root, a-\-l-}-c-\-d. . . ., is arranged according to the powers of some letter, the square will also be arranged according to powers of the same letter. We may now derive the root from the square, in the following manner : SQUARE ROOT OF POLYNOMIALS. 165 OPERATION. |rt-j_54-e-fd5 , root a* +2ab+2ac+2ad +b*+2bc+2bd a* ' 2ab+2ac+2ad +b*+2bc+2bd. . & 2ab + Z* +2ac + 2ad +2bc+2bd 2ac ,-2b+2c+d 2ad +2bd. . . . +2cd +d* 2ad +2bd +2cd +d? We find a as in the former example, and take its square from the whole expression. We then divide the first term of the remainder by 2a, and write the quotient, b, in the root, and also in the divisor. We then multiply the complete divisor by b, subtract the product from the first remainder, and thus obtain a new dividend. Then writing 2a-f-26 for a partial divisor, we find c in the same manner as we found b ', and thus we continue till the work is finished. If we examine the several subtrahends, taking the terms diag- onally in the operation, we shall find a 2 , 2ab, 2ac, 2c7, etc. ; &*, 2&c, 2bd, etc. ; c', led, etc. ; d 1 , etc. That is, we have, in the ope- ration, the square of each term of the root, together with twice the product of each term by all the terms ichlch follow it. Thus we have exactly reversed the process of forming a polynomial square. Hence the following general RULE. I. Arrange the terms according to the powers of some letter, and write the square root of the first term in the quotient. II. Subtract the square of the root thus found from the given quantity, and bring down two or more terms for a dividend. III. Divide the first term of the dividend by twice the root already found, and write the result both in the root and in the divisor. IV. Multiply the divisor, thus completed, by the term of the root last found, subtract the product from the dividend, and proceed with the remainder, if any, as before. NOTE According to the law of signs in evolution, every square root obtained will still be a root, if the signs of all its terms be changed. 166 EVOLUTION. EXAMPLES FOR PRACTICE. 1. What is the square root of a a -|-2a&-|-2ac-f-6 9 -f-26c-|-c 9 ? Ans. a-f-6-j-c. 2. What is the square root of a* 6a 2 6+4a 3 + 96" 126 + 4 \ Am. a 2 36+2. 3. What is the square root of x*+ 4x*-f2z 4 2x'-f-5x 9 2x 4. What is the square root of 1 2o-j-3a a 4a'-f 3a 4 2a'-f- Ans. 1 a-f-a a a\ 5. What is the square root of 4a*l 9 1 2a'& 9 -f8a*&'-|-9a 9 & a 6. What is the square root of 9x" 30x 6 7. What is the square root of a 4 6a*6c-|-4aW 2a 2 Ans. a a 8. What is the square root of a 4 a'6-|- -^- -- - -f ? Ans , a _._ + _ 9. What is the square root of x* Ans. 10. What is the square root of a*l~* 10ai- 1 +27 Ans. al~ l S+a^ft. 11. What is the square root of a 4m +6a f ~c"4-lla am c 8n +6a'' l c 8w -f- ? Ans. a' m -f3a m c n -|-c an . SQUARE ROOT OF NUMBERS. . In order to discover the process of extracting the squaro root of a number, it is necessary to determine 1st. The relative number of places in a number and its square root. 2d. The local relations of the several figures of the root to the periods of the number. SQUARE ROOT OF NUMBERS. 167 3d. The law by which the parts of a number are combined in the formation of its square. 222. The relative number of places in a given number and its square root may be shown by illustrations, as follows : I 3 = 1 I 1 = 1 9'= 81 10'= 1,00 99'= 98,01 100' = 1,00,00 999' = 99,80,01 1000' = 1,00,00,00 From these examples we perceive that a root consisting of 1 place may have 1 or 2 places in the square j and that in all cases the addi- tion of 1 place to the root adds 2 places to the square. Hence, If we point off a number into tico-fiyure periods, commencing at the right hand, the number of periods will indicate the number of places in the square root. 233. If any number, as 2345, be decomposed at pleasure, the squares of the parts, beginning with the highest order, will be rela- ted in local value as follows : 2000' = 4 00 00 00 2300 2 = 5 29 00 00 2340' = 5 47 56 00 2345' = 5 49 90 25. Hence The square of the first fyure of the root is contained wholly in the first period of the power ; the square of the first tico Jiynres of the root is contained wholly in the first two periods of the power ; and so on. 234. If the figures of a number bo separated into two parts, and written with their local value, we may then form the square of the number by the formula for a binomial square. Thus, 76 =70 -|- 6. And if we put a = 70 and 6 = 6, then a-\-b = 76 ; and we shall have a' = 4900 ' 2ab = 840 V = 36 a *+2ab-\-b*= 5776=76' Hence, the binomial square may be used as a formula for extracting the square root of a number. 168 EVOLUTION. 1. Let it, be required to extract the square root of 5776. There are two periods in the num- OPERATION. her, indicating that there will be two places in the root. As the square of 5 / < 6 [76 the tens is contained wholly in the a a , 40 00 first period, (233), we first seek the 2, 140 8 76 greatest perfect square in 57. This 2a-\-b, 146 8 76 we find to be 49, the root of which is 7, the first figure of the root sought. Hence we have a = 70, and subtracting a 2 , or 4900, from the entire number, we have 876 for a remainder, which must be equal to (Za-\-l)b ; (23O). Dividing the remainder by the partial divisor, 2, or 140, we have b = 6, the second figure of the root. Completing the divisor, we have 2a-\-b = 146 ; whence (2a-f-^)X b = 876, and the work is complete. It is obvious that we may omit ciphers, and still employ the figures with their proper local values, in the operation. It will not then be necessary to form the partial divisor separate from the com- plete divisor. If the given number consists of more than two periods, we may extract the two superior figures of the root from the first two periods, (233), bringing down another period to the remainder. Then a in the binomial formula will represent the part of the root already found, considered as tens of the next inferior order ; and so on. 2. Required the square root of 22657C. OPERATION. ft 22 6*5 76 | 476, Ant. 16 87 665 609 946 5676 5676 Having found 47, the square root of the first two periods, we bring down the last period, and have 5676 for a new dividend. "We SQUARE ROOT CI- NUMBERS. 169 then take :."a = 47x2 94 for a partial divisor, whence we obtain I 6, the last figure of the root. We should observe that by pimply doubling the 7 in the 87. we may obtain 94, the new trial divisor. From these principles and illustrations, we have' the following RULE. I. Point off the given number into periods of two figures each, counting from unit 1 s place toward the left and right. II. Find the greatest square number in the left-hand period, and write its root for the first figure in the root sought ; subtract the square number from the left-hafid period, and to the remainder bring down the next period for a dividend. III. Al the left of the dividend write twice the first figure of the root, for a trill divisor divide the dividend, exclusive of its right- hand figure, by the trial divisor, and write the quotient for a trial figure in the root. IV. Annex the trial figure of the root to the trial divisor for a complete divisor ; multiply the complete divisor by the trial figure in the root, subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. V. Take the last comjrtcte divisor, doubling its right-hand figure, for a new trial divisor, with which proceed as before, tilt the work is finished. NOTES. 1. If there is a remainder after all the periods have been brought down, annex periods of ciphers, and continue the operation to as many decimal places as are required. 2. If the denominator of a fraction is not a perfect square, the fraction may be first reduced to a decimal, and its root then taken. EXAMPLES FOR PRACTICE. 1. What is the square root of 7225 ? Ans. 85. 2. What is the square root of 108241 ? Ans. 329. 3. What is the square root of 651249 ? Ans. 807. 4. What is the square root of 9741 CO ? Ana. 987. :>. What is the square root of 50985G1? Ans. 2258. 15 170 EVOLUTION. 6. What is the square root of 6634.1025 ? Ann. 81.45. 7. What is the square root of 1812886084? Ans. 42578. 8. What is the square root of .339889 ? Ans. .583. 9. What is the square root of .00524176? Ans. .0724. 10. What is the square root of 477 ? Ans. 21.8403-f . 11. What is the square root of 11.09 ? Ans. 3.3301G+. 12. Required the square root of 7 1 , 3 gsV ^' ts - 7 3 oV 13. Required the square root of 73^50^7. Ans. TT j Tn7 . 14. Required the square root of %$$ Ans. j7 T . 15. Required the square root of 5$. Ans. 2.3604+. CONTRACTED METHOD. 2SI. When the required root is a surd, the work may be abridged by the method of contracted decimal multiplication. To insure a correct result, each contracted divisor should contain at least one redundant pl'.u -e that is, one place more than is necessary to produce the required order of units in the product. This figure should be multiplied mentally, and the tens (increased by 1 when the units are 5 or more) carried to the product of the next figure. To illustrate this principle, let it be required to divide 28337 by 53194, correct to 3 decimal places. In multiplying the first divisor, of 53194) 28337 (.5327 which the last figure, 4, is treated as re- 26597 dundant, we say 5 times 4 are 20, and 5319 1740 reserve the 2 tens for the next partial 1595 product; then, 5 times 9 are 45, and 2 532 145 tens added make 47, and we write the 106 unit figure of this result for the first in 53 39 the contracted product. In multiplying 37 the second divisor, 5319, we have 9x3 = 27 ; hence there will be 3 tens to carry, because 27 is nearer 30 than 20. The third divisor is 532, one unit being carried to 531 of the preceding divisor, because the rejected figure, 9, is greater than 5. SQUARE ROOT OF NUMBERS. 171 1. Required the square root of 7.12 correct to six decimal places. We continue the operation as usual until we have obtained the dividend, 1776. At this point we omit the pe- riod of ciphers, and consider 533 as the divisor ; and in multiplying by 3, the new root figure, we carry the 1 ten 46 from the product of the redundant figure 6, and 1 also from the 8 units 526 3600 in this product, making 1601 for the 3156 first contracted product. After this 5328 44400 we drop one figure from the right, to 42624 form each successive divisor, and thus 5336 1776* continue till the work is finished. 1601 OPERATION. 12.668333 Am. ?;120000 4 312 276 534 160 53 15 16 It will be observed that there are as many figures in the root thus obtained, as there are in the assumed number. From this illustration, we have the following RULE. I. If necessary, annex periods of ciphers to the (/wen number, and assume as many figures as there are places required in the root ; then proceed in the usual manner until all the assumed figures have been brought down. II. form the next trial divisor as usual, but omit to annex to it the trial figure of the root, reject one figure from the right to form each subsequent divisor, and in multiply ing regard the right-hand figure of each contracted divisor as redundant. NOTE. If a rejected figure is 5 or more, increase the next figure at the left by 1. EXAMPLES. 1. Find the square root of 56 correct to 7 decimal places. Ans. 7.4833147-f . 2. Find the square root of 14 correct to 7 decimal places. Ans. 3. 7416573 + . 172 EVOLUTION. 3. Find the square root of 18 correct to 4 decimal places. Ans. 4.2426+. 4. Find the square root of 19 correct to 6 decimal places. Ans. 4.358898+. 5. Find the square root of 52.463 correct to 7 decimal places. Ana. 7.2431346+. 6. Find the square root of 7 correct to 8 decimal places. Ans. 2.64575131 -f. 7. Find the value of 5 3 correct to 5 decimal places. Ans. 11.18034+. CUBE ROOT OF POLYNOMIALS. 236. We may deduce a rule for extracting the cube root ol a polynomial in a manner similar to that pursued in square root, Ly analyzing the combination of terms in the binomial cube. If the binomial, a+6, be cubed, we have We will now consider how the process may be reversed, and the root extracted from the power. We observe 1st. That the first term of the root may be obtained by taking the cube root of the first term of the power. Thus, Va* = a. 2d. The second term of the root may be found by dividing the second term of the power by three times the square of the first term of the root. Thus, = b. 3d. The last three terms of the power may be factored, and written as follows : or Thus we see that if to the trial divisor, 3 a 2 , we add a correction, 3a&+&' 2 . or (3a+6)6, the result will be a complete divisor, which multiplied by A, will give the last three terms of the power. Hence, the whole operation of extracting the root, a+6, from the cube, a s +3a 2 6+3& 2 +6 3 , may be written as follows : CUBE BOOT OF POLYNOMIALS. 173 OPERATION. 3a 8 Having found a, the first term of the root, we take its cube from the whole expression, and obtain 3a*6-|-3a& a -|-& 8 . Dividing the first term of this remainder by 3 a 2 , we obtain 6, the second term of the root. To complete the divisor, we first write the quantity 3a-|-&; and multiplying this by 6, we have 3aZ>-j-& 9 , which added to the trial divisor, gives 3a 2 -f-3a&-{-6 2 , the complete divisor. Multiplying this by />, and subtracting the product from the dividend, there is no remainder, and the work is complete. 297. To recapitulate, -we may designate the quantities employed in the foregoing operation, as follows : Trial divisor, 3a* "\ First factor of correction, 3a-\-b ( ^ ( G) Correction of trial divisor, 3ab-\-b* C Complete divisor, 3 a _j_3 a &_j_i a J 338. Next, suppose there are three terms in the root, as a-j-6 +<-' Assume s =. a-\-b; then s-\-c = a-j-6-j-c; and we have (s+c) 3 = s 3 +3s a c+3sc s +c 8 . If we proceed as in the last example, we shall obtain a-\-b, or that part of the root represented by s, and subtract its cube from the whole expression. There will then be left 3s a c-j-3$c a -|-c 3 , which may be factored and written (3.s a -f3sc-fc a )r* or {3s a -j-(3s-f-c)c}c. And we perceive that 3x 2 will be the new trial divisor to obtain c, and that (3s-f-c)c will be the new correction. The value of 3s a , or 3(a-|-&) a , may be obtained by multiplication. It will be more convenient, however, to derive it by the addition of three quantities already used in the operation. Thus, 15* 174 EVOLUTION. Last complete divisor, 3a*-f-3a&-{-& a -\ Last correction, Sab-\-l* \. n\ Square of last term of the root, 6* J 3.s 3 = 3(a-j-) a = 3a 3 -f 6ai-|-36 a Let it now be required to find the cube root of the polynomial OPEBATION. la^-faj2, root. x* 3x*+x 3z 6 -3.z 4 -llz 8 +62: 2 +12.i:-8 x 3 Having arranged the polynomial according to the exponents of x, we proceed as in the former example, and obtain x 2 , the first term of the root, 3x* 3x 4 Ilx 3 -j-6x 8 -f- 12x 8 the first remainder, 3x 4 the trial divisor, and x the second term of the root. To com- plete the trial divisor according to formula (a), we write three times the first term of the root plus the second, or 3x 3 -|-a:, for the first factor of the correction. Whence we have (3x 8 -|-x)x, or Sx'-j-cc 2 , for the correction; 3x 4 -f-3x*-j-x* for the complete divisor; (3x 4 -j- Sos'-f a; 1 )*, or 3x B -f-3z 4 -fV, for the product; and 6x 4 12x s -f- Gx a -|-12x 8 for the new dividend. To form the new trial divisor according to formula (&), we have (3x 4 -|-3x 8 -fx 9 ) -f- (3x'-|-a; 3 ) -j-* 2 = 3x 4 -f 6x s -f 3.x 2 ; whence, by division, we obtain 2 for the third term of the root. To com- plete the new trial divisor, we have for the first factor of the cor- rection, 3(x a -|-.T) 2 = 3x a -f-8^ 2. This may be obtained in the operation from the former factor 3# a -j-:r, by simply multiplying its second term by 3, and annexing the 2. We now find the correc- tion, complete divisor, and product as before, and the work is finished. It is evident that three or more terms of the root will sustain the same relation to the next succeeding term, that the first sustains to the second, or the first and second to the third. CUBE ROOT OF POLYNOMIALS. 175 23O. From the foregoing analysis we derive the following RULE. I. Arrange the polynomial according to the, powers of some letter, and write the cube root of the first term in the quotient ; sub- tract the cube of the root thus found from the polynomial, and ar- range the remainder for a dividend. II. At the left of the dividend write three times the square of the root already found, for a trial divisor ; divide the first term of the dividend by this divisor, and write the quotient for the next term of the root. III. To three times the first term of the root annex the last term, and icrite the result at the left, and one line below, the trial divisor ; multiply this result by the last term of the roof, for a correction of the trial divisor ; add the correction, and the result will be the com- plete divisor. IV. Multiply the complete divisor by the last term of the root, subtract the jtrotluct from the dividend, and arrange the remainder for a new dicidend. V. Add together the last complete divisor, the last correction, and the square of the last term of the root, for a new trial divisor ; and by division obtain another term of the root. VI. Take the first factor of the last correction with its last term multiplied by 3, and annex to it the last term of the root, for the first factor of the new correction ; with which proceed as before, till the work i$ finished. EXAMPLES FOR PRACTICE. 1. What is the cube root of 27 8 -f 108a s -{-144a-f 64 ? Ans. 3a-}-4. 2. What is the cube root of a 6 -j-6x 6 40x s 4-96x 64 ? Ans. x*-\-1x 4. 3. What is the cube root of Sx e 36x 6 -f66.r 4 63x 8 -f33x 2 9x -f-1? Ans. 2x a Sx+l. 4. What is the cube root of a fl -f 9a 8 &+24aV/ l -|-9a > &' 24a 2 Z> 4 -f9^ s Z/'? Ans. a*+3abb*. 5. What is the cube root of a 9 6a 8 -|-27 7 74a fl +159a 6 234a 4 -j-257a 8 174a 8 -j-60 8 ? Ans. a 3 2a a +5a 2. 176 EVOLUTION. - G. What is the cube root of *' 3x 8 +6.t T 10;c'-f 12x 12x -flOx 3 G.r'+Sx 1? ^l s . a; 1 *'-f.r 1. 7. What is the cube root of 8 a 12 s i -f SGa'ic -f 6a*& s SGaVc aV4-54a&V+9a*6 3 c 27a6V+27V? -Aws. 2a rti-f SJc. 195x 4 195 r 1 8. What is the cube root of *' 12**+ j 70;c'-J- -^j- 3* 1 ? 1 -T^64 ? 4z +r 9. What is the cube root of o; 8 -}-Gx 8 64o: e 962r 6 -j-192jc*4- Root*. Cubes. Roots. 1 r 1 9 729 10 99 970,299 100 999 997,002,999 1000 CUBE ROOT OF NUMBERS, 24O. To establish a rule for extracting the cube root of a num- ber, we must first ascertain the relative number of places in a cube and its root. This relation is exhibited in the following examples : Cubes. 1 1,000 1,000,000 1,000,000,000 Thus we perceive that a number consisting of one place, may have from one to three places in its cube; and that in all cases the addition of one place to the root adds three places to the cube. Hence, ]f a number le pointed off into three-figure periods, commencing at units place, the number of periods icill indicate the numb** of places in the root. 341. To ascertain how the several figures of the root arc related in local value to the periods of the power, we may decompose any number, as 5423, and form the cubes of its several parts, as follows : 5000 3 = 125 000 000 000 5400 s = 157 464 000 000 5420 3 = 159 220 088 000 5423 3 = 159 484 621 967. Hence, CUBE ROOT OF NUMBERS. 177 The cube of the first figure of the root is contained wholly in the first period of the power j the cube of the first two figures of the root is contained wholly in the first two periods of the p wer ; and so on. 242. To employ the binomial cube as a formula for extracting the cube root of a number, we must represent the first figure or figures of the root, taken with their local value, by a, and the re- maining figures by b. The operation will then be the same, in form and principle, as that employed in extracting the cube root of alge- braic quantities. 1. Let it be required to find the cube root of 164,206,490,176. OPERATION. 1642064901 76|5476 125 154 616 7500 39206 8116 32464 1627 11389 874800 6742490 886189 6203323 16416 98496 89762700 539167176 89861196 539167176 There are four periods in the given number, indicating that there will be four figures in the root. As the cube of the first figure will be contained wholly in the first period, (241), we seek the greatest perfect cube in 164. This we find to be 125 ; its root is 5, which we write as the first figure of the root sought. We may now consider the 5 as tens of the next inferior order in the root, and let a = 50, and b represent the next figure. A.nd since the cube of a-\-b will be contained wholly in the first two periods of the number. (241), we subtract a 3 , or 125. from 164, and to the remainder bring down the next period, making 39206. Then this result must contain at least 3a a 6-f-3a6 s -}-& s , (236), and we there- fore divide it by 3 a 9 , or 7500, as a trial divisor, and obtain 4 for the value of 1>. or the second figure of the root. To complete the divisor, we have 3a-\-b = 154 for the first factor of the correction, and (3a-f- 6)6 = 616 for the first correction ; M 178 EVOLUTION. whence by addition we obtain 8116, the complete divisor. Multiply- ing this by 4, and subtracting the product from the dividend, we have, after bringing down the next period, 6742490 for a new dividend. We may now form a new trial divisor, according to (388, I). We shall have 8116-f 616-J-16 = 8748 j or 874800, if we give to the figures their local value with respect to the lowest order in the dividend. By division, we have 7 for the next figure of the root. To find a correction for the new trial divisor, we annex the last figure, 7, to 3 times the former figures of the root, and obtain 1627 for the first factor j and we then continue the operation, repeating the former steps, till the work is finished. Hence we have the following RULE. I. Point off the given number into periods of three figures each, counting from units' place toward the left and right. IT. Find the greatest cube in the left hand period, and place its root for the first figure of the required, root ; subtract this cube from the first period, and to tlie remainder bring down the next period for a dividend. III. At the left of the dividend write three times the square of the root already found, and annex two ciphers, for a trial divisor ; divide the dividend, and write the quotient for the next figure of the root. IV. To three times the first figure of the root annex the last; multiply this result by the last root figure, for a correction to the trial divisor; add the correction, and the result will be the complete divisor. V. Multiply the complete divisor by the last figure of the root, subtract the product from the dividend ', and to the remainder bring down another period for a, new dividend. VI. Add together the last complete divisor, the last correction, awl the square of the last figure of the root, and*annex two ciphers, for a new trial divisor ; then l>y division obtain another figure of the root. VII. Take the first factor of the last correction, multiplying its right hand figure by 3, and annex the last figure of the root, for the first /actor of the new correction ; with which proceed as in the former steps, till the work is finished. CUBE ROOT OF i\ UMBERS. 179 EXAMPLES FOR PRACTICE. 1. Find the cube root of 148877. Ans. 53. 2. Find the cube root of 571787. Ant. 83. 3. Find the cube root of 256047875. Ans. 635. 4. Find the cube root of 354894912. ' Ans. 708. 5. Find the cube root of 11852.352. Ans. 22.8. G. Find the cube root of 144125083907. Ans. 5243. 7. Find the cube root of 128100283921. Ans. 5041. 8. Find the cube root of 105555,569176. Ans. 4726. 9. Find the cube root of 731189187729. Ans. 9009. 10. Find the cube root of 1762.790912. Ans. 12.08. 11. Find the cube root of 1061520150G01. Ant. 10201. 12. Find the cube root of 33212361:641984. Ans. 321.44. 13. Find the cube root of 1371737997260631 . Ans. 111111. 14. Find the cube root of .171467. Ans. .55555-f. 15. Find the cube root of .004235801032. Ans. .1618. CONTRACTED METHOD. 243. In applying the method of contracted decimal division to the extraction of the cube root of a number, we observe, 1st. For each new figure in the root, the terms in the operation extend to the right 3 places in the column of dividends, 2 places in the column of divisors, and 1 place in the extreme left-hand column- Hence, 2d. If at any point in the operation we omit to bring down new periods in the dividend, we must shorten each succeeding divisor 1 place, and each succeeding term in the left-hand column 2 places. 3d. If, however, for the -first contraction in the column of divisors, and in the left hand column, we simply omit the ea-tcinlcd part^ and afterward contract according to the precept just given, each con- tracted multiplicand will have one redundant figure. 180 EVOLUTION. 1. Find the cube root of 850 correct to 8 decimal places. OPERATION. [9.472682374-, root. 850.000000 ~ 729 274- 1096 24300 121000 25396 101584 2327 19789 2650800 19416000 2070589 18694123 2841 568 2690427 721877* 2690995 538199 28 17 269156 183678 269173 161504 26919 22174 21535 2692 639 538 269 101 81 27 20 19 We proceed in the usual manner until we reach the first contract- ed dividend, 721877, which is obtained in the common way, the period of ciphers being omitted. The corresponding trial divisor, with the ciphers at the right omitted, is 2690427, the right hand figure of which is redundant, being of an order lower than is required to obtain a product corresponding in local value to the contracted dividend. By division, we have 2 for a new figure in the root. To obtain a correction whose lowest figure shall be of the same order as the lowest in the trial divisor, we form the term 2841 in the common way, but omit to annex 2, the Ir.st figure in the root. Then 2841 X 2 = 5682, of which 568 is the part required for the correction. We then have 2090995 for a complete divisor, 538199 for a product, and 183678 for the new dividend. For the next trial divisor, we add 2690995 and 568, and reject one figure, thus obtaining 269156. The square of 2, the last root figure, is of CUBE HOOT OF NUMBERS. 181 course rejected, on account of its inferior local value. The remain- ing part of the operation requires no further explanation. It will be seen that the number of places obtained in the root is equal to the number of places assumed in the power. Hence wo have the following RULE. I. Assume as many places in tJie "power as there are places required in the root, and proceed in the usual manner till all the assumed figures have been brought doicn. II. Form the next trial divisor as usual, omitting the ciphers at the right; and reject one place in forming each subsequent trial divisor. III. In completing the first contracted divisor, omit to annex the new figure of the root to the corresponding term in the left-hand column, and reject two places in funning each succeeding term in. this column. IV. In multiplying, treat the right-hand figure of each contract- ed term as redundant, both in the column at the left, and in the col- umn of divisors. NOTE. To avoid complicating the process of contracting, it is better to use none but full periods, even if the root is thereby carried beyond the required number of places. EXAMPLES FOB PRACTICE. 1. Find the cube root of 3 correct to 6 decimal places. Ans. 1.442249-f . 2. Find the cube root of 7 correct to 6 decimal places. Ans. 1.912931-f . 3. Find the cube root of 156 correct to 8 decimal places. Ans. 5.38321261+. 4. Find the cube root of 34786 correct to 6 decimal places. Ans. 32.643859-f . 5. Find the cube root of 10.973937 correct to 6 decimal places. Ans. 2.222222-f-. 6. Find the cube root of 1500.101520125 correct to 8 decimal places. Ans. 11.44740066-}-. 7. Find the cube root of 1.164132 correct to 6 decimal places. Ana. 1.051963 -r 16 182 RADICAL QUANTITIES. SECTION IV. RADICAL QUANTITIES. 344. A Eadical Quantity is a root merely indicated, cither by the radical sign or by a fractional exponent ; as 3|/a, $ 'a 6, c(a-f &)*, m ^x 1 y*. A radical quantity may be either surd or rational. The quantity or factor placed before a radical is its coefficient. Thus in the examples just given, 3, 1, c, andm are the coefficients of the radicals. 24>. The Degree of a radical quantity is denoted by the radical index, or by the denominator of the fractional exponent. Thus, i I/a, (a 6)2 are radicals of the 2d degree ; ~^x* y, a*l* are radicals of the 3d degree; vac, (x-f-y) are radicals of the nth degree. 24 6. Similar Radicals are those in which the same quantity is affected by radical signs having the same index. Thus, 4 vV-fi, ftf+b, and 7 (a a -f-&)^ are similar radicals. REDUCTION OF RADICALS. CASE I. 947. To reduce a radical to its simplest form. A radical is in its simplest form when it contains no perfect power corresponding to the degree of the radical. 1. Reduce l/48a"x* to its simplest form. We have seen that the nth root of a quantity is equal to the prod- uct of the nth roots of its component factors; (227). Hence we have REDUCTION. 188 It will be seen that we first separate the quantity under the rad- ical sign into two factors, one of which is a perfect square. Then according to the principle of evolution just adduced, we have the product of two radicals, one of which, 1/lGaV 2 , is rational, and the other, l/3.r, is a surd. The expression is then reduced by extract- ing the root of the rational part, and multiplying it by the surd. 2. Reduce 3F&*y Sx*t/ 4 to its simplest form. Factoring as before, we have Y = s x i^y x 3 X 2ry X Hence the following RULE. I. Separate the factors of the quantity under the radical sign into two groups, one of which shall contain all tlie perfect pow- ers corresponding in degree to the radical. II. Extract the root of the rational part, multiply this root by the given coefficient, and prefix the product to the surd or irrational part. EXAMPLES FOR PRACTICE. Reduce the following radicals to their simplest form : 184 RADICAL QUANTITES. 13. (2a'i 5 3a 6 i 7 ) 5 . Ans. a&(2a* 31*] 15. l/8a 2 "x 4 '". 16. if cTV^. 17. (2x 9w 18. a- m c(a m V n a^V 1 )*". ^ns. c a (c n a mn )~ n ". 248. When the quantity under the radical sign is a fraction, we may transform it in such a manner that the denominator shall be a perfect power corresponding in degree to the indicated root. Then after simplifying, the quantity remaining under the radical sign will be entire. It will generally be expedient to separate the given fraction into two factors, one of which shall be a perfect power } we may then operate upon the surd part separately. 1. Reduce 1/4 i to its simplest form. OPERATION. l/4f = T/^XV = l/,ftxV = VAX 4X33 = -& 1/33, An*. 2. Reduce v -fa to its simplest form. OPERATION. In like manner reduce the following : 3. ^-i. Ans. 4. 5. G. : Ti- REDUCTION. . 185 CASE II. 249. To reduce a rational quantity to a radical, or to introduce a coefficient of a radical under the radical sign. Since involution and evolution are the converse of each other, we have a = ytf = fa* = Va 4 , etc. Whence, we have also, a yb = i/a*Xi/b = V~c?b. We have, therefore, the following RULE. I. To reduce a rational quantity to a radical : Involve it to a power denoted by the degree of the required radical, and write the result under the radical sign. II. To introduce the coefficient of a radical quantity under tho radical sign : Involve it to a power denoted ly the degree of the rad- ical f and multiply the quantity under the radical by the power thus obtained. EXAMPLES FOB PRACTICE. 1. Reduce ab* to a radical of the second degree. An*. \/a*b*. 2. Reduce 5a a .ry 3 to a radical of the 3d degree. AM. ^ / 125(t.ry. 3. Reduce a cz to a radical of the 4th degree. Ans. (a 4 4a f +6aW 4<*cV+cV)i Introduce the coefficients of the following radicals under the rad- ical sign : 4. 4ul/27yT Ans. 1/32^.7^" 5. 3.t 2 Vx y. Ans. T V 27x 7 6. (_2i) 1 /2a, Ans. l / 2a s 16* 186 RADICAL QUANTITIES. CASE III. 2oO. To reduce radicals to a common index. m mr It may be shown that a n = a, r being any integer -whatever. Let x = a n ; (1) involving (1) to the th power, x n = a m , (2) (2) rth " af=a'", (3) mr taking the nrth root of (3), x = a~, (4) m mr equating values of x in (1) and (4), a n = a nr . Hence, 1 If Loth terms of a fractional exponent be multiplied or di- vided by the same number, the value of the expression will not be altered. From (233) we have and d^ssrVa-r. Hence, 2. If both the index of a radical and the exponent of the quan- tity under the radical sign be multiplied or divided by the same number, the value of the expression will not be altered. 1. Reduce (ab)" and (a a x) to a common index. (a&)* = (a6)* = (a f &')*l i 2 i fAns. (aVp = (aV)S = (aV) J 2. Reduce "^a'c and VxV to a common index. V..., V35- Ilence, we have the following RULE. I. When the quantities are affected by fractional expo- rents : Reduce the yiven exponents to their least common denomi- nator ; then raise each quantitifto a power denoted by the numera- tor of its new exponent, and affect each result with a fractional index equal to the reciprocal of the common denominator. ADDITION. 187 II. When the quantities are affected by radical signs : Find the least common multiple of the given indices for the common index required ; and raise the quantity under each radical sign to a pow* er indicated Ly tJie quotient of the new index divided l)ij the given index. EXAMPLES TOR PRACTICE. 1. Reduce a 2 , (cd) ^,and (a'c)* to a common index. Ant. a^, (cd*) A, (aV)A. 2. Reduce (3a'x)^, (2ax*)7,and (5aV)& to a common index. AM. (81aV)T2, (8aV)T2, (25aV)T2. 1 2 3. Reduce (a &) 2 and (a-J-i) 3 to a common index. , or (a 1 3a 2 Am. , or (a 4 -f 4a' 4. Reduce a, 1/ac, v a*x, and ^2ac* to a common index. 5. Reduce -j/2, V2, and v/2 to a common index. Am. VlUa, V82, VTO. C. Reduce a*, V 5x, K^2ax, and ^4a^x to a common index. 7. Reduce v x y, v x-\-y^ and v x 2 ^ a to a common index. 8. Reduce V ax, *Vxy, and ^ ex to a common index. *t.TlS* Cb & v } ADDITION OF RADICALS. 51. When the quantities to be added are similar radicals, it is evident that the common radical part may be made the unit of addi- tion ; the result will then be a single radical whose coefficient is the sum of the coefficients of the given radicals. Radicals which do 188 KADICAL QUANTITIES. not appear to be similar, may become similar when reduced to their simplest forms. 1. What is the sum of 71/oe, Sl/oc, and 51/oc? ==15l/^r Ant. 2. What is the sum of 1^8^ 1^27Vc, and OPERATION. Sum = (5a-|-4c)^ / a a c, If the given radicals are dissimilar, the addition can only be indi- cated. Hence the following RULE. I. Reduce each radical to its simplest form. II. If the resulting radicals are similar, add their coefficients, and to the sum annex the common radical ; if dissimilar, indicate the addition by the proper signs. EXAMPLES FOR PRACTICE. 1. Find the sum of ViQa*x and l/4a a x. -4ns. Qa^/x. 2. Find the sum of 1/32^ Vll, and 1/128. Ans. 18]/2. 3. Find the sum of &W, V f35, and 1^625. Ans. 10^/5. 4. Find the sum of 1^108, 9^4, and 1^1372. Ans. 19^4. 5. Find the sum of -j/4, i/f, and i/yg. Ans. -j/2. 6. Find the sum of f^b ^IJ, an 1 T V |f|. Ans. |J^3. 8 ~ 7. Find the sum of j^J, |1/^ and Jl/j J. ^ln. |-J|/'G. 8. Find the sum of Sl/abm*, wl/4o6, and V25abni\ Ans. 10ml/6. 9. Find the sum of 2aT/c 2 x- c 2 ^, 3cl/a a ic a>, and 5 l/aV.r a'c'. J.TW. lOacl/ a^y SUBTRACTION. 189 10. Find the sum of l/20a f w 2Uoc7-f 5mc and l/20ni a 60aci-j-45a*i. Ans. (c a 11. Find the sum of S^'a?, ^a?, and 2^ x d -An*. 12. Find the sum of 5a(cx*dx*^ and 2* Ans. %ax(c cQ 3 . 13. Find the sum of \ ^~ ' > \f ' ^, \ and ' a-j-6 * a 6 . 21/a a 6- 14. Find the sum of V(l+ a)' 1 , I/a 2 (1 -fa)- 1 , and <>&=SF>. ' VWa SUBTRACTION OF RADICALS. . When the radicals are similar, it is evident that we may make the common radical the unit of subtraction. Hence the following RULE. I. Reduce each radical to its simplest form. II. If the resulting radicals are similar , find the difference of the coefficients, and to the result annex the common radical part j if dissimilar, indicate the subtraction by the proper sign. - EXAMPLES FOB PRACTICE. 1. From 41/135 take 21/60L Ans. 8l/lK 2. From 1/75 take 1/50. Ans. 5(|/3 !/2). 3. From 3l/16^> take 3l/^T Ans. (12a 9 4 From \y ~ '.? take ' P' STT 3 Ans i-i/11 8 1^5 4 2 /490a 888 190 RADICAL QUANTITIES. 6. From (aV 3c'x)* take 2(aV 3d a x^. Am. (c 2d) (a* 7. From (a 1 a&'+a'fc &)* take (a' 3a'&+3a6 a ')*. 8. From o\/- take 6\/- ~n ^- ns - -- fl/a 1. X - 1 * X-j-1 X 3 - 1 MULTIPLICATION OF RADICALS. . It has already been shown (S3 7) that the wth root of the product of two or more factors is equal to the product of the nth roots of those factors. And since the converse of this pro- position is true, we shall have Va X V& = V ob- it' the radicals have coefficients, the product of the coefficients may be taken separately. Thus, IF the radicals have not a common index, they must first be reduced to the same decree. O Let it be required to find the product of a^/x and OPERATION. ay'x = a \ Product, ab y = Hence the following RULE. I. If necessary, reduce the given radicals to a common indrx. II. Multiply the quantities in the radical parts together, and place Jie product under the common radical sign ; to this result prefix the product of the given coefficients, and reduce the whole to its simplest form. DIVISION.^. >> 1 91 EXAMPLES FOR PRACTICE. Find the following indicated products: 1. 5/5x3/8. Ans. 301/10. 2. 41/I2X3/2. Ans. 24/6. 3. 3/2X2/8. Ans. 24. 4. 2/5x2l / 10x3/6. Ans. 120/3. 5. 21^4x3/4. Ans. 12 1 6. 5c 7. 9. i^ISXl/ia ^ws. V225000 a l x y s /^ 2 3 1^ 5 10. - 7 J- X- -\KX\f~i Ans. Z>\y x \a 2 V/j/ 9 OF RADICALS. 2> involving (3) to the power denoted by q, x*' = a"; (4) extracting the root whose index is qs, hence, by equating values of x in (1) and (5), p ~ - m. (<*)' = a, which was to be proved. We conclude, therefore, that in multiplication, division, involution and evolution, the same rule will apply, whether the exponents are positive or negative, integral or fractional THEORY OF EXPONENTS. 199 EXAMPLES. 1221 1. Multiply a~b* by a 3 6 3 , and simplify the product 2. Simplify the expression 3. Multiply x4_3x2-j- by OPERATION. 3x4+9x^3x1 4. Divide x 5Vx 5^x s 5^/x 6^x by ^x f OPERATION. 9|/x 5. Multiply x^ by x^. ^n. x^. 6. Multiply a'fti by aM. J.ns. a'l/ail 7. Find the product of a*, a*, a^, and a~*. Ans. T^a*. 8. Divide aM by a^c^ ^4n5. (-\ *. 200 RADICAL QUANTITIES. 9. Divide (**)* by (*')*. 10. Multiply a- a 4 by a^-j-1. Ans. a' a*. 11. Multiply 2^x 8 -f l/^" by 3^* T/ayl .4ns. C.r 4- Vaty 3 QVafytxy. 12. Multiply a*- 2a~i-}-a- 1 by a* a~*. Ans. a^ 3-}-3a~J ~-. 13. Divide a Z> by i/a +i/b. Ans. -|/a i/b. 14. Divide a^ 2a2-j-a^ by a^ 1. -4ns. a^ a 15. Multiply J+a'^+a^-fai'-f a^^-f ^ by a^ ^. Ans. a'i 1 . 16. Divide ^-f^a^-fa^ by a^-f-z^-f al ^Ins. x s x^a^-^d 3 . 17. Simplify (a^-a^) TT . -4ns a 3 . 18. Simplify ( ^-H ^- 2v/5. i i {KfWCO}' 19. Simplify \ r ^ 20. Si 3^72-3(3)3 21. Simplify -t-. Jut. |(2 Cl/i3+3)(Vi3-f 1 /3)(Vl3 1/3) J ^jw. IMAGINARY QUANTITIES. 201 IMAGINARY QUANTITIES. 969. It has been shown (998 ? 3) that an even root of a nega- tive quantity is imaginary, an expression for such a root being a symbol of an impossible operation. Thus if w take a a , which is numerically a perfect square, and affect it with the minus sign, we can not obtain the square soot of the result. For we have (+a) = -fa', (-ay = + a'. Hence the indicated root, V a a , is not real but imaginary. Such expressions are, however, of frequent occurrence in analysis and its application to physical science, and conclusions of the highest impor- tance depend upon their use and proper interpretation. We there- fore proceed to investigate the rules to be observed in operating with such quantities. 263* When a real and an imaginary quantity are connected in a single expression, the whole is considered imaginary on account of the presence of the imaginary part. Thus the binomial, 4-j-F 3, considered as a single quantity, is imaginary. 964. According to (927'), we may have 1/HTa = 1/X( 1) = i/o ' Vl ; also, V a* 6 3 +2aft = V (a &)'X( 1) = (a V)V~\. Hence, if we regard only quadratic expressions, every imaginary quantity may be reduced to the form, in which a is the real part, b the coefficient of the imaginary part, and v 1 the imaginary factor. Thus we may employ only the single symbol, V 1, to indicate that a quantity is imaginary. 965. For convenience in multiplication and division of imagi- nary quantities, we will now obtain some of the successive powers of the symbol v 1, and deduce the law of their formation. =+v'=i, -i)' = o/-i) x o 7 -!) == -i, ; -i)' = (-1) X (V 7 -!) = -V~\, -!) 4 = (-V=i) x (i/ 1) = 4-1, 202 RADICAL QUANTITIES. Multiplying these powers, in their order, by the 4th, we shal . obtain the 5th, 6th, 7th and 8th, the same as the 1st, 2d, 3d, and 4th ; and so on. 2GG. The common rules for multiplication and division of radicals will apply to imaginary quantities, with a simple modifica- tion respecting the law of signs. Let it be required to find the product of I/ a and I/ b. To obtain the true result, we must separate the imaginary symbol V 1 from each factor. Thus l/IT a X V~l) == i/a -1/L X j/fc !/ L = Vai x X ( 1) [From a real quantity, and negative. But if we multiply by the common rule for radicals, (253), we shall have a result erroneous with respect to the sign before the radical. Proceeding as in the first operation we find that (_l/H^)x( V~b) =- -f-T/o6 (1) = 1/ofc ; Thus, like signs produce , and unlike signs produce -j-. Hence, 1. The product of two imaginary terms will be real, and the sign before the radical will be determined by the common rule re- versed. We may operate in like manner in division of imaginary quanti- ties. Thus, -\-V~ab - V\ _T/_ ff fc Val - 1/ 1 That is, like sipis produce -j- and unlike signs produce . Hence, IMAGINABY QUANTITIES. 203 2. The quotient of one imaginary term divided "by another will bf real, and (lie sign before the radical will be governed by the com- mon rule. 2G7. Let us assume the equation a-f bVl = a'+b'Vl, (1) in which a and a! are real. By transposition, a a! = (b f &)!/ L (2) Now it is evident that in this equation a = a'. For, if a > o! or a a rational quantity. 2. Rationalize x*. The factor is x ; for, xX^ = &* = * 279. To find a factor which will rationalize a bino- mial in the form of aV6, or VaV&, m and n being each some power of 2. The product of the sum and difference of two quantities is equal to the difference of their squares. Hence, if we multiply the given binomial in this case, by the same terms connected by the opposite sign, the indices of the product must be respectively the halves of the given indices ; and a repetition of the process will ultimately ra- tionalize the quantity, provided m and n arc any powers of 2. 1. Rationalize a-j-|/ c - The factor is a ^/c ; for we have 2. Rationalize j/a \/x. d/a 4 vAr) d/a-f-V^) = a -/a; ; (a -j/x) (a-f !/*) = a' x, a rational result; and the complete multiplier is 18* 210 RADICAL QUANTITIES. A trinomial may be treated in a similar manner, when it con- tains only radicals of the 2d degree. 3. Rationalize T/5+J/2 -v/3 1/5+1/2 -f y/3 J/15 1/10+ 2 j/6 4 2V/10 16 40= .24 a rational result ; and the complete multiplier is ( 1 /5+ 1 /2-|- 1 /3) (4 2 1 /10). Thus we perceive that it is necessary simply to change the sign of one of the terms of the trinomial, and multiply by the result, repeating the process with the product. SSO. To find a factor which will rationalize any binomial surd whatever. Let the binomial be represented by the general form, a 7 I*. Put x = a r and y = b* ; and let n be the least common multiple of r and s. Then x* y* will be rational. But x-\-y will exactly divide x*-{-y n when n is odd, and x n y n when n is even ; and x y will exactly divide x n y n whether n is odd or even; (89). These quotients will therefore be the factors that will rationalize the respective divisors. Hence, let q represent the required factor; then CO q = -jry-' for a r -j-6 '", when n is odd ; x*i/* - L. Ov for a r Z> ' , i being odd or even. RATIONALIZATION. 211 1. Rationalize the binomial a*-f-6*. Since n = 6, an even number, we have from (2), 9 = ~= ~jj | the factor required ; and the rational result. The foregoing methods may be applied in the solution of the fol- lowing EXAMPLES. 1. Reduce to a fraction whose denominator shall be rational. Ans. - 2. Reduce - to a fraction whoso denominator shall be rational. 21/1531/1 An.. - g - 3. Reduce to a fraction whose denominator shall be rational. 1/2 4. Reduce T/Q to a fraction whose denominator shall be rational. V72 Ans. 3 5. Reduce - to a fraction whose denominator shall be 1/7+1/3 rational. Am 5d/7 1/3) 4 6. Reduce }L to a fraction whose denominator shall be I/ a i/c rational. a-f-l/oc 212 KADICAL QUANTITIES. 7. Reduce !_ JLk__ to a rational denominator. 1/11 v/5 Q 8. Reduce /i]i /"g to its simplest form. .4ns. l/ll 9. Reduce , -- -^ to its simplest form. Ans. 4-f- ^ 15. 10. Reduce -- -, - to its simplest form. " S * - - 11. Simplify (8+1/8) (8+T/5XT/5-2) (5-1/5) (v/8+1) 12 . Simplify _ 1-fa 1/1 a' a 13. Find the factor which will rationalize -j/5 V2. Ans. , 14. Reduce . - _ to a rational denominator. Ans g '^+ aa RADICAL EQUATIONS. 281. A Radical Equation is one in which the unknown quan- tity is affected by the radical sign. 282. In order to solve a radical equation, it is necessary in the first place to rationalize the terms containing the unknown quantity. In case of fractional terms, this may be effected in part by methods already explained. But the process is commonly one of involution. The following are examples of simple equations containing radical quantities. RADICAL EQUATIONS. 213 1. Given 1/x-j-ll-fl/o; 4 = 5 to find x. OPERATION. y ^+114-1/^=4 = 5 ; by transposition, 1/x-j-ll = 5 V x 4 ; squaring and reducing, x-|-ll = x-f 21 lOl/x A ; transposing and reducing, l/x 4 = 1 ; squaring and reducing, x = 5. T- 2. Given r - - = - r to find x. OPERATION. x 5 __ Multiplying both terms of 2 x 5 2l/?=5]c ' 4cr 35 the fraction by the nume- > - - - = - - - > rator, (279), J ' _ clearing of fractions, etc., 2Vx*5x = 2x 30 ; dividing by 2, l/x 5x = x 15 ; squaring and reducing, 25x = 225 ; whence, a; = 9. c m-t/a m 3. Given -j, T -\ -- - = -7 - 7 to find x. * xa /x/a The least common multiple of the denominators is x a = (^/x-t-i/a) (i/x i/a) } and the solution will be as in the fol- lowing OPERATION. p rr-> (c m) 1 214 RADICAL QUANTITIES. From these illustrations, we derive the following precepts for the solution of radical equations : 1. It is sometimes advantageous to rationalize the denominate! of a fractional term, before transposition or involution. 2. An equation should be simplified as much as possible before involution ; and care should be taken so to dispose the terms in the two members as to secure the simplest results after involution. EXAMPLES FOR PRACTICE. Find the values of the unknown quantity in each of the follow- ing equations : /*=7. 2. x+3 = 1/V 4x4-5~9*. Ans. x = 5. 3. ^i,/ T-TO ,_ *, ^ ns = 16. a 9 4a 4 - Ans - x = 5. _?L I? n -. An*, x = \x , - 6. -f - - =- - . Ans. x = |. T/l-fa: 1/1 x* V\ *' / -\/a-\-x* 8 7. T/c-f-arr: 1 -/^ Vc+x 8 - x ^9-f xx*^S = 3. 9. 2 1 / 21/x 32 = 1/327 ^in*. x = 50. a a-fc c /a-fc\* 0- 7^=2-^=4 = 7^2* 11. a 3 3*x -j- x*K 3a a; ~ a x. ^Lns. x = 3a 1. EADICAL EQUATIONS. 215 c * 13. x4- Vc* ax = . Ans. x = l/c a ax 23- rVPST- *" - -II 15. 17. V 5+x+F 5 x = F 10. Ans. x = 5. I~, = 2a -. -4ns. a = * 1/i+i 3 19. x-\-a = ^ Itf+xV-V+aP. Ans. x = 4a 2 41/6^9 20. -/= - = j= -- Ans. x = 6. 1/6x4-2 41/6x4-6 4-4- c 21. / - ^w*. = 4. rrrr - . -4n*. x = - . Vax+b 3Vax+5b 4 25. -- -^ = 9. .. ~. 97 27. J - = - . AM. x = a 216 QUADRATIC EQUATIONS. SECTION V. QUADRATIC EQUATIONS. 983. A Quadratic Equation is an equation of the second degree, or one which contains the second power of the unknown quantity, and no higher power ; as 3x a = 48, and ax* 2bx = c. That term of the equation which does not contain the unknown quantity, is called the absolute term. 984. Quadratic equations are divided into two classes Pure Quadratics, and Affected Quadratics. PURE QUADRATICS. . A Pnre Quadratic Equation is one which contains tho second power only, of the unknown quantity ; as 3.r a 7 = 20. NOTE. A pure equation, in general, is one which contains but a single power of the unknown quantity. 986. It is evident that by the rule for solving simple equations? every pure quadratic may be reduced to the form of x* = a, in which a may be any quantity, real or imaginary, positive or negative. Extracting the square root of both members of this equation, we have, x = -}-j/a or |/a Hence, Evdry pure quadratic equation has two roots, equal in numerical value, but of opposite signs. x * _ 24 1. Given - --- j = - 32, to find the values of x. " H PURE QUADRATICS. 217 I - ^ OPERATION. a__4 ^ 3 24 x 9 6 " 4 = 2" " 2 ' clearing of fractions, 2x' 8 3x'-f72 = 6x* 384 ; collecting terms, ~ "X 2 ^ ^ '% . 7x a = 448 j 3 ^ dividing by 7, x a = 64 ; by evolution, x 8, -4ns. We have therefore, for the solution of pure quadratics, the following _^A <~~ o ~ >' "~ , that 2a, the coefficient of x, becomes fractional, thus rendering the solution a little complicated. In such cases it will be sufficient to reduce the first member to the simplest entire terms. The equation will then be in the form ax*+bx = c, (1) in which a and b are integral in form, and prime to each other, and c is entire or fractional. To render the first term of equation (1) a perfect square, multiply both members by a; thus, a*x*-\-abx = ac. (2) Adding -^ to both members, we have V+ ate+l" = ac+?, (3) * \ &" where the first member is a complete square. Now if b is even, will be entire; but if 6 is odd, - will be fractional, a result which 19* 222 QUADRATIC EQUATIONS. we wisli to avoid. To modify the rule to suit the latter case, sup- pose equation (3) to be multiplied by 4 ; thus, 4aV-f4a&x-f& a = 4ac-f&'. (4) The first member is now a complete square, and its terms are entire. Moreover, we observe that (4) may be obtained directly from (1) by multiplying (1) by 4a, and adding l>* to both sides of the result. Hence, for the second method of completing the square in the first member, we have the following RULE. I. Reduce the equation to the form o/ax a -|-bx = c, where a and b are prime to each other. II. If b is even, multiply the equation by the coefficient of x 1 , and add the square of one half the coefficient of x to Loth members. III. If 1) is odd, multiply the equation by 4 times the coefficient of x a , and add the square of the coefficient of x to loth members. The above rule may be considered as more general than the first ; for if applied to equations in the form of x a -f-2ax = I, the opera- tion will be the same as by the first rule, with the simple modifica- tion of avoiding fractions in the first member, when 2a is fractional. 1. Given 5x a Qx = 8, to find the values of x. OPERATION. 5z' Gx = 8. Multiplying by 5, and ad- ") ding 3 a , or 9, to both mem- [ 25z' T -30;c-}-9 = 49 ; bers, 3 by evolution, 5x 3 = 7 ; whence, 5x = 10 or 4 ; or, x = 2 or J. 2. Given ISa;' 55x = 350, to find x. OPERATION. 15x*_55z = 350. Dividing the given equation by 5, 3x* lice = 70 ; multiplying by 12, and ad- j 36x ,_ 132;c+12l = 961 ding 121 to both members, ) by evolution, 6x 11 = 31 ; whence, reducing, x = 7 or J^. AFFECTED QUADRATICS. 223 "We will now apply this rule to an equation in the form of x*-\-2ax = b, where 2a is odd. 3. Given x 2 7x = 44, to find x. OPERATION. x' Ix = 44. Multiplying by 4 times 1, ) 4 x *_28x-|-49 = 225 and adding T to both sides, ) by evolution, 2x 7 = 15 ; whence, x = 11 or 1. Thus we may always operate in such a manner as to avoid frac- tions in the first member ; and indeed in the second member, if we first reduce both members of the equation to entire terms. EXAMPLES FOR PRACTICE. Solve each of the following equations : 1. 5x'-j-4x = 204. Ans. x = 6 or J. 2. 5x'-f 4x = 273. Ans. x = 7 or M. 3. 7 x* 20x = 32. Ans. x = 4 or f . 4. 6x'-f 15x = 9. Ans. x = % or 3. 5. 2x f 5x = 117. Ans. x = 9 or *. 6. 21x* 292x = 500. Ans. x = 11 j$ or 2. 7. 6x' 13x-f 6 = 0. Ant. x = f or j. 8. 7x* 3x ;= 160. Ans. x = 5 or 3 ^. 9. 3x 53x = 34. Ans. x = 17 or f . 10. x-{-13x 140 = 0. Ans. x = 7 or 20. 12. af+llx 80 = 0. Ans. x = 5 or 16. 13. 7x-f- '"^ = 50. ^ns. a; = 2 or W- iU ox 14. - - -j =. =x + 14. Ans. x = 28 or 9. 224 QUADRATIC EQUATIONS. A/2rr _ 1 1 ^ 15. ; =26 4s. ^. *=8 or 5 3x 2 2x 5 10 TREATMENT OP SPECIAL CASES. 292. Either of the two preceding rules is sufficient for the solution of any quadratic equation whatever. There are certain cases, however, where the solution may be much simplified, cither by a -modification of one of the common rules, or by a special prep- aration of the example. 293. When the coefficient of the highest power of the unknown quantity is a perfect square. In this case the equation will be in the form of aV+fcc = c. (1) Let the quantity to be added to complete the square in the first member, be represented by t*. Then oV+fcH-^ = c+t\ (2) Now in any binomial square, the middle term is twice the product of the square roots of the extremes. Hence, -= And equation (2) becomes 4 which may be used as the formula for completing the square, ir this case. Or we may proceed according to the following RULE. Divide the coefficient of x by twice the square root of tA coefficient of x 2 , and add the square of this result to both memo**"* AFFECTED QUADRATICS. 225 1. Given 25x* 20x = 8, to find the values of x. In this example the number to be added to complete the square is and the whole operation will be 25x* 20x = 3, 25x 8 20x-f4 = l, 5x 2=1, 4x 9 x 51 2. Given = ^-, to find the values of x. For this example we have MW)'- &)'=' and the solution is as follows : 49 ~~2 = 63' If! f 49 100 49^ Sf ' 64 ~ 64 ' 2x 7 10 T'~~s~' ~s t x = *f or f J, ^ln. EXAMPLES FOR PRACTICE. 1. 16z' fl2x = 10. Ans. x = ^ or 2. 36x' 5x = T 3 ^. Ans. x=^ or 3. Six 3 12x = J. Ans. x = | or 49x a 6x 40 , 841x 9 58x 5 - --- - = 11. Ans. x= 6. = 2 or - 226 QUADRATIC EQUATIONS. 294. When the equation is in the form of x 2 +2ax=(2a+m)m. (1) In this case we may avoid tedious numerical operations, by the use of the auxiliary quantity, 2a. 1. Given x* 5x = 6, to find the values of x. Put 2a = 5 ; then 2a+l = 6; and the equation becomes x* 2ax = 2a-j-l ; whence, x 3 2ax-\- a 9 = a a -f-2a-f 1, x a=(a+l), x = 2a-|-l or 1, x = 6 or 1, Ans. 2. Given x a -{-19x = 92, to find x. Assume 2a = 19 ; then 4(2a-|-4) = 8a-f 16 = 92 ; and the solution will be as follows : x 3 +2ax = 8a-fl6, x'+2ax4-a a = a'+8a+16, x+a = (a+4), x = 4 or 23, Ans. Let it be observed that we always put the coefficient of x equal to 2a. Then the method will apply, provided the second member is or 4a-}-4, or 6a+9, or 8a-f-16 ; or, in general, 2am-}-m 9 ; that is, any multiple of 2a plus the square of the multiplier. EXAMPLES FOE PRACTICE. Solve the following equations : 1. x 2 Ix = 8. Ans. x = 8 or 1 . 2. x*+llx = 26. Ans. x = 2 or 13. 3. x 8 17 = 60. Ans. x = 20 or 3. 4. .c'-f 21x = 46. ute. x = 2 or 23. AFFECTED QUADRATICS. 227 5. x 9 75x = 76. Ans. x = 76 or 1. 6. x*+12x = 385. Ans. x = 5 or 77. 7. x 8 325x = 3350. Ans. x = 335 or 10. 29o. When large numerals may be avoided in the operation, by the use of an auxiliary quantity. As all the examples of this kind can not be included in any general classification, we give the following illustrations : 1. Given x-f-9984x = 160000, to find the values of x. Put 2a = 10000 ; then 2a 16 = 9984, and 32a = 160000. Whence, x f -f (2a 16)x = 32a, x *_j.(2a 16)*+(a 8) 3 = a a +16a+64, x+(o-8) = (a+8), x = 16 or 10000, Ant. 2. Given x*-f 45z = 9000, to find x. Put a = 45 ; then 200a = 9000. 4z'-f 4az-fa a = a a +800a, 2x+a = l/a(a+800) = 1/45x845. Multiply one of the factors under the radical by 5, and divide the other by 5 ; then we have 2x+a = 1/225x169, 2x+l5 3 = 15 13, 2x = 15 -10 or 15 16, x = 75 or 120, Ans. 3. Given 16x f 225* = 225, to find x. Put 15 = a ; then 16 = a-f-1 ; whence, 4(a+l)V 4 2 O- 2(a-f 1) = 2a f -f 2a or 2o, = a(a-f-l) or a, x = 15 or j, 228 QUADRATIC EQUATIONS. EQUATIONS IN THE QUADRATIC FORM. 29G. There are many equations which, though not really quad- ratic, or of the second degree, may be solved by methods similar to those employed in quadratics. All such equations are reducible to the following form : x* n +2ax = b ; in which x represents a simple or a compound quantity, and n is positive or negative, integral or fractional. It is always necessary that the greater exponent should be twice tJie less. 1. Given x 4 16z a = 28, to find the values of x. OPERATION. ^__16a; a = 28. Adding 8 a , or 64, x 4 16x'-|-64 = 36 ; extracting the square root, x 9 8 = 6 ; by transposition, x 9 = 14 or 2 ; whence, x = 1/14 or zhj/2, Arts. 2. Given x 6x 2 = 5, to find the values of x. OPERATION. x 6z* = 5. Completing the square, x 6a;*-j-9 = 4 ; by evolution, x~ 3 = 2; or, x 3 = 5 or 1 ; involving to the 2d power, x = 25 or 1, -4ns. 3. Given or-S-j-lOar- 1 = 24, to find the values of x. OPERATION. arS-flOz- 1 =24, Completing square, or 2 -j-lOor * -j-25 = 49 ; extracting square root, x~ l -\-b = 7; transposing, x~ l = 2 or 12 ; taking the reciprocals, x = ^ or y^, Am. AFFECTED QUADRATICS. 229 4. Given (x 8 f 2x) a 23(> 3 -f 2x) = 120, to find the values of x. This equation is in the quadratic form, for it contains the first and second powers of the compound quantity, x a -|-2x. For convenience, let us assume y = x a -f 2x. Then by substitution in the given equation, wo have if 2Zy = 120 j whence, y 23y+AJA = y, y-V = 1 y = 15 or 8. We have now the two equations, x*+ 2z = 15 and z'-f 2x = 8, which are solved as follows : x*+2x = 15, x*+2x = 8, x a +2x-f-l = 16, +2aj+l = 9, z+1 = 4, as+l = 3, x = 3 or 5. x = 2 or 4. Thus the equation has four roots, 3, -5, 2, -4; ami it will be found by trial that any one of these four values will satisfy the given equation. Equations of the third and fourth degrees may often be solved like quadratics, even if they do not, at first, present themselves in the quadratic form, like the last equation. If any equation is sus- ceptible of such a solution, it will be found to contain the first and second powers of some compound quantity, with known coefficients. To determine whether this be the fact in any particular case, we may proceed as follows : Transpose all the terms to the first member; and if the highest power of the unknown quantity is not even, multiply the equation through by the unknown letter, to render it even. Then extract the square root to two or more terms, as the case may require ; and if at any time a remainder occurs, which, with or without the abso- lute term, is a multiple, or an aliquot part of the root already ob- tained, a reduction to the quadratic form may be effected. Other- wise it will be impossible. 20 230 QUADRATIC EQUATIONS. 5 Given x 4 4x' 14x a -f-36x-f 45 = 0, to find x. SOLUTION. ' 14x'+36x-f-45 = 0|x*-2x 2x-2x 4x-14x f 4x'-f 4x* 18x a -f36x-|-45 Or by factoring, _18(x 8 2x)-f 45 Hence the given equation may be written thus : (x* 2x) a 18(x a 2x)+45 = 0. Without substituting any letter for the compound quantity, x f 2x, the remaining part of the operation will be as follows : (x 2x)' 18(x a 2x) = 45, (a? 2x) f 18(x'-2x)+81 = 36, (x 8 2x) 9 = 6, whence, re* 2x = 15, (1) or, x* 2x = 3. (2) From (1) we obtain a: = 5 or 3 ) * (2) x = 3 or 1 f 6. Given x*-f-4ax > -f-2a*x 4a* = 0, to find x. As the highest exponent of x is odd, we multiply the equation by x, and obtain x 4 -f-4ax 3 -f 2aV-4a'a; = 0. Taking the square root of two terms, and factoring the remainder, (x'+2axV 2a*(x'-f2ax) = 0. Assume y=x*-f2ax ; then y*2a*y = 0, = a 4 y = 2a* or 0. Hence we have two equations, solved as follows : x*+2ax = 2a a , x s -f-2ezx = 0, x-j-o == dbi/3, x a = 2ax, x =r a;a|/3. x = 2a. Hence, x = a(l y/3), (l-{- 1 /3), or 2a, AFFECTED QUADRATICS. 231 7. Given 25x* 6-f ^ = |, to find x. The two extremes in the first member are perfect squares. We will therefore seek for a middle term which will render the square complete. This will be twice the product of the square roots of the extremes} or We therefore add 1 to both members, and solve as follows : 10z f 1 = 100*:p30.r-ff = 10-fj = y, lOxiF! = 1 10-c = 5 or 2, or 2 or *= ^or :pi, Ans. 8. Given x-\-yx= 21, to find the values of *. OPERATION. x+ tyx+4 = 25, . -i/cc+2 = 5, l /x = B or 7, as = 9 or 49, Ans. It should be observed here that when the equation contains a radical, as in the last example, it cannot be satisfied by the roots obtained, without a trial of signs. The roots found in the last solu- tion are 9 and 49 j but we have !/9 = +3 or 3, ^49"= +7 or 7. Now we may verify the given equation, if we take j/ie =-|-3 or 7 j but not otherwise. 232 QUADRATIC EQUATIONS. Thus, putting x = 9 and ^/x = 3 -in the given equation, we have 9+12 = 21; also with x = 49 and i/x = 7, we have 4928 = 21 ; and the equation is satisfied in both cases. But putting x = 9 and -j/x = 3, we have 912 = 21; also with x = 49 and -j/x = -f-7, we have 49+28=21; both of which are false. In general, it will be found that a radical equation can be satisfied by each of the roots of solution, under at least one of the possible combinations of signs. 2 9. Given 2i/x-f - = 5, to find x. l/x We have here a radical equation which is not in the quadratic form. In such cases, it is generally better to clear the equation of radical signs, either entirely or partially. Thus, 2x+2 = 2x 5!/x = 2, 16aj 40j/x+25 = 9, 4V/X 5 = 3, 4j/x = 8 or 2, l/x = 2 or j, x = 4 or |, Ans EXAMPLES OF EQUATIONS SOLVED LIKE QUADRATICS. 1. x 4 34x 3 = 225. Ans. x = 5 or 3. 2. x 6 35x 3 +216 = 0. Ans. x = 2 or 3. 3. x fl 4x 8 621 = 0. Ans. x = 3 or ^^23. 4. x l +31x 6 32 = 0. ^Tw. x =*2 or 1 . AFFECTED QUADRATICS. 233 5. *_ a = 56. Ans. x = 4 or 6. a* 1 23" = 8. ^n,. *=V4 cr VIT. ? L 7. 2Qx 31x" = 12. .4ns. x = (f ) or (f/. 8. 3 lOrr = 3. ^Lns. x = 27 or 9. X _|_5_if5 = 6. Ans. x = 4 or 1. 10. O-f-12)i-Kse-{-12)l == 6. -4ns. x = 4 or 69. 11. (z+a)^-f2i(:e-f a)? = 3Z S . -4ns. x = i 4 a or 81^ 4 a. 12. x-f l/5x+10 = 8. .4ns. x = 18 or 3. 13. 9x+4-f-21/9x-f4 = 15. Ans. x = | or |. 14. 1/iO+a VicTfi = 2. Ans. x = 6 or 9. 15. (x 5) s 3(> 5)* = 40. Ans. x = 9 or 5-ff / ( 5)*. 16. 2(1+ a x 1 ) (1+oj a; 1 )^-}-^ = 0. Ans. a?=||l/4r or JJVTT 17. z_}_l6_31/a;-f-16 = 10. ^Ins. x = 9 or 12. 18. 81* 2 +17+-\ = 99. Ans. x= 1 or J. 19. 25* 3 + 6+- = -. ^ns. * = 2 or 20. a: 4 +2x 8 7x 9 8a+12 = 0. Ans. x = 1, 2, 2, or 3. 21. x 3 8x a +19rr 12 = 0. ^ns. x = 1, 3, or 4. 22. a 4 10x 8 -f 35x 8 50x-f24 = 0. ^Ins. x = 1, 2, 3, or 4. 23. x* 8az 8 -f 8aV-f32a 3 a; 9a 4 = 0. ^ns. a; = a(2;l/T3) or a(2i/3). 24. y_2cy-l-(c 9 -2y+2^=c. An*. y = 20* QUADRATIC EQUATIONS. PROMISCUOUS EXAMPLES IN QUADRATICS. 1. *+!!* = 80. Ans.x = 5 or -16. 2 3 3# 3 3x 6 4 3 1 1 2{V 1) 4(>+l) = 8* ^ ns - * = 3 or 5 _ "" = 5 or -J. " An*. x = I or 28. 7. a'- 8. ^-^ax+V = 0. . x = 10 _i_ 8a! 20 ' 49 + 2l ^ f ^ ns - = 7 or 11 . ,, ^ a 12r, " 361~~l9" = An *' ^ = 152 or 76. 19 8 16 ' a " *" *' Ans ' x ~ 3 or 1 ' 13. ^+li+l/?+li =42. ^ MS . cc = 5 or 1/38. 14. ^2^61/^=2^+6 = 11. ^tn*. x = 1 or 1 21/15. 17#* 15. a: 4 + -= 34x+16. Ans. x = 2, 2, 8, or . 16. *-l=: . x x = 4 or 7J. 18 . x Y 3|/2- AFFECTED QUADRATICS. ~^~ _2+*' "2|/2~"V8 ' 235 (2 1 /2)4. 19 V* - -Wl -=s ,4ns. x = 1 = 1/?. l+l/l.^ a 1 x = 22. 23. 24. 25. 26. 27. 28. 2a Ofl J " 30 81. 32. = x 2. J.TW. a; = 5 or 3. 9< = 756. 13x w = 6. 243 or = c(l-*). = 352aV . x = V| or Vf c 1 2 ' (c+2) or sfcar 7. a 6 a; a; = a or 5. Ans. x = o. x Ans. x = a-4-# a x a-fx-fl/^ax-f-x* . x = it V X 26 6 s 236 QUADRATIC EQUATIONS. SIMULTANEOUS EQUATIONS CONTAINING QUADRATICS. 3O7. Having treated of quadratics containing one unknown quantity, we will now consider certain cases of simultaneous equations where one or more is of a degree higher than the first. 2O8. In general, the solution of two quadratic equations, involv- ing two unknown quantities, depends upon the solution of a single equation of the fourth degree, containing one unknown quantity. To show this, let us represent the two equations under a general form, as follows : = 0, (1) c' = 0, (2) in which the coefficients a, b, c, etc., and a', I', c r , etc., may have any value, positive or negative, integral or fractional. Arranging the terms in these equations with reference to cc, and factoring, we have a *+(ay+c)x+by'+dy+c = 0, (3) Subtracting (4) from (3), we have l(a-a')y+c-c'lx+(l-b'W+(d-d'^+(c--c>) =- 0. By transposition, we have [(a a'>+c c'] x = (6' 6y-f(rf' whence we obtain (l' This value of x substituted in either equation, (3) or (4), will give a final equation involving only y. Without actually making this substitution, which would lead to a complicated expression, it is obvious that the resulting equation would be of the fourth degree. For the value of x is in the form of ri/+s in which y is involved to the second power. Therefore the term containing cc a in (3), or (4), must involve y to the fourth power. TWO UNKNOWN QUANTITIES. 237 Hence, two equations, essentially quadratic, and containing two unknown quantities, can not, in general, be solved by the rules for quadratics. 299. There are certain cases where simultaneous equations, involving one or more of the unknown quantities to a higher degree than the first, may be solved by means of a final equation in the quadratic form. Most of the examples of this kind are embraced in these three cases : 1st. Where one of the equations is simple, and the other quadratic. 2d. Where both of the equations are quadratic, but homogeneous. 3d. Where one or both of the equations are symmetrical, involving the different letters in a similar manner with respect to coefficients and exponents. The following are illustrations : 1st. SIMPLB AND QUADRATIC EQUATIONS. The solution is effected in this case by the ordinary methods of elimination. 1. Given { 5 *'~ Qxy = 8 I to find x and y. \ 3x 2 = 6 J From equation (2), x = from( l), OPERATION. 5*' Qxy = 8, (1) Sx 2y = 6. (3) 72, I2y = 108 ; 3 21 4y-^ = T ; 4y = 12 or 9; whence, y = 3 or J. and x = 4 or 238 QUADRATIC EQUATIONS. 2d. HOMOGENEOUS EQUATIONS. In the case of homogeneous equations, an auxiliary quantity is employed in the elimination. 2. Given j ^f^ IT } to find the values of x and y ' SOLUTION. Put x = vy } then the given equations become - LI) Y-vy> = 6, or y' = -JL (1) - 6 8 when ec, __ = __, 6+90 = 8e' 4t>, St.* 13 = G, = 2 or -|. Taking v = 2, equation (2) gives y = 1 ; whence a; = 2. Taking v = |, the same equation gives 8 3 ^ = 7^, ', whence x = qi . % = ^ l 7 ^ 1/7 It may be observed here, that in this example, as in all other examples of simultaneous equations, the different sets of values which are capable of satisfying the equations, will be found by tak- ing the signs in their order ; that is, the upper signs should be taken throughout, and the lower signs throughout. Thus, when y = +1, x = -{-2 ; 8 3 7/ _ I ~ _ _ . *' "' TWO UNKNOWN QUANTITIES. 239 3d. SYMMETRICAL EQUATIONS. When the equations are of this description, they may be solved by taking advantage of multiple forms, and of the necessary rela- tions existing between the sum, difference, and product of two quantities. 3. Given \ ^ ~ , t to find the values of x and y. ( xy = 21 ) OPERATION. x+y = 10, (1) xy := 21. (2) Squaring (1), x*+2xy+i/* = 100 ; subtracting 4 times (2), x* 1xy-\-y* = 16 ; by evolution, x y = 4; but in (1), cc-fy = 10; whence, x = 1 or 3 ; y = 3 or 7. 4. Given { ^"^ = ^l to find the values of x andy. ( x*+xy+y* = 133 J OPERATION. Put x ~{~y = s ) an( ^ V^cry =p. Then the given equations become -{-p=:19, (1) '- p = 133. (2) Dividing (2) by (1), s_p = 7; (3) whence from (1) and (3), s = 13, and p = 6 ; that is, x+y = 13, and xy = 36. Proceeding now as in the third example, we have = 169 > xy= 5, x = 9 or 4, = 4 or 9 240 QUADRATIC EQUATIONS. ' 5 Given } X 4 ' ^ 2 = 3 v to find x and y. OPERATION. Put o= then a* = P\ and y? = Q* . Substituting these values in the given equations, From (1) P'-f 2P<2+ ' = 36 j (3) taking (2) from (3), 2PQ = 16 ; (4) taking (4) from-(2), P* 2P-f (? a = 4 ; by evolution, P ^ = 2 ; P=4or2; Q = 2 or 4. whence we have x$ = 4 or 2, y* = 2 or 4 ; x 8 = 64 or 8, y = 32 or 1024. x = 8 or 2j/2. In this example, the auxiliary letters were used to avoid frac- tional exponents in the operation. This practice, however, is not a necessity, but only a convenience. The auxiliary letters should be made to represent the loivest powers of the unknown quantities. 6. Given \ X '+ X V = 208 / ^ find x &nd y (y+x^s^ 1053 ) OPERATION. Assume x 5 = P, y 3 = Q ; then, x^ = P a , ^ = Q* ; Substituting these values in the given equations, and factoring, P(P+Q)= 208 = 13-16, (1) C"(e-fP) = 1053 = 13-81. (2) TWO UNKNOWN QUANTITIES. 2-11 Dividing (2) by (1), 7n=IJj-> (3) J-J- op 4 f) or, C = 'T' and />= ~. = a: . =:6 4 ; from(T), e'=y=:729; ~/ whence, x = 8, and #==27. If we take tlic minus sign in the second member of equation (*), - , we shall obtain /-/?./ *=Sl/=y ; y = 27T/:K'. VX/*/U ( XJ .,, = 8 ) 7. Given ] , , - r \ to find the values of x and y. /- [' s. / t a; -py = loL ) . OP EnATIO ,. - v -+ 1 . f * = 152. (2) Cubing (1), , x'-f 3*yf 3^/ 2 -f^ 8 = 512 j (3) taking (2) from (3), SxV+3-ry* = 360, (4) ay(*H-y) = 120; (5) dividing (5) by (1), xij = 15. (6) Whence, by combining (1) and (6) as in the 3d example, x = 5 or 3, y = 3 or 5. >- - V ^ 3OO. For examples of more than two unknown quantities, ru> additional illustrations arc necessary. The few cases which lead to a final equation in the quadratic form are to be treated by the same y - methods that apply to the preceding. And skill in the management of this whole class of examples, must depend less upon precept than upon practice. . 21 Q 242 QUADRATIC EQUATIOXS. SOS. As ixiliary to the solution of certain questions, particu- larly in geometrical progression, we give the following PROBLEM. Given x-\-y = * and :ry = p, to find the values of x* "HiA J ' 3 -\-y*i x '-\~#*> an d x *-\-y"i expressed in terms of s aud/>. SOLUTION. x+y = *, (1) xy=p. (2) Squaring the first, x*-\-'xy-\-y* =. s a ; .-toy =2;,; 1st result, x*+*. (C7) ) b Multiplying (A) by subtracting 4th result. The following example will illustrate the use of these formulas. 1 . Given \ ' ^ ~ (-to find the values of x and ?/. U<_^-2417) In this example we have s = 9, = 81, s 4 = 6561. Hence, from (C), we have 6561 324p+2p f = 2417 ; p '_162p = 2072, p 9 162p-|-6561 = 4489, p 81 = 67, sry =p = 148 or 14. If we take :ry = 148, the values of x and y will be imaginary, Taking xy = 14, with the equation ^-(-y = 9, we readily obtain x = 7 or 2, y = 2 or 7. 3 or -21. 2. TWO UNKNOWN QUANTITIES. 243 EXAMPLES OP SIMULTANEOUS EQUATIONS. Find the values of tho unknown quantities in the following groups of equations : ! (x-y =15) jx=18or 121, \x-2f = Of Ans - ty = = 1207. * = 22) = 35) = 25 f = 48) a = 3) = 336 ) =40) '( = 72or 8. 7. j-y+Jf- =130) (5x- = 7x i A = x- y) = ^ ' 9. ( 5x - = 25 ' ( = 10 or 45. f Sx'+xy = 336 ) A (x= 28 or 12, '( = 72 = 6. =161) 10. - 3 -2^-^ = ll. (x= l /| f ^ =2 =50) Ans , or i 40. 42. 43. \ ' ( x = 2744 or 8, | y 9604 or 4. = 1009 ) = 582193 \ ' ( x = 81 or 16, An8 ' ) y == 16 or 81. i '-8x =64) , , } y-2x%5= 4J jxV-f^ = 30| u - \ tr jlns. irt.+i . ~< t +( 45. 46. 47. 48 = 3093 ) x-y = 3J- = 7 _ = 2 or 5. I J>_ X ~ 2 ' 26' 1 15 / = -Q or T^' o lo Z = -j Or -r-r. 4 44 THEORY OF QUADRATICS. 247 THEORY OF QUADRATICS. . Having treated of the practical methods of solving quad- ratic equations, we will now proceed to consider certain general principles relating to quadratics. 3O3. Let us resume the general equation, x*+2ax = b. (A) If we solve this equation, and represent one root by r and the other by r 1 ', we shall have + 6, (1) = aV~a*+b; (2) By adding these equations, and also multiplying them together, we obtain r+r' = 2, (3) rr' = -b. (4) That is, 1. The sum of the two roots is equal to the coefficient of x taken with the contrary sign. 2 The product of the two roots is equal to the absolute term taken with the co.itrary sign. 3O4. From equations (3) and (4) in the last article, we have 2a = (r-j-r'), and b = rr'. Substituting these values in (J.), and transposing the absolute term, we have x* (r-f r')x+rr' = j or by factoring, (xr)(xr') = 0. Hence, If all the terms of a quadratic equation be transposed to the first member, the result will consist of two binomial factors, formed by annexing the two roots with their opposite signs to the unknown quantity. 3O>. A Quadratic Expression is one which contains the first and second powers of some letter or quantity. By the principle established in the preceding article, any quadratic expression whatever may be resolved into simple factors. ^ ' 248 QUADRATIC EQUATIONS. 1. Let it be required to resolve the expression, :t*-f-12.r 45, in- to simple factors. Assume rr a -f- I2x 45 = , This equation readily gives c4 f x = 3, x = 15. Hence, a a -{-12.r 45 = (x 3) (x-f-15), Ans. -j 2. Separate 5x a 8x-{-3 into simple factors s We first separate the factor 5 ; thus, r* We may now factor the quantity within the parenthesis, a*s iu tiie last example; thus, i\ 4_ 1 ~5 -5' *> \ ^ . 3 = lor-. ^j -^, v. . r , And the given quantity is factored as follows : x d Sx'-^+S = 5(x-l) (^-J), ^,. O Q OJI ^| h ii EXAMPLES. ^ 1. Resolve x a -|-2a; 120 into simple factors. -'Z- -4iw. (a: 10) (x+ 12.) 2. Resolve x^Qx+ll into simple factors. V* Jf * " "7 x s - ^~ / -*- ^' " ^4s. (x 15) (x 20). 5. Rcsolvcx* into simple factors. .-j,-+.r s J-,, K _j X - r f i- THEORY, OT QUADRATICS. 249 V _JJ-4-l-^_.JL__Jt / ~ ,30 p -** "" S~ X 6. Resolve 15x*-f-19x-{-6 into simple factors. Am. 15(aH-) (x+I). 7. Resolve ex 7 2ax-\-c*x 2ac 3 into simple factors. / * 2a \ ^4ns. c(x -- -J(x-fc 3 ). 3OO. The same principle also enables us to construct an equa- tion whose roots shall be any given quantities. This is done by multiplying together the two binomial factors, which, according to the principle in question, the required equation must contain. 1. Find the equation whose roots shall be \ and \. Factoia, or, Gx'-f-z 1 = 0, Ans. EXAMPLES. 1. Find the equation whoso roots shall be 6 and 15. Ans. x*Sx 90 = 0. 2. Find the equation whose roots shall be 3 and 15. Ans. x a --12x 45 = 0. 3. Find the equation whose roots shall be 16 and 9. AM. a; 8 25x-f 144 = 0. 4. Find the equation whose roots shall be 84 and 1. (V - Ans. x 9 83x 84 = 0. 5. Find the equation whose roots shall be | and . - i 6. Find the equation whose roots shall be | and ^. (Y- , IT* i An , ,.____ =0 . 7. Find the equation whose roots shall be ^ and \. Ans. 8x 9 G.r-j-1 = 0. 8. Find the equation whose roots shall be 2a and c. l^^^l - ' Ana. x* tfac^xfiZac ^= 0. 250 QUADRATIC EQUATIONS. DISCUSSION OF THE FOUR FORMS. 3O7. In the general equation -t 2 -f 2ax = b, the coefficient of x, as well as the absolute term, may be either positive or negative. Hence, to represent all the varieties, with respect to signs, we must employ the four forms, as follows: x*+2ax = -f-fe, (1) x a 2ax = +6, (2) x*+2ax = b, (3) x *2ax = -b. (4) From these equations we obtain x = a+i/'cf+b, (1) x = -j-al/a a -h&, (2) a; = al/a 9 ^ ( 3 ) x = -fartl/a 2 6. (4) "We may now consider what conditions will render these roots real or imaginary, positive or negative, equal or unequal. SO8. Heal and imaginary roots. In the first and second forms, the quantity a a -f-6, under the rad ical, is positive, and the radical quantity is therefore real. But in the third and fourth forms, the quantity a 2 b, under the radical, will be negative when b is numerically greater than a' ; in which case the radical quantity is imaginary. Hence, 1. In ea.ch of the first and second forms, both roots are always real. 2. In each of the third and fourth forms, both roots are imagi- nary when the absolute term is numerically greater than the square of one half the coefficient of x ; otherwise they are real. f ; KO9. Positive and negative roots. Since a~-\-b > a 3 and a 2 b < a 2 , we have vV-f-6 > a and l/a a b < a It follows, therefore, that the signs of the roots in the first and second forms will correspond to the signs of the radical ; but the THEORY OF QUADRATICS. 1 signs of the roots in the third and fourth forms will correspond to the signs of the rational parts. Hence, 1. In each of the first and second forms, one root is positive and the other negative. 2. In the third form both roots are negative, and in the fourth form loth roots are positive. 31O. Equal and unequal roots. It is obvious that in the first and second forms the two roots are always unequal j for in each of these forms, one root is numerically the sum of a rational and a radical part, and the other the difference of the same parts. The same may be said of the third and fourth forms, if we ex- cept the case where a 2 = b ; in which case the roots are eqnal, and we have, for the third form, x = a0 = a or a, and for the fourth form, x = -f-a0 = -fa or -\-a. Hence, 1. In each of the first and second forms, the two roots are always unequal. 2. In each of the third and fourth forms, the roots will be equal tchen the absolute term is numerically equal to (he square of one half the coefficient of x ; otherwise they will be unequal. In the first and third forms, the negative root consists of the sum of the rational and radical parts ; while in the second and fourth forms, the positive root consists of the sum of the two parts. Hence, if we exclude the case of equal roots, 3. In the first and third forms the negative root is numerically greater tlian the positive. 4 In the second and fourth forms, the positive root is numerical- ly greater than the negative. The principles which we have now established, respecting the roots of quadratic equations, are all that are of importance, either theoretically or practically. 252 QUADRATIC EQUATIONS. DISCUSSION OF PROBLEMS. 311. In the solution of particular problems involving quadrat- ics, we shall find that in certain cases both roots of the equation will answer the conditions of the problem, while in other cases only one of the roots is admissible. The reason is, that the algebraic expression is more general in its meaning than ordinary language ; and thus the equation which rep- resents the conditions of the given problem, will sometimes be found to represent the conditions of other analogous problems. 1. A man bought a horse for a certain price. Now if he sells him for $24, he will lose as much per cent, as the horse cost ; re- quired the price of the horse. Let x denote the price. Then x X -777- ? or T7"7T> will be the loss, if 1UU 1UU he sells him for $2-4. Hence, we have or, a: a -100.r = 2400; a' lOOjc-f 2500 = 100 j whence, x 50 = zblO; or, x = GO or 40. Both values of x fulfill the conditions. For, 60X-60 = 36 ; and 6036 = 24. 40X-40 = 16 ; and 4016 = 24. 2. A person bought a number of sheep for $240 if he haa bought 8 more for the same sum, each sheep would have cost $1 less. How many sheep did he purchase ? 240 Let x = the number of sheep purchased ; then - = cost of one. Had he purchased 8 sheep more, the cost of one would havo 240 240 240 been. Hence, - 1 = _ ; reducing, x*'+Sx = 1920 ; a+8;c+16= 193( 5; or, 3+4= 44; whence. x = 40 or ? 48. DISCUSSION OF PROBLEMS. 253 In this case only the first value of x is admissible. The negative result, 48, is numerically the answer to the problem which would be formed by substituting in the above, the word more for the word less, and the word less for the word more. INTEKPRETATION OF IMAGINARY 'RESULTS. Ilti. We have seen that when the absolute term of a quadratic is negative, and numerically greater than the square of one half the coefficient of the second power of the unknown quantity, the roots of the equation will be imaginary. Now the imaginary roots will always satisfy the equation, and it is necessary to ascertain what they indicate respecting the conditions of the problem which the equation represents. 1. Let it be required to divide 20 into two such parts, that their product shall be 140. Let x = one part ; then 20 x = the other. Hence. x(2Qx) = 140 ; or, x 9 20x = 140, x a 20*+100 = 40, x 10 = x = The result is imaginary ; how shall it be interpreted ? Recurring to the problem, we find that the greatest possible product that can be formed by multiplying together two parts of 20, is 10x10 = 100, the product of the halves of 20. Thus we find that the prob- lem is impossible. 2. A farmer would inclose 50 square rods in rectangular form, by a fence whose entire length shall be 24 rods. Required the length and breadth of the inclosurc. Let x = the length, and y = the breadth } then x-\-y 12, and xy = 50 ; x*+2xy+y> = 144, ** 2xy+y* = 56, x y = whence, x = 61/14, y = G^V^ 22 254 QUADRATIC EQU>\TION T 3." Thus, again, the results are imaginary. The problem, however, is impossible. For, if any given area is to be inclosed in rectangular form, the perimeter will be the least when the figure is a square. But the square root of 50 exceeds 7 j hence the field will have a perimeter of more than 28 rods, and can not be inclosed by a fence 24 rods long. We conclude, therefore, That imaginary roots indicate impossible conditions in the problem. PROBLEM OF THE LIGHTS. 313. To illustrate more fully the rules of algebraic interpreta- tion, we present for discussion the following general PROBLEM. Find upon the line which joins two lights, A and B, the point which is equally illuminated by them ; admitting that the intensity of a light at any given distance, is equal to its intensity at the distance 1/divided by the square of the given distance. -^ jr A c B b. In this case, both values of x are positive; therefore, both points of equal illumination are situated to the right of A. The first value of x is less than c ; because -7- - = is less than unity, being a proper fraction. This value of x is also greater than one half of c ; for we have l/a=|/a; (1) and since a > 6, -j/a-f|/6 < 2^/a. (2) Dividing (1) by (2), we have, 256 QUADRATIC EQUATIONS. and consequently, 1 Hence, the first point of equal illumination is at C, between A and 13, but nearer B than A. The second value of x is greater than r for, - is greater I/a yb than unity, being an improper fraction. Hence, the second point is at C', in the prolqngation of the line beyond B. These conclusions arc evidently correct. For, the supposition that a is greater than b, implies that B is the feebler light ; both points should therefore be nearer B than A. d. Suppose a < b. The first value of x is positive. It is, moreover, less tlian one lialf of c. For we have l /a = v /a; (1) and since a < 5, i/-f ;/& > 2j/a. (2) Dividing (1) by (2), we obtain, and therefore, Hence, the first point of equal illumination falls between A and B, and nearer A than B, as it should, because A is the lesser light. The second value of x is negative, since the denominator,y / cr j/Z>, is negative. Now in the statement of this problem, we considered distances reckoned from A toward the right as positive j hencs, according to the rule for interpreting negative results, previously established, (183), we must consider the negative result in this case, as a distance to be reckoned from A toward the left. Heiice, the second point will be situated to the left of A, at C". And this is as it should be, because A, under the present supposition, is the lesser light. PROBLEM OF THE LIGHTS. 257 3d. Suppose a b. In this case, the first value of x is positive, and equal to -^ u Hence the first point of equal illumination is midway between A andB. The second value of x is -~ = oo. This result indicates that there is no other point of equal illumination in the line AB, or in AB produced, at a finite distance from A. These conclusions are obviously correct. For, under the present supposition, the two lights are equally intense. Hence any point, to be equally illuminated by them, must be equally distant from them ; and the only point which fulfills this condition is the point midway between them. If, however, we consider a and b as two varying quantities, at first unequal, but continually approaching equality, then the second value of x will become greater and greater by degrees, until it reaches infinity. Under these conditions, the second point of equal illumi- nation will continually recede from A, moving toward the right or toward the left, according as a is greater, or less than b, until it is finally removed to an infinite distance. In this view of the case, it is sometimes said that there are two points of equal illumination, under the hypothesis, a =. b' } one point being at an infinite distance from A. 4 tli. Suppose a = b and c 0. The first value of x reduces to = : hence the first point Zj/a is situated at A. The second value of x is - , the symbol of indetermination ; (188, 4). This result shows that there arc an infinite number of other points equally illuminated by the two lights. These interpretations are evidently correct. For, as the lights, under the present hypothesis, are equally intense, and both situated at A, every point in space must be equally illuminated by them 5. Suppose c = 0, and a > b or a < b. Both values of x now reduce to j and the common rule for interpreting zero might lead us to suppose that the two points of 22* R 258 QUADRATIC EQUATIONS. equal illumination coincide with the point A. But this conclusion is not strictly correct; for it is obvious that when two lights, of unequal intensities, occupy the same place, there is no point in space equally illuminated by them ; not even the point in which they are both situated. Let us return to the original equation (m), which truly represents the conditions of the problem. If we put c = 0, the result is a _ b x a - oT 8 '^ ^-2..C + > - an equation which can not be satisfied by any value of x whatever, while a > I or a < b. For by substituting any value for x we shall always obtain two unequal fractions. If x = 0, the two mem- bers are two unequal infinities. We conclude, therefore, that under the supposition, c = 0, while a and b are unequal, the problem fails altogether, and is impossible. Thus we learn that zero may be the answer to a possible, or an impossible problem. And whenever we obtain this symbol as the result of a solution, we must not interpret it on the assumption that the thing required in the problem is possible ; but we must first determine whether the conditions are rational or absurd, by consid- ering the nature of the problem, or by substituting zero in the original equation. PROBLEMS PRODUCING QUADRATIC EQUATIONS. . It will be found that some of the following problems may be solved by a single unknown quantity, while others require two. Still others may be conveniently solved by means of either one or two letters. It is left to the judgment and skill of the learner to discover the mode of solution, in each example, which is most simple. 1. It is required to divide the number 14 into two such parts, that 9 times the quotient of the greater divided by the less, may be equal to 16 times the quotient of the less divided by the greater. Ans. 8 and 6. "-- PROBLEMS PRODUCING QUADRATICS. 259 2. A company, dining at an inn, agreed to pay $3.50 for the ""entertainment; but before the bill was presented, two of the party j. 3 <- left, in consequence of which each of the others had to pay 20 cents more than if all had been present. How many persons dined ? Ans. 7. .4' ^ 3. There is a certain number, which being subtracted from 22, and the remainder multiplied by the number, the product will be 117. What is the number ? An*. 13 or 9. 4. It is required to divide the number 18 into two such parts, that the squares of these parts may be to each other as 25 to 16. I .// / t~ ', '-o Y / i ' - Ans. 10 and 8. 5. The difference of two numbers is 4, and their sum multiplied by the difference of their second powers, is 1600. What are the f ~ Q (3 numbers ? 2-- , - '^) - / - u-G Ans. 12 and 8. ? 6. What two numbers are those whose difference is to the less as 4 to 3, and whose product multiplied by the less is equal to 504 ? Ans. 14 and 6. 7. A man purchased a field, whose length was to its breadth as 8 to 5. The number of dollars paid per acre was equal to the number of rods in the length of the field ; and the number of dollars given for the whole was equal to 13 times the number of rods round the field. Required the length and breadth of the field. Ans. Length, 1 04 rods ; breadth, 65 rods. 7 y. 8. There is a stack of hay, whose length is to its breadth as 5 to 4, and whose height is to its breadth as 7 to 8. It is worth as many cents per cubic foot as it is feet in breadth ; and the whole is worth at that rate 224 times as many cents as there are square feet on the 'ritV bottom. Required the dimensions of the stack. Ans. Length, 20 feet ; breadth, 16 feet ; height, 14 feet. 9. There is a number, to which if you add 7 and extract the square root of the sum, and to which if you add 16 and extract the square root of the sum, the sum of the two roots will be 9. What is the number? >/L -f- t l^? ~tf Ans. 9. NOTE. Represent the number by ar 7. y ' 10. A and B together carried 100 eggs to market, and each received the same sum. If A had carried as many as B, he would "Y 260 QUADRATIC EQUATIONS. have received 18 pence for them ; and if B had taken as many as A, he would have received 8 pence. How many had each ? Ans. A 40, and B 60. 11. The sum of two numbers is 6 ; and the sum of their cubes is 72. What are the numbers ? Ans. 4 and 2. 12. A man traveled 36 miles in a certain number of hours; if he had traveled one mile more per hour, he would have required 3 hours less to perform his journey. How many miles did he travel per hour ? | * - ^ -t 3 )d ^-t Y- ~ / V Ans. 3 miles. ^ ' ; 13. The sum of two numbers is 100, the difference of their I/H ~ 'Square roots is two ; what are the numbers? Ans. 36 and 64. 14. A gentleman bought a number of pieces of cloth for 675 dollars, which he sold again at 48 dollars a piece, and gained by the bargain as much as one piece cost him. What was the number of pieces? ^ V * /rw * Ans. 15. r ? q 15. A merchant sold a piece of cloth for 39 dollars, and gained is much per cent, as it cost him. What did he pay for it ? Ans. $30. 16. A merchant sent for a piece of goods and paid a certain sum for it, besides 4 per cent, for carriage ; he sold it for $390, and and thus gained as much per cent, on the cost and carriage as the 12th part of the purchase money amounted to. For how much did lie buy it? 3$- Ans. 8300. <017. ^ Ans. A, 36 ; B, 30. _ 18. Divide the number 60 into two such parts that their product shall be 704. ^ , ~ * s - 7 V Ans. 44 and 16. 19. A vintner sells 7 dozen of sherry and 12 dozen of claret for ( 50, and finds that he has sold 3 dozen more of sherry for 10 than he has of claret for 6. Required the price of each. Ans. Sherry, 2 per dozen: claret, 3. y */ T.M c PROBLEMS PRODUCING QUADRATICS. 261 20. A set out from C towards D, and traveled 7 miles a day. After he had gone 32 miles, B set out from D towards C, and went t+j l-^vt/y^day T ' g of the whole journey; and after he had traveled as many days as he went miles in a day, he met A. Required the' distance from C to D. Ans, 76 or 152 miles. , 21. A farmer received $24 for a certain quantity of wheat, and an equal sum at a price 25 cents less per bushel for a quantity of '' barley, which exceeded the quantity of wheat by 16 bushels. How many bushels were there of each ? Ans. 32 bushels of wheat and 48 of barley. .22. Two travelers, A and B, set out to meet each other, A leaving 2, and whose sum, multiplied by the sum of their squares, is 272^$} Ans. 5 and 3. 'y 7 ;/- i?"^) 24. A and B hired a pasture at a certain rate per week, agreeing A.C*- 3that each should pay according to the number of animals he should , , , have in the pasture. At first A put in 4 horses, and B as many as cost/ him 18 shillings a week ; afterward B put in 2 additional 1 houses, 'and found that he must pay 20 shillings a week. At what rate was the pasture hired ? Ans. 30 shillings per week. + y 25. If a certain number be divided by the product of its two aigits, the quotient will be 2 ; and if 27 be added to the number, the dibits will be inverted. What is the number ? Ans. 36. 26. It is required to find three numbers, such that the difference of the fi~st and second shall exceed the difference of the second and third by 6, the sum of the numbers shall be 08, and the sum of the squares 441. Ans. 4, 13, and 16. 27. What two numbers are those whose product is 24, and whose sum added to the sum of their squares is 62 ? Ans. 4 and 6. 262 QUADRATIC EQUATIONS. 28. It is required to find two numbers, such that if their product be added to their sum, the result shall be 47 ; and if their sum be taken from the sum of their squares, the remainder shall be 62. \. , ^ -_ L -j i t, j , 7 /, V V -^ ^* - fJ& Ans. 7 and 5. t "*2r^_^~> f tn> 7 NOTE. In many examples of two unknown quantities, giving rise to symmetrical equations, it will be found convenient to denote one of the unknown quantities by x+y, and the other by x y. 29. The sum of two numbers is 27, and the sum of their cubes is 5103. What are the numbers ? An*. 12 and 15. 30. The sum of two numbers is 9, and the sum of their fourth powers is 2417. What are the numbers ? Am. 1 and 2. N _ Cl. The product of two numbers multiplied by the sum of their squares, is 1248 ; and the difference of their squares is 20. What are the numbers ? Ans. 6 and 4. L 32. Two men are employed to do a piece of work, which they can finish in 12 days. In how many days could each do the work I , p rov id e( i it would take one 10 days longer than the other Ans. One in 20 days ; the other in 30 days. - The joint stock of two partners was $1000 ; A's money was in trade 9 months, and B's 6 months ; when they shared stock and ^' gain/A' receive^ $1,140 and B $640. What was each man's stock ? Ans. A's, $600 ; B's, $400. 34. A speculator, going out to buy cattle, met with four droves. In the second were 4 more than 4 times the square root of one half I 2the number in the first ; the third contained three times as many /as the first and second ; the fourth was one half the number in the third, and 10 more; and the whole number in the four droves was 1121. How many were in each drove ? ^ ^"^ 2. y i. Ans. 1st, 162; 2d, 40 ; 3d, 606 ; 4tli, 313. f 35. Find two numbers, such that if the sum of their squares be subtracted from three times their product, 11 will remain; and if the difference of their squares be subtracted from twice their prod- uct, the remainder will be 14. Aim. 3 and 5. 36. Divide the number 20 into two such parts, that the product of their squares shall be 9216. Atm. 12 and 8. n y * 96 PROBLEMS PRODUCING QUADRATICS. 263 37. Divide the number a into two such parts, that the product "^ of their squares shall be b. ( Greater part, | + - (a 9 Ans. 1 . i I Less part, | ^ * "" ^ 38 The greater of two numbers is a 9 times the less, and the /M 6 product of the two is i 1 . Find the numbers. Ans. - , and ab. 39. A certain number is formed by the product of three consec- utive numbers; and if it be divided by each of them in turn, the sum of the quotients will be 74. What is the number ? . j 120; that is, 4-5'6; or ( 120 ; that is, (4) (5) (6). 40. An engraving, whose length was twice its breadth was mounted frJt* on Bristol board, so as to have a margin 3 inches wide, and equal in area t'o the engraving, lacking 36 inches. Find the width of t+JC the engiaving. '^tf* Ans. 12 inches. '41. A man has two 'square" lots of unequal dimensions, containing ', together 25 A. 100 P. If the lots were contiguous to each other, it would require 28 J rods of fence to embrace them in a single in- 7 ~ closure of six sides. Required the dimensions of the two lots. -^ Ans. 62 rods and 16 rods, 50 rods and 40 rods. - '- 42. A person has 1300, which he divides into two portions, and lends at different rates of interest. He finds that the incomes from the two portions are equal ; but if the first portion had been lent at the second rate of interest it would have produced 36, and if the second portion had been lent at the first rate of interest it would have produced 49. Find the rates of interest. A^s. 7 -and 6 per cent. ' ^ se * s out fr m Condon to York, and B at the same time YL 2-/ from York to London, both traveling uniformly. A reaches York 25 hours, and B reaches London 36 hours, after they have met on the road. Find in what time each has performed the journey. Ans. A, 55 hours ; B, 66 hours. 44. A owns a village lot, in square form, containing 36 square rods ; B owns the adjacent lot on the same street, which is also a M **- 264 QUADRATIC EQUATIONS. square, but greater than A's. Now if A should purchase all the front of B's lot, so as to extend the rear boundary line of his own through B's lot, parallel to the street, the two neighbors would pos- sess equal quantities of land. Find the length of one side of B's lot. Ans. 6(l-f-|/2) rods. 45. There are three numerical quantities having the following relations to each other ; the sum of the squares of the first and second, added to the first and second, is 32 ; the sum of the squares of the first and third, added to the first and third, is 42 ; and the sum of the squares of the second and third, added to the second and third, is 50. Required the quantities. f 1st, 3 or 4 ; Ans. J 2d, 4 or 5 : (3d,5or-G. 46. "What is'llic side of that cube -which contains as many solid units as there arc linear units in the diagonal through its opposite corners. Ans. V3. 47. It is required to find two quantities such that their sum, their product, and the sum of their squares, shall all be equal to each other. Ans. (3+1/^3), and i(3q:l/^). 48. Find those two numbers whose sum, product, and difference of their squares, are all equal to each other. Ans. J(l3d:|/5) f and j(l^v^5). 49. Find two numbers, such that their product shall be equal to the difference of their' squares, and the sum of their squares shall be equal to the difference of their cubes. Ans. i|/5, and ^1/5). J PROPORTION. 265 SECTION VI. PROPORTION. AND THE THEORY OF PERMUTATIONS AND COMBINATIONS PROPORTION. 31*>. Two quantities of the same kind may be compared, and their numerical relation determined, by finding how many times one contains the other. This mode of comparison gives rise to ratio and proportion. 31G. Ratio is the quotient of one quantity divided by another of the same kind regarded as a standard of comparison. There are two methods of indicating the ratio of two quantities. 1st. By writing the divisor before the dividend, with two dots between them; thus, a : b indicates the ratio of a to b, where a is the divisor and b the dividend. 2d. In the form of a fraction ; thus, the ratio of a to I may bo written b a 317. A Compound Ratio is the product of two or more ratios. Thus, Simple ratios, I '* . The Extremes in a proportion are the first and fourth terms. 326. The Means in a proportion are the second and third terms. 327. When the first of a series of quantities has the same ratio to the second, as the second has to the third, as the third to the fourth, and so on, the several quantities are said to be in continued proportion, and any one of them is a mean proportional between the two adjacent ones. Thus, if a : I b : c = c : d = d : e, then tf, b, c, d, and e are in continued proportion, and b is a mean proportional between a and c, c a mean proportional between b and d; and so on. 328. One quantity is said to vary directly as another when the two quantities, by reason of their mutual dependence, have always a constant ratio, so that if one be changed the other will be changed in the same proportion. Thus, for illustration, suppose, in the purchase of a commodity, a certain quantity, A, costs a certain sum, B. Now if the price of unity remain the same, it is evident that 2 A will cost 2B ; 3 A will 267 cost 3j8 ; and in general, mA will cost mJS. In this case the cost is said to vary directly as the quantity. 32O. One quantity is said to vary inversely as another when the first has a constant ratio to the reciprocal of the other. 330. One quantity is said to vary as two others jointly, when it has a constant ratio to the product of the two. 331. The Sign of Variation is the symbol oc ; thus, the expres- sion, A oc J3, signifies that A varies as B. From the definition of variation, it is evident that the expression, A oc B, is equivalent to the proportion, A: = m:l, where m is a constant. This proportion gives A = mB. Hence the general truth, If A vary as J5, then A is equal to B multiplied by some constant guantity. PROPOSITIONS IN PROPORTION. 339. A Proposition is the statement of a truth to be demon- strated, or of a problem to be solved. 333. A Scholium is a remark showing the application or limit- ation of a preceding proposition. 334. If in the proportion a : b =. c : d, the second method of indicating ratio be employed, we have -:-t which is the fundamental equation of proportion ; and any proposi- tion relating to proportion will be proved, when shown to be con- sistent with this equation. PROPOSITION I. In every proportion, the product of the ex- tremes is equal to the product of the means. 268 PROPORTION. Let a : b == c : d represent any proportion j then by formula (A\ - = - ; a c clearing of fractions, be = ad. That is, the product of b and c, the means, is equal to the prod- uct of a and d, the extremes. SCHOLIUM. From the last equation, we have The first mean, c r The second mean, The first extreme, The second extreme, Hence, 1st. Either mean is equal to the product of the extremes divided by the other mean. (1) 2d. Either extreme is equal to tJie product of the means divided by the other extreme. (2) PROPOSITION II. Conversely : If the product of two quantities is equal to the product of two others, then two of them may be taken for the means, and the other two for the extremes of a proportion. Let be = ad. Dividing by ac, - = - > a c hence by formula (A), a : b = c : d, in which the factors of the first product, be, arc the means, and the factors of the second product, ad, are the extremes. PROPOSITION III. If four quantities be in proportion, they will be in proportion by ALTERNATION ; that is, the antecedents will be to each other as the consequents. Let a : b =c: ?; then by formula (A\ - = - J (1) a c PROPERTIES. 269 be ^multiplying (1) by c, = d ; (2) dividing (2) by b t - = j (3) hence, a : c = b : d, in which a and c, the antecedents of the given proportion, are pro- portional to b and d, the consequents of the given proportion. PROPOSITION IV. If four quantities be in proportion, they wittbe in proportion by INVERSION; that is, the second will be to the fast, as the fourth to the third. Let a : b = c : d ; b d then by formula (A), = > clearing of fractions, bc = ad; hence by Prop. II, 6 : a = d : c. SCHOLIUM. The last two propositions axe but modifications of Prop. II. Thus we learn that from every equation three different forms of proportion may be derived. Let i ad = be ; then, > J a : b = c : d} or, * a : c =b : d; ^ or, b : a = d : c. , PROPOSITION V. Quantities which are proportional to the same quantities are proportional to each other. If a : b m : n, (1) and c : d = m : n, (2) we are to prove that a : b = c id. From (1), - = - ; a m from (2), - = -; c m b d, hence, = ; a c or, a : b = c : d. 23* 270 PROPORTION. PROPOSITION VI. If four magnitudes be in proportion, they must be in proportion by COMPOSITION or DIVISION; that is, the first is to the sum or difference of the first and second, as the third is to the iwm or difference of ike third and fourth. If a : b = c : d, we are to prove that a : a+b = c : c+d. By formula U), - = -: (1) whence, 1-^ 1-1 , 0J) a c 1-1=1-1; (3) a c from (2), ^ZL _ a c f /ox ft c ^ from (o), = j a c hence from (4), a : a-f-6 = c : c-\-d; and from (5), a : a 6 = c : c d. SCHOLIUM. In 4 like manner, it may he sho - <+ ^PROPOSITION VII. If four quantities be in proportion, the sum of the first and second is to their difference, as the sum of the third and fourth is to their difference. If a : b = c : d, we are to prove that a-\-b : a b = c-\-d : c d. By Prop. VI, a : a-\-b = c : c-fd; (1) also, a : a 6 = c : c d ; (2) from(1)> _ = . a c a b c d from (2), (4) C a b c d dividing (4) by (3), = ; whence, a-j-6 : a b = c-\-d : c d. PROPERTIES. 271 PROPOSITION VIII. If there be a proportion, consisting of three, or more equal ratios, then either antecedent will be to its consequent, as the sum of all the antecedents is to the sum of all the consequents. Suppose a : b = c : d = e : f= g : h =, etc. Then by comparing the ratio, a : b, first with itself, and after- ward with each of the following ratios in succession, we obtain ab = ba, ad = bc y af=be, ah =-; by, etc. ; whence, a(b+d+f-\-h+ etc.) = b(a-\-c+e+g+ etc.), or, a:b = a+c+e+g+ etc. : b+d-\-f+h+ etc. PROPOSITION IX. If four quantities be in proportion, the term** of eitlier couplet may be multiplied or divided by any number, and the results will be proportional. Let a : b = c : d ; b d then, - = - And since the value of a fraction is not changed by multiplying or dividin" 1 both of its terms by the same number, we have =* -: in which n may be either integral or fractional. If n be integral, we have, from (1) and (2), na : nb = c : d, (3) a : b =. nc : nd / (4) in which the terms of the given couplets are multiplied. But put n = ; then (3) and (4) become m' "- ;-=*. .d, (5) mm (6) m m in which the given terms are divided. 272 PROPORTION. PROPOSITION X. If four quantities be in proportion, either the antecedents or the consequents may be multiplied or divided by any number, and the results iu cccry case will be proportional* Let a : b = c : d ; then, l - = * e . (1) nit nd whence, = ; (2) y (3) in which to, may be cither integral or fractional. If n be integral, we have from (2) and (3), a : nb = c : nd; (4) na : b = nc : d} 0) in which the given antecedents and consequents arc multiplied. Put n = ; then (4) and (o) become m b_ d ' m ~~ " in a c : b = : ; 77i m in which the given antecedents and consequents are divided. PROPOSITION XI. I/four quantities icliich are in proportion, be multiplied or divided, term by term, by four other quantities also in proportion, the products, or quotients, taken in order, will be propor- tional. If a:b = c:d, (1) and x : y = m : n, (2) then we are to prove that ax : by t= cm : dn, a b c d and := : x y m n From (1) and (2), we obtain ad = be ', (3) xn = ym ; (4) PROPERTIES. 273 multiplying (3) by (4), (ax)(dn) = (%)(cm) ; (5) whence, from (5), ax :by =. cm : d/i ; , P ,, a & c cZ and trom (o) _._.-._:_. x y m n PROPOSITION XII. If four quantities be in proportion, Wee powers or roots of the same quantities will be in proportion. Let a : b = c : d', b d then = m a c W Raising (1) to the nth power, also taking tho th root of the same, i i b d* (3) Hence from (2), a* : b* = c n : d* ; JL L LI. and from (3), a" : b n = c n : d* . PROPOSITION XIII. If three quantities be in continued propor- tion, the product of the extremes is equal to the square of the mean. Let a : b = b : c ; then by Prop. I, ac = bb = b*. SCIIOLIUM. Taking the square root of the last equation, we have b = V ac ; hence, The mean proportional between two quantities is equal to the square root of their product. PROPOSITION XIV. If three quantities be in continued propor- tion, the first is to the third, as the square of the first is to the square of the second ; that is, in the duplicate ratio of the first and second. Let a : b = b : c ; then b 9 = ac ; multiplying by a, ab* = a*c ; whence, by Prop. II, a : e = a 9 : 6*. S 274 PROPORTION. PROPOSITION XV. If four quantities be in continued propor- tion, the first is to the fourth, as the cube of the first is to the cube of the second; that is, in the triplicate ratio of the first and second. Let a : b = b : c = c : d', then ac = b 9 , (1) and c* = bd; (2) multiplying (1) by (2), ac 8 b*d ; whence, by Prop. II, a : d = b 9 : c* ; or, a : d = a 8 : b*. PROBLEMS IN PROPORTION. To show some of the applications of the preceding principles, we give the following problems : 1. Find two numbers, the greater of which shall be to the less as their sum to 42, and as their difference to 6. Let x = the greater, and y = the less. By the conditions, \ * C x : y = x y : 6. (2) Prop. V, x+y : 42 = x y : 6, (3) Prop. Ill, x+y : xy = 42 : 6, (4) Prop. VII, 2x : Zy = 48 : 36, (5) Prop. IX, x:y=4:3, (6) From (1) and (6), Prop. V, 4 : 3 = x+y : 42, (7) (2) (6), 4:3 = a: ; y : 6, (8) " (7), Prop. I, x+y = 56, " (8) " " a: y = 8; whence, x = n . Ans. and y = 24 j 2. Divide the number 14 into two such parts that the quotient of the greater divided by the less, shall be to the quotient of the less divided by the greater, as 16 to 9. PROBLEMS. 275 Let x = the greater, then 14 x = the less. x 14. _ x By the conditions. : - = 16 : 9. J 14 x x Multiplying terms, Prop. IX, x* : (14 z) a = 16 : 9, extracting square root, x : 14 x = 4 : 3, by composition, Prop. VI, x : 14 = 4 : 7, dividing consequents, x : 2 =. 4 : 1, x = S] whence, a c Ans. 14 x = 6 ) 3. There are three numbers in continued proportion; their sum is 52, and the sum of the extremes is to the mean as 10 to 3. Re- quired the numbers. Three numbers in continued proportion may be represented by x, xy, xy* ; for we observe that the product of the extremes will then be equal to the square of the mean. Hence, * 52 (1) = 10 ' : 3. (2) From (2), y f +l:y = 10:3, (3) or, y-fl : 2y = 10 : 6, (4) by Prop. VH, #'H-2y+l : y' 2^+1 = 16 : 4, taking square root, #-|-l : y 1 = 4:2, by Prop. VII, 2y : 2 = 6 : 2, or, y:l = 3:l, whence, y = 3 | and from (1), ic = 4 j 4. The product of two numbers is 112 ; and the difference of their cubes is to the cube of their difference, as 31 to 3. What are the numbers ? By the conditions, \ ay = 1 12, U- * : x- s = 31 : 3. 2) t x-L By the cond.Uons, { - y* : (x-y) s = 31 : 3. (2) From (2), Prop. IX, x'-fry-f^ 9 : tfZxy+y* = 31 : 3, (3) by Prop. VI, 3;ry : (x yj = 28 : 3, (4) by substitution, 336 : f (xt/y = 28 : 3, (5) whence, / ^ \x~yy = 3^/ * (6) or, x y = 6. (7) From (1) and (7), we obtain x = 14, y = 8. y -- 276 PROPORTION. 5. What two numbers are those whose difference is to their sum as 2 to 9, and whose sum is to their product as 18 to 77 ? Let x and y represent the numbers. By the conditions, \ x ~^ : X +V = 2: > < 1 > ( x+y : xy = 18 : 77. (2) From (1), Prop. VII, 2x : 2y 11 : 7, (3) From (2), by substitution, -y : -- = 18 : 77, (5) by Prop. IX, Ify : 1 V = 18 : 77, (C) or, 18 : lly = 18 : 77, (7) or, 1 : y = 1 : 7, (3) whence, y 7. y. // 6. Two numbers have such a relation to each other, that if 4 be added to each, they will be in proportion as 3 to 4 ; i.nd if 4 be / subtracted from each, they will be to each other as 1 to 4. What are the numbers ? Ans. 5 and 8. 7. Divide the number 27 into two such parts, that their product shall be to the sum of their squares as 20 to 41. Ans. 12 and 15. 8. In a mixture of rum and brandy, the difference between the quantities of each is to the quantity of brandy, as 100 is to the number of gallons of rum; and the same difference is to the quan- tity of rum, as 4 to the number of gallons of brandy. How many gallons are there of each? Ans. 25 of rum, and 5 of brandy. 9. There are two numbers whose product is 320 ; and the differ- ence of their cubes is to the cube of their difference as 61 to 1 . What are the numbers? Ans. 20 aj:d 16. NOTE. In the last example, put x+y the greater, and x y = the less. 10. Divide 60 into two such parts, that their product shall be to the sum of their squares as 2 to 5. Ans. 40 and 20. 11. There are two numbers which are to each other as 3 to 2. , v If 6 be added to the greater and subtracted from the less, the sum p / f PROBLEMS. 277 and the remainder will be to each other as 3 to' 1. What are the numbers ? Ans. 24 and 16. 12. There are two numbers which are to each other as 16 to 9, and 24 is a mean proportional between them. What are the num- bers ? 4. Ans. 82 and 18. 13. The sum of two numbers is to their difference as 4 to 1, and the sum of their squares is to the greater as 102 to 5. What are the numbers ? Ans. 15 and 9. \ ^^;; 0}4y The number 20 is divided into two parts, which are to each oilier in the duplicate ratio of 3 to 1. Find the mean proportional - L, between these parts. Ans. 6. 15. There are two numbers in the proportion of 3 to 2 ; and if 6 be L v added to the greater, and subtracted from the less, the results will be as 9 to 4. What nrc llio numbers ? Ans. 39 and 26. ^ 16. There arc three numbers in continued proportion. The I product of the first and second is to the product of the second and third, as the first is to twice the second ; and the sum of the first and third is 300. What are the numbers ? Ans. 60, 120 and 240. 17. The sum of the cubes of two numbers is to the difference of their cubes, as 559 to 127 ; and the square of the first, multiplied by the second, is equal to 294. What arc the numbers ? Ans. 7 and 6. 18. The cube of the first of two numbers is to the square of the second as 3 to 1, and the cube of the second is to the square of the first as 96 to 1. What arc the numbers ? Ans. 12 and 24. 19. Given the proportion (a;-f- 1) 4 : (x1) 4 = 2(x+l) 9 : (x 1), to find the value of x. i/24-l 20. Prove that a : b = c : d, when (o+6-|-c+rf) (a I c+d) = (a 6-f-c d) (a-j-& c - \i -pa./ Now Z? is greater or less than Z, according as is greater or less than unity. That is, when r+l" the number of combinations will be increased by giving to r its succeeding value ; but when n r the number of combinations will be diminished by giving to r its succeeding value. ^+> / r \ ? * - ^~ AND COMBINATIONS. 283 But if And if Hence, that value of r which will give the greatest number of combinations, must not be less than , or greater than - -- [-1, , . A . r f Al Al . j hence, it will have one of the three values, S. 'V - //'*/' t^/^v 1st. Suppose n even,. Then the first and third values will be fractionalj and therefore impossible for r ; hence in this case 2d. Suppose n odd. Then the second value will be fractional, and consequently impossible for r ; hence, in this case r must have at least one of the other values. We will show that it may have either of them. For, suppose ' By (34O), the number of combinations will be the same, if n+l That is, when n is odd, the greatest number of combinations will be obtained by making > Ti1 n-4-l r= or r = , the two values of r giving the same result. EXAMPLES OF PERMUTATIONS AND COMBINATIONS. 1. How many different permutations may be formed of 10 letters taken four at a time ? Ans. 5040. 2. How many different permutations may be made of 6 things taken all together in a set ? Ans. 720. 284 PERMUTATIONS AND COMBINATIONS. 3. How many different permutations may be made of 10 things taken all together ? Ans. 3628800. 4. How many different numbers can be formed with the five Arabic characters, 4, 3, 2, 1, ; each of the characters occurring once, and only once in each number ? Ans. 90. 5. How many different combinations may be formed of 8 things taken 4 at a time ? Ans. 70. 6. How many different combinations may be made of 16 things taken 5 at a time ? Ans. 4368. 7. How many different parties of 6 men each can be formed from a company of 20 men ? Ans. 38760. 8. In how many different ways can a class of 6 boys be placed in line, one boy being denied the privilege of the head ? Ans. 60CT. 9. Find the greatest number of different products that can be formed with the prime numbers under 40, the products being all composed of the same number of factors. Ans. 1716. 10. The number of permutations of n things taken 5 at a time, is equal to 120 times the number of combinations of the n things taken 3 at a time ; find n. Ans. n 8. 11. At a certain house there were 8 regular boarders ; and one of them agreed with the landlord to pay $35 for his board so long as he could select from the company different parties, equal in number, to sit each for one day on a certain side of the table. At what price per day did he secure his board ? Ans. $.50. 12. A and B have each the same number of horses j and A can make up twice as many different teams by tj.kiug 3 horses together, as B can by taking 2 together. Required the number of horses that each has. Ans. 8. 13. There are 12 points in a plane, no three of which are in the same straight line with the exception of five, which are all in the same straight line. How many different straight lines can be formed by joining the points. Ans. 57. ARITHMETICAL PROGRESSION. 285 SECTION VII. *- OF SERIES. 342. A Series consists of a number of terms following one another, but so related that each may be derived from one or more of the preceding, by a fixed law. A scries may be finite or infinite, converging or diverging. 343. A Finite Series is one which by its law of development must terminate, or have only a finite number of terms. 344. An Infinite Series is one which by the law of its devel- opment can never terminate, but may have an infinite number of terms. 345. A Converging Series is one whose successive terms con tinually diminish in numerical value. 346. A Diverging Series is one whose successive terms con- tinually increase in numerical value. ARITHMETICAL PROGRESSION. 347. An Arithmetical Progression is a series of numbers or quantities increasing or decreasing from term to term by a common difference. We may consider the common difference as a quantity continually added, in the algebraic sense ; hence, it will be positive in an in- creasing series, and negative in a decreasing series. Thus, 1, 3, 5, 7, 9,.... is an increasing arithmetical progression, in which the common dif- ference is -f-2 j and 20, 18, 16, 14, 12,.... is a decreasing arithmetical progression, in which the common dif- ference is 2. 286 SERIES. 348. To investigate the properties of an arithmetical progres- sion, we may suppose the series to terminate ; there will then be five parts or elements; the first term, the last term, the number of terms, the common difference, and the sum of the terms. The first term and last term are called the extremes, and all the terms between the extremes are called arithmetical means. $4:O. In an Arithmetical Progression, the last term is equal to the first term plus the common difference multiplied by the number of terms less 1. Let a denote the first term, I the last term, d the common differ- ence, and n the number of terms j then the series will be represented thus : a, (a+rf), (a+2d), (a+3O. In an arithmetical progression the sum of any two terms equidistant from the extremes is equal to the sum of the extremes. Let t denote a term of the series which has r terms before it, and /' a term which has r terms after it ; then the terms, t and t', will be equidistant from the extremes. Suppose the series to be increas- ing ; then from the nature of the series, * = a+rd ; (1) t' = lrd', (2) whence, by addition, t+t = a+L 51. The sum of the terms of an Arithmetical Progression is 'qua! lo one half the sum of the two extremes, multiplied by the num- ber of terms. Represent the sum of the series by S, then we have S = a-f(a+d)+(a+2d)-f . . . . +1. (1) By writing the series in the reverse order, we have also S = ** d)+J- t 2d+. ...+. (2) ARITHMETICAL PROGRESSION. 287 Therefore, by addition, Now equation (3) expresses the sum of n terms, each equal to (a-|~0 ; hence, and dividing by 2, we obtain the formula, n . Jb insert any number of arithmetical means between two given terms. Let n f denote the number of means to be inserted. Then the number of terms in the completed series will be n'-f-2 j and we shall have n = w'-f-2. This value of n substituted in formula (4), (34L0), gives , l-a whence, = ^ 7 -p 1 . ((7) Having the common difference, the means are readily obtained. APPLICATION OF THE FORMULAS. 353. The two formulas, I = a+(nl)d, (A) contain in all five quantities, a, /, n } rf, S, four of which enter eacK equation. Now if any three of these quantities be given, the other two may be found ; for, if the values of the three given quantities be substituted in the formulas, there will result two equations con- taining only two unknown quantities. 1. The first term of an arithmetical series is 5, the common dif- ference 3, and the number of terms 24. Find the last term, and *the sum of the series. We have given, a 5, tf=3,w = 24; hence, by formula (A), I 5+(24 1)3 = 74 ) and bj formula (B), S = ^ (5^.74^ _ 943 } 288 SEEIES. 2. Given a = 15, d = 2, and S = 60, to find the number of terms. Substituting the given values in (A) and (Z), we have 1= 15 2(7i-l), (1) 60 = ? (15+0; ( 2 ) whence, from (1), I = 17 2n, 120 Ion and from (2), I = n 120 15 . =n-2, 120 15?i =47>i 2n a n_16n= 60, n = 10 or 6. Both values of n arc possible ; for there are two scries answering to the given conditions, one having 6 terms, and the other 10 ; these are 15, 13, 11, 9, 7, 5, 3, 1, -1, -3, and 15, 13, 11, 9, 7, 5. The sum of cither scries is 60. EXAMPLES FOE PRACTICE. 1. The first term of an arithmetical scries is 7, the common difference 3, and the number of terms 36 ; find the last term. Ana. 112. 2. The first term of an arithmetical series is 275,. the last term 5, and the number of terms 46 ; required the sum of the terms. Ans. 6440. 3. The sum of an arithmetical series is 156, the number of terms 8, and the common difference 5. Required the two extremes. 4. Find the sum of the terms in an arithmetical progression, knowing that the first term is 1, the common difference f, and the number of terms 101. Ans. 2626. ARITHMETICAL PROGRESSION. 289 5. Required to find four arithmetical means between 7 and 37. Ans. 13, 19, 25, 31. 6. The first term of an arithmetical series is 3, the number of terms 60, and the sum of the terms 3720; required the common difference, and the last term. Ans. *d= 2, I -_ 121. -"* 7. What will be the sum of the series if 9 arithmetical means bo inserted between 9 and 109 1 Ans, 649. 8. If three arithmetical means be inserted between and A, what will be the common difference? / Am. 5 ' ? . 9. What debt can be discharged in a'year by paying 1 cent the first day, 3 cents the second, 5 cents the third, and so on, increasing the payment each day by 2 cents ? Ans. 1332 dollars 25 cents. 10. A footman travels the first day 20 miles, 23 the second, 26 the third, and so on, increasing the distance each day 3 miles. How many da}*s must he travel at this rate to go 43 miles ? Ans. 12. 11. Find the sum of n terms of the progression of 1, 2, 3, 4, 5 ' 6 ' ..... - An*. S = 12. Find the sum of n terms of the progression 1, 3, 5, 7, ..... Ans. S = n\ 13. The sum of the terms of an arithmetical series is 950, the common difference is 3, and the number of terms 25. What is the first term ? Ans. 2. > 14. A man bought a certain number of acres cf land, paying for the first $ j ; for the second, $ : j ; and so on. When he came to settle he had to pay $3775. How many acres did he purchase ? Ans. 150 acres. 1 15. The 14th, 134th, and last terms of an arithmetical progres- sion are 66, 666, and 6666, respectively. Ilequired the number of terms. Ans. 1334. (a - 66 6 - (>(, . f-0 -4- t 1L0 V / ^ i. / 3 ? 25 T 290 SERIES. THE TEN CASES. I. Given any three, of the quantities, a, 1, n, d, S, to find the other two. This problem will present ten cases, each giving rise to two for- mulas, making in all twenty different formulas, or four values for each letter. The results in each case may be obtained directly from the two fundamental equations, or those of any particular case my be derived from some preceding case, as is most convenient. The whole wril be left as an exercise for the student. Given. a, d', n I, d, n , n, I d,n,S l,n,S a, d,l a,l,S l,d,S To Find. a, 8 d,S a, I 1, S n, d n,a A _ Formulas. = x+o- 2Sn(nV}d 2^ w !>/ n(n / 2S Z = a. n ~ T~\~ w( 1) S = 2^ (/-j-a) ' -', J=a+(n-lX AKITHMETICAL PROGRESSION. 291 PROBLEMS IN ARITHMETICAL PROGRESSION 10 WHICH THE FORMULAS DO NOT IMMEDIATELY APPLY. When in the conditions of a problem no three of the five parts, a, /, ?i, d, S, are directly given, the general formulas will not directly apply. *It is usually necessary in such instances to represent the several terms of the series by means of two or more unknown quantities ; and for this purpose there are two methods of notation. 1st. Let x denote the first term and y the common difference ; thus, This method of notation, however, is seldom the most expedient. 2d. When the number of terms is odd, denote the middle term by .r, and the common difference by y ; then we shall have, for three terms, (x y), x, (x-\-y) ; for five terras, (x2y), (x y), x, (x+y), (*+2y). And when the number of terms is even, represent the two middle terms by x y and x-\-y respectively, 2y being the common differ- ence ; thus, (x-3y), (x-y), (x+y), (*+%) The advantage of the second method is, that the sum of all the terms, or the sum and difference of two terms equidistant from the extremes, will each contain but a single unknown quantity. 1 . There are three numbers in arithmetical progression ; the sum of these numbers is 18, and the sum of their squares is 158. What are the numbers? Ans. 1, 6, 11. 2. There are five numbers in arithmetical progression ; their sum / is 65, and the sum of their squares 1005. What are the numbers ? Ans. 5,9, 13, 17,21. 3. Tt is required to find four numbers in arithmetical progression, such that their common difference shall be 4, and their continued product 176985. Ans. 15, 19, 23, 27. \~t r ' 1 " ^'7-f'-~ 292 SERIES. 4. There are four numbers in arithmetical progression ; the sum of the extremes is 8, and the product of the means 15. What are the numbers? Ans. 1, 3, 5, 7. 5. A person starts from a certain place and goes 1 mile the first day, 2 the second, 3 the third, and so on , in six days after, another sets out from the same place in pursuit, and travels uniformly 15 miles a day. How many days after the second starts before they are together ? ), we have r = Vigi = Vyi = 3. Therefore, the series is 6, 18, 54, 162, 486, Ans. 4. Find the sum of the series G, 2, ,}... .to infinity. We have given, a = 6, r =. \; hence, by formula ((7), ffs-p-SB*, Am. 5. Find the exact value of the decimal .454545. . . .to infinity. This is a circulating decimal, and may be expressed thus : 45 45 45 UK) ~*~ 10000 ~*~ 1000000 ~1 In all such cases, the repetend, taken with its local value, will bo the first term of a geometrical series, of which the ratio will be 10 or some power of 10. In the present example we have = rfo' r = m' 9 hcncc ' e_ 45 (i l \_ 45 100 _ 5 "lOO^V 1 ~ 100/ 100 X 99 -If ***' 6. Find the value of 1 1 . . 4- to infinity. a a a We have a = 1, r = ', Lcnco, ]= , Ans. x +x L-V- %i y ^ -w^ A -r EXAMPLES FOR PRACTICE. 1. Find the sum of 9 terms of the series 1, 2, 4, 8, .... Ans. 511. 2. Find the 8th term of the progression 2, 6, 18, 54, Ans. 4374. 3 Find the sum of 10 terms of the series 1, |, $, 2 8 f , j a GEOMETRICAL PROGRESSION. 297 4. Required two geometrical means between 24 and 192. .4ns. 48, 96. 5. Required 7 geometrical means between 3 and 768. Am. 6, 12, 24, 48, 96, 192, G. Find the value of 1 -f- }.+ T 9 g + f f + . .* , to infinity. AM. 4. 7. Find the value of f + 1 + f + T& -f ---- to infinity. -4ns. 4. 8. Find the value of 5 -j- | -j- f -f- 2 \ -f- . . . . to infinity. Ana. 7f 9. Find the value of the decimal .323232. . . . to infinity. Aus. -. 10. Find the value of the decimal .212121 ____ to infinity. Ans. sV 11. Find the value of 4 j + J ^ -f & ---- to infinity. 12. Find the value of \ ^ -f- 3 ^ T c 5 -f ---- to infinity. a; x* x* 13. Find the value of 1 -\ --- f- -, -j- ^ -f ---- to infinity. a Am. - a x 1 CC* X* X* 14. Find the value of --- , -j- - -. -- ; + * infinity. a a* ' a* a T a'-J-x 3 15. The sum of a geometrical series is 1785, the ratio 2, and the number of terms 8 ; find the first term. Ans. 7. 16. The sum of a geometrical series is 7812, the ratio 5, and the number of terms 6 ; find the last term. Ans. 6250. 17. The first term of a geometrical series is 5, the last term 1215, and the number of terms 6. What is the ratio ? Ans. 3. 18. A man purchased a house with ten doors, giving $1 for the first door, $2 for the second, $4 for the third, and so on. What did the house cost him ? Ans. $1023. 298 SERIES. PROBLEMS IN GEOMETRICAL PROGRESSION TO WHICH THE FORMULAS DO NOT IMMEDIATELY APPLY. 9 3G5. The terms of a geometrical progression are represented in a general manner as follows : x, xy, xy*, xy\ ---- In the solution of problems, however, the following notation is gen- erally preferable : 1st. When the number of terms is odd, the series may be repre- sented thus : V-^ x \ xy, y* ; * *' . \j -' x > x ^ v> -*> 2d. When the number of terms is even, the series may be ex- pressed thus : * * y* r~ p f^* We may also represent three terms as follows : y* ,/ r * ~fe x ' IJ * y 1. 'The sum of three numbers in geometrical progression is 26, and the sum of their squares 364. What are the numbers 1 Let the numbers be denoted by x, V ' xy, y. Then x -\-T/xy+y = 26 = a, (1) and 3 -|-xy-t-y 2 = 364 = b. (2) Transposing V xy in (1), squaring and reducing, * a +^-f-y = a 3 2al/^. (3) From (2) and (3), a 8 2al/xy = b ; / a *b whence, V xy =. = 6. From (1) and (2) x = 2, and y = 18. Hence, the numbers are, 2, 6, 18, Ans. GEOMETRICAL PROGRESSION. 299 2. The sum of four numbers in geometrical progression is 15 or a, and the sum of their squares 85 or b. What are the numbers ? Taking the proper notation for an even number of terms, we have " and, ^ +x ' +y > + ^ ==6 .> " (2) ty Assume &-\-y = s, and xy=p-, then by (3O1), cc a -f if = s*2pi ; x*+y*. = s 8 3p. Substituting the values of (z-fy) and O 2 -fy 2 ), in (1) and (3), Squaring (3), and then transposing 2;cy, or 2p, whence, from (4) and (5), ( s) a 2p = 6 s*+2p ; or, a 9 2as-| 7 2s 2 4p == b. , (6) Clearing (o) of fractions, and putting xy =p in second member, x * -\~y* = a ^ 1>* j or > s * 3sp = ap ps j whence, p = -_i_. (7) Substituting this value of p in (6), and reducing, we have a_2 a .s 2 = ab+2bs ; or, as*-}- 5s = (a* 6). Restoring the numerical values of a and 6, 15s'+85s = 70x15,- whence, * = 6. Substituting the values of a and s in (9), and we obtain P = 8; that is, x ~\~y = 6, xy = 8 ; whence, x =. 2, y = 4. Therefore, the required numbers are 1, 2, 4, 8, Ans. SERIES. 3. There are three numbers in geometrical progression; their sum is 21, and the sum of their squares is 189. Find the numbers. Ans. 3, 6, 12. 4. Divide the number 210 into three parts, so that the last shall exceed the first by 90, and the parts be in geometrical progression. * , *y f Y y *- Ans. 30, 60, and 120. 5. The sum of four numbers in geometrical progression is 30 ; and the last term divided by the sum of the mean terms is 1^. What are the numbers ? K fy- *^/~*Y J Ans. 2, 4, 8, and 16. 6. The sum of the first and third of four numbers in geometrical progression is 148, and the sum of the second and fourth is 888. What are the numbers ?X/Yy'Vy**y?te. 4, 24, 144, and 864. 7. It is required to find three numbers in geometrical progression, such that their sum shall be 14, and the sum of their squares 84. Ans. 2, 4, and 8. S. There are four numbers in geometrical progression, the sec- ond of which is less than the fourth by 24 ; and the sum of the extremes is to the sum of the means as 7 to 3. What are the num- bers? Y" >y * y % *y 3 Ans. 1, 3, 9, and 27. _, 9. There are three numbers in geometrical progression ; the sum of the first and second is 20, and the difference of the second and third is 30. What are the numbers ? X,*y* *y3lns. 5, 15, 45. 10. The continued product of three numbers in geometrical pro- gression is 216, and the sum of the squares of the extremes is 328. What are the numbers ? }L , Y-y , * y *" Ans. 2, 6, 18. 11. The sum of three numbers in geometrical progression is 13, and the sum of the extremes being multiplied by the mean, the product is 30. What are the numbers^^ Ans. 1, 3, and 9. 12. There are three numbers in geometrical progression ; their continued product is 64, and the sum of their cubes is 584. What are the numbers ? / / / */ * *" Ans. 2, 4, 8. 13. There are three numbers in geometrical progression; their continued product is 1, and the difference of the first and second is IDENTICAL EQUATIONS. 301 to the difference of the second and third as 5 to one. What aro the numbers? '+* * ^ Ans ' l 5 ' of 120 dollars was divided between four persons in fJL0 such a manner that the shares were in arithmetical progression ; if the second and third had each received 12 dollars less, and the fourth 24 dollars more, the shares would have been in geometrical ""^''p^gfesiJn/^Jind the shares. Ans. $3, $21, $39, and $57. /*-*:. / 16. The sum of six numbers in geometrical progression is 189, and the sum of the second and fifth is 54. What are the numbers ? */ *yr*Y%Xy 2 f *-*f'~ Am. 3, 6, 12, 24, 48, and 96. 17. The sum of six numbers in geometrical progression is 189, and the sum of the two means is 36. What are the numbers ? AUK. 3, 6, 12, 24, 48, and 96. 18. A man borrowed p dollars ; what sum must he pay yearly in order to cancel the debt in n years, interest being allowed on the unpaid parts of the principal at r cents per annum on a dollar ? IDENTICAL EQUATIONS. An Identical Equation is one in which the two members are either the same algebraic expression, or the one member is merely another form for the other. In every case, either the one member may be reduced to the other directly, or the two members may be reduced to some expression different from either, from which both members may be supposed to originate. Thus, ax-{-b = ax-\-b, , , __ _ __ _ ~l+x x ~ * are identical equations. In the first, tne two members have exactly 26 302 SERIES. the same form. In "the second, the second member may be reduced to the form of the first, by performing the multiplication indicated. In the-third, each member may be reduced to the fraction, --- . 1-j-x 367. There are certain properties of identical equations, which are of great importance in the further treatment of series, and iu the general theory of equations. In order to investigate these properties, let us first consider what any term containing the variable x, as ax n , will become when x = 0, under the various conditions of the exponent. 1 Suppose n to be positive ; then if x =. 0, we have ax* = a ' 0* == 0. 2. Suppose n to be negative ; then if x = 0, we have a a ax =^=o =GO - 3. Suppose n to be nothing or zero ; then if x = 0, we have ax* = a - () = a 1 = a. Q* * -j I 368. We are now prepared to demonstrate the following propo- sitions : I. An identical equation is satisfied for any value whatever of the unknown quantity. The truth of this proposition follows directly from the definition of an identical equation. It is implied in all algebraic transforma- tions, that the value of a function is not changed by changing its form, whatever quantities the symbols represent. Hence, if the two members of an equation are the same in form, or reducible to the same expression, they must be equal, whatever value be substi- tuted for the unknown quantity. To illustrate this principle, we will take the following identical equation, where the form is such that the identity of the two members is not apparent from inspection. IDENTICAL EQUATIONS. 303 If in this equation we make x equal to 1, 2, 3, 4, 5, etc. succes- sively, we shall have, {4+1+1}' = 2(16+1+ 1}, {1+1+0}' = 2} 1+1+0}, {0+1+1}*:= 2{ 0+1+*!}, { 1+1+4 f = 2{ 1+1+16}, { 4+1+9 }' = 2{ 16+1+81 f, etc., every result being a true equation. II. Conversely : Every equation which is satisfied for any value whatever of the unknown quantity, is an identical equation. Suppose the given equation to be cleared of fractions, and each member arranged according to the ascending powers of the unknown quantity. Then the equation may be represented thus : Atf+a*+ <7x e + . . . . = 4V+.BV+ CV'+ .... (D in which, by hypothesis, we have a < b < c . . . , and a ' < b' < c' It is implied, also, that the coefficients, A, B, (7, etc., and A', B'j (7, etc., are all finite quantities greater than zero, and independent of*', and the number of terms may be limited or unlimited. Divide both members of equation (1) by x a ' } we have A+x b - a + OB+ . . . . = A'x u '-~+'x*'+ C'x"+ . . . , (2) in which the exponents, b a, c a, etc., in the first member, are all positive, because a < b < c Now by hypothesis, the given equation is true for all values of x j hence every modification of it will be true for all values of x. Make x = ; then in the first member of equation (2), every term after the first will reduce to zero, (307, 1), and we shall have A = A V + J?'**' + C'x e '-*+ .... (3) Now since a' < b r < c' . . . . , we must have (a' a) < (b f a) < (c' a) < Hence, in equation (3), the first exponent, a a', is the least of aU> But we observe, 1st. The exponent, a' a, can not be a positive quantity; for in 304 SERIES. that case the term containing it would reduce to zero when x = 0, (367, 1), and we should have A = 0, which is contrary to the implied conditions of the proposition, 2d. The exponent, a' a, can not be a negative quantity ; for in that case the term containing it would reduce to infinity when x = 0, (367, 2), and we should have A oc, which is also con- trary to the implied conditions of the proposition. Now since a' a can neither be a positive nor a negative quantity, it must be nothing or zero ; that is, a' a =. 0, or a f = a. It follows also that each of the other exponents, b' a, c' a, etc., in equation (3), is positive, being algebraically greater than zero ; hence all the terms after the first in the second member of this equation, must disappear when x =r 0, (367, 1), and we shall have, A = A'x = A'. Now since A and A' are independent of x, we shall have Ax a A'x af , whatever be the value of x. We may therefore suppress these terms in equation (1). There will result Bx*+ Cx c -\- .... = B'y?'+ CV'+ . . . . , whence, by reasoning as before, we shall find that b = //, c = c', etc. B = B' C= C', etc. That is, equation (1) is an identical equation, the two members having the same form. Hence, the given equation is also identical, and the proposition is proved. It is obvious -that the preceding demonstration will apply if one or more of the exponents, a, l>, c, .... are negative ; or if a = 0, in which case each member will contain an absolute term. III. In every equation which is satisfied for any value whatever of the unknown quantity, and which involves like powers of this quantity in the two members, the coefficients of the corresponding powers will be equal, each to each. Let us assume the equation, Aar+Bx*+ Cx*-\- .... = J V-f^V-f <7'V+ the number of terms bein<: either limited or unlimited. IDENTICAL EQUATIONS. Now if this equation is capable of being satisfied for any value of x, then according to the preceding demonstration, not only must the exponents of x in the two members be equal respectively, but the coefficients also must be equal, each to each ; that is, A = A', B = B', 0= C", ;etc. Every such equation is obviously identical, though it is not neces- sary that A, B, C, etc. should be of the same form respectively, as A', B 1 , C', etc. IY. In every equation which is satisfied for any value whatever of the unknown quantity, and which has zero for one of its members, the coefficients of the different powers of the unknown quantity are separately equal to zero. Let Ax*+Bx*+ Cx c +Dx*+ . . . . = 0, (1) represent the equation, arranged according to the ascending powers of x. The coefficients, J., B, C, D, etc., are supposed to be inde- pendent of x, and consequently the same for all values of x. Divide every term in this equation by x a ; we shall have A+ Bx*- a -{- Cx*-*+Dx d + =0. (2) In this equation make x = ; then since the exponents, b a, c a, d a, etc., are all positive, every term after the first will reduce to zero, (3G7, 1), and we shall have A = 0. Suppressing Ax 9 in equation (1), and then dividing through by x*, we obtain B+ Cx*-*+Dx*-*+ . . . . = 0. (3) In this equation make x = 0, and we have ^=0. In like manner we may prove that each of the other coefficients is equal to zero. It is important to observe in this connection that the coefficients, A, B, C, D, etc., must be supposed to represent polynomial exj)res- sions, ichich reduce to zero in consequence of having positive ana rrfji+ive parts that are respectively equal to each other. 26* ' u 306 SERIES. A DECOMPOSITION OF RATIONAL FRACTIONS. 3O9. By means of the properties of identical equations, a fraction may often be separated into two or more partial fractions, whose denominators shall be simpler than the given denominator. In every such case, the given fraction is the sum of the partial fractions ; hence its denominator will be a common multiple of the denominators of the partial fractions. O q-1 1. Separate - into partial fractions. By inspection, we perceive that x 2 7z+10 = (x5)(x 2). Now assume 8 *-3L JL_ JB_ ( (x 5)O 2) ~x 5 "t" x 2" Since the first member is simply the sum of the two fractions in the second member, this is obviously an identical equation. Clearing of fractions and uniting terins, we have . ._ Sx 31 = A+)X (24+55), (2) in which 31 in the first member, and (24 +55) in the second, may be considered as coefficients of x. Now according to (3G8, III), the coefficients of the like powers of x in the two members must be equal ; and we have, therefore, 4+5 = 8, (3) 24+55=31. (4) From these equations we readily obtain whence from equation (1), we have 3 An *- X _7 X _|_10 - x 5 7 x It should be observed that equations (3) and (4) are the equations of condition, which must exist in order that equation (1) shall be true for all values of x. DECOMPOSITION OF FRACTIONS. 307 1x* I x 2. Separate - -- into partial fractions. (*+l)(2x 1) Suppose, if possible, lx*+x A clearing of fractions, we obtain 7x*+x = (2A+J3)x+(B A); transposing all the terms to the first member, we have JB = 0. (2) If this equation be possible, it must be an identical equation ; and as one member is zero, the coefficients of the different powers of x must be separately equal to zero (368, IV) ; and we shall have 7 = 0, which is absurd. Hence, we infer that the fraction can not be sep- arated into partial fractions, having numerators independent o/*x. Again, assume 7x'-\-x Ax Bx clearing, of fractions and collecting terms, 7*'+ x = (2A -f B}x*+(B equating the coefficients of like powers of x, 2A+J3 = 7, B A = 1 ; whence we obtain A = 2, j5 = 3 ; and by substitution in equation (1), 2x 3x Ans - From this example we learn that if we assume an impossible form for the partial fractions, the fact will be made apparent by some absurdity in the equations of condition. NOTE. If the given denominator consists of three or more factors, there will be three or more partial fractions. But there will always be as many equations of condition as there are numerators to be determined. 308 SERIES. EXAMPLES FOR PRACTICE. 7/y 24 1. Resolve a j into partial fractions. X JJC j L-x. 5 2 Ans. -J . x 1 x2 2. Resolve a , >> 9 n ^ nto partial fractions. - , Gar 2 22ar+18 . Itcsolve . vx-i - -- rr into partial fractions. x 1 2 4. Resolve p^-- into partial fractions. C vC 1 ! - 5. Resolve Q7r into partial fractions. CC - ioJC -J-oO -1- ^_ " THE RESIDUAL FOR3IULA. 3 TO. It lias been shown in (89, 4) that x m y m is exactly divisible by x y, if m is a positive whole number. The form of the quotient is as follows : the number of terms in the quotient being equal to m. Now suppose x=y; then each term will become a 1 "- 1 , and since there are ?n. terms, we have the formula, The subscript equation, y = x, is used to indicate the condition un- der which the first member of (A] will be equal to the second. RESIDUAL FORMULA. 309 371. We will now show that this formula is true, whatever bo the value of m. There will be two cases : 1st. When m is positive and fractional. r r . r - Assume m = - then X M y m = x' y'. _ Let x ' = z j then x* = z r , and x =. *. J_ lso let y = u } then y By proper substitutions we have J_ r_ Also let y = u } then y* = u r , and y = u*. X y sfu' Z'U' z u Now suppose xy, then z = u ; and since r and s are positive whole numbers, we have from (1) j - " v u / r r i .- r) z u rz- r r i = Hence, the formula is true when the exponent is positive and frac- tional. 2d. Wlien m is negative, and either integral or fractional. Suppose the exponent of x and y to be m; we shall have ^ Now suppose x = y ; then whether m be integral or fractional, we shall have, from the principles already established, hence, =(-") X (m.-) = Hence, the formula holds true universally. 310 SERIES. BINOMIAL THEOREM. 372. The Binomial Theorem has for its object the develop- ment of a binomial with any exponent, into a series. This theorem is expressed by an equation, called the Binomial Formula. 373. It is required to expand (a-j-x) n into a series, n beiny any real quantity , positive or negative, entire or fractional. We observe that a+x = aYl -|- - } >, therefore (a-j-z) 11 = a* ( 1 -f -\ . (x \ n 1 _j J , and then multiply the result by a n , we shall have the expansion of (a-\-x) n . Put z = - then (l + - Y = (1+ *)". , | % Let us UQW assume the equation, (!+*) = A+Bz+ Cz*+Dz*+Ez*+ .... (I) in which A, B, C, D, etc., are independent of z. "We are to find the values of these coefficients which will render equation (1) true for all possible values of z. Suppose z = ; then from equation (1), we have A = 1. Hence, the assumed development becomes (l+z) w = 1-hflH- Cfe f +0* i +J5k 4 +.... (2) for^ll- values of z. Put z = u . ; then . \+.(\+uy = l+Bu+Cu' l +Du*+Eu'+.... (3) - Subtracting (3) from (2), and dividing the result by z , we obtain Vi + .y.-(i ) != /.^x ^N (4) 2 - U \ Z - U I \ Z - U I Let P = 1-j-z, and Q = 1 -f?* ; then P Q = 2 w. Equation (4) now becomes ' .... (5) Now suppose z = u ; then P Q. And by the Residual Forn> ula (37O), we shall have i BINOMIAL THEOREM. 311 ) =*>> / z'u* \ - ) = 42*, etc. z u /=* Substituting these values in equation (5), we have Multiplying both members of equation (6) by (l-j-2), gives -f -f-26 r Multiplying both members of equation (2) by n, gives CO ____ (8) Now by equating the second members of (7) and (8) we shall have an identical equation, because it may be satisfied for any value of z. Therefore the coefficients of the like powers of z in equations (7) and (8) are equal, each to each (3O8, III), and we shall have =n, or C = B -, ZD+2C=nC, or 2>= = wD, or E = Therefore, the values of the coefficients arc ; etc. n(n l)(n 2) 23* 312 SERIES. Substituting these values in (1) we have o and by restoring the value of z, which is - , or. finally, multiplying both members of (&) by a n , Equation (c) is the binomial formula, as it is usually -written. It will be observed, however, that in the three equations, (), (*), ( c \ the coefficients, or the factors depending on n, are the same ] and in practice, either (#), (6), or (<$ may be employed, according to the form of the binomial to be expanded. 31 74:. By inspecting the general formula (c\ we perceive that in the expansion of a binomial in the form of a-\-x, the law of the exponents is as follows : 1. The exponents of the leading letter in flie successive terms form a scries, commencing in the first term with the exponent of the binomial, and diminishing by 1 to the right. 2. The exponents of the second letter form a scries, commencing in the second term with unity, and increasing by 1 to the right. And the law of the coefficients is as follows : 3. The coefficient of the first term is unity, and that of the second term is the exponent of the required power. 4. If the coefficient of crny term be multiplied by the exponent of the leading letter in that term ; and divided by the exponent of the second letter plus 1, the remit will be the coefficient of the fol- lowing term. 37>. If we take the least factor in each of the successive coef- ficients of the expansion, commencing at the second, we have a de- creasing series , (]), (Ti2), (n 3), etc., in which the common difference is unity. BINOMIAL THEOREM. 313 Suppose n to be a positive integer, then the least factor in the nu- merator in the (w+2)d term will be (H ), or 0, and this term will disappear. But if n is negative or fractional, then no one of the factors, (M 1), ( 2), (M 3), etc., can be zero, and the expan- sion may be continued indefinitely. Hence, ^ 1. When n is a positice intfyer, the expansion of the binomial will be a finite series, the number of terms being n-j-1. 2. When n is negative or fractional, the expansion of the bino- mial will be an- -infinite series. APPLICATION OF THE BINOMIAL FORMULA. 370. Let us resume the equation, 7i(. 1) 7/(M !)( 2) (a+x)* = a*+na-*x+ -- - 4 y V~V -f- - . 8 a n ~V + .... (e) If n be entire and positive, this formula will be an expression of involution, denoting some power of the binomial. If n be fractional and positive, the formula will be an expression of evolution, denoting some root of the binomial. If n be negative, the formula will express the reciprocal of some power or root of the binomial. 377. Since the binomial coefficients depend entirely upon the exponent ?*, they may be formed independently. To do this, we have simply to commence with unity and multiply by n] > TT~ } o etc., continually. 1. Expand (a j-) 6 into a scries. Here n = G j hence, The first coefficient is 1=1 " second " " lyC = G third " " GX:] = 15 " fourth " 16X3 = 20 fifth " 20 XJ = 15 " sixth " " 15 X? = G " seventh" " GX = 1 Since the odd powers of x arc negative, we have for the literal factors of the terais, SERIES. ffl , a'x, Therefore the expansion will he (a a;)' = a 6 6a^+15a.V 20V-f 15aV C^'-f x. 2. Expand (a-j-a:)- into a series. In this example n = i. Keprcsent the coefficients by A, J5, C, D ---- ; then A= + 1 Ji = AX n =+ The literal factors of the terms will be a?, a~v x , : vJ or by clearing of negative exponents, We might have obtained this last result directly, by putting the binomial in the form of a~ ( 1-| J 5 . It is well, however, to note the transformations made above. BINOMIAL THEOREM. 315 3. Expand a into a series. Observe that Whence, by expanding the factor f 1 -j ) we obtain 4. Expand (a' x*)* into a series. If we take the descending powers of a*, commencing with the . 5th, and the ascending powers of x a , commencing with the first, we have for the literal factors of the terms, 16 n^nr* n*-r* n*Y* /r 8 ?- 8 f l * CL , Cp C i tt.t/j Ui JC , U JL j C-.- Hence, with the coeflicients the development becomes V-flOaV 10aV-j-5aV- EXAMPLES FOB PRACTICE. . . 1. Find the fifth power of a b. * t Ans. a' 5a6+10ai 7 lOa'^ 2. Find the sixth power of 1+c. A ^cW^/V Ans. !-|-6c4-15c a 4-20c 8 -fl5c 4 -f6c'-fc a . 3. Find the seventh power of x-\-y. Ans. x 7 +7^V-h- n ^y+35xy+35xy-f21xy+7xy+y'. 4. Find the eighth power of a 3 1. Ans. a 16 8a u +28a" 56a 10 -j-70a 56a'+28a 4 8a a +l. 5. Find the ninth power of a c. . a 9 J)a 8 c -j- 36a'c 2 -r84aV -f 126a 5 c 4 126aV + 84aV 6. Expand \ 7. Expand (a a a:') 6 . 316 SEBIES. 8. Expand (* z 4 )'. Ans. z 10 SxV+KW 10xV f +5:cV - 9. Expand (o'^-f cfy 9 ) 6 . . a 1 V-f 6a" a: 3 i ^j 10. Expand (a a:) 3 into a scries. C^-^JL *^wt */ # t #, --h-i^-i . / /Y_^ x_ __ x 9 . 3x* - 3-5a* i**-rf V ""^a 2-4a 8 ~"2-4-6^"~"2-4-G-8a 4 """ ir'^'r 1 i 11. Expand (1 x)* into a series. ^ 3 3-6 3-6-9 3-6-9-12 f. , ^^.J s ^fl2. Expand (a+1)* into a series. 13. Expand (a-J-i) 3 into a series. J/! , 1 jg_ , .2-Sy 2-6-Sf X \ ^3a 3-Oa' ^ 3-G-Oa 1 3-G-9-l:V^ "/ -f 14. Expand -L , a -4t- LX ^4^ y.i 1 ^-, Expand _ 1nt6 a scri XTr 1 "f)<-T*-4 <^ix < / J/A 16. Expand / a --^ a2 into a series. _ ~~ , , 2^1 l1 ' Expand (a c 2 )^ into a series. ^_^ ^ ^ ^.. '" f /. 2c a 2c 4 2-4c a (J2 ' 4 7c a \ jSSi a \ ' ~ 3^ ~ 3 T 6^~ a " ' 3 6 9 a 3 ~~ 3 6 9 1 2 a 4 " "J 18. Expand ^c'-j- x a )~2 into a series. ^ ** An, l--A - ' / ? , ' j c\ 2c'" 1 "2-4c 4 2-4- 6c 8 BINOMIAL THEOREM.^./ tf^-J ^J*A = 19. Expand (1 a)~ 3 into a series. - _^ t,*-^ ~ -JQ Am. ; ^20. Expand^ (a 3 x 3 )* into a series. ^-/ ^-^r < ^V^V'^-'^ry "TT^ * f*f&*~ 3x 9 3x 4 3-5x' 3^-9x* C^"**- A. Am. ya [a - ^ - ^j^ - 4 . 8 .' 1:2 . 16tt f j , ,^ 21. Expand (a+y)~ 4 into a series. -{-12a& a SI 9 . 2. Find the fourth power of 2a+3x. AM. 16a 4 +96a'a:+ 216a^ a 3. Find the fourth power of 1 \a. -/-4{j Ans. 1 Sa+fa' ^a' 4. Find the fourth power of a* ax-fa*. Ans. 8 -4a 7 .i- + lOaV ICaV + 19aV L6aV -f- lOoV " 5. Expand (4a a 3x)? into a scri -Ji.- .. . _ \ 16a 9 512a 4 24576a' "V FRENCH'S THEOREM. 3 TO. When a binomial having numerical coefficients is to bo raised to any power, the coefficients of the expansion may be ob- tained with great facility by means of a simple modification of the binomial formula. We have (z-f-*0 w = n(nV) n(n IVn 2) s+n*-'iH- ir a^VH- >, A o ^ 3 ^+ - 6 A i) In this equation make z = ax, and u=.ly } then BINOMIAL THEOREM. 319 in which a and b may represent the numerical coefficients of x and y. Now denote the numerical coefficients of the expansion by C 19 C 2) C$, etc. We shall then have in which C l = c c- n - b C * * I a' r r "- 1 - b * * ~2~ a' n - 2b c 64 (7 5 4 a 1. Find the fourth power of ba-\-ox. In this example we have n = 4, a = 5, 6 = 3, and the coefficients of the expansion will be C , = 5* =625 C 2 = 625 - f | = 1500 C 3 = 1500 -f | = 1350 <7 4 = 1350 f |= 540 C 5 = 540 i - |= 81 Hcnco, (5a-f 3x) 4 = 625a 4 -fl500az+1350aV+540az'-f81.r 4 , Ans. 2. Find the fourth power of 2 4 .6436 We have n = 4, a = , 6 = , and - = = - o 5 a 5 2 5 Hence the coefficients of the expansion are c, (!)' = if c-z= if-f-f = t3i t- Y 3 = ill ' 1 ' f = W <^ = w i f = m u. &Y+ 5?'V+ a a ^ f -f20/V > 4-J.s^/- 4 + i ^' r *+ W' 1 '- G. Find the fifth power of . 4 5 m* 7U 1 7H 8 7?r m 1 /I JO I _ I _ '1024 250 ^ 1GO 200^600 3125' 7. Find the eighth power of . 2 '2m m* m c 7wi 4 7??^ a 85 7 7 yl j) o I __ I _____ * 250 32 J 04 32 '*" 128 ~~ 32m 1 "" Gi/u* "" 1 1 S2w? ' 250m 6 ' DEVELOP^IEXT OF SURD ROOTS IXTO SERIES. SI8O. The approximate value of a surd root mny lie oLtaincd with much facility by expanding the root into a series. Let u n represent that perfect ?>th power, which is next less or next greater than the given number, and let I denote the difference between this power and the ijivcn number. Then a n +b, or a n b, will express the given number. But we have DEVELOPMENT OF SUED BOOTS. 321 a" Developing the radical parts into series, we have 2 n rf 2;i a 3 * " n 2 3/i a 8 " The second members of these equations contain no radicals ; hence, Any Kurd may l>e developed into a series of rational terms; whence ly summing the series, we may obtain approximately the indicated root. It should be observed that the smaller the fraction n is, the more rapidly will the series converge. 1. Find the cube root of 76, to six decimal places. The smallest fraction will result by taking the cube which is next less than 76, or 64 ; thus, We may now develop the radical part by equation (1), in which q A ^ _ To form the binomial coefficients we have the factors, 1 _1 n ~~3 l n 1 1 4 11 2n 3 bn 15 1 2?i_ 5 1 5?i_ 7 3 ~~9 Qn ~~jf 1 3n 2 1 6n 17 4n 3 7??- 21 We represent the successive terms by A, B, C, etc.; and to secure accuracy in the final result to the 6th place of decimals, wo should V 322 SERIES. carry the computation in each term to the 7th place. Thus we have A *= + C = # = ^7 = .F 7 = = //= Algebraic sum, A -T 3 S A T 3 * c D E F G 1.0000000 . .0625000 .0039062 .0004069 .0000508 .0000069 .0000010 .0000001 1.0589559 = ^I+A 4 4.235824, Am. Whence, 2. Find the 5th root of 25, to 6 decimal places. The most convenient fraction will result by taking that 5th power which is next greater then 25, or 32 thus, Equation (2) will now apply ; and the operation will be as follows : L_ L n~5~ 1-n 2 l4n 19 2n ' 5 5?i 25 l2n 3 1 5n 4 3n 5 6/1 5 1371 7 1 Qn 29 4n 10 - 77i 35 A = + 1.0000000 B = J ' & = _ 437500 (7 = 1 ' A ' & = 38281 D = 1 ' A * = 5024 E T 7 7T >' D = 769 IT 4S A ' E = 128 G = 1 'h' F = 22 A G = Algebraic sum, Whence, .9518272 = 2 = 1.903654-f, EXPANSION OF FRACTIONS. 323 EXAMPLES FOR PRACTICE. Find tlio values of the following indicated roots, to the 6th lecimal place : 1. V9. * Ans. 2.080084. 2. VsT Ans. 3.141381. 3. VETO. Ans. 4.641589. 4. VTTO. Ans. 4.791420. 5. V597. Ans. 3.122851. 6. v'oa Ans. 1.978602. 7. 1/4. Ans. 1.319508. 8. ^8275. Ans. 5.047104. 9. VT25. Ans. 1.993235. EXPANSION OF FRACTIONS INTO SERIES. 381. An irreducible fraction may always be converted into a series, by dividing the numerator by the denominator. 1. Convert into a series. 1+a Observe that 1-f a ~ a-fl Hence, there may be two ways of dividing j 1st. 2d. C-i+i a 1 1 a a' 824 SERIES. The law of expansion is obvious in both quotients, and we havo from the same fraction two series; thus, Y^ a = 1 -f a 2 a 3 -}- a 4 = 3 0; E= 36; F=+ 36; '=0, " G = +288, etc. Substituting these values in the assumed development (1), and observing that the term containing C will disappear because (7=0, we have , = - +318x 2 X -+OX X KOTE. It is not necessary to transpose the terms to one member ; for if neither member is zero we have simply to equate the coefficients of the like powers of x in the two members, according to the third property of identical equations. The method of Indeterminate Coefficients is applicable to a great variety of examples, but always with this provision, viz. : That we determine by inspection what power of the variable will be contained in the first term of the expansion, and make the first term of the as- sumed development correspond to the knoicn fact. If the assumed development commence with a power of the variable higher than it should, the fact will be indicated by an absurdity in one of the resulting equations. If, however, the assumed development commence with a power of the variable lower than is INDETERMINATE COEFFICIENTS. 327 necessary, no absurdity will arise ; but the redundant terms will disappear by reason of the coefficients reducing to zero. EXAMPLES FOE PRACTICE. _ 1. Develop - - into a scries. 1 ox Ans. l 2. Develop - - - - into a series. r 1 x x 3 Ans. 3. Develop - - - - into a scries. 1 ox Lx Ans. l-J-2z-|-8z > -|-28x a +100z 4 -f-356:e*+ .... .r) . 4. Develop 5 r- t into a series. Ans. 2 5. Develop -^ - ^-j- into a series. ox x 2 4 8x 16x 32x' AnS ' ^ + 9 + 2T + -gT + 243" +" 6. Develop . , , , . into a series. v 34 nto a sees. 2a)x (2a 3a'X-|-(3a 4a 4 )x 8. Develop \f\ x into a series. x x* ox* ~ " ~~ "~ 2 ~~ 24 "~ G : 2-4-6-S NOTE. Assume Vl x = A+Bx+ Ox > +Dz*+ ---- ; then square both members, and the equations for the coefficients will be readily obtained. 328 SERIKS. 9. Develop l/l-f-3x-|-5x 8 -f 7x 3 -j-9x 4 -|- into a series of rational terms. - 11*' 23*' 179x*_ I i 10. Develop alent scries. + *'+ x 4 Ans. mtoancquiv- REVERSION OF SERIES. 383. The Reversion of a Series is the process of finding the value of the unknown quantity in the scries, expressed in terms of another unknown quantity. 1. Given y ax-{-lx' 2 -\-cx*-\-dx*-}-cx*-\- , to find the valuo of x in terms containing the ascending powers of y. In this equation, x and y arc two indeterminate quantities, and either may have any value whatever without altering the form of the series. We may therefore apply the method of Indeterminate Coefficients. Assume We may now find by involution the values of *, cc s , x*, x*, etc., carrying each result only to the term containing y*. Then substitut- ing for T, x a , x 3 , etc., in the given equations we shall have, after transposing y, oA I y-\- aB II I A 1 4-21AB +21AC + IB 9 + cA + dA* -f This is an identical equation, being true for all values of y. And if we place the coefficients of the different powers of y sepa- rately equal to zero, (368. IV), and reduce the resulting REVERSION OP SERIES. 329 tions, wo shall obtain the values of the assumed coefficients as follows : A= L a If wo substitute these values of A, E, (7, etc., in equation (t), wo shall have the value of x in terms of a, 2, <...., and the powers of y ; that is, the given series will be reversed. 2. Given y = a .x-\-lx*-\-cx* -\-djc 1 e of x in terms of y. Assume x = Ay+fy 9 + Proceeding as before, we shall obtain, , to find the valuo .. (1) In the preceding examples the letters a, I, c, . . . ., represent any coefficients whatever. Hence, in reverting any series in cither of these forms, we may determine the values of the assumed coefficients by an application of formula (#"), or (G). 3. Revert the series y x-4-2x 2 -}-4x 3 -f-8x 4 + Assume, x = Ay+By*+ C/+ Dy'+ .... 28* 330 SERIES. If we now substitute in formula (F) } a = 1, b = 2, c = 4, d = 8, we shall obtain ^1 = 1, B = 2, (7 =4, D = 8. Hence, x = y 2y f + 4y 8 8^ 4 -|- . . . . , ^n. EXAMPLES FOR PRACTICE. 1. Revert the series y = x+x*-\- x*--x 4 -\- x s + .... Am. X = yy*+y*y*+y* 2. Revert the series y = z+Sx'-f 5x*-f7z 4 -f 9x'-f . . . . -An*. *= 3. Revert the series x = y ^_L^ _ ?. _L ?. _ 2 ^3 4 + 5 S ^ =x+ i~2 + r^3 + rl^4 + r^ 4. Revert the scries y =x x'+x* x 7 -f-x 9 x n -|- .... 5. Revert the series y = 2x-j-3x 8 -j-4x 6 -f-5^ T -f ____ 6. Revert the scries x = 2^+4y f -f6y 8 +8/-f-10y ft -f- -^- _____.... 384. One of the principal objects in reverting a series is, to obtain the approximate value of the unknown quantity when the sum of the series is known. Thus, 1 . Given ^ = 2x -- 3~ + ~^ --- f~+ ____ to find the a P~ proximate value of x. 4z* 6^ 8 8x 4 Letusput s = 2x + -- -j- ---- , (i) and consider x and s as variables. Reverting (l) y by formula (^), 8 S 138 8 = ---- Now if we put s = in this equation, the result will bo a con REVERSION OF SERIES. 331 verging series ; and we may find the approximate value of x, by computing the values of the terms separately. Thus, 2 1 = ' =' 125000 1512 ~ 256 ' 1512 ,.000013 Hence, x = .135993, EXAMPLES. 1. Given | = 5x 20^'-f80x s 320x 4 +1280x 6 ....,to find the approximate value of x. Ans. x = .117647. =*+++++ ..... tofind the approximate value of x. Am. .454620. 1 x* x* x 1 3. Given - = x Q |Q jp> f- ---- > to find the approxi mate value of x. Ans. x = .201369. 4. Given -= * ^ + ^-^g +-> to find thc a P" proximate value of x. Ans. x = .274655. SUMMATION OF INFINITE SERIES. 385. The Summation of a Series is the process of obtaining a finite expression equivalent to the series. 38G. The method of summing a given series must always depend upon the nature of the series, or the law governing its development. Formulas have already been given for the summa- tion of arithmetical and geometrical series. We will now investigate the methods of summing a variety of other series. 332 SERIES. RECURRING SERIES. 387. A Recurring Series is one in which a certain number of consecutive terms, taken in any part of the series, sustain a fixed relation to the term which immediately succeeds. Thus, is a recurring series, in which if any two consecutive terms be taken, the product of the first by 3x* plus the product of the second by 2.r, will be equal to the next succeeding term. The coefficients of these multipliers, or 3, 2, arc called the scale of relation. In the recurring series the scale of relation is 8, 2, 1. 388. A recurring series is said to be of the first order, when each term after the first depends upon the term which immediately precedes it. The scale of relation will consist of a single part, and the series will be a geometrical progression. A recurring series is said to be of the second order, when each term after the second depends upon the two preceding terms; the scale of relation consists of two parts. A recurring scries is said to be of the third order, when each term after the third depends upon the three preceding terms ; the scale of relation consists of three parts. 389. To find the scale of relation and the sum of a recurring series of the second order. 1st. Let a, 1), c, d, represent the coefficients of any four consec- utive terms; and let in, n f denote the scale of relation. Then from the nature of the scries, we have ma -\-nb = c ) ml-{-nc =. d\ These two equations will determine the values of m and n. 2d. To find the sum of the scries, denote the terms of the series* by A, B, C, etc., and let S'= A+B+C+D+E+ ... (1) RECURRING SERIES. 333 TLc scries is supposed to contain the ascending powers of x, the first power occurring in either A or B. Then because the series is of the second order, we have C = mAx*+nBx D = mBx*+nCx , v E = mCx*+nDx etc. etc. etc. Adding these equations, and observing the value of S in (1), we have SAB = mx*S+nx(S A). Whence we obtain 39O. To find the scale of relation and the sum of a recurring series of tlie third order. 1st. Let a, b, c, rf, e,J\ represent the coefficients of any six con- secutive terms ; also represent the scale of relation by m, 72, r. Then we have, ma-\-nb-\~rc = d \ ml)~\-nc-\-rd = e > (7*) mc-\-nd-\-re = f ) These three equations will determine the values of m, n and r. 2d. To find the sum of the series, represent the terms by, A, B, C, etc., and put S=A+B+C+D+E+F+ ... (1) Then because the series is of the third order we shall have D = mAx*+nBx*+rCx'\ E = m&tf+nCx*+rDxf F = mCx*+nDx*+rEx{ etc., etc., etc., etc. y By addition, observing the value of S in (1), we have SABC = vnx'S+nx\S A)+rx(S A J5) we obtain A++C\ -- 1 rx IMC* nix* In Hkv manner formulas may be obtained for the summation of recurring series of higher orders. 334 SERIES. 391. To apply these formulas in the summation of any given series, we must first determine the scale of relation by (P) or (I 7 ), and then we may obtain the sum of the series from (Q) or (V). If the order of the series is not known, we should first deter- mine the values of m and n by formula (P). and ascertain by trial whether the scale of relation thus found will apply to the givon series. If it will not apply, we may determine the values of m, //, and V from formula (T\ and ascertain by trial whether the series can be developed by the new scale thus obtained. If this also fail, we must establish other formulas corresponding to still higher de- grees, and continue the trials. If, however, we resort in the first place to a formula corresponding to an order higher than that of the given series, then one or more of the quantities, m, n, r,' etc., will prove to be zero, and the re- maining numbers may be taken as the scale of relation, without further trial. 1. Find the sum of the infinite series, 1 -f- 4x -f- 10x f -j- 22x* -f 46x 4 -f .... To determine the scale of relation, we have a = 1, I = 4, c = 10, d = 22. These values substituted in formula (/*), give m+4 = 10, 4w+10 = 22, from which we readily obtain m 2, n = 3 These numbers form the true scale of relation ; for we perceive that any coefficient after the second in the given series, is equal to three times the first preceding coefficient, minus twice the second preceding coefficient. To find the sum of the series, we have A = 1, B = 4*. Whence by formula (Q) +x ~ - ' We have "thus obtained the sum of the series in the form of an RECURRING SERIES algebraic fraction. Conversely, the given series may be developed from this fraction, either by division, or by the method of indeter- minate coefficients. Indeed, it will be found that the sum of every recurring scries is an irreducible fraction, from which the series may be supposed to originate. The fraction from which any partic- ular series is supposed to arise, is called the generating fraction for that series ; it is the same as the sum of the series. EXAMPLES FOR PRACTICE. 1. Find the sum of l+3*+4*+7*'+ll* 4 + .... 1+2* Ans. 2. Find the sum of l+6*+12*'+48*'+120* 4 + .... 1+5* Ans. -r- 3. Find the sum of 1+2* 5*'+2G*' 119* 4 + .... + 6.r Ans. 4. Find the sum of 1 -f 4o; + 3*' 2x* + 4 x< + 1 7ar* + 3 *" + ^Jl* 5. Find the sum of l+3*+5* f +7*'+9* 4 + Ans. (I * 6. Find the sum of l+*+5**+13* > +41.r 4 +121* B + Am. 12* 3** 7. Find the sum of l+4*+6**+ll* a +28* 4 +63* 6 + Ans. x 7r* fil r* 8. Find the sum of - + x'+ ' + 10*'+ -^- +91*"+ . . . iu a u Ans. - r-=- 336 SEEIES. DIFFERENTIAL METHOD. 393. Tlic Differential Method is the process of finding any term of a regular scries, or the sum of any number of terms, by means of the successive differences of the terms. 303. To find any term of a series ly the differential method. If we subtract cacli term of a series from the next succeeding term, we shall obtain a new scries called t^Q first order of differences. If we subtract each term of this new scries from the succeeding term, we shall obtain a series called the second order of difference* ; and so on. Let r/, &, c, d, c, .... represent a regular series, the successive terms being formed according to any fixed law. We will write the given terms in a vertical column, and proceed by actual subtraction to form the several orders of differences, placing each order in a separate column, and each difference at the right of the subtrahend. The result is as follows : .Series. 1st order of differences. 2'1 onler of differences. 3d order of differences. 4th orler of differences. a I & a c c I c _2i+a d dc d2 c-f-6 rf_3 c _|_3i_ a e cd c 2d+c c3d+Scl e_ 4- d. 2 + V - . To find the sum of any number of terms of a series, by ike. differential method. Represent the given series by a, b, c, d : e, . . . . (1) And denote the sum of n terms by S. We are to find the values of S in functions of a, d 19 d ZJ <7 3 , etc. ,Lct us assume the auxiliary series, 0, ., -f-&, tt-f6-|-c, a-f6+c+rf, ---- (2) it is ob^.ious that the (n-(-l)th term of this series is the same as the sum of n terms of the given series, (1), and may be placed equal to S. Now let c?V . The use of formulas (A) and (_) may be illustrated by the following examples : 1. Find the 12th term of the series, 1, 5, 15, 35, 70, 126, etc. We first form the successive orders of differences, as follows : 1, 5, 4, 15, 10, 6, 35, 20, 10, 4, 70, 35, 15, 5, 1, 126, 56, 21, 6, 1, 0. Thus we have n = 12, and - ** = !, d,=4, V 15. Sum the series I 4 , 2 4 , 3 4 , 4 4 , 5 4 , etc., to n terms. n* ?i 4 n* n ^ ^+2~ +3 ~ sir 16. Sum the series (m+1), 2(m+2), 3(m+3), 4(m-|-4), etc., to n terms. INTERPOLATION. 3OG. Interpolation is the process of introducing between the terms of a series, intermediate terms which shall conform to the law of the series. It is of great use in the construction of mathe- matical tables, and in the calculations of Astronomy. 397. The interpolation of terms in a series is effected by the dif- ferential method. In any series, the value of a term which has n terms before it is expressed by formula (wi), (393), which is , n(n 1) T nn lYn - 2 n - -- ; d , If in this formula we make n a fraction, then the resulting equa- tion will give the value of a term intermediate between two of the given terms, and related to the others by the law of the series. If n is less than unity, the intermediate term will lie between the first and second of the given terms ; if n is greater than 1 and less than 2. the intermediate term will lie between the second and third of the given terms \ and so on. INTERPOLATION. 341 = 2.758924 , '22 = 2.802039 / to find the cube roots Given ^23 = 2.843807 / of intermediate nuin- '24 = 2 88450 Abers } by, interpolation. . '25 = 2.924018/ 1. Required the cube root of 21.75. We have No. Cube Boots. d, d, ** d 4 21 2-758924 22 2.802039 +.043115 23 2.843867 -[-.041828 .001287 24 2.884501 +040632 .001196 +.000091 25 2.924018 +.039519 .001113 +.000083 .000008 Hence, to find the cube root of 21.75 by the formula, we have a = 2.758924, n = .75, d l = +.043115, d. 2 - .001287, d 3 = +.000091, etc. These values substituted in the formula, give 1st term, +2.758924 " + .032336 " + .000121 2d 3d 4th Whence, + .000004 2.791885, Arts. If it were required to find the cube root qf any number between 22 and 23, we might put n equal to the excess of the number above 21, and employ the same values for c/,, c? 2 , t? 3 , etc., as before. But greater accuracy will be attained by making 22 the first term of the series, and employing the corresponding differences ; in which case n will be a proper fraction. EXAMPLES FOR PRACTICE. Find by interpolation, 1. The cube root of 21.325. 2. The cube root of 21.875. 29* Ans. 2.773083. An*. 2.796722. 342 SERIES. 3. The cube root of 21.4568. Ans. 2.778785. 4. The cube root of 22.25. Ans. 2.812613. 5. The cube root of 22.684. Ans. 2.830784. 6. The cube root of 22.75. Ans. 2.833525. 398. On three consecutive days, the angular distances of tho sun from the moon, as seen from the earth, were as follows : 1st day, noon, 66 6' 38". " " midnight, 72 24' 5". 2d " noon, 78 34' 48". " " midnight, 84 39' 4". 3d " noon, 90 37' 18". " " midnight, 96 29' 57". In the data here given, the interval of time is 12 hours. Hence, to find the distance of the sun from the moon at intermediate times, ft must always be some fractional part of 12. Thus, for the distance at 3 o'clock p. M. of the first day we have n = -fy = J, and a =. 66 6' 38" ; for the distance at 6 o'clock A. M. of the second day, n = T 2 = A, and a = 72 24' 5". For the distance at 3 o'clock p. M. of the second day, n = T \ = |, and a = 78 34' 48". EXAMPLES FOR PRACTICE. Find by interpolation the distance of the sun from the moon, 1. At 3 o'clock p. M. of the first day. Ans. 67 41' 38" 2. At 6 o'clock p. M. of the first day. Ans. 69 16' 13" 3. At 9 o'clock p. M. of the first day. Ans. 70 50' 21" 4. At 3 o'clock A. M. of the second day. Ans. 73 57' 23" 5. At 6 o'clock A. M. of the second day. Ans. 75 30' 16" 6. At 9 o'clock A. M. of the second day. Ans. 77 2' 44" 7. At 3 o'clock p. M. of the second day. Ans. 80 6' 27", 8. At 6 o'clock p. M. of the second day. Ans. 81 37' 43", 9. At 9 o'clock p. M. of the second day. Ans. 83 8' 35" LOGARITHMS. LOGARITHMS. The Logarithm of a number is the exponent of the power to which a certain other number, called the Lage, must be raised, in order to produce the given number. Thus, in the expression, to the base a. An equation in this form is called an exponential equation. If in this equation we suppose a to be constant, while b is made equal to every possible number in succession, the corresponding values of x will constitute a system of logarithms : hence, 4OO. A System of Logarithms consists of the logarithms of all possible numbers, according to a given base. Any positive number greater than unity may be made the base of a system of logarithms. For, by giving to x suitable values, the equation a x = b will be true for all possible values of &, provided a is positive and greater than 1. Hence, There ma./ be an indefinite number of systems of logarithms. SOB. If in the equation a* = b, we suppose b to represent a perfect power of a, then x will be some integer but if /> is not a perfect power of , then x will be some fraction. Hence, A logarithm may consist of an integral and a fractional part. 4O2. The Index or Characteristic of a logarithm is the integral part 5 and <1OS. The Mantissa is the fractional part of a logarithm. For illustration, let 5 be the base of a system ; then we have 5 2 ' 25 = 5+ = VV = 37.384. Thus, the logarithm of 37.384 to the base 5, is 2.25 ;Vthe index of this logarithm is 2, and the mantissa .25. PROPERTIES OF LOGARITHMS. 4LO4. There are certain properties of logarithms, which are common to all systems. To investigate these general properties, let 344 SERIES. a denote the base of the system ; also, designate the logarithm of a quantity by log., written before the quantity. 1. In am/ system, the logarithm of unity is 0. For, let a* =. 1 ] then x ==. log. 1. But by (88), if a* = 1, then x = 0, or log. 1 = 0. 2. In any system, the logarithm of the base itself is unity. For, let a* = a ; then x r= log. a. But by (88), if a* = a, then x = 1, or log. a = 1. 3. The logarithm of the product of two numbers is equal to the sum of the logarithms of the two numbers. For, let MI = a z , n = a* ; then x = log. m, z = log. n. But by multiplication we have mn = a*+' ; therefore, log. mn = x-\-z = log. m-j-log. n. 4. The logarithm of a quotient is equal to the logarithm of the dividend diminished by the logarithm of the divisor. For, let m = a x y n = a* ; then x = log. m, z = log. .. By division we have _ a *-* n. therefore, log. ( j = x z = log. m log. 5. The logarithm of any power of a number is equal to tht logarithm of the number multiplied by the exponent of the power. For, let m = a* ; then x = log. m. By involution we have m r = a n ; therefore, log. (m c ) = rx = r log. m. 6. The logarithm of any root of a number is equal to the logarithm of the numbei' divided by the index of the root. For, let m = a* ; then x log. m. By evolution we have LOGARITHMS. . The principal use of logarithms is to facilitate arithmet- ical computations. By means of the last four properties, we may avoid the ordinary labor of multiplication, division, involution, and evolution, these operations being practically performed by addition and subtraction. For this purpose, it is necessary to have a ToUe of Logarithms, so constructed that we may readily obtain the logarithm of any numbqr within a certain limit, or the number corresponding to any logarithm, to a certain degree of approximation. The common tables give the logarithms of numbers from 1 to 10,000, correct to 6 decimal places. With a table of this kind, we have the following obvious RULES FOR COMPUTATION. I. To multiply one number by another : Find the logarithms of the given numbers ; add these logarithms, and find the number corresponding to the sum ; this number will be the required product ; (404, 3). II. To divide one number by another : Find the logarithms of the given numbers ; subtract the logarithm of the divisor from that of the dividend, and find the number corresponding to the difference ; thi* number icill be the required quotient ; (4O4, 4). III. To raise a number to any power : Find the logarithm of the given number, and multiply it by the exponent of the required poirer ; then find the number corresponding to this product, and it will be the required power ; (4O4. 5)- IV. To extract any root of a number : Find the logarithm of the given number, and divide it by the index of the root ; then find the number corresponding to the quotient, and it trill be the required ror>t; (4O4, 6). NOTE. From (4OO), we infer that negative numbers, as such, have no logarithms. But we may always employ logarithms in calculations where negative factors are involved, by disregarding signs until the absolute value of the product or quotient is obtained. 346 SERIES. THE COMMON SYSTEM. <1O6. Any positive number except unity may be made tho base of a system of logarithms. But the only base used in practic- al calculations, is 10. The logarithms of numbers according to this base, form what is called the Common System of logarithms. NOTE. Besides the common system, there is another, called the No- perian System, from Baron Napier, the inventor of logarithms. This system is of great theoretical importance, and its relation to other systems will be shown in a subsequent article. 4O7. The peculiarities which constitute the advantage of tho common system, may be shown as follows : Since 10 is the base of the system, log. 1 = log. 10, = 0, log. 10 = log. 10 1 = 1, log. 100 = log. 10 3 = 2, log. 1000 = log. 10 3 = 3, log. 10000 = log. 10* = 4. Now it is obvious that if any number, integral or mixed, be great- er than 1 and less than 10, its logarithm will be entirely decimal; if the number be greater than 10 and less than 100, its logarithm will be 1 plus a decimal ; if greater than 100 and less than 1000, its logarithm will be 2 plus a decimal ; and so on. Hence, 1. The common logarithm of an integer or a mixed number icill have a positive index, equal to the number of integral places minus 1 . Again, since the logarithm of 10 is 1, it follows that if a number be divided by 10 continually, the logarithm will be diminished by 1 continually, the decimal part remaining unchanged. Let us take any number, as 5468, and denote the mantissa, or the decimal part of its logarithm, by m. Then we have (1.) (2.) log. 5468 = 3-fm, log. .5468 = 1-j-m, log. 546.8 = 2-f w, log. .05468 = 2-fm, log. 54.68 = 1-fm, log. .005468 = 3-f-m, log. 5.468 Q-fm; log. .0005468 = 44m; LOGARITHMS. ?>41 in which 3, 2, 1, 0, are the indices of the logarithms in column (1); and 1, 2, 8, 4, are the indices of the logarithms in column (2) and m, the decimal part in all. Hence, 2. If two numbers consist of the same figures, and differ only in ike position of the decimal point, their logarithms, in the common sys- tem, will have the same decimal part, and will differ only in the val- ues of the index. 3. The common logarithm of a decimal fraction will have -a negative index ; if the significant part of the decimal commence at the tenths' place, the index of the logarithm will be 1 ; but if ci- phers occur between the decimal point and the first significant figure, the index of the logarithm will be numerically equal, to the number of intervening ciphers, plus 1. 4O8. In writing the logarithm of a decimal fraction, the minus sign is placed before the index, and the decimal or positive part an- nexed without any intervening sign. Thus, from a table of loga- rithms, we have log. .0546 = 2.737193, in which the minus sign must be understood as affecting only the index 2. This logarithm is therefore equivalent to 2+.737193. COMPUTATION OF LOGARITHMS. 4OO. Since the rules for computing by logarithms require a logarithmic table, it becomes necessary to calculate the logarithms of an extended series of "numbers. The only practical method of doing this, is by means of a converging series, expressing the value of any logarithm in known terms. Let us resume the fundamental equation, a* = b, (1) in which x is the logarithm of b, to the base a. i Assume a == 1 -j-c, b 1-j-p ; then (1-M" = l+P, (2) where x is the logarithm of 1-f-p, to the base a. SERIES. Raise both members of equation (2) to the nth power; then (1+e)"* = (1-f />)" Expanding both members by the Binomial Theorem, we have C-l)(-2) O-1)0 2)Q 3) Dropping unity irom both Tnembers, and dividing by ?i, we obtain / (na-1) (ng-l)(ng-2) (wa-l)(na?-2Xg-8) , , \ aj ^ c " 1 2 2-3 2.3.4 ") = f + fi^+ ( -^?/+ ( -^^^-V+. - - This equation is true for all values of n ; it will* be true, there- , fore when n = 0. Making this supposition, the eouation reduces to +*' '- 2 + if -4 + 5 - -f + - +-- ( From equation (2), we perceive that ' * ^ \ rr --= log. (1+p). Hence, if we place 1 equation (3) will become Thus, we have obtained an expression Tor the logarithm of the number l-}-p, or b. This expression consists of two factors; namely, the quantity in the parenthesis, which depends upon the number, .and the quantity Jl/, which depends upon the base of the system. -41O. It is obvious that if a definite value be given to .If, the base of the system will be fixed and determinate. Baron Napier arbitrarily assumed M= 1. To determine the base of the system, according to this assumption, substitute 1 for M in equation (4) ; (4rOO). We shall have, after reducing, LOGARITHMS. 349 Putting s = 1, we shall have c 9 8 4 * =C " + 3 .Reverting the series, we obtain s' Restoring the value of s, = l + F2 + F^"3 + 1-2-3-4 + 1-2-3-4-5 + * * ' ' By taking 12 terms of this series, we find the approximate value of c to be 1.7182818. But the base ^as 1-f-c; hence, adding 1 to the resultyand representing thexsum.by e, the- usual symbol for the Naperian base, we have e = 2.T182818, which is the base of the Naperian system. 411. In the general formula, (-4), the quantity Jf, which depends upon the base, is called the modulus of the system. Thus, the modulus of the Naperian system is unity. Let us here designate Naperian logarithms by nap. log., and log- arithms in any other system by log., simply. Then, Dividing (1) by (2), we obtain M = I g-( 1 +^) . (3) nap. log. (l+j>) ' or, {nap. log. (1+p) JX^= log. (l+j>), (4) where M is the modulus of the system in which the logarithm of the second member is taken. Hence, The modulus of any particular system i the constant multiplier which will convert Naperian logarithms into the logarithms of that system. 30 350 SERIES. . Formula (A~) can be employed for the computation of logarithms, only when p is less than unity; for if p be greater than unity, the series will be dwergintj. The series, however, may be transformed into another which will be always converyiny. Let us resume the logarithmic series, If in this equation we substitute p for p, we shall If we subtract equation (2) from equation (1), observing that we sliall Assume These values substitu ; whence we obtain = 1 p x ^- 1 6 + 72*!' + ' ' / ' 4- 1) 6 7(2*+!) ' ' ' The first member of this equation is equivalent to log. (z-f-1) log. z. Hence, finally, we have log.3 = +l) 7+ " / : This series is rapidly converging, and may be employed with facility for the computation of logarithms, in the Naperian, or in the common system. To commence the construction of a table, first make z = 1 ; then log. = 0, and the formula will give the value of log. (z-f-1), or log. 2. Next make z = 2 ; then the formula will give the value of log. (z-f 1), or log. 3 ; and so on. LOGARITHMS. 353 It is necessary to compute directly the logarithms of prime num bers only, in any system; for, according to (404, 3), the loga rithm of any composite number may be obtained, by adding the logarithms of its several factors. 413. We will now illustrate the use of formula (J?) } by com- puting the Nuperian, logarithms of 2, 4, 5, and^lO. Make 2 = 1; then nap. log. 2 = 0, and nap. log. (z-j-1) = nap log. 2 ; and since M= 1, we have 3 _ _ _.. We first form a column of numbers, by dividing | by 3 s , or 9, continually; then dividing the first of these members by 1, the second by 3, the third by 5, and so on, we obtain the several terms of the series. 32 JL .66666666 -r- 1 = .66666666 7407407 ^ 3 2469136 823045 -f- 5 == 164609 91449 -4- 7 = 13064 10161 -v- 9 = 1129 1129 -4-11 = 103 125 -j-13 = 10 14 -^15 = 1 .69314718 = nap. log. 2. 2 Whence, by (4O4, 5), 1.38629436 = nap. log. 4. Next make 2 = 4; then 2-j-l = 5 ; and 22-j-l = 9 ; and we have nap. io S .5= 2 ( i L + 1 L. + ^L. + T i- + ....) +nap . lo , ; .4. 9 = 81 81 81 2 .22222222 274348 3387 42 + 1 3 5 7 = .22222222 91449 677 6 .22311354, sum of series 352 SERIES. To .22314354 Add nap. log. 4 = 1.38629436 1.60943790 = nap log. 5. Add nap. log. 2= .69314718 Whence, by (4O4, 3), 2.8025? 508 = nap. log. 10. 414. In order to compute common logarithms, we must first determine the modulus of the common system. From (411), equation (3), we have "" nap. log.(l+p) In this equation, make 1+p = 10, the base of the common system. Then we have the value of the modulus sought. Substituting this value in formula (*), we obtain the formula for common logarithms, as follows : log. (2+1) log. z = (O To apply this formula, assume z = 10 ; then log. 3 = 1, and 23+1 = 21. .86858896 ^04136138 -f- 1 = .04136138 > 9379 + 3 = 3126 21 + 5 = 4 21 441 .04139268, sum of scries. Add log. z = 1.0 log. (2+1) = 1.04139268 = log. 11. If we make z = 99, then z+1 = 100, and 2*+l = 199. In this case, the formula will give the logarithm of 99; for, log. (2+!) log. z = log. 100 log. 99 = 2 log. 99. 199 199 s = 39601 .80858896 436477-7-1 = .00436477 11 -j- 3 = 4 .00436481, sum of series. LOGARITHMS. 353 Therefore, we have 2 log. 99= .00436481, whence, 1.99563519 = log. 99. Subtract log. 11 = 1.04139268 .95424251 ''= log. 9 And by (4O4, 6), 4 log. 9 = .47712126 = log. 3. Thus we may compute logarithms with great facility, using the formula for prime numbers only. USE Or TABLES. S 175. The following contracted tables will illustrate the princi- ples of logarithms, and the methods of using the larger tables. The logarithms are taken in the common system. TABLE I. LOGARITHMS FROM 1 TO 100. H, Log. i N< Log. N. Log. i N. Log. 1 OWOOO 26 1 414973 51 1 707570 i 76 880814 2 301030 27 1 431364 52 1 716003 77 886491 3 477121 28 1 447158 53 1 724276 78 892095 4 602060 29 1 462398 54 1 732394 79 897627 5 698970 30 1 477121 55 1 740363 80 903090 6 778151 i 31 491362 56 1 748188 81 908485 7 845098 32 505150 57 1 755875 82 913814 8 908090 33 518514 58 1 763428 83 919078 9 954243 i 34 531479 59 1 770852 84 924279 10 1 000000 35 544068 60 1 778151 85 1 929419 11 1 041393 36 1 55f5303 61 1 785330 86 1 934498 12 1 079181 37 1 568202 62 1 792392 87 1 939519 13 1 113943 38 1 579784 63 1 799341 88 1 944483 14 1 146128 39 1 591065 64 1 806180 89 1 949390 15 1 176091 40 1 602080 65 1 812913 90 1 954243 10 1 204120 41 1 612784 66 1 819544 91 1 959041 17 1 230449 42 1 623249 67 1 826075 92 1 963788 18 1 255273 43 1 633468 68 1 832509 93 1 968483 I 19 1 278754 44 1 643453 ! 69 1 838849 94 1 973128 20 1 301030 T 45 1 653213 70 1 845098 95 1 977724 i 21 1 322219 ! 46 1 602758 ' 71 1 851258 08 1 982271 ! 23 T 342423 47 1 672098 72 1 857333 97 1 986772 i 23 1 -J 6 1 728 48 1 681241 73 1 863323 98 1 991226 24 1 380211 49 1 690196 74 1 869232 99 1 995635 25 1 397940 50 1 698970 75 1 875061 100 2 000000 30* x 854 SEBIES. TABLE II. LoGAiUTinis OF LEADING NUMBERS WITHOUT INDICES. JS. 0. 1. o 3. 1 4. ,5. 6. 7. 8. 9. 100 000000 000434 000868 001301:001734 002160 002598 003029 003461 008891 101 0043'21 004750 005181 005609 006038 006466 006894 007321 007748 008174 102 008600 C09026 00945$ 00987ti 0103CO 010724 011147 011570 011993 012415 108 012837 013251; 013680 014100 014521 01494C 015300 015779 016197 016616 104 01703J-- 017451 01780K 018284 0187CO 01911(1 019532 019947 020o6.1 020775 105 021189 021603 0220 Hi 022428 022841 023252 02301)4 024075 0244 8C 024896 100 025306 02571,1 02G125 020538 026942 027351 027757 028164 028571 028978 107 02938, 02978H 030195 080600 031004 03140fc 031812 Oo2216 032;; if OS3021 108 033424 033826 03-122^ 034628 035029 03543C 035830 036230 036621 037028 109 037420 037825 038223 038620 039017 039414 039811 040207 04060!: 040998 In table I, the logarithms are given, with indices, in columns adjacent to the columns of numbers. In table II, each figure in the row at the top may be annexed to any number in the left-hand column ; the logarithm of any number thus formed, will be found at the right of the number in the column, and beneath the figure at the top. The proper index may be supplie 1 in any case, according to the theory of logarithms. Thus, to obtain the logarithm of 1023 by this table, we find 102 in the left-hand column, and 3 in the top row j and opposite the former, and under the latter, we 'find 009876, the decimal part of the logarithm. Hence, log. 1023 = 3.009876. In like manner, we find log. 104.2 = 2.017868, log. .1078 = 1.032619. CASE I. 416. To find the logarithms of numbers when their factors are in the tables. RULE. Take out from the tables the logarithms of the factors, and find their sum ; the result will be the logarithm required. EXAMPLES FOR PRACTICE. 1. Required the logarithm of 533.5. Observe that 533.5 = 106.7x5 ; hence, log. 106.7 = 2.028164 log. 5 = .698970 2.727134, AM. LOGARITHMS. 355 2. Find tho logarithm of 520. 'tfjAm. 2.710003 3. Find the logarithm of 146. f 3 X ?- Ans. 2.164353. 4. Find the logarithm of 1450. if+^ + f* Am. 3.161368. 5. Find the logarithm of 1.59. '^l^^^l^Ans. .201397. 6. Find the logarithm of 2034. (ifj y ^ Ans. 3.308351. 7. Find the logarithm of 76.37. SO- ?/ *7 Ans. 1.882923. 8. Find the logarithm of .0201..^**^ *- .4ns. 2.303196.^*7* 9. Find the logarithm of .3822. .f/ A-4 2- Ans. 1.582290. /."//' v 10. Find the logarithm of 16995./0-3/ 33+f- Ans. 4.230321. 417. To find the logarithms of numbers intermediate between the numbers on the table. Since the logarithms in any table form a regular series, we may interpolate for intermediate logarithms, by the usual formula, If the logarithm of the given number is intermediate between the logarithms of table I, it will be necessary to take account of the first and second differences. But we may always employ table II, where the logarithms increase so slowly that two terms of the formula will give the result accurately. The first four figures of a number, counting from the left, will be called the four superior figures ; and the others, the inferior figures. To apply the formula, a will represent that logarithm of the table which is next less than the required logarithm, and n' will denote the inferior- figures of the number, regarded as a decimal. Hence the following RULE. Take out the logarithm of the four superior figures of the given number ; multiply the difference between this logarithm and the next greater in the table, by the inferior places of the num- ber, considered as a decimal; add this product to the former resulf, cde. . ..p^m-odt etc., ^1= ^fabcde. . .pq rH ^ l ^bcdef. . .pq n ^ l + etc., U ' = abed. . . -pq m j the subscript expressions m 2, m 1, m, denoting the number of literal factors which enter each term. We thus have the identical equation, ) j ~ -f U and placing the second member of this equal to zero we have x m +^x m - 1 -^Ex"*-- -f- ____ Sx*+ Tx+ U = (2) an equation of which a, 6, c, . . . . jp, ^ are the roots, since these values substituted in succession for x in the first member of the eq. (1) will cause this first member, and consequently the second member, to vanish. The relations between the coefficients A t B, C, etc., and the roots of eq. (2), may be expressed as follows: 1 The coefficient of the second term is equal to the algebraic xum of all the roots, with the signs changed. 2. The coefficient of the third term is equal to the algebraic sum of all the different products formed by multiplying the roots, two and two. 3. The coefficient of the fourth term is equal to the algebraic mm of all the different products formed by multiplying the roots with their signs changed, three and three. 364 PROPERTIES OF EQUATIONS. 4, And in general ; The coefficient of the term having n terms before it, is equal to the algebraic sum of all the different products formed by multiplying the roots, with their signs changed if n ?'.-: odd, n and n. Hence, 5. The absolute term is the continued product of all the roots, with their signs changed when the number denoting the degree of the equation is odd. This principle will enable us to construct an equation, the roots of which are given, and the composition of eq. (1) shows that eq. (2) thus constructed can have no other than the assumed roots ; for there is no value of x differing from one of these roots which can cause the first member of eq. (1) to disappear. From this we might conclude that every equation involving but one unknown quantity, has as many roots as there are units in the exponent of its degree, and can have no more. 4^*5. Admitting that every equation containing but one unknown quantity has at least one root, real or imaginary, it may be demonstrated that the first member of every equation of the with degree, the second member being zero, may be regarded as the continued product of m binomial factors of the first degree with respect to the unknown quantity. We will first prove that, If & is a root of an equation of the form x^Ax^ 1 -^- B.x~~*+ .... Tx+U= 0, (1) its first member can be exactly divided by x a. For if we apply the rule for division, we shall finally arrive at a remainder which will not contain x j since for each quotient term obtained, the new dividend is at least one degree lower than that which precedes. Calling the entire quotient Q and the remainder R, we shall have an identical equation. The substitution of a for x causes the first member, and also the first term in the second member of this equa- tion, to vanish. Hence, R = 0. But by hypothesis R does not contain x ; it is therefore equal to zero whatever value be attributed to x, and the division is exact PROPERTIES OF EQUATIONS. 365 4 2G. The converse of the last principle is also true; that is, Jf the first member of the equation, x m +Ax m - l -\-x m -*-}- ____ Tx+ 7=0, can Ic exactly divided ~by x a, then a is a root of the equation. For, suppose the division performed, and that the quotient is Q; then we shall have the identical equation, x m -\-Ax m - 1 +JBx m -* + ____ Tx-\- U = Q(x a). But x = a causes the second member of this equation to vanish ; it will therefore cause the first member to vanish, and consequently satisfy the given equation. 42 7 Every equation containing but one unknown, quantity has a number of roots denoted l>y the exponent of its degree, and no more. Resuming the equation, and admitting that it has one root, o, x a must be a factor of its first member ; (425). The quotient which arises from the division of the polynomial, x m +Ax m ~ 1 -}- ____ Tx-\- U, by x a, will be of the form ' we shall therefore have the identical equation, Now the second member of this equation will vanish for any value of x which reduces the second factor to zero. If then the assumed root of the equation, x- l +A'x~-*+ .... Fx+ U' = 0, be denoted by &, we shall have r~ l +A'x~-*+ I , 7 , (x m ~^A"x^ '+ .... T"x+ U"\ ...T'x+U' \ A third equation may be formed in the same way, and then a fourth, and so on, until the (in l)th equation is finally reached, in which the second factor in the second member is of the first degree with respect to x. 35* 366 PBOPEliTIES OF EQUATIONS. Taking this last equation, and substituting for its first member the second, in the next preceding equation, and thus continuing the process of substitution until the first equation of the scries is arrived at, the result will be the following identical equation : ) (.r-)(x-^)(x c).... J = = ....Tx+V - ( (*-?)(*- 2) The second member of this equation vanishes for any one of the in values, x = a, x = 6, x = c, . . . . x = p, x = q, and consequently these values are severally roots of the equation, Moreover, no value of x that differs from some one of these values, can satisfy the equation ; for no such value will cause any one of the factors in the second member of the identical equation to be zero, a condition requisite to make the product zero. The equation therefore has ra roots and no more. 428. From the foregoing principles we conclude, 1. That in an equation in which the second term does not ap- pear, that is, the term containing the next to the highest power of the unknown quantity, the algebraic sum of the roots is 0. 2. If an equation has no absolute term, at least one of its roots is 0. 3. The absolute term being the continued product of all the roots of an equation, it must be exactly divisible by each of them. 4. An equation may be constructed, which shall have any assumed roots. 5. The degree of an equation may be reduced by 1 for each of its known roots. EXAMPLES. 1. What is the equation having -f-2, 3 for its roots ? Ans. x*-\-x 6 = 0. 2. What is the equation having the roots -(-1? 2, 4 ? Ans. oj'-f-Sx'H- 2x 8 = 0. 3. What is the equation having for its roots -j-3, 2, 1, -|-5 ? Ans. x 4 5x 7x a +29x+30 = 0. PROPERTIES OF EQUATIONS. 367 4. What is the equation of which the roots are 1-fV 5, 11/^5, -f v'5, !/5 ? .4/zs. x- 4 2x 3 -fx a 4-10.r 30 = 0. 5. What is the equation of which the roots are 1, 2, -|-3, 2+1/ZT3, 2 1/H3 ? 4n. r 6 4* 4 + 22** 25x 42 = 0. 6. One root of the equation x_5;r 8 -f 13x 21 = is -j-3 ; what is the reduced equation ? Ans. x* 2x-{-7 = 0. 7. One root of the -equation Z 4 -{-2x 3 34.r 2 4-12*4-35 rrr is 7 ; what is the depressed equation ? Ans. x 3 5x*+x-\-5 = 0. 8. Two of the roots of the equation x'Sx* 4*"+30a- 36 = are -f-2, 3 ; what is the depressed equation, and what are its roots ? r The depressed equation is Ans. J x* kt'4-6 = ; (and its roots are 2+l/~2, 21/^2. 4t2O. -4w,?/ equation having fractional coefficients can be trans- formed into another in which the coefficients are entire, that of the first term being unity. If the coefficient of the first term of the given equation is not unity, make it so by dividing through by this coefficient. Then the equation will be of the form. x m +Ax m - l -{-Bx m r*-\- ____ Tx-\- U = 0, in which it is supposed that some or all the coefficients, A, B, etc., are fractional. Assume x = , (/ being entirely arbitrary, and substitute this a value of x in the equation j it then becomes _ M *ri + j: .7-^+^7=0. a m ~ a" 1 - 1 n a" 1 - 2 ~ a ' Whence, by multiplying through by a TO , '-*+ .... T(j-*y+ Ua m = 0. Now since a is arbitrary, its value may be so selected that it and its 368 PROPERTIES OF EQUATIONS. powers will contain the denominators of tlic fractional coefficients of the original equations. We present the following examples for illus- tration. 1. Transform the equation, x . + - + - + - = o, in u ' p into another which shall have no fractional coefficients, and which shall have unity for its first coefficient. Make x = ; substituting this value of a*, the equation bc- ninj> ' comes 7/ s ml* l>i/ c ' I J \ ' [ _ _._ Q ?}i 8 y< 3 ^ 3 flu 3 //*/? 9 nih*p p ~ Multiplying every term of this by m s n 3 j> 3 , we havo y*-\-cuipy*-\-l>ni l np !l y-}-cm*>*p t =. 0. When the denominators of the coefficients have common factors, we may make x equal to y divided by the least common multiple of the denominators. ax 9 bx c 2. Transform the equation x -] -J -I = 0, into another 1 2> })l m p which shall have no fractional coefficients, and that of the first term be unity. To effect this it is sufficient to put x = . With this value of pm x the equation becomes jf__ , Wf_ , _^_ _, 1 _ Q 2> 3 iu* jni? ' pm* p Multiplying every term by^ s ??i 3 , we obtain f+ftf+lp^my^fm* = for the transformed equation required. 8. Transform the equation x*-\- ~ -\- -f- - -[-- = into another having no fractional coefficients. AM. ,y 4 -r-20/-|-18-24^-f-7(24)V+2(21) 8 = 0. PROPERTIES OF EQUATIONS. 360 In transforming an equation having fractional, into another with entire coefficients, in terms of another unknown quantity, it is im- portant to have the transformed equation in the lowest possible terms. The least common multiple of the denominators will not necessarily be the least value of a that will give the required equa- tion. If, in each case, the denominators be resolved into their prime factors, it will be easy to decide upon the powers of these factors to be taken as the factors of a. The following illustration will render further explanation unnec- essary. 4. Transform the equation, 3 13 17 ^.3 _ __ ~2 I _ _ "^ _ 35 "2450 68600" into another of the same form with the smallest possible entire coefficients. Writing y for x and multiplying the second, third and fourth terms, by a, a 5 , a 9 , respectively, we have o I q I "T y' _ & - a y+ __ a . x _ = 0. The denominators, resolved into their prime factors, are 7-5, 7 a -5 3 -2, 7 3 -5 a -2 3 ; and assuming a = 7 * 5 2, the equation may be written 3 -7 -5 -2 13 -r-tf'2* 17-7 3 -5 3 -2 3 which reduces to y_6y'+26y 85 = 0. In this example, the least common multiple of the denominators is 7 8 5 2 2 8 } and had this value been taken for a, instead of 7*5-2, the coefficients of the transformed equation would have been much larger than they are, as found above. When a root of the transformed equation is known, the corres- ponding root of the original equation will be given by the relation 8TO PROPERTIES OP EQUATIONS. COMMENSURABLE ROOTS. 4SO. A number is commensurable with unity when it can be expressed by an exact number of units or parts of a unit; a num- ber which can not be so expressed is incommensurable with unity. 431. Every equation having unity for the coefficient of the fir^t term, and for nil the other coefficients, whole numbers, can have only ichole numbers for its commensurable roots. This being one of the most important principles in the theory of equations, its enunciation should be clearly understood. Such equations may have other roots than whole numbers ; but its roots can not be among the definite and irreducible fractions, such as ~, J, ip, etc. Its other roots must be among the incommensurable quantities, such as j/2, (S)" 5 , etc.; i. e., surds, indeterminate deci- mals, or imaginary quantities. To prove the proposition, let us suppose , a commensurable but irreducible fraction, to be a root of the equation, x m -\- Ax^+Bx-* .... Tx+ U=Q, A, B, etc., being whole numbers. Substituting this supposed value of x, we have Transpose all the terms but the first, and multiply by b m ~ l y and we have Now, as a and b are prime to each other, b can not divide a, or any number of times that a may be taken as a factor ; for being irreducible, X & is also irreducible, as the multiplier a will not be measured by the divisor b ; therefore can not be expressed a m in whole numbers. Continuing the same mode of reasoning, - COMMENSURABLE ROOTS. 371 can not express a whole number, but every term in the other mem- ber of the equation expresses a whole number. Hence, the supposition that the irreducible fraction is a root of the equation, leads to this absurdity, that a series of whole numbi^s is equal to an irreducible fraction. Therefore, we conclude that any equation corresponding to these conditions can not have a definite commensurable fraction among its roots. 43^2. It has been shown (42O) that an equation having fraction- al coefficients, that of the first term being unity, can be changed in- to another of the same form, with entire coefficients. The expres- sion entire must there be understood in its algebraic sense ; that is, the new coefficients being entire merely in algebraic form, may be irrational or imaginary. In the preceding article it is proved that if these coefficients are whole numbers, all the commensurable roots of the equation are also whole numbers '; moreover, these roots must be found among the divisors of the absolute term ; (428). If the divisors of the absolute term are few and obvious, those answering to the roots may be found by trial substitutions ; but in most cases the labor will be abridged by the rule suggested by the following investigation : Suppose a to be a commensurable root of the equation, z" t -}-J.x m - 1 4- -f^x 3 -f Sj?+Tjc+U= 0. Writing a for -ce, transposing 'all the terms except the last to the second member, and dividing through by a, we have - = a"- 1 - Aa m -*. . . . RcfSaT. a But, since a is a root of the equation, is an entire number; trans pose T^to the first member of the last equation, make \- T d N } , and divide both members of the resulting equation by a ; it then becomes N 1 = a m ~z Aa~-* RaS. a The second member of this equation is a whole number; the first 372 PROPERTIES OF EQUATIONS. member is therefore entire ; and if S be transposed to this mem- ber, and - -f S be denoted by .A 7 ".,, we shall again have, after di- viding through by a, the equation, = -}(x c) .... (x ni)(x n) ; and since the sum of all the products that can be formed by multi- plying m factors in sets of m 1 and m 1, is the same as the sum of all the quotients which can be obtained by dividing the continued product of tho factors, by each factor separately, it follows that X X X X A 1= - -H -_{-...._}- j-- x a x b x m x n So likewise the sum of the products of the binomial factors taken m 2 and m 2, is the same as the sum of all the quotients obtained by dividing the continued product by all the different products of the binomial factors taken 2 and 2 ; that is, JL'a X_ X_ X_ o / ._ _. \ f / \ I / _.. > ( . ,.\ I * " I By like reasoning it may be shown that *. , , 2-8 - (xa\xV)(x-c) 1 t (*-a) (x-wi) (x *i x A so for the next coefficient in order, etc., etc. EQUAL ROOTS. 43*3. It has been seen (427) that if a, 6, c,. . . .,m, n arc the roots of the equation, X = x m + Ax n ~ l +Bx'-*+ ; . . . + 3Tx+ U = 0, it may be written, X= (xa}(xl~}(xc] . . .( x m}(xn) 0. Now if a number p of these roots are each equal to , a number q equal to b, and a number r equal to c, the last equation becomes X ( x a}P(xb^(xc) r (xm}(xn} = 0. But since A" contains p factors equal to x a, q factors equal to x i } r factors equal to x c, its first derived polynomial will con- X X tain the term p times, the term ^. a times, the term x.~~ ' x b EQUAL ROOTS.- 377 Y Y X r times, besides the terms > , etc., corrcspond- a: c x m x u ing to the single roots, (434) ; that is, P x ,.^ + j^ + __ + _^. + ^_. The factor (x a) p is found in every term of this expression for A", except the first, from which one of the p equal factors, x , has been suppressed by division. Hence, (x a} p ~ l is the highest power of x a, which is a factor common to all the terms of A",. For like reasons (x &)*" 1 , (x c) r ~* are the highest powers of the factors x &, x c, which arc common to all the terms of X 1 ; hence, is the greatest common divisor which exists between the first member of the proposed equation audits first derived polynomial. The supposition that the given equation contains one or more sets or species of equal roots, necessarily leads to the existence of this greatest common divisor. Conversely : if there be a common di- visor between A" and A", there must be one or more sets of equal roots belonging to the equation. For, if (x a)' be a factor of the greatest common divisor, then the composition of A", shows that (x- a)'* 1 is a factor of A*, and that a is therefore t-^-l times a root of the equation A"= 0. Hence the conclusions : 1. An equation involving but one unknown quantity, x, and of which the second member is zero, has equal roots ii there he between its first member, X, and its first derived polynomial A",, a common divisor containing x. 2. The greatest common divisor, D, of X and AT, , is the product of those binomial factors of A r , of the first degree with respect to x, which correspond to the equal roots, each raised to a power whose ex- ponent is one less than that with which it enters A". Therefore, To determine whether an equation has equal roots, and if so, to find them, if possible, we have the following E.ULE. I. S'-ek the greatest common divisor between the first member of the proposed equation and its first derived polynomial. 32* 878 PROPERTIES OF EQUATIONS. If no common divisor be found, there are no equal roots; but if one l>e found, there are equal roots ; in which case II. Make an equation by placing the greatest common devisor, D, equal to zero ; then any quantity which is once a root of D = ic ill be twice a root of X = 0j any Quantity which is twice a root of D = will be three times the root of X ; and so on. It will at once be seen that, if D contains a factor of the form (x a) 1 , t being a positive whole number greater than unity, and we denote the greatest common divisor which exists between D and its first derived polynomial D 11 by D r , then D' will contain the factor (x a)*" 1 . And, again, denoting by Z>', the first derived polynomial of Z>', and by D" their greatest common divisor, (x a)'~ 2 will be a factor of D". This process being continued, as the exponent of (x a), and consequently, the degree of the greatest common divisor, diminishes by one for each operation, it is plain that when the degree of the equation, #=;i>; is too high to be solved, we may in certain cases make the determin- ation of the equal roots depend upon the solution of equations of lower degrees, until finally one is obtained which can be solved. To illustrate, suppose that for the equation, JT=0, it is found that D = (x a)\x b) n (x c) ; then D' = (x a)"-i(a>- &)*-, D" = (x a y-*(x &)*-, The equation, /* > = (x a)(x 6) = 0, may be solved, giving the roots x = a, x = b, and (*-a)*M, (*-&)+', (x c)', are factors of X, or a and b are each n-}-l times roots, and c twice a root, of the equation, X=0. Dividing the given equation by the product, (x )"+ (x 6)"+ (x c) 3 , its degree will be depressed 2n-{-4 units. EQUAL KOOTS. 379 EXAMPLES. 1. Docs the equation x 4 2x 8 7x a -j-20x 12 = 0, contain equal roots, and if so, what are they ? The first derived polynomial of the first member is 4x 3 6x 2 14.T+20. The greatest common divisor between this and the first member of this equation is x 2 ; therefore x = 2 is twice a root of the equa- tion, and x*2x 3 7z'-f20.r 12 n:ay be divided twice by x 2, or once by (x 2) 2 = x 9 4./--J-4. Performing the division, we find the quotient to be x a -j-2x 3, and the original equation may now be written (x s 4x-|-4)(a; a +2a? 3) = 0. This equation will be satisfied by the values of x found by placing each of these factors equal to zero. From the first we get x = 2, x = 2, and from the second x = 1 x = 3 ; hence the four roots of the given equation are 1 , 2, 2, 3. 2. Find the equal roots of the equation x'+^-llx* 8x'+20x+16==0. Ans. 3. What arc the equal roots of the equation x_2x 4 -f3^ 8 7x a -f-8x 3 = 0? Ans. It has three roots, each equal to 1. 4. What are the roots of the equation x*2x* lla?+ 12o:+36 = ? Ans. Its roots are ,3, 3, 2, 2. 5 What are the equal roots of the equation X= x 1 5o; 8 2z 5 -f-3Sx 4 31x 8 61#'-f96;c 36 = 0? We find D = x* 3* 8 x'-j-llz 0, D f = xl. Hence or- = 1 is twice a root of the equation D = 0, and three times a root of the given equation. 880 PKOPERTIES OF EQUATIONS. Dividing D x' 3x z x*+IIx 6 by Z)' 2 = x* 2x+ fiad for the quotient x 2 x 6 = (x 3)(x- r 2). Therefore, and X= (x 3) 2 (x-f2) 2 0r 1). Hence the roots of the given equation are Ans. 3, 3, -2, -2, +1, +1, +1. 436. Having an equation involving Lut one unknoicn quantity, to transform it into another, the roots of which shall differ from those, of the proposed equation l)y a constant quantity. Assume x m -\-Ax m - l -\-x m -*+ Cx m - 3 + ____ -j- TJC+ U=Q, and denote the new unknown quantity by y, and by x' the arbitrary but fixed difference which is to exist between the corresponding values of x and y\ we shall then have x = y-\-x'. Substituting this value of x in the given equation, it becomes Developing the terms separately, by the binomial formula, and ranging the aggregate o ing powors of y, we have arranging the aggregate of the results with reference to the ascend- + TX' y/-j- nix -f .4(??i IX" 1 - 2 C(m 3X m-l 2) -- x'' 2 ml x n = 0. (1) An examination of this developed first mciubcr leads to these conclusions : 1. The absolute term of the transformed equation, or the coeffi- cient of y, is what the first member of the given equation becomes when x' is substituted for x. TRANSFORMATIONS. 381 3. The coefficient of y, the first power of the unknown quantity, is what the first derived polynomial of the first member of the given equation becomes, when in it x' takes the place of x. 3. The coefficient of' y* is what the second .derived polynomial of the first member of the given equation becomes when it is divided by 2, and x' takes the place of x. 4- And in general, the coefficient of y n is what the nth derived polynomial of the first member of the given equation becomes when it is divided by the product of the natural numbers from 1 to n in- clusive, and x is replaced by x'. Representing the first member of the given equation, and its suc- cessive derived polynomials, after x' has been substituted for x, by A 7 , A' 7 ,, A*' 3 , A' 3 , etc. respectively, the transformed equation may be written Or, by inverting the order of terms, 437. By comparing eqs. (1) and (!') of the preceding article it is shown that, jrt _ -i ttDd 1-2 . . (m-5) = m ~2~ *"+^0-lX+* ; (3) the degree of the coefficients of the equation, with respect to cc', increasing by at least one from term to term as we pass from left to right, the absolute term being of the Twth degree. Now, since x' is an arbitrary quantity, such a value may be as- sumed for it as will cause it to satisfy any reasonable condition. We may therefore form an equation, by placing any one of these coeffi- cients equal to zero, regarding ./' as the unknown quantity, and any root of this equation will cause the corresponding term of the transformed equation (I'), (43G), to disappear. 882 PEOPERTIES OF EQUATIONS. Suppose mx'-\- A = ; -whence x' = --- m If this value of x' be substituted in the equation just referred to, it takes the form Hence, to transform an equation into another which shall be in- complete in respect to the second term : Substitute for the unknown quantity another, minus the coefficient of the second term divided by the exponent of the degree of the equation. 438. The third term will disappear from the transformed equa- tion when x' is made equal to either of the roots of the equation, m _ 1 m ^ x '*+ A(m l)x'+ JB = 0. 2 But there may exist such a relation between m, A, and B, that the A value, :c' = -- , will satisfy this equation : in which case the van- m ishing of the second term of the transformed equation will involve that of the third. To find what this relation is, substitute this value of x' in the above equation, and it becomes 2 m a v J m This, reduced as follows, ^Lz. 1 . '_ (m _i) +jB =o, 2 m v ' m ( Wl -_l)^[ 9 _2(m 1) J.'-f 2m.g = 0, (m 1)^L 2 = gives finally = When the values m, A, and B, will satisfy this equation, the third term of the transformed equation will disappear with the second. In general, to find the value of x f which will free the transformed equation of the third term, an equation of the second degree must be solved ; and to free it of the fourth term, the equation* to be solved would be of the third degree; and finally, to make the abso- lute term disappear, would require the solution of the original equation. TRANSFORMATIONS. 383 EXAMPLES. 1. Transform the equation x* -\-2px q = 0, into another which which shall not contain the second term. 2 This is done by making x = y -- ~ = y p (437) ; Lt whence, by (436), *' = 00 f 2x^-2 X' 1 =- ; 2( J p)-2p = 0; Therefore, the required equation 13 = 0, from which we find y = V q-\-p* ; and since cc = y -p, the values of x are given by the formula, x = the same as that found by the rule for quadratics. 2. Transform the equation x*--px? -\-qx-\-r = 0, into one not having the second term. Make x = y ^- ; then * -(-fM-fM- _i 2-3 ~ 2-3 Hence, the equation sought is or, bj making jl gr = m, and 9: ^| +r = n, 384 PROPERTIES OF EQUATIONS. 3. Transform the equation x 4 12a; 3 -f-17x 9 Qx+ 7 = 0, into another which shall not contain the 3d power of the unknown quantity. 1 By (437), put x = y+ ; or x == B+y. Here x r = 3 and m = 4. X' = (3) 4 12(3)'-{-17(3)'-9(3)+7,or A 7 =-110. A7j = 4(3)' 3G(3) a -f-34(3) 9, or A", = 123. 4^ = 6(3)-SO(3)+17, or = - 37. Therefore the transformed equations must be y 4 37y a 123y 110 = 0. 4. Transform the equation a; 3 6x'-J-13x 12 = 0, into another wanting its second term. x = 2-fy; then = (2)' G(2) 3 -}-13(2) 12, or A' 7 =2. = 3(2)' 12(2)+13, or A", = +1. or ?-= 0. Therefore, the transformed equation must bo -2 = 0. 5. Transform the equation x *__4. x _8x+32 = 0, into another whose roots shall be less by 2. Put x - 2+y. Ans. y 4 +4y*24 y = 0. As this transformed equation has no term independent of y. t y is one of its roots, and x = 2 is therefore a root of the original equation. TllANSFOBMATIOXS, 385 6. Transform the equation, = 0, into another whose roots shall be greater by 3. Ans. y-f-4/4-9,y 9 42y = 0. 7. Transform the equation, x _Sx 3 -f z'+82x 60 = 0, into one incomplete in respect to its second term. Ans. #* 23,y 8 -f 22y+69 = 0. 439. Resuming the transformed equation (!'), (436), which is = 0, and replacing y by its value, y = x x', it becomes Now it is evident that, by developing the first member of this equation and arranging the result with reference to the descending powers of .r, the first member of the original equation will be repro- duced } for, by this operation we will have merely retraced the steps by which eq. (10 was derived from eq. (1) in the article referred to Hence we have the identical equation, ....Sx*+fx+U The quotients and remainders obtained by the division of the first member of this equation by any quantity, will not differ from those arising from the division of the second member by the same quan- tity. Dividing the second member by x x r , the first remainder is A'', and the quotient, (*-*')'+ 386- PROPERTIES OF EQUATIONS. and this divided again by x x f , will give for the second remainder JTj and the quotient, X 9 X 7 X 9 It is unnecessary to continue this process further, to see that these successive remainders are the coefficients of the transformed equation (!') beginning with the absolute term, or the coefficient of //. The divisor to be employed is x x' if the roots of the trans- formed equation are to be less, in value, than those of the given equation by the constant difference x' j if greater, the divisor must be .r-j-.r. Hence, an equation may be transformed into another of which the roots are greater, or less, than of the given equation by by the following RULE. I. Divide the first member of the given equation (the sec- ond member being zero} by x plus the constant difference between the roots of the two equations, continuing the operation until a remain- der is obtained which is independent of x; then divide the quotient of this division by the same divisor, and so on, until m divisions have been performed. II. Write the transformed equation, making these successive re- mainders the coefficients of the different powers of the unknown quantity, beginning with the zero power. It must be borne in mind that the term plus in this rule is used in its algebraic sense. By a little reflection, it will seem that the mth quotient will be the coefficient of x m in the original equation, and that this will also be the coefficient of the highest power of the unknown quantity in the transformed equation. EXAMPLES. 1. Transform the equation, x 4 4x 9 8;e-h32 = t), into another of which the roots shall be less by 2. This is example 5 of the last article. Make x = 2-j-^, or y = x 2 ; TRANSFORMATIONS. 387 then the operation is as follows : -2)* 4 4x 3 8*4-32(*' 2* a 4* 16 x'2x> 2*' 8* _-2*+4* a 4* 3 _8* a; 2)*' 2* a 4* 16(V 4 4* a 4-8* a;'-- 2* a 16.^+32 4x 16 16jc-f32 4*+ 8 2X 4(* __ 2.r 4 2-3 Hence, the transformed equation is y 4 +4y 0^24^4-0=0; or, y*+4y* 24y = 0, as before. 2. Transform the equation, x* 12x 8 +16-c 9 9*4-7 = 0,into one having roots less by 3. Here * = y4~3> ory = x 3. OPERATION. >_9 X _|_7(V_9* 10* 39 x 4 3*' l(Kr a -f30.;c 39.1-4- 7 39*4-117 110 = A 17 , 1st remainder. PROPERTIES OF EQUATIONS. 28 * 3x a 28* 39 123 = , , 2d remainder. a 3)* 3(1 a 3 28 9 ^ a 37 = -~> 3d remainder. Hence, #*:0y 37,/ 123^ 110 = 0, is 'the transformed equation. We shall have 4 remainders, if we operate on an equation of the 4th degree ; 5 remainders with an equation of the 5th degree ; andf in general, n remainders with an equation of the nth degree. The transformation of equations by division, treated of in this article, if performed by the ordinary rule, would be too laborious for practical application ; but by a modified method of division, called Synthetic Division, it becomes expeditious and easy. As preliminary to the explanation of this method of division, we must explain the process of MULTIPLICATION AND DIVISION BY DETACHED COEFFICIENTS. * 4L4O. It has been seen that when two polynomials are homoge- neous their product is also homogeneous, and the number which de- notes its degree is the sum of the'numbers denoting the degrees of the factors. It is evident that if the polynomials contain but two letters, and both are arranged with reference to the same letter, the product will be arranged with reference to that letter. Since, in the operation of multiplying the terms of the multiplicand by the terms of the multiplier, the products of the coefficients are not DETACHED COEFFICIENTS. 889 affected by the literal parts to which they are prefixed, these coeffi- cients may be detached and written down with their signs in their proper order, and the multiplication performed as with polynomials. The partial products, numerical or literal, being carefully arranged as if undetached, are then reduced and the literal parts annexed. EXAMPLES. 1. Multiply a'+2ax-j-o: 2 by a+x. OPERATION. l-j-2-j-l, Detached coefficients of multiplicand. 1-|-1 " " multiplier. 1+2+7 1+2+1 1+3+3 + 1 Product of coefficients. Now by annexing the proper literal parts to the several v terms thus obtained, we have a 3 +3a a x+3a;r a +z 8 ; Ans. This method of multiplication may be employed when the two polynomials contain but one letter. 2. Multiply 3x a 2;c 1 by OPERATION. 3 2 1 3+2 9 G 3 +642 9+0 7 2 whence, or, 9x 3 1x 2, Ans. When any of the powers of the letters, between the highest ana lowest, do not appear in either factor, the terms corresponding to such powers must be supplied, with the coefficient 0. 33* 390 PROPERTIES OF EQUATIONS. 3. Multiply x*+2x> I by a a -f2. The factors completed are x'-\-2x*+Qx 1 and cc'-f-Oz-f 2. Hence the operation i.s l-l-2-j-Ol 1+2+0-1 2+4+0-2 1+2+2+3+02 and the product, or, ^+2x 4 +2x 3 +3x 3 2. 4. Multiply 3x a 2x 1 by 4x+2. Ans. 12x s 2x a 8x 2. 5. Multiply 3x a 5x 10 by 2x 1. Ans. Qx 3 22.c 3 6. Multiply o^+iry+y by x a xy-\-y*. Ans. x 4 +x a 7. Multiply x s 4x a +5x 2 by x'+4x 3. Ans. x 6 14x*+30x a 23x+6. 441. Now, if detached coefficients can be used in multiplication, so in like cases, they may be employed for division. When the divi- dend and divisor contain but two letters and are homogeneous, the degree of the quotient will be the excess of the degree of the div- idend over that of the divisor. EXAMPLES. 1. Divide a 4 3a'z 8aV+18ax'+16x 4 by a' 2ax2x\ OPERATION. 1_3_8+18+16|1 2 2 122 118 __!_G-fl8-fl6 -1+2+ 2 _8-fl6-j-lG Hence the quotient is DETACHED COEFFICIENTS. 391 2. Divide a 6 5a 8 Z> 8 +aV+6a 4 W by a 1 Sa&'+JA In this example we must supply the term Oa*& in the dividend, and the term Q-a?l> in the divisor. The operation then is, fl+6 2 l+Q3 -fl _ 1+0 02+0+62 _2+Q+6 2 Therefore we have, for the quotient, or, a 3. Divide x> 4x 4 17a, 3 13x a llz 10 by cc'+3 OPERATION. 14 17 13 II 10 |l+3+2 1+3+ 2 __ i_7+2 5 _7_19_13-11 10 7 21 14 + 2+ 11110 + 2+6+4 Hence the quotient is When the dividend and divisor contain but a single letter, absent terms in either, answering to powers of this letter between the highest and lowest, must be inserted with the coefficient 0. In the examples we have wrought to illustrate the method of division by detached coefficients, the coefficients have been taken entire, that of the first term of the divisor, in each case, being unity ; the process, however, will be the same whatever these coefficients may be. When the coefficient of the first term of the divisor is not unity, it may be made so by dividing both dividend and divisor by this coefficient. The quotient term will then be the first term of the corresponding dividend, as is seen in all the abova examples. 392 PROPERTIES OF EQUATIONS. SYNTHETIC DIVISION. . To explain what synthetic division is, and to deduce a rule for executing it, let us take the first example in the preceding article. If the signs of the second and third terras of the divisor be changed, each remainder will be. found, by adding the terms of the product of these two terms by the term of the quotient, to the corresponding terms of the dividend ; observing that by the nature of the operation, the product of the first term of the divisor by the term of the quotie-nt, cancels the first term of the dividend. Be- sides, since the first term of the divisor is unity, any quotient term is the same as the first term of the partial dividend to which it belongs. The process may now be indicated as follows : 2216 2 216 Quotient, 118 Hence the quotient is a 2 ax 8x', as before found. The dividend and divisor are written in the usual way, after changing the signs of the last two terms of the latter; and a hori- zontal line is drawn far enough beneath the dividend for two inter- vening rows of figures. Bring down the first term of the dividend for the first term of the quotient. The products of the second and third terms of the divisor by the first term of the quotient are written, the first in the first row under the second term of the div- idend, and the second in the second row under the third term of the dividend. The sum of the second vertical column is ther written for the second term of the quotient. The next step is multiply the second and third terms of the divisor by the second term of the quotient, placing the first product in the first row under the third term of the dividend, and the second in the second row under the fourth term of the dividend. The sum of the third vertical column is the third term of the quotient. The sums of the fourth and fifth columns each reduce to zero. The operation for the last example in the preceding article is SYNTHETIC DIVISION. 393 14 17 13 -111011 3 2 _3_|_21 64-15 _ 2+14 4+10 17-f- 2 5 and for the quotient we have x _7x_|_2x 5. No difficulty will now be experienced in understanding this general RULE. I. If the coefficient of the first term of the arranged divisor is not unity, make it so by dividing loth dividend and divisor by this coefficient. II. Write down the detached coefficients of the dividend and di- visor in the usual way, changing the signs of all the terms of the of the latter except the first, and draw a line far enough lelow the dividend for as many intervening rows of figures as there ore terms, less one, in the divisor, and bring down the first term of the dividend, regarded as forming a vertical column, for the first term of the quotient. III. Write the products of the second, third, etc., terms of the di- visor by the first term of the quotient, beneath the second, third, etc., terms of the dividend in their order, and in the first, second, etc., rows of figures ; and bring down the sum of the second vertical column for the second term of the quotient. IY. Multiply the terms of the divisor, exclusive of the first, as before, by the second term of the quotient, and write the products in their respective rows, beneath the terms of the dividend beginning at the third j bring down the sum of the third vertical column for the third term of the quotient. V. Continue this process until a vertical column is found of which the sum is zero, the sums of all the following aho being zero when the division is exact ; otherwise continue the operation until the de- sired degree of approximation is attained. Having thus found the coefficients of the quotient, annex to them the proper literal parts. In applying this method of division it is unnecessary to write the first term of fhe divisor, since it is unity and is not used in the oper. ation. 35 394 PKOPERTIES OF EQUATIONS. EXAMPLES. 1. Divide 1 x by 1+x. Ans. 1 2x+2z a 2x s +ctc. 2. Divide 1 by 1+z. Ans. 1 x-f x a x-'+z 4 etc. 3. Divide a 6 5a 4 x+10aV lOaV-f 5ax 4 x 6 by a 8 2ax+x*, Ans. a 3 3a a z-|-3ax 2 z 3 . 4. Divide x 9 bx'+lbx' 24x-j-27x a 13x+5 by * 4 2x 3 + 4x a 2-r-f-l. J.?is. a; 3 3x--f5. 5. Divide # T y T by x y. Ans. x*+tfy+xyj&f+a*y*^?y*+?> 4 4- 3. The transformation of an equation into another having roots less or greater than those of the given equation by a fixed quantity, may now be expeditiously made by the method of synthetic division. i . Transform the equation x* 4x 8 8^-|-^^ = i n ^ another whose roots shall be less by two. The second power of x not appearing in this equation, it must bo introduced with for its coefficient. FIRST OPERATION. 1_4 8+32j2_ 24832' 12416, = SECOND OPERATION. 1_2 4 16|2 2-M) 8'" lH-04, 24 = X\ THIRD OPERATION. FOURTH OPERATION. 1-4-0 4|2 1+212 2+4" 2~ 1_L9 ^ 1 J-4 * 2 ~ 2-3" Bence the transformed equation is 24 = 0. SYNTHETIC DIVISION. W> Instead of keeping the above operations separated, they may be united and arranged as follows : 1_4 -4-0 - 8+32|2_ 2 4 832 2 4 16, = X 2 08 4, 24 = ^', 2_+4 V7 2, -0=^ 2_ i-5* " 2-3 To understand this, it is only to be borne in mind that the divisor is the same throughout, and that the first term, 1, of the successive dividends, \vhich if written would all fall in the vertical column at the left, is omitted. Transform the equation x* 12x 8 -f-17x a 9x-|-7 = 0, into an- other whose root shall be 3 less. OPERATION. 1 _12 -f 17 - 9 -|- 7 (3 _j_ 3 _27 _- 30 117 _ 9 _io _ 39^110 = X* 4. 3 18 84 _ 6 28, 123=^7, 4- 3 -- 9 o 07 -^a - 3, -37 = -g J = ^ 3 2-3 Hence the transformed equation -is y*+) m j dbC^dz etc., that the sign it is to be used before those only which have odd numbers for their exponents ; when the exponent is even, the plus sign is to be understood. If the root of equation (1) be -f-|/6, the aggregate of these developments will be composed of two parts, the one rational and the otherurd. The rational part will be the algebraic sum of those the even powers of -\/b for factors, the zero pow- er bei^l * Represent this part by M. le other^urd. Th irnfl BMt^" ^ rbe^H ifeS? PROPERTIES OF EQUATIONS. 399 The irrational part will be the algebraic sum of the terms having the odd powers of y'b for factors. But since (j/&) 3 = b^/b, (j/&) 6 = ^> 2 |/&, etc., the different terms of this part can be repre- sented by a single term of the form N^/b^ N being the algebraic sum of the coefficients of y/7>. Hence equation (2), under the sup- position that a-f- j/6 is a root of equation (1), becomes which can be true only when we have separately M = 0, N= 0; In reducing equation (2) to equation (3), the upper signs in the expansions of the terms of equation (2) were used. If the lower signs in the equation and the expansions of its terms be used, which is equivalent to supposing a i/b to be a root of equation (1), tho reduced equation will be MN^/b = 0. (4) in which M and N are evidently the same as in equation (3). Hence if equation (1) has a root, a-\-]/b, ^ ^ as a ^ so ^ e r00 ^ a V & Now let us suppose that a-\-V i is a root of eq. (1); then since the even powers of V b are real and the odd powers imaginary, the developed first member of eq. (2) will be composed of two parts, the one real and the other imaginary. Represent the real part by M'. The imaginary part is the algebraic sum of the terms having the odd powers of 1/b for factors. But since (I/ 6) 3 = V b* ( b) = bV b, (V 7 ^)) 6 = V b T (^b) = b*V~b, etc., the different terms of this part can be reduced to a single term of the form N'v b. Hence, under the supposition that a-\-V b is a root of equation (1), equation (2) becomes M'-\-N'V^b = 0, (5) which requires that we have separately M' = 0, N' = ; (367). By using the lower signs of the terms and their expansions in equa- tion (2), which supposes a V b to be a root of eq. (1), we find and by a simple inspection of the expanded terms of equation (2), we see that M' and N' in equations (5) and (6) are the same. Whence we conclude that if equation (1) have a root, a J- K b, it has also the root, a 1 b. 400 PKOPERTIES OF EQUATIONS. RULE OF DES CARTES. 446. An equation can not have a greater number of positive roofs than there are variations in the signs of its terms, nor a greater number of negative roots than there are permanences of signs. NOTE. A variation is a change of sign in passing from one term to another ; a permanence occurs when two successive terms have the same sign. It is obvious that the number of variations and permanences taken together must be equal to the number of terms, less 1. Let the signs of the terms of an equation be + + + + + -> the second, sixth, and eighth terms giving permanences, and the other terms variations. To introduce a new positive root into the equation, we must mul- tiply the equation by some binomial factor in the form of x a ; and the signs of the partial and final products will be as follows : 1st. 2d. 3d. 4th. 5th. 6th. 7th. 8th. 9th. Now we observe, in the final result, that the 2d, 6th, and 8th terms, or the terms which give permanences in the proposed equa- tion, are ambiguous ; consequently, let these ambiguous signs be ta- ken as they may, the number of permanences has not been increas- ed. But the number of terms has been increased by 1 ; hence, the number of variations has been increased by 1, at least. Again, to introduce a new negative root into the proposed equa- tion, we must multiply by some binomial factor in the form of x-\-a' and the partial and final products will be as follows : 1st. 2d. 3d. 4th. 5th. 6th. 7th. 8th. 9th. + + + + 4- _ +4-- + -- + + - + + + Her.c, in the final result, the 3d, 4th, 5th, 7th and 9th terms, 01 the terms which give variations in the proposed equation, are am- V CARDAN'S RULE. 401 biguous ; consequently, the number of variations has not been in- creased. But the number of terms has been increased by 1 ; hence, the number of permanences has been increased by 1, at least. Thus we have shown that the introduction of each positive root must give at least one additional variation, arid the introduction of each negative root must give at least one additional permanence. Hence the whole number of positive roots can not exceed the num- ber of variations, and the whole number of negative roots can not exceed the number of permanences; the proposition is therefore proved. 44 7. Although the introduction of a positive root -will always give an ; dditional variation of signs, it is not true that a. variation 9f signs in the terms of an equation necessarily implies the presence of a real positive root. Thus, the equation, x 3 x 9 T.r-f-15 = has 2 variations of signs, and 1 permanence. But its roots are 2-fi/Hl, 2 l/~l, and 3, no one being positive and real. But ichen the roots are all real, the number of positive roots is equal to the numler of variations, and the number of negative roots is equal to the number of permanences. CARDAN'S RULE FOR CUBIC EQUATIONS. . It has been shown, (437"), that any equation can be transformed into another which shall be deficient of its second term That is, every cubic equation can be reduced to the form of . x*+3px = 2q> (1) and the solution of this equation must involve the general solution of cubics. We make 3/> the coefficient of x, and 2q the absolute term, in order to avoid fractions in the following investigations: Assume x = v-\-y ; then cq. (1) becomes Expanding and reducing, we have * 3 + 34* 402 PROPERTIES OF EQUATIONS. Now as the division of x into two parts is entirely arbitrary, we are permitted to assume that v y+p = o > (4) whence, from eq. (3), v*-{-y* = 2q. (5) If we obtain the value of y from (4), and substitute it in (5), we shall have, after reducing, whence, v* = q3/3. If P denote the numerical value of the greatest negative coefficient in an equation, and n the number of terms which precede the first negative coefficient, then VP-|-1 will be a superior limit of the positive roots of this equation. Let ./ m -h A^-'-f Bx m ~*-\- Or" 1 - 8 -}- -f Tx+ U 0. (1) If we omit those positive terms, if any, which occur between x m and the first negative term, and then put P for the coefficient of every other term after x m , we shall have f*3\ , t _(p.<+ J R c + .... ^Px-fty^ 0. (2) Now it is evident that any value which, substituted for x, will give a positive result in eq. (2) will give a positive result also in eq. (1). For, the sum of all the negative terms in (1) can not possibly be greater than the negative part of (2) ; besides, there may be one or more positive terms in (1) which are omitted in (2). LIMITS OF REAL ROOTS. Dividing every term of (2) by x m , we obtain Make x \/P-}-l r-fl, where r is put in the place of '(/P for the sake of simplicity. Remembering that P = r n , we have Summing the geometrical series found in the parenthesis, by ^ (-??')] the equation becomes r \ n ~ l H) (4) Now since , is less than unity, the expression which consti- tutes the first member of (4) is positive. Moreover, the negative (r \ n ~ l 1 , must always be less than unity, whatever be the value of r ; hence, no value of r, however great, will render the first member of (4) negative. Thus we have shown that if we substitute for x the quantity VP + 1, or any greater value, the result will be positive in equation (3) ; the result will therefore be positive in cq. (2), and also in eq. (1). Hence, by (4r*IS ? 2\ \AP-r-l is a superior limit of the positive roots in any equation, which wr.s to be proved. In applying the principle just established, the absolute term must be regarded as the coefficient of x 9 ; and if the equation is incom- plete, the deficient terms must be counted, in finding n. It should be observed also, that an equation having no negative term can have no positive roots. For, every positive number sub- sti+uted for x will render the first member positive. That is, no positive value of x can reduce the first member to zero. EXAMPLES. 1. Find the superior limit of the positive roots of the equation _f fu;_}_2x 8 _ 14x 3 2Cxr-f 10 = 0. llcre n = 3 and P = 26. Hence we have, in whole numbers, = 4, An*. 408 NUMERICAL EQUATIONS OF HIGHER DEGREES. 2. Find the superior limit of the positive roots of the equation x 4 5x 3 25x a 12a:-68==0. Ans. 6. 3. Find the superior limit of the positive roots of the equation a: 4 5.*' 93+12 = 0. Ans. 4. 4. Find the superior limit of the positive roots of the equation x *+ x *+3x8 0. Ans. 3. 454. To determine the superior limit of the negative roots of an equation, numerically considered, Change the signs of the alternate terms, c mnting the deficient terms when the equation is incomplete; then, apply the, preceding rule. For, according to (444), the positive roots in the new equation will be numerically the negative roots in the given equation. EXAMPLES. 1. Find the superior limit of the negative roots of the equation a' 3. c +5:e+7 = 0. Ans. ^7-f 1 = 3, in whole numbers. 2. Find the superior limit of the negative roots of the equation rr 4 15.*; 2 lO.r+24 = 0. Am. 5. 3. Find the superior limit of the negative roots of the equation X __3 x *_j-2.r 4 +27x s 4x a 1 = 0. Ans. 4. LIMITING EQUATION. 45t5. If there be one equation whose roots, taken in the order of their values, are intermediate between the roots of another, the former is said to be the limiting equation of the latter. /JoG. Any equation being given, its limiting equation may be formed Ly putting its first derived polynomial equal to zero. If a, 6, c,. . . .&, I are the roots of the given equation X = 0, and a', &', c', . . . . k r are the roots of the derived polynomial X^ = 0, each set being arranged in the order of their values, then we are to show that all these roots, taken together, and arranged in the order of their values, will be as follows : In both equations, put x = z'+w, developing the terms, and ar- LIMITING EQUATION. 409 ranging the results according to the ascending powers of u. Ob- serve that JTo is the first derived polynomial of X l ; hence, adopt- ing the same notation as in (43O), we have, from the two equations, x = x* -MV-f -^'-f- * 3 + . . . = o, (i) X l = X' 1 4- A' 2 + -:_3w a + -*i*'+ . . . . = ; (2) where, it will be observed, A 7 , A' 13 A T ' 2 , etc., represent what X, ATj, Ao, etc., become, when x' takes the place of x. Now suppose x' = r ; that is, a; i= r-|-w, r being any rootf o/ the git- e n equation. Then X' = ; and as A r ' u A 7 ^, A / 3 , now receive definite values, the values of X and A\ may, or may not become zero by giving a particular value to u. Dropping X' from (1), and factoring the result, we have X = u ( X'.+ ^u+^u*4- ....}, (3) (4) where the different terms may be essentially positive or negative, according to the values or r and w, upon which they depend. Now, it is evident that by causing u to diminish numerically, each term after the first, in the parenthesis, may be made as small as we please ; and by making u sufficiently small, the sum of the terms containing u, in each parenthesis, may be made less, than the first term X' , ; in which case the essential sign of the quantity in either parenthesis will depend upon the sign of X' j . Thus, when u is indefinitely small, the signs of the functions, A" and A,, will depend upon the signs of u (A 7 j) and X 7 n respectively. Hence, when u is negative, A^and A\ will have opposite signs; but when u is pos- itive, X and A' 1 will have the same signs. / 157'. Thus we have shown, that if we substitute in a given equa- tion X = 0, and its first derived polynomial Xj = 0, a quantity r u. which is insensibly less than the root r, the results will have op- posite signs ; but if we substitute the quantity r-f-u, which is insensibly greater than the root r, the results will have the same sign. 408. Consider the quantity substituted in the two functions to be insensibly less than a, the least root of X = 0, and let it increase 35 410 NUMERICAL EQUATIONS OF HIGHER DEGREES. till it is insensibly greater than a. In passing the root a. the func- tion X will change sign, (452 9 3) ; hence the signs of the func- tions will be as follows : X X, X X l rx au -f , -f, For }'x $sa , or else -J-, \x=a+u , + +. Now let the substituted quantity increase from x = a-\-u to x = b , a value insensibly near to b, the next root of X =. 0. According to the principle already established (4-57), X and X l must now have opposite signs. And since X can not have changed its sign during the change of x from . We have seen, (435), that-the equal roots of an equa- tion may always be found and suppressed. Now let X = xr+Ax^+Bx-*-^ ____ Tr-\-u = represent any equation having no equal roots, and X 1 its first derived polynomial, or its limiting equation. V\ 7 e will now apply to the functions, JCand.JT n a process similar to that required for finding their greatest common divisor (1O5), but with this modification, namely; that ice change the signs of the siii'i-wicr remainders, and neither introduce nor reject a negatibe factor, in pn paring for division. Denote the successive remainders, with their signs changed, by STURM'S THEOREM. 411 \ R, R IJ R. 2 ,. . . .J^n-n ^n- Since the given equation has no equal roots, there can be no common divisor between X and X l , (435); hence, if the process of division be continued sufficiently far, tho last remainder, R n , must be different from zero, and independent o/x. Now in the several functions, ., X l> R, M\^ R*,- R n \y RM let us substitute for x any number, as A, and having arranged the sig;i:i of the results in a row, note the number of variations of signs. Next substitute for x a number, 7t', greater than 7i, and again note the number of variations of signs. The difference in the number of variations of signs, resulliny from the two substitutions, will -be equal to the number of real roots comprised between h and h'. This is Sturm's Theorem, which we will now demonstrate. Let Q, Qi, (?;>> On-i Qn denote the quotients in the succes- sive divisions. Now in every case, the dividend will be equal to tho product of the divisor and quotient, plus the true remainder, or the remainder with its sign changed. Hence, (1) X = X l Q R (2) X } = R Q, - (3) R =^C a - a v (4) R = R.Q ' From these equations, it follows, 1 If any number be substituted for x in the functions X, X n R 7 RJ,. . . .R n , no two oj them can become zero at the same time. For, if possible, let such a value of h be substituted for x as will render X l and R zero at the same time. Then the second equation of (A) will give R , = ; whence, the third equation will become R. 2 = ; and tracing the series through, we shall have, finally, R n = 0, which is impossible. 2. If any one of the functions become zero by substituting a particular value for x, the adjacent functions will have contrary signs far the same value. For, suppose R l in the third equation to become zero ; then this equation will reduce to R = R. 2 . That is, R and /?._, have con- trary signs. Having established these principles, suppose the quantity h, which 412 NUMERICAL EQUATIONS OF HIGHER DEGREES. is to be substituted simultaneously in all the functions, to be a vari- able, changing by insensible degrees from a less to a greater value. As it passes any of the roots, the function to which this root be- longs will reduce to zero, and change sign, (45S, 3). Let p be a little less than a certain root of 7?.,, and q a little grea'tcr than the same root, the two values being so taken, however, that no root of R, or R 3 shall be comprised between them,. As h changes fromjp to q, R 2 will reduce to zero, and change sign. But neither JK l nor E 3 will change sign; and since, according to the second principle, these functions have opposite signs when R% = 0, they must have opposite signs also when h = p or h q. Now when h =p, the arrangement of signs must be R l R. 2 R 3 R^ R 2 R 3 + - or ,- +; giving one variation and one permanence, whichever way the double sign be taken. When h q the signs must become R l 7? 2 R 3 R 1 R. 2 R 3 + T , or =F +; giving, as before, one variation and one permanence, so that tho whole number of variations is neither increased nor diminished. This reasoning obviously applies to any function which is situated between tico other functions. Hence, 3. When h passes a root of any function intermediate between X and Rn , the number of variation^ of signs will not be altered. As the last function, R m is independent of x, its sign will not be changed by any substitution for x. It follows, therefore, that if any change is produced, in the number of variations of sign's, it must re- sult from the alternation of signs in the original function X. Let a,b,c,d, .... I be the roots of JT, taken in the order of their values. Then the roots of X l will be found, the first between a and 6, the second between b and r, and so on ; (458). The de- gree of X l is less by 1 than the degree of X; hence, if the degree of X is odd the degree of X l will be even, arid if the degree of X is even the degree of X^ will be odd. Now take h less than a ; ac- cording to (4:512, 1), the signs of X and X l will be unlike, giving a variation. Let h increase till it is insensibly greater than a ; X will change sign, and the variation between X and X l will be lost. STURM'S THEOREM. 413 Now let h increase till it is insensibly loss than It. Tt will pnss the first root of Xi, causing the signs of X and X l to be again unlike; but by (3;, this change in the sign of X l will not alter the whole number of variations in the signs of the functions. Again let h increase till it is insensibly greater than b ; X will again change sign, and another variation will be lost. In like manner it may be shown that the number of variations will be diminished by 1 evert/ time h p ses a root f X; hence the truth, of the theorem. 4O1. If we substitute for x in the several functions h . =. QQ and h'*= -\- oo, successively, we shall determine at once the whole number of real roots in the given equation. To ascertain the signs of the functions resulting from these substitutions, we require the following principle : If in am/ polynomial involving the descending powers of x, in- finity be substituted for x, the sign of the whole exf>ression will de- pend upon the sign of the first term. Let Ax m -\-Bx m - l -\-Cx m -^Dx m - y -j- j^e"- 4 -}- . . . . bo the given polynomial, if x = GO, then because every term in the second member is less than any assigna- ble quantity, or zero, (188, 2). Multiplying both members of (1) by x m j we have Ax m > Bx n - l + Cx^+Dx^+Ex-*-}- ... (2) That is, when x GO, the first term of the given polynomial is num;rio:iHy greater than the sum of all the other terms. Hence the sign of the whole will be the same as % the sign of the first term. 4LG2. In the application of Sturm's Theorem, we may always suppress any numerical factor in any of the functions X lt R, It lt etc. j for this will not affect the sign of the result. 1. Given the equation x* 3x 2 12x-|-24 = 0, to find the nurn bcr and situation of the real roots. Suppressing monomial factors, we have for tho several functions* X = x 3 3x a I2x -f- 24, X I = x a 2x \ R = x 2, R, =4. 35* 4J4 NUMERICAL EQUATIONS Oi 1 HIGH Ell DEGREES. Substituting in these functions x == co and ;r = -j-ce succes- sively, we obtain the following results, in respect to signs : X X l R R, For $ x = &>> + +> 3 variations. _ Hence, the given equation has 3 real roots. Since the signs in the given equation present two variations and one permanence, two of the roots must be positive, and the other negative, (446). To ascertain the situation of the positive roots, let us substitute in the functions, x = 0, x 1, x = 2, etc., suc- cessively, noting the variations of signs in t]ie results. x = 0, signs, -f- -J-, 2 variations. x = l, -f -f, 2 x = 2j " 4- 1 variation. * = 8, + +, 1 a; = 4, -f + +, 1 a; = 5, -f -j- + +, " Since one variation is lost in passing from x = 1 to x = 2, and ono also in passing from x = 4 to x = 5, one positive root must be situ- ated between 1 and 2, and the other between 4 and 5. To ascertain the situation of the negative root, substitute x = 0, x = 1, x = 2, etc. j the signs are as follows : x = 0, signs, -f- -{-, 2 variations. * = i, * + +, 2 " For / 2, -f- 4- -f, 2 -3, + + +, 2 -4, - -j- - +, 3 Hence, the negative root is situated between 3 and 4. The initial figures of the several roots will be 1, 4, and 3. 2. Given the equation, x- 4 _2x 8 7. 2 -flOx-j-10 = 0, to find the number and situation of its real roots. In this example, we have X = x 4 2x 3 7x s -flO^+10, ^ = 2x 9 3x a 7x-f 5, ^ = 17.r 2 23x 45, ^ = 152x 305, R. 2 = 524785. Let x = oo; we have -f -f- -f-, 4 variations. -f + + -f +, STURM'S THEOREM. 415 Hence, the roots are all real. And since the signs of the given equation give 2 variations and 2 permanences, two of the roots tire positive and two are negative. To find the situation of the roots, , 2 variations. , 2 , 2 , i 3 , 3 , 4 Hence, there are two roots situated between 2 and 3, one between and 1, and one between 2 and 3. We wish now to find limits which will separate the two roots that lie between 2 and 3. Let us transform the given equation into another whose roots shall be less by 2. By (443), the operation will be as follows : 1 2 7 +10 +10(2 +2 -14 - 8 7 4, + 2 == X' 2+46 Let x = x = x = ; we have + + 1; " + 9 1 - J x = 3; " + + + + x = -1; + + X =. -2; " + X = O . ti + + 2 3, 10 = X' t 2 +8 4 -+5 = ^ 2 2.3 The transformed equation is therefore F=y+6y'+5.y' l(ty+2 = 0. Now since the two positive roots of the original equation are found between 2 and 3, the two corresponding roots of the transformed equation must lie between and 1. The situation of these roots may be found from V alone, by trials as follows : Substitute x = 0, .1, .2, .3, .4, .5, .6, .7, .8. The signs of Fare _|_ _j_ _[_ _j.. 416 NUMERICAL EQUATIONS OF HIGHER DEGREES. Hence, one root of V lies between .2 and .3, and one between .7 and .8. Consequently the initial figures of the two positive roots in the original equation, are 2.2 and 2.7. NOTE. If we had found the initial figures of the two positive roots of V to be the same, we should have proceeded to transform F", and make similar trials with the result. We are now prepared to find the roots of an equation to any de- gree of accuracy, by IIORNER'S METHOD OF APPROXIMATION. 463. In the year 1819, W. G. Homer Esq., an English math- ematician, published a most elegant and concise method of approxi- mating to the roots of a numerical equation of any degree. The process consists in a series of transformations, the routs of each suc- cessive equation being less than the roots of the preceding equation by the initial figures of the preceding roots. But in making the several transformations, the initial figures are obtained by trial division, as in square and cube root, and not by substitutions, as in the last article. 464. Let us take an equation in the general form, thus : X = x M +Ax m - l +x m - 9 + ____ -f TJ-+ U = (1) Let r represent the initial figure or figures of one of the real roots of this equation, as found by Sturm's Theorem, or otherwise. Now let the equation be transformed into another whose roots shall be less by r. Put x r-j-y j we shall have V = y +4y- '4-7?y- 2 4- . . . . -f- T'y+ U ' = (2) In this equation y is supposed to represent a decimal, since r in- cludes at least the entire part of the required root. Hence, the terms containing the higher powers of y are comparatively small ; neglecting these, we have, approximately, Denote the first figure of this quotient by s ; put y = s-f-z. Trans- forming eq. (2) into another whose roots shall be less by s, we have V = -z + A" z-*+B" -.""-'* + . . . , + r'z-f U" = 0. HORNER'S METHOD. 417 Whence we have, as before, z = where t is another figure of the required root. This process may be continued at pleasure, and we shall have, finally, Hence, to solve a numerical equation of any degree, we first Jind by Sturm's Theorem, or otkericisfo the number- of real roots, and also the first figure or figures of each. We may then approximate to the value of any root by the following RULE. I. Transform the given equation into another whose roots shall be Itss by the initial figure or figures of the required root. II. Divide the absolute term of the transformed equation by the penultimate < o efficient, as n trial iff visor, find write the first figure of the quotient as the neoct jigure of the root sought. III. Transform (he last equation into another whose, roots shall be less than those of the previous equation by the figure last found ; and thus continue till the root be obtained to the required degree of accuracy. NOTES. 1. The successive transformations required in obtaining any root may all be made in a single operation ; and for the sake of perspi- cuity, the coefficients obtained in each transformation may be marked or numbered. 2. If a trial figure of the root, obtained by any division, reduces the absolute term X', and the penultimate coefficient J"'j,to the same sign, this figure is not the true one, and must be diminished. 3. To obtain the negative roots, it will be most convenient to change the signs of the alternate terms of the given equation, and find the. positive roots of the result; these, with their signs changed, will be the negative roots required. 4. If the penultimate coefficient, T', should reduce to zero in the ope- ration, the next figure of the root may be obtained by dividing the ubanlnte term, X', by the coefficient which precedes T', and extracting the square root tf the quotient. For, if T' vanishes, we have, in the transformed equation, = 0; or y = , B* NUMERICAL EQUATIONS OF HIGHER DEGREES. EXAMPLES. 1. Given x' 2.x 3 2Qx 40, to find the approximate value of x. By Sturm's Theorem we find that this equation has only one real root, the initial figure being 6. We now obtain the decimal part, to 2 places, as follows : OPERATION. 20 4016.23 24 24 ~ -f4 <> 1G GO 13.448 10 <> 64 <*> 2.552 G 3.24 67.24 3.23 1G.2 > 70.52 16.4 .2 > 16.6 EXPLANATION. We first transform the given equation into an- other whose roots are less i y 6, using the method of Synthetic Di- vision, explained in (44:3). The coefficients of the transformed equation are 16, 64 and 16, marked (1) in the operation. Divid- ing the absolute term 16, taken with the contrary sign, by the penultimate coefficient 64, we obtain .2, the next figure of the root. We next transform the equation whose coefficients are marked (1), into another whose roots are less by .2, the resulting coefficients be- ing marked (2). Dividing 2.552 by 70.52, we obtain .03, the next figure of the root. The operation may thus be continued till the root is obtained to any required degree of accuracy. 2. Given x'+x* 30x 2 20x 20 = 0, to find one value of x. By Sturm's Theorem we find the initial figures of the two real HORNEK'S METHOD. 419 roots to be 5 and 5. Changing the signs of the alternate terms of the equation, we obtain the decimal part of the negative root, by the following OPERATION. IV. III. II. I. 1 -1 5 30 20 -f- 20 5p 20|5.731574 150 -f-4 10 30 10.2929 5 """ \ 83.153 9.7727 14 (1) 105.00 228.153 <> .5202" 5 13.79 93.149 .3304 <" 19.0 118.79 <*> 321.302 < 4 > .1898 0.7 14.28 4.455 .1653 19.7 133.07 325.757 < 5> 245 . i 14.77 > 147.84 4.474 232 20.4 (3) 330.23 .7 .65 .15 13 21.1 148.49 330.38 0- 7 .7 .65 .15 < 2 > 21.8 149.14 <> 330.5 .65 .1 > 150. * 330.6 .1 <4> 2 <> 331. (6) 38 Am. 5.731574. EXPLANATION. "We proceed as in the preceding example till we obtain the terras marked (2), in the operation. Dividing 10.2929 by 321.302, we obtain .03 for the next figure of the root. At this point we commence to apply decimal contractions, accord ing to the principles employed in the contracted method of cube root; -(24:3). Let it be observed, that ear Ji contracted term in the operation contains one redundant figure at the right. 420 NUMERICAL EQUATIONS OF HIGHER DEGREES. Commencing with column IV, we have 21. 8 X -03 = .65, which added to column III gives 148.49. Then 148.49X-03 = 4.455, which added to column II gives 325.757. Then 325.757 X -03 = 9.7727, which added to column I gives .5202. Again adding .65 to column III, we have 149.14. Then 149.14X-03 = 4.474, which added to column II gives 330.23, after dropping one place. Again, adding .65 to column III gives 150 after dropping two places. In like manner we continue till the work is finished. NOTE. Observe, as a general rule, to contract the several columns, for each root figure, as follows : Column I, place ; column II, 1 place ; column III, 2 places ; column IV, 3 places ; and so on. Find the real roots of the following equations : Ans. 5.1345787253. 3...T 8 +2x 2 23x 70 = 0. 4. x'+70x 300 = 0. 5. 6. a 500 = 0. = 0. 7. rc 4z 9 24x+ 48 = 0. 8. z 4 -f-z'4-z a x 500 = 0. Ans. Ans Ans. 3.7387936782. Ans. 7.6172797559. ( 3.3792053825, 4.5875359541, 6.9667413367. ( 1.7191292611, . < 6.5461457261. 14.2652749871. Ans. ( 4.4604168201, } 4.9296646474. 9. x 4 9*' llz a 10. a 4 = 0. Ans . f = 0. Ans. .1796840250, 10.2586086356. 2.8580833082, .6060183069, .4432769396, 3.9073785547. 11. x __ = 0. Ans. < /-3.0653157918, \_ .6915762805, - .1756747993. ) .8795087084, \ 3.0530581627, NOTE. Full solutions of the examples above may be found in the Key. p ; ' >>\v> UNIVERSITY OF CALIFORNIA LIBRARY This book is DUE on the last date stamped below. OCT 21 1947 I6*r ; -100w-12,'46(A2012sl6)4120 TU esit