Digitized by tine Internet Arciiive in 2008 witii funding from IVIicrosoft Corporation http://www.arcliive.org/details/elementsofalgebrOOIoomrich PREFACE. The stereotype plates of my Treatise on Algebra having be- come so much worn in the printing of more than 60,000 copies that it had become necessary to cast them aside, I decided to improve the opportunity to make a thorough revision of the work. I therefore solicited criticisms from several college pro- fessors who had had much experience in the use of this book, and in reply have received numerous suggestions. The book has been almost entirely rewritten, nearly every page of it having been given to the printer in manuscript. The general plan of the original work has not been materially altered, but the changes of arrangement and of execution are numerous. In the former editions, in place of abstruse demonstrations, I some- times employed numerical illustrations, or deductions from par- ticular examples. In the present edition such methods have been discarded, and I have aimed to demonstrate with concise- ness and elegance every principle which is propounded. This book therefore aims to exhibit in logical order all those principles of Algebra which are most important as a prepara- tion for the subsequent branches of a college course of mathe- matics. I have retained, with but slight alteration, a feature which was made prominent in the former editions, that of stating each problem twice: first as a restricted numerical problem, and then in a more general form, aiming thereby to lead the student to cultivate the faculty of generalization. At the same time I have very much increased the number of examples in- corporated with each chapter of the book, and at the close have given a large collection of examples, to which the teacher may resort whenever occasion may require. The proofs of the work have all been examined by Prof H. A. Newton, to whom I am indebted for numerous and im- portant suofprestions. 541 H21 CONTENTS. CHAPTER I. DEFINITIONS AND NOTATION. ^^^^ General Definitions 9 Symbols which denote Quantities 10 Symbols which indicate Operations 11 Symbols which indicate Relation 14: Combination of Algebraic Quantities 16 CHAPTER 11. ADDITION 21 CHAPTER III. gUBTBAGTION 25 CHAPTER IV. MULTIPLICATION. Case of Monomials — Case of Polynomials 32 General Theorems proved 38 CHAPTER V. DIVISION. Case of Monomials — Case of Polynomials 41 When a^ii" is exactly divisible 47 To resolve a Polynomial into Factors 50 CHAPTER VI. GREATEST COMMON DIVISOR. — LEAST COMMON MULTIPLE. How the Greatest Common Divisor is found 52 Rule applied to Polynomials 55 How the Least Common Multiple is found 58 CHAPTER VII. FRACTIONS. Definitions and General Principles 62 Reduction of Fractions 64 Addition and Subtraction of Fractions..... 69 Multiplication and Division of Fractions 72 I VI CONTENTS. CHAPTER Vin. • • • • • • •• • • •«• • • "****' • JS^QTUtlONS OF THE FIRST DEGREB. * 'l^efimtions — Axioms employed 79 Transformation of Equations 81 Solution of Equations 84 Solution of Problems 87 CHAPTER IX. EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITT. Different Methods of Elimination „ 99 Equations containing three or more Unknown Quantities 105 Problems involving several Unknown Quantities 109 CHAPTER X. DISCUSSION OF PROBLEMS. Positive Values of a; — ^Negative Values 114 Zero and Infinity 118 Indeterminate Solutions 121 Inequalities 122 CHAPTER XI. INVOLUTION. Powers of Monomials 127 Powers of Polynomials 130 CHAPTER XII. EVOLUTION. Roots of Monomials 133 Square Root of Polynomials 136 Square Root of Numbers 139 Cube Root of Polynomials 143 Cube Root of Numbers 146 CHAPTER XIII. RADICAL QUANTITIES. Transformation of Radical Quantities 150 Addition and Subtraction of Radical Quantities 157 Multiplication and Division of Radical Quantities 159 Powers and Roots of Radical Quantities 164 Operations on Imaginary Quantities 166 To find Multipliers whidi shall render Surds rational 168 Square Root of a Binomial Surd 171 Equations containing liadical Quantities 174 CONTENTS. . Vll CHAPTER XIV. y' , , , , , . , EQUATIONS OF THE SECOND DSOR^Si, ^ , , ^ '* \ T * '? * \" ' Incomplete Equations of the Second Degree " 177 Complete Equations of the Second Degree 182 Equations which may be solved like Quadratics 187 Problems producing Equations of the Second Degree 192 Equations containing two unknown Quantities 196 General Properties of Equations of the Second Degree 203 Discussion of the general Equation of the Second Degree 206 Discussion of particular Problems 209 Geometrical Construction of Equations 211 CHAPTER XV. RATIO AND PROPORTION. Definitions 214 Properties of Proportions 216 Variation 222 Problems in Proportion 224 CHAPTER XVI. PROGRESSIONS. Arithmetical Progression — General Principles 227 Examples 230 Geometrical Progression — General Principles 233 Examples 237 CHAPTER XVII. CONTINUED FRACTIONS. — PERMUTATIONS AND COMBINATIONS. Properties of Continued Fractions 240 Permutations and Combinations 244 CHAPTER XVin. BINOMIAL THEOREM. Powers of a Binomial 248 Sir Isaac Newton's Binomial Theorem 251 Powers and Roots of Polynomials 255 To extract any Root of aNumber 257 CHAPTER XIX. SERIES. Definitions — Method of Difibrences 259 Interpolation 263 Development of Algebraic Expressions into Series 265 vm CONTENTS. {^tAod of UVidfet^-mihad CoeflBcicnts 267 To re^qlvQ a^ Krecticn into Simpler Fractions 272 IkVer^gibKi (?f Series .t.OA^.^~ *'. 273 (jlenerai Demonstration of the Binomial Theorem 275 Applications of the Binomial Theorem ^.. 278 CHAPTER XX. LOOASITHMS. Properties of Logarithms 284 The common System of Logarithms 287 Multiplication and Division by Logarithms 293 Involution and Evolution by Logarithms 294 Exponential Equations 296 Compound Interest and Annuities 297 Computation of Logarithms 301 CHAPTER XXI. GENERAL THEORY OF EQUATIONS. Definitions and general Properties 306 An Equation of the nth Degree has n Roots 308 Law of the Coefficients of every Equation 310 Number of imaginary Roots 315 Transformation of Equations 316 Descartes's Rule of Signs 320 Composition of derived Polynomials 321 Equal Roots 322 Sturm's Theorem 323 Solution of simultaneous Equations of any Degree 331 CHAPTER XXn. NUMERICAL EQUATIONS OF HIGHER DEGREES. Commensurable Roots 834 Incommensurable Roots — Horner's Method 338 Equations of the fourth and higher Degrees 347 Newton's Method of Approximation 350 Method of double Position 861 Miscellaneous Examples .'. 865 ALGEBRA. CHAPTEE I. DEFINITIONS AND NOTATION. /l. Quantity is any thing that can be increased or diminished, and that can be measured. A line, a surface, a solid, a weight, etc., are quantities; but the operations of the mind, such as memory, imagination, judg- ment, etc., are not quantities. A quantity is measured by finding how many times it con- tains some other quantity of the same kind taken as a stand- ard. The assumed standard is called the unit of measure, 2. Mathematics is the science of quantity, or the science which treats of the properties and relations of quantities. It employs a variety of symbols to express the values and relations of quantities, and the operations to be performed upon these quan- tities, or upon the numbers which represent these quantities. 3. Mathematics is divided into pure and mixed. Pure math- ematics comprehends all inquiries into the relations of quan- tity in the abstract, and without reference to material bodies. It embraces numerous subdivisions, such as Arithmetic, Al- gebra, Geometry, etc. In the mixed mathematics, these abstract principles are ap- plied to various questions which occur in nature. Thus, in Sur- veying, the abstract principles of Geometry are applied to the measurement of land; in Navigation, the same principles are applied to the determination of a ship's place at sea ; in Optics, they are employed to investigate the properties of light; and in Astronomy, to determine the distances of the heavenly bodies. A 2 lO ALGEBRA. ■^4. Algebra is'tHat.b.ranch of mathematics in which quantities "aj'C'H^^Ve&ditted b^^ letters, and their relations to each other, as 'well as the operations to be performed upon them, are indi- cated by signs or symbols. The object of algebraic notation is to abridge and generalize the reasoning employed in the so- lution of all questions relating to numbers. Algebra may there- fore be called a species of Universal Arithmetic. *5. The symbols employed in Algebra may be divided into iJiree classes : 1st. Symbols which denote quantities. 2d. Symbols which indicate operations to be performed upon quantities. 8d. Symbols which indicate the relations subsisting between different quantities, with respect to their magnitudes, etc. Symbols which denote Quantities. 6. In order to generalize our reasoning respecting numbers, we represent them by letters, as a, &, c, or cc, y, z, etc., and these may represent any numbers whatever. The quantities thus represented may be either Jcnoivn quantities — that is, quantities whose values are given ; or unknown quantities — that is, quan- tities whose values are to be determined. Known quantities are generally represented by the first let- ters of the alphabet, as a, 6, c, c?, etc., and unknown quantities by the last letters of the alphabet, as cc, y, 2, u, etc. This, how- ever, is not a necessary rule, and is not always observed. 7. Sometimes several quantities are represented by a single letter, repeated with different accents, ns a', a'\ a'", a"", etc., which are read a prime, a second, a third, etc. ; or by a letter repeated with different subscript figures, as a^, a^, 03, a^, etc., which may be read a one sub, a two sub, a three sub, etc. All these symbols represent different quantities, but the ac- cents or numerals are emplo3'ed to indicate some important re- lation between the quantities represented. / DEFINITIONS AND NOTATION. 11 8. Sometimes quantities are represented by the initial letters of their names. Thus s may represent sum; d, difference or di- ameier ; r, radius or ratio ; c, circumference ; h, height, etc. All these letters may be used with accents. Thus, in a problem relating to two circles, d may represent the diameter of one cir- cle, and d' the diameter of the other ; c the circumference of one, and c' the circumference of the other, etc. -k^ Symbols which indicate Operations. -§. The sign of addition is an erect cross, +, called plus, and when placed between two quantities it indicates that the second is to be added to the first. Thus, 5 + 3 indicates that we must add 3 to the number 5, in which case the result is 8. We also make use of the same sign to connect several numbers togeth- er. Thus, 7 + 5 + 9 indicates that to the number 7 we must add 5 and also 9, which make 21. So, also, 8 + 5 + 13 + 11 + 1 + 3 + 10 is equal to 51. The expression a + 5 indicates the sum of two numbers, which we represent by a and h. In the same manner, m+?2+x+^ indicates the sum of the numbers represented by these four let- ters. If we knew, therefore, the numbers represented by the letters, we could easily find by arithmetic the value of such ex- pressions. / - 10. The sign of subtraction is a short horizontal line, — , called minus. When placed between two quantities, it indicates that the second is to be subtracted from the first. Thus, 8 — 5 indi- cates that the number 5 is to be taken from the number 8, which leaves a remainder of 3. In like manner, 12 — 7 is equal to 5, etc. Sometimes we may have several numbers to subtract from a single one. Thus, 16 — 5—4 indicates that 5 is to be subtracted from 16, and this remainder is to be further diminished by 4, leaving 7 for the result. In the same manner, 50 — 1—5 — 3 — 9 — 7 is equal to 25. The expression a-^h indicates that the number designated by a is to be diminished by the number designated by h. 12 ALGEBRA. 11. The double sign ± is sometimes written before a quan* tity to indicate that in certain cases it is to be added, and in others it is to be subtracted. Thus, 6 db c is read h j^l^^ or minus c, and denotes either the sum or the difference of these two quantities. 12. The sign of multiplication is an inclined cross, x. When placed between two quantities, it indicates that the first is to be multiplied by the second. Thus, 3x5 indicates that 3 is to be multiplied by 5, making 15. In like manner, axh indi- cates that a is to be multiplied by h; and axbxc indicates the continued product of the numbers designated by a, ft, and c, and so on for any number of quantities. Multiplication is also frequently indicated by placing a point between the successive letters. Thus, a.h.c.d signifies the same thing as axbxcxd. Generally, however, when numbers are represented by let- ters, their multiplication is indicated by writing them in suc- cession without any intervening sign. Thus, abc signifies the same as axbxc^ or a.b.c. The notation a.b or ab is seldom employed except when the numbers are designated by letters. If, for example, we attempt to represent in this manner the product of the num- bers 5 and 6, 5.6 might be confounded with 6-^\ and 56 would be read fifty-six, instead of five times six. The multiplication of numbers may, however, be denoted by placing a point between them in cases where no ambiguity can arise from the use of this symbol. Thus, 1.2.3.4.5 is sometimes used to represent the continued product of the numbers 1, 2, 3, 4, 5. X 13. When two or more quantities are multiplied together, each of them is called a factor. Thus, in the expression 7x5, 7 is a factor, and so is 5. In the product abc there are three factors, a, b, c. ^ When a quantity is represented by a letter, it is called a literal factor. When it is represented by a figure or figures, it DEFINITIONS AND NOTATION. 13 is called a numerical factor. Thus, in the expression 5a5, 5 is a numerical factor, while a and h are literal ^tors. 14. The sign of division is a short horizontal line with a point above and one below, -:-. When placed between two quantities, it indicates that the first is to be divided by the second. Thus, 24-^6 indicates that 24 is to be divided by 6, making 4. So, also, a^h indicates that a is to be divided by h. Generally, however, the division of two numbers is indi- cated by writing the divisor under the dividend, and drawing a line between them. Thus, 24 h- 6 and a^h are usually writ- , 24 . a ten -TT and 7. 15. The products formed by the successive multiplication of the same number by itself, are called the powers of that number. Thus, 2 X 2=4, the second pT5wer or square of 2. 2x2x2 = 8, the third power or cube of 2. 2 X 2 X 2 X 2 = 16, the fourth power of 2, etc. So, also, 3x3 = 9, the second power of 3. 3x3 X 3 = 27, the third power of 3, etc. Also, axa = aa, the second power of a. ax a xa=aaa, the third power of a, etc. In general, any power of a quantity is designated by the number of equal factors which form the product. -. \My(A.'^iC -'■' ' ''"- ■^' ^ 16. The sign of involution is a number written above a quan- tit}'-, at the right hand, to indicate how many times the quan- tity is to be taken as a factor. ^A root of a quantity is a factor which, multiplied by itself a certain number of times, will produce the given quantity. XThe figure which indicates how many times the root or fac- tor is taken, is called the exponent of the power. Thus, instead of aa, we write a^, where 2 is the exponent of the power; instead of aaa, we write a^, where 3 is the expo- 14 ALGEBRA. nent of the power ; instead of aaaaa, we write a®, where 5 is the exponent of the power, ^^Q-i^rruu^ju - Ir h^' (r\j ^d, or <. When placed between two quantities, it indicates that they are "un- equal, the opening of the angle being turned toward the greater number. When the opening is toward the left, it is read great- er than ; when the opening is toward the right, it is read less than. Thus, 5>3 denotes that 5 is greater than 3 ; and 6<11 denotes that 6 is less than 11. So, also, a>5 denotes that a is greater than h ; and x as c is to d; and this is express- ^.d by writing them thus : a : b=c:dy or a:b::c:d. X 23. The sign of variation is the character od. When written between two quantities, it denotes that both increase or diminish together, and in the same ratio. Thus the expression scotv de- notes that s varies in the same ratio as the product of t and v. 16 ALGEBRA. X J4. Three dots .*. are sometimes employed to denote there- fore, or consequently. A few other symbols aVe employed in Algebra, in addition to those here enumerated, which will be explained as they occur. Combination of Algebraic Quantities. ^ 25. Every number written in algebraic language — that is, '^by aid of algebraic symbols — is called an algebraic quantity, or an algebraic expression. Thus, 3a^ is the algebraic expression for three times the square of the number a. 7a^Z>* is the algebraic expression for seven times the third power of a, multiplied by the fourth power of b. ^ 26. An algebraic quantity, not composed of parts which are separated from each other by the sign of addition or subtrac- tion, is called a monomial, or a quantity of one term, or simply a term. Thus, 3a, bbc, and Ixy"^, are monomials. Positive terms are those which are preceded by the sign plus, and negative terms are those which are preceded by the sign minus. When the first term of an algebraic quantity is posi- tive, the sign is generally omitted. Thus a-{-b—c is the same as -\-a-\-b—c. The sign of a negative term should never be r 27. The coefficient of a quantity is the number or letter pre- fixed to it, showing how often the quantity is to be taken, c^' < Thus, instead of writing a + a-f-a-f-a-f-a, which represents 5 a's added together, we write 6a, where 5 is the coefficient of a. In ^{x-\-y), 6 is the coefficient oi x-{-y. When no coeffi- cient is expressed, 1 is always to be understood. Thus, la and a denote the same thing. The coefficient may be a letter as well as a figure. In the ex- pression nx, n may be considered as the coefficient of x, be- cause X is to be taken as many times as there are units in n. If n stands for 5, then nx is 5 times x. When the coefficient ' DEFINITIONS AND NOTATION. 17 is a number, it may be called a numerical coefficient ; and when it is a letter, a literal coefficient. In 7aa7, 7 may be regarded as the coefficient of ax^ or 7a may be regarded as the coefficient of x. 28 The coefficient of a positive term shows how many times the quantity is taken positively^ and the coefficient of a nega- tive term shows how many times the quantity is taken nega- tively. Thus, -\-4:X—-\-x-\-x-[-x-\-x; — 4x— —x—x—x—x. ^29. Similar terms are terms composed of the same letters, affected with the same exponents. The signs and coefficients may differ, and the terms still be similar. Thus, 2>ah and lab are similar terms. Also, ba\ and —Sa^c are similar terms. X.30. Dissimilar terms are those which have different letters or exponents. Thus, axy and axz are dissimilar terms. Also, Sab"^ and 4a^5 are dissimilar terms. ▼*31. A polynomial is an algebraic expression consisting of more than one term; as, a-\-b ; or a-^2b—bc-{-x. A polynomial consisting of two terms only is usually called a binomial; and one consisting of three terms only is called a trinomial. Thus, Sa-\-6b is a binomial; and 6a — Sbc-\-xy is a trinomial. y 32. The degree of a term is the number of its literal factors. ThusJ^ 8a is a term of the first degree. 6ab " second " 6a^bc^ " sixth " In general, the degree of a term is found by taking the sum of the exponents of all the letters contained in the term. Thus the degree of the term bab'^ccP is 1+2 4-1-/- 8, or 7; that is, this term is of the seventh degree. .18 ALGEBRA. X 33. A polynomial is said to be homogeneou s_yi\iQn all its terms are of the same degree. Thus, Sa^— 4a6 -f 62 is of the second degree, and homogeneous 2a3-f3a2c-4c2c^ " third But ba^—2ah-\-c is not homogeneous. 34. The reciprocal of a quantity is the quotient arising from dividing a unit by that quantity. Thus the reciprocal of 2 is ^ ; the reciprocal of a is ^ 35. A function of a quantity is any expression coutainmg that quantity. Thus, ax'^-\-h is a function of a?. ay^-\-cy-\-cl is a function of?/. ax^—hy^ is a function of x and y. Exercises in Algebraic Notation. 36. In the following examples the pupil is simply required to express given relations in algebraic language. Ex. 1. Give the algebraic expression for the following state- ment: The second power of a, increased by twice the product of a and 6, diminished by c, and increased by c?, is equal to fifteen times x. Ans. a'^-{-2ah—c-\-d=\bx. Ex. 2. The quotient of three divided by the sum of x and four, is equal to twice h diminished by eight Ex. 3. One third of the difference between six times x and four, is equal to the quotient of five divided by the sum of a and h. Ex. 4. Three quarters of x increased by five, is equal to three sevenths of b diminished by seventeen. Ex. 5. One ninth of the sum of six times x and five, added to one third of the sum of twice x and four, is equal to the product of a, 6, and c. Ex. 6. The quotient arising from dividing the sum of a and b by the product of c and (i,is greater than four times the sum of m, 72, cc, and y. DEFINITIONS AND NOTATION. 19 37. In the following examples the pupil is required to trans- late the algebraic symbols into common language. Ex.1. -T- — h h c a-\-b' Ans. The quotient arising from dividing the sum of a and X by &, increased by the quotient of x divided by c, is equal to the quotient of m divided by the sum of a and h. Ex.2. 7a2 + (i-c)x(cZ+e)=a7+2/- How should the preceding example be read when the first parenthesis is omitted? d + 6— c 3 a-}-7 6a — d Ex. b. 2a ^fb^ —ac— 5(a -f- m + cc). \t. . VSb^-^Vi 1 ^^•^- l + 2a ^^^+? Computation of Nuraerical Values. " 38. The numerical value of an algebraic expression is the re- sult obtained when we assign particular values to all the let- ters, and perform the operations indicated. Suppose the expression is 20^1). If we make a = 2 and h — 2>^ the value of this expression will be 2x2x2x8 = 24. If we make (7.=4 and ?; = 3, the value of the same expression will be 2x4x4x3 = 96. The numerical value of a polynomial is not affected by changing the order of the terms, provided we preserve their re- spective signs. The expressions a^-\-2ah-\-h'^^ a"^ -\-h'^ -\-2ab^ and h'^-\-2ab^a'^y have all the same numerical value. Find the numerical values of the following expressions, in which a = Q^ b = 6, c=4, m=8, and n-=2. Ex.1. a^+Sab-c^ J.n5. 36 + 90-16 = 110. Ex.2. a'x{a + b)-2abc, Ans. 156. 20 ALGEBRA. Ex. 3. 5- +c«. Ans. 28. a + dc V2ac4-c* ^ Sc.5. V^^^^c+ V2ac-\-c\ Ux.e. 3v^+2aV2a+I+2c. ^07.7. (3v^+2a)V2a+&+2c. ^05.9. a 6 m— ?i m+n a2Z)VmH3a6cm4-2 abcm-{-l Find the numerical values of the following expressions, in which a=S, Z>=5, c=2, m=4, w=6, and x=9. jEc. 10. ^^ 5 ^. ^?25. 8. ox—a^—c jE'a:. 11. 5a:-7Vx. ^715.24. Ux. 12. 2a;2 4- V2^H^. -Atw. 175. '^x. 13. vTo+n-VlO+n. -^^ ^ic. 14. 6(m2H-n2)4-4aaj. '^'' ^ic. 15. a* - 4a3 + 7a2 - 6a. Ex. 16. V5Vm+5'v/^+ Vm-f Vx. Ex. 17. m—n-\-b m-^2n 2a4-m Ex. 19. / ADDITION. 2J CHAPTER 11. ADDITION. 39. Addition^ in Algebra, is the connecting of quantities ta getlier bj means of their proper signs, and incorporating such as can be united into one sum. When the Quantities are similar and have the same Signs. 40. The sum of 8a, 4a, and 5a, is obviously 12a. That is, 3a 4- 4a 4- 5a = 12a. So, also, ^^a^ --4a, and —5a, make —12a; for the minus sign before each of the terms shows that they are to be sub- tracted, not from each other, but from some quantity which is not here expressed ; and if 2>a^ 4a, and 5a are to be subtracted successively from the same quantity, it is the same as subtract- ing at once 12a. Hence we deduce the following RULE. Add the coefficients of the several quantities together^ and to their sum annex the common letter or letters^ prefixing the common sign, EXAMPLES. Sa -Sab 2b-\-Sx a-2x'^ 2a -h 2/' 6a -6ab 6b-\-7x 4:a-Bx^ 5a-f2?/2 7a - ab b-{-2x 3a— 5a;2 9a+3y2 a -lab 46+307 7a- x"" 4a+6?/2 16a -17 ab The pupil must continually bear in mind the remark of A^i. 26, that, when no sign is prefixed to a quantity, plus is always to be understood. When the Quantities are similar, but have different Signs. 41. The expression 7a — 4a denotes that 4a is to be sub- tracted from 7a, and the result is obviously 3a. That is, 7a— 4a = 3a. 22 ALGEBRA. The expression 5a — 2((-\-Sa — a denotes that we are to sub- tract 2a from 5a, add 3a to the remainder, and then subtract a from the last sum, the result of which operation is 5a. That is, 5a— 2a+ 8a— a— oa. It is generally most convenient to take the sum of the posi- tive quantities, which in the preceding case is 8a; then take the sum of the negative quantities, which in this case is Sa^, and we have 8a— 3a, or 5a, the same result as before. Hence we deduce the following IIULE. Add all the i^osilive coefficients together^ and also all those that are negative; subtract the least of these results from the greater; to the difference annex the common letter or letters^ and prefix the sign of the greater sum. EXAMPLES. — 2>a 6x+5a?/ lay- 7 -2a^x -6«H2Z/ 4- 7a -3.c+2a?/ - ay-{- 8 a^x 2a2-3Z> + 8a x—^ay 2ay— 9 -Za^x -5a2-8^ — a 2:r+ ay 8r,7/- 11 na^x 4a2-2/> 4- 11a fox-^lay WJien some of the Quantities are dissimilar. 42. Dissimilar terms can not be united into one term by ad- dition. Thus 2a and Zb neither make ba nor oh. Their sum can therefore only be indicated by connecting them by their proper signs, thus, 2a4-3Z/. In adding together polynomials which contain several groups of similar quantities, it is most convenient to write thejn in such a manner that each group of similar quantities may occu py a column by itself. Ilcncc we deduce the following RULE. Write the quantities to he added so that the similar terms may be arranged in the same column. Add up each column separately^ and connect the several results hy /heir projuT signs. ADDITION. 23 EXAMPLES. 1. Add 2a+3&4-4c, a-{-2b-{-5c, Sa—b + 2c, and _a+4&_6r, Avs. 5a -j- 86 + 5c. 2. Add 2x^/~2xH?/^ Sx-^-f x?/-f 4z/2, a;2_^.^^3^2^ and 4^^ — 2y^—Sxy. Ans. 6x'^—xy-\-6y^. 8. Add 5aV — 2x7j^ 2>ax — 4:xy^ 7xy — 4:ax, aV-{-6xy^ and 2aic— 8x?/. Ans. 6a'^x'^-{-Sxy-\-ax. 4. Add 2a2-3ac + 3&-cf^, 4:a''-ac + 2cd-b, 3a'^ + 2ac-'ib -i-3cd, and a^— 2rtc+5c+6cr —^x^-^-bax^+l^ ^ab 9. From 3a4-5Z>— 2c subtract 2a— b. Ans. a + 66— 2c. 10. From babc-2b-^ subtract ^abc-2b-\-\. Alls. 2abc—7. 11. From 4a2— 7a-l-3x subtract a'^-\-Sa—2x. Ans. 3a2— 10a+5x. 12. From a—b-{-2m—x subtract 3a:-f-m— 4&+a. 13. From 2x^—x'^y-{-6xy^ subtract x^—2xy"-\-y^. 14. From m-\-n subtract m-^n. 15. From m-\-n-\-x subtract —m—n—x. 16. From Sa^-Sa— 7 subtract -2a2_4a+10. 17. From 77i'* + 37?^^— 4m2— 2;?i+l subtract 7n* — 2m^-{-m^ -3m4-5. 18. From x^-6x^-\-10x^-B subtract x^-^-^x^-lOx^+S. 19. From da^-^ax+2x'^—14^a^x subtract x^ — Wa^x -^ 2a^ —4:aoc 20. From 6abx—4:mn-\-6ax subtract Smn-{-6ax—Sabx. Subtraction may be proved, as in Arithmetic, by adding the remainder to the subtrahend. The sum should be equal to the minuend. SUBTRACTION. 27 K 47. It will be perceived that the term subtraction is used in a more general sense in Algebra than in Arithmetic. In Arith- metic, where all quantities are regarded as positive, a number is always diminished hy subtraction. But in Algebra the dif- ference between two quantities may be numerically greater than either. Thus the difference between + a and — 6 is a + h. The distinction between positive and negative quantities may be illustrated by the scale of a thermometer. The de- grees above zero are considered positive, and those below zero negative. From five degrees above zero to five degrees below zero, the numbers stand thus, + 5, +4, 4-3, +2, +1, 0, -1, -2, -3, -4, -5. The difference between a temperature five degrees above zero and one which is five degrees helow zero, is ten degrees, which is numerically the sum of the two quantities. Ten is said to be the algebraic difference between +5 and —5. 48. When dissimilar terms have a common literal part^ the difference of the terms may be expressed, as in Art. 44, by in- closing the difference of the coefl&cients in a parenthesis, and prefixing it to the common letter or letters. Thus the difference between ax^ and bx^ may be written (a—b)x'^. EXAMPLES. 1. Fiom ax^if subtract — Si^y. Ans. {ci-\-Z)x'^y'^, 2. From 2ax-\-Zy subtract bbx—y. Ans. {2a—6b)x-\-4:7/. 8. From mx-^-ny subtract 2>ax—2y. Ans. {in — Za)x-\-{n-\-2)y, 4. From 4mV^-|-3 subtract 2aVx—l. 5. From 2ax* + 35x^—7 subtract ^inx^—nx^-{-2. 6. From amx^-^-bnx^ ■\-cx subtract bmx^—anx^-\-ax, 7. From m-\-am-{-bm subtract am-\-bm-\-cm. 8. From l^Zax^-^-ba^x^^la^x* subtract x^^2>ax^—ba^x\_ 28 ALGEBRA. 49. Use of the Parenthesis. — If we wish to indicate that one polynomial is to be subtracted from another, we may inclose it in a parenthesis, and prefix the sign minus. Thus the expres- sion , , , . indicates that the polynomial m—n-\-x is to be subtracted from the polynomial a—b. Performing the operation indicated, we ^^^® a—b—m+n^x. The expression a—b-\-(m—n-{-x) indicates that the polynomial m—7i-\-x is to be added to the polynomial a— 5, and the result is a—b-{-m—n~{-x. Hence we see that a parenthesis preceded by the plus sign may be removed without changing the signs of the inclosed terms ; and, conversely, any number of terms, with their prop- er signs, may be inclosed in a parenthesis, and the plus sign written before the whole. ^ But if the parenthesis is preceded by the minus sign, the signs of all the inclosed terms must be changed when the pa- renthesis is removed ; and, conversely, any number of terms may be inclosed in a parenthesis, and preceded by the minus sign, provided the signs of all the inclosed terms are changed. 50. According to the preceding principle, polynomials may- be written in a variety of forms. Thus, a—b—c-\-d is equivalent to a—{b-\-c—d\ or to a—b—{c—d), or to a+d—{h-^c). These expressions are all equivalent, the first form being the simplest. EXAMPLEa Keduce the following expressions to their simplest forms. 1. 2a^-6a^b+3ab^^{a^-\-¥-ab''), Ans, a^-6a^b-\-4ab^-b^ SUBTEACTION. 29 2. a-\-h-\-c-{a~h)-{b-c). Ans. h^2c. 3. 4a2-^)-(2a-8Z>+l) + 3a. Ans. 4:a?-\-a-]-2h-l. 4. a + 25-(3?7i-2a+8x2)^ 6. 3a3-2a2-j-a4-l-(2a3-a2-a+5)-(a3_a2-5a-4). 6. a + h-{2a-'6h)-{pa + n)-{-Ua-\-2h). 7. Sa^x?/ — 55^2?/ -f- "Jcxy"^ — Sy^ — {a?xy + Zhx^y — 4(^?/2 -f 9^^). 8. 7aa;2_ lOa^xH lla2ic-5a* - {^ax^-\-l2aV-Qa'x-^a^). V 51. Hence we see tliat when an expression is inclosed in a parenthesis, the essential sign of a term depends not merely upon the sign which immediately precedes it, but also upon the sign preceding the parenthesis. Thus m-\-{-\-n) is equivalent to m+n, and m+(— n) " m—n. But m— (4-n) " m—n^ and m—{—n) " m-\-n. •^ The sign immediately preceding n is called the sign of the quantity ; the sign preceding the parenthesis may be called the sign of the operation ; while the sign resulting from the opera- tion is called the essential sign of the term. We perceive that when the sign of the quantity is the same as the sign of opera- tion, the essential sign of the term is positive ; but when the sign of the quantity is different from the sign of operation, the essential sign of the term is negative. X 52. Use of Negative Quantities. — The introduction of nega- tive quantities into Algebra enables us not only to compare the magnitude^ but also to indicate the relation or quality of the objects about which we are reasoning. This peculiarity will be understood from a few examples : 1st. Gain and Loss in Trade. — Suppose a merchant to gain in one year a certain sum, and in the following year to lose a certain sum ; we are required to determine what change has taken place in his capital. This may be indicated algebraical- ly by regarding the gains as positive quantities, and the losses as negative quantities. Thus, suppose a merchant, with a cap- ital of 10,000 dollars, loses 3000 dollars, afterward gains 1000 80 ALGEBRA. dollars, and then loses again 4000 dollars, the whole may be expressed algebraically thus, 10,000 - 3000 + 1000 - 4000, which reduces to +4000. The + sign of the result indicates that he has now 4000 dollars remaining in his possession. Suppose he further gains 500 dollars, and then loses 7000 dol- lars. The whole may now be expressed thus, 10,000-3000+1000-4000+500-7000, which reduces to —2500. The — sign of the result indicates that his losses exceed the sum of all his gains and the property originally in his possession ; that is, he owes 2500 dollars more than he can pay. 53. 2d. Motion in Contrary Directions. — Suppose a ship to sail alternately northward and southward, and we are required to determine the last position of the ship. This may be indi> cated algebraically, if we agree to consider motion in one direc- tion as a positive quantity, and motion in the opposite direction- as a negative quantity. Suppose a ship, setting out from the equator, sails north- ward 50 miles, then southward 30 miles, then northward 10 miles, then southward again 20 miles, and we wish to determ- ine the last position of the ship. If we call the northerly mo- tion + , the whole may be expressed algebraically thus, 50-30 + 10-20, which reduces to + 10. The positive sign of the result indi- cates that the ship was north of the equator by 10 miles. Suppose the same ship sails again 40 miles north, then 70 miles south, the whole may be expressed thus, 50-30+10-20+40-70, which reduces to —20. The negative sign of the result indi- cates that the ship was now south of the equator by 20 miles. We have here regarded the northerly motion as + , and the SUBTRACTION. 31 southerly motion as — ; but we might with equal propriety have regarded the northerly motion as — , and the southerly motion as +. It is, however, indispensable that we adhere to the same system throughout, and retain the proper sign of the result, since this sign shows whether the ship was at any time north or south of the equator. In the same manner, if we regafd westerly motion as +, we must regard easterly motion as — , and vice versa; and, gen- erally, when quantities which are estimated in different direc- tions enter into the same algebraic expression, those which are measured in one direction bemg treated as +, those which are measured in the opposite direction must be regarded as — . 54. The same principle is applicable to a great variety of examples in Geography, Astronom}^, etc. Thus, north latitude is generally indicated by the sign +, and south latitude by the sign — . West longitude is indicated by the sign +, and east longitude by the sign — . Degrees of a thermometer above zero are indicated by the sign -h, while degrees below zero are indicated by the sign — . A variation of the magnetic needle to the west is indicated by the sign +, while a variation to the east is indicated by the sign -. The date of an event since the birth of Christ is indicated by the sign -|- ; the date of an event before the birth of Christ, by the sign — ; and the same distinction is observed in a great variety of cases which occur in the application of the mathe- matics to practical problems. In all such cases the positive and negative signs enable us not merely to compare the mag- nitude^ but also to indicate the relation of the quantities con- sidered. 32 ALGEBRA. CHAPTER ly. . MULTIPLICATION. 56. Multiplication is the operation of repeating one quantity as many times as there are units in another. The quantity to be multiplied is called the multiplicand; and that by which it is to be multiplied is called the multiplier. 56. When several quantities are to be multiplied together, the result will be the same in whatever order the multiplica- tion is performed. In order to demonstrate this principle, let unity be repeated five times upon a horizontal line, and let there be formed four such parallel lines, thus, Then it is plain that the number of units in the table is equal to the five units of the horizontal line repeated as many times as there are units in a vertical column ; that is, to the product of 5 by 4. But this sum is also equal to the four units of a vertical line repeated as many times as there are units in a horizontal line ; that is, to the product of 4 by 5. Therefore the product of 5 by 4 is equal to the product of 4 by 5. For the same reason, 2x3x4 is equal to 2 x 4 x 3, or 4 x 3 x 2, or 3x4x2, the product in each case being 24. So, also, if a, /^ and c represent any three numbers, we shall have ahc equal tu hca or cab. CASE I. When both ilie factors are monomials. 57. Suppose it is required to multiply 5a by 4Z>. The prod- uct may be indicated thus, 5a x 41), MULTIPLICATION. 33 But since the order of the factors may be changed without affecting the value of the product, the factors of the same kind may be written together thus, 4 X bah ; or, simphfying the expression, we have w Wah. Hence we see that the coefficient of the product is equal to the product of the coefficients of the multiplicand and multiplier. 58. The Law of Exponents.^W e have seen, in Art. 16, that when the same letter appears several times as a factor in a product, this is briefly expressed by means of an exponent. Thus, aaa is written a^, the number 3 showing that a enters three times as a factor. Hence, if the same letters are found in two monomials which are to be multiplied together, the ex- pression for the product may be abbreviated by adding the exponents of the same letter. Thus, if we are to multiply a^ by a^, we find a^ equivalent to aaa, and a^ to aa. Therefore the product will be aaaaa, which may be written a^, a result which is obtained by adding together 3 and 2, the exponents of the common letter a. XHence we see that the exponent of any letter in the product is equal to the sum of the exponents of this letter in the multiplicand and multiplier. 59. Hence, for the multiplication of monomials, we have the following EULE. Multiply together the coefficients of the two terms for the coeffi- cient of the product. Write after this all the letters in the two monomials, giving to lach letter an exponent equal to the sum of its exponents in the twc factors. EXAMPLES. 1. 2. 3. 4. Multiply Idbc 5a^Pc 9amx*y 6a^h^c* by 6mn Sah'^c 4:am^xy^ 8a^bc^ Product S6ahcrm ISa^Z^V SQa'^m^xY 48a^Z>V B2 ^ 34 ALGEBRA. 5. Multiply 9a^x by la^y. 6. Multiply 12a'b*c' by lla^iV. ^ 7. Multiply a"^ by a**. ^rw. a"^+*». a Multiply 8a"* by Ga**. 9. Multiply a'" by a'". J-tw. a^. ^ 10. Multiply 9a"* by 12a"*. " 11. Multiply a'^b by aZ)"*. Ans. a'^-^^h'^+K 12. Multiply 6a"*x» by 6a"*cc. ^??5. 30a2"*a:»+i. 13. Multiply 3a2"*ic» by 4a"*a;3« J.n5. 12a3"*a:*^ 14. What is the continued product of 6a, 4m^x, and 9a^m^x? Ans. 180a^m^x^. 15. What is the continued product of 7a^b, ab^, and 4ac^.^ 16. What is the continued product of Sa'^cc, 5ab^, and 7abx? 17. What is the continued product of a, ab, abc, abed, and abcdx ?J 18. What is the continued product of a^, a^h^, o?bc, and CASE 11. I When one or both of the factors are polynomials. ^ 60. Kepresent the sum of the positive terms of any polyno- mial by a, and the sum of the negative terms by b. Then a—b will represent any polynomial whatever. In like man- ner, c—d will represent any other polynomial whatever. It is required to find the product of a—b by c—d. In the first place, let us multiply a—b by c. This implies that the difference of the units in a and b is to be repeated c times. If 4- a be repeated as many times as there are units in c, the result will be -\-ac. Also, if —5 be repeated as many times as there are units in c, the result will be —be, for —6 taken twice is —2b, taken three times is —36, etc.; and if it be repeated c times, the result will be —cb or —be The entire operation may be exhibited thus : a — b ac—bc. Next let us multiply a—b by c—d. When we multiply MULTIPLICATION. 35 a—h by c, we obtain ac—bc. But a — b was only to De taken c—d times; therefore, in this first operation, we have repeated it d times inore than was required. Hence, to have the true product, we must subtract d times a—b from ac—bc. But d times a— 6 is equal to ad-bd, which, subtracted from ac—bc^ ° ac—bc— ad-\-bd. If the pupil does not perceive the force of this reasoning, it will be best to repeat the argument with numbers, thus: Let it be proposed to multiply 8—5 by 6—2; that is, the quantity 8—5 is to be repeated as many times as there are units in 6—2. If we multiply 8—5 by 6, we obtain 48—80; that is, we have repeated 8—5 six times. But it was only required to repeat the multiplicand /oz^r times, or 6—2. We must there- fore diminish this product by twice 8 — 5, which is 16—10; and this subtraction is performed by changing the signs of the subtrahend. Hence we have 48-30-16 + 10, which is equal to 12. This result is obviously correct; for 8— 5 is equal to 3, and 6— 2 is equal to 4; that is, it was re- quired to multiply 8 by 4, the result of which is 12, as found above. We have thus obtained the following results : +ax(+5)=+a5, -\-ax{—b)=—ab^ —ax{+b) = —ab, r-ax{—b)=-{-ab, from which we perceive that when the two factors have like signs, the product is positive; and when the two factors have un- like signs, the product is negative. 61. Hence, for the multiplication of polynomials, we have the following general RULE. '^Multiply each term of the multiplir.and by each term of the mul- Q6 ALGEBRA. iipltevj and add together all the partial products^ observing thai like signs require + i'^^ the product, and unlike signs — . EXAMPLEa 1. 2. Multiply 2a + 36 a^'-^-^ah+W by 4a— 5Z> a-\-h Partial (8a2+12a6 a^^-2a%-\- a\? products ( -\^ah--\m a%-[-2aJ/-\-h^ Result 8aH 2aZ>-1562 a^+Sa^Z^+Sai^+i^ ' w ^ It is immaterial in what order the terms of a polynomial are arranged, or in what order the letters of a term are ar- ranged. It is, however, generally most convenient to arrange the letters of a term alphabetically, and to arrange the terms of a polynomial in the order of the powers of some common letter. 3. Multiply a'^—ab+b^ by a + b. Ans. a^-^b^. 4. Multiply a^—2ab-\-b^ hy a—b. 5. Multiply 3a2-2a+5 by a -4. 6. Multiply a^-ab-^-b'^ by a'' + ab-\-b\ Ans. a*-^a%^-{-b\ 7. Multiply 2a2_3a&+4 by a''-\-2ab-S. 8. Multiply a^ + a'^b-]-ah'^-\-b^ by a-b. 9. Multiply a-\-7nb by a-{-nb. 10. Multiply Sa-\-2bx—Sx^ by Sa—2bx+Sx\ 11. Multiply together cc— 6, x-{-2, and x-\-S. 12. Multiply together x—S, x—4:, x-\-6, and x—6. 13. Multiply together a'^+ab+b^, a'^—ab+b^, and a^^b\ 14. Multiply together a+a;, b-\-x, and c+a-. 16. Multiply a'^-{-a^b-\-a%^-\-ab^-[-b* by a— Z>. 16. Multiply a3-3a2 + 7a-12 by a^+Sa+2. 17. Multiply ir^ + 2a:3 + 8x2 + 2a: + l by a:2-2a:4-l. 18. Multiply 14:a^x—6a'^bx+x^ by 14a3ic+6a2Z;x— x^. 19. Multiply x^—x^y-\-xy^ by x^—xg^—y^. ' 20. Multiply 8a;2+8xy— 5 by 4x^—7xy-\-9. 62. Degree of a Product. — Since, in the multiplication of two monomials, every factor of both quantities appears in the prod MULTIPLICATION. 37 uct; it is "obvious that tbe degree of the product will be equal to the sum of the degrees of the multiplier and multiplicand. Hence, also, if two polynomials are homogeneous^ their product will be homogeneous. Thus, in the first example of the preceding article, each term of the multiplicand is of the first degree, and also each term of the multiplier; hence each term of the product is of the second degree. For a similar reason, in the second exam- ple, each term of the product is of the third degree ; and in the sixth example, each term of the product is of the fourth degree. This principle will assist us in guarding against er- rors in the multiplication of polynomials, so far as concerns the exponents. . 63. Number of Terms in a Product. — When the product aris- ing from the multiplication of two polynomials does not admit of any reduction of similar terms, the whole number of terms in the product is ecLual to the product of tlie numbers of the terms in the two polynomials. Thus, if we have five terms in the multiplicand, and four terms in the multiplier, the whole number of terms in the product will be 5x4, or 20. In gen- eral, if there be m terms in the multiplicand, and n terms in the multiplier, the whole number of terms in the product will be mxn. 64. Least Number of Terms in a Product. — If the product of two polynomials contains similar fer-ms, the number of terms in the product, when reduced, may be much less than mn; but it is important to observe that among the different terms of the product there are always two which can not be combined with any others. These are, 1st. The term arising from the multiplication of the two terms affected with the highest exponent of the same letter. 2d. The term arising from the multiplication of the two terms affected with the lowest exponent of the same letter. For it is evident, from the rule of exponents, that these iwo partial products must involve the letter in question, the one 88 ALGEBRA. with a higher^ and the other with a lower exponent than any of the other partial products, and therefore can not be similar to any of them.-t'Hence the product of two polynomials can never contain less than two terms. ^ y,^ 65. For many purposes it is sufficient merely to indicate the multiplication of two polynomials, without actually perform- ing the multiplication. This is effected by inclosing the poly- nomials in parentheses, and writing them in succession, with or without the sign x. When the indicated multiplication has been actually performed, the expression is said to be fa> panded.^ EXAMPLES. 1. Expand (a +5) (c+rf). Ans. ac-^hc-\-ad-\-bd. 2. Expand 9a-7{h-c). 3. Expand and reduce 14(12^a-6-c)+13(44-a-c)-15(7-a-c). 4. Expand and reduce 28(a-&+c)+24(a+Z>-c)-13(6-a-c). 5. Expand and reduce 24a^66-9(a+i)-f-25a-19(Z>-c)-17(a+5-c). 6. Expand and reduce 53(a_Z>+c)-27(a+5-c)-26(a-6-c). 66. The three following theorems have very important ap- plications. ,,. ' The square of the sum of two nmaJbrn^s is equal to the square of the firsts plus twice the product of the first hy the second^ plus Hie square of the second. Thus, if we multiply a-\-h by a-\-h a'^-f ah , ah^V' we obtain the product \a2 4-2a64-R Hence, if we wish to obtain the square of a binomial, we can, according to this theorem, write out at once the term.« MULTIPLICATION. 39 of the result, without the necessity of performing an actual multiplication. EXAMPLES. 1. (8a4-5)2^ ?^V^^/w-fi''' 6. (oa2 4-7aZ/)2=:ivA7»(;t'irf;^ 8. {6a + Shf= -^^i^O^lr^fo- 8. (2a+-^)2=.^i^vL('. • / 67. T/ie square of the difference of two numbers is equal to iht square of the first, minus twice the product of the first by the sec- ond, plus the square of the second. Thus, if we multiply a—b by a—b d^— ah ^ ab^h^ we obtain the product a'^—2ah-\-b'^. ^ EXAMPLES, 1. (2a-^8/>)2 = ^>^>l'-//rc.^ ^ 6. {la'-nabY^ 2. (5a-4i)2=i6A-X(;^(r-f ^^^ 7. [l d'b'' -12abf ^^ ■ 4. (6a2^8x)^-=^i^^'*-5fef'-^*'^^''- 9. (2-^)2:= ^rUi ' ' 5. {x-\yf=^^^^' 'V- 10. (4-^)2= ;^.| , 68. Meaning of the sign ±. Since (a+&)2r=a2-|-2aZ>+Z'', and (a-bf^a^^2ab + b\ we may write both formulas in the following abbreviated form, {a-±bf = o?±2ab + b''; which indicates that, if we use the + sign of b in the root, we must use the + sign of 2ab in the square ; but if we use the — sign of b in the root, we must use the — sign of 2ah in the square. By this notation we are enabled to express two dis- tinct theorems by one formula. 40 ALGEBRA. 69. The product of the sum and difference of two nw m he r B - is equal to tJie difference of their squares. •Thus, if we multiply a-\-h by a—h oFTaE -ah-y^ Are obtain the product a^ —h\ EXAMPLES. 1. (3a4-26) (3ci-25) = 2. (7a5 + x) \lah—x)= ^ / 8. (8aH-7Z>c)(8a-75c)= QHe>^^ - ^<^ fi' 4. (5aH6Z;^)(5a2-663)=: it^n - - "^ 5. (4a2+37r2ic)(4a2— 3mx)=:/5'.'?^ 6. (3a26+a') (3a2^>-a3)= 7. (m + 1) (m— 1) = • 8. (4+i)(4-i)= ,,'^.^^ The pupil should be drilled upon examples like the pre- ceding until he can produce the results mentally with as great facility as he could read them if exhibited upon paper, and without committing the common mistake of making the square of a + /^ equal to a^-^J)^^ or the square of a— & equal to a'^—h\ The utility of these theorems will be the more apparent when they are applied to very complicated expressions. Fre- quent examples of their application will be seen hereafter. DIVISION. y-y CHAPTER Y. DIVISION. '^ 70. Division is the converse of multiplication. In multipli- cation we determine the product arising from two given fac- tors. In division we have the product and one of the factors given, and we are required to determine the other factor. The dividend is the product of the di'V.isor and quotient, the divisor is the given factor, and the quotient is the factor re- quired to be found. .' ^ ^ When the divisor and dividend are both monomials. • 71. Since the product of the numbers denoted by a and b is denoted bj a6, the quotient of ab divided by a is ^ ; that iS", ab^a=h. Similarly, we have abc-^a=bc, abc-^c=ab, abc-^ab =c, etc. The division is more commonly denoted thus: abc 7 aba abc , ' a ,0 ' c ' abc abc ^ abc be ~ ^ ac~ '' ah~ ' So, also, 12m?i divided by 8m gives ^n; for Zm multiplied by 4?^ makes Vlmn. 72. Rule of Exponents in Division. — Suppose we have a^ to be divided by a^. We must find a quantity which, multiplied by a^, will produce a^. We perceive that a^ is such a quanti- ty ; for, according to Art. 58, in order to multiply a^ by a2,*we add the exponents 2 and 8, making 5 ; that is, the exponent 8 of the quotient is found by subtracting 2, the exponent of the divisor, from 5, the exponent of the dividend. ^Hence, in order to divide one power of any quantity by an- other power of the same quantity, subtract the exponent of the divisor from the exponent of the dividend. 42 ALGEBRA. 73. Propel' Si^n of the Quotient. — The proper sign to be pre- fixed to a quotient may be deduced from the principles al- ready established for multiplication. The product of the di- visor and quotient must be equal to the dividend. Hence, because -\-ax{^-h)=-\-ahA (^ah^{-\-h)=i-\-a. -ax{-^h)=-ahX ^^^^^^^^^ \-ah-r{+h)=-a. -\-ax{—h)=—ahA j—ab^{—h)=-\-a. —ax{—b)=-{-abj [-{-ab^{—b)=—a. Hence, if the dividend and divisor have like signs, the quotient will be positive ; but if they have unlike signs, the quotient will be negative, 74. Hence, for dividing one monomial by another, we have the following . RULE. > 1. Divide the coefficient of the dividend by the coefficient of tlie divisor, for a new coefficient. 2. To this result annex all the letters of the dividend, giving to each an exponent equal to the excess of its exponent in the dividend above that in the divisor. 3. If the dividend and divisor have like signs, prefix the plus sign to the quotient; but if they have unlike signs, prefix the minu^ sign. EXAMPLES ( 1. Divide 20ax^ by 4cc. Ans. bax\ 2. Divide 2ba^xy'^ by —5ay^. Ans. -^ba^xy^. 3. Divide —12ab^x^ by IWx. Ans. —6ab'^x. 4. Divide -77 a^b^'c^ by -lla¥c\ Ans. 7a^b^c\ 5. Divide 48aWc''d by -Uab'^c. 6. Divide -IdOa^iW by BOa^b'd\ 7. Divide -260a''bV by -6abx\ 8. Divide 272a^b'c'x^ by -17a''b^cx\ lL> 9. Divide -^2a^b^c hy^2\aPc. 10. Divide -mOa^b'^x by -bObx. rp^Z aj^^j DIVISION. 43 ^5 / 75. Value of me Symbol a^. — The rule given in Art. 72 con- ducts us in some cases to an expression of the form a°. Le,t it be required to divide a^ by 0/^ Accordingrtt^.the rule, tte quotient will be a^-^^ or a°. 'Now every number Is contained in itself once: lieace the value of the quotient must be unitj; that is, /.^~" ^^'''^^^^^ "" ^' ^''^>"(^j,. To demonstrate this principle generally, let a represent any quantity, and m the exponent of any power whatever. Then, •• by the rule of divisiou, , ..^,,;/,,; .«,vJA/>^ But the quotient obtained by dividing any quantify by if- * self is unity; that is, a°=l, or any quantity having a cipher for its exponent is equal to unity. 76. Signification of Negative Exponents. — The rule given in Art. 72 conducts us in some cases to negative exponents. ThuSj^et it be required to divide a^ by a^. We are directed to subtract the exponent of the divisor from the exponent of the- dividend. We thus obtain a^~^, or a~\ . a^ But a^ divided by a^ may be written -g ; and, since the value of a fraction is not altered by dividing both numerator i and denominator by the same quantity, this expression is eq„ivalenttoiV;-J^/^-:;:^P^^-^' Hence a~^ is equivalent to *^^ ^ ■- - ,-^ So, also, if a^ is to be divided by a^, this i^nay be written (^_l_ _3 '> a^~'a'^~ In the same manner, we find .+ . ." ' that is, any quantity having a negative exponent is equal to the reciprocal of that quantity with an equal positive exponent.^ 44 ALGEBRA. 77. Hence any factor may be transferred from the numera- tor to the denominator of a fraction, or from the denominator to the numerator, by changing the sign of its exponent. Thus, ^ may be written a5~\ that is, the denominator of a fraction may be entirely re- moved, and an inkgral form be given to any fractional ex- pression. This use of negative exponents must be understood simply as a convenient notation, and not as a method of actually de- stroying the denominator of a fraction. 78. To divide a Polynomial hy a Monomial. — We have seen, Art. 60, that when a single term is multiplied into a polyno- mial, the former enters into every term of the latter. Thus, {a-\-h)m=am-\-hm; therefore {am + hmi) -^m^za-^-h. Hence, to divide a polynomial by a monomisJ, we have the following RULE. Divide each term of the dividend hy the divisor^ and <:onnect the quotients hy their proper signs. EXAMPLES. 1. Divide Sx^+Qx^+Sax—Wx by Sx. Ans. a;'4-2r-f ".— 5. Z Divide Sahc+12ahx—9a^h by Sah. Ans. c4-4x— 3a- i8. Divide ^OaW + QOa^h^ -17 ah by -ah. k. Divide 16a'^hc—10acx^-{-6ac'^d^ by —6a'^c. 5. Divide 20x^-35x*-15a;3+V5a;2 by -6x\ 6. Divide 6«V/-12aV/+15aV?/3 by Sa'^x^. 7. Divide x^+^—x'"+^-^x'^-^^—x'"+* by x". 8. Divide 12aY-16aV+20ay-28aY by -4aV' DIVISION. i6 79. To divide one polynomial hy another. Let it be required to divide The object of this operation is to find a third polynomial which, multiplied by the second, will reproduce the first. It is evident that the dividend is composed of all the partial products arising from the multiplication of each term of the divisor by each term of the quotient, these products being added together and reduced. Hence, if we can discover a term of the dividend which is derived without reduction from the multiplication of a term of the divisor by a term of the quotient, then dividing this term by the corresponding term of the divisor, we shall be sure to obtain a term of the quo- tient. But, from Art. 64, it appears that the term a^, which con- tains the highest exponent of the letter a, is derived without re- duction from the multiplication of the two terms of the divisor and quotient which are affected with the highest exponent of the same letter. Dividing the term a^ by the term a of the divisor, we obtain a, which we are sure must be one term of the quotient sought. Multiplying each term of the divisor by a, and subtracting this product from the proposed dividend, the remainder may be regarded as the product of the divisor oy the remaining terms of the quotient. We shall then ob- tain another term of the quotient by dividing that term of the remainder which is affected with the highest exponent of a by the term a of the divisor, and so on. Thus we perceive that at each step we are obliged to search for that term of the dividend which is affected with the high- est exponent of one of the letters, and divide it by that term of the divisor which is affected with the highest exponent of the same letter. We may avoid the necessity of searching for this term by arranging the terms of the divisor and dividend in ike order of the powers of one of the letters. The operation will then proceed as follows: 46 ALGEBRA. a-\-h, the divisor. The arranged dividend is a^-^2ah-\-b'^ a2+ ah a -1-6, the quotient. ab-\-b^j the first remainder. ab + b^ 0, remainder. For convenience of multiplication, the divisor is written on the right of the dividend, and the quotient under the divisor. 80. ITencc, to divide one polynomial by another, we have the following RULE. 1. Arrange both polynomials in the order of the powers of thr same letter. 2. Divide the first term of the dividend by the first term of the divisor^ for the first term of the quotient. 8. Multiply the whole divisor by this term, and subtract the product from the dividend. 4. Divide the first term of the remainder by the first term, of the divisor J for the second term of the quotient. * 5. Multiply the whole divisor by this term, and subtract tJie product from the last remainder. 6. Continue the same operation until a remainder is found equal to zero^ or one whose first term is not divisible by the fijrst term of the divisor. When a remainder is found equal to zero, the division is said to be exact When a remainder is found whose first term is not divisible by the first term of the divisor, the exact division is impossible. In such a case, the last remainder must be placed over the divisor in tbc form of a fraction, and an- nexed to the quotient. EXAMPLES. 1. Divide 2a'^b-^b^-\-2ab'^^-a^ by a^ -^b"- -\- ab. Ans. a-\-h. 2. Divide x^ — a^-\-S(t'^x—Bax? by x^a. Ans. a^^2ax-^a\ 5. Divide a^-\-x^-^2a^x^ by a'^-^ax-{-x\ A ns. a* -I- a\L + ax^-\- x\ DIVISION. 47 4. Divide a^ —16a^x^ -\- 64:X^ by a'^—4:ax-\-^x\ 5. Divide a'^-\-Qa^x^—4:a^x-\-x'^—4:ax^ by a'^—2ax+x'^. Ans. a^^2ax-^x\ ^. Divide S2x'+y' by 2a?+y. i'^^^V />-^^^'^L^ :r ., V V' 7. Divide x^ + ^y + ^3/* — ^V " ^^2/^ "~ 2/^ ^X ' ^^—^•/r ^ < • ' ^ ^^ 8. Divide ic^+x^ + Sx— 4^2— 3 by x^—2x—6. 9. Divide a^-^^ by o?-^2o?h^-2aW'-^h\ ^C Divide a;H 1-2x3 ^^^ x'^-l-^x. 11. Divide a:;* + 2/*+a^^y^ by x^-^-y^-^-xy. 12. Divide 12x^-192 by 8x-6. ^ ^ns. 4x3 + 8x2+16x+82. vl3. Divide ^x^-^^ by 2x2-2^/2. S^^M. Divide a^ ^Za'^y -Za'^lP- -W by a^-Sa^^^ + Sa^^.^a^ \h. Divide x6-6x* + 9x^-4 by x^-\: \6. Divide aHa35-8a2Z^2_^19«Z>3_ 15^,4 by o?-^^al-bJ)\ 17. Divide x^ + i/^ + Sx?/— 1 by x+?/— 1. 18. Divide a252 4-2a5c2— aV— 5V by ah-^ac—hc. 19. Divide a^-i^ by a-h, 20. Divide a*-5* by a-J. 81. Hitherto we have supposed the terms of the quotient to be obtained by dividing that term of the dividend which is aflPected with the highest exponent of a certain letter. But, from Art. 64, it appears that the term of the dividend affected with the lowest exponent of any letter is derived without re- duction from the multiplication of a term of the divisor by a term of the quotient. Hence we may obtain a term of the quotient by dividing the term of the dividend affected with the lowest exponent of any letter by the term of the divisor containing the lowest exponent of the same letter; and we may even operate upon the highest and lowest exponents of a certain letter alternately in the same example. 82. a^—b^ is always divisible by a—b. From the examples of Art. 80 we perceive that a^—b^ is divisible by a—b ; and a*—b* is divisible by a—b. We shall find the same to hold 48 ALGEBRA. true, whatever may be the value of the exponents of the two letters ; that is, ilie difference of the same powers of any two quart' titles is always divisible hy (he difference of the qnxintities. Thus, let us divide a^ — h^ by a — h: 7/ a—h, divisor. -a'h a*, partial quotient. a'h-h\ The first term of the quotient is «*, and the first remainder is a*h—h^j which may be written Now if, after a division has been partially performed, the remainder is divisible by the divisor, it is obvious that the dividend is completely divisible by the divisor. But we have already found thata'' — Z;* is divisible by a — b; therefore a^—h^ is also divisible by a—b ; and, in the same manner, it may be proved that a^—h^ is divisible by a— 6, and so on. 83. To exhibit this reasoning in a more general form, let n represent any positive whole number whatever, and let us at- tempt to divide a'^—b^ by a—b. The operation will be as follows ; .b» ■ba^-'^ a—b, divisor. a"-^, quotient. The first remainder is ba'^~^—b\ Dividing a" by a, we have, by the rule of exponents, a"^-' for the quotient. Multiplying a—b by this quantity, and sub- tracting the product from the dividend, we have for the first remainder Z>a"-^— Z>", which may be written Now, if this remainder is divisible by a—b, it is obvious chat the dividend is divisible by a—b; that is, if the difference of the same powers of two quantities is divisible by the difference of the quantities^ then will the difference of the powers of the next higher degree be divisible by that difference. Therefore, since a*—b^ is divisible by a-^b, a^—b'' must be «^ ^ DIVISION. ' . 49 divisible hy a—b ; also a^-^W^ and so on for any positive value of n. The quotients obtained by dividing the difference of the same powers of two quantities by the difference of the quan- tities follow a simple law. Thus, (a2_52)-f-(a-Z))=a+5. (a5_65)-f-(a-&)=aHa36+a2Z>24.a^>3+M, etc. etc. etc. The exponents of a decrease by unity, while those of h in- crease by unity, v 84. It may also be proved that the difference of like even pow- ers of any two quantities is always divisible by the sum of ihc quantities. Thus, {o?-b'')^{a-\-b) = a-b. (a* _ h^) ^\a-\-b)=:a^- a^b + ab^ - b\ la^_l/)^(a+b) = a'-a'b-\-a'b^-a^b^ + ab^-b', etc. etc. etc. Also, the sum of like odd powers of any two quantities is always divisible by the sum of the quantities. Thus, {a^ 4- 53) ^ (a + Z^) = a2 - a5 + b\ la'^b^)^la + b) = a''-a^b-^a^b^-ab^-\-b\ {a?^b')^(a-^b) = a^-a'b + a'b^-a^P + a^b^-ab^'^b% etc. etc. etc. The exponents of a and b follow the same law as in Art. 83, but the signs of the terms are alternately plus and minus. 85. When exact division is impossible. — One polynomial can not be divided by another polynomial containing a letter which is not found in the dividend; for it is impossible that one quantity multiplied by another which contains a certain letter should give a product not containing that letter. C 50 ALGEBRA. A monomial is never divisible by a polynomial, because every polynomial multiplied by another quantity gives a prod- uct containing at least two terms not susceptible of reduction. Yet a binomial may be divided by a polynomial containing ^ny number of terms. Thus, a*— i* is divisible by a^-^a^h-^-ab^-^-h^^ and gives for quotient a—h. To resolve a Polynomial into Factors. 86. When a polynomial is capable of being resolved into factors, the factors can generally be discovered by inspection, w^ or from the law of formation. ^ ^ li all the terms of a polynomial have a common factor, that factor is a factor of the polynomial ; and the other factor may be found by dividing the polynomial by the common factor. EXAMPLES. 1. Eesolve Zo'l'^ -^2,ah'^ -\-Zah'^c into factors. Ans. 3a62(a-f 6+c). 2. Eesolve ba'^h'^-^lOaW—ba%'^—ba%'^ into factors. Ans. ba%\a'-2ah-h''-l), 3. Eesolve M¥c^ —\2ab'^c^mx—lSah'^c^y into factors. Ans. 6ab'^c\a—2mx—Sy), ^ 4. Eesolve 1 a^h"^ —1 a^lP —1 a^V^c into factors. Ys. Eesolve Sa'^hc-\-\2ah'^c—l%a'bc'^ into factors. ^6. Eesolve lOaJj^cmx—bali^cy ■\-bah'^c into factors. 87. When two terms of a trinomial are perfect squares, and .t\ie third term is twice the product of their square roots, the trinomial will be the square of the sum or difference of these roots, Arts. %^ and 67, and may be resolved into factors ac- cordingly. EXAMPLES. 1. Eesolve j^—^ah^y^ into factors. Ans. {a — h) {a — h). 2. Eesolve a'^-\-A:ah-\-4J? into factors, Ans. (a + 25)(a-}-2i). 3. Eesolve (L'^—^ah + %^ into factors. -- -) DIVISION. 51 Kesolve Oa^— 24a5 + 16&2 into factors. -^C^v-^vV /3-:5'. ■ Kesolve 25a*— GOa^^^ + 3666 j^to factors. 6. Kesolve 4:mV—4:mn-{-l into factors. -I ^>^-'>v-'>V'''''''''^^"^ 7. Eesolve 4:da'b^-16SaW-{-U4:aW into factors. V "—- ^ -* 8. Eesolve 7i^+2n2+7i into three factors. (-/v^S ( ''^ "^ "9. Eesolve 16a'^b'^—24:a'^bmx-{-9mV into factors. 10. Eesolve mV-\-2mV-\-mhi^ into three factors. 88. 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 two terms, Art. 69, and may be resolved into factors accordingly. EXAMPLES. 1. Eesolve 4:a^—%^ into factors. Ans.{2a+3b){2a—Bb). 2. Eesolve da^b^—Wa^c^ into factors. 8. Eesolve a^x—9ax^ into three factors. 4. Eesolve a*— 6* into three factors. 5. Kesolve a^—W into its factors. 6. Eesolve a^—b^ into four factors. 7. Eesolve 1— ^^V into two factors. 8. Eesolve 4 — ^V into two factors. 89. If the two terms of a binomial are both powers of the same degree, it may generally be resolved into factors accord- ing to the principles of Arts. 82-84. EXAMPLES. 1. Eesolve a^—W into its factors. A7is.{o?-\-ab-\-b'^){a~-b). 2. Eesolve a^-\-b^ into its factors. 8. Eesolve a^—W into four factors. 4. Eesolve a^—^W into its factors. 6. Eesolve 8a^ — 1 into its factors. 6. Eesolve 8a^— 86^ into three factors. . 7. Eesolve 1 + 276^ into its factors. 8. Eesolve 8a^ + 276^ into its factors. 9. Eesolve a^^ — h'^^ into five factors. 52 ALGEBRA. V CHAPTER YL GREATEST COMMON DIVISOR. — LEAST COMMON MULTIPLE. '90. A common divisor of two quantities is a quantity which will divide them both without a remainder. Thus 2ab is a common divisor of Wh^x and lOd^h^y. y 91, A prime factor is one that can not be resolved into any other factors. It is, therefore, divisible only by itself and uni- ty. Thus the quantity ^a^—^ah is the product of the three prime factors 2, a, and a—h. A 92. The greatest common divisor of two quantities is the greatest quantity which will divide each of them without a remainder. It is the continued product of all the prime fac- tors which are common to both. The term greatest here refers to the degree of a quantity, or of its leading term, and not to its arithmetical value. 93. When both quantities can be resolved into prime fac- tors by methods already explained, the greatest common di- visor may be found by the following RULE. Resolve both quantities into their prime factors. The continued product of all those factors which are common to hothy will he the greatest common divisor required. EXAMPLES. 1. Find the greatest common divisor of 4^'^hx and Qah^j^, Resolving into factors, we have ^a^hx = 2a xja xhxx. ^ah'^x^ = 2a xSbxhxxxxxx. The common factors arc 2a, 6, and x. Ilence the greatest common divisor is 2abx. GREATEST COMMON DIVISOR. 53 2. Find the greatest common divisor of 4am^4-45m^ and Ban + 3bn, Kesolving into factors, we have 4:am'^ + ^bm? = 2m x 2 w (a 4- 5). San-\-Sbn=Bn{a-{-b). Hence a-\-b is the greatest common divisor. S. Find the greatest common divisor of x^—y^ and x^—y"^. x^—f={x—y){x^-\-xy-\-y'^y x^-y^ = {x-y){x+y). Hence x—y is the greatest common divisor. ^ 4. Find the gre^atest common divisor of Sda^bmx'^ and ^2amV, ra^ru/. "^J^^Jp ' 5. Find the greatest common divisor of %a^x—^abx-{-W^x and ^a}y—W-y. ;/ - - ^ 6. Find the greatest common divisor of dmx'^—Qmx-^-m and 'dnx^—n, ^, ^ 7. Find the greatest coi^mon divisor of 12a2— 36aZ>-|-27^'* and 8a2-18i2. ^(X - i\r -^ 94. When the given quantities can not be resolved into prime factors by inspection, the greatest common divisor may be found by applying the following principle : The greatest common divisor of two quantities is the same with the greatest common divisor of the least quantity^ and their remain- der after division. To prove this principle, let the greatest of the two quanti- ties be represented by J., and the least by B. Divide A by B; let the entire part of the quotient be represented by Q^ and the remainder by R. Then, since the dividend must be equal to the product of the divisor by the quotient, plus the remainder, we shall have A^QB-'tR. Now every number which will divide B will divide QB ; and every number which will divide R and QB will divide R-{- QB, or A. That is, every number which is a common divisor of B and i? is a common divisor of A and B. Again ; every number which will divide A and B will di- 54 ALGEBBA. '^ vide A and QB; it will also divide A-^QB, or R. That is, every number which is a common divisor of A and B is also a common divisor of B and R. Hence the greatest common divisor of A and B must be the same as the greatest common divisor of B and R. 95. To find^ tlien^ the greatest common divisor of two quanti- ties, we divide the greater by the less; and the remainder, which is necessarily less than either of the given quantities, is, by the last article, divisible by the greatest common divisor. Dividing the preceding divisor by the last remainder, a still smaller remainder will be found, which is divisible by the great- est common divisor; and by continuing this process with each remainder and the preceding divisor, quantities smaller and smaller are found, which are all divisible by the greatest com- mon divisor, until at length the greatest common divisor must be obtained. Ilence we have the following ^ RULE. Divide the greater quantity hy the less, and the preceding divisor hy the last remainder, till nothing remains ; the last divisor will he the greatest common divisor. When the remainders decrease to unity, the given quanti- ties have no common divisor greater than unity, and are said to be incommensurahle, or prime to each other. EXAMPLES. 1. What is the greatest common divisor of 372 and 246? 872 246 246 246 126 126 120 126, the first remainder. 120, the second remainder. 120 120 6, the third remainder. 20 Here we have continued the operation of division until wo GREATEST COMMON DIVISOR. 55 obtain for a remainder; the last divisor (6) is the greatest common divisor. Thus, 246 and 872, being each divided by 6, give the quotients 41 and 62, and these numbers are prime with respect to each other; that is, have no common divisor greater than unity. 2: What is the greatest common divisor of 336 and 720 ? niM){lx3 Ans. 48. 8. What is the greatest common divisor of 918 and 522 ? Ans. 18. 96. In applying this rule to polynomials some modification may become necessary. It may happen that the first term of the arranged dividend is not divisible by the first term of the divisor.' This may arise from the presence of a factor in the divisor which is not found in the dividend, and may therefore be suppressed. For, since the greatest common divisor of two quantities is only the product of their common factors, it can not be affected by a factor of the one quantity which is not found in the other. We may therefore suppress in the first polynomial all the factors common to each of its terms. We do the same with the second polynomial ; and if any factor suppressed is com- mon to the two polynomials, we reserve it as one factor of the common divisor sought. But if, after this reduction, the first term of the dividend, when arranged according to the powers of some letter, is not divisible by the first term of the arranged divisor, we may mul- tiply the dividend by any monomial factor which ivill render its first term divisible by the first term of the divisor. This multiplication will not affect the greatest common di- visor, because we introduce into the dividend a factor which belongs only to a part of the terms of the divisor ; for, by sup- position, every factor common to all the terms has been sup- 97. The preceding principles are embodied in the following general 56 ALGEBRA. RULE. 1. Arrange the two polynomials according to the powers of some letter; suppress all the monomial factors of each ; and if any fac- tor suppressed is common to the two polynomials^ reserve it as one factor of the common divisor sought. 2. Multiply the first polynomial hy such a monomial factor as will render its first term divisible hy the first term of the second polynomial; then divide this result hy the second polynomial^ and continue the division till the first term of the remainder is of a lower degree than the first term of the divisor. 3. Take the second polynomial as a dividend^ and the final re- mainder in the first operation as a divisor ^ and proceed as before^ and so on till a remainder is found that will divide the preceding divisor. This remainder^ multiplied hy the common factors^ if any^ reserved at the beginning^ will give the greatest common divisor. EXAMPLEa 1. Find the greatest common divisor of x^+4cc^H-5x-f 2 and a:3+4x2+5a:-f2 X^-\-bx-\-4: x-l - cc2+ cc+2 — x^—6x—4: ^[6x^6_ Suppressing the factor 6 inM;nis remainder, we have jc+l for the next divisor. x+l X-\-4: 4x+4 4x+4 Here the division is exact; hence, by the rule, x-\-l is the greatest common divisor sought 2. Find the greatest common divisor of 6x^—7ax^—20a^x and 6x^-\'2ax—8a\ Suppressing the factor 2 in the second polynomial, we pro- ceed thus: GREATEST COMMON DIVISOR. 57 6x^-7ax''-20a^x 6x^-{-2ax^- 8a^x 8a;2+acc— 4a* 2x-Sa -dax^-na'x - 9a^x-12a^ Suppressing the factor — 8a^, 8^2+ ax—4ca^ I 8a;+4a 8x2-|-4ax I cc— a —Sax— 4:0^ —Sax— 4:0^ Hence Sx-\-4a is the greatest common divisor. 8. Find the greatest common divisor of 4:a^—2a'^—Sa-\-l and 8a2-2a-l. We first multiply the greater polynomial by 8, to render its first term divisible by the first term of the other polyno- mial. 12a3-6a2-9cx+3 12a^-Sa^-4ta 8a2_2a-l 4a, +2 2a2-5a+3 6a2-15a+ 9 6a^- 4a- 2 -lla+11 Here we multiply the first remainder by 8, to render the first term divisible by the first term of the divisor. As the two partial quotients 4a and 2 have no connection, they are separated by a comma. Eejecting the factor —11 from the second remainder, we proceed as follows : 8a2_2a-l Sa^-Sa Sa-\-l a-1 a-1 Hence a— 1 is the greatest common divisor. V 4. Find the greatest common divisor of a^—Sah-{-2h'^ and a^-ah-2h\ Ans, a-2b. 02 58 ALGEBRA. 5. Find the greatest common divisor of a^—a^b-\-Sab^—Sb^ and a^—5ab-\-4:b\ Ans. a—b. \\ 6. Find the greatest common divisor of Sx-^— 13a;2-f-232;— 21 and 6x3+a72-44a:+21. Ans. 3a;-7. PT. Find the greatest common divisor of x*-7x3 4-8a;2+28x-48 and x^ -^x" -\-l^x-\4u Ans. x—2. 98. To find the greatest common divisor of three quantities. — Find the greatest common divisor of the first and second, and then the greatest common divisor of this result and the third quantity. The last will be the greatest common divisor re- quired. EXAMPLES. 1. Find the greatest common divisor of Sahn^, Qb'^m}^ and 12m^x. Ans. Sm^. 2. Find the greatest common divisor of 4x^—21x2+15x4-20, K— 6x-h8, and x^— x— 12. Aiis. x— 4. 3. Find the greatest common divisor of 6x* 4- x^ — x, ^4x^— 6x2— 4x4-3, and 2x^-^x^-\-x—l, Ans. 2x— 1. ^4. Find the greatest common divisorof4x*4-9x^4-2x2— 2x— 4, 3x3 4-5x2— x4- 2, and xHx2-x4-2. Ans. x4-2. LEAST COMMON MULTIPLE. 99. One quantity is a multiple of another when it can be di- vided by it without a remainder^ Thus bah is a multiple of 5, also of a and of h \ When one quantity is a multiple of an- other, the former must be equal to the product of the latter by some entire factorJ Thus, if a is a multiple of i, then a=vib, where m is an entire number. 100. A common multiple of two or more quantities is one which can be divided by each separately without a remainder. Thus 20a^h'^ is a common multiple of 4a6 and ba^b^ 101. The least common multiple of two or more quantities ia the least quantity that can be divided by each without a re* LEAST COMMON MULTIPLE. 59 mainderT] Thus 12a2 is the least common multiple of Sa^ and 4a. 102Jjt is obvious that the least common multiple of two or more quantities must contain all the factors of each of the quan- tities, and no other factors. Hence, when the given quantities can be resolved into prime factors, the least common multiple may be found by the following RULE. Resolve each of the quantities into its prime factors ; taJce each factor the greatest number of times it enters any of the quantities; multiply together the factors thus obtained^ and the product will he the least common multiple required. EXAMPLES. 1. Find the least common multiple of ^x^y and Vlxy"^. Eesolving into factors, we have 9x'^yz=zSxSxxyj and 12xy'^=Sx2x2xyy. The factor 3 enters twice in the first quantity, also the fac- tor 2 enters twice in the second; x twice in the first, and y twice in the second. Hence the least common multiple is 2 X 2 X 3 X Sxxyyj or SQx'^y^. 2. Find the least common multiple of 4:a%^, 6a%j and lOa^oc^, We have 4.a'^b^ = 2 x 2aahb, 6a'^b=2xSaab, lOa^x^ =2x baaaxx. Hence the least common multiple is 2 X 2 X 3 X 6aaabbxx, or QOa^V. 8. Find the least common multiple of a''^x—2ahx-\'h'^x and ii^y^b^y. Here we have a^x-^2abx-\-b'^x={a—b) {a—b)x, a^y-b''y^{a-{-b)(a^b)y. Hence the least common multiple is (a — b) {a — b)(a-\- b) xy, or a^xy — ab^xy — a%xy + h^xy. 60 ALGEBRA. '4 Find the least common multiple of 6a%'^j 10aZ>*, and 2abx> Ans. lOa^b^x, ' 5. Find the least common multiple of ^tol!^^ 4aar^, 5fe, and Wx\ Alls. QOa^b^x^. 6. Find the least common multiple of a;'— 3cc+2 and x^—1. Ans, {x+l){x-l){x-2), or x^-2x''^x-\-2. ^ 7. Find the least common multiple of a^x-\-b^x and Sa^— 56^, Ans. 6x{a-\-b){a—b){a^—ab-^h^l or 6a'^x—6a^bx-\-5ab^x—5b*x. 103. When the quantities can not be resolved into factors by any of the preceding methods, the least common multiple may be found by applying the following principles : ! If two polynomials have no common divisor, their product must be their least common multiple ;^ feut if they have a com- mon divisor, their product must contain the second power of this common divisor^ Their least common multiple will therefore be obtained by dividing their product by their greatest com- mon divisor. Hence, to find the least common multiple of two quantities, we have the following RULE. i Divide the product of the two polynomials by their greatest com- mon divisor ; or divide one of the polynomials by the greatest com' mon divisor J and multiply the other by tlie quotvent. EXAMPLES. 1. Find the least common multiple of Gx^ — x — l and 2x2+8ic-2. The greatest common divisor of the given quantities is 2x— 1. j^ence the least common multiple is ^ (6^'-^-l)y +3^-2) ,, (2..+3._2)(3.+ l). v' 2. Find the least common multiple of cr^— 1 and x'^-f ^— 2. //\ ^ ' ^?^^- (.T'-l)0r4-2> Y ^ 3. Find the least common multiple of a;^— 9x2-f-23x-15 aiid a5»-8aJ4-7. Ans. (a;3-9a;»+23a;-15)(cc-7). LEAST COMMON MULTIPLE. 61 104. When there are more than two polynomials^ find the least common multiple of any two of them ; then find the least common multipla of this result, and a third polynomial ; and so on to the last. ' 4. Find the least common multiple of a2+2a — 3, a^—l^ and a-1. -^ ',._,-... Ans. (a2-l)(a+3). 6. Find the least common multiple of 4:0^ -\-l^ 4a2— 1, and 2a-l. Ans. 16a*-l. ^ 6. Find the least common multiple of a^—a, a^+1, and a^—1. Ans. a{a^—l). / 7. Find the least common multiple of (a:+2a)^, (j^— 2a)^, and ic^— 4a^. - Ans, (x^— 4a^)^. hm^h Kh 'm komhn/Jih^ , JlfW^ 62 ALGEBRA. CHAPTER YII. FRACTIONS. 105. A fraction is a quotient expressed [^ described in Art Tybj writing the divisor under the dividend with a line be- tween them. Thus t is a fraction, and is read a divided by h. 106. Every fraction is composed of two parts: the divisor, which is called the denominator^ and the dividend, which is called the numerator, \ 107. An entire quantity is an algebraic expression which has no fractional part, as a^—^ah. An entire quantity may be regarded as a fraction whose de- nominator is unity. Thus, a'^=^. 108. A mixed quantity is an expression which has both en- tire and fractional parts. Thus a^ + - is a mixed quantity. 109. General Principles of Fractions. — The following princi- ples form the basis of most of the operations upon fractions : \st. In order to multiply a fraction hy any number^ we must multiply its numerator or divide its denominaim hy that number. Thus the value of the fraction ~ is h. If we multiply the numerator by a, we obtain ^ or ah; and if we divide the de- nominator of the same fraction by a, we obtain also ah; that is, the original value of the fraction, 6, has been multiplied by a. 2d. In order to divide a fraction hy any number^ ive must divide its numerator or multiply its denominator hy that number. Thus the value of the fraction ^ is ab. If we divide the FRACTIONS. 63 numerator by a, we obtain ^, or h ; and if we multiply the de- nominator of the same fraction by a, we obtain ^, or h ; that is, the original value of the fraction ah has been divided by a. TSd The value of a fraction is not changed if we multiply or di- vide both numerator and denominator hy the same number, ^, ah abm abmx , Thus, — = = =b, ' a am amx 110. The proper Sign of a Frojction. — ^5]ach term in the numer- ator and denominator of a fraction has its own particular sign, and a sign is also written before the dividing line of a fraction. The relation of these signs to each other is determined by tlie principles already established for division. The sign prefixed to the numerator of a fraction affects merely the dividend; the sign ^ prefixed to the denominator affects merely the divisor ; but the sign prefixed to the dividing line of a fraction affects the quotient. ; The latter sign may be called the apparent sign of the fracti'on, while the real sign of the fraction is the sign of its numerical value when reduced. The real sign of a fraction depends not merely upon its a]> parent sign, but also upon the signs of the numerator and de- noniinator. From Art. 73, it follows that ah —ah —= = -}-h, a —a , —ah ah . and ■ = — = —h. a —a Also, since a minus sign before the dividing line of a frac- tion shows that the quotient is to be subtracted, which is done by changing its sign, it follows that ah —ah ^ a ~ —a ~ ' J —ah ah , and — =+5. a —a Hence we see that of the three signs belonging to the numer- 64 ALGEBRA. ator, denominator, and dividing line of a fraction, any two ma^ be changed from -^ to —, or from — to -[-, without affecting the real sign of the fraction. 111. When the numerator or denominator of a fraction is a polynomial, it must be observed that by the sign of the numer- ator is to be understood the sign of the entire numei-ator^ as dis tinguished from the sign of any one of its terms taken singly, ^, a + ^ + c . • 1 . X , —a—h—c Thus, — IS equivalent to -\ . When no sign is prefixed either to the terms of a fraction or to its dividing line, plus is always to be understood. Reduction of Fractions. 112. To reduce a Fraction to its Lowest Terms. — A fraction Is in its lowest terms. when the numerator and denominator con- tain no common factor; and since the value of a fraction is not changed if we divide both numerator and denominator by the same number {Art. 109), we have the following RULE. Divide both numerator and denominator by their greatest com mon divisor. Or, Cancel all those factors which are common to both numeratoi and denominator. EXAMPLES. a^bc 1. Reduce ^ ^.^ to its lowest terms. „^ - a^bc cxa^b Wc have 6a^b^ bb x a^b' Canceling the common factors a^i, we have a'^bc _ c 2. Reduce ^ ^ to its lowest terms. a^c+a^x ■^ , cx-Ji-x^ x{c-\-x) X We have -5— — r= 2) . \ =~i' a^c + a^x a^(c + x) a* FRACTIONS. 65 Eeduce 777 p^— to its lowest terras. Ans. --. lOac—obc be a 4. Reduce -^ . to its lowest terms. Ans. — >. 5. Keduce -^-^ — ^r-r- — ^ to its lowest terms. Ans. |. ^ 6. Reduce » ? . /> rij-? to its lowest terms. An5. — ^• 7. Reduce -5 — ^ , . ,^ to its lowest terms. 8. Reduce .^ 1 — ^ to its lowest terms. . ^r^> ( ^^^j^/fi cp2— 16 9. Reduce -^; ^^7: to its lowest terms. .n ID ^ 80^3-160:2+23:^-6 - T*f-^^^^-^ J ^^- ^'^^"" 2x3_nc.2+17x-6 *^ f^pwest terms. ^^:V'. ' V 11. Reduce 2x^~x^-x+2 *^ ^*^ ^^^^^* *®^"^^- '^^hf^ to T> A 2cc3 + 9x24-7x-3 , ., , ^^ ^w • 12. Reduce » -^ . ^ 9 — r^ — r— r to its lowest terms. 3x^+5x2- 15ic+ 4 - 113. To reduce a Fraction to an Entire or Mixed Quantity. — • When any term of the numerator is divisible by some term in the denominator, the division indicated by a fraction may be at least partially performed. Hence we have the following RULE. Divide the numerator hy the denominator^ continuing the opera- tion as far as possible ; then write the remainder^ if any^ over the denominator^ and annex the fraction thus formed to the entire part EXAMPLES. 1. Reduce to an entire quantity. - (90 ^ 'Y^ q]j 2a^ . /(^^ 2. Reduce — ^ to a mixed quantity. (X ^ 6Q ALGEBRA. 3. Reduce to a mixed quantity. Ans. a+x-\ . (I — X Cl — X X^ — ti J" 1^ 4. Reduce ^ to an entire quantity. rjOtnCA^'t"^ if " 5. Reduce ^ to a mixed quantity. x!- ^' / '^ ^oc 6. Reduce ^r to a mixed quantity. & ^ ^ 1 7. Reduce ^ — ^ — —^ — to an entire quantity. ■ -■ '^ x^—Sx-\-2'^.^^zjijyC-/j ^ "^ 0. Reduce 9,^1.10 to an entire quantity. 114. To reduce a Mixed Quantity to the Form of a Fraction. — This problem is the converse of the last, and we may proceed :^'m. by the following RULE. Multiply the entire part hy the deyiominator of the fraction ; to the product add the numerator with its proper sign^ and write the result over the denominator. EXAMPLES. q2 ^2 ^2 1. Reduce x-\ to the form of a fraction. Ans. — . X X ax I Cfy 2. Reduce x-\-—;z to the form of a fraction. Ans. 2.T-7 /^^^TAY 8. Reduce 5h — ^— — to the form of a fraction. 6x . d^ 4. Reduce 1+ "~ "~ to the form of a fraction. ^J^ - a cu cc— 3 6. Reduce 1 + 2x4--^ — to the form of a fraction.? ^x -^iJJjiJL bx ^.^ 6. Reduce 7+ ^ ?, to the form of a fraction. ^^_3JZi-. < V A /r>.^ c 4a2_50 7. Eeduce 2a— 7 — ^ ^ to tlie form of a fraction. 8. Reduce (a— 1)^— ^^ to the form of a fraction. 115. 71? reduce Fractions having Different Denominators to Equivalent Fractions having a Common Denominator: n c rfh Suppose it is required to reduce the fractions ^, -7, and — to a common denominator. Since, by Art 109, both terms of a fraction may be multiplied by the same quantity without changing its value, we may multiply both terms of each frac- tion by the product of the denominators of the other fractions, and we shall have a_adn c _hcn , m_hdm h hdn d hdri n hdn The resulting fractions have the same value as the proposed fractions, and they have the common denominator hdn. Hence we have the following RULE. Multiply each numerator into all the denominators^ except its own^for a new numerator^ and all the denominators together for the common denominator. EXAMPLES. 1. Reduce t and to equivalent fractions having a common denommator. Ans. — , — ^ — . be be 2. Reduce ^, ^,and - to equivalent fractions having a common denominator. 3 2cc 4:X 8. Reduce -r, -^, and a+-^ to equivalent fractions having a common denominator. ■ / 08 ALGEBRA. 4. Reduce ^> -=-i and to equivalent fractions having a common denominator. 5. Reduce -' — ^— > and zr- — to equivalent fractions hay« ing a common denominator. 116. Fractions may always be reduced to a common denom inator by the preceding rule ; but if the denominators have any common factors, it will not be the least common denominator. The least common denominator of two or more fractions must be the least common multiple of their denominators. Suppose it is required to reduce the fractions ^-g and — t6 ox ^x equivalent fractions having the least common denominator. The least common multiple of the denominators is 12a::^ Mul- I2x'^ tiply both terms of the first fraction by -Q-y, or 4, and both 12cc2 terms of the second fraction by -r—, or Sx, and we shall have 8a , 16hx 12^2 '^"d 12^' which are equivalent to the given fractions, and have the least common denominator. Hence we deduce the following RULE. Find the least common multiple of all the denominators^ and use Hiis as the common denominator. Divide this common denominator hy each of the given denomina- tors separately^ and m^ultiply each numerator hy Hie corresponding quotient. The products will he the new numerators. 2a bx 6. Reduce ^ry- and 7775-5 to equivalent fractions having the Sbc Wc^ A 4ac X least common denominator. Ans. -7^ and -^r^' 7. Reduce r and -^ — j^ to equivalent fractions having the least common denonunator. Ans. ^ — yt and -^ — j^. ~ 8. Eeduc FRACTIONS. 69 \ 8. Eeduce -3, — , and - to equivalent fractions having the least common denominator. ^o T> J ^« -^^ 11^ J '^(^ + ^) ^ • 1 * i> 9. Keduce 77-j ^77^—, ^777—? and — ^^^j to equivalent frac- 87?^ Som 28m 4m ^ tions having the least common denominatorSH^jXy:— //jj^ 2 3 2cc— 3 'i;<:'^'>>i^^'!^>n.>'^V/4^.^ ^ 10. Eeduce -j ^r -, and -r^ — 7 to equivalent fraction^ cc 2x— 1 4x^—1 ^ ■: 2aL<*?^ having the least common denominator. ^^T'W^y^ fliTi?'' Idditid df Fractions. .7. The denominator of a fraction shows into how manj parts a unit is to be divided, and the numerator shows how many of those parts are to be taken. Fractions can only be added when they are like parts of unity ; that is, when they have a common denominator. In that case, the numerator of each fraction will indicate how many times the common frac- tional unit is repeated in that fraction, and the sum of the nu- merators will indicate how many times this result is repeated in the sum of the fractions. Hence we have the following RULE. Reduce the fractions to a common denominator ; then add the numerators together^ and write their sum over the common denomi- nator. If there are mixed quantities, we may add the entire and fractional parts separately. EXAMPLES. 1. What is the sum of and o ? Eeducing to a common denominator, the fractions become Zx , 2x -g- and -^. Adding the numerators, we obtain -^. D It is plain that three sixths of x and two sixths of x make five sixths of x. 70 ALGEBRA. 2. What is the sum of -r, -, and — ? .. adn-\-hcn-\-hdm 3. What is the sum of j and r? a-\-b a—b 4. What is the sum of 5£c, 7r^» and — ^ — ? 5. What is the sum of 2a, Sa-f— » and a4-pr? 2a , a+2a:, lajc-*^^^ 6-' ^^^ ^+-9r Ans. oa-f -r^. 4o 6. What is the sum of a+cc, , and -^-? a—x a ^ Atis. a+a:+24--^ . a4-b —b a^—ax 7. What is the sum of —^ and —^r— ? Ans. a. o -vm- ^ • xT_ p ^ a— 2m J a+2m« 8. What IS the sum oi ^, — j — , and — j — ? n TtTT. X • xT- r* ^« — ^ -, na + Z)„ 9. What IS the sum of and — ? m-fn m-\-n 10. What is the sum of -4^-^-, ^""^ , and -^t:?-? /- "11. What is the sum of |4^ a^d ^4^? /■•owu*-*;, .13a-295 7Z>-21a , 9^>-lla, '^ 12. What IS the sum of -^7 jr-, —-^-7 ^, and — -7 r^? 6(a-Z)) b{a—by 6{a—b) A71S. 9. / 13. What IS the sum of ^ , — — -, :p- — z— , ^ — ^I— -, 1— cc 1 + ic l-l-x^ 1— a;J* and -1? Subtraction of Fractions. 118. Fractions can only be subtracted when they are like parts of unity ; that is, when they have a common denomina- tor. In that case, the difference of the numerators will indi- cate how many times the common fractional unit is repeated in the difference of the fractions. Iloiice we have the following FRACTIONS. 71 RULE. Reduce the fractions to a common denominator ; then suhtract i]ie numerator of the subtrahend from the numei^ator of the minu- endj and write the result over the common denominator, EXAMPLES. 1. From -77- subtract -— . o Eeducing to a common denominator, the fractions become lOx , 9« 1^ ^^^ 15- __r , lOx ^x X Hence we have -^p — ^H — rE? and it is plain that ten fifteenths of cc, diminished by nine fif- teenths of X, equals one fifteenth of x, 2. From -=- subtract ^. 7 _ ^ 9a— 4aj , ^ ^ 5a— 3x 3. From — = — subtract ^ — . It must be remembered that a minus sign before the divid- ing line of a fraction affects the quotient {Art. Ill); and since a quantity is subtracted by changing its sign, the result of the subtraction in this case is 9a— 4x 6a — Sx which fractions may be reduced to a common denominator, ai>d the like terms united as in addition. . T^ OX . ^ ^ ax J, 2acx 4. 1 rom X — subtract ^ . Ans. yr — -„. b—c b~\-c b^—d^ _ _, _ , 2-^7x . ^ bx-^ . S55X-6 5. From 2x-\ 5 — subtract x ^rr— . Ans. o ZL Ut^UU^zJ^ 6. From Sec 4-^ subtract x . : ^ JLVc 7. From ^^-r- subtract -tt-. - h / 72 ALGEBRA. ^ ^ 13a-56 . , 7a^2b . 2ba-llb 8. From -. subtract — -^ — . Ans. ^Ti . 4 V2i Qa-lb , , , 5a , eUb-lba'-eSb^ ^' ^^^"^ 3^326 ^^^'^""' W ^^- —27^^:1186^-^ ^ 10. From ^^^ subtract unity.'-^ll^^^' ^ 11. From ^ subtract |^:=|^. 11?/ Ix—by ax X 12. From — ^ subtract -f^.--^ . Multiplication of Frojctions. 119. Let it be required to multiply ^ by ^. First let us multiply t by c. According to the first princi- pie of Art. 109, tbe product must be -r. But the proposed multiplier was J ; that is, we have used a multiplier d times too great. "We must therefore divide the result by d; and, according to the second principle of Art. 109, we obtain ^^ ^ ^^ Hence we have the following RULE. Multiply the numerators together for a new numerator^ and the denominators for a new denominator. Entire and mixed quantities should first be reduced to frac« tional forms. Also, if there are any factors common to the nu- merator and denominator of the product, they should be caa celed. EXAMPLES. X ziX iT^ 1. Multiply ^ by -5-. Ans. ^ 2T 2. Multiply I by g|. FRACTIONS. 1^ y _^ 73 Multiply })-{ by -. 4. Multiply -^ by -^^. An^. j^-^,. 0. Multiply -^ — j^ by — —r. Am. -, — -yr^. 6. Multiply together ^, — , and -^r^. 7. Multiply together — , — , and -^. Qi C AO X + 1 T X — 1 Ynf' I 1 Q^ I 8. Multiply together x^ , and 7. Ans x^^x a + b' ' a^-\-ab' 9. Multiply by ^w?n, 10. Multiply . ^,„ ,^ , ,,„ „^ - -,„ .. by a^^V. ^ -^ (a2Z>V) {d^bh^) iaWc^) -^ 11. Multiply together ^^^, ^^^, -^^, and -^. . 2m^n Ans. --3-. 12. Multiply together -=^ -S ^ — ^, and -^^^ { ^ *^ ° l{m—n 39(a— 0) 66{x»—y t 13. Multiply -i — = -^rr- by ^-77-5 — ./JLX^^ - ^ , ,f . . , , , 3ax a^— ^2 bc-\-bx ^ c—x 14. Multiply together ^, ^,-^,, ^,^^, and ^-^. , 3x Ans. -J—. 4y 15. Multiply top^ether -1 — ^, ^, and 1 + - — ^ -^ ° 1 + 2/ 03 + ^2 1—; ■X Ans. 1-^ \ 16 M It' 1 ^(q^ — ^) ^ a(« + a^), ^ I) 74 . Aa^'^ ALGEBRA. V ' 18. Multiply cc^-jr+l by -2+- + I. Ans. (r^+l + A ^ 19. Mulfply _^-_^_^-^^ by -^. ^„.. 2. ^20. Multiply ^-,--+1 by J+?+l. ^„. |+ J,+l. 120. Multiplication of Quantities affected with Negative Expo- nents. — Suppose it is required to multiply -3 by — . According to the preceding article, the result must be —^. 1 1 ^ But, according to Art. 76, -3 may be written a-^ ; — may be -i fl a written a-^; and -g may be written a-*. Uence we see that a-^xa-'^=.a-^\ that is, the rule of Art. 58 is general, and applies to negative as well as positive exponents. 1 a*' EXAMPLES. 1. Multiply —x-'^ by x~\ Ans. — ic-*, c 2. Multiply a-2 by -a\ ^^ 8. Multiply fi-3 by «3. ^•. ; 4. Multiply a-*" by a^ ex 5. Multiply a-"* by a-\ f' 6. Multiply (a-Z>)s by {a-b)-^ ' ^^'i Division of Fractions. 121. If the two fractions have the same denominator, then the quotient of the fractions will be the same as the quotient of their numerators. Thus it is plain that | is contained in -J as often as 3 is contained in 9. If the two fractions have Tioi the same denominator, we may perform the division after hav- ing first reduced them to a common denominator. Let it be required to divide ? by ^ FRACTIONS. 75 Keducing to a common denominator, we have ^ to be di- vided by 0. It is now plain that the quotient must be repre- sented by the division of ad by fee, which gives j- ; a result which might have been obtained by inverting the terms of the divisor and multiplying by the resulting frac- tion ; that is, ^ j b ' d~ b c~ be' Hence we have the following RULE. Invert the terms of the divisor^ and multiply the dividend by the resulting fraction. Entire and mixed quantities should first be reduced to frac- tional forms. EXAMPLES. 1. Divide ^ by -q-. Ans. IJ. 2. Divide -j- ^J -^' 3. Divide -^-- — ^ by a'^-\-x-^ '^ x-\-a 4. Divide —^ by -^. 6. Divide —3 3- by . Ans. c^-\~cx-\-x^ 7. Divide — ^ H — ~ by -^ ^. ^4725. Unity. a+o a — fe '^ a—b a-\-b ^ a 8. Divide 7a''-Sx-\-- by fe^-^. . 2 la%— 9/10:4-3771 J.715. ^j^ . 6o^n—an ^\ ALGEBRA, 9. Divide loic^^-— - by a; . Ans. = = ^, be '^ c OCX— 5a4-56 10. Divide x^H 7 by 7 — w. ^W5. -7 n— . a— 6 "^ a—b ao—am-\-om 82(m+6) ^ 128n(m+6y 12. Divide -^ + - by —^ h-. ^?i5. -, y^ X '' y^ y X y ^ 13. Divide ^±^ + ? by ^±^- -^. ^r^. Unity. x+2/ y y ^+y 1 1 x^-i-i ^14. Divide x^^-— 4-2 by x-f-. ^rw. . x^ -^ X X a-f 64-c 15. Divide a}-h^-c'^-2bc by 122. Division of Quantities affected with Negative Exponents. — Suppose it is required to divide -^ by -3. According to the preceding article, we have X But, according to Art 76, -3 may be written a-* ; -3 may be written a~^ ; and — may be written a-^. Ilence we see that that is, the rule of Art. 72 is general, and applies to negative as well as positive exponents. EXAMPLES. 1. Divide a~* by — a"^. Ans. — a"^, or — i 2. Divide — a^ by a~\ ' ^ a Divide 1 by a-\ T-*^ 4. Divide 6rt** bv — 2a-^ 'iO^ FRACTIONS. 77 5. Divide h'^-'' by h^, ^ ^^ 6. Divide 12x-22/-* by -^xy\ -l^ ^ 7. Divide {x—y)-"^ by (cc— ?/)-«. (^ "^ . 123. jTAe Reciprocal of a Fraction. — According to the defini- tion in Art 34, the reciprocal of a quantity is the quotient aris- ing from dividing a unit by that quantity. Hence the recipro- cal of ^ is . a ^ h h a a' that is, the reciprocal of a fraction is the fraction iiiverted. Thus the reciprocal of j—— is ; and the reciprocal of -J o~\-x a bTi is b+c. It is obvious that to divide by any quantity is the same as to multiply by its reciprocal^ and to multiply by any quantity is the same as to divide by its reciprocal. 124. How to simplify Fractional Expressions. — The numerator or denominator of a fraction may be itself a fraction or a mixed quantity, as -^. In such cases we may regard the quantity above the line as a dividend, and the quantity below it as a divisor, and proceed according to Art. 121. Thus, 2i-|=4x 1=^=31. The most complex fractions may be simplified by the appli- cation of similar principles. EXAMPLES. 1. Simplify the fraction -, 1+1 rr.! . . . , h-\-a a-\-h This expression is equivalent to —7 — : > 01- to — :; — X T) which is equal to r, Ans, b a-\-b ^ o 78 ALGEBRA. 2. Simplify a—m a-\-m Arts. 24 8. Find the value of the fraction ||. 4. Find the value of the fraction 5. Simplify 22a5c %^mnx llab ' 3mx 6. Simplify 7. Simplify 8. Simplify a+h a-h c+d ^ c-d a+b a-h' c-d a-\-x ' c-^d a-x • a—x ' a-\-x a+x a—x a—x a-\-x ,2 n wi'-n^ 1 n a m'-\-n^' Ans. Ans. a-\-m a—m Ans. \. LP ac—hd ac-\-bd' a^-\-x^ 2ax ' h+ d-\- m Ans. Ans. m. adn-\-am bdn-\-hm,-\-C7C EQUATIONS OF THE FIRST DEGREE. 7^ r, CHAPTEK VIII. EQUATIONS OF THE FIRST DEGREE. 125. An equation is an expression of equality between two algebraic quantities. Thus 2>x=2ab is an equation denoting that three times the quantity x is equal to twice the product of the quantities a and 6. 126. l^hQ first member of the equation is the quantity on the left side of the sign of equality, and the second memher is the quantity on the right of the sign of equality. Thus, in the preceding equation, Zx is the first member, and 2ah the second member. 127. The two members of an equation are not only equal numerically, but must have the same essential sign. If, in the preceding equation, x represents a negative quantity, then the first member is essentially negative, and the second member must also be negative ; that is, either a oy h must represent a negative quantity. 128. Equations are usually composed of certain quantities which are known^ and others which are unknown. The known quantities are represented either by numbers, or by the first letters of the alphabet; the unknown quantities are usually represented by the last letters of the alphabet. 129. A root of an equation is the value of the unknown quantity in the equation ; or it is any value which, being sub- stituted for the unknown quantity, will satisfy the equation. For example, in the equation 8cc-4=24-cc, suppose x=l. Substituting 7 for x^ the first member becomes 3x7—4; that is, 21-- 4, or 17; and the second member be- 80 ALGEBRA. comes 24—7 ; that is, 17. Hence 7 is a root of the equation, because when substituted for x the two members are found to be equal. 130. A numerical equation is one in which all the kno"^'n ^quantities are represented oy figures; as, x^-\-^£- = Zx-\-Vl. 131. A literal equation is one in which the known quantities are represented by letters, or by letters and figures. Thus x^-\-adi?-\-hx=m. \ ^^. ^ and x'_3ax^+5Sa;» = 16 [ ""'^ ^'^'""^ equafons. 132. The degree of an equation is denoted by the greatest number of unknown factors occurring in any term. If the equation involves but one unknown quantity, its de- gree is denoted by the exponent of the highest power of this quantity in any term. If the equation involves more than one unknown quantity, its degree is denoted by the greatest sum of the exponents of the unknown quantities in any term. Thus ax-{-h=cx-\-d is an equation of the^?*5^ degree^ and is sometimes called a simple equation. 4x2— 203=5— x^ r^^^ ^xy—4:X-\-y=4:0 are equations of the second degree^ and are frequently called quadratic equations, x^-^ax^=2h and x^-{-^xif-\-y=m are equations of the third degree^ and are frequently called cubic equations. So also we have equations of the fourth degree, sometimes called hi-quadratic equations ; equations of the fifth degree, etc., up to the nih. degree. Thus x'^+ax^-'^—h is an equation of the ?ith degree. 133. To solve an equation is to find the value of the unknown quantity, or to find a number which, being substituted for the unknown quantity in the equiation, renders the first member identical with the second. The difl[iculty of solving equations depends upon their de- gree, and the number of unknown quantities they contain. A EQUATIONS OF THE FIRST DEGREE. 81 134. Axioms. — The various operations which we perform upon equations, in order to deduce the value of the unknown quantities, are founded upon the following principles, which are regarded as self-evident. "^1. If to two equal quantities the same quantity be added, the sums will be equal. 2. If from two equal quantities the same quantity be sub- tracted, the remainders will be equal. 3. If two equal quantities be multiplied by the same quanti- ty, the products will be equal. 4. If two equal quantities be divided by the same quantity, the quotients will be equal. 135. Transposition. — Transposition is the process of changing ^fv^- a term from one member of an equation to the other without 4t/fv« destroying the equality of the members. ^ Let it be required to solve the equation x-^az=h. ^j^ ,, ,,[,:■ . v.r ■ If from the two equal quantities x-\-a and h we subtract the ' same quantity a, the remainders will be equal, according to the last article, and we shall have x-\-a—a = b—a, or x = h — a. Let it be required to solve the equation x—a — h. If to the two equal quantities x—a and h the same quantity a be added, the sums will be equal, according to the last arti- cle, and we have x—a-\-a=h-^aj or x=h-\-a. ---J 136. Hence we perceive that we may transpose any term of an equation from one merriber of the equation to the other, provided we change its sign. It is also evident that we may change the sign of every term of an equation without destroying the equality ; for this is, in fact, the same thing as transposing every term in each member of the equation. D2 82 ALGEBRA. EXAMPLES. In the following examples, transpose the unknown terms to the first member and the known terms to the second member. 1. 5ir4-12 = 8a;+18. Ans. 6x~3a:=18-12. 2. 4x--7=21 — 8x. Ans. Ax -\-Sx=21-\-7. 3. 2x-15 = -7ic-f30. Ans. 2cc+7a7=304-15. 4. ax-{-bc=m—2x. Ans, ax-\-2x=m—hc, 5. 4:ax—b-\-2c=Sx—2ab^STnx. Ans. 4:aX'—Sx-\-Smx=b—2c—2ab. 6. 4:ah—ax—2c=hx—Sm. Ans. ax-{-bx=4:ab—2c+3m. 7. ab^cx—2mx=Sax—4:b. Ans. Sax-\-cx-\-2mx=ab-\-4b. 137. To clear an Equation of Fractions. — Let the equation be ^=b. If we multiply each of the equal quantities - and b by the same quantity a, the products will be equal by Art. 134, and we shall have x=ab. Suppose the equation is — \-j^=m. If we multiply each of the members of the equation by a, we shall have . ax If we multiply each of the members of this equation by 6, we shall have bx-\-ax=zabm. Hence, to clear an equation of fractions, we have the follow- ing RULE. Multiply each member of the equation by all the denominators, EXAMPLES. £C cc 3 1. Clear the equation -— -=- of fractions. Ans. 20x-12a:=45. 3.r 2x 3 2. Clear the equation — ^— — == of fractions. ^ Ans. 63a;-70a'=45. EQUATIONS OF THE FIRST DEGREE. 83 8. Clear tlie equation y — ^+^=:6 of fractions. Ans. 40ic-105a^+28a:=840. Cf, CP, cc 4. Clear the equation 0+1;+^ = 10 of fractions. 138. An equation may always be cleared of fractions by muL tiplying each member into all the denominators; but some- times the same result may be attained by a less amount of mul- tiplication. Thus, in the last example, the equation may be cleared of fractions by multiplying each term by 12 instead of 6x4x2, and it is important to avoid all useless multiplica- tion. In general, an equation may be cleared of fractions by multiplying each member by the least common multi;ple of all the denominators. 2x Sx 7 5. Clear the equation "c' + x~To ^^ fractions The least common multiple of all the denominators is 20. If we multiply each member of the equation by 20, we obtain 8x+ 15^=14. The operation is effected by dividing the least common mul- tiple by each of the denominators, and then multiplying the corresponding numerator, dropping the denominator. 4:X Sx 8 6. Clear the equation y~~TZ~9T ^^ ^^^^^^^^^' X — 4: 1 7. Clear the equation Sx——t—=j^ of fractions. It should be remembered that when a fraction has the mi- nus sign before it, this indicates that the fraction is to be sub- tracted^ and the signs of the terms derived from its numerator must be changed, Art. 118. ^^^ 86c.-3x+12:=l. ^ ^ , . a—x Sx—2h x-{-ab 8. Clear the equation — r r — = — ^— • Ans. a^—a^x—Sax-^2ah=hx-{-ah\ 84 ALGEBRA. 139. Solution of Equations. — An equation of the first degree containing but one unknown quantity may be solved by trans- forming it in such a manner that the unknown quantity shall stand alone, constituting one member of an equation ; the other member will then denote the value of the unknown quantity. Let it be required to find the value of x in the equation 4x— 2 bx 3x ^ Clearing of fractions, we have 32x-16 + 25x=80a;+200. By transposition we obtain 32x+25x-30:r=200+16. Uniting similar terms, 27£c=216. Dividing each member by 27, according to Art. 134, we have To verify this value of a;, substitute it for x in the original equation, and we shall have 32-2 40 24 ^ or 6+5 = 6 + 5; that is, 11 = 11, an identical equation, which proves that we have found the correct value of x. , 140. Hence we deduce the following RULE. 1. Clear the equation of fractions^ and perform all the opera- tions indicated. 2. IVanspose all the terms containing the unJcnoivn quantity to one side^ and all the remaining terms to Hie other side of the equa- tion^ and redujce each member to its most simple form. 8. Divide each member by the coefficient of tJie unknown quan tity. There are various artifices which may sometimes be em* EQUATIONS OF THE FIRST DEGREE. 85 ployed, by which the labor of solving an equation may be con- siderably abridged. These artifices can not always be reduced to general rules. If, however, any reductions can be made be- fore clearing of fractions, it is generally best to make them ; and if the equation contains several denominators, it is often best to multiply by the simpler denominators first, and then to effect any reductions which may be possible before getting rid of the remaining denominators. Sometimes considerable labor may be saved by simply indicating a multiplication during the first steps of the reduction, as we can thus more readily detect the presence of common factors (if there are any), which may be canceled. The discovery of these artifices will prove one of the most useful exercises to the pupil. EXAMPLES. 1. Solve the equation _ 8x4-1 2:r-f 9 , , 8x ^ = -^+4. Clearing of fractions, 63x-24cc-3 = 14a:H- 63 + 84. Transposing and reducing, 25x=150. Dividing by 25, cc=:6. To verify this result, put 6 in the place of x in the original equation. Solve the following equations: \^. Sax—4:ab = 2ax—6ac. 3. Sx''-10x^Sx-\-x''. ^ aid^+x"^) ax ^ . x-5 , , 284-03 6. "^=bc+d+l X X 7. 3a; H =^ = 5H t: . ins . X = 4:b- .6c. Ans. X-- = 9. Ans. X-. _d ~c' Ans. X- =9. Arts. X — ab-\ bc + d' Ans. ■ X-. = 7. 00 ALGEBRA. c 8. 6ax^2b+4:hx=2x-\-bc. Ans. x= ^ ba-\-4:b-2' "^^ 9. x-\ — - — =12 ^ — . Am, x=5, 11 3x ^-^ i- ^^+U 1 ^ 11. IJx- 4^-3 -12' / 12. 3x— a4-cx=— ;t . -4.m. x=- ^ -. da 8a + 8ac— 3 \^^^ 8x X , 2x . abed 13. c+r=4x+-T. Ans. x=-—-— — - — — — — — -, (f b a 3bd-\-ad—4:abd—2ab 14. {a-{-x){b-{-x)—a{b-\-c)=— — \-x^. Am. ^=t-. 15. Ilr3x_4^=5-6x+^. ^ 5 o o 3ir-3 , 20-cc 6a:-8 4ic-4 Z 16. 0.-- ^+4 = -2 -^+-g-. 7:c+16 a:+8 x 18. 21 4x-ll~3* 6x-[-7 7a;-13 _ 2a-H-4 /j. 9 ^6x+S~~ 3 • w^.-»7'± zi o ^7/, i 70aZ>— 3ac 19. '^,ab-\-zac—^cx=^ac-\-2ao—6cx. Am. x— — ^j-r-; — . 6 5 3 4 o20c 20. 1(.-|)+1(.-|)4(.-|). ^n..=|. 21. 7x4-9 / 1x—\\ _ . J c ^ {x 9-) = 7. ^ns. a:=5. 22. -2-+-3-=-4- + ~6~"'" '^^^ 23. l-l'+4-|:=7-|-+10-,4^. ^«.. x=^. / 3&^*-te-'-9i-^^"-12x -'"••^-432- 24. —7^ \ T-^ — i^* Am. ic=ll. 85 6X-101 5 76 EQUATIONS OF THE FIRST DEGREE. 87 9x+4 4x-19 6x^S2 llx+13 . .^^ 25 — ^^H r:; — = — T^ ^1 • Ans. X =100. 5x-48^ 51 17 ol 4 7 87 26 \ = —, — ^^ n. ^^^5- ic=l. a;+2^x + 8 x'-{-bx-{-^ 27. ^ 1 — n ?;. Ans. x=6. x-2 x-4: x-6 x-S ^o X X a . _ a^{h—a) 29. (x+|)(x-|)+|=(x+5)(a:-8). ^ns. a:=12. Solution of Problems. 141. A problem in Algebra is a question proposed requiring us to determine the value of one or more unknown quantities from given conditions. 142. The solution of a problem is the process of finding the value of the unknown quantity or quantities that will satisfy the given conditions. 143. The solution of a problem consists of two parts : 1st. The statement, which consists in expressing the condi- tions of the problem algebraically ; that is, in translating the conditions of the problem from common into algebraic lan- guage, ov forming the equation. 2d. The solution of the equation. The second operation has already been explained, but the first is often more embarrassing to beginners than the second. Sometimes the conditions of a problem are expressed in a dis- tinct and formal manner, and sometimes they are only implied, or are left to be inferred from other conditions. The former are called explicit conditions, and the latter implicit conditions. 144. It is impossible to give a general rule which will enable 88 ALGEBRA. US to translate every problem into algebraic language, since the conditions of a problem may be varied indefinitely. The fol- lowing directions may be found of some service : Represent one of the unknown quantities hy some Utter or sym- bol^ and then from the given conditions find an expression for each of the other unknown quantities^ if any^ involved in the problem. Express in algebraic language the relations which subsist between the unknown quantities and the given quantities; or, by means of the algebraic signs, indicate the operations necessary to verify the value of the unknown quantity, if it was already known. PROBLEMS. Prob. 1. What number is that, to the double of which if 16 be added, the sum is equal to four times the required number? Let X represent the number required. The double of this will be 2x. This increased by 16 should equal 4x. Hence, by the conditions, ^x-\-\^—\x. The problem is now translated into algebraic language, and it only remains to solve the equation in the usual way. Transposing, we obtain 16=4a:-2a;=:2cc, and 8=x, or x—^. To verify this number, we have but to double 8, and add 16 to the result ; the sum is 32, which is equal to four times 8, according to the conditions of the problem. Prob. 2. What number is that, the double of which exceeds its half by 6? Let cc=the number required. Then, by the conditions, 2x-|=6. Cleari n g of fractions, 4ic — a: = 12, or 3x=12. ITence ' a: = 4. To verify this result, double 4, which makes 8, and diminish EQUATIONS OF THE FIRST DEGREE. 89 it by the half of 4, or 2 ; the result is 6, according to the con- ditions of the problem. Prob. 3. The sum of two numbers is 8, and their difference 2. What are those numbers? Let a:=the least number. Then x-f 2 will be the greater number. ■ The sum of these is 2a;+2, which is required to equal 8. Hence we have 2a;+2=:8. By transposition, 2x=8 — 2 = 6, and x=S, the least number. Also, x-\-2 = bj the greater number. Verification. 5 + 3 = 8) -,. ^ . ... p; Q_9 ( accordmg to the conditions. The following is a generalization of the preceding Problem. Prob. 4.^ The sum of two numbers is a, and their difference b. What are those numbers? Let X represent the least number. Then x-\-b will represent the greater number. The sum of these is 2x+5, which is required to equal a. Hence we have 2x-\-b=a. By transposition, 2x=a—b,_ or ^=-i^ = o~o^ ^^^® ^^^^ number. Hence x-\-b—- — --\rb—-^-\--^, the greater number. As these results are independent of any particular value at- tributed to the letters a and i, it follows that Half the difference of tiuo quantities^ added to half their sum, is equal to the greater ; and Half the difference subtracted from half the sum is equal to the less. The expressions f + l ^^^ f— | ^^^ called /ormt/Zas, because they may be regarded as comprehending the solution of all questions of the same kind ; that is, of all problems in which we have given the sum and difference of two quantities. 90 ALGEBRA. Thus, let a=8 ) . ,, ,. , , 7j_o f ^ 1° t^® preceding problem. Then --|--=— ^r— =5, the greater number. And -—-=—^=3, the less number. 0) S P O 10; r 6; 12 2; 23 11; 100 100 their difference - 50 ; required the numbers. 1; 5 4; 10 I 4; Prob. 5. From two towns which are 54 miles distant, two travelers set out at the same time with an intention of meet- ing. One of them goes 4 miles and the other 5 miles per hour. In how many hours will they meet? Let X represent the required number of hours. Then 4x will represent the number of miles one traveled, and bx the number the other traveled; and since they meet, they must together have traveled the whole distance. Consequently, 4a:+5x=:54. Hence 9^=54, or x=Q. Proof. In 6 hours, at 4 miles an hour, one would travel 24 miles; the other, at 5 miles an hour, would travel 30 miles. The sum of 24 and 30 is 54 miles, which is the whole distance. This Problem may be generalized as follows : Prob. 6. From two points which are a miles apart, two bod- ies move toward each other, the one at the rate of m miles per hour, the other at the rate of n miles per hour. In how many hours will they meet? Let X represent the required number of hours. Then tux will represent the number of miles one body moves, and nx the miles the other body moves, and we shall obvious- ly have mx-\-nx=a. EQUATIONS OF THE FIKST DEGREE. 91 __ a Hence x— — ; — . This is a general formula, comprehending the solution of all problems of this kind. Thus, Let the 90 one body distances ^-^3^. moves 210; Eequired the time of meeting. We see that an infinite number of problems may be pro- posed, all similar to Prob. 5 ; but they are all solved by the formula of Prob. 6. We also see what is necessary in order that the answers may be obtained in whole numbers. The given distance (a) must be exactly divisible by m-\-n. Prob. 7. A gentleman, meeting three poor persons, divided 60 cents among them ; to the second he gave twice, and to the third three times as much as to the first. What did he give to each? Let a:=the sum given to the first; then 2x=the sum given to the second, and 3x=the sum given to the third. Then, by the conditions. That is, 6x=60, or 05=10. Therefore he gave 10, 20, and 80 cents to them respectively. The learner should verify this, and all the subsequent results. The same problem generalized: Prob. 8. Divide the number a into three such parts that the second may be m times, and the third n times as great as the first. . a ma na Ans. l + m-|-?2' l+m + ?2' l-\-m-\-n What is necessary in order that the preceding values may be expressed in whole numbers? Prob. 9. A bookseller sold 10 books at a certain price, and afterward 15 more at the same rate. Now at the last sale he 92 ALGEBRA. received 25 dollars more than at the first. What did he re- ceive for each book? Ans. Five dollars. The same Problem generalized: Prob. 10. Find a number such that when multiplied success- ively by m and by n, the difference of the products shall be a. ,»«-, Ans. . 772 — 72. Prob. 11. A gentleman, dying, bequeathed 1000 dollars to three servants. A was to have twice as much as B, and B three times as much as C. What were their respective shares? Ans. A received $600, B $300, and C $100. Prob. 12. Divide the number a into three such parts that the second may be m times as great as the first, and the third n times as great as the second. . a ma mna 1 + m+mn' l-\-m-\-mn'' l-\-m-\-mn Prob. 13. A hogshead which held 120 gallons was filled with ii mixture of brandy, wine, and water. There were 10 gallons of wine more than there were of brand)', and as much water as both wine and brandy. What quantity was there of each? Ans. Brandy 25 gallons, wine 35, and water 60 gallons. Prob. 14. Divide the number a into three such parts that the second shall exceed the first by m, and the third shall be equal to the sum of the first and second. . a— 2771 a + 2771 a Prob. 15. A penson employed four workmen, to the first of whom he gave 2 shillings more than to the second ; to the sec- ond 3 shillings more than to the third ; and to the third 4 shil- lings more than to the fourth. Their wages amount to 32 sliil- lings. What did each receive ? Ans. They received 12, 10, 7, and 3 shillings respectively. "^ Prob. 16. Divide the number a into four such parts that the second shall exceed the first by m, the third shall exceed the second by ??, and the fourth shall exceed the third by p. Ans. The first, -. ; the second, -r ^ ; EQUATIONS OF THE FIRST DEGREE. 93 the third, ^ ; the fourth, j -. Problems which involve several unknown quantities may oft- en be solved by the use of a single unknown letter. Most of the preceding examples are of this kind. In general, when we have given the sum or difference of two quantities, both of them may be expressed by means of the same letter. For the diffei^ ence of two quantities added to the less must be equal lo the greater ; and if one of two quantities be subtracted from their sum, the remainder will be equal to the other. Prob. 17. At a certain election 86,000 votes were polled, and the candidate chosen wanted but 8000 of having twice as many votes as his opponent. How many voted for each ? Let x=:the number of votes for the unsuccessful candidate; then 86,000 — x = the number the successful one had, and 86,000-^+8000 = 2^. Ans. 18,000 and 28,000. "^ Prob. 18. Divide the number a into two such parts that one part increased by h shall be equal to m times the other part. . ma — h a-\-h 772 + 1 ' m + 1 Prob. 19. A train of cars, moving at the rate of 20 miles per hour, had been gone 3 hours, when a second train followed at the rate of 25 miles per hour. In what time will the second train overtake the first? Let x=:the number of hours the second train is in motion, and cc+8=the time of the first train. Then 25x=the number of miles traveled by the second train, and 20(cc+8)=:the miles traveled by the first train. But at the time of meeting they must both have traveled the same distance. Therefore 25cc = 20x + 60. B}^ transposition, 5cc=60, and a: =12. Proof. In 12 hours, at 25 miles per hour, the second train goes 800 miles ; and in 15 hours, at 20 miles per hour, the first train also goes 300 miles; that is, it is overtaken by the sec- ond train. 94: ALGEBRA. Prob. 20. Two bodies move in the same direction from two places at a distance of a miles apart ; the one at the rate of n miles per hour, the other pursuing at the rate of m miles per hour. When will they meet ? Ans. In — — hours. m—n This Problem, it will be seen, is essentially the same as X prob. 10. Prob. 21. Divide the number 197 into two such parts that four times the greater may exceed five times the less by 50. \ Ans. 82 and 115. Prob. 22. Divide the number a into two such parts that m times the greater may exceed n times the less by b. ma—h na-\-b Ans. m-\-n ' m-\-n When 71=1, this Problem reduces to Problem 18. When 5=0, this Problem reduces to Problem 24. Prob. 23. A prize of 2329 dollars was divided between two persons, A and B, whose shares were in the ratio of 5 to 12. What was the share of each ? Beginners almost invariably put x to represent one of the quantities sought in a problem; but a solution may often be very much simplified by pursuing a different method. Thus, in the preceding problem, we may put x to represent one fifth of A's share. Then bx will be A's share, and 12a; will be B's, and we shall have the equation 5ic+12jc=2329, and hence a; =137; consequently their shares were 685 and 1644 dollars. V Prob. 24. Divide the number a into two such parts that the first part may be to the second as m to n. . ma na A71S. vi-\-n^ 7n-{-n Prob. 25. What number is that whose third part exceeds its fourth part by 16? Let 12ic=the number. Then 4a:-3u;=16. EQUATIONS OF THE FIRST DEGREE. 95 or x=zl6. Therefore tlie number =:= 12 x 16 = 192. Prob. 26. Find a number such that when it is divided suc- cessively by m and by n, the difference of the quotients shall be a. amn Ans. 71 — 711 Prob. 27. A gentleman has just 8 hours at his disposal ; how far__may he ride in a coach which travels 9 miles an hour, so as to return home in time, walking back at the rate of 3 miles an hour ? Ans. ISjuilfiSL Prob. 28. A gentleman has just a hours at his disposal ; how far may he ride in a coach which travels m miles an hour, so as to return home in time, walking back at the rate of n miles an hour? , arriTi ., Atis. miles. Prob. 29. A gentleman divides a dollar among 12 children, giving to some 9 cents each, and to the rest 7 cents. How many were there of each class ? Prob. 80. Divide the number a into two such parts that if the first is multiplied by m and the second by n, the sum of the products shall be b. . h — na ma—h Ans. — Prob. 31. If the sun moves every day 1 degree, and the moon 13, and the sun is now 60 degrees in advance of the moon, when will they be in conjunction for the first time, sec- ond time, and so on? Prob. 32. If two bodies move in the same direction upon the circumference of a circle which measures a miles, the one at the rate of n miles per day, the other pursuing at the rate of m miles per day, when will they be together for the first time, sec< end time, etc., supposing them to be h miles apart at starting? Ans. In --i^ i^ ?^±*, etc., days. m — 7L 7n — n nn—n It will be seen that this Problem includes Prob. 20. 96 ALGEBRA. Prob. 33. Divide the number 12 into two sucb parts that the difference of their squares may be 48. Prob. 34. Divide the number a into two such parts that the difference of their squares may be 6. , d^—h (j?-\-h ^'^' ~%r ^ "2^- Prob. 35. The estate of a bankrupt, valued at 21,000 dollars, is to be divided among three creditors according to their re- spective claims. The debts due to A and B are as 2 to 3, while B's claims and C's are in the ratio of 4 to 5. What sum must each receive? Prob. 36. Divide the number a into three parts, which shall be to each other as m : ?i : ^. . ma na pa 'm-f-w-|-^' m-f-7i-f-^' m-\-n-{-p' When p=l, Prob. 36 reduces to the same form as Prob. 8. Prob. 37. A grocer has two kinds of tea, one worth 72 cents per pound, the other 40 cents. How many pounds of each must be taken to form a chest of 80 pounds, which shall, be worth 60 cents? Ans. 50 pounds at 72 cents, and 30 pounds at 40 cents. Prob. 38. A grocer has two kinds of tea, one worth a cents per pound, the other h cents. How many pounds of each must be taken to form a mixture of n pounds, which shall be worth c cents? . n(c—h) , Ans, — ^^ — j-^ pounds at a cents, and — ^^ — 7—^ pounds at b cents. a—b ^ Prob. 89. A can perform a piece of work in 6 days ; B can perform the same work in 8 days; and C can perform the same work in 24 days. In what time will they finish it if all work together? Prob. 40. A can perform a piece of work in a days, B in 6 days, and Cine days. In what time will they perform it if all work together? . abc , ° A71S. —J -y- days. ab-{-ac-\-bc "^ Prob. 41. There are three workmen, A, B, and C. A and B together can perform a piece of work in 27 days; A and G EQUATIONS OF THE FIRST DEGREE. 97 together in S6 days; and B and C together in 54 days. In what time could they finish it if all worked together ? A and B together can perform -^ of the work in one day. AandC " A B and C " 6^ " " Therefore, adding these three results, 2A+2B + 2C can perform i^r + -5V 4- "sV in one day, =: -xV in one day. Therefore, A, B, and C together can perform ^ of the work in one day ; that is, they can finish it in 24 days. If we put X to represent the time in which they would all finish it, then they would together perform ^ part of the work in one day, And we should have i ■ i , i _ 2 Prob. 42. A and B can perform a piece of labor in a days ; A and C together in b days ; and B and C together in c days. In what time could they finish it if all work together ? . 2ahc , Ajis. —^ J- days. ao-{-ac-{-oc This result, it will be seen, is of the same form as that of Problem 40. Prob. 43. A broker has two kinds of change. It takes 20 pieces of the first to make a dollar, and 4 pieces of the second to make the same. Now a person wishes to have 8 pieces for a dollar. How many of each kind must the broker give him ? Prob. 44. A has two kinds of change ; there must be a pieces of the first to make a dollar, and b pieces of the second to make the same. Now B wishes to have c pieces for a dollar. How many pieces of each kind must A give him ? V Ans, — — r- of the first kind ; — ^ — ^ of the second. a — b a — b Prob. 45. Divide the number 45 into four such parts that the first increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, shall all be equal. In solving examples of this kind, several unknown quantities are usually introduced, but this practice is worse than super- E 98 ALGEBRA. fluous. The four parts into which 45 is to be divided maj be represented thus: The first =a;-2, the second =ix-\-2y the third =f, the fourth =2x] for if the first expression be increased by 2, the second dimin- ished by 2, the third multiplied by 2, and the fourth divided by 2, the result in each case will be x. The sum of the four parte is 4-^x, which must equal 45. Hence £c=10. Therefore the parts are 8, 12, 5, and 20. Prob. 46. Divide the number a into four such parts that the first increased by m, the second diminished by m, the third multiplied by m, and the fourth divided by Tn, shall all be equal. . ma ma , a m^a (m + 1)' ' (m+l)2' ' (m + l)2' (m+lf Prob. 47. A merchant maintained himself for three years at an expense of $500 a year, and each year augmented that part of his stock which was not thus expended by one third thereof. At the end of the third year his original stock was doubled. What was that stock? Prob. 48. A merchant supported himself for three years at an expense of a dollars per year, and each year augmented that part of his stock which was not thus expended by one third thereof At the end of the third year his original stock was doubled. What was that stock ? 148a Ans. -^. Prob. 49. A father, nged 54 years, has a son aged 9 yeai-s. In how many years will the age of the father be four times that of the son ? Prob. 50. The age of a father is represented by a, the age of his son by h. In how many years will the age of the father be n times that of the son ? a-^nh Ans. Z-, n— 1 EQUATIONS WITH MORE THAN ONE UNOTOiv^^'. QlTAJSTTITY.'-, 99 CHAPTER IX. EQUATIONS OF THE FIRST DEGREE CONTAINING MORE THAN ONE UNKNOWN QUANTITY. 145. If we have a single equation containing two unknown quantities, then for every value which we please to ascribe to one of the unknown quantities, we can determine the corre- sponding value of the other, and thus find as many pairs of values as we please which will satisfy the equation. Thus, let 2x-{-4y=l6. (1.) If y=lj we find x=6] if y=2, we find cc=4, and so on; and each of these pairs of values, 1 and 6, 2 and 4, etc., sub- stituted in equation (1), will satisfy it. Suppose that we have another equation of the same kind, as, for example, 5a: +83/ =19. (2.) We can also find as many pairs of values as we please which will satisfy this equation. But suppose we are required to satisfy both equations with the same set of values for x and y ; we shall find that there is only one value of x and one value of y. For, multiply equa- tion (1) by 3, and equation (2) by 4, Axiom 3, and we have 6x+12^=48, (3.) 20x+12.y=76. (4.) Subtracting equation (3) from equation (4), Axiom 2, we have 14cc=:28; (5.) whence a: =2. (6.) Substituting this value of x in equation (1), we have 4+42/=16; (7.) whence 2/ =3. (8.) Thus we see that if both equations are to be satisfied, x must equal 2, and y viust equal 3. Equations thus related arc called simoltaneous equations 100 ALGEBRA. l^.f^'i * SimuUmeov^ equatums are those which must be satisfied hf tb^ 'sartie Vakes of the unknown quantities. When two or more simultaneous equations are given for so- lution, we must endeavor to deduce from them a single equation containing only one unknown quantity. We must therefore make one of the unknown quantities disappear, or, as it is termed, we must eliminate it. 147. Elimination is the operation of combining two or more equations in such a manner as to cause one of the unknown quantities contained in them to disappear. There are three principal methods of elimination : 1st, by ad- dition or subtraction ; 2d, by substitution ; 8d, by comparison. 148. Elimination ly Addition or Subtraction. — Let it be pro- posed to solve the system of equations 5x+43/=35, (1.) 7x-3?/=6. (2.) Multiplying equation (1) by 8, and equation (2) by 4, we have 15a;+12y=105, (8.) 28x-123/=24. (4.) Adding (3) and (4), member to member (Axiom 1), we have 48^=129 ; (5.) whence cc=:3. (6.) We may now deduce the value of y by substituting the value of X in one of the original equations. Taking the first for ex- ample, we have 15 +4^/= 85 ; whence 4y=20, and y=5. 149. In the same way, an unknown quantity may be elimi- nated from any two simultaneous equations. This method is expressed in the following RULE. Multiply or divide the equatiojis^ if necessari/, in such a manner that ouc of the unknown quantities ahall hai^e the same coefficient in EQUATIONS WITH MOBE THAN ONE UNKNOWN QUA^S^TITY. 101 both. Then subtrdct one equation from :ihe (kite^ if^ihosigvs of these coefficients are alihe^ or odd them logeit^eri/ihe&igiyixirt unlike. In solving the preceding equations, we multiplied both mem- bers of each by the coefficient of the quantity to be eliminated in the other equation ; but if the coefficients of the letter to be eliminated have any common factor, we may accomplish the same object by the use of smaller multipliers. In such cases, 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 multipliers which we can employ. 150. Elimination by Substitution. — Take the same equations as before: 5x-h4y=35, (1.) 7x-32/=6. (2.) Finding from (1) the value of y in terms of x, we have .=?^. (3.) Substituting this value of y in (3), we have Ix J = 6. Clearing of fractions, 28ic-105 + 15x=24; whence x=3. Substituting this value of x in (3), we have The method thus exemplified is expressed in the following RULE. Find an expression for the value of one of the unknown quan- tities in one of the equations ; then substitute this valuje for iliat quantity in the other equation. 151. Elimination by Comparison. — Take the same equations as before: 5x4-42/ =35, (1.) 7x-Sy=6. (2.) 102 . ALGEBRA. , . iPejoiye froirt eftcK equation an expression for y in terms of a^ ^.iSd-WeWe " 35— 5x 2/ = -^f— , (3.) and 2/=—^. (4.) Placing these two values equal to each other, we have 7a;-6 _ 35-5a; 3 ~ 4 • Clearing of fractions, 28ir-24=:.105-16x; whence 43aj=129, and £c=3. Substituting this value of x in (3), The method thus exemplified is expressed in the following RULE. Find an expression for the value of the same unknown quantity in each of the equations^ and form a new equation by placing these values equal to each othe7\ In the solution of simultaneous equations, either of the pre- ceding methods can be used, as may be most convenient, and each method has its advantages in particular cases. General- ly, however, the last two methods give rise to fractional expres- sions, which occasion inconvenience in practice, while the first method is not liable to this objection. When the coefficient of one of the unknown quantities in one of the equations is equal to unity, this inconvenience does not occur, and the method ^by substitution may be preferable ; the first will, however, com- imonly be found most convenient EXAMPLES. 1. Given j a ^n~^A \^^ ^"^ *^^ values of x and y, Ans. ic=8; y=4. ICQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY. 103 1+1=7 2. Given to find a: and y, 5. Given < ^ > to find cc and y. ^"f",.— ^ K^ . be— ad be— ad ^ y ' Ans. x=—j ■ • y= . no— ma rac—na 6. Given | g^J^/^'l^^ | to find x and y. \ ^' ^^^^^ 1 Sx+SylsS 1 '^ ^^^ ^ "^^ 2/. ^^ J fl 1 r\ \ 8. Given I V to find cc and y, y 11= n\ . 2 2 \ y X J Ans. x= ; y=- m-\-n^ ^ m—n C\ 9. Given < > to find x and v- 1 12^±97_ ( ^ 115?/-17~ J Ans. x=2^', y=Z\. x-\-a , « \ -L^-\-y^b=2a I ] L, 10. Given to find x and y. x-{-a+^- = l+na j ^725. x=na—a; y—a-^h. 104 ALGEBRA. ^ ^11. Given {i^Z\6y=-S^^ 1 ^^ ^^^ ^ ^^^ y- 12. Given ^^^_^ ^^ V to find a: and 3. 5y 4:y-l9_x 20-2y 13. Given < ^ ^ o . 01 f 13 _ ^ ) .. r- Ja^+23/+3- 4:^-52/4-6/ 14. Given < „ -. q Mo find x and y. ( 6x—by+4:~Sx-\-2y+l ] Ans. x=7 ] y=S, {nr — ?y^ — (j. 1 ^ , [• to find a; and y. x—y=o ) ^ Ar^.x=^^; 2,^__. jy^+2^ 4+12 16. Given to find x and y. ^725. a:=2; y=7. 4x-Sy-7 _Sx 2y 6 j 5 ~10 15 6 /to find X '^ I ^_^2/ -t_ y-^ I ^ I -^ \ ^"^^ y- 17. Given y 8 "^2 20 15 "^6"^ 10 Ans. x=S; y=2. EQUATIONS WITH MORE THAN ONE UNK^N^OWN QUANTITY. 105 Equations of the First Degree containing more than Two Unknown Quantities. 152. If we have three simultaneous equations containing three unknown quantities, we may, by the preceding methods, reduce two of the equations to one containing only two of the unknown quantities; then reduce the third equation and either of the former two to one containing the same two unknown quantities ; and from the two equations thus obtained, the un- known quantities which they involve may be found. The third quantity may then be found by substituting these values in either of the proposed equations. Take the system of equations 2x+3y-f42;=16, (1.) ^x+2y-bz= 8, (2.) 5;r-6z/-f32= 6. (3.) Multiplying (1) by 3, and (2) by 2, we have 6a; 4- %-f 122=^48, (4.) 6a;+4^-102=16. (5.) Subtracting (5) from (4), 5z/+22z=32. (6.) Multiplying (1) by 5, and (3) by 2, we have 10:c+15?/-f20z=80, (7.) 10x-12z/-h62r=12. (8.) Subtracting (8) from (7), 27^ -|- 142= 68. (9.) Multiplying (6) by 27, and (9) by 5, we have 135z/+5942=864. (10.) 135y+70z = 340. (11.) Subtracting (11) from (10), 5242=624 ; whence 2=1. Substituting this value of z in (6), 5y+22=32; whence 2/ =2. Substituting the values of y and 2 in (1), 2x4-64-4=16; whence x=3. E2 106 ALGEBRA. 153. Hence, to solve three equations containing three un known quantities, we have the following RULE. From the three equations deduce two containing only two un» known quantities ; then from ifiese tvjo deduce one containing onl§ one unknown quantity. 154. If we hadybwr simultaneous equations containing four unknown quantities, we might, by the methods already ex- plained, eliminate one of the unknown quantities. We should thus obtain three equations between three unknown quanti- ties, which might be solved according to Art. 152. So, also, if we had/ve equations containing five unknown quantities, we might, by the same process, reduce them to four equations containing four unknown quantities, then to three, and so on. By following the same method, we might resolve a system of any number of equations of the first degree. Hence, if we have m equations containing m unknown quantities, we proceed by the following RULE. \st. Combine successively any one of the equations with each of the others^ so as to eliminate the same unknown quantity; there will result m—1 new equations^ containing m—1 unknown quan- tities. 2d. Combine any one of these new equations with the others^ so as to eliminate a second unknoiun quantity ; thei^e tuill result m — 2 equations^ containing m—2 unknown quantities. Sd. Continue this series of operations until there residts a single equation containing but one unknown quantity^ from ivhich the value of this unknoiun quantity is easily deduced. 4:th. Substitute this value for its equal in one of the equations con- iaining two unknown quantities, and thus find the value of a second unhwwn quantity \ substitute these values in an equation contain- ing three unknown quantities^ and find the value of a third; and so on, till the values of all are determined. Either of the unknown quantities may be selected as the one EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY. 107 W to be first eliminated. It is, however, generally best to begin with that which has the smallest coefficients ; and if each of the unknown quantities is not contained in all the proposed equations, it is generally best to begin with that which is found in the least number of equations. Sometimes a solution may be very much abridged by the use of peculiar artifices, for which no general rules can be given. EXAMPLES. Solve the following groups of simultaneous equations : (2x+^y-Sz = 22\ [x=S. ^I. <4:X-2y-\-5z=ls[ Ans, \y=7. [x-\-y=za\ ~-2. } x-\-z=bl ( y-\-z=c ) Note. Take the sum of the three preceding equations. / x-{-y+z=:29\ Y S. ■] ^+2y+8z=:62V Ans, ilxA-iy+iz=10) ( x+^y-^^z=S2\ (ix-\-iy+h^=12) { x-\-y-z=lS20\ I — 5. -^ x—y+z= 654 V Ans. ■< ir 3.987. 2/ =654. 2 = 821. 6. 7. x—y-^z= 6 8ix-4|z/+5^2=t:82 10^;z;-9|2/+ll2=71 Ans. }y:=:6S0. {z=94:5. 108 ALGEBBA. — 8. ^ X y 1+1=5 X z - \.y ^ J - + ^a X y z 1-1+1.5 X y z - . ~H-^i=' , r 12 7^ 2a; + 3?/ 8a;+4z 30 37 z 3x+4;^ ' 5?/+a 222 8 [,63/4-9z 2x4-3 y c2x+by-1z=-2^^ 11. \ 6x-y-\-Sz= 227 L 7x4-6y4-^= 297 X y 2z ^^ } 4'^6'^3 76 2+ 8"+5- ^^ ^4-^4-^=248 J 7a;-2z4-8w=17 4:y—2z-\-v=ll 5y—Sx-^2u= 8 47/_8?^4-2y= 9 8«4-8w=33 ^n^. X: y= Z = 2 a—b+c 2 2 J.7W. - a:;= 2/^ 2r = 2 2 -4/15. {S ic=13. 24. 62. -Arw. x= 12. y= 30. 2 = 168. Lv= 60. J.7W. a:=2, z = S, EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY. lOS — 14. x-\-y-{-z-\-u + v=26 x-{-y + z+t-{-v=27 x-{-y-\-t-{-u-^v=:2S x-{-z+i-^u+v^29 y-\-z+t-]-u-\-v=30 Note. Take the sum of these six equations. Am. x=S. y=4u z=6. t=7. v=8. ,3 V Problems involving Equations of the First Degree with several Unknown Quantities. Prob. 1. Find two numbers such that if the first be added to four times the second, the sum is 29 ; and if the second be added to six times the first, the sum is %Q. Prob. 2. If A's money were increased by ZQ shillings, he would have three times as much as B ; but if B's money were diminished by 5 shillings, he would have half as much as A. Find the sum possessed by each. Prob. 3. A pound of tea and three pounds of sugar cost six shillings ; but if sugar were to rise 50 per cent, and tea 10 per cent., they would cost seven shillings. Find the price of tea and sugar. Ans. Tea, 55. per pound ; Sugar, 4 pence. "iProb. 4. What fraction is that to the numerator of which if 4 be added the value is one half; but if 7 be added to the de- nominator, its value is one fifth ? Ans. t^. ^ Prob. 5. A certain sum of money, put ont at simple interest, amounts in 8 months to $1488, and in 15 months it amounts to $1530. What is the sum and rate per cent. ? Prob. 6. A sum of money put out at simple interest amounts in m months to a dollars, and in n months to h dollars. Re- quired the sum and rate per cent. h—a Ans. The sum is ; the rate is 1200 x - n — m ' na—mb' Prob. 7. Tlvcre is a number consisting of two digits, the sec- ond of which is greater than the first ; and if the number be divided by the sum of its digits, the quotient is 4 ; but if the digits be inverted, and that number be divided by a number 110 ALGEBRA. greater by two than the diflference of the digits, the quotient is 14. Required the number. Let X represent the left-hand digit, and y the right-hand digit. Then, since x stands in the place of tens, the number will be represented by \^x-^y. Hence, by the first condition, lOo^-hy ^^. x-\-y ' by the second condition, y-x-^r'l' Whence a: =4, ?/=8, and the required number is 48. *^ Prob. 8. A boy expends thirty pence in apples and pears, buying his apples at 4 and his pears at 6 for a penny, and aft- erward accommodates his friend with half his apples and one third of his pears for 13 pence. How many did he buy of each? Prob. 9. A father leaves a sum of money to be divided among his children as follows: the first is to receive $300 and the sixth part of the remainder ; the second, $600 and the sixth part of the remainder; and, generally, each succeeding one receives $300 more than the one immediately preceding, together with the sixth part of what remains. At last it is found that all the children receive the same sum. What was the fortune left, and the number of children ? Ans. The fortune was $7500, and the number of children 5. Prob. 10. A sum of money is to be divided among several persons as follows: the first receives a dollars, together with the nth part of the remainder; the second, 2a, together with the nth part of the remainder; and each succeeding one a dollars more than the preceding, together with the n\\\ part of the re- mainder ; and it is found at last that all have received the same Bum. What was the amount divided, and the number of per- sons? Ans. The amount was a(?i — 1)*; the number of persons =?? — 1. Prob. 11. A wine-dealer has two kinds of wine. If ho mixes ^ EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY. Ill 9 quarts of the poorer with 7 quarts of the better, he can sell the mixture at 65 cents per quart ; but if he mixes 3 quarts of the poorer with 5 quarts of the better, he can sell the mixture at 58 cents per quart. What was the cost of a quart of each kind of wine? Ans. 48 cents for the poorer, and 64 for the better. Prob. 12. A person owes a certain sum to two creditors. At one time he pays them $530, giving to one four elevenths of the sum which is due, and to the other $30 more than one sixth of his debt to him. At a second time he pays them $420, giving to the first three sevenths of what remains due to him, and to the other one third of what remains due to him. What were the debts? Prob. 13. If A and B together can perform a piece of work in 12 days, A and C together in 15 days, and B and C in 20 days, how many days will it take each person to perform the same work alone? This problem is readily solved by first finding in what time they could finish it if all worked together. Prob. 14. If A and B together can perform a piece of work in a days, A and C together in b days, and B and C in c days, how many days will it take each person to perform the same work alone? ^ , ^ , . . . 2aoc T ^ . 2aoc , Ans. A in — —^ 7 days* 13 m —. — -, days; ac-^bc—ab "^ ab-{-bc—ac "^ ^ . 2abc , C in -1 7- days. ab-\-ac—bc Prob. 15. A merchant has two casks, each containing a cer- tain quantity of wine. In order to have an equal quantity in each, he pours out of the first cask into the second as much as the second contained at first; then he pours from the second into the first as much as was left in the first ; and then again from the first into the second as much as was left in the second, when there are found to be a gallons in each cask. How manj gallons did each cask contain at first ? , 11a -, 5a Ans, -TT— and — . 112 ALGEBRA. Prob. 16. A laborer is engaged for a days on condition that he receives p pence for every day he works, and pays q pence for every day he is idle. At the end of the time he receives a pence. How many days did he work, and how many was he idle? Ans, He worked —^ — days, and was idle -^ days. p + q -^ p + q ^ Prob. 17. A certain number consisting of two digits contains the sum of its digits four times, and their product three times. What is the number? Prob. 18. A father says to his two sons, of whom one was four years older than the other. In two years my age will be double the sum of your ages ; but 6 years ago my age was 6 times the sum of your ages. How old was the father and each of the sons ? Ans. The father was 42, one son 11, and the other 7 years old. Prob. 19. It is required to divide the number 96 into three parts such that if we divide the first by the second the quo- tient shall be 2, with 8 for a remainder ; but if we divide the second by the third, the quotient shall be 4, with 5 for a re- mainder. What are the three parts? Ans. 61, 29, and 6. Prob. 20. Each of seven baskets contains a certain number of apples. I transfer from the first basket to each of the other six as many apples as it previously contained ; I next trans- fer from the second basket to each of the other six as many apples as it previously contained, and so on to the last basket, when it appeared that each basket contained the same number of apples, viz., 128. How many apples did each basket contain before the distribution? Ans. The first 449, the second 225, the third 113, the fourth 67 the fifth 29, the sixth 16, and the seventh 8 apples. 155. When we have only one equation containing more than one unknown quantity, we can generally solve the equation in an infinite number of ways. For example, if a problem involv- ing two unknown quantities {x and y) leads to the singh equa- tion ax-{-by=Cj EQUATIONS WITH MORE THAN ONE UNKNOWN QUANTITY. US we may ascribe any value we please to ic, and then determine the corresponding value of y. Such a problem is called inde- terminate. An indeterminate problem is one which admits of an indefinite number of solutions. 156. If we had two equations containing three unknown quan- tities, we could, in the first place, eliminate one of the unknown quantities by means of the proposed equations, and thus obtain one equation containing two unknown quantities, which would be satisfied by an infinite number of systems of values. There- fore, in order that a problem may be determinate^ its enuncia- tion must contain as many different conditions as there are un- known quantities, and each of these conditions must be express- ed by an independent equation. 157. Equations are said to be independent when they express conditions essentially different^ and dependent when they express the same conditions under different forms. Thus ] Q^ _in [ ^^® independent equations. But lo^ cf^-|j.[ ^^® ^^^ independent, because the one may be deduced from the other. 158. If, on the contrary, the number of independent equa- tions exceeds the number of unknown quantities, these equa- tions will be contradictory. For example, let it be required to find two numbers such that their sum shall be 8, their difference 2, and their product 20. From these conditions we derive the following equations : a:+2/=8, x-y=2, xy=z20. From the first two equations we find x=b and 2/=3. Hence the third condition, which requires that their product ehall be equal to 20, can not he fidjilled. 114 ALGEBBA. CHAPTER X. DISCUSSION OF PROBLEMS INVOLVING SIMPLE EQUATIONS.— INEQUALITIES. 159. To discuss a 'problem or an equation is to determine the values which the unknown quantities assume for particular hy- potheses made upon the values of the given quantities, and to interpret the peculiar results obtained. We have seen that if the sum of two numbers is represented by a, and their differ- ence by &, the greater number will be expressed by —5—, and the less by — ^. Here a and h may have any values whatever, and still these formulas will always hold true. It frequently happens that, by attributing different values to the letters which represent known quantities, the values of the unknown quanti- ties assume peculiar forms, which deserve consideration. 160. We may obtain jive species of values for the unknown quantity in a problem of the first degree : Ist. Positive values. 2d. Negative values. ^ 3d. Values of the form of zero, or -r. A '^ 4th. Values of the form of -r-. 5th. Values of the form of ^. We will consider these five cases in succession. 161. 1st. Positive valves are generally answers to problems in the sense in which they are proposed. Nevertheless, all posi- tive values will not always satisfy the enunciation of a prob- lem. For example, a problem may require an answer in whoU DISCUSSION OF PROBLEMS. 115 numbers^ in which case a fractional value of the unknown quan- tity is inadmissible. Thus, in Prob. 17, page 93, it is implied that the value of cc must be a whole number, although this con- dition is not expressed in the equations. We might change the data of the problem, so as to obtain a fractional value of cc, which would indicate an impossibility in the problem pro- posed. Problem 43, page 97, is of the same kind ; also Prob, 7, page 109. If the value obtained for the unknown quantity, even when positive, does not satisfy all the conditions of the problem, the problem is impossible in the form proposed. 162. 2d. Negative values. Let it be proposed to find a number which, added to the number 5, gives for a sum the number a. Let x denote the re- quired number; then, by the conditions of the problem, h-{-x=a; whence x = a—b. This formula will give the value of x corresponding to any assigned values of a and b. For example, if a=7 and 5=4> then tc = 7— 4 = 3, a result which satisfies the conditions. But suppose that a=6 and 6=8, then x=5—8=—B. We thus obtain for x a negative value. How is it to be in- terpreted ? By referring to the problem, we see that it now reads thus : What number must be added to 8 in order that the sum may be 5? It is obvious that if the word added and the word sum are to retain their arithmetical meanings, the proposed problem is impossible. Nevertheless, if in the equation 8 4- a; = 5 we substitute for -fee its value —3, it becomes 8-3 = 5, an identical equation; that is, 8 diminished by 3 is equal to 5, or 5 may be regarded as the algebraic sum of 8 and —3. The negative result, x=—3, indicates that the problem, in a 116 ALGEBRA. strictly arithmetical sense, is impossible ; but, taking this value of X with a contrary sign, we see that it satisfies the enunciation when modified as follows: What number must be svhtrax^ted from 8 in order that the difference may be 6 ? The second enunciation differs from the first only in this, that we put sub- iract for add, and difference for sum. If we wish to solve this new equation directly, we shall have whence x=S—5, or 3. 163. For another example, take Problem 50, page 98. The age of the father being represented by a, and that of the son by 5, then will represent the number of years before the age of the father will be n times that of the son. Thus, suppose a =54, 6=9, and ?2 =4; ^, 54-86 18 ^ then x= — ^ — = — = 6. o o This value of x satisfies the conditions understood arithmet- ically ; for if the father was 54 years old, and the son 9 years, then in 6 years more the age of the father will be 60 and the son 15 ; and we see that 60 is 4 times 15. But suppose a =45, 6=15, and 7i=4; ^, 45-60 -15 ^ then x= — - — =——-=— 5. o o Here again we obtain a negative result How are we to in- terpret it ? By referring to the problem, we see that the age of the son is already more than one fourth that of the father, so that the time required is already past by five years. The problem, if taken in a strictly arithmetical meaning, is impossible. But let us modify the enunciation as follows: The age of the father is 45 years ; the son's age is 15 years; how many years since the age of the father was four times that of his son ? DISCUSSION OF PROBLEMS. 11? The equation corresponding to this new enunciation is whence 60—4:x=4^5—x; and a: ==5, a result which satisfies the modified problem taken in its arith- metical sense. From this discussion we derive the following general prin- ciples : l5^. A negative result found for the unknown quantity in a proh lem of the first degree indicates that the problem is imjpossiblej if understood in its strict arithmetical sense. 2d. TJiis negative value, taken with a contrary sign, may he re- garded as the answer to a problem whose enunciation only differs from that of the proposed problem in this, that certain quantities which were added should have been subtracted, and vice versa. 164. In what case would the value of the unknown quantity? in Prob. 20, page 94, be negative? Ans. When ri>m. Thus, let m = 20, n = 26, and a = 60 miles; To interpret this result, observe that it is impossible that the second train, which moves the slowest, should overtake the first. At the time of starting, the distance between them was 60 miles, and each subsequent hour the distance increases. If, however, we suppose the two trains to have been moving uniformly along an endless road, it is obvious that at some former time they must have been together. This negative result indicates that the problem is impossible if understood in its strict arithmetical sense. But if the prob- lem had been stated thus : Two trains of cars, 60 miles apart, are moving in the same direction, the forward one 25 miles per hour, the other 20. How long since they were together ? The problem would have furnished the equation 25ic=20:r + 60; whenoe a:=-fl2. 118 ALGEBRA. If we wish to include both of these cases in the same enun- ciation, the question should be, Required the time of Hmr heiiig together ^iQdivmg it uncertain whether the time ^2^ past or future^ EXAMPLES. 1. What number is that whose fourth part exceeds its third; part by 16? Ans. -192. How should the enunciation be modified in order that the result may be positive? 2. The sum of two numbers is 2, and their difference 8. What are those numbers? Ans. —3 and +5. How should the enunciation be modified in order that both results may be positive ? 3. What fraction is that from the numerator of which if 4 be subtracted the value is one half, but if 7 be subtracted from the denominator its value is one fifth ? , —^ . . ^'^^ How should the enunciation be modified in order that the problem may be possible in its arithmetical sense? 4. Find two numbers whose difference is 6, such that four times the less may exceed five times the greater by 12. Ans. -42 and -36. Change the enunciation of the problem so that these num- bers, taken with the contrary sign, may be the answers to the modified problem. 165. 8d. We may obtain for the unknown quantity valms of the form of zero, or -7. In what case would the value of the unknown quantity in Prob. 20, page 94, become zero, and what would this value signify ? Ans. This value becomes zero when a = 0, which signifies that the two trains are together at the outset. In what case would the value of the unknown quantity in Prob. 50, page 98, become zero, and what would this value signify ? DISCUSSION OF PKOBLEMS. 119 Ans. When a=nb, which signifies that the age of the father is now n times that of the son. In what case would the values of the unknown quantities in Prob. 88, page 96, become zero, and what would these values signify ? When a problem gives zero for the value of the unknown quantity, this value is sometimes applicable to the problem, and sometimes it indicates an impossibility in the proposed question. 166. 4th. We may obtain for the unknown quantity values of the form of -^. In what case does the value of the unknown quantity in Prob. 20, page 94, reduce to -^, and how shall we interpret this result? Ans. When m = n. On referring to the enunciation of the problem, we see that it is absolutely impossible to satisfy it ; that is, there can be no point of meeting ; for the two trains, being separated by the dis- tance a,. and moving equally fast, will always continue at the same distance from each other. The result ^ may then be re- garded as indicating an impossibility. The symbol q is sometimes employed to represent infinity^ and for the following reason : If the denominator of a fraction is made to diminish^ while the numerator remains unchanged, the value of the fraction must increase. For example, let m— 72 = 0.01 ; then x=-^=4; = 100a. m — n .01 Let m-n = 0.0001] then a:=-^ = -^ = 10,000«. m^n .0001 Hence, if the difference in the rates of motion is not zero, the 120 ALGEBRA. two trains must meet, and the time will become greater and greater as this difference is diminished. If, then, we suppose this difference to be less than any assignable quantity^ the time represented by will be greater than any assignable quantity, 7Yi-^n Ilence we infer that every expression of the form j found for the unknown quantity indicates the impossibility of satis- fying the problem, at least in Jiriiie numbers. In what case would the value of the unknown quantity in Prob. 10, page 92, reduce to the form ^r? and how shall wc in- terpret this result? 167. The symbol 0, called zero, is sometimes used to denote the absence of value, and sometimes to denote a quantity less than any assignable value. The symbol oo , called infinity, is used to denote a quantify greater than any assignable value. A line produced beyond any assignable limit is said to be of infinite length ; and time ex- tended beyond any assignable limit is called infinite duration. We have seen that when the denominator of the fraction —3- becomes less than any assignable quantity, the value of the fraction becomes greater than auy assignable quantity. Hence we conclude that ^ = co\ that is, a finite quantity divided by tero is an expression for in- finity. Also, if the denominator of a fraction be made to increase while the numerator remains unchanged, the value of the frac- tion must diminish; and when the denominator becomes greater than any assignable quantity, the value of the fraction must be- come less than any assignable quantity. Hence we conclude that ci 00 that is, a finite quantity divided by infiriiiy is an expression fyr zero mSCUSSION OF PKOBLEMS. 121 168. 5th. We may obtain for the unknown quantity values of the form of ^. In what case does the value of the unknown quantity in Prob. 20, page 94, reduce to ^, and how shall we interpret this lesult? ^?75. When a = 0, and m — n. To interpret this result, let us recur to the enunciation, and observe that, since a is zero, both trains start from the same point; and since they both travel at the same rate, they will always remain together ; and, therefore, the required point of meeting will be any where in the road traveled over. The problem, then, is entirely indeterminate^ or admits of an infinite number of solutions ; and the expression ^ may represent any finite quantity. We infer, therefore, that an expression of the form of q found for the unknown quantity generally indicates that it may have any value whatever. In some cases, however, this value is subject to limitations. In what case would the values of the unknown quantities in Prob. 44, page 97, reduce to ^, and how would they satisfy the conditions of the problem? Ans. When a=b=c, "which indicates that the coins are all of the same value. B might therefore be paid in either kind of coin ; but there is a "limitation, viz., that the value of the coins must be one dollar. In what case do the values of the unknown quantities in Prob. 38, page 96, reduce to -, and how shall we interpret his result? 169. The expression ^ may be conceived to result from a fraction whose numerator and denominator both diminish si- multaneously, but in such a manner as to preserve the same relative value. If both numerator and denominator of a frac- tion are divided by the same quantity, its value remains un- F 122 ALGEBRA. changed. Hence, if j represent any fraction, we may conceive both numerator and denominator to be divided by 10, 100, 1000, etc., until each becomes less than any assignable quanti- ty, or 0. The fraction then reduces to the form of J=^, but the ralue of the fraction has throughout remained unchanged. For example, we may suppose the numerator to represent the circumference of a circle, and the denominator to represent its diameter. The value of the fraction in this case is known to be 3.1416. If now we suppose the circle to diminish until it becomes a mere point, the circumference and diameter both become zero, but the value of the fraction has throughout re- mained the same. Hence, in this case, we have 5 = 3.1416. Again, suppose the numerator to represent the area of a cir- cle, and the denominator the area of the circumscribed square; then the value of the fraction becomes .7854. But this value remains unchanged, although the circle may be supposed to diminish until it becomes a mere point Hence, in this case, ^° '^"^^ g=.7854. Hence we conclude that the symbol ^ may represent any finite quantity. So, also, we may conceive both numerator and denominator of a fraction to be multiplied by 10, 100, 1000, etc., until each becomes greater than any assignable quantity; the fraction then reduces to the form of ^. Hence we conclude that the 00 ^ symbol — may also rej^resent any finite quantity, INEQUALITIES. 170. An inequality is an expression denoting that one quan- tity is greater or less than another. Thus Zx > 2ah denotes that three times the quantity x is greater than twice the prod« act of the quantities a and h. INEQUALITIES. 123 171. In treating of inequalities, the terms greater and less must be understood in their algebraic sense ; that is, a negative quantity standing alone is regarded as less than zero ; and of two negative quantities, that which is numerically the greatest is considered as the least; for if from the same number we sub- tract successively numbers larger and larger, the remainders must continually diminish. Take any number, 5 for example, and from it subtract successively 1, 2, 8, 4, 5, 6, 7, 8, 9, etc., we obtain 5-1, 5-2, 5-3, 5-4, 5-5, 5-6, 5-7, 5-8, 5-9, etc., or, reducing, we have 4, 3, 2, 1, 0, -1, -2, -8, -4, etc. Hence we see that —1 should be regarded as less than zero; —2 less than —1; —3 less than —2, etc. 172. Two inequalities are said to subsist in the same sense when the greater quantity stands at the left in both, or at the right in both ; and in a contrary sense when the greater quanti- ty stands at the right in one and at the left in the other. Thus 9>7 and 7>6, or 5<8 and 3<4, are inequalities which sub- sist in the same sense; but the inequalities 10>6 and 3<7 subsist in a contrary sense. 173. Properties of Inequalities. — 1st. If the same quantity he added to or subtracted from each memher of ayi inequality, the re suiting inequality will always subsist in the same sense. Thus, 8 > 3. Adding 5 to each member, we have 8-f-5>8 + 5, and subtracting 5 from each member, we have 8-5 > 3-5. Again, take the inequality -3<-2. Adding 6 to each member, we have ^3-h6<-2 + 6, or 3<4; and subtracting 6 from each member, _3-6<-2-6, or -9<-a 124 ALGEBRA. 174. Hence we conclude that we may transpose a term from one member of an inequality to the other, provided we changa its sign. Thus, suppose ct'^h''>W-^2a'. Adding 2a2 to each member of the inequality, it becomes Subtracting U^ from each member, we have a2+2a2>3Z.2_^2^ or 3a2>262. ( 175. 2d. If we add together the corresponding members of two or more inequalities which subsist in the same sense^ the residiing in- equality will always subsist in the same sense. Thus, b>4: 4>2 7>3 Adding, we obtain 16 > 9. 176. 8d. If one inequality be subtracted from another which sufh sists in the same sense^ the result will not always be an inequality subsisting in the same sense. Take the two inequalities 4<7 and 2<3 ' Subtracting, we have 4— 2<7— 3, or 2<4, where the result is an inequality subsisting in the same sense. But take 9<10 and 6< 8 Subtracting, we have 9 — 6 > 10— 8, or 3>2, where the result is an inequality subsisting in the contrary sense. We should therefore avoid as much as possible the use of this transformation, or, when we employ it, determine in what sense the resulting inequality subsists. 177. 4th. Ifive multiply or divide each member of an inequality by the same positive quantity, the resulting inequality ivill subsist in the same sense. m INEQUALITIES. Thus, if a— 5, then — may—mb. and a b 125 Hence an inequality may be cleared of fractions. Thus, sup- pose we have a^—h^ &—d'^ Multiplying each member by 6ac?, it becomes 3a(a2-52)>2cZ(c2-cZ2). 178. 5th. If loe multiply or divide each member of an inequality by tJie same negative number^ the resulting inequality will subsist in the coyitrary sense. Take, for example, 8>7. Multiplying each member by —3, we have the opposite in- equality — 24<-21. So, also, 15>12. Dividing each member by — 3, we have -5<-4. Therefore, if we multiply or divide the two members of an inequality by an algebraic quantity, it is necessary to ascertain whether the multiplier or divisor is negative, for in this case the resulting inequality subsists in a contrary sense. 179. 6th. If the signs of all the terms of an inequality be changed, the sign of inequality must be reversed. For to change all the signs is equivalent to multiplying each member of the inequality by —1. 180. Reduction of Inequalities. — The principles now establish- ed enable us to reduce an inequality so that the unknown quan- tity may stand alone as one member of the inequality. The 126 ALGEBRA. other member will then denote one limit of the unknown quantity. EXAMPLES. 1. Find a limit of x in the inequality 7 5cc 95 ^ Multiplying each member by 12, we have U-15x<96-24:x. Transposing, 9a;<81. Dividing, x<9. 2. 2x4- f-8<6. Ans. x<6 o 8. 8a^-2>^-t Ans.x>'^ 2 5 /y> rv* rf* /y» /y» ^- 2 + 3+4+6 + 12 -^>^- ^ cc cc— 3 ^ 4cc+l . . 5. 7j ^-^^<^ E — • -47W. cc>4. o 15 5 ^+4a;-8>8 6. Given <( k i k /" ^o find the limits of x, 6a:+5^^<18 . (a;>2. Ans. \ U<3. 7. A man, being asked how many dollars he gave for his watch, replied, If you multiply the price by 4, and to the prod- uct add 60, the sum will exceed 256; but if you multiply the price by 8, and from the product subtract 40, the remainder •will be less than 118. Eequired the price of the watch. 8. What number is that whose half and third part added together are less than 105 ; but its half diminished by its fifth part is greater than %Z ? 9. The double of a number diminished by 6 is greater than 22, and triple the number diminished by 6 is less than double the number increased by 10. Required the number. INVOLUTION. 127 CHAPTER XI. INVOLUTION. 181. A power of a quantity is the product obtained by tak- ing that quantity any number of times as a factor. Thus the first power of 3 is 8 ; the second power of 3 is 3 X 8, or 9 ; the fourth power of 3 is 8x3x8x8, or 81, etc. Involution is the process of raising a quantity to any power. 182. A power is indicated by means of an exponent. The exponent is a number or letter written a little above a quantity to the right, and shows how many times that quantity is taken as a factor. Thus the first power of a is a\ where the exponent is 1, which, however, is commonly omitted. The second power of a is a x a, or a^, where the exponent 2 denotes that a is taken twice as a factor to produce the power aa. The third power of a is axaxa^ or a^, where the exponent 3 denotes that a is taken three times as a factor to produce the power aaa. The fourth power of a is axaxaxa^ or a*. Also the 72th power of a is axaxaxa^ etc., or a repeated as a factor n times, and is written a". The second power is commonly called the square^ and the third power the cube. 183. Exponents may be applied to polynomials as well as to monomials. Thus {a-\-h-\-cY is the same as (a+2>+c)x(a+64-c)x(rt + Z) + c), or the third power of the entire expression a'\-h~\-c. 128 ALGEBRA. Powers of Monomials. 184. Let it' be required to find the third power or cube of According to the rule for multiplication, we have {2d'b''f=2aW X 2aW x 2aW = 2 x 2 x2a^a^aWb^b^=.8a^b\ In a similar manner any monomial may be raised to any power. Hence, to raise a monomial to any power, we have the fol- lowing RULE. Baise the numerical coefficient to the required power^ and multi- ply the exponent of each of the letters by the exponent of the re- quired power, 185. Sign of the Power. — With respect to the signs, it is ob- vious from the rules for multiplication that if the given mono* mial be positive, all of its powers are positive ; but if the mo- nomial be negative, its square is positive, its cube negative, its fourth power positive, and so on. Thus —ax—a=-{-a^^ — ax —ax —a=—a^j — ax—ax—ax—a=-\-a*, — ax —ax —ax —ax —a=—a^, etc. In general, any even power of a negative quantity is positive^ and every odd power negative ; but all powers of a positive quaU' iity are positive. EXAMPLES. 1. Find the square of IWbcd'^. Ans. \21a%''cH\ 2. Find the square of —ISx^yz^. ^ 8. Find the cube of 1ab'^x\ 4. Find the cube of —Sxy'^z^. 5. Find the fourth power of 4aiV. 6. Find the fourth power of —baWx. 7. Find the fifth power of 2aWx\ 8. Find the fifth power of -SaZ^V. 9. Find the sixth power of Zab^x^, INVOLUTION. 129 ^ lOPFind the sixth power of -1d?h^x'', U "( ^ 11. Find the seventh power of 2a^x^y, 12. Find the mth power of abV. cC' 186. Powers of Fractions. — Let it be required to find the third power of %^. "l 6c From the rule for the multiplication of fractions, we have /2ahy_2ah^ 2ab^ 2a^_8a^ \Sc ) ~ Sc ^ Sc ^ 3c ~27c^' In a similar manner any fraction may be raised to any power. Hence, to raise a fraction to any power, we have the following RULE. Baise both numerator and deno7ninator to the required power. EXAMPLES. X.^j.' . ,. r.Sab-'c^ . 9a25V ^Otl. i^md the square of -= — 5. Ans. ,^ ., , . Jj 2. 1^ md the square of — j^— . A71S. H 3. Find the cube of Z:|^ omn 4:bc ' ' IQh^c^' ^ ax ^^ 4. Find the cube of 5. Find the fourth power of i^. .JiAi^' 6. Find the fourth power of - ^-^. M£^\^ 7. Find the fifth power of - ^^'. _ ..A5^£-^IiC:l 8. Find the fifth power of ^^^' —^ ^ 9. Find the sixth power of - 5^!^. > ^"i^tki ^ 2mn^ -~- F 2 ^- " 180 ALGEBRA. 187. Negative Exponents. — The rule of Art. 184, for raising a monomial to any power, holds true when the exponents of any of the letters are negative, and also when the exponent of the re- quired power is negative. Let it be required to find the square of a'^. This expres- sion may be written -3, which, raised to the second power, be- 1 " comes -g, or a~®, the same result as would be obtained by multiplying the exponent —3 by 2. Also, let it be required to find that power of 2am^ whose exponent is —3. The expression {2am'^)~^ may be written -^ — ^, which 1 equals 3 g . Transferring the factors to the numerator, we A a m have 2~^a'^%"®, or \a-hn~^. EXAMPLES. Find the value of each of the following expressions. 1. {Za'h-y. Ans. Mh-\ 2. {1a-%''c-''xf. Ans. AQa-'h^c-^x^. 3. (3a52x-V-2)-2. Ans. la-^-^xY^ 4. (— 4a^x~y)~^ Ans. -^~*x^y~*. 5. {-6ab-'x-^y. ^juiioJ^yC'^ 6. (-3a^x-V)-3. _-^,e;L-'o^^r' 7. {ia-'bx-*)-\ ^^ l'*lr'' 8. {^Za-nh-xy. ±IL^Z<'*' 9. (-4a-3i2x-r*. ^\Xi,^^.o. 10. (-2a5-Vic-*y)*. -i^^ ^ C < r 188. Powers of Polynomials. — A polynomial may be raised to any power by the process of continued multiplication. If the quantity be multiplied by itself, the product will be the second power; if the second power be multiplied by the orig- inal quantity, the product will be the third power, and so on. Hence we have the following INVOLUTION". 131 RULE. %J^iL^^^-^^ Midi^-^^^-qiia/)^^^ h^-4(sdf-^'mMl-tt has-been^ iaken tos a factor as many times as there are units in the exponent of the required jpower. EXAMPLES. 1. Find the square of 2a+852. Ans. ^a^ ^VlalP' -^-W , 2. Find the square of a-\-m—n. 3. Find the cube of 1a^-^Za-\. 4. Find the cube of a-{-\. 6. Find the cube of a^lh-\-Zx. 6. Find the fourth power of a—h. 7. Find the fourth power of 2« — 85. 8. Find the fourth power of (]?-\-}?. 9. Find the fifth power of a—h. 10. Find the square of "^. Ans. ^ rfl^^tK ^ ay— OX a^y^ —zaoxy -\- oV 11. Find the cube of ?^±5?. 12. Find the cube of m—n a^-h a-W 189. Square of a Polynomial. — We have seen, Art. QQ^ that the square of a binomial may be formed without the labor of actual multiplication. The same principle- may be extended to polynomials of any number of terms. By actual multipli- cation, we find the square of a+J+c to be a2-|-52_^c2-f2a&+2acH-2&c; that is, the square of a trinomial consists of the square of each term J together with twice the product of all the terms multiplied to- gether two and two. In the same manner we find the square of a 4-5+ c+c? to be a2 4.i3_|_c2_|.(^2_^_2a&4-2ac-f2ac?+25c+2M+2cc^; that is, the square of any polynomial consists of the square of each term, together with twice the sum of the products of all the terms multiplied together two and two. 132 ALGEBRA. EXAMPLES. 1. Find the square of a-\-h-\-c-\-d-{-x, 2. Find the square of a—h-\-c. 3. Find the square of 1 + 2a; +3x2. 4. Find the square of 1— a: +ir2_x^ / 5. Find the square of a — 2h-\-Sab—m. 6. Find the square of l — Sx-\-Sx'^—x^. 7. Find the square of a— 2?>+3c— 4c?. In Chapter XVITI. will be given a method by which any power of a binomial may be obtained without the labor of multiplication. i r ^ EVOLUTION. 133 CHAPTEK XII EVOLUTION. 190. A root of a quantity is one of the equal factors which, multiplied together, will produce that quantity. If a quantity be resolved into two equal factors, one of them is called the square root. If a quantity be resolved into three equal factors, one of them is called the cube root. If a quantity be resolved into four equal factors, one of them is called the fourth root., and so on. 191. Evolution is the process of extracting any root of a given quantity. Evolution is indicated by the radical sign ^f. Thus, -y/a denotes the square root of a. Ya denotes the cube root of a. yH denotes the nth root of a. 192. Surds. — When a root of an algebraic quantity which is required can not be exactly obtained, it is called an irration- al or surd quantity. Thus, v/o^ is called a surd. VS is also a surd, because the square root of 3 can not be expressed in numbers with perfect exactness. A rational quantity is one which can be expressed in finite terms, and without any radical sign ; as, a, ba^^ etc. 193. An imaginary root is one which can not be extracted on account of the sign of the given quantity. Thus the square root of —4 is impossible, because no quantity raised to an even power can produce a negative result. A root which is not imaginary is said to be real. 134 ALGEBRA. Roots of Monomials. 194. According to Art. 184, in order to raise a monomial to any power, we raise the numerical coefficient to the required power, and multiply the exponent of each of the letters by the exponent of the power required. Hence, conversely, to ex- tract any root of a monomial, we extract the root of the nu- merical coefficient, and divide the exponent of each letter by the index of the required root Thus the cube root of 64a^6^ is 4a^6. 195. Sign of the Root. — We have seen. Art. 186, that all pow« ers of a positive quantity are positive ; but the even powers of a negative quantity are positive, while the odd powers are negative. Thus +a, when raised to different powers in succession, will give j^a, +a\ H-a^, +a*, +«', +< +a\ etc. and —a, in like manner, will give —a, H-a^, — a^, -|-«*j — o^^ +tt^, — a^, etc. Hence it appears that if the root to be extracted be express- ed by an odd number, the sign of the root will be the same as the sign of the proposed quantity. Thus, \/ — a^— _a; and If the root to be extracted be expressed by an even number, and the quantity proposed be positive^ the root may be either positive or negative. Thus, y/o}= ±a. If the root proposed to be extracted be expressed by an even number, and the sign of the proposed quantity be negative, the root can not be extracted, because no quantity raised to an even power can produce a negative result. 196. Hence, to extract any root of a monomial, we have the following RULE. \st. Extract the required root of the numei^ical coefficient. 2d. Divide the exponent of each literal factor hy the index of Oie required root. EVOLUTION. 135 Sd. Every even root of a positive quantity must have the double sign ± , and every odd root of any quantity must have the same sign as that quantity. From Art. 186, it is obvious that to extract any root of a fraction, we must divide the root of the numerator by the root of the denominator. Thus, 3 /o^ « T 3 / ~c? a EXAMPLES. 1. Find the square root of 64a^Z)*. Ans. :i^8a^b\ 2. Find the square root of WQa'^b^c^x^. Ans. ±14:ab'^c^x*. 3. Find the square root of 226a^H^x^ 4. Find the cube root of GAa^^b^x^. Ans. 4:ahV. 5. Find the cube root of —12ba^x^y^. Ans. —bax^y'^. 6. Find the cube root of -343a^^V2. 7. Find the fourth root of 81a*^>l Ans. ±Zab\ 8. Find the fourth root of 256a*Z>iVl 9. Find the fifth root of -32a^Z>^V^ Ans. -2abV. 10. -bind the square root oi ^ ^ ^ , . Ans. ±- — r-^. Y 11. Find the square root of fff^'f' i^JLjL 12. Fmd the cube root of ^-^r-s- -4^- fi — 5. bm'^x^ zmx^ 13. Find the cube root of — 216cV • ~ ~'^^f 14. Find the fourth root of -ktt^-^- , r 81^ V^ '^^V'^ 15. Find the square root of 64a-2Z>-*x*. Ans. ±.'8a-^)-'^x\ 16. Find the cube root of ^hVla-^-'^x^. 17. Find the fourth root of 256a-'^5- V. 18. Find the fifth root of •^Z1a-^%-^H\ Ans. -2a-^b-^x 19. Find the square root of {a — byx^. Ans. dr:{a—'b)x^ 20. Find the cube root of {a-^bylx-^yf. 186 ALGEBRA. Square Root of Polynomials. 197. In order to discover a rule for extracting the square root of a polynomial, let us consider the square of a+Z>, which is a^-\-2ah-{-h'^. If we arrange the terms of the square accord- ing to the dimensions of one^etter, a, the first term will be the square of the first term of the root; and since, in the present case, the first term of the square is a^, the first term of the root must be a. Having found the first term of the root, we must consider the rest of the square, namely, 2db-\-h'^^ to see how we can de- rive from it the second term of the root. Now this remainder may be put under the form {2a-\-h)h; whence it appears that we shall find the second term of the root if we divide the re- mainder by 2a-\-h. The first part of this divisor, 2a, is double of the first term already determined ; the second part, 5, is yet unknown, and it is necessary at present to leave its place empty. Nevertheless, we may commence the division, employing only the term 2a ; but as soon as the quotient is found, which in the present case is 5, we must put it in the vacant place, and thus render the divisor complete. The whole process, therefore, may be represented as fol- lows: a2-f2a&+&2(a+6 a2 2a+h)2ab-{-h'' 2ah-\-h'' If the square contained additional terms, we might continue the process in a similar manner. We may represent the first two terms of the root, a+Z;, by a single letter, m, and the re- maining terms by c. The square of m+c will be m'^-\-2mc-\-c\ The square of the first two terms has already been subtracted from the given polynomial. If we divide the remainder by 2m as a partial divisor, we shall obtain c, which we place in the root, and also at the right of 2m, to complete the divisor. We then multiply the complete divisor by c, and subtract the y D^OLUTIOK 137 product from the dividend, and thus we continue until all the terms of the root have been obtained. 198. Hence we derive the following RULE. 1st. Arrange the terms according to the powers of some one let- ter ; take the square root of the first term for the first term of the required root, and subtract its square from the given polynomial, 2d. Divide the first term of the remainder by twice the root al- ready found, and annex the residt both to the root and the divisor. Multiply the divisor thus completed by the last term of the root, and subtract the product from the last remainder. 3d. Double the entire root already found for a second divisor. Divide the first term of the last remainder by the first term of the second divisor for the third term of the root, and annex the result both to the root and to the second divisor, and proceed as before until all the terms of the root have been obtained. If the given polynomial be an exact square, we shall at last find a remainder equal to zero. EXAMPLES. 1. Extract the square root of a^ —2a^x-\-3aV —^ax^ -{-x^ a* — 2a^x + 2>aV — 2ax^ + x* ( a^ — ax + x^ 2a'^-ax) -2a^x-{-SaV — 2a^x-\- c^x? 2a^-2ax+x')2aV-2ax^-\-x'- 2a^x^-2ax^^-x*' For verification, multiply the root a^^ax-\-x^ by itself, and we shall obtain the original polynomial. 2. Extract the square root of a^-^2ab-\-2ac^b'^^2bc-^c^, 3. Extract the square root of 10x^-10a;3-12a^H5x-2 + 9cc6-2a;+l. 4. Extract the square root of 8ax=»+4a2a;2_|.4ic* + 16Z>2ic2-hl6Z>*-hl6a52x. Ans. 2x'^-^2ax+4.b\ 138 ALGEBRA. 5. Extract the square root of -\I>a'b'' + a« - 6a'b - 20a'b^ +b^-^ Ida'^b' - 6ab\ ""6. Extract the square root of Sa^/^-ha*— 4a^6+4Z;*. ^7. Extract the square root of 4:X^-\-12x^-\-5x^—6x-\-l. Am. 2x2 4-3ic-l. — 8. Extract the square root of 4x* - 12aa? + Iba^x^-^^o^x -h 16a*. Ans. 2^2— 3aa:+4a2. 9. Extract the square root of 25x* - 30ax3 + ^^a?x'' - 24:a^x + 1 6a*. Ans. 5x^— 3ax+4a*'*. x^ 4 10. Extract the square root of x'^—x^-\--j-\-4:X—2-\- -^. ^ X Ans. x^—^-\ — . 2 X 199. When a Trinomial is a Perfect Square. — The square of a-\-b is a^-\-2ab-{-b'^^ and the square of a — b is a^— 2aZ>4-&^. Hence the square root of a'^±2ab-\-b'^ is a±b; that is, a trino- mial is a perfect square when two of its terms are squares, and the third is the double product of the roots of these squares. Whenever, therefore, we meet with a quantity of this de- scription, we may know that its square root is a binomial ; and the root may be found by extracting the roots of the two terms which are complete squares, and connecting them by the sigp of the other term. EXAMPT>ES. 1. Find the square root of 4a2^12a/; + 9/>2. Ans, 2a + Sb. 2. Find the square root of 9a^—24ab-\-16b\ 3. Find the square root of 9a* - 30^/'/^ + 25o2Z,^ 4. Find the square root of 4a2-f 14aZ>+16Z>2, i/j^ossible. No algebraic binomial can be a iperfvci square^ for the square of a monomial is a monomial, and the square of a binomial necessarily consists of three distinct terms. EVOLUTION. 139 Square Root of Numbers. 200. The preceding rule is applicable to the extraction of the square root of numbers; for every number may be re- garded as an algebraic polynomial, or as composed of a cer- tain number of units, tens, hundreds, etc. Thus 529 is equivalent to 500 + 20 + 9; also, 841 " 800+40 + 1. If, then, 841 is the square of a number composed of tens and units, it must contain the square of the tens^ plus twice the product of the tens by the units, plus the square of the units. But these three terms are blended together in 841, and hence arises the peculiar difficulty in determining its root. The following principles will, however, enable us to separate these terms, and thus detect the root. 201. 1st. For every two figures of the square there will he one figure in the root, and also on([for any odd figure.\ Thus the square of 1 is 1 10 " 1,00 100 " 1,00,00 " 1000 " 1,00,00,00 the square u of 1 9 is 1 81 u u 999 u 9§.01 99,8?),01' The smallest number consisting of two figures is 10, and its square is the smallest number of three figures. The smallest number of three figures is 100, and its square is the smallest number of five figures, and so on. Therefore the square root of every number composed of one or two figures will contain one figure ; the square root of every number composed of three or four figures will contain two figures ; of a number from five to six figures will contain three figures, and so on. Hence, if we divide the number into periods of two figures, commencing at the units' place, the number of periods will indicate the number of figures in the square root. 202. 2c?. Tlie first figure of the root will be the square root of ilce greatest square number contained in the first period on the left. 140 ALGEBRA.. For the square of tens can give no significant figure in tbe first right hand period, the square of hundreds can give no fig- ure in the first two periods on the right, and the square of the highest figure in the root can give no figure except in the first period on the left. Let it be required to extract the square root of 5329. This number contains two periods, indicating that there 53,29 (70 + 3, the root will be two places in the 49 00 root. Let a + h denote the 140-f3)~429 root, where a is the value of 4 29 the figure in the tens' place, and h of that in the units' place. Then a must be the great- est multiple of 10, which has its square less than 5300 ; this is found to be 70. Subtract a^, that is the square of 70, from the given number, and the remainder is 429, which must be equal to {2a-\-h)h. Divide this remainder bj 2a, that is by 140, and the quotient is 3, which is the value of h. Com- pleting the divisor, we have 2a + J = 143; whence (2a-|-i)Z), that is 143 X 3, or 429, is the quantity to be subtracted ; and as there is now no remainder, we conclude that 70+3, or 73, is the required square root. For the sake of brevity, the ciphers may be omitted, pro- vided we retain the proper local values of the figures. If the root consists of three places of figures, let a represent the hundreds, and h the tens; then, having obtained a and h as before, let the hundreds and tens together be considered as a new value of a, and find a new value of h for the units. Kequired the square root of 568516. Having found 75, the square root of 56,85,16(754 ohe greatest square number contained in 49 the first two periods, we bring down the 145) 785 last period, and have 6016 for a new div- 725 idend. We then take 2a, or 150, for a 1504) 6016 partial divisor, whence we obtain i=4 6016 for the last figure of the root. The en- tire root is therefore 754. EVOLUTION. 141 203. Hence, for the extraction of the square root of num- bers, we derive the following RULE. 1st. /Separate the given number into periods of two figures each^ beginning from the units'' place. 2d. Find the greatest number luhose square is contained in the left-hand period ; this is the first figure of the required root. Sub- tract its square from the first period^ and to the remainder bring down the second period for a dividend. 3o?. Double the root already found for a divisor^ and find how mavy times it is contained in the dividend^ exclusive of its right- hand figure ; annex the 7'esult both to the root a7id the divisor. 4:th. Multiply the divisor thus increased by the last figure of the root^ subtract the product from the dividend^ and to the remainder bring down the next period for a new dividend. 6th. Double the whole root now found for a new divisor^ and proceed as before^ continuing the operation until all the periods are brought down. In applying the preceding rule, it may happen that the prod- uct of the complete divisor by. the last figure of the root is greater than the dividend. This indicates that the last figure of the root was taken too large, and this happens because the divisor is at first incomplete — that is, is too small. In such a case, we must diminish the last figure of the root by unity until we obtain a product which is not greater than the dividend. EXAMPLES. 1. What is the square root of 294849 ? Ans. 543. 2. What is the square root of 840889 ? 3. What is the square root of 1142761 ? 4. What is the square root of 32239684? 5. What is the square root of 72777961 ? 6. What is the square root of 3518743761? 204. Square Root of Fractions. — We have seen that the root of a fraction is equal to the root of its numerator di- 142 ALGEBRA. vided by the root of its denominator Hence the square root The number 5.29 may be written -— -:, and its square root oo 1 OfiCOA is — , or 2.3. So, also, 18.6624 may be written , and 432 its square root is --- , or 4.32. That is, the square root of a decimal fraction, or of a whole number followed by a decimal fraction, may be found in the same manner as that of a whole number, if we divide it into periods commencing with the deci- mal point. In the extraction of the square root of an integer, if there is still a remainder after we have obtained the units' figure of the root, it indicates that the proposed number has not an exact square root. We may, if we please, proceed with the approxi- mation to any desired extent by supposing a decimal point at the end of the proposed number, and annexing any even num- ber of ciphers, and continuing the operation. We thus obtain a decimal part to be added to the integral part already found. So, also, if a decimal number has no exact square root, we may annex ciphers and proceed with the approximation to any desired extent. EXAMPLES. 1. What IS the square root of ^r^^-^ c Ans. ^^. Aovv bo 2. What is the square root of ^^^^,,,, ? ^ 1^0569 8. What is the square root of " 67551961 4. What is the square root of 9.878449 ? 6. What is the square root of 58.614336 ? 6. What is the square root of .558009 ? 7. What is the square root of .03478225 ? Find the square roots of the following numbers to five deci mal places. EVOLUTION. 8. Of 2. Ans. 1.41421. 11. Of 4| 9. Of 10. Arts. 8.16227. 12. Of tV. 0. Of 9.1. 13. Of A 143 Cube Boot of a Polynomial. 205. We already know that the cube of a-{-h is a^ + 8a^5 ■\-2>ah'^-\-h^. If, then, the cuhe ^nqyq given, and we were re q^uired to find its root^ it might be done by the following method. When the terms are arranged according to the powers of one letter, a, we at once know, from the first term, a^, that a must be one term of the root. If, then, we subtract its cube from the proposed polynomial, we obtain the remainder dto^h + ^aW '■)- ^^, which must furnish the second term of the root. Now this remainder may be put under the form (3a2+3aZ^ + ^/2)Z,; whence it appears that we shall find the second term of the root if we divide the remainder by Za'^-\-^ah-\-h'^. But, as this second term is supposed to be unknown, the divisor can not be completed. Nevertheless, we know the first term, Sa^, that is thrice the square of the first term already found, and by means of this we can find the other part, h; viz., by dividing the first term of the remainder by Sa^. We then complete the divisor by adding to it ^ab-^-h^. If this complete divisor be multi- plied by 6, it will give the last three terms of the power. Let it be required to find the cube root of 8a^+36a2^-l-54aZ^2 + 276^. 8aH36a2Z>+54a&2_|_27Z'3(2a+3& 8a3 36fl^Z> + 54q/>H27^^^ Having found the first term of the root, 2a, and subtracted its cube, we divide the first term of the remainder, 36a2Z>, by three times the square of 2a, that is 12a^, and we obtain ^b for the second term of the root. We then complete the divi- sor by adding to it three times the product of the two terms of 144 ALGEBRA. the root, which is 18a5, together with the square of the last term 3^, which is 96^ Multiplying then the complete divisoi by 3^, and subtracting the product from the last remainder, nothing is left. Hence the required cube root is 2a-{-Sb. This result may be easily verified by multiplication. 206. If the root contains three terms, as a4-6 + c, we may ^ut a-\-b = m. Then {a-^h-\-cy = (m+cy=m^-{-Sm'^c-{-Smc'^-\-c^. If we proceed as in the last example, we shall find a-\-h, and we subtract its cube from the given polynomial. There will then remain 3m'^c-\'Smc^-\-c^, which may be written (3m2 + 3mc+c2)c. We perceive that Sm^ will be the new trial divisor to obtain c. We then complete the divisor by adding to it Smc-\-c^. Let it be required to find the cube root of Sa^—S6ha^-\- 66h^a' - QBb^a^ + SSb'a" - %'a + b\ 8a« - 366a'' + 666=a* - Q^h^a" + 336*a' - 96*a + h\2a} - 36a + 6' 8ae 1 2a* - 1 86a' + 96 V ) - 3660^ + 666='a* - Q'Sb^a" -366a»+546V-276'a' ''"*"^^^"' t'66V-963a + 6* ) 126V-3663a3+336*a^-96^«+6« 126V-366V+336«a''-96^a + 6« The first term of the root is 2a2, and subtracting its cube, the first term of the remainder is — 36/>a*, which, divided by 3 times the square of 2a^, gives —2>ba for the second term of the root. Complete the divisor as in the last example, and multi- ply it by —Bba. Subtracting the product from the last remain- der, the first term of the second remainder is 12b'^a^. To form the new trial divisor, we take three times the square of the part of the root already found, viz., 2o^—Zba. Divide the first term of the remainder by 12«*, and we obtain b"^ for the last term of the root. We now complete the divisor by add- ing to it three times the product of the third term by the sum of the first two terms, and also the square of the last term. Multiplying the divisor thus completed by i^, we find the prod- EVOLUTION. 145 uct equal to the last remainder. Hence tlie required cube root 207. Hence, for extracting the cube root of a polynomial, we derive the following RULE. 1st. Arrange the terms according to the powers of some one let- ter ; take the cube root of tJie first term^ and subtract the cube from the given polynomial. 2o?. Divide the first term of the remainder by three times the sqxiare of the root already found ; the quotient will be the second term of the root. Sd. Complete the divisor by adding to it three times the product of the two terms of Uie root and the square of the second term. 4tth. Multiply the divisor thus increased by the last term of the rootj and subtract the product from the last remainder. 6th. Take three times the square of the part of the root already found for a new trial divisor, and proceed by division to find an- other term of the root. 6th. Complete the divisor by adding to it three times the product of the last term by the sum of the first tivo terms, and also the square of the last term, with which proceed as before till the entire root has been obtained. We may dispense with forming the complete divisor accord- ing to the rule if each time that we find a new term of the root we raise the entire root already found to the third power, and subtract the cube from the given polynomial. EXAMPLES. "1. What is the cube root of a^-6a^-\-15a^—20a^-]-16a'^- 6a + l? Ans. a2-2a + l. 2. What is the cube root of 6:^^-40x3+ ^^ + 9 6a: -64? Ans. 8. What is the cube root of 18;r* + 86ccH24ic+8 + 82xH x^-\-6x'^7 Ans. 4. What is the cube root o^b' + b^-bP-l + Sb? An^. G 146 ALGEBRA- ' 6. What is the cube root of 8x^ -S6a^-\- 66x^ - GSx^ -\- 33x2 ^ 9x+l? ^ 6. What is the cube root of 8x^+^Saa^-\-60a^a^—80a^a^— 90a'x''^108a'x-27a^? 7. What is the cube root of 8ic^-36ax*+102aV— 171aV f 204aV- 144^^0;+ 64a« ? (7z^Z>e i^ooi of Numbers, 208. The preceding rule is applicable to the extraction of the cube root of numbers ; but a difficulty in applying it arises from the fact that the terms of the powers are all blended to- gether in the given number. They may, however, be separated by attending to the following principles : 1st. For every three figures of the cube there will be one figure in the root, and also one for any additional figure or figures. Thus, the cube of 1 is 1 " 9 " 729 " 99 " 970,299 " 999 " 997,002,999 the cube of 1 is 1 10 " 1,000 100 " 1,000,000 " 1000 " 1,000,000,000 Hence we see that the cube root of a number consisting of from one to three figures will contain one figure ; the cube root of a number consisting of from four to six figures will contain two figures; of a number from seven to nine figures will con- tain three figures, and so on. Hence, if we divide the number into periods of three figures, commencing at units' place, the number of periods will indi- cate the number of figures in the cube root 209. 2c?. The first figure of the root will be the cube root of the greatest cube number contained in the first period on die left. For the cube of tens can give no significant figure in the first right-hand period ; the cube of hundreds can give no figure in the first two periods on the right ; and the cube of the high- est figure in the root can give no figures except in the first period on the left. Let it be required to extract the cube root of 438976. This number contains two periods, indicating that there will EVOLUTION. 147 be two places in 438,976 ( 70 f 6, the root, the root. Let a be 343,000 .1 1 e ^\. 70»x3 = 14700-— — — — • the value of the 70x6x3=: 1260 95976 figure in the tens' 6^= ?? 95976 place, and 6 of that complete divisor, 15996 in the units' place. Then a must be the greatest multiple of 10 which has its cube less than 438000 ; that is, a must be TO. Subtract the cube of 70 from the given number, and the re- mainder is 95976. This remainder corresponds to Zc^h-\-Za}?' -f 5^, which may be written Divide this remainder by Zd?-^ that is, by 14700, and the quo- tient is 6, which is the value of h. Complete the divisor by adding to it Zah^ or 1260, and IP-^ or 36. The complete divisor is thus found to be 15996, which, multiplied by 6, gives 95976. Subtracting, the remainder is zero, and we conclude that 70 + 6, or 76, is the required cube root. For the sake of brevity the ciphers may be omitted, provided "we retain the proper local values of the figures. If the root consists of more than two places of figures, the method will be substantially the same. Let it be required to extract the cube root of 279,726,264. 279,726,264(654 . ^^^^ found 65, the cube root of ' ' ^ the greatest cube contained in the first ^-*-^ two periods, we bring down the last period, and have 5101264 for a new dividend. We then take three times the square of the root already found, or 12675, for a partial divisor, whence we obtain 4 for the last figure of the root. We then complete the divisor by adding to it three times the product of 4 by 65, and the square of 4, regard being paid to the proper local values of the figures. The complete divisor is thus found to be 1275316, which, multiplied by 4, gives 5101264. Hence 654 is the rC' quired cube root. 108 90 25 11725 63726 58625 12675 780 16 1275316 5101 264 5101 264 148 ALGEBRA. 210. Hence, for the extraction of the cube root of numbers^ we derive the following RULE. l5^. Separate the given number into periods of three figures each, "beginning at the units^ place. 2d. Find the greatest cube contained in the left-hand period; its cube root is the first figure of the required root. Subtract the cube from the first period, and to the remainder bring down the second period for a dividend. 3d Talce three hundred times the square of the root already found for a trial divisor; find how many times it is contained in tlie dividend, and write the quotient for the second figure of the root. 4:ih. Complete the divisor by adding to it thirty times the product of the two figures of the root, and the square of the second figure. bth. Midtiply the divisor thu^ increased by the last figure of the root; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. 6ih. Take three hundred times the square of the whole root now found for a new trial divisor, and by division obtain another figure of Hie root. 7th. Complete the divisor by adding to it thirty times the product of the last figure by the former figures, and also the square of iJie last figure, with ivhich proceed as before, continuing the operation until all the p)eriods are brought down. It will be observed that three times the square of the tens, when their local value is regarded, is the same as three hund- red times the square of this digit, not regarding its local value. In applying the preceding rule, it may happen that the prod- uct of the complete divisor by the last figure of the root is greater than the dividend. This indicates that the last figure of the root was taken too large, and this happens because the divisor is at first incomplete, that is, too small. In such a case we must diminish the last figure of the root by unity, until we obtain a product which is not greater than the dividend. EXAMPLES. V • 1. Find the cube root of 163667323. Ans. 547- EVOLUTION. 149 2. Find the cube root of 89651821. Ans 341. ^ S. Find the cube root of 4019679. Ans. 159. -4. Find the cube root of 12^895213625. -5. Find the cube root of 188.05'6,925752. -6. Find the cube root of 759299,343g67. 211. Cube Boot of Fractions. — The cube root of a fraction is .5qual to the root of its numerator divided bj the root of its de- nominator. Hence the cube root of -rrrw ^s tV The number 12.167 may be written -t-V^tV^? ^^^ its cube root is fl", or 2.3. That is, the cube root of a decimal fraction, or of a whole number followed by a decimal fraction, may be found in the same manner as that of a whole number, if we divide it into periods commencing with the decimal point. In the extraction of the cube root of an integer, if there is still a remainder after we have obtained the units' figure of the root, it indicates that the proposed number has not an exact cube root. We may, if we please, proceed with the approxi- mation to any desired extent, by supposing a decimal point at the end of the proposed number, and annexing any number of periods of three ciphers each, and continuing the operation. We thus obtain a decimal part to be added to the integral part already found. So, also, if a decimal number has no exact cube root, we may annex ciphers, and proceed with the approximation to any de- sired extent EXAMPLES. 1. Find the cube root of ttVA- ^'tis. ^% 2. Find the cube root of 14^. ^ Ans, 2f . 8. Find the cube root of 18.312058. 4. Find the cube root of 1892.819053. 5. Find the cube root of .001879080904. Find the cube roots of the following numbers to 5 decimal places : 6. 15.25. Ans. 2.47984. 7. 8.7. Ans. 1.54668. 8. 100.1. ^n5. 4.64314. a 4. Ans. 1.58740. 10. 11. 11. i 12. f. 150 ALGEBRA. CHAPTER XIII. RADICAL QUANTITIES. 212. A radical quantity is an indicated root of a quantity: as y^, y/a, etc. Radical quantities may be either surd or ra- tional. Radical quantities are divided into degrees^ the degree being denoted by the index of the root. Thus, V3 is a radical of the second degree ; 1^5 is a radical of the third degree, etc. 213. The coefficient of a radical is the number or letter pre- fixed to it, showing how often the radical is to be taken. Thus, in the expression 2 -y/^j 2 is the coefficient of the radical. Similar radicals are those which have the same index and the same quantity under the radical sign. Thus, Sy/a and 5/5 are similar radicals. Also TVh and lOv/6 are similar radicals. 214. Use of fractional Exponents. — We have seen. Art. 196, that in order to extract any root of a monomial, we must di- vide the exponent of each literal factor by the index of the re- quired root Thus the square root of a'^ is a^, and in the same manner the square root of a^ may be written a^, that of a* will 6 1 be a^, and that of a, or a\ is a . Whence we see that a is equivalent to y a, a^ " ^/a\ ±a^63^-2. .jij^^^. 10. V48a^6V. -fifi'^//:. <^6..o 11. Vl92?6^.-tT7r75W^ 12. ^y^la'Wx. "^V^VJaV" 13. Vrt'-a^a;. J.rzs. 2aZ)2\/7a^ Ans. Zahyiac^. Ans. ^ah^cS/^ac^, i^~ . . - 216. When the quantity under the radical sign is Vi fraction^ it is often convenient to multiply both its terms by such a quantity as will make the denominator a perfect power of the degree indicated. Then, after simplifying, the factor remain- ing under the radical sign will be entire. 14. SVf. 15. #V;+6V3i. ,16. V81+2-v/|. 17. ^/U+^^f^-. 18. |.V|+3V|. 19. IS-v/A+SVA- Ans. 3V|=3Vil=|.VT0. Ans. -11/2 + 31/14. ^715. 4 a/34-1 Vis. Ans. 4 V5+^-v/5. ^W5. |V'304-^V14. ^ns. iy'154-f ViO. 20. a/Uals/l. ^725. ^-^ah. 21- (-^)\^- Ans, ^Jd'-V\ - '!^ Ans. Y^abc. b 23. 21/44-314. Ans. v/4-f v/9. 24. 2l/2| + 7l/7||. Ans. |i/9+2^\/l8. 217. The following principle can frequently be employed in simplifying radicals : TJie mnih root of any quantity is equal to the mth root of the nth root of thai quantity. That is, G 2 154 ALGEBRA. For, if we raise each expression to the mth power, it becomes y/S Thus, the fourth root = the square root of the square root, the sixth root =the square root of the cube root, or the cube root of the square root the eighth root = the square root of the fourth root, or the fourth root of the square root, the ninth root =the cube root of the cube root. Hence, when the index of a root is the product of two or more factors^ we may obtain the root required hy extracting in succession the roots denoted hy those factors. Ex. 1. Let it be required to extract the sixth root of 64. The square root of 64 is 8, and the cube root of 8 is 2. Hence the sixth root of 64 is 2. Ex. 2. Extract the eighth root of 256. Ans. 2. -Ex. 3. Find the fourth root of 1874161. Ans. 37. -"Ex. 4. Find the sixth root of 148085889. Ans. 23. "Ex. 5. Find the ninth root of 887420489. Ans. 9. Ex. 6. Find the eighth root of ^^89,0,6^5. Ans. 15. 218. When the index of a root is the product of two or more factors, and one of the roots can be extracted, while the other can not, a radical may he simplified hy extracting one of the roots. Thus, v/9=V3. Keduce the following radicals to their simplest forms : Ex. 1. V&. Ans. y%i. Ex.2. VzWh''. Ex. 8. V^. Ex.4. \/25a*62c«. 6 p. Ex.5. ^25a2 6462- Ex.6. \/^. Ans.^\^ ^ m^n^ m'n V ^^v ' RADICAL QUANTITIES. 155 To introduce a Factor under the Radical Sign, 219. The square root of the square of a is obviously a, and the cube root of the cube of a is a, etc. That is, a=^\/a:^=Vo^=:\/a^^ etc. Whence, also, ay/h = y^ x\/T>= y/o^. Hence, to introduce a factor under the radical sign, we have the following RULE. liaise the factor to a 'power denoted hy the index of the required root, and write it as a factor under the radical sign. EXAMPLES. 1. Reduce ax^ to a radical of the second degree. ^ Ans. Vcl'^x*, 2. Reduce 2d^hx to a radical of the third degree. Ans. VSM^. 8. Reduce 5 + 6 to a radical of the second degree. Ans. V25-\-10b-\-h\ Transform the following radicals by introducing the coe£&> cients as factors under the radical sign : 4 eVSi. Ans^ VT26. 5. 4-v/i+3i-v/8. Ans. -\/2 + V98. 6. ay -^aiy —. Ans. 2Vab. 8. 2l/2 + 7v/5. Ans. VU+VmE. 9. 7^v/7j|. ■ Ans. l^^:^ To change the Index of a Radical. 220. From Art. 219, it follows that v/a = \/a^=V^=V«*, etc.; V^ = v/^ = Va' = Va\ etc. ; V^ = Va^ = Va' = r < etc. 156 ALGEBRA. Hence we see that the index of any radical may he multiplied hy any number, provided we raise the quantity under the radical sign to a power whose exponent is the same number ; or the index of any radical may be divided by any number, provided we extract that root of the quantity under the radical sign whose index is die same number. If, instead of the radical sign, we employ fractional expo- nents, we shall have 12 3 4 a^ =0^ =za^ =0^ , etc. 12 3 4 0^ =aj^ =a^ z^a^ , etc. Hence we see that we may multiply or divide both terms of a fractional exponent by the same number, without changing tJie value of Hie expression. EXAMPLES. Yerify the following equations : 1. v^+v/g+Vi^v/i+v/e+VIe. 8. 2V^i?^^2ab^T^ = 2^/7^. 4. 2 V^^ = 2'/a3-3a26+3a62_63. 5. 8 V^x^ - l^x'y + 6x2/2 _ y3 ^ 3 V2^^. To reduce Radicals to a Common Index, 221. Let it be required to reduce Va and Va to equivalent radicals having a common index. Substituting for the radical signs fractional exponents, the given quantities are 1 1 a^ and a^. Reducing the exponents to a common denominator, the ex pressions are a 2 a^ and a^, or Va^ and Va^, RADICAL QUANTITIES. 157 which are of the same value as the given quantities, and have a common index 6. Hence we derive the following RULE. Reditce the fractional exponents to a common denominator^ raise each quantity to the power denoted hy the numerator of its new ex- ponentj and take the root denoted hy the common denominator, EXAMPLES. 11 1 1. Eeduce a , a^, and a^ to a common index. Ans. a^ a^ and a^. 2. Reduce a , a , and b to a common index. Ans. ar^ a^ ^ and h^. 11 1 8. Reduce 2^, 8^, and 5^ to a common index. Ans. V6i, 'v^, and 'v/l25. 12 3 4. Reduce 3^, 2^, and 2^ to a common index. Ans. 'v/729, "v/256, and "v/512. 1 _i 1 5. Reduce a^, a*^, and a" to a common index. mn 2>i 2ni An^. a^^j a^^j and a^^, 6. Reduce v/8, V5, and '/7 to a common index. 7. Reduce V2aZ), \/8a6^, and v/5a^ to a common index. 8. Reduce Va-\-b, \/~a—h^ and S/a^—W- to a common index. To add Radical Quantities together. 222. "When the radical quantities are similar^ the common radical part may be regarded as the unit^ and the coefficient shows how many times this unit is repeated. The sum of the coefficients of the given radicals will then denote how many times this unit is to be repeated in the required sum. If the radicals are not similar they can not be added, because they have no common unit. In such a case, the addition can 158 ALGEBRA. only be indicated by the algebraic sign. Kadicals which are apparently dissimilar may become similar when reduced to their simplest forms. Hence we have the following RULE. Reduce each radical to its simplest form. If the resulting radi- cals are similar^ add their coefficients^ and to their sum annex Hie common radical. If they are dissimilar ^ connect them by the sign of addition, EXAMPLES. 1. Find the sum of V27, ViS, and Vlb. Ans. 12 VI. 2. Find the sum of 4Vli7, 3 V75, and VT92. Ans. 61 \^. 8. Find the sum of V72, Vl28, and V 162. Ans. 23 V2. 4. Find the sum of VT80, Vi05, and V320. Ans. 23 V5. 5. Find the sum of 3^1, 2 V^, and 4VA. Ans. VTO. 6. Find the sum of V^500, Vl08, and v/256. Ans. 12 '/i. 7. Find the sum of ^40, 1/135, and v/320. Ans. 91^5. 8. Find the sum of 2 V|, V60, Vl5, and V|. Ans. 4|Vl5. 9. Find the sum of Vi5?, ^/SO?, and VbaFc. Ans. {a-\-*Jc)y/bc, 10. Find the sum of Vl8^ + ^/W^. Ans. {^a%-^bab)V2ab, ll.Bnd,l..™ofV^,V/g%nav/g. 12. Find the sum of V-^a^ V^oab^, and 5i Vah. Ans.{2a-\-10h)Vd), RADICAL QUANTITIES. 169 To find the Difference of Radical Quantities. 223. When the radicals are similar, it is evident that the Bubtraction may be performed in the same manner as addition, except that the signs in the subtrahend are to be changed. Hence we have the following RULE. Reduce each radical to its simplest fi)rm. If the resulting radi- cals are similar^ find the difference of the coefficients^ and to the re- sult annex the common radical part. If they are dissimilar, thfi subtraction can only be indicated. EXAMPLES. 1. From V58 take Vll2. Ans. 4i/7. 2. From 5 V26 take 3 Vi5. Ans. VE, 8. From 2\/50 take VI8. Ans. 7^2. 4. From VSOaFx take V20a^\ 6. From 2^72^ take Vl62a\ 6. From VT92 take ^24. Ans. 2v/3. 7. From 2^320 take 8\/40. 8. From V ^^ take \/ -^. Ans. {Sa-1)\/ -^. To multiply Radical Quantities together. 224. We have found, Art. 215, that Va multiplied by Vb is equal to Vctb. Hence, V2xVS = V6. If the radicals have coefficients, the product of the coeffi' cients may be taken separately. Thus, aVxxbVy=axbxVxxVy=ahVxy; also, 8^/8x5\/2 = 15Vl6=:60. If the radicals have not a common index, they must first be reduced to a common index. Hence we have the following 160 ALGEBRA. RULE. If necessary^ reduce the given radicals to a common index. Mul- tiply the coefficients together for a new coeffijcient ; also multiply the quantities under the radical signs together^ and place this product under the common radical sign. Then reduce the result to its sim- plest form. EXAMPLES. Find the value of the following expressions : . 1. 3V'8x2\/6. Ans. 24 V^. 2. 6V8x3V5. Ans. SOVTO. 3. V2 X v/3. Ans. Vn. 4 5 VS X 7 Vl X V2. Ans. 140. 5. c^/axdVa. Ans. acd. 6. iV^xbVl. Ans. 70 V^. 7. kVlx-^VVJ. 8. iv/l8x5\/20. 9. i/ix7l/6xiv/5. Aris.lV^. 225. We have seen, Art 58, that the exponent of any letter in a product is equal to the sum of the exponents of this letter in the multiplicand and multiplier. That is, a'"xa''=a'"+'', where m and n are supposed to be positive whole numbers. When one or both of the exponents are negative^ we must take the algebraic sum of the exponents. For, suppose n is negative. Then 1 a"* a"* X a"" = a"* X — , by Art. 76, = — — a*^"*. The same relation holds true when m and n are fractional ; p r p^r that is, a^xaf=a^ '. For a« X a}=Va^ x Va% Art. 214, = V^^ X Va'^', Art. 220, =7^3^m¥, Art.224,=a «* =a^ \ Hence we conclude that the exponent of any letter in a product is equal to Hie algebraic sum of Hie exponents of this letter in ihi RADICAL QUANTITIES. 161 multiplicand and multiplier^ whether the exponents are positive or negative^ integral or fractional. EXAMPLES. 1. Multiply 5a* by 3a* Ans. Iba^. 2. Multiply 21a* by 3a^ Ans. 63a^. 8. Multiply 3x*2/* by 4ccV 1 ■T7 4. Find the product of a^, a , a , and a 6. Find the product of a^, a , a^, and a^^. Multiplication of Polynomial Radicals. 226. By combining the preceding rules with that for the multiplication of polynomials, Art. 61, we may multiply togeth* er radical expressions consisting of any number of terms. Ex. 1. Let it be required to multiply a*+2a*-a*by a*-3a*+2. a*+2a*-a* a*^3a*+2 5 3 a'^+2a -a^ -3a -6a*+3a* 4-2a*+4a*-2a* T ^ k3 I ^73 O^i a — a —6a -{-7a —2a^ Ans. Ex. 2. Multiply 3 + V5 by 2 - VB. ^7i5. 1 - VB. Ex. 3. Multiply 7 + 2^6 by 9-6Ve. Ans. 3-17 V'6. Ex. 4. Multiply 9 + 2^10 by 9-2 VlO. Ans. 41. Ex. 5. Multiply 3 V45-7\/5 by Vi|+2\/9i. ^ns. 34. Ex. 6. Multiply cVa + cZ/^ by cVa—dVb. Ex. 7. Multiply a^-aHa'^-a'+a^-« + «^-l by a^+1 -4ri5. ac^—hd"^. -Iby a'^+l Ans. a'^— 1. 162 ALGEBKA. To divide one Radical Quantity hy another. 227. The division of radical quantities depends upon the fol lowing principle : The quotient of the nth roots of two quantities is eqval to the nth root of their quotient ; or, Va _n fa for the nth power of each of these expressions is ^, Art. 186. Let it be required to divide 4:a'^V6by by 2aV3b. ;^=7r-\/-77f =2aV2v, Ans, Hence we have the following EULE. If necessary, reduce the given radicals to a common index. Di- vide the coefficient of the dividend by that of the divisor for a new coefficient ; also the quantity under the radical sign in the dividend hy that in the divisor, and place this quotient under the common radical sign. Then redux:e the result to its simplest form. EXAMPLES. 1. Divide 8VT08 by 2V6. Am. 12 V2. 2. Divide 8 \/612 by 4 V2. Ans. 8 Vi 3. Divide 6 VU by 3 V2. Ans. 6. 4. Divide 4v/72 by 2\/l8. 5. Divide 4V6a^?/ by 2VSy. \ i '^ ^ 6. Divide 16{a^b)'^ by S{acy. 7. Divide 4 \/l2 by 2 \/3. Am. 2 V^. 8. Divide VOi by 2. Ans. \/2. 9. Divide VMc by Vcib'^c^. Ans. \ j-^- 228. We have seen, Art. 72, that the exponent of any letter in RADICAL QUANTITIES. 163 a quotient is equal to the difference between the exponents of this let- ter in the divisor and dividend. The same relation holds true whether the exponents are posi- tive or negative^ integral or fractional , that is, universally, p r p r a~^^a'=a^"\ For the quotient must be a quantity which, multiplied by the divisor, shall produce the dividend ; and, according to Art. 225, the exponent of any letter in a product is in all cases equal to the algebraic sum of the exponents of this letter in the multiplicand and multiplier. Hence this relation must hold true universally in division. EXAMPLES. 1. Divide {abf by {ah)^. Ans. {ahf, 2. Divide a^ by a*. 8. Divide ^Vab by 2 Voh. Ans. 2 Vd). 4. Divide ^m\a-h)^ by Zm{a-h^, Ans. 2>m{a-h)'^. 1111 42 5. Divide a^lr by o^li^. Ans. a^F^. 6. Divide 4* by 2*. Ans, 4- ^ V2 Division of Polynomial Radicals. 229. By combining the preceding rules with that for the di- vision of polynomials, Art. 80, we may divide one radical ex- pression by another containing any number of terms. Ex. 1. Let it be required to divide d^—a^—a'^ -j- a^ by a^— 1. a^-a^-a^+a^ a^-a^ a^—a* Ans, -a* 4- a* :.2. Divide 8a-5 by 2a*-Z>* Ans. ^a^+2ah^-\-h^. 164 ALGEBRA. Ex.3. Divide a-41a'^-120 by a^+4a^ + 5. Ans. a^ — 4a ^ -hlla^ — 24 Ex. 4. Divide a^ + Ub^ by a*+4i*. u4ns. a^-4:ah^'\-16hi. Ex. 5. Divide cc"^— ccz/^+a:;'^!/— ?/^ by x^ — y^. Ans. x-\-y. To involve a Radical Quantity to any power. 230. Let it be required to raise a^ to the nth power. Ill l_|_i 1. The square of a"* is a'^xa'''=a'" '"''=a''\ 1 1111 11 ?i The cube of a^ is a^xa'"'xa'^=a'^ '"' "'=«"*. l 1 1 i 1+i+Ietc » The wth power of a"* is a"* x a*" x a'", etc., =:a"' "^ "^^ "=a''\ Hence, to involve a radical quantity to any power, we have the following RULE. Multiply the fractional exponent of the quantity hy the exponent of the required power. If the radical has a coefficient^ let this be in- volved separately ; then reduce the result to its simplest form. If the quantity is under the radical sign, it is generally most convenient to substitute for this sign the equivalent fractional exponent; but if we choose to retain the radical sign, we must raise the quantity under it to the required power. EXAMPLES. 1 4 1. Required the fourth power of fa^. Ans. ^^^. 2. Required the cube of f-ZS. Ans. f v^. 3. Required the square of 3 V%. 4. Required the cube of 17 V2T. 5. Required the fourth power of \VQ. Ans. ^. 6. Required the fourth power of 2 V3a*6. Ans. 16a2v/9«^. 7. Required the fourth power of ahVab. Ans. a^b^. RADICAL QUANTITIES. 165 A71S. a''-\-2ab-\-h\ 8. Kequired the sixth power of (a +5)* 9. Kequired the. value of V(51y x VCHf- ^ns. -gV- 10. Kequired the value of ViM^^VJ^^^'- Ans. {2ahf, To Extract any Root of a Radical Quantity. 231. A root of a quantity is a factor which, multiplied by itself a certain number of times, will produce the given quanti- 1 n ty. But we have seen that the ?ith power of a"" is a*". There- n ^ fore the nth root of a^ is a'"\ Hence we derive the following EULE. Divide the fractional exponent of the quantity by the index of the required root. Jf the radical has a coefficient^ extract its root sep- arately if possible; otherwise introduce it under the radical sign. Then reduce the result to its simplest form. If the quantity is under the radical sign, and we choose to retain the sign, we must, if possible, extract the required root of the quantityunder the radical sign ; otherwise we must mul- tiply the index of the radical by the index of the required root. EXAMPLES. 1. Find the square root of 9(3)^ 2. Find the cube root of ^V2. 3. Find the square root of 10^. 4. Find the cube root of -i^a^. 5. Find the fourth root of ff-a^. 6. Find the cube root of yVs^^^A ^^ 7. Find the cube root of ^\/|. 8. Find the square root of 3 Vb. Ans. VT35. 9. Find the fourth root of J v/|. Ans. ^1/12. Ans. 3 v/3. Ans.^V^. in Ans. xfi. 166 ALGEBRA. Operations on Imaginary Quantities, 232. It has been shown, Art 195, that an even root of a neg- ative quantity is impossible. Thus, V— 4, V — 9, V— 5a are algebraic symbols representing operations which it is impossi- ble to execute ; for the square of every quantity, whether posi- tive or negative, is necessarily positive. Quantities of this na* ture are called imaginary or impossible quantities. Neverthe- less, such expressions do frequently occur, and it is necessary to establish proper rules for operating upon them, 233. The square root of a negative quantity may always he rep- resented hy the square root of a positive quantity multiplied by the square root of —1. Thus, V'^=V4:X -1 = 2V^, -/i:3=: V3 X -1 = V3 V^l, V^^z=z\/ax —l = VaV — l. The factor V — 1 is called the imaginary factor, and the other factor is called its coefficient. 234. When several imaginary factors are to be multiplied together, it is best to resolve each of them into two factors, of which one is the square root of a positive quantity, and the other V — 1. We can then multiply together the coefficients of the imaginary factor by methods already explained. It only remains to deduce a rule for multiplying the imaginary factor into itself; that is, for raising the imaginary factor to a power whose exponent is equal to the number of factors. The first power of V — l is V— 1. The second power, by the definition of square root, is —1. The third power is the product of the first and second pow- ers, or —1 X V— 1 = — V — l. The fourth power is the square of the second, or +1. The fifth is the product of the first and fourth ; that is, it is the same as the first; the sixth is the same as the second, and RADICAL QUANTITIES. 167 so on ; so that all the powers of V — 1 form a repeating cycle of the following terms : ~" /^, +1. + -1, -1, EXAMPLES. 1. Multiply V^ by V— 4. 2. Multiply l-\-V^ by 1-V^. 3. Multiply Vl8 by V^. 4 Multiply 5 + 2-/^ by 2-i/^. 5. Multiply aV^ by cV—d. 6. Multiply 1-^^^ by itself 7. Multiply 2Vl~V^ by 4^3-2 6, Ajis. Ans. 2. Ans. -^2V^. 8. Multiply a+Vb- Ans. 14-8V-I5. 1 by a— v^-/^. Ans. a^+h. 9. Multiply aV-aW by V-M\ 10. Multiply V3^4-V^ by VZ:^— yCl,. 11. Multiply i/3l7+V^^39 by V-119-V^=T83. ^?25. 2%/7. 235. Division of Imaginary Quantities. ^—The quotient of one imaginary term divided by another is easily found by resolv- ing both terms into factors, as in the preceding article. Ex. 1. Let it be required to divide V~^ah by V — a. V—ab VahV^ nr . — ■ = V 6, Ans* V — a yay — 1 Ex. 2. Divide l^^^ by cV^. Ex. 3. Divide unity by V — 1. Ex. 4. Divide a by hV^. Ex. 5. Divide a by Va-/^. Ans. ^. c Ans. — V — 1. 168 ALGEBRA. Ex.6. Divide V_12 + V-6+ ^-9 by V-S. Ex. 7. Divide 2V8--/^^30 by -V^, Am. V6+4V^. To find Multipliers which shall cause Surds to become BationaL 236. 1st. When the surd is a monomial The quantity Va is rendered rational by multiplying it by Va. 1 1 For ^aX\^=a^xa^=a. 1 9 So, also, a^ is rendered rational by multiplying it by a^. 1 • . . . ^ 1 Also, a^ is rendered rational by multiplying it by a^; and a" by multiplying it by a ~«. Hence we deduce the following RULE. Multiply the surd by the same quantity having such an exponent as^ when added to the exponent of the given surdj shall make unit?/. 237. 2d., When the surd is a binomial If the binomial contains only the square root, multiply the given binomial by the same terms connected by the opposite sign^ and it will give a rational product. Thus the expression Va-^Vb multiplied by Va— V^ gives for a product a—b. Also the expression Va+Vb multiplied by \/a—Vi gives for a product Va—Vb, which may be rendered rational by multiplying it by Va-^Vb. In general, V« =*= yi may be rendered rational by success- ive multiplications whenever m and n denote any power of 2. When m and n are not powers of 2, the binomial may still be rendered rational by multiplication, but the process becomes more complicated. Ex. 1. Find a multiplier which shall render V'b+VS ration- al, and determine the product. RADICAL QUANTITIES. 169 Ex. 2. Find a multiplier which shall render VS— Vx ration- al, and determine the product. Ex. 3. Find multipliers which shall render VS — V^ ration- al, and determine the product. 238. 3d. When the surd is a trinomial When a trinomial surd contains only radicals of the second degree, we may reduce it to a binomial surd by multiplying it by the same expression, with the sign of one of the terms changed. Thus, Va-^Vb-^Vc multiplied by Va-\-Vb—Vc gives for a product a-\-b—c-{-2Vabj which may be put under the form of m-{-2Vab. Ex.1. Find multipliers that shall make V6-\-VS—V2 ra- tional, and determine the product. Ex. 2. Find multipliers that shall make 1 -}- V2 -\- VB ration- al, and determine the product. To transform a Fraction whose Denominator is a Surd in such a Manner that the Denominator shall be Rational. 239. If we have a radical expression of the form a Vb-\-Vc or -7= y=, it may be transformed into an equivalent expres- sion in which the denominator is rational by multiplying both terms of the fraction by v^i V~c. Hence the RULE. Multiply both numerator and denominator by a factor which will render the denominator rational. EXAMPLES. Reduce the following fractions to equivalent fractions having a rational denominator : 1 2 A 2V3 1. —7=. Ans. — r— . V3 • _ ^ 9 1 A V5+V2 ^. -7= 7=. Arts. . V5-V2 ■ 3 H 170 ALGEBRA. 3 ^^ ^' 3>V2- . 3\/24-2 ^7Z5. ^ — . a-Vb a+Vb • 6. ,_ ^- . VS + V2+1 Ans. 2+-/2-V6. 6. % 7. ^ V6. 8 ^2 Am. V2. ^ l + a+Vl-a» l4-a-Vl-a2 l + Vl-a^ a 240. The utility of the preceding transformations will be seen if we attempt to compute the numerical value of a fractional surd. Ex. 1. Let it be required to find the square root of ^ ; that is, Vs to find the value of the fraction — — . ^^ V2-1 Making the denominator rational, we have ' „ , and the value of the fraction is found to be 0.6546. 7V6 Ex. 2. Compute the value of the fraction Ex. 3. Compute the value of the fraction Vn-\-Vs A71S. 3.1003. V6 V7-\-VB Ans. 0.5595. Vs Ex. 4. Compute the value of the fraction — — ;= ;=> ^ 2V84-3\/6-7\/2 Ans. 0.7025. RADICAL QUANTITIES. 171 9 + 2VT0 Ex. 5. Compute the value of the fraction 9-2 VIO" Ans. 5.7278. Square Boot of a Binomial Surd, 241. A binomial surd is a binomial, one or both of whose terms are surds, as 2 + VS and Vb — V2. A quadratic surd is the square root of an imperfect square. If we square the binomial surd 2+'v/8, we shall obtain 7 + 4^/3. Hence the square root of 7+4V3 is 2-\-V3] that is, a binomial surd of the form a± Vb may sometimes be a perfect square. 242. The method of extracting the square root of an expres- sion of the form a±.Vb is founded upon the following princi- ples: 1st. The sum or difference of two quadratic surds can not be equal to a rational quantity. Let Va and Vb denote two surd quantities, and, if possible, suppose V-a^Vb=c, where c denotes a rational quantity. By transposing Vb and squaring both members, we obtain 2c The second member of the equation contains only rational quantities, while Vb was supposed to be irrational; that is, we have an irrational quantity equal to a rational one, which is impossible. Hence the sum or difference of two quadratic surds can not be equal to a rational quantity. 243. 2d. In any equation which involves both rational quanti- ties and quadratic swxis, the rational parts in the two members are equal^ and also the irrational parts. Suppose we have x-^-Vy^a-'tVh, 172 ALGEBRA. Then, if x be not equal to a, suppose it to be equal to a+m; 60 that m=Vb—Vy; that is, a rational quantity is equal to the difference of two quadratic surds, which, by the last article, is impossible. There- fore x=a, and consequently Vy=Vb. 244. To find an expression for the square root of a±: Vh. Let us assume Va-f V^= V^+ Vy. (1.) By squaring, a-\-Vb=x-\-2Vxy-{-y, (2.) By Art. 243, a=x+y, (3.) and Vh=2Vxy. (4.) Subtracting (4) from (3), we have a—Vh=x—2Vxy-\-y. (5.) By evolution, Va-Vb = v^- v^. (6.) Multiplying (1) by (6), we have V^^~h=x—y. (7.) Adding (3) and (7), a+V^^h^^x, (8.) Hence, ic= ^ . (9.) Subtracting (7) from (3), a-^ftf-b y= — 2 • ^^^'^ It is obvious from these equations that x and y will be ra- tional when a^—b is a perfect square. If a^ — b be not a perfect square, the values of Vx and Vy will be complex surds. Hence, to obtain the square root of a binomial surd, we pro- ceed as follows : Let a represent the rational part, and Vb the radical part, and find the values of x and y in equations (9) and (10). Then, if the binomial is of the form a-f v^, its square root will be Vx-[-Vy. If the binomial is of the form a—Vby its square root will be Vx— Vy. RADICAL QUANTITIES. 175^ EXAMPLES. 1. Kequired the square root of 4-+-2'\/3. Here a=4 and ^^h=l^/Z\ or 5=12. 4-|-vT6iri2 Hence aj= ^ — *^» 4_V16-12 , y^ 2 =^- Hence Vx + Vy = -/S 4- 1, ^^5. Verification. The square of -v/S + l is 3-|-2V34-l=4+2V3. 2. Eequired the square root of 114-6V2. Here a=ll and V^=6V2; or 5 = 72, cc=9 and 2/=2. 3. Required the square root of 11— 2 V30. Ans. VQ—VE. 4. Required the square root of 2 + ^3. Ans. V^+V^. 5. Required the square root of 7 + 2VT0. Ans. V6-\-V2. 6. Required the square root of 18 + 8 VS. Ans. VIO + 2V2. 245. This method is applicable even when the binomial con tains imaginary quantities. 7. Required the square root of 14-4V — 3. Here a=l and V5=4V— 3. Hence 5=— 48 and a^— 5=49. Therefore £c=4 and y=—S. The required square root is therefore 2 + '/— 3, J.n5. 8. Required the square root of — ^+-iV — 3. 9. Required the square root of 2^—1 or 0+2V^^. ^n5. i+-/iri. 174 ALGEBRA. 10. Eequired the value of the expression V6-I-2-V/5- V6-2\/5. IL Eequired the value of the expression V4+3-v/i:20"+ V4-3V^^. Ans. 6. 12. Eequired the square root of — 3+V — 16. ^725. 1 + 2 \/^^. 13. Eequired the square root of 8 V — 1- Ans. 2 + 2-/^. Simple Equations containing Radical Quantities. 246. When the unknown quantity is aifected by the radical sign, we must first render the terms containing the unknown quantity rational. This may generally be done by successive involutions. For this purpose we first free the equation from fractions. If there is but one radical expression, we bring that to stand alone on one side of the equation, and involve both members to a power denoted by the index of the radical. Ex. 1. Given x-{-Vx'^-Sx-{-60 = 12 to find x. Transposing x and squaring each member, we have a;2-3x+60=144-24a;+ic2; whence x=4:. Ex.2. Given Vx-\-Va-\-x= = to find x. IT-S-v/S ^^^- ^=i' Ex.3. Given n """^ ^^ ^^^ ^- Ex 4. Given V^x^—7x—6 = 9-2x to find x. 247. If the equation contains two radical expressions com- bined with other terms which are rational, it is generally best to bring one of the radicals to stand alone on one side of the equation before involution. One of the radicals will thus be made to disappear, and, by repeating the operation, the remain- ing radical may be exterminated. Ex.6. Given Va;+19-h Va;+10 = 9 to find x. KADICAL QUANTITIES. 175 By transposition, Vx-\-19 — 9—Vx-{-10. Squaring, ir4-19=:91+ic-18Vx+10. Transposing and reducing, Squaring, cc + 1 = 1 6 ; whence x = 6. Ex. 6. Given V36-\-x=lS+Vx to find x. SEx. 7. Given Vx-{-4:ab = 2b-{-Vx to find x. -^Ex. 8. Given x- Va^^xVb^+x^—a^-i-a to find x. 248. When an equation contains a fraction involving radical quantities in both numerator and denominator, it is sometimes best to render the denominator rational by Art. 239 ; but the best method can only be determined by trial. Ex. 9. Given Vj+V^ ^ 3 ^^ ^^^ ^^ Multiply both terms of the first fraction by Vx-\- Vx—S^ and we have {Vx-[-V^^f _ 3 x-{x-3) ~ic-3' or {V^-{-Vx^y ^ cc-3 Extracting the square root, Vx-hVx—S^- Clearing of fractions, Vx^-Sx-\-x—S=S] whence x=4, Ans. \/ T? 1A n- V9CC4-13+V9^ 1Q . n A W Ex.10. Given ■ , ;==13 to find x. ^ V9x+13-\/9^ Ex. 11. Given '\/ax-{-b='\/cx-\-d to find x. Ex. 12. Given — Z =^ to find oc 3+5'v^^+24 176 ALGEBRA. Ex. 13. Given ^^^ r.:l + |(V^~l) to find x. Ans. x=S. - fn -finrl o' Vx-\-Sn Vx-{-n Ex. 14. Given — — = — :J to find x. Ans. X: ( mn y \m—n) ' Ex.15. Given (V^-6) (v^4-25)=(5 + 3v^-) (v^+3) to find X. Ans. x=9. ^ Ex. 16. Given V2x-3n=BVn--V2x to find x. Ans. x=2n, Ex.17. Given -—= =—^ to find x. Vx+2 Vx+^0 Ex. 18. Given — = — = — ;= to find x. •n ^^ ^. 6x—9 ^ V6x—S ^ , Ex.19. Given --= 1 = — to find x. -v/6^+3 2 Ex.20. Given V^a+x=2Vb+x— Vx to find x. _(a-by Am, X— „ , , 2a— o EQUATIONS OF THE SECOND DEGREE. 177 CHAPTER XIY. EQUATIONS OF THE SECOND DEGREE. 249. An equation of the second degree^ or a quadratic equation with one unknown quantity, is one in which the highest power of the unknown quantity is a square. 250. Every equation of the second degree containing but one unknown quantity can be reduced to three terms ; one con- taining the second power of the unknown quantity, another the first power ^ and the third a known quantity ; that is, it can be reduced to the form _ , x^-{-px=q. Suppose we have the equation ^+^+^+18|=(»+3) («.-!). Clearing of fractions and expanding, we have 9cc2-f7cc-6+4x-8 + 164=12a;2+24^-36. Transposing and uniting similar terms, we have Dividing by the coefficient of cc^, that is, by —3, we have ccHy^=62, which is of the form above given, p in this case being equal to ^, and q being equal to 62. 251. In order to reduce a quadratic equation to three terms^ we must first clear it of fractions, and perform all the operations indicated. We then transpose all the terms which contain the unknown quantity to the first member of tne equation, and the known quantities to the second member ; unite similar terms, and divide each term of the resulting equation by the coeffi- cient of x\ H 2 178 ALGEBRA. 252. An equation of the form x^-^px=q is called a completi equation of the second degree, because it contains each class of terms of which the general equation is susceptible. 253. The coefficient of the first power of the unknown quan- tity may reduce to zero, in which case the equation is said to be incomplete. An incomplete equation of the second degree, when reduced, contains but two terms : one containing the square of the un- known quantity, and the other a known term. Incomplete Equations of the Second Degree. 254. An incomplete equation of the second degree may be reduced to the form d^=.q. Extracting the square root of each member, we have If g- be a positive number, either integral or fractional, we can extract its square root, either exactly or approximately, by the rules of Chap. XII. Hence, to solve an incomplete equa- tion of the second degree, we have the RULE. Reduce the equatioji to the form x^=q, and extract the square root of each member of the equation. 255. Since the square of both -f m and —m is -|-77i^, the square of -^Vq and that of —Vq are both -\-q. Hence the above equation is susceptible of two solutions, or has two roots; that is, there are two quantities, which, when substituted for x in the original equation, will render the two members identical. These are + Vq and — Vq. Hence, Every incomplete equation of the second degree has two roots^ equal in numerical value, hut with opposite signs. EXAMPLEa Find the values of x in each of the following equations : 1. 4:X^-7 = Sx^-\-9. Ans. x=zt4. EQUATIONS OF THE SECOND DEGREE. 179 Show that each of these values will satisfy the equation. 2. cc2_17==130-2x2. Ans.x=±:7. ±Va6 4. x'^-\-ab=6x\ Ans. x= — ^ — . 2^2 . . a 6. x-k-Va^ + x^^ ,—^ =. Ans. x=±——. Va^+x"" V3 6. ax'^—oc=hx^—Sc-{-d. ^- a~'^+12~24~^ + 24- 8. 12a6+a:2=4a2 4-9^'. 9. ii_^^3-^. 10. X+\/x'^— 17= , ^. ^715. X=db44. Vcc2-17 a:; + a cc-a 2(a2 4-l) 11. 1 ; =^ /l , \/1 V ■^^^- ^=±1. cc— a ic + a (l + a)(l — a) 12. T^ = l' J.7Z5. a;=±3. X Note. Clearing of fractions and transposing, we find in eacli member of this equation a binomial factor, which being cancel* ed, the equation is easily solved. PROBLEMS. Prob. 1. What two numbers are those whose sum is to the greater as 10 to 7, and whose sum, multiplied by the less, pro- duces 270? , Let 10a:j:= their sum. ^ Then 7x = the greater number, and 3x = the less. Whence 80^2 = 270, and cc2 = 9; therefore x=±S, and the numbers are ±21 and ±9. 180 ALGEBRA. Prob. 2. What two numbers arc those whose sum is to the greater as m to w, and whose sum, multiplied by the less, is equal to a f ♦ Arts. ^J-^—- and dbv^ ^ m{m—n) v [m—n) m Prob. 3. What number is that, the third part of whose square being subtracted from 20, leaves a remainder equal to 8? Prob. 4. What number is that, the mth part of whose square being subtracted from a, leaves a remainder equal to h f Ans. ±Vin{a—b). 12 3 Prob. 5. Find three numbers in the ratio of ^, q, and -j, the sum of whose squares is 724. Prob. 6. Find three numbers in the ratio of m, n, and p, the sum of whose squares is equal to a, Ans. I cum?' I CLTV^ I dtp' Prob. 7. Divide the number 49 into two such parts that the quotient of the greater divided by the less may be to the quo- tient of the less divided by the greater as | to ^. Ans. 21 and 28. Note. In solving this Problem, it is necessary to assume a principle employed in Arithmetic, viz.. If four quantities are proportional, the product of the extremes is equal to the product of the means. Thus, if a:h\\c : c?, then ad=bc. Prob. 8. Bivide the number a into two such parts that the quotient of the greater divided by the less may be to the quo- tient of the less divided by the greater as m to n. . aVm , aVn A71S. —=z = and Vm 4- Vn Vm -\- Vn Prob. 9. There are two square grass-plats, a side of one of EQUATIONS OF THE SECOND DEGREE. 181 which is 10 yards longer than a side of the other, and their areas are as 25 to 9. What are the lengths of the sides ? Prob. 10. There are two squares whose areas are as m to n, and a side of one exceeds a side of the other by a. What are the lengths of the sides ? . aVm T aVn Ans. —7=z ~ and V Vm — Vn Vm — Vn Prob. 11. Two travelers, A and B, set out to meet each other, A leaving Hartford at the same time that B left New York. On meeting, it appeared that A had traveled 18 miles more than B, and that A could have gone B's journey in 15| hours, but B would have been 28 hours in performing A's journey. What was the distance between Hartford and New York ? Ans, 126 miles. Prob. 12. From two places at an unknown distance, two bodies, A and B, move toward each other, A going a miles more than B. A would have described B's distance in n hours, and B would have described A's distance in m hours. What was the distance of the two places from each other ? -y/m-^-Vn Ans. ax— prr —. ym — yn Prob. 13. A vintner draws a certain quantity of wine out of a full vessel that holds 256 gallons, and then, filling the vessel with water, draws off the same quantity of liquor as before, and so on for four draughts, when there were only 81 gallons of pure wine left. How much wine did he draw each time ? Ans. 64, 48, ^Q, and 27 gallons. Note. Suppose - part is drawn each time. Then 256 —— — ^^ remains after the first draught. Similarly, —^ — - remains after the second draught, and so on. Hence ^^^(--^^^ =81. 182 ALGEBRA. Prob. 14. A number a is diminished by the nth part of it- self, this remainder is diminished by the nth part of itself, and so on to the fourth remainder, which is equal to h. Required the value of n. Ans. —= — -^. Va-Vb Prob. 15. Two workmen, A and B, were engaged to work for a certain number of days at different rates. At the end of the time. A, who had played 4 of those days, received 75 shil- lings, but B, who had played 7 of those days, received only 48 shillings. Now had B only played 4 days and A played 7 days, they would have received the same sum. For how many days were they engaged? Ans. 19 days. Prob. 16. A person employed two laborers, allowing them different wages. At the end of a certain number of days, the first, who had played a days, received m shillings, and the second, who had played h days, received n shillings. Now if the second had played a days, and the other h days, they would both have received the same sum. For how many days were they engaged? hVm—aVn , Ans. — -= ^=- days. Complete Equations of the Second Degree. 256. In order to solve a complete equation of the second de- gree, let the equation be reduced to the form If we can by any transformation render the first member of this equation the perfect square of a binomial, we can reduce the equation to one of the first degree by extracting its square root. Now we know that the square of a binomial, T+a, or ar*4- 2ax-|-^/'*, is composed of the square of the first term, plus twice the product of the first term by the second, plus the square of the second term. Hence, considering x^-\-px as the first two terms of the EQUATIONS OF THE SECOND DEGREE. 185 square of a binomial, and consequently px as being twice the product of the first term of the binomial by the second, it is evident that the second term of this binomial must be ^. 257. In order, therefore, that the expression x^-\-px- may be rendered a perfect square, we must add to it the square of this V second term ^ ; and in order that the equality of the two mem- bers may not be destroyed, we must add the same quantity to the second member of the equation. We shall then have Taking the square root of each member, we have whence, by transposition, xz= _^rb y ^+^. Thus the equation has two roots: one corresponding to the plus sign of the radical, and the other to the minus sign. These two roots are 258. Hence, for solving a complete equation of the second degree, we have the following RULE. 1st. Reduce the given equation to the form ofx?-\-px=q. 2d. Add to each member of the equation the square of half the co- ^cient of the first power of x. 8d. Extract the square root ofhoth members^ and the equation will he reduced to one of the first degree^ which may he solved in the usual manner. 259. When the equation has been reduced to the form x^^ px=q, its two roots will be equal to half the coefficient of the sec- 184 ALGEBRA. ond term, taken with a contrary sign, plus or minus the square root of the second member , increased by the square of half the coeffi- cient of the second term. Ex. 1. Let it be required to solve the equation a;2-10ic=-16. Completing the square by adding to each member the square of half the coefficient of the second term, we have a;2_i0x+25=26-16=9. Extracting the root, a:— 5= ±3. Whence x=6±S= 8 or 2, ^n5. To verify these values of x, substitute them in the original equation, and we shall have 82_i0x8 = 64-80=-16. Also, 22-10x2= 4-20= -16. Ex. 2. Solve the equation 2aj2_|_3^_20 = 70. Ans. x= -f 6 or —9. Ex. 3. Solve the equation Sx'^—Sx-{-6=6l. Eeducing, x'^—x=—^. Completing the square, x'^—x-{-^=\—^=-^. Hence x=^±i=-| or -J, An^, Second Method of completing the Square, 260. The preceding method of completing the square is al- ways applicable ; nevertheless, it sometimes gives rise to incon- venient fractions. In such cases the following method may be preferred. Let the equation be reduced to the form ax'^-\-bx=^c, in which a and b are whole numbers, and prime to each other, but c may be either entire or fractional. Multiply each member of this equation by 4a, and it becomes 4:a^x'^-[-4:abx=4:ac. Adding h^ to each member, we have 4:aV + 4:abx ■i-b^= 4ac + b^, where the first member is a complete square, and its terms are entire. Extracting the square root, we have 2ax-^d=±:V4:ac+b\ EQUATIONS OF THE SECOND DEGREE. 185 Transposing 5, and dividing by 2a, _ -5db V4 ac+6'^ ""- 2^ ' which is the same result as would be obtained by the former rule ; but by this method we have avoided the introduction of fractions in completing the square. If h is an even number, - will be an entire number ; and it would have been sufficient to multiply each member by a, and add J to each member. Hence we have the following RULE. 1st. Reduce the equation to the form ax^-{-hx=c, where a and b are prime to each other. 2d. ^ b is an odd number ^ multiply the equation by four times the coefficient ofx^^ and add to each member the square of the coeffi- cient of X. 3d. If b is an even number , multiply the equation by the coeffi- cient of x^^ and add to each member the square of half the coefficient ofx. Ex. 4. Solve the equation Gx^— 13x=— 6. Multiplying by 4x6, and adding 13^ to each member, we have 144x2-312x+169=rl69-144=25. Extracting the root, 12x— 13 = ± 5. Whence 12x^18 or 8, and x—% or f. Ex. 5. Solve the equation llOa:^— 21cc=— 1. Multiplying by 440, and adding 21^ to each member, we have 48400x2-9240^+441 = 1. Extracting the root, 220x- 21 = ± 1. Whence x=^ ot -^. Ex. 6. Solve the equation 7x2~3x=160. Ans. x=5 or — ?;2.. 261. Modification of the preceding Method. — The preceding method sometimes gives rise to numbers which are unneces- 186 ALGEBRA. sarily large. When the equation has been reduced to the form ax'^-\-bx—c, it is sufficient to multiply it by any number which will render the first term a perfect square. Let the resulting equation be mV-\-nx=:q. The first member will become a complete square by the ad- dition of (h— ) ) ^^^ t^® equation will then be This method is expressed in the following RULE. Having reduced the equation to the form ax^-\-hx=Cj multiply the equation by any quantity {the least possible) which will render the first term a perfect square. Divide the coefficient of x in this new equation by twice the square root of the coefficient of a?^ and add the square of this residt to both members. Ex.7. Solve the equation 8ic2 4-9:r=99. ^9\2 8249 16x2+18a;+Q' 16 4 4' cc=3 or — -— , Ans. o Ex. 8. Solve the equation 16a:^-15:c=34. 16.-15x+(^)' 2401 64 403=8 or — -r, 4' x=z or — tt;, Ans, Ex.9. Solve the equation 12x^-=2l-{-^, , Solve the following equations: 4 Ex. 10. ia;2_^aj + 20^=421. Ans. x=7 or -6^. Ex. 11. cc2-x-40 = 170. Ans. x=l6 or -14. Ex.12. 3a:2 4-2a:-9 = 76. EQUATIONS OF THE SECOND DEGREE. 187 Ex.13. Ja;2— Jx4-7f=8. This equation reduces to x^— 15^0;=— 46. Ans. a;=llf or 4. Ex.15. 1?_I^=^:=2. Ana. x=^ or Q. Ex.16. ^0-^-^=0. cc cc+l cc+2 Ex.17. a;2_xV3=x-^V3. , 1/3 + 3 a/3-1 ^n^. a:= — - — or — ^ — . Ex.18. ^ -=— ^. J.725. a:=5 or 1. x—2 X—4: 3 Ex.19. — ; 1 = -. Ans. x—a or —2a. a+ic X 2 Ex. 20. x'^—(a-\-b)x-{-ab=0. Ans. x—a or &. Ex.21. (3x-25)(7a; + 29)=:0. Ans. x^%\ oy -^\. ^ __ 3.T-2 , 2x-5 10 , 13 1 ^^- 22- 2^35 + 3^-2 = ¥ ^^^- ^=¥ "^ f Ex.23. (ic-l)(a;-2)+(a;-2)(x-4)=6(2a:-5). J.725. a? =8 or ^. „ ^. 170 170 51 , .12 Ex.24. ~ — -. Ans. x—^ or —If. ■CI o^ a^ + aa:: -I- x^ . a^—ax-\-x^ ah hjx. 25. h- a-\-x a—x Sa—4:h-\-x' Ans. x=—Sa or 3a — r-. 6 Equations which may be solved like Quadratics. 262. There are many equations of a higher degree than the second, which may be solved by methods similar to those em- ployed for quadratics. To this class belong all equations which contain only (wo powers of the unknoivn quantity^ and in which 188 ALGEBRA. the greater exponent is double the less. Such equations are of the where n may be either integral •or fractional. For if we assume y=a;**, then y^—x^^ and this equation becomes ^2_|_py_^ . whence £c'*=?/=— ^iy 5'+^. Extracting the nth root of each member, we have Ex.1. Solve the equation cc*-13a;2=-36. Assuming ^=y^ the above becomes 2/2-13?/= -36; whence y=9 or 4. But, since x^=y^ x=±.Vy. Therefore a:= ± V9 or ± Vi. Thus X has four values, viz., +3, —3, +2, —2. To verify these values: 1st value, (4-3)*-13(+3)^=-36, i. e., 81-117=-36. 2d value, (-3)*- 13(-3)2=-36, i.e., 81-117= -36. 3d value, (+2)^-13( + 2)2= -36, i.e., 16- 52= -36. 4th value, (-2)*-13(-2)2= -36,* i.e., 16- 52 = -36. Ex. 2. Solve the equation cr^- 35^3 =—216. Assuming x^=y^ the above becomes 2/'-352/=-216; whence 2/= 27 or 8. Hence x= \/^=3 or 2. This equation has four other roots which can not be de- termined by this process. Ex. 3. Solve the equation a;H-4v^=21. Assuming Vx=y^ we have 2/^+43/=21; EQUATIONS OF THE SECOND DEGREE. 189 whence y=S or —7. Therefore cc=:9 or 49. Although the square toot of 9 is generally ambiguous, and may be either -f 3 or —8, still, in verifying the preceding val- ues, -y/i can not be taken equal to —8, because 9 was obtained by multiplying- + 8 by itself For a like reason, -y^ can not be taken equal to + 7. A similar remark is applicable to sev- eral of the following examples. 263. The same method of solution may often be extended to equations in which any algebraic expression occurs with two exponents^ one of which is double the other. Ex.4. Solve the equation {x'^xf-2^{x''-^x)=-V10, Assuming x^-\-x=y^ this equation becomes 2/^-262/= -120; whence y=20 or 6. We have now the two equations, cc^ -1-03=20, and x'^-\-x=% the first of which gives 03= —5 or +4, and the second gives cc= — 8 or +2. Thus the equation has four roots, -5, +4, -8, +2, and any one of these four values will satisfy the given equa tion. Ex.5. Solve the equation Vic+12+ \/a3+12 = 6. We find ic+12 = 16 or 81. Hence cc=:4 or 69. ^ Ex.6. Solve the equation 2x^^y/2x^-\-l = ll, This equation may be written 2a^2-|-l + V2x2-fl = 12. Hence '2x^ + 1 = ^ or 16; therefore ic=4-2, -2, +V7|, -Vl\. 190 ALGEBRA. 264. Equations of the Fourth Degree. — An equation of the fourth degree may often be reduced to an equation contain- ing the first and second powers of some compound quantity, with known coefficients, in the following manner* Transpose all the terms to the first member; then extract the square root to two terms, and see if the remainder (with or without^ the absolute term) is a multiple of the root already obtained.* Ex.7. Solve the equation ir*-12xH44a;2-.48x=9009. We may proceed as follows : x*-12a;34.44a;2-48cc-9009=0(x2-6a; - 12^:3 + 36x2 8x2-48x-9009, or 8(x2-6a:)-9009. Hence the given equation may be expressed as follows; (x2-6x)24-8(a:2-6a;) = 9009. Ans. x=l% or -7, or 3±3V-10. Ex.8. Solve the equation ic*— 2x^+3; =132. Ans. x=4: or —3, or ^±^'v/_43. Ex.9. Solve the equation x*-\-4:a^ = 12. Ans. x= ± V2 or ± V — 6. Ex. 10. Solve the equation a;®— 83^^=513. *Ans. x=S or — v/l9. 6 3 Ex.11. Solve the equation x'^-\-x'^=766. Ans. a: =243 or -v/28*. Ex.12. Solve the equation Jcc^— ia^= — A- Ans. x=i\/2. Ex.13. Solve the equation 2x^+307^=2. Ans. x=\ or —8. Ex. 14. Solve the equation ^a?— i-v^=22^. Ans. 37=49 or -^. Ex.15. Solve the equation Vl0-{-x-Vl0-^x=2. Ans. x=6 or —9. EQUATIONS OF THE SECOND DEGREE. 191 Ex. 16. Solve the equation a;^-l- 20x^-10 = 59. Ans. x=V^ or v/^=23. Ex. 17. Solve the equation Sx^''-2x''-\-S = ll. Ans, x= v/2 or V—^' Ex.18. Solve the equation Vl+x—x'^—2{l-{-x—x'^) = i. Am. x=^±iV^ or -^i^VII. Ex.19. Solve the equation V^4-Vx=20. Ans. x=:256 or 625. Ex.20. Solve the equation x'^—^x^ + 7x'^-Qx=18. Ans. x=3 or —1, or liV— 5. Ex.21. Solve the equation x'^-\-6x-^4:=5Vx^-}-6x-{'28. Ans. x—4: or —9, or —^±^V—ol, Ex. 22. Solve the equation x'^+B=2Vx^-2x-\-2 + 2x. An^. x—1. Ex.23. Solve the equation {x-{-Vxy-{x-{-Vxf =20692, Ans. ic = 9 or 16? Ex. 24. Solve the equation ic+ V25-|-cc=157. Ans. cc=l44 or 171. Ex.25. Solve the equation Vx—l=x—l. Ans. x=l or 2. 265. We have seen that every equation of the second de- gree has two roots; that is, there are two quantities which, when substituted for x in the original equation, will render the two members identical. In like manner, we shall find that every equation of the third degree has th7'ee roots^ an equa- tion of the fourth degree has four roots, and, in general, an equation of the mth degree has m roots. Before determining the degree of an equation, it should be freed from fractions, from negative exponents, and from the radical signs which affect its unknown quantities. Several of the preceding examples are thus found to furnish equations 192 ALGEBRA. of the fourth degree, while others furnish equations of the second degree. The above method of solving the equation x'^^-\-px^—q will not always give us all of the roots, and we must have recourse to different processes to discover the remaining roots. The subject will be more fully treated in Chapter XXI. Problems producing Equations of iJie Second Degree. Prob. 1. It is required to find two numbers such that their difference shall be 8 and their product 240. Let ir=:the least number. Then will a:+8 = the greater. And by the question, x{x-^8)=x^-\-8x=2'^0. Therefore cc=12, the less number, cc -1-8=20, the greater. Proof. 20—12 = 8, the first condition. 20x12 = 240, the second condition. Prob. 2. The Keceiving Keservoir at Yorkville is a rectan- gle, 60 rods longer than it is broad, and its area is 5500 square rods. Eequired its length and breadth. Prob. 3. What two numbers are those whose difference is 2a, and product hf . Ans. a±ya^-\-o, and —a±Va^+b. Prob. 4. It is required to divide the number 60 into two such parts that their product shall be 864. Let ic=one of the parts. Then will 60— a? = the other part And by the question, x{60—x)-—60x—x^=S64:. The parts are 36 and 24, Ans. Prob. 5. In a parcel which contains 52 coins of silver and copper, each silver coin is worth as many cents as there are copper coins, and each copper coin is worth as many cents as there are silver coins, and the whole are worth two dollars. How many are there of each ? Prob. 6. What two numbers are those whose sum is 2a and Vroduct b? Ans. a-\-VaF—b and a-Va^—b. EQUATIONS O^ THE SECOND DEGEEE. 193 Prob. 7. There is a number consisting of two digits whose sum is 10, and the sum of their squares is 58. Kequired the number. Let ic=:the first digit. Then will 10— cc=the second digit. And a;2_|_(io_^)2^2a^2_20a7+100=58; that is, ic2_iQ^^_2i^ a;2-10a^+25 = 4, ic=5±2 = 7 or 3. Hence the number is 73 or 37. The two values of x are the required digits whose sum is 10. It will be observed that we put x to represent the first digit, whereas we find it may equal the second as well as the first. The reason is, that we have here imposed a condition which does not enter into the equation. If x represent either of the required digits, then 10— a? will represent the other ^ and hence the values of x found by solving the equation should give both digits. Beginners are very apt thus, in the state- ment of a problem, to impose conditions which do not appear in the equation. The preceding example, and all others of the same class, may be solved without completing the square. Thus, Let X represent the half difierence of the two digits. Then, according to the principle on page 89, h-\-x will rep- resent the greater of the two digits, and 5— a? the less. The square of 5+07 is 25 + 10a:+ a?^, 5-07 " 25-1007+ 00" . The sum is 50 +207^, which, according to the problem, =58. Hence 207^= 8, or a?^= 4, and X =±2. Therefore, 5 4- a? =7, the greater digit, and 5—0? =3, the less digit. Prob. 8. Find two numbers such that the product of their sum and difference may be 5, and the product of the sum of their squares and the difference of their squares may be 65. 194 ALGEBRA. Prob. 9. Find two numbers such that the product of theii sum and difference may be a, and the product of the sum of their squares and the difference of their squares may be ma. Ans.^;^l 111 — a Prob. 10. A laborer dug two trenches, whose united length was 26 yards, for 356 shillings, and the digging of each of them cost as many shillings per yard as there were yards in its length. What was the length of each ? A71S. 10, or 16 yards. Prob. 11. What two numbers are those whose sum is 2a, and the sum of their squares is 2h? Ans. a+Vb—d^, and a—Vb—a\ Prob. 12. A farmer bought a number of sheep for 80 dollars, and if he had bought four more for the same money, he would have paid one dollar less for each. How many did he buy ? Let X represent the number of sheep. 80 Then will — be the price of each. 80 And 2 would be the price of each if he had bought four more for the same money. But by the question we have 80^ 80 ^ X ~X + 4: Solving this equation, we obtain a:: =16, Ans. Prob. 13. A person bought a number of articles for a dol- lars. If he had bought 26 more for the same money, he would have paid c dollars less for each. How many did he buy ? Ans. -6±^M±^ Prob. 14. It is required to find three numbers such that the product of the first and second may be 15, the product of the first and third 21, and the sum of the squares ofthe second and third 74. Ans. 8, 5, and 7. EQUATIONS OF THE SECOND DEGREE. 195 Prob. 15. It is required to find three numbers such that the product of the first and second may be a, the product of the first and third Z>, and the sum of the squares of the second and third c. /o^j^^ / c , / A71S. v ; aSj — — ^^ ; o\/ - Prob. 16. The sum of two numbers is 16, and the sum of their cubes 1072. What are the numbers? Ans. 7 and 9. Prob. 17. The sum of two numbers is 2a, and the sum of their cubes is 2b. What are the numbers ? Ans.a + \/^ ^ and «— v/' 3a V 3a • Prob. 18. Two magnets, whose powers of attraction are as 4 to 9, are placed at a distance of 20 inches from each other. It is required to find, on the line which joins their centres, the point where a needle would be equally attracted by both, ad- mitting that the intensity of magnetic attraction varies inverse- ly as the square of the distance. . j 8 inches from the weakest magnet, I or —40 inches from the weakest magnet Prob. 19. Two magnets, whose powers are as m to t?, are placed at a distance of a feet from each other. It is required to find, on the line which joins their centres, the point which is equally attracted by both. The distance from the magnet m is — ^:= Ans The distance from the magnet n is VmdzVn Prob. 20. A set out from C toward D, and traveled 6 miles an hour. After he had gone 45 miles, B set out from D to- ward C, and went every hour -^V of the entire distance ; and after he had traveled as many hours as he went miles in one hour, he met A. Eequired the distance between the places and D, Ans. Either 100 miles, or 180 miles. 196 ALGEBRA. Prob. 21. A set out from C toward D, and traveled a miles per hour. After he had gone h miles, B set out from D toward C, and went every hour ^th of the entire distance ,• and after he had traveled as many hours as he went miles in one hour, he met A. Kequired the distance between the places C and D. ^-"(^-vTW^- Prob. 22. By selling my horse for 24 dollars, I lose as much per cent, as the horse cost me. What was the first cost of the horse? Ans. 40 or 60 dollars. Prob. 23. A fruit-dealer receives an order to buy 18 melons provided they can be bought at 18 cents a piece ; but if they should be dearer or cheaper than 18 cents, he is to buy as many less or more than 18 as each costs more or less than 18 cents. He paid in all $3.15. How many melons did he buy? Ans. Either 15 or 21. Prob. 24. A line of given length (a) is bisected and pro- duced ; find the length of the produced part, so that the rect- angle contained by half the line, and the line made up of the half and the produced part, may be equal to the square on the produced part. Ans.^{l + Vb), Equations of the Second Degree containing Two Unknown Quantities. 266. An equation containing two unknown quantities is said to be of the second degree when the highest sum of the expo- nents of the unknown quaiitities in any term is two. Thus 0^2+2/2 = 13, (1.) and . x-\-xy+y=ll, (2.) are equations of the second degree. 267. The solution of two equations of the second degree containing two unknown quantities generally involves the so- lution of an equation of the fourth degree containing one un- EQUATIONS OF THE SECOND DEGREE. 19T known quantity. Thus, from equation (2), we find Substituting this value for y in equation (1) and reducing, we have an equation which can not be solved by the preceding methods. Yet there are particular cases in which simultaneous equations of a degree higher than the first may be solved by the rules for quadratic equations. The following are the principal cases of this kind : 268. l5^. When one of the equations is of the first degree and the other of the second, — We find an expression for the value of one of the unknown quantities in the former equation, and substi- tute this value for its equal in the other equation. Ex. 1. Given \ ^' + S^^-2/'=23 ) ^^ ^^^ ^ ^^^ From the second equation we find Substituting this value for x in the first equation, we have 49-28?/ 4-%'^+21?/-6y2_2/2^23, which may be solved in the usual manner. Ans. \ ^ Ex. 2. Given { ^""'XZ^^yCf^ \ to find x and y. |?/=3 or -Y= [ 10x-{-y _ \ Ex. 3. Given \ 3 ~^^ [• to find x and y. (92/-9a;=18i . (cc=2 or —\, (2/=4 or |. For equations of this class there are in general two sets of values of x and y. 198 ALGEBRA. 269. 2c?. When both of the equations are of the second degree^ and homogeneous. — Substitute for one of the unknown quanti- ties the product of the other bj a third unknown quantity. Ex. 4. Given j 9 Z_i [ *^ ^^^ ^ ^^^ V- If we assume x^vy^ we shall have 12 y2y2 _^ yy2 _ -[^^ whencc 2/2 Therefore v'^-\-v v—2' From which we obtain v = 8or3. Substituting either of these values in one of the preceding expressions for 2/^ we shall obtain the values of ?/; and since x=vyj we may easily obtain the values of x. o x=±S or ±-— . Ans.i ^ il—±\ or =t-T=. ■ V6 Ex. 5. Given \ ^'■^^^"' J^ \ to find aj and y. Assuming x=vy^ we find ^=2; ^^ "o"- a:=±7 or ±--. Ans.< y= ±4 or ±— p' Ex. 6. Given { '^JiXtS' " 1 ^ ^^^ ^ -^ V' Assuming x=vy^ we find v=- or — . a:= ±1 or ± EQUATIONS OF THE SECOND DEGEEE. 199 For equations of this class there are in general four sets of values of x and y. It should be borne in mind that to any one of the four values of x there corresponds only one of the four values of y. Thus, when x in the 6th example is +1, y must be -4-8, and can not be one of the other three values given above. 270. Sd When the unknown quantities entei' each equation sym- metrically. — Substitute for the unknown quantities the sum and difference of two other quantities, or the sum and product of two other quantities. Ex. 7. Given I y x >• to find x and y. { x+y=12 ) Let us assume x=z-\-v, y = z-'V. Then x-\-y=2z = 12 or z = 6. That is, X—6 + V and y=Q—v. But from the first equation we find x^-\-y^ = 18xy. Substituting the preceding values of ic and y in this equation, and reducing, we have 432+36?;2=648-18i;2, whence v=±2. Therefore cc= 4 or 8, and 3/ =8 or 4. Ex. 8. Given \ ^ r to find x and ?/. x=S or 5. y=^5 or 3. Ex. 9. Given ] f . ^ 7 non C t^ ^^^ ^ 3,nd y. ix^y+xy^ = SSO) ^ 5 or 6. '=6 or 5. Ans. I Ans. } 271. 4:th. When the same algebraic expression is involved to different powers^ it is sometimes best to regard this expression as the unknown quantity. 200 ALGEBRA. Ex.10. Given \ ^ ^^^ -« o > tofindxandy. The first equation may be written (x+2/)'4-2(x+i/) = 120. Regarding x-\-y as a single quantity, we find its value to be either 10 or -12. Proceeding now as in Art. 268, we find x=Q or 9, or — QifVS; y=4 or 1, or — SiVS. Ex. 11. Given \ ^V^J^-ft \ to find x and y. Regarding xy as the unknown quantity, its value from the first equation is found to be either 8 or -12. j x=2 or 4, or 3± V2T. (y=4 or 2, or 3=fV21. /^ 4x_85 \ Ex. 12. Given Xy"^ 3/ ~~ 9 >• to find ic and y, ( a;-2/=2 ) Regarding - as the unknown quantity, we find its value to if be either 5 17 [2/=3or-T^. For several of these examples there are other roots, some of which can not be obtained by the processes heretofore ex- plained. The roots of two simultaneous equations are some- times infinite, as in the equations a?^— a?y=:8, and 3:^—^^ = 12, since two quantities that are infinitely great may difier by a finite quantity. Solve the following groups of simultaneous equations : x^^2xy-f^\\ . (a:=±-v/|. -4-^ — 1 ) ( — " Ex. 13. EQUATIONS OF THE SECOND DEGREE. 201 U+2/4-V^=14:f (2/=2 or 8. cc=6 or 30. 30 or 6. [xy 18 ) Ex 17 i(^^+2/^)^¥=468) . \x ^^'^^' \ {x^.y)xy=^0 \ ^""''Xy x=2, 3. Ex.18. I rn\\ ^n..i^=^*^^*^- Ex.19. \ ^ ^ [ ^^-{^=64 018. Ex. 20. ^ 1.15 5- Arts, j cc=l or 4 3/=4 or 1. (r=±2. Ex.21, o , i^iy ^ns. I x—y=S ) \y=^ or —5. 4-ccy +2/* = 931 [ . j x= ±5 or ±3 Ex.23. T:^^^~:r Ans Ex. 24. a;2+a:?/+2/2=:49 j * (^=±3 or ±5 ((7+a;)(6+2/)-80) ^^ p-l or 3. ( x-^y=6 ) ' \y-4: or 2. JVbiJtf. Put 7+07=2!, 6-\-y=v, PROBLEMS. 1. Divide tlie number 100 into two such parts that the sum of their square roots may be 14. Ans. 64 and 36. 2. Divide the number a into two such parts that the sum of their square roots may be b. , Ans. 2^2'^2a—b\ 8. The sum of two numbers is 8, and the sum of their fourth powers is 706. What are the numbers ? Ans. 3 and 5. 12 202 ALGEBRA. 4. The sum of two numbers is 2a, and the sum of their fourth powers is 2b. What are the numbers? Ans, a ± V—Sa^±VSa*-{-b. 5. The sum of two numbers is 6, and the suni of their fifth powers is 1056. What are the numbers? Ans. 2 and 4. 6. The sum of two numbers is 2a, and the sum of their fifth powers is b. What are the numbers ? 10a 5 7. What two numbers are those whose product is 120 ; and if the greater be increased by 8 and the less by 6, the product of the two numbers thus obtained shall be 800 ? Ans. 12 and 10, or 16 and 7.5. 8. What two numbers are those whose product is a; and if the greater be increased by b and the less by c, the product of the two numbers thus obtained shall be df m . /m^ ah Ans. 7r±\/-: , and 2^4 c r?? /m^ ah VT-7' where m = 2 V 4 c d—a—bc c 9. Find two numbers such that their sum, their product, and the difference of their squares may be all equal to one another. 4„. 1+4 and 1 + 4 that is, 2.618, and 1.618, nearly. 10. Divide the number 100 into two such parts that their product may be equal to the difference of their squares. Ans. 88.197, and 61.803. 11. Divide the number a into two such parts that thcii product may be equal to the difference of their squares. . Sa±ay^ , —a::^aV6 Am. ?j and ^r . EQUATIONS OF THE SECOND DEGREE. 203 12. The sum of two numbers is a, and the sum of their re- ciprocals is h. Eequired the numbers. An. l^vfTl General Properties of Equations of the Second Degree. 272. Every equation of the second degree containing hut one unknown quantity has two roots, and only two. We have seen, Art. 250, that every equation of the second degree containing but one unknown quantity can be reduced to the form x^-{-px=q. We have also found, Art. 257, that this equation has two roots, viz., ^=-f +\/^+|-, and _|_y/^+|-. This equation can not have more than two roots ; for, if pos- sible, suppose it to have three roots, and represent these roots by x', x'\ and x'^\ Then, since a root of an equation is such a number as, substituted for the unknown quantity, will satisfy the equation, we must have x'^+px'=:q, (1) x"''^px"^q, (2.) x"'''-\-px"'^q. (3.) Subtracting (2) from (1), we have x"'-x"''^p{x'-x'')^(^. Dividing by x' —x" , we have (cc'+cc")+^=0. (4.) In the same manner, we find {x'-\-x"')\p=.^. (5.) Subtracting (5) from (4), we have that is, the third supposed root is identical with the second ; hence there can not be three different roots to a quadratic equation. 273. The algebraic sum of the two roots is equal to the coeffi- cient of the second term of the equation taken with the contrary sign. 204 ALGEBRA. If we add together tlie two values of x in the general equa- tion, the radical parts having opposite signs disappear, and we obtain ^ _ :2 — _ ■~2~2~ ^' Thus, in the equation cc^— 10x=— 16, the two roots are 8 and 2, whose sum is + 10, the coefficient of x taken with the contrary sign. If the two roots are equal numerically, but have opposite signs, their sum is zero, and the second term of the equation vanishes. Thus the two roots of the equation x^=lQ are -f4: and —4, whose sum is zero. This equation may be written ic2+0x=16. 274. The product of the two roots is equal to the second member of the equation taken with the contrary sign. If we multiply together the two values of x (observing that the product of the sum and difference of two quantities is equal to the difference of their squares), we obtain Thus, in the equation cc^— 10x=:— 16, the product of the two roots 8 and 2 is +16, which is equal to the second mem- ber of the equation taken with the contrary sign. 275. The last two principles enable ns to form an equation whose roots shall be any given quantities. Ex. 1. Find the equation whose roots are 3 and 6. According to Art. 273, the coefficient of the second term of the equation must be — 8 ; and, according to Art. 274, the sec- ond member of the equation must be —15. Hence the equa- tion is a;2— 8a:=— 15. Ex. 2. Find the equation whose roots are —4 and —7. Ex. 3. Find the equation whose roots are 5 and —9. Ex.4. Find the equation whose roots are —6 and 4-11. Ex. 5. Find the equation whose roots are 1 and -^2. Ex. 6. Find the equation whose roots are — ^ and + J. Ex. 7. Find the equation whose roots arc —-J and +i. EQUATIONS OF THE SECOND DEGREE. 205 Ex. 8. Find the equation whose roots are 1 ± \/5. Ex. 9. Find the equation whose roots are 1 ± V^^. 276. Every equation of the second degree whose roots are a and bjinay be reduced to the form (a:— a)(cc— 6)=0. Take the general equation x'^-\-px=q, and write it x^-\-px—q—0. Then, by Art. 273, p=-{a-j-b)] and by Art. 2 74, ^ = — «^- Hence, by substitution, x'^—{a-\-b)x-{-ab=0] or, resolving into factors, {x-a){x-h)=0. Thus the equation cc^— 10a:=— 16, whose roots are 8 and 2, may be resolved into the factors cc— 8 = and a:— 2 = 0. It is also obvious that if a is a root of an equation of the second degree, the equation must be divisible by x—a. Thus the preceding equation is divisible by a;— 8, giving the quo- tient x—2, Ex. 1. The roots of the equation x^-{-6x-\-8 = are —2 and —4. Eesolve the first member into its factors. Ex. 2.' The roots of the equation cc2_j_6a;—27 = are -f3 and —9. Resolve the first member into its factors. Ex. 3. The roots of the equation cc^— 2a:— 24 = are +6 and —4. Resolve the first member into its factors. Ex.4, Resolve the equation a;2-f 73x4-780 = into simple factors. A71S. {x-\-60) {x+ 13) = 0. Ex.5. Resolve the equation a;'^—88a:+ 1612 = into simple factors, Ans.{x—62){x-26)=0. Ex. 6. Resolve the equation 20-^+3:— 6 = into simple fac- tors. Ans. 2{x+2){x-^)=0. Ex.7. Resolve the equation 3a:2— 10a:— 25 = into simple factors. A ns. 3 (a: — 5) ^x + f ) = 0. 206 ALGEBRA. Discussion of the General Equation of the Second Degree. 277. In the general equation of the second degree x'^-\-px=q. the coefficient of x^ as well as the absolute term, may be either positive or negative. We may therefore have the four fol- lowing forms: First form. Second form, Third form, Fourth form. x^-^-pxrzzq. x^—px — q. x^-\-px=—q, x^—px——q. From these equations we obtain — f-v?-?- We will now consider what conditions will render these roots positive or negative, equal or Unequal, real or imaginary. 278. Positive and negative roots. Since X~^^ ^^ greater than ^, s/ P I ^ ^.,^4. T 4. 41 P ~-f 5' must be greater than ^. For the same reason, y 4: —9. niust be less than ^. Therefore, in the first and second forms, the sign of the roota will correspond to the sign of the radicals; but in the third and fourth forms the sign of the roots will correspond to the sign of the rational parts. Hence, in the first form^ one root is positive and the other negative^ and the negative root is numerical- ly the greatest; as in the equation x^-\-x=zQ^ whose roots are -4-2 and -3. EQUATIONS OF THE SECOND DEGREE. 207 In the second form one root is positive and the other negative^ and the positive root is numerically the greatest^ as in the equa- tion x^—x=Q^ whose roots are —2 and +3. In the third form both roots are negative, as in the equation x2+5x— — 6, whose roots are —2 and —3. In the fourth form both roots are positive, as in the equatioq. 0-2— 5x=— 6, whose roots are +2 and +3. 279. Equal and unequal roots. In the first and second forms the tiuo roots are always unequal; but in the third and fourth forms, when q is numerically equal to ^, the radical part of both values of x becomes zero, and the two roots are then said to be equal. In this case the third form gives x=- and the fourth form gives the third form gives x= — ~±{)= — ^, Thus, in the equation a:^+6x=— 9, the two roots are —3 and —3. We say that in this case the equation has two roots, because it is the product of the two factors a:: 4-3 = and a:-h3 = 0. So, also, in the equation x^—^x=^^^, the two roots are +3 and +3. 280. Ileal and imaginary roots. nrp> Since ^, being a square, is positive for all real values of ^, it follows that the expression ^^rq can only be rendered neg- ative by the sign of q ; that is, the quantity under the radical sign can only be negative when q is negative and numerically 092 greater than ■!—. Hence, in the first ayid second forms, both roots are always real; but in the third and fourth forms both roots are imaginary when q is numerically greater than ^. 208 ALGEBRA. Thus, in the equation x^-\-4:X=—6^ the two roots are and in the equation x'^—4:x=—6, the two roots are +2±i/:i2. It will be observed that when one of the roots is imaginary, the other is imaginary also. 281. Imaginary roots indicate impossible conditions in the pro- posed question which furnished the equation. The demonstration of this principle depends upon the fol- lowing proposition : the greatest product which can he obtained by dividing a number into two parts and multiplying them iogetJi' er is the square of half that number. Let p represent the given number, and c? the difference of the parts. , Then, from page 89, ^ + - = the greater part, and f ~ 2 ~ *^^ ^^^ ^^^* and A~~~I~ *^® product of the parts. Now, since ^ is a given quantity, it is plain that the prod- uct will be the greatest possible when c?=0; that is, the great- est product is the square of |-, half the given number. For example, let 10 be the number to be divided. We have 10 = 1 + 9; and 9x1= 9. 10 = 2 + 8; and 8x2 = 16. 10=3 + 7; and 7x3=21. 10=4 + 6; and 6x4 = 24. 10 = 5+5; and 5x5 = 25. Thus we see that the smaller the difference of the two parts, the greater is their product; and this product is greatest when the two parts are equal. Now, in the equation x^-~px——q^ p is the sum of the two roots, and q is their product. Therefore q can never be great- er than -v- 4 EQUATIONS OF THE SECOND DEGREE. 209 Ifj then, any problem furnislies an equation in which q is negative, and numerically greater than ^, we infer that the conditions of the question are incompatible with each other. Suppose it is required to divide 6 into two parts such that their product shall be 10. Let X represent one of the parts, and 6— x the other part Then, by the conditions, (c(6-cc) = 10; whence cc^ — 6cc = — 10, and cc = 3±V — 1. The imaginary value of x indicates that it is impossible to find two numbers whose sum is 6 and product 10. From the preceding proposition, it appears that 9 is the greatest product which can be obtained by dividing 6 into two parts and multi- plying them together. Discussion of Particular Prohlems, 282. In discussing particular problems which involve equa- tions of the second degree, we meet with all the different cases which are presented by equations of the first degree, and some peculiarities besides. All the different cases enumerated in Chapter X. are presented by Prob. 19, page 195, when we make different suppositions upon the values of a, m, and n; but we need not dwell upon them. The peculiarities exhibited by equations of the second de- gree are double values of x, and imaginary values. 283. Double Values of the Unhnown Quantity. — We have seen that every equation of the second degree has two roots. Some- times both of these values are applicable to the problem which furnishes the equation. Thus, in Prob. 20, page 195, we obtain either 100 or 180 miles for the distance between the places and D. C E D I L^ — i Let E represent the position of A when B sets out on hia 210 ALGEBRA. journey. Then, if- we suppose CD equals 100 miles, ED will equal 65 miles, of which A will travel 80 miles (being 6 miles an hour for 5 hours), and B will travel 25 miles (being 5 miles an hour for 5 hours). If we suppose CD equals 180 miles, ED will equal 135 miles, of which A will travel 54 miles (being 6 miles an hour for 9 hours), and B will travel 81 miles (being 9 miles an hour for 9 hours). This problem, therefore, admits of two positive answers, both equally applicable to the question. Problems 22 and 28, page 196, are of the same kind. In Problem 18, page 195, one of the values of x is positive and the other negative. C A C B » Let the weaker magnet be placed at A, and the stronger at B ; then C will represent the position of a needle equally at- tracted by both magnets. According to the first value, the distance AC = 8 inches, and CB = 12 inches. Now, at the dis- tance of 8 inches, the attraction of the weaker magnet will be 4 represented by ^ ; and at the distance of 12 inches, the at- 9 traction of the other magnet will be represented by z-92) and these two powers are equal ; for 4__9_ 32" 122" But there is another point, »C', which equally satisfies the conditions of the question ; and this point is 40 inches to the left of A, and therefore 60 inches to the left of B ; for 402 ■"602* 284. Imaginary Values of the Unknown Quantity. — We have seen that an imaginary root indicates impossible conditions in the proposed question which furnished the equation. In sev- eral of the preceding problems the values of x become imag- inary in particular cases. EQUATIONS OF THE SECOND DEGREE. 211 When will the values of x in Prob. 6, page 192, be imag- inary ? Alls. When h > o?. What is the impossibility involved in this supposition? Ans. It is impossible that the product of two numbers can be greater than the square of half their sum. When will the values of x in Prob. 11, page 194, be imag- inary ? Ans. When a?>b; or {2af > 4J). What is the impossibility involved in this supposition? Ans. The square of the sura of two numbers can not be greater than twice the sum of their squares. When will the values of x in Prob. 17, page 195, be imag- inary ? Ans. When a^>h; or {2af > 85. What is the impossibility of this supposition ? Ans. The cube of the sum of two numbers can not be great- er than four times the sum of their cubes. When will the values of x in Prob. 4, page 180, be imag- inary, and what is the impossibility of this supposition ? 285. Geometrical Construction of Equations of the Second De- gree. — The roots of an equation of the second degree may be represented by a simple geometrical figure. This may be done for each of the four forms : First form. — The first form gives for x the two values ^=-f +V4H' and x^-^-\/^q. Draw the line AB, and make it equal to y^. From B draw ^ .^ BC perpendicular to AB, and make it equal to ^. Join A and C ; then AC will repre- sent the value of\J^-\-q. For AC^^ AB2+BC2 (Geom., Prop. 11, Bk. lY.). With C as a centre, and CB as a radius, describe a circle cutting AC in D, and AC produced in E. For the first value of x the radical is positive, and is set off from A toward C ; then — ^ is set off from C to D ; and AD. estimated from A to D, represents the first value of cc. \ 212 ALGEBRA. For the second value of x we begin at E, and set off EC equal to — ^ ; we then set off the minus radical from C to A j and EA, estimated from E to A, represents the second value of X. JSecond form. — The second form gives for x the two values The first value of x is represented by AE estimated from A to E. The second value is +DC — CA, the latter being esti- mated from C to A. Hence the second value is represented by DA estimated from D to A. Third form. — The third form gives for x the two values Draw an indefinite line FA, and from any point, as A, set off a distance AB=— ^. We set off its , value to the left, because ^ is negative. IP D- -'E A At B draw BC perpendicular to FA, and make it equal to ^/~q. With C as a centre, and a radius equal to ■^, describe an arc of a circle cutting FA in D and E. Now the value of Vx"^ "^^ ^® ^-^ ^^ ■^^- "^^^ ^^* ^^^^ ^^ X will be represented by — AB+BE, which is equal to — AE. The second will be represented by — AB— BD, which is equal to —AD; so that both of the roots are negative, and are esti- mated in the same direction, from A toward the left. Fourth form. — The fourth form gives for x the two values Construct the radical part of the value of a; as in the last case. Then, since ^ is positive, we set off its value AB from EQUATIONS OF THE SECOND DEGREE. 213 Q A toward the right. To AB we add BD, / which gives AD for the first value of x; / , and from AB we subtract BE, which leaves ^ ^- ...'^ -^ AE for the second value of x. Both val- ues are positive, and are estimated in the same direction, from A toward the right. Equal Roots. — If the radius CE be taken equal to CB, that is, if Vq is equal to -^j the arc described with the centre C will be tangent to AF, the two points D and E will unite, and the two values of x become equal to each other. In this case the radical part of the value of x becomes zero. Imaginary Roots. — If the radius of the circle described with the centre C be taken less than CB, it will not / meet the line AF. In this case q is nnmerical- ly greater than ^, and the radical part of the V value of X is imaginary. 214 ALGEBRA. CHAPTER XV. RATIO AND PROPORTION. 286. Ratio is the relation whicli one quantity bears to an- other with respect to magnitude. Ratio is denoted by two points like the colon ( : ) placed between the quantities com- pared. Thus the ratio of a to 6 is written a : 6. The first quantity is called the antecedent of the ratio, and the second the corisequent. The two quantities compared are called the terms of the ratio, and together they form a couplet The quantities compared may be polynomials; nevertheless, each quantity is called one term of the ratio. 287. A ratio is measured by the fraction whose numerator is the antecedent and whose denominator is the consequent of the ratio. Thus the ratio of a to 6 is measured by 7. 288. A compound ratio is the ratio arising from multiplying together the corresponding terms of two or more simple ratios. Thus the ratio of a to & compounded with the ratio of c to 0? becomes ac to bd. The ratio compounded of the ratios 3 to 5 and 7 to 9 is 21 to 45. 289. The duplicate ratio of two quantities is the ratio of their squares. Thus the duplicate ratio of 2 to 3 is 4 to 9 ; the duplicate ratio of a to & is a^ to h^. 290. The triplicate ratio of two quantities is the ratio of their cubes. Thus the triplicate ratio of a to 6 is a^ to h^. 291. If the terms of a ratio arc both multiplied or both divided RATIO AND PROPORTION. 215 hy the same quantity^ the value of the ratio remains unchanged. The ratio of a to & is represented by the fraction 7, and the value of a fraction is not changed if we multiply or divide both numerator and denominator by the same quantity. Thus , a _ma OT a\h—ma'.mh=z-:-. n n PROPORTION. 292. Proportion is an equality of ratios. Thus, if a, h^ c, d are four quantities such that a when divided by h gives the same quotient as c when divided by c?, these four quantities are called proportionals. This proportion may be written thus, a : 6 : : c : c?, or a:b=c.dj or ^-- In either case the proportion is read a is to b as c is to d. 293. The terms of a proportion are the four quantities which are compared. The first and fourth terms are called the ex- tremes^ the second and third the means. The first term is called ihQ first antecedent^ the second term the^r^^ consequent^ the third term the second antecedent^ and the fourth term the second con- sequent. 294. When the first of a series of quantities has the same ratio to the second that the second has to the third, or the third to the fourth, and so on, these quantities are said to be in continued proportion^ and any one of them is a mean propor- tional between the two adjacent ones. Thus, if a : b i\b : c y, c : d :\ d : e^ then a, 5, c, c?, and e are in continued proportion, and 5 is a mean proportional between a and c, c is a mean proportional between b and c?, and so on. 216 ALGEBRA. 296. Alternaiion is when antecedent is compared with ante- cedent and consequent with consequent. Thus, if a:b:: c : dj then, by alternation, a:c::b:d. See Art. 801. 296. Inversion is when antecedents are made consequents, and consequents are made antecedents. Thus, if a:b :: c: dj then, inversely, b: a::d: c. See Art. 802. 297. Composition is when the sum of antecedent and conse- quent is compared with either antecedent or consequent. Thus, if a b:: c: dj then, by composition, a-\-b: a:: c-\-d: c, and a-\-b:b::c-\-d:d. See Art. 304. 298. Division is when the difference of antecedent and con- sequent is compared with either antecedent or consequent Thus, if a:b::c:dj then, by division, a—b a:: c—d : c, and a—b :b:: c—d : d. See Art. 805. 299. If four quantities are in proportion, the product of the ex- tremes is equal to the product of the means. Let a:b::c:d. Then I = J, Art. 292. Multiplying each of these equals by 5c?, we have ad=bc, 300. Conversely, if the product of two quantities is equal to the product of two other quantities^ the first two may be made Vie ex- tremes^ and the other two the means of a proportion. Let ad=bc. Dividing each of these equals by bd, we have q_ c b~d' or a:b::c:d, Art. 292. RATIO AND PROPORTION. 217 EXAMPLES. 1. Given the first three terms of a proportion, 24, 15, and 40, to find the fourth term. 2. Given the first three terms of a proportion, Sab^, A^aW^ and 9a^5, to find the fourth term. 3. Given the last three terms of a proportion, 4a^5^, 2>aW^ and 2a% to find the first term. 4. Given the first, second, and fourth terms of a proportion, 6/, 7x2^^, and 21x^?/, to find the third term. 5. Given the first, third, and fourth terms of a proportion, a-j-J, a^—b^j and {a — bf^ to find the second term. Which of the following proportions are correct, and which are incorrect? 6. 3a+46:9a+8J::a-26:8a-45. 7. da^-^W : 16a'-25ab+8b^ ::16a^+26ah-\-8b^: 25a''-16b^, 8. a^->rb^ : a'+b^ :: a^-b^ : a-b, 9. a^-\-b^ : a+b:: a'-a^b-^-aW-a^b^-i-a^-b' : a^-b\ 301. If four quantities are in proportion^ they will be in pro- portion when taken alternately. Let a:b::c:d; then a_c h~d' Multiplying by 6, ^ = ^- Dividing by c, ^ = ^, or a:c:'.h'.d. 302. If four quantities are in proportion^ ifiey will be in pro- portion when taken inversely. Let a:b::c:d; then 1 = ^. Divide unity by each of these equal quantities, and we hare b^d a c' OT b: a:: d: c. K 218 ALGEBRA. 303. Ratios that are equal to the same ratio are equal to each other. If a:h::m:n, (1.) and c:d::m:n, (2.) then a:b::c:cL a _m From proportion (1), From proportion (2), ^ = "" c m Hence a _c or a:h\:c:d. 304. If four quantities are proportimial^ they wiU be propor- tional by composition. Let a:b::c:d; ^x. a c then T = J- b d Add unity to each of these equals, and we have b^ d^"-' . ^ . a+5 c-\-d that IS, -r = -^» or a-\-b:b:: c-\-d: d. 305. If four quantities are proportional^ they will he propor- iional by division. Let a:b::c:d; then -j = ^. Subtract unity from each of these equals, and we have b ^-d ^' ,, , . a—h c—d that 18, -r = -d-' or a — b -.b:: c—d: d. KATIO AND PROPORTION. 219 306. If four quantities are proportional^ the sum of the first and second is to their difference^ as the sum of the third and fourth is to their difference. Let a:b::c:d. By composition, Art. 804, a-\-b:b:: c-{-d: d. By alternation, Art. 301, a-{-b : c-j-c?:: b: d. Also by division, Art. 805, a — b :b:: c—d: d; by alternation, a—b:c—d::b:d. By equality of ratios. Art. 808, a-\-b: c-{-d:: a—b: c—d, or a-\-b: a—b:'.c-{-d\ c—d. 307. If four quantities are in proportion^ any equimultiples of the first couplet will be proportional to any equimultiples of the second couplet. Let a:b::c: d; then T = ^• b d Multiply both terms of the first fraction by m, and both terms of the second fraction by n, and we have ma _ nc mb ~ nd'' or ma :mb:: nc: nd, 308. If four quantities are in proportion, any equimultiples of the antecedents will be proportional to any equimultiples of the con- sequents. Let a:b::c:d; then a _ c b~d' Multiply each of these equals by m, and we have ma _ mc ~F~~d' 220 ALGEBRA. Divide each of these equals by w, ma _ mc nb nd* or ma :nb:: mc: nd. 309. If any number of quantities are proportional, any one an- tecedent is to its consequent as the sum of all the antecedents is to the sum of all the consequents. Let a:b :: c: d :: e:f; then, since a:b::c:dj ad = bc; (1.) and, since a:b::e:f af=be; (2.) also ab = ba. (3.) Adding (1), (2), and (3), ab-{-ad-\-af= ba-{-bc-\-be; that is, a {b-\-d-{-f) = b{a-\-c+e). Hence, ^r^. 300, a:b:: a-{-c+e:b-\-d+f 310. If there are two sets of proportional quantiAies^ the prod- ucts of the corresponding terms will be proporti(mt>L Let a'.bwc'.d, and e:f::g'.h; then Qje\bf'.\cg\dk, *-i a c and -c — T- Multiplying together these equal quantities, we have ae _cg bf~dh: or ae:bf::cg:dh. 311. If four quantities are in proportion^ like powers or roots pf these quantities will also be iri proportion. RATIO AND PROPORTION. 2! Let a:b: :c:d; then a c Eaising each of these equals to the nth power, ''dJ^' we obtain that is, a" : h^ : : c"" : c?". In the same manner, we find 1 1 1. 1 : c» : dn. 312. If three quantities are in continued proportion^ the product of the extremes is equal to the square of the mean. If a:h::b: Cj then, bj Art. 299, ac=bb=h\ 313. If three quantities are in continued proportion, the first is to the third in the duplicate ratio of the first to the second. Let a:b::b: c ; then i = i c Multiply each of these equals by y, and we have a a a b that is, or 314. If four quantities are in continued proportion, the first ia to the fourth in the triplicate ratio of the first to the second. Let a:b::b: c:: c: d; then 1 = 1 (1.) and 1 = '- (2.) V '^b' b c' a a^ c ~62' a : c : : a^ : b\ h~ d' a _a h~b' 1 a a ,- . also ^ = T. (3.) 222 ' ALGEBRA. Multiplying together (1), (2), and (3), we have a? _ dbc __ a nence a:d::a^:b\ VARIATION. 315. Proportions are often expressed in an abridged form. Thus, if A and B represent two sums of money put out for one year at the same rate of interest, then A : B : : interest of A : interest of B. This is briefly expressed by saying that the interest varies as the principal. A peculiar character ( a) is used to denote this relation. Thus interest a principal denotes that the interest varies as the principal. 316. One quantity is said to vary directly as another when the two quantities increase or decrease together in the same ratio. Thus, in the above example, A varies directly as the interest of A. In such a case, either quantity is equal to the other multiplied by some constant number. Thus, if the interest varies as the principal, then the interest equals the product of the principal by some constant number, which is the rate of interest. If A a B, then A = mB. If the space (S) described by a falling body varies as the square of the time (T), then S = mT2, where m represents a constant multiplier. 317. One quantity may vary directly as the product of sev- eral others. Thus, if a body moves with uniform velocity, the space de- scribed is measured by the product of the time by the velocity. If we put S to represent the space described, T the time of motion, and V the uniform velocity, then we shall have S a T X V. RATIO AND PKOPORTION. 228 Also the area of a rectangalar figure varies as the product of its length and breadth. The weight of a stick of timber varies as the product of its length X its breadth x its depth x its density. 318. One quantity is said to vary inversely as another when the first varies as the reciprocal of the second. Thus, if the area of a triangle be invariable, the altitude varies inversely as the base. If the product of two quantities is constant, then one varies inversely as the other. In uniform motion the space described is measured by the product of the time by the velocity ; that is, SaTxY; S whence '^ ^ v* If the space be supposed to remain constant, then Ta —' that is, the time required to travel a given distance varies in- versely as the velocity. Conversely^ if one quantity varies inversely as another, the product of the two quantities is constant Thus, if T a ^, then the product of T by Y is equal to a constant quantity. 319. One quantity is said to vary directly as a second^ and inversely as a third^ when it varies as the product of the second by the reciprocal of the third. Thus, according to the New- tonian law of gravitation, the attraction (G) of any heavenly body varies directly as the quantity of matter (Q), and inverse- ly as the square of the distance (D). That is, ^^% 224 ALGEBRA. EXAMPLES. * — fi ' \ ^^ ^"^ ^ ^^^ y* Since x+?/: ic:: 5 : 3, bj division, -4r^. 805, y : x ::2 :S. 2x Hence 3?/=:2cc, and y=-^. Substituting this value of y in the second equation, we ob- tain 2^2 Therefore x = ± 8, and 2/ =±2. 2. Given j ^3 ~_V^ ' [ to find x and y. From the first equation, Art. 306, 2x : 2y : : 4 : 2 ; whence x = 2?/. Substituting this value of x in the second equation, we find ?/=2, and cc=4. 3. Given -^ ^ ^ ^^ ^ ^J V to find a; and y, { xy—K>o ) '' By Art 311, cc+y : x—y : : 8 : 1. By ^r^. 306, 2aj : 2^/ : : 9 : 7. Hence x = -^. Substituting this value of x in the second equation, we find ?/=±7, and ir=±9. (a;='_2/3:(x-y)3::61:l) , . , , 4. Given i ^ ^ ^^^ }• to find x and y, { a:?/=320 ) From the first equation, by division. Art. 805, Zxy{x^y)\{x-yY:'.m'.l, Hence 960 : {x--yf : : 60 : 1, or 16 ; {x-yf : : 1 : 1. KATIO AND PROPORTION. 226 Therefore x—y=±4:. Hence x^—2xy-^y^ = 16, and 4xy=12S0. By addition, x^-h2xy+y^=12m. Hence x+y=dtS6. Therefore cc=: ±20 or ±16, and 2/=±16or±20. ( X -— TJ ' X 1/ — Xl/ ' ' I ' 2 I 5. Given \ ^ • ^ 2 * ' ' r to find x and y. . i x=4: or 2. Ans. { :2 or 4. 6. Given -j "^ — i to find x and y. Vy—x-\- ya—x : ya—x : : 5 : 2 ) 4a 7. Given ic+ V^ : x—Vx :: SVx-{-6 : 2Vx to find x. Ans. x=9 or 4. 8. What number is that to which if 1, 5, and 13 be severally added, the first sum shall be to the second as the second to the third? Ans.S. 9. What number is that to which if a, b, and c be severally added, the first sum shall be to the second as the second to the third? . h'^—ac Ans. ■ a — 2h^-c 10. What two numbers are as 2 to 3, to each of which if 4 be added, the sums will be as 6 to 7? y 11. What two numbers are as m to n^ to each of which if a be added, the sums will be as p to 5- .^ Ans. ^-^(>^-y) . (in{p-q) ^ mq—np ' mq — np ' 12. What two numbers are those whose difierence, sum, and product are as the numbers 2, 3, and 5 respectively ? Ans. 2 and 10. K2 226 ALGEBRA. 13. What two numbers are those whose difference, sum, and product are as the numbers m, n, and p ? Arts, — — and — —. n-\-m n—m 14. Find two numbers, the greater of which shall be to the less as their sum to 42, and as their difference to 6. Arts. 32 and 24. 15. Find two numbers, the greater of which shall be to the less as their sum to a and their difference to h. Ans. g±^, and ^* 2{a—b) 2 16. There are two numbers which are in the ratio of 3 to 2, the difference of whose fourth powers is to the sum of their cubes as 26 to 7. Kequired the numbers. Ans. 6 and 4. 17. What two numbers are in the ratio of m to tz, the differ- ence of whose fourth powers is to the sum of their cubes as V ^ Q^ 3,3 3.3 Ans, —^ X— 7 i, and -^ x— ^ .. q m* — n^' q rn^ — ii* 18. Two circular metallic plates, each an inch thick, whose diameters are 6 and 8 inches respectively, are melted and form- ed into a single circular plate 1 inch thick. Find its diameter, admitting that the area of a circle varies as the square of its diameter. 19. Find the radius of a sphere whose volume is equal to the sum of the volumes of three spheres whose radii are 3, 4, and 5 inches, admitting that the volume of a sphere varies as the cube of its radius. 20. Find the radius of a sphere whose volume is equal to the sum of the volumes of three spheres whose radii are r, r', and r". PROGRESSIONS. 227 CHAPTER XYI PROGRESSIONS. ARITHMETICAL PROGRESSION. 3^0. An arithmetical progression is a series of quantities which increase or decrease by a common difference. Thus the fol- lowing series are in arithmetical progression : 1, 8, 5, 7, 9, . . . 20, 17, 14, 11, 8, . . . a, a-\-dj a-^2dj a + 3c?, . . . a, a—d, a — 2d, a— So?, . . . In the first example the common difference is -f2, and the series forms an increasing arithmetical progression ; in the sec- ond example the common difference is —3, and the series forms a decreasing arithmetical progression. In the third example the common difference is 4-c/, and in the fourth example it is —d. 321. In an arithmetical progression having a finite number of terms, there are five quantities to be considered, viz., the first term, the last term, the number of terms, the common differ- ence, and the sum of the terms. When any three of them are given, the other two may be found. We will denote , the first term by a, the last term by ?, the number of terms by n, the common difference by cf, and the sum of the terms by s. The first term and the last term are called the extremes j and all the other terms are called arithmetical means. 322. In an arithmetical progression the last term is equal to the first term plus the product of the common difference hy the number of terms less one. 228 ALGEBRA. Let the terms of the series be represented bj a, a-fo?, a + 2dj a-\-Sdj aH-4(i, etc. Since the coefficient of c? in the second term is 1, in the third term 2, in the fourth term 3, and so on, the nth term of the series will be a-\-{n—l)dj or l=a-\-{n—l)dj in which d is positive or negative according as the series is an increasing or a decreasing one. 323. The sum of any number of terms in arithmetical progres' sion is equal to one half the sum of the two extremes multiplied by the number of terms. The term preceding the last will be l—d^ the term preceding that Z— 2(i, and so on. If the terms of the series be written in the reverse order, the sura will be the same as when written in the direct order. Hence we have 5=a-h(a-f cZ)-}-(a+2c^)-f (a4-3cZ)4- .... +?, s=l J^ll-d)^{l -2d)^{l -%d)+ +a. Adding these equations term by term, we have 2s={a^-l) + {a^T) + {a-\-l)+ 4-(a+0- Here a+Z is taken n times; hence 25=.?2(a4-Z), or 5=:-(a + 0- 324. Tn an arithmetical p^'^^ogression the sum of the extremes is equal to the sum of any tivo terms equidistant from the extremes. This principle follows from the preceding demonstration. It may also be shown independently as follows : The mth term from the beginning is a-\-{m — V)d. The mth term from the end is l—{m—\)d. And the sum of these terms is a + Z. 325. To insert any number of arithmetical means between two given terms. The whole number of terms in the series consists of the two PROGRESSIONS. 229 extremes and all the intermediate terms. If, then, m repre- sents the number of means, m + 2 will be the whole number of terms. Substituting m4-2 for n in the formula, Art. 322, we have Z=a + (m-f l)c?, the common difference, or d= w + 1 whence the required means are easily obtained by addition. 326. The two equations contain five quantities, a, Z, w, c?, 5, of which any three being given, the other two can be found. We may therefore have ten different cases, each requiring the determination of two dif- ferent formulae. These formulae are exhibited in the following table, and should be verified by the student. No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Given, a, C?, W, ?, d, n, a, Z, ??, a, ??, 5, 72, C?, 5, Z, 72, 5, a, cZ, Z, a, ?, 5, a, £?, 5, Ij d, 5, Re- quired. a, s, cZ, 5, dj, a, I a,d, 72,5, 72, C?, / =ra + (72 — l)cZ; a = Z— (^ — l)c?; Z — a 771:1' 2.s-2a?2 n(72-l)' s (77-iyi 72 2 25 , arr Z; 72 I— a , ^ Formulae. 5=-J?2[2a + 0?-iy]. 5zz=^72[2Z-(72-l)6Z]. s=^2(a + Z). , 25 I = a. n , _s {n-V)d ^ -^ + "~2~- cZ= 27iZ-25 72(72—1)* 72 = d 25 a + Z' _ (Z+q)(Z— g-t-f?) ^ ~ 2(^ c?= 2s-a-l' Z, 72, /^-i.f±V 2c/.+Ca-^c^)» ; ,^^-2a±V(2a-cf)' + 8^. 2c« 230 ALGEBKA. EXAMPLES. 1. The first term of an arithmetical progression is 2, and the common difference is 4 ; what is the 10th term ? Ans. 38. 2. The first term is 40, and the common difierence —3 ; what is the 10th term? 3. The first term is 1, and the common difference f ; what is the 10th term? 4. The first term is 1, and the oemmon difference —J; what is the 10th term ? 5. The first term is 5, the common difference is 10, and the number of terms is 60 ; what is their sum ? Alts. 18000. 6. The first term is 116, the common difference is —4, and the number of terms is 25 ; what is their sum ? 7. The first term is 1, the common difference is f , and the number of terms is 12 ; what is their sum ? 8. The first term is 1%, the common difference is — | ; and the number of terms is 10 ; what is their sum ? 9. Required the number of terms of a progression whose sum is 442, whose first term is 2, and common difference 8. Am. 17. 10. Required the first term of a progression whose sum is 99, whose last term is 19, and common difference 2. 11. The sum of a progression is 1465, the first term 5, and the last term 92 ; what is the common difference? 12. Required the sum of 101 terms of the series 1, 3, 5, 7, 9, etc. Aiis. 10201. 13. Find the nth term of the series 1, 3, 5, 7, 9, etc. Ans. 27i-l. 14. Find the sum of n terms of the series 1, 3, 5, 7, 9, etc. Aiis. n\ 15. Find the sum of n terms of the series of numbers 1, 2, 3, 4, 5, etc. n(M-l) Ans. Q *^ PROGRESSIONS. 231 16. Find the sum of n terms of the series 2, 4, 6, 8, etc. Arts. n(n+l). 17. Find 6 arithmetical means between 1 and 50. 18. Find 7 arithmetical means between -J and 8. 19. A body falls 16 feet during the first second, and in each, succeeding second 82 feet more than in the one immediately preceding ; if it continue falling for 20 seconds, how many feet will it pass over in the last second, and how many in the whole time? Ans. 624 feet in the last second, and 6400 feet in the whole time. 20. One hundred stones being placed on the ground in a straight line at the distance of two yards from each other, how far will a person travel who shall bring them one by one to a basket which is placed two yards from the first stone ? Ans. 20200 yards. PROBLEMS. 327. When of the five quantities a, Z, ??, c?, s, no three are directly given, it may be necessary to represent the series by the use of two unknown quantities. The form of the series which will be found most convenient will depend upon the conditions of the problem. If x denote the first term and y the common difference, then a?, a?-j-?/, a: +2?/, a? +3?/, etc., will represent a series in arithmetical progression. It will, however, generally be found most convenient to rep- resent the series in such a manner that the common difference may disappear in taking the sum of the terms. Thus a pro- gression of three terms may be represented by one of four terms by x—Sy, x—y, x-\-y, x-]-Sy; one of five terms by x—2y^ ^—Vi ^i ^+2/? ^ + 2i/. Prob. 1. A number consisting of three digits which are in arithmetical progression, being divided by the sum of its digits, gives a quotient 26 ; and if 198 be added to itj the digits will be inverted ; required the number. Ans. 284. 232 ALGEBRA. Prob. 2. Find three numbers in arithmetical progression the sum of whose squares shall be 1232, and the square of the mean greater than the product of the two extremes by 16. Ans. 16, 20, and 24. Prob. 3. Find three numbers in arithmetical progression the sum of whose squares shall be a, and the square of the mean greater than the product of the two extremes by b. /a -2b rr /a -2b , /a-2b ,, Ans. V— g Vb; V — 3-5 ^^^ V— 3- + ^^- Prob. 4. Find four numbers in arithmetical progression whose sum is 28, and continued product 685. Ans. 1, 5, 9, 13. Prob. 5. A sets out for a certain place, and travels 1 mile the first day, 2 the second, 3 the third, and so on. In five days afterward B sets out, and travels 12 miles a day. How long will A travel before he is overtaken by B ? Ans. 8 or 15 days. This is another example of an equation of the second de- gree, in which the two roots are both positive. The following diagram exhibits the daily progress of each traveler. The di- visions above the horizontal line represent the distances trav- eled each day by A ; those below the line the distances trav- eled by B. A. 12345 6 7 8 9 10 11 12 13 14 15 I NI I I I I I I I I I I I I \ \ T \ \ i i \ i \ I B. 123456789 10 It is readily seen from the figure that A is in advance of B until the end of his 8th day, when B overtakes and passes him. After the 12th day, A gains upon B, and passes him on the 15th day, after which he is continually gaining upon B, and could not be again overtaken. Prob. 6. A goes 1 mile the first day, 2 the second, and so on. B starts a days later, and travels b miles per day. How long will A travel before he is overtaken by B ? . 2b^\±V{2b-lf-^ab . Ans, — ^ — days. PROGRESSIONS. 238 In what case would B never overtake A ? (25-1)2 Am. W hen a > ^ — ^, • . oo For instance, in the preceding example, if B had started one day later, he could never have overtaken A. Prob. 7. A traveler set out from a certain place and went 1 mile the first day, 8 the second, 5 the third, and so on. After he had been gone three days, a second traveler sets out, and goes 12 miles the first day, 13 the second, and so on. After how many days will they be together ? Ans. In 2 or 9 days. Let the student illustrate this example by a diagram like the preceding. Prob. 8. A and B, 165 miles distant from each other, set out with a design to meet. A travels 1 mile the first day, 2 the second, 3 the third, and so on. B travels 20 miles the first day, 18 the second, 16 the third, and so on. In how many days will they meet ? Ans. 10 or 33 days. GEOMETRICAL PROGRESSION. 328. J. geometrical progression is a series of quantities each of which is equal to the product of the preceding one hy a constant factor. The constant factor is called the ratio of the series. 329. When the first term is positive, and the ratio greater than unity, the series forms an increasing geometrical progres- sion, as 2, 4, 8, 16, 32, etc., in which the ratio is 2. When the ratio is less than unity, the series forms a decreas- ing geometrical progression, as 81, 27, 9, 3, etc., in which the ratio is -J. 330. In a geometrical progression having a finite number of terms, there are five quantities to be considered, viz., the first 234 ALGEBRA. term, tlie last term, the number of terms, the ratio, and the sum of the terms. When any three of these are given, the other two may be found. We will denote the first term by a, the last term by i, the number of terms by w, the ratio by r, and the sum of the terms by s. The first term and the last term are called the extremes^ and all the other terms are called geometrical means. 331. In a geometrical progression^ the hst term is equal to the product of the first term by that power of the ratio whose exponent is one less than the number of terms. According to the definition, the second term is equal to the first multiplied by r, that is, it is equal to ar ; the third term is equal to the second multiplied by r, that is, it is equal to wr^'^ the fourth term is equal to the third multiplied by r, that is, it is equal to ar^ ; and so on. Hence the nth term of the series will be equal to ar^"^ ; hence we shall have l=ar'^-^. 332. To find the sum of any number of terms in geometrical progression, multiply the last term by the ratio^ subtract the first term^ and divide the remainder by the ratio less one. From the definition, we have s=a-\-ar-\-ar^-\- , . . . -l-ar^-^H-ar'*-^ Multiplying this equation by r, we have rs=ar-{-ar^-^ .... -f ar^-^+a^""^-!- «^**- Subtracting the first equation from the second, member from member, we have rs—s=ar^—a. ar^ — a Hence s= r— ; r— 1 or, substituting the value of I already found, we have rl—a PROGRESSIONS. 235 If we had subtracted the second equation from the first, we should have found a—rl which is the most convenient formula when r is less than unity, and the series is, therefore, a decreasing one. 333. To find the sum of a decreasing geometrical series when the number of terms is infinite, divide the first term hy unity diminished hy the ratio. The sum of the terms of a decreasing series may be repre- sented by the formula a—rl Now, in a decreasing series, each term is less than the pre- ceding, and the greater the number of terms, the smaller will be the last term of the series. If the number of terms be in- finite, the last term of the series will be less than any assigna- ble number, and rl may be neglected in comparison with a. In this case the formula reduces to 1-r 334. To find any number of geometrical means between two given terms. In order to solve this problem, it is necessary to know the ratio. If m represent the number of means, 772 + 2 will be the whole number of terms. Hence, putting m-{-2 for n in the formula, Art. 331, we have l=ar'^+^] 1 /Am+l whence we obtain ^— W That is, to find the ratio, divide the last term by the first term, and extract the root which is denoted hy the number of means plus one. Having found the ratio, the required means may be ob- tained by continued multiplication. 236 ALGEBRA. 335. The two equations l=ar^-\ s: ar'^—a contain five quantities, a, I, n, r, 5, of which any three being given, the other two can be found. We may therefore have ten different cases, each requiring the determination of two quan- tities, thus giving rise to twenty different formulae. The first four of the following cases are readily solved. The fifth and sixth cases involve the solution of equations of a higher degree than the second. When n is not large, the value of the un- known quantity can generally be found by a few trials. The four remaining cases, when n is the quantity sought, involve the solution of an exponential equation. See Art. 416. These different cases are all exhibited in the following table for con- venient reference. No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Given. a,r, n, a, Z, ?i, a, w, 5, Z, ri,5, a, Z, 5, a,r,s, I, r, 5, Re- quired. Z, 5, a, I, r,?, r, n, Z, 71, I =ar'^-'^ s =- Formulse. ar'^—a a= _Z_^ y.n-1' (r-Vys s =■ r-1 Ir^'—l »n— 1* p-l_^n-l 7*1—1 ^n-\ t — a ar'^—rs = a—s; ?(5— Z)'»-^ = a(5 — a)*»-\ a(5-a)"-i=rZ(5-Z)»»-i ; (5-Z)?-"-5?'"-i= -Z. 5 = Ir—a s— a ^37' ioff. Z— losf. a log. ?' 4-1. log. Z— log. a log. (.9-a)-log. (.9-Z) + 1. a-\.{r-l)s ^ ^^ log. [a-f-(/'-l>-]-log. a r ' log. r a=Zr— (r— l)s,- n: log. Z— log. [Zr—(r— !).-?] log. r + 1. PROGRESSIONS. 287 EXAMPLES. 1. Find the 12th term of the series 1, 3, 9, 27, etc. We have 7=ar«-i = 3^^ = 177147, Ans. 2. Given the first term 2, the ratio 3, and the number of terms 10 ; to find the last term. Ans. 39366. 3. Find the sum of 14 terms of the series 1, 2, 4, 8, 16, etc. 5=^^^^=2^^-1 = 16383, Ans. r — 1 4. Find the sum of 12 terms of the series 1, 3, 9, 27, etc. Ans. 265,720. 5. Given the first term 1, the last term 512, and the sum of the terms 1023 ; to find the ratio. 6. Given the last term 2048, the number of terms 12, and the ratio 2 ; to find the first term. 7. Find the sum of 6 terms of the series 6, 4f, 3f , etc. Ans. 19|i|. 8. Find the sum of 15 terms of the series 8, 4, 2, 1, etc. Ans. lo-jj^jT^-g^. 9. Find three geometrical means between 2 and 162. 10. Find two geometrical means between 4 and 256. 11. Find three geometrical means between a and b. Ans. Va^b, \/ab, Vab^. 12. Find the value of 1+J-|-J+-|-+, etc., to infinity. =- — T=2, Ans. 1-r 1-i 13. Find the value of l4--J+-^4--2V+, ^^c, to infinity. Ans. I". 14. Find the value ofl+i+-iV4-A+, etc., to infinity, 15. Find the ratio of an infinite progression whose first term^ is 1, and the sum of the series |-. Ans. ^. 16. Find the first term of an infinite progression whose ratio is tVj and the sum f . Ans. f. 17. Find the first term of an infinite progression of which the ratio is -, and the sum 7l' 71 — 1' 18. Find the value of the series 34-2-}-|+, etc., to infinity 238 ALGEBRA. 19. Find the value of the series 3^+l + f +, etc., to infinity. 20. A gentleman, being asked to dispose of his horse, said he would sell him on condition of receiving one cent for the first nail in his shoes, two cents for the second, and so on, doubling the price of every nail to 82, the number of nails in his four shoes. What would the horse cost at that rate ? Ans. $42,949,672.95. PROBLEMS. Prob. 1. Find three numbers in geometrical progression such that their sum shall be 21, and the sum of their squares 189. Denote the first term by x and the ratio by y ; then X -{• xy ^ xy^ =21, (1.) x''-\-xhf-{-xY = 189. (2.) Transposing xy in Eq. (1), squaring, and reducing, we have x^ -h xy + xY = 441 - 42a;?/. (8.) Comparing (2) and (8), xy=6j or ir=-. Substituting this value of x in Eq. (1), and reducing, we have Whence y=2 or ^, and x=3 or 12. The terms are therefore 3, 6, and 12, or 12, 6, and 3. Prob. 2. Find four numbers in geometrical progression such that the sum of the first and second shall be 15, and the sum of the third and fourth 60. By the conditions, x-\-xy=zl5, (1.) xy''-\-xy^ = 60. (2.) Multiplying Eq. (1) by 3/^, we have xy^ -h xy^ = 16y'^ = 60. Therefore 2/^ = 4, and y = ± 2. Also £c±2a;=15; therefore x=6 or —15. Taking the first value of x and the corresponding value of y, we obtain the series 5, 10, 20, and 40. Taking the second value of x and the corresponding value oft/, we obtain the series —15, 4-80, —60, and 4-120; PKOGRESSIONS. 239 which numbers also perfectly satisfy the problem understood algebraically. If, however, it is required that the terms of the progression be positive, the last value of x would be inapplica- ble to the problem, though satisfying the algebraic equation. Several of the following problems also have two solutions, if we admit negative values. Prob. 3. Find three numbers in geometrical progression such that their sum shall be 210, and the last shall exceed the first by 90. Ans. 30, 60, and 120. Prob. 4. Find three numbers in geometrical progression such that their sum shall be 42, and the sum of the first and last shall be 34. Ans. 2, 8, and 32. Prob. 5. Find three numbers in geometrical progression such that their continued product may be 64, and the sum of their cubes 584. Ans. 2, 4, and 8. Prob. 6. Find four numbers in geometrical progression such that the difference between the first and second may be 4, and the difference between the third and fourth 36. Ans. 2, 6, 18, and 54. Prob. 7. Find four numbers in geometrical progression such that the sum of the first and third may be a, and the sum of the second and fourth may be h. Ans. -:— — rr, — 77,, -^ r^, aUQ Prob. 8. Find four numbers in geometrical progression such that the fourth shall exceed the second by 24, and the sum of the extremes shall be to the sum of the means as 7 to 3. Ans. 1, 3, 9, and 27. Prob. 9. The sum of $700 was divided among four persons, whose shares were in geometrical progression, and the differ- ence between the greatest and least was to the difference be- tween the means as 37 to 12. What were their respective shares ? Ans. 108, 144, 192, and 256. Prob. 10. Find six numbers in geometrical progression such that their sum shall be 1365, and the sum of the third and fourth shall be 80. Ans. 1, 4, 16, 64, 256, and 1024. 240 ALGEBBA. CHAPTER XYII. CONTINUED FRACTIONS.— PERMUTATIONS AND COMBINATIONS.^ 336. A continued fraction is one whose numerator is unity, and its denominator an integer plus a fraction, whose numera- tor is likewise unity, and its denominator an integer plus a fraction, and so on. The general form of a continued fraction is 1 a-fl b+1 c+1 d-\-j etc. When the number of terms a, h, c, etc., \s> finite^ the continued fraction is said to be terminating ; such a continued fraction may be reduced to an ordinary fraction by performing the op- erations indicated. 337. To convert any given fraction into a continued fraction. Let - be the given fraction ; divide m by n; let A be the n quotient, and^ the remainder: thus, nun P Divide nhjp; let a be the quotient, and q the remainder: thus, ^ . ? .1 — =a4---=aH — . P P P ? Similarly, £=6+^=^+^? CONTIlSrUED FRACTIONS. 241 and so on, so that we have' :=A+1 'J m n a + 1 & + , etc. We see, then, that to convert a given fraction into a contin- ued fraction, we proceed as if we were finding the greatest com- mon divisor of the numerator and denominator; and we must, therefore, at last arrive at a point where the remainder is zero, and the operation terminates ; hence every rational fraction can he converted into a terminating continued fraction. Ex. 1. Transform -g-jy into a continued fraction. 1 Ans. 3 + 1 22 + 1 1+i. Ex. 2. Transform -144 '^^'^^ a continued fraction. 1 Ans. 1 + 1 1 + 1 1+1^. Ex. 3. Transform ^|4 i^^to a continued fraction. 1 Ans. 2 + 1 1 + 1 2 + 1 -L-r-g-T* Ex. 4. Transform |44- i^^^o a continued fraction. Ex. 5. Transform f|-^ into a continued fraction. Ex. 6. Transform \^ into a continued fraction. 338. To find the value of a terminating continued fra/^tion, Ex. 1. Find the value of the continued fraction 1 2 + 1 3 + i- Beginning with the last fraction, we have L 242 ALGEBBA. 3+i=J#. Hence 1 _4 3+i~"- Therefore 2+3+i=- and 1 a+i-^'^"*- Ex.2. Find the value of the continued fraction 1 3+1 2+1 4+^ Ex.8. Find the value of the continued fraction 2 + 1 3 + 1 2+1 2 + i Ex.4. Find the value of the continued fraction 1 1 + 1 ^ 2+1 1 + 1 1 + 1 l+^Sj. 339. To find the value of an infinite continued fraetion. Let the fraction be 1 a+1 i + 1 c+, etc. An approximate value of this fraction is obtained by omit- ting all its terms be3'^ond any assumed fraction, and obtaining the value of the resulting fraction, as in the previous article. CONTINUED FBACTIONS. 243 Thus we obtain 1st approximate value, - ; 2d approximate value, a + 1 = ; b 3d approximate value, a-\-l , ^ {ab-\-l)c+a' 4th approximate value, ^ ^ iab-{-l)cd+ad-{-ab + lj etc. 340. The fractions formed by taking one, two, three, etc., of the quotients of the continued fraction are called converging fractions, or conver gents. The convergents, taken in order, are alternately less and great- er than the continued fraction. The first convergent - is too great, because th& denominator a , a is too small ; the second convergent , is too small, be- cause a-f T is too great, and so on. 341. When a fraction has been transformed into a continued fraction, its approximate value may be found by taking a few of the first terms of the continued fraction. Thus an approximate value of -^^r ^s -J, which is the first term of its continued fraction. By taking two terms, we obtain f^, which is a nearer ap- proximation ; and three terms would give a still more accu- rate value. Ex. 1. Find approximate values of the fraction iW^ . AnS. f, -g-, -y-f. Ex. 2. Find approximate values of the fraction ■|44- Ex. 3. Find approximate values of the fraction H^. 244 ALGEBRA. 342. By the preceding method we are enabled to discover the approximate value of a fraction expressed in large num- bers, and this principle has some important applications, par- ticularly in Astronomy. Ex. 4. The ratio of the circumference of a circle to its diam^ ter is 3.1415926. Find approximate values for this ratio. AnQ 22 333 35S Ex. 5. The length of the tropical year is 365d 5A. 48m. 48& Find approximate values for the ratio of bh. 48m. 48^. to 24 hours. Ans. J, "5^, ^, yVt- Ex. 6. In 87969 years the earth makes 277287 conjunctions with Mercury. Find approximate values for the ratio of 87969 to 277287. ArtQ X fi 7 13 33 JlllS. -J, Yy^, Tj-g, ^fTj TIFT- Ex. 7. In 57551 years the earth makes 36000 conjunctions with Yenus. Find approximate values for the i:atio of 57551 to 36000. i 11 o 8 235 Ex. 8. In 295306 years the moon makes 3652422 synodical revolutions. Find an approximate value for the ratio of 295306 to 3652422. Ans. ^. Ex. 9. One French metre is equal to 3.2809 English feet. Find approximate values for the ratio of a metre to a foot. Ex. 10. One French kilogramme is equal to 2.2046 pounds avoirdupois. Find approximate values for the ratio of a kilo- gramme to a pound. Ex. 11. One French litre is equal to 0.2201 English gallons. Find approximate values for the ratio of a litre to a gallon. PERMUTATIONS AND COMBINATIONS. 343. The different orders in which things can be arranged are called their perrmdations. In forming permutations, all of the things or a part only may be taken at a time. Thus the permutations of the three letters a, Z>, c, taken all together, are ahc, acb. hac. hca, cab, cha. PERMUTATIONS AND COMBINATIONS. 245 The permutations of the same letters taken two at a time are ahj aCj ha^ be, ca, cb. The permutations of the same letters taken one at a time are a, 5, e. 344. Tlie number of permutations of n things taken m at a time is equal to the continued piroduct of the natural series of numbers from n down to 7i — m-\-l, Suppose the things to be n letters, a^b, c, d The number of permutations of n letters, taken singly or one at a time, is evidently equal to the number of letters, or to n. If we wish to form all the permutations of n letters taken two at a time, we must write after each letter each of the 71—1 remaining letters. We shall thus obtain 7z(n— 1) permu- tations. If we wish to form all the permutations of n letters taken three at a time, we must write after each of the permutations of n letters taken two at a time each of the n — 2 remaining let- ters. We shall thus obtain n{n—l){n — 2) permutations. In the same manner we shall find that the number of permu- tations of n letters taken four at a time is n{n-l){n-2){n-S). Hence we may conclude that the number of permutations of n letters taken m at a time is n{n-l){n-2)(n-S) (n-m + 1). 345. The number of permutations of n things taken all together is equal to the continued product of the natural series of ^lumbers from 1 to n. If we suppose that each permutation comprehends all the n letters; that is, if m=:?2, the preceding formula becomes n{n-l){n-2) 8x2x1; or, inverting the order of the factors, 1.2.3.4 (n-V)n, which expresses the number of permutations of ri things taken all together. 246 ALGEBRA. For the sake of brevity, 1.2.3.4.... {n—l)n is often de- noted by \n; that is, \n denotes the product of the natural num- bers from 1 to 71 inclusive. 346. The combinations of things are the different collections which can be formed out of them without regarding the orc^er in which the things are placed. Thus the three letters a, b, c, taken all together, form but one combination, abc. Taken two and two, they form three combinations, ab, ac, be. 347. 77ie numher of combinations ofn things^ taken m at a timCj is equal to the continued product of the natural series of numbers from n down to n—m-\-l divided by the continued product of the natural series of numbers from 1 to m. The number of combinations of n letters taken separately, or one at a time, is evidently n. The number of combinations of n lett,ers taken two at a time IS \ . For the number of permutations of n letters taken two at a time is n{n — l\ and there are two permutations (ai, ba) corre- sponding to one combination of two letters ; therefore the num- ber of combinations will be found by dividing the number of permutations by 2. The number of combinations of n letters taken three at a . n(n-l)(n-2) time IS — i^ — ^\ ^. For the number of permutations of 7i letters taken three at a time is n{n — l){n — 2\ and there are 1.2.3 permutations for one combination of these letters ; therefore the number of com- binations will be found by dividing the number of permuta- tions by 1.2.3. In the same manner we shall find the number of combina- tions ofn letters taken m at a time to be y?(?i-l)(n-2) .... (?i— m+1) 1X3 m PERMUTATIONS AND COMBINATIONS. 247 EXAMPLES. 1. How many different permutations may be formed of 8 letters taken 5 at a time ? Ans. 8.7.6.5.4= 6720. 2. How many different permutations may be formed of the 26 letters of the alphabet taken 4 at a time ? Ans. 858800. 3. How many different permutations may be formed of 12 letters taken 6 at a time ? Ans. 665280. 4. How many different permutations may be formed of 8 things taken all together? ^725.1.2.8.4.5.6.7.8=40320. 5. How many different permutations may be made of the letters in the word Roma taken all together ? 6. How many different permutations may be made of the letters in the word virtue taken all together? 7. What is the number of different arrangements which can be formed of 12 persons at a dinner-table ? Ans. 479001600. 8. How many different combinations may be formed of 6 letters taken 3 at a time ? Ans. = 20. L.Jj.O 9. How many different combinations may be formed of 8 letters taken 4 at a time ? Ans. 70. 10. How many different combinations may be formed of 10 letters taken 6 at a time ? Ans. 210. 11. A telegraph has m arms, and each arm is capable of 72 distinct positions; find the total number of signals which can be made with the telegraph. 12. How many different numbers can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, each of these digits occurring once, and only once, in each number ? 248 ALGEBKA. CHAPTER XYIII. BINOMIAL THEOREM. 348. The binomial theorem, or binomial formula^ is a formula discovered by Newton, by means of which we may obtain any power of a binomial x+a, without obtaining the preceding powers. 349. By actual multiplication, we find the successive powers of ic-f a to be as follows: {x -\-ay=x^-{-2ax-\- a^, {x + af =x^ + Sax^-^ Sa'^x + a^, {x + ay = a;* + 4:ax^ + Qa^x"^ -^4:a^x+ a*, {x-{-a.y=x^-\-6ax'' + 10aV + 10a^x'-\-5a*x-\-a^. The powers of x — a, found in the same manner, are as fol- lows: {x — ay =x^— 2ax + a'^, (a?— a)^ — a;^ — 3ax2 4- 8a^a?— a^, {x--ay=x*—4:ax^-\-6aV—4:a^x-\-a*, {x-ay=x^-6ax^+10a''x^-10a^x^+6a''x-~a\ On comparing the powers of x-\-a with those of x—a, we perceive that they only differ in the signs of certain terms. In the powers of cc+a, all the terms are positive. In the powers of x—a, the terms containing the odd powers of a have the sign minus, while the terms containing the even powers have the sign plus. The reason of this is obvious; for, since —a is the only negative term of the root, the terms of the power can only be rendered negative by a. A term which contains the factor —a an even number of times will therefore be positive; if it contain it an odd number of times it must be negative. Hence it appears that it is only necessary to seek for a method of ob- taining the powers of ir+a, for those will become the powers of a:— a by simply changing the signs of the alternate terms. BINOMIAL THEOREM. 249 350, Law of the Exponents. — The exponents of cc and of a in the different powers follow a simple law. In the first term of each power, x is raised to the required power of the binomial ; and in the following terms the exponents of cc continually de- crease by unity to zero, while the exponents of a increase by unity from zero up to the required power of the binomial. 351. Law of the Coefficients. — The coefficient of the first term is unity ; that of the second term is the exponent of the power ; and the coefficients of terms equidistant from the extremes are equal to each other ; but after the first two terms it is not ob- vious how to obtain the coefficients of the fourth and higher powers. In order to discover the law of the coefficients, we will form the product of several binomial factors whose second terms are all different; thus, {x-{-a){x^h) = x^-\-a x-\-ah. {x^a){x-\-h){x-^c). (x-\-a){x-^h){x-\-c){x-{'d). •.x^-{-a x^-^-ah -\-ac -\-hc x-^abc. x* + a x'+ab x'^-\-ahc ■\-h -\-ac -\-ahd 4-c + hc -^acd ^d -\-ad -^hd -i-cd + hcd x-\-ohccL In each of these products the exponent of cc in the first term fs equal to the number of binomial factors, and in the follow- ng terms continually decreases by one. The coefficient of the first term is unity ; the coefficient of the second term is the sum of the second terms of the binomial factors ; the coefficient of the third term is the sum of all their products taken two and two, and so on. T7ie last term, is the product of the second terms of the binomial factors. L 2 250 ALGEBRA. 352 We will now prove that if the laws of formatwn jusi stated are true for any power^ they will also hold true for the forma- tion of the next higher power. Suppose that we have found the product of m binomials x-\-a^ x+bj .... x-\-k. Let Pj denote the sum of the second terms of the binomials, Pg the sum of the different products of these second terms taken two and two, P3 the sum of their products taken three and three, and so on ; and let Fm denote the product of all these second terms. The product of the given binomials will then be x'"-\-F^x'^-'^ + 'P^x'^-^-]-F^x^-^ 4-Pm. Multiplying this polynomial by a new binomial, x-\-l, we ot)* tain the following product: ^+HPi + 1 .m-l_,_p^ + ZP + ^P,n-l X + ^Pr + P2 + ^P, The law of the exponents of x remains the same. The co- efficient of the first term is still equal to unity, and that of the second term is the sum of the second terms of the -m + l bino- mials. The coefficient of the third term consists of the sum of the products of the second terms of the m binomial factors taken two and two, increased by the sum of the same second terms multiplied by Z, which is equivalent to the sum of the products of the second terms of the m+1 binomials taken two and two. The coefficient of the fourth term consists of the sum of the products of the second terms of the m factors of the first product taken three and three, increased by the sum of the products of their second terms taken two and two multiplied by ?, which is equivalent to the sum of the products of the sec- ond terms of the m-f-l binomials taken three and three, and so on. The last term is equal to the product of the second terms of the m binomial factors multiplied by I, which is equiv- alent to the product of the second terms of the m-f 1 binomials. Hence tlic law which was supposed true for m factors is true for m + l factors; and therefore, since it has been verified for two factors, it is true for three; being true for three fiictors, it is also true for four, and so on ; therefore the law is general. BINOMIAL THEOREM. 251 353. Powers of a Binomial. — If now, in the preceding bino- mial factors, we suppose the second terms to be all equal to a, the product of these binomials will become the mth power of x-\-a. The coefficient of the second term of the product becomes equal to a multiplied by the number of factors ; that is, it is equal to ma. The coefficient of the third term reduces to a^ repeated as many times as there are different combinations of m letters m^iifYi 1^ taken two and two ; that is, to — \r-^ — -a^. The coefficient of the fourth term reduces to a^ repeated as many times as there are different combinations of m letters 1 • m(m—l){m—2) „ , taken three and three; that is — ^ — Too ^^ ) ^^^ ^^ o^* The last term will be a^. Hence the m\h power of a?H-a may be expressed as follows: {x + ay = x"^ H- max^-^ + ^Yo" ' 1.2 / ^3^m-3 1.2:3 m(m— l)(m— 2) , ^ „ . , 354. We perceive that if the coefficient of any term he multi- plied hy the exponent of x in that term^ and the product he divided hy the exponent of a in that term increased hy unity ^ it will give the coefficient of the succeeding term. Forming thus the seventh power of cc+a, we obtain (ir+ay=a^^ + 7a:rH21aV4-35a3xH35aV + 21aV+7a«a;+a^ We have thus deduced Sir Isaac Newton^s Binomial Theorem. 355. In any power of a binomial jr + a, the exponent of x hegins, tn the first term with the exponent of the povjer, and in the follow- ing terms continually decreases hy one. Tlie exponent of a com mences with one in the second term of the power ^ and continually increases hy one. The coefficient of the first term is one^ that of the second is the ex 252 ALGEBRA. jponent of the power ; and if the coefficient of any term he muUi- plied hy the exponent ofx in that term^ and divided by the exponent of a increased hy one^ it will give the coefficient of the succeeding term. 356. The coefficient of the nth term from the beginning is equal to the coefficient of the nth term from the end. If we change the places of cc and a, we shall have, by the law of formation, (a + xY =a^+ mxa'"'-^ + ^^^^7^ x^a-^-^ + m(wi— l)(m— 2) „ o L.A.O The second member of this equation is the same as the sec- ond member of the equation in Art. 353, but taken in a reverse order. Comparing the two, we see that the coefficient of the second term from the beginning is equal to the coefficient of the second term from the end ; the coefficient of the third from the beginning is equal to that of the third from the end, and so on. Hence, in forming any power of a binomial, it is only necessary to compute the coefficients for half the terms ; we then repeat the same numbers in a reverse order. 357. The mth power of x-\-a contains m-\-\ terms. This ap- pears from the law of formation of the powers of a binomial developed in Art. 352. Thus the fourth power of x-\-a con- tains five terms ; the sixth power contains seven terms, etc. 358. The sum of the coefficients of the terms in the nth power of x-\-a is equal to the nth power of 2. For, suppose x—\ and a = l, then each term of the formula without the coefficients reduces to unity, and the sum of the terms is simply the sum of the coefficients. In this case (x-fa)"^ becomes (1 + 1)"*, or 2^^. Thus the coefficients of the second power are 1 + 24-1= 4=2', third " 1-1-3 + 34.1= 8 = 2^ fourth *' 1+4 + 6 + 4 + 1 = 16 = 2^ eta BINOMIAL THEOREM. 253 359. To obtain the development of (x— a)^, it is sufficient to change -\-a into —a in the development of (a: + a)™. In con- sequence of this substitution, the terms which contain the odd powers of a will have the minus sign, while the signs of the re- maining terms will be unchanged. We shall therefore have J- • iu m(m—l)(m — 2) , ^ -^ 2 3 ^^'^"*~H EXAMPLES. 1. Find the sixth power of a + b. The terms without the coefficients are a^, a% a'b^, a^W, a^h\ ab', W. The coefficients are 1 (K ^^ 15x4 20x3 15x2 6x1 ' ' 2 ' 3 ' 4 '~"6~'~6~' that is, 1, 6, 15, 20, 15, 6, 1. Prefixing the coefficients, we obtain (a + bf = «« _^ Qw>b + 15«^52 + 20aW -f 15a=Z.* + Qah^J^h^. 2. Find the ninth power of a—b. The terms without the coefficients are a^, a%, cCb^ a'h^, a'b\ a'b\ a%\ o}b\ ab\ b\ The coefficients are 9x^ 36x7 84x6 126x5 126x4 84x3 36x2 9x1 ' ' 2 ' 3 ' 4 ' 5 '~6 '~T~'"~8~'~9~^ that is, 1,9, 36, 84, 126, 126, 84, 36, 9, 1. Prefixing the coefficients, we obtain ia^bY:=a^-^cH-\-ZMb''-^Ma%^+12Qa'b'-12^a'b'' + ^4:a%- It should be remembered that it is only necessary to com- pute the coefficients of AaZ/*the terms independently. 3. Find the seventh power of a— x. 4. Find the third term of {a-^-bf^. 5. Find the forty-ninth term of {a~-xf\ 6. Find the middle term of (a + a:)^o. 264 ALGEBRA. 360. A Binomial with Coefficients. — If the terms of the given binomial have coefficients or exponents, we maj^ obtain any power of it by means of the binomial formula. For this pur- pose, each term must be raised to its proper power denoted by the exponents in the binomial formula. 7. Find the fourth power of 2x-\-^a. For convenience, let us substitute y for 2x and h for 8a. Then {y-\- hf = ?/* + ^^h + ^y'^h' -f ^yh' 4- h\ Kestoring the values of y and 6, the first term will be (2x)*=16ic*, the second term will be 4(2cc)^x3a = 96a;^a, the third term will be %{2xf x {2>af = 2lhx''a\ the fourth term will be 4,{2x)x{%af=2lQxa^, the fifth term will be (3a)*=81a*. Therefore (2x + 3a)* = 1 6x* + 9 6a;3a+ 2 IG^r^aH 21 6ira3+81o^ It is recommended to write the three factors of each term in a vertical column, and then perform the multiplication as indi- cated below : Coefficients, 1+4 +6 +4+1 Powers of 2cc, 16a:* + ^x^ + ^tx^ + 2x -\- \ Powers of 3a, 1 + 3a + Oa'^ + 27a^ +81a* (2a;+3a)*=T6^*+96a:3a+216a:'^a2+2I6a:a3+8Ta*. 8. Find the fifth power of 2aa:-3&. Coefficients, 1+5 +10 +10 +5+1 Powers of 2ax, 32a5.T*+ 16a*x* + Sn'ar' + 4a»x» + 2ax -f 1 Powers of -36, 1 - 36 +96' - 276» + 81 6« -2436' (2ax - 36)* = 32a»x» - 240a*a:*6 + 720a='a:='6» - 1080a=a:»6» + 8 10ax6* - 2436*. 9. Find the fourth power of 2a:+5rt^. Ans. 1 6x* + 1 mx'^a' + GOOx^a* + 1000a;a« + 6250^. 10. Find the fourth power of 7?-\-4:y'^. 11. Find the sixth power of a^+3aft. Ans. a^«+-18a^^Z> + 135a'*/;2 4.540ai2i3 + 1215a^oZ>*+ 1458a«^H 729a«Z)« 12. Find the seventh power of 2a — 3/^. Ans. 128a' - 1344a«/; + mASa'V^ - 15120a*Z»3 + 22680a3Z>*-20412a^^H10206a^.«-21876l 13. Find the fifth power oiha^^^xhj. BINOMIAL THEOREM. 255 14. Find the sixth power of o?x-\-hy'^. 15. Find the fifth power of ax—1. 16. Find the fifth term of{a^-by^. 17. Find the fifth term of {Sx^-4:y^y, 18. Find the sixth power of 5 — ^. Powers and Boots of Polynomials. 361 If it is required to raise a polynomial to any power, we riiay, by substituting other letters, reduce it to the form of a binomial. We obtain the power of this binomial by the gen- eral formula ; then, restoring the original letters, and perform- ing the operations indicated, we obtain the required power of the proposed polynomial. Ex. 1. Let it be required to raise a + 5+c to the third power. If we put &-f c=m, we shall have {a-\-b + cy=:{a-\- my = a^ + Sa'^m + Sam"^ + m^, or a^ + Sa%b -\.c)-{-Sa{h-\- cf + {h + cy. Developing the powers of the binomial b-\-c, and performing the operations indicated, we obtain (a _j_ 5 4- c)3 :zr a^ + Sa^Z) + Sa^c + 80^2 + 6a6c + 3ac2 + 53 4. 3^;2c + 36c2+cl Ex. 2. Find the fifth power of x-\-a-{-b. Ex. 3. Find the fourth power of a'^—ab-\-b'^. Ans. a^-4:a'b + 10a'b'^-Ua'b^-{-l9a'h* -16a:'b'-{-10a'^b'-4:ab'' + b\ Ex. 4. Find the fifth power of l-\-2x-^Sx^ Ex. 5. Find the sixth power of a-^b+c. A71S. a«+6a^6+6«5c+15a*Z>2 4_ SOa^Z'c+lSa^c^-f 20a3J3 + eOa^b^c 4- eOa^bc'' -f 20q-V + 15a2^* + QOa'^b^c + 90a262c2+ 60a2Z>c3 + 16aV+ 6ab'-\- SOab'c+ 60ab^c^ 4- eOab^c^ + SOabc* + Qac' + 6« + 6b'c + 16b'c^ + 20^^03 + 15Z^V+ 66cHc'. 362. The binomial theorem will inform us how to extract any root of a polynomial We know "that the mth power of 256 ALGEBRA. x-\-a is x^ -\- max"^~'^ -}- other terms. The first term of the root is, therefore, the mth root of the first term of the polynomial. Also the second term of the root may be found by dividing the second term of the polynomial by mx^~^ ; that is, the first term of the root raised to the next inferior power, and multiplied by the exponent of the given power. Hence, for extracting iny root of a polynomial, we have the following RULE. Arrange the terms according to the poivers of one of the letters^ and take the mth root of the first term for the first term of the re- quired root. Subtract the mth power of this term of the root from the given polynomial, and divide the first term of the remainder hy m times the (m — l) power of this root ; the quotient will he the second term of the root. Subtract the mth power of the terms already found from the given 2)olynomial, and, using the same divisor, proceed in like mannei' to find the remaining terms of the root. Ex. 1. Find the fourth root of 16a*-96a3a;4-216a2x2-216aa:3_^3l3^, 16a*-96rt3x+216a2^2_216acc3+81x*(2a-3x 16a*____ 82a3)__-96tt^ 16a*-96a3a;+216a2x2-216ax3^31a^, Here we take the fourth root of 16a*, which is 2a, for the first term of the required root, subtract its fourth power, and bring down the first term of the remainder, — 96a^a;. For a divisor, we raise the first term of the root to the third power and multiply it by 4, making 82a^. Dividing, we obtain — 3.t for the second term of the root. The quantity 2a— Sx, being raised to the fourth power, is found to be equal to the proposed polynomial. Ex. 2. Find the fifth root of 80x3^32a;5_80a;*-40a;2 f lOx-1. Ans. 2a:— L BINOMIAL THEOREM. 257 Ex. 8. Find the fourth root of Ans. 8x2-2x-2. 363. To extract any Boot of a Number. — The preceding meth- od may be applied to the extraction of any root of a number. Let n be the index of the root, n being any whole number. For a reason similar to that given for the square and cube roots, we must first divide the number into periods of n figures each, beginning at the right. The left-hand period may contain less than n figures. Then the first figure of the required root will be the nth root of the greatest nth. power contained in the first period on the left. If we subtract the ?2th power of this root from the given number, and divide the remainder by n times the (n— l)th power of the first figure, regarding its local value, the quotient will be the second figure of the root, or possibly a figure too large. The result may be tested by raising the whole root now found to the nth. power ; and if there are other figures they may be found in the same manner. Tn the extraction of the nth root of an integer, if there is still a remainder after we have obtained the units' figure of the root, it indicates that the proposed number has not an exact nth root. We may, if we please, proceed with the approximation to any desired extent by annexing any number of periods of n ciphers each, and continuing the operation. We thus ob- tain a decimal part to be added to the integral part already found. So, also, if a decimal number has no exact nth root, we may annex ciphers, and proceed with the approximation to any de- sired extent, dividing the number into periods commencing with the decimal point. Ex, 1. Find the fifth root of 33554432. 335.54432(32 243 5.3*=r405) 925 32^= 33554432. Ex. 2. Find the fifth root of 4984209207. 268 ALGEBRA. Ex. 3. Find the fifth root of 10. Ex. 4. Find the fifth root off. Ans. .922. 364. When the index of the required root is composed of two factors^ we may obtain the root required by the successive extraction of simpler roots, Art. 217. For the mnXh root of any number is equal to the mih. root of the nth root of that number. Thus we may obtain the fourth root by extracting the square root of the square root. We may obtain the sixth root by extracting the cube root of the square root, or the square root of the cube root. It is, however, best to extract the roots of the lowest degrees first, because the operation is less laborious. We may obtain the eighth root by extracting the square root three times successively. We may obtain the 7ii7ith root by ex- tracting the cube root twice successively. Ex. 1. Find the fourth root of Qa'^b^ -f. a* - 4:a% - 4:ah^ + b\ Ex. 2. Find the sixth root of 6a'b + Iba'b'' + a^ + 20a^b^ + Ida'^b* -{-b^-h 6ah\ Ex. 3. Find the eighth root of 1024x'^-hl792xy+256a;«+1120a:y+1792icy4-448x3y* + ?/«-t-112xy-hl6x/. SEBIES. 259 CHAPTER XIX. SERIES. 365. A series is a succession of terms each of which is de- rived from one or more of the preceding ones by a fixed law. This law is called the law of the series. The number of terms of the series is generally unlimited. Arithmetical and geomet- rical progressions afford examples of series. 366. A converging series is one in which the sum of the first n terms can not numerically exceed some finite quantity, how- ever great n may be. Thus, 1, ^, J, I", -^, etc., is a converging series. 367. A diverging series is one in which n can be taken so large that the sum of the first n terms is numerically greater than any finite quantity. Thus, 1, 2, 3, 4, 5, 6, etc., is a diverging series. 368. When a certain number of terms are given, and the law of the series is known, we may find any term of the series, or the sum of any number of terms. This may generally be done by the method of differences. 369. To find the several orders of differences fiyr any series: Subtract the first term from the second, the second from the third, the third from the fourth, etc. ; we shall thus form a new series, which is called the ^irs^ order of differences. Subtract the first term of this new series from the second, the second from the third, etc. ; we shall thus form a third series, called the second order of differences. Proceed in like manner for the third, fourth, etc., orders of differences, and so on till they terminate, or are carried as far as may be thought necessary. ^60 ALGEBRA. Ex. 1. Find the several orders of differences of the series ol square numbers 1, 4, 9, 16, etc. Squares. 1 4 9 16 25 letDifC 2d Diflf. 3 5 2 7 2 9 2 2 SdDifi: Ex. 2. Find the several orders of differences of the series of cube numbers 1, 8, 27, etc. Cubes, 1 8 27 64 125 216 1st DiflF. 2d Diflf. SdDiff: 7 12 19 18 6 87 24 6 61 80 6 91 S6 6 4th DiA Ex. 8. Find the several orders of differences of the series of fourth powers 1, 16, 81, 256, 625, 1296, etc. Ex. 4, Find the several orders of differences of the series of fifth powers 1, 82, 248, 1024, 8125, 7776, 16807, etc. Ex. 5. Find the several orders of differences of the series of numbers 1, 8, 6, 10, 15, 21, etc. 370. To find the nth term of any series: Let a, 5, c, c?, e, etc., represent the proposed series. If we subtract each term from the next succeeding one, we shall ob- tain the first order of differences ; if we subtract each term of this new series from the succeeding term, we shall obtain the second order of differences, and so on, as exhibited in the fol- lowing table : Scries. l8t Diff. a h h-a c-b c d d-c e-d e 2d DiflTerences. C— 2Z>+a d-2c+h e—2d-\-c 3d Order of Differences. d-Sc-^Sb- e-3d-\-Sc- 4th Order of DiffereDccs. c— 4c?+6c— 4/;+a SERIES. 261 Let D; D'', D''', D'''', etc., represent the first terms of the several orders of differences. Then we shall have D^=b—aj whence ^=: a +D'. I)''=c-2b+a, " c=a-]-2J)'-\-D'\ W=zd-Sc+Sb-a, '' d^a+SD'-\~SJ)''-]-W. etc., etc. The coefficients of the value of c, the third term of the pro- posed series, are 1, 2, 1, which are the coefficients of the second power of a binomial ; the coefficients of the value of d, the fourth term, are 1, 3, 3, 1, which are the coefficients of the third power of a binomial, and so on. Hence we infer that the co- efficients of the 7ith term of the series are the coefficients of the (n— l)th power of a binomial. If we denote the nth term of the series by Tn, we shall have Z 2t.o + , etc. Ex. 1. Find the 12th term of the series 2, 6, 12, 20, 80, etc. The first order of differences, 4, 6, 8, 10, etc. " second order of differences, 2, 2, 2, etc. " third order of differences, 0, 0. Here D' = 4, J)" ^% and D^'^=:0. Also a = 2 and n^Vh. Hence Tj2=2 + llD^ + 55D'' = 24-44-hll0 = 156, Ans, Ex. 2. Find the twentieth term of the series 1, 3, 6, 10, 15, 21, etc. Here D' = 2, D'' = l, a=rl, and w = 20. Therefore T2o-:l + 19D'-fl71D'^ = l +38 + 171=210, ^tw?. : Ex. 3. Find the thirteenth term of the series 1, 5, 14, 30, 55, 91, etc. Ex. 4. Find the fifteenth term of the series 1, 4, 9, 16, 25, 36, etc. Ex. 5. Find the twentieth term of the series 1, 8, 27, 64, 125, etc. Ex. 6. Find the nth term of the series 1, 3, 6, 10, 15, 21, etc. . 7?(n + l) Arts. -^^ — '-, 262 ALGEBRA. Ex. 7. Find the nth term of the series 1, 4, 10, 20, 35, etc. Anc. ^(^+^)(^+2) ^ 6 Ex. 8. Find the nth term of the series 1, 6, 15, 35, 70, 126, eta 371. To find the sum ofn terms of any series: Let us assume the series 0, a, a+6, a+Z>-|-c, a-\-h-\-c-\-d^ etc. (1.) Subtracting each term from the next succeeding, we obtain the first order of differences, a, &, c, c?, etc. (2.) Kow it is clear that the sum of n terms of the series (2) is equal to the (nH-l)th term of series (1); and the nth order of differences in series (2) is the (?2 4-l)th order in series (1). If, then, we denote the sum ofn terms of series (2) by S, which is the same as the (n + l)th term of (1), we may obtain the value of S from the formula of the preceding article by substituting for a, n-|-l for n, a for D; D' for D", etc. Hence S_„a+— 2— D + -g-g D + ^^^-^ D , etc. When any one of the successive orders of differences be- comes zero, this formula gives the exact sum of the terms. When no order of differences becomes zero, the formula may still give approximate results, which will, in general, be nearer the truth the greater the number of terms employed. EXAMPLES. 1. Find the sum of 15 terms of the series 1, 3, 6, 10, 15, 21, etc. Here a=l, T>'=% D'' = l, D'''=0. Therefore 8=15 + 15.14+5.7.13 = 680, Am, SERIES. 263 2. Find the sum of 20 terms of the series 1, 4, 10, 20, 35, etc. 8. Find the sum of n terms of the series 1, 2, 3, 4, 5, 6, etc. Ans. ^(^+^1 4. Find the sum of n terms of the series 1\ 2^ 32, 42, 52, etc. ^^,, ^(^ + I)(2n4l) ^ 6 6. Find the sum of n terms of the series 1^ 23, 3^ 4^ 5=*, etc. Ans. fc^. 4 6. Find the sum of n terras of the series 1, 3, 6, 10, 15, etc. . n(72+l)(n+2) 7. Find the sum of n terms of the series 1.2, 2.3, 3.4, 4.5, 5.6, etc. . ?i(n+l)(n+2) Ans. — — —(r^ — ■ — k o 8. Find the sum of n terms of the series 1, 4, 10, 20, 35, etc. n(72 + l)(;7 + 2)(n+3) ^'''' 2X4 • Interpolatwn. 372. Interpolation is the process by which, when we have given a certain number of terms of a series, we compute inter- mediate terms which conform to the law of the series. Interpolation may, in most cases, be effected by the use of the formula of Art. 370. If in this formula we substitute n-\-l for n, we shall have T„+.=a+nD-+^(!^D"+ "("-^iy-^) D'-'+, etc., which expresses the value of that term of the series which has n terms before it When n is a fraction less than unity ^ ^n-\-i 264 . ALGEBRA. stands for a term between the first and second of the given terms. When n is greater than 1 and less than 2, the inter- mediate term will lie between the second and third of the given terms, and so on. In general, the preceding formula will give the value of such intermediate terms. EXAMPLES. 1. Given the cube root of 60 equal to 3.914868, " " " 61 " 3.936497, " " " 62 " 3.957891, " " " 63 " 3.979057, « " " 64 " 4.000000, to find the cube root of 60.25. Here D'= +.021629, D''= -.000235, D"'== +.000007, etc a = 3.914868, and n = .25. Substituting the value of n in the formula, we have T„+i=a+iD'-^D''+^D--, etc. The value of the 1st term is +3.914868, u u 2d " + .005407, u u 3(J u + .000022, u u 4tii » + .000000. Ilence the cube root of 60.25 is 1 5.920297. 2. Find the cube root of 60.5. Alls. 3.925712. 3. Find the cube root of 60.75. ^725. 3.931112. 4. Find the cube root of 60.6. Arts. 3.927874. 5. Find the cube root of 60.33. Alls. 3.922031. 6. Given the square root of 30 equal to 5.477226, » " " 31 " 5.567764, " " " 32 " 5.656854, " " " 33 " 5.744563, *' " " 34 " 5.830952, to find the square root of 30.3 Ans. 5.504544, 7. Find the square root of 30.4. Ans. 5.513619. 8. Find the square root of 30.5. Ans. 5.522681. 9. Find the square root of 30.6. -^725. 5.531727. 10. Find the square root of 30.8. Ans. 5.549775. SERIES. 265 Development of Algebraic Expressions into Series, 373. An irreducible fraction may be converted into an in- finite series by dividing the numerator by the denominator, ac- cording to the usual method of division. Ex. 1. Expand ^ — - into an infinite series. J. — x 1— a;)l (l + a;+a;2+x3-l-^*4-, etc. \ — x X X- -x^ x" x'- -x' x^ x'- -X* Hence z=zl-]-x-\-x'^-{-x^-\-x^-{-x^-{-. etc., to infinity. 1—x "^ Suppose x—^, we shall then have J^ = J^=2::::l+i + i + i+A + , CtC. Suppose cc=J, we shall then have Y^=Y^=-|=l+i+i+A+A+, etc. Ex. 2. Convert :; into an infinite series. l-\-x Ans. l—x+x'^—x^-\-x'^ — x^-{-j etCc Suppose x — i^ we shall then have ^-x=l-l-i+i-i+TV-A+, etc. Ex. 3. Convert into an infinite series. a+x rp ry" /y>3 ^4 Ans. 1 \-- — 3+-^— 7 etc. a a^ a^ a* Ex. 4. Convert into an infinite series. a—x rp rp£i rp"^ rrA Ans. l+- + -;+^4-— . + , etc. ■jyi" a o? cr a* 266 ALGEBRA. 1 4-cc Ex. 5. Convert —^— into an infinite series. 1—x Ex. 6. Convert — i— into an infinite series, a— x Ans. 14- — H — 2-+^-+-r+) 6*^ a a^ a^ or Ex. 7. Convert ~ into an infinite seriea X—x-^-x? Ans. I4-X— cc^— x*4-cc®+cc'— , etc. Ans. l—x^—x^-\-x^-\-x^—x^—^ etc. 1— a? Ex. 8. Convert :; r into an infinite series. l—x-\-x^ Ex. 9. Convert = 5 into an infinite series. 1—x—x^ Ans, l+2x-f3a;2+5a;^+8a;*H-13a:^-f-, etc 374. An algebraic expression whicli is not a perfect square may be developed into an infinite series by extracting its square root according to the method of Art. 198. Ex. 1. Develop the square root of 1 4- a: into an infinite series. 1 . (^ ,x x^ x^ 5a;* , l+x(^l+2-g+jg-j28+.etc 1 »-l) ' , -+4 ^«-a -f X^ a;3 ic* 4" "8 "^64 ^«-f4) x^ cc* 8 64 a? cc* cc* x^ 8*^16 64 "^256 5.T* x^ x^ 64 ' 64 256' SERIES. 267 Hence the square root of 1 +x is equal to l+2-« +16-128 + ''^- Suppose x=lj we shall have -v/2 = l+i-i+yV-TlH+, etc. Ex. 2. Develop the square root of ct^+x into an infinite seriea Ans. a^-- -— +_— ^-— — _+ etc. Ex. 8. Develop the square root of a* + cc into an infinite series. Ex. 4. Develop the square root of a^—x into an infinite seriea Ex.5. Develop the square root of a^-\-x^ into an infinite series. Method of Undetermined Coefficients, 375. One of the most useful methods of deviiloping algebraic expressions into series is the method of undetermined coefficients. It consists in assuming the required development in the form of a series with unknown coefficients, and afterward finding the value of these coefficients. This method is founded upon the properties of identical equations. 376. An identical equation is one in which the two members are identical, or may be reduced to identity by performing the operations indicated in them. As ax-\-h=ax+h^ tt2. -x^ ax a — X a l-\-x 1+x 377. It follows from the definition that an identical equation is satisfied hy each and every value which may he assigned to a let- ter which it contains^ provided that value is the same in both members of the equation. Every identical equation containing but one unknown quan- tity can be reduced to the form of A + Bx4-Cx2+Dx3 + , etc. = A' + B'cc + C'iK2^DV + , etc. 268 ALGEBRA. 378. If an equation of the form K-\-^^x^-^J?^\-, etc. = A' + B'a: + C':r2 + , etc., must he satisfied for each and every value given to Xj tJien the co- efficients of the like powers of x in the two members are equal each to each. For, since this equation must be satisfied for every value of £c, it must be satisfied when x=0. But upon this supposition all the terms vanish except two, and we have A=A'. Suppressing these two equal terms, we have Bx + Ca;2 + , etc. = B'a: + C 'it^ + , etc. Dividing each term by re, we obtain B + Ca: + , etc.r=:B' + Cx + , etc. Since this equation must be satisfied for every value of x, it must be satisfied when x=0. But upon this supposition B = B'. In the same manner w€ can prove that D=D', etc. 379. Wlienever we have an equation of the form M -f Nrr -f- P^2 + Q.r' -+- , etc. = 0, which is true for every value of x^ all the coefficients of x are equal to zero. For, if we transpose all the terms of the equation in the last article to the left-hand member, we shall have A-A' + (B-B>H-(C-COx2 + (D-D')a;3 4-,etc.=0. But it has been shown that A = A', B = B', etc.; whence A- A' = 0, B-B' = 0, etc. If we substitute M for A -A', and N for B — B', etc., the equation will be M+Nx + Pa:2-f Qa:3 + , etc.=0. whence M=:0, ]Sr = 0, P = 0, etc. Ex. 1. Expand the fraction - — -- into an infinite seriea 1 — &x It is plain that this development is possible, for we may divide the numerator by the denominator, as explained in Art 373. Let us, then, assume the identical equation SERIES. 269 l±^=A+Bx+Cx^-{-J)x'-{-Ex' + , etc., i. — oX where the coefficients A, B, C, D are supposed to be independ- ent ofx, but dependent on the known terms of the fraction. In order to obtain the values of these coefficients, let us clear this equation of fractions, and we shall have l + 2x=A + (B-3A):r+(C-3B)x2+(D-8C)xH (E-8D)cc*+, etc. Now, since this is supposed to be an identical equation, the coefficients of the like powers of x in the two members are equal each to each. Therefore A = l. B-3A = 2, whence B=5; C-3B=0, '' C = 15; D-3C=0, " D=45; E_3D=0, " E = 135, etc. Substituting these values of the coefficients in the assumed series, we obtain l^:^=l + 6x-\-16x''-\-4:5x^-{-lS6x'-^, etc., l — 6x where the coefficient of each term after the second is three times the coefficient of the preceding term. 380. The method thus exemplified is expressed in the fol- lowing RULE. Assume the proposed expression equal to a series of the form A + Ba^-f Ca:;^4-, etc. ; clear the equation of fractions^ or raise it to its joroper power ^ and place the coefficients of the like powers of x in the two members equal each to each. Then find from these equa- tions the values of A, B, C, etc., and substitute these values in the assumed development. Ex. 2. Expand the fraction - — into an infinite seriea l — 2x-j-x^ Assume - — — - = A-i[-'Bx-^Cx'^-^'Dx^+Ex'^+, etc. J. — ^iX -p X Clearing of fractions, we have 270 ALGEBRA. l = A4-(B-2A)x4-(C-2B-fA)x2 + (D-2C + B)a:* + (E_2D + C)a;H, etc. Therefore we must have A=l, B-2A=0, whence B = 2A=:.2; C-2B + A = 0, " C = 2B-A = 3; D-2C + B = 0, " D:=2C-B=4; E_2D + C = 0, " E = 2D-C=5, etc. Therefore ^-— ^-__^=l+ 2^+3x2 +4^3 + 5x*H-, etc. Ex. 3. Expand the fraction ^ "^ ^ , into an infinite 1— cc— cc^ Ans. l + Sx-{-4x'-{-7x^-{-llx'-\-lSa^+29x^-{-, etc., where the coefficient of each term is equal to the sum of the coefficients of the two preceding terms. 1—x Ex. 4. Expand ^j — — into an infinite series. Ans. l+x-\-5x^-\-lSx^+4:lx*+121x^-^, etc. What is the law of the coefficients in this series? series. Ex. 5. Expand Vl—x into an infinite series. ty /yi2 ry^S P^/Y»4 ^O^ Ans. l_-_-______-, etc. 1—X Ex. 6. Expand .. ^ into an infinite series. X ~j~ X -J" X Ans. l—2x-\-ocy^-\-x^—2x'^-\-x^+x^—, eta Ex. 7. Expand Va^—x^ into an infinite series. 381. Proper Form of the assumed Series.-^ln applying the method of undetermined coefficients to develop algebraic ex- pressions into series, we should determine what power of the variable will be contained in the first term of the development, and assume a corresponding series of terms. Generally the first term of the development is constant, or contains cc°; but the first term of the series may contain x with any exponent either positive or negative. If the assumed development com- mences with a power of cc lower than is necessary, no error will SERIES. 271 result, for the coefficients of the redundant terms will reduce to zero. But if the assumed development commences with a power of X higher than it should, the fact will be indicated by an absurdity in one of the resulting equations. The form of the series which should be adopted in each case may be determined by putting a; = 0, and observing the nature of the result. If in this case the proposed expression becomes equal to a finite quantity, the first term of the series will not contain x. If the expression reduces to zero, the first term will A contain x; and if the expression reduces to the form — , then the first term of the development must contain x with a nega- tive exponent. Let it be required to develop ^ ^"^o a series. Assume -=K-\-'Bx-\-Qx'^-\-T)x^-[-, etc. 6x—x'^ Clearing of fractions, we have l=3Aaj-l-(3B — A)x2+, etc., whence, according to Art. 378, we obtain 1=^0, which is ab- surd, and shows that the assumed form is not applicable in the present case. Let us, however, assume r = Aa?-i-hB4-Ca7+B:i;H, etc. 6x — x^ ' Clearing of fractions, we have l=^K+{2>']^-A)x-^r{^o-^)x^^-{^J)-Q)x'+,e\JQ. Therefore 3A=:1, whence A=r-|-; 3B-A = 0, '' B=|; 3C-B=z:0, '' C- 1 TT» 3D-C=0, " D=A- Substituting these values, we find 1 _^~^ x^ X ^' , X 272 ALGEBRA. To Resolve a Fraction into Simpler Fractions. 382. When the denominator of a fraction can be resolved into factors, the principles now developed enable us to resolve the fraction itself into two or more simpler fractions^ having these factors for denominators. In such a case, the given fraction is the sum of the partial fractions. g^ ][2 Ex. 1. Kesolve the fraction -^ — ^ into partial fractions. We perceive that x^— 5x+6 = (cc— 2)(cc— 3). . 5x-12 A B Assume -5 — = -= rr4- x^—bx+io x—2 0^-3' in which the values of A and B are to be determined. Clearing of fractions, we have 5x-12 = (A + B):r-(3A4-2B). By the principle of Art. 378, A + B=5, and 3A+2B = 12. From which we obtain A =2 and B = 3. Substituting in the assumed equation, we have bx-12 _ 2 3 x^-bx-\-(6~ x-2 x—Z' Ex. 2. Eesolve — ^ — j into partial fractions. sc — i Ans, 07+1 iC— 1 6x— 19 Ex. 3. Eesolve —z — ^ ^^ into partial fractions. Ans. ;r--|- a:—3* Zx'^—1 Ex. 4. Kesolve —3 into partial fractions. Xr ^—30 Ans. ~—r-{- r + -. X+l X—1 X Ex. 6. Resolve -, r^-^ — ^^r^-^ — • into partial fractions. \X — L) \X — Zj [X — Oj A 1 2,3 Ans. --f x—l x—2 x— 3* SERIES. 273 Ex. 6. Eesolve 5^2+2^-1 (x4-l)(x-l)(2x+l) Ans. into partial fractions. 02+1 X— 1 2.T+r Ex. 7. Eesolve -:; — ^ ^ . . . into partial fractions. 1-5x2+4^* Ans. l+x 1 + 16 1+2^ ' l-2x' Reversion of Series. 383. TJie reversion of a series is the finding the value of the unknown quantity contained in an infinite series by means of another series involving the powers of some other quantity. This may be accomplished by the method of undetermined coefficients in a mode similar to that employed in Art. 879. Ex. 1. Given the series tj^x-^-x'^-^qi?-^-^ etc., to find the value of cc in terms of y. Assume a:;= A?/+B?/2 + C?/^+D^* + . etc. Find, by involution, the values of x^^ x^, x\ and a:^, carrying each result only to the term containing y^. Then, substituting these values for x^ x^^ x^^ etc., in the given equation, we shall have 2/*+E 2/'+, etc. 4-2AD + 2BC + 3A2C + 8AB2 4-4A3B -f A^ Since thia is an identical equation, we place the coefficients of the like powers of y in the two members equal to each other, and we obtain A=+l, B=-l, C=+l, D=-l, E=r+1, etc. Hence we have x—y—y'^-\-y^—y'^-\-y^~^ etc., Ans. nr'2 /ytO qA Ex. 2. Given the series y—x— — -\-- — — + , etc., to find the ^ 4: O y-Ay+B f+C jr'+D +A' +2AB +2AC +A' +B^ +3A'B +A* value of X in terms of y. yl ^,3 y\ Ans. a;=2/+|-+|- + | + , M? etc. 274 ALGEBRA. Ex. 3. Given the series 2/=^?— -n+'q—x+'H' — j ^^-j to find the value of x in terms of y. Ans. .=2,+|V|L + ^+^^+, eta Ex. 4. Given the series y—x-{-x^+x^-{-x''-]-x^-{-j etc., to find the value of x in terms of y. Alls. x=y—y^-{-2y^—5y'' + 14:7/—, etc. Ex. 5. Given the series y=x+Sx^-{-5x^-\-7x*-\-9x^-\-j eta, to find the value of a; in terms of?/. Ans. x=y-d7f-{-lSy'-67y*-\-SSlf-, etc. 384. When the sum of a series is known, we may sometimes obtain the approximate value of the unknown quantity by re- verting the series. Ex.1. Given ix-\-^^-\-^-\-^x'^-\-^s^x^-^, etc.=i, to find the value of x. If we call s the sum of the series, and proceed as in the last article, we shall have £C=:25-52-f |s3-is* + -|s^-, etc. Substituting the value of 5, we find ^=i— A+uV— -m+irATr— J etc., or a?=0.446354 nearly. Ex. 2. Given 2a:+3x3H-4x*+5xH, etc.=^, to find the value ^^^- , 5 353 195^ 1525^ or a;=i-T-^4-Tiw-i«Vri44-, etc. =0.2300 nearly. , and the quotient is a"-i-f-a"-2Z>-f a"-3^2_^ 4-^>«-i. The number of terms in this quotient is equal to n; for h is contained in all the terms except the first, and the exponents of b are 1, 2, 8, etc., to 71—1, so that the number of terms containing h is ti— 1, and the whole number of terms is equal to n. Now, when a=h^ each term of the above quotient becomes a"-^, and, since there are n terms in the quotient, this quotient reduces to na^-^. Second. Suppose n to be a positive frax:tion^ or n=—, where p and q are positive whole numbers. 1 P Let a'^=Xj whence a^—x^^ and a—x'i, \ V. Also, let ^^=2/, whence h'i—y'P^ and h—y[i. Then, substituting, we have a^^ln ^a'i — hi _xP—yP _ x—y a—h a—h x'^—yi x'i—y'i x—y But^ and q are positive integers ; therefore, when a =5, and, consequently, x=:y^ according to case first, the numerator of the last fraction becomes pxP-'^^ and the denominator becomes qxfi-^ ; that is, the fraction reduces to ^ or —xP-1. qx9 ■1 ' 276 ALGEBRA. 1 Substituting for x its value a*?, the fraction reduces to --a 3 or — a^ , or na)^~^. q ' q ' Third. Suppose n to be negative^ and either integral or frac- tional ; or let n=—m. Then we shall have a"— /;»_a-^— J-»'^_a"^ h^ _ 1 lf»'—a^_ 1 a'^—b^ a—h ~ a—b ~ a—b ~ a"'b"''' a — b ~'~d"'b"'' a — b' Now, when a=^bj the first factor of the last expression re- duces to — 2^^, or — a-2"*, and the second factor (bj one of the preceding cases) reduces to ma'^-~'^. Hence the expression be- comes — a~^^xma'^-'^j or — ma~'^~\ or 7?a"~^. 387. It is required to obtain a general formula expressing the value of{x-\-ay^j whether m be positive or negative, integral or fractional. JSrowa:+a=a:/l4--); therefore (a:4-«)"''^ic"'(l4--) • / r?\"* If then we obtain the development of (1+-1 , we have only to multiply it by x^ to obtain that of (a;+a)"*. Let -=z; then, to develop (l + 2y^ assume {l+zy=A-\-Bz+Cz-'-\-Dz^-\-, etc., (1.) in which A, B, C, D, etc., are coefficients independent of z, and we are to determine their values. Now this equation must be true for any value of z; it must therefore be true when 2 = 0,' in which case A = l. Substituting this value of A in Eq. (1), it becomes (H.2)'«=:l + Bz4-C22 + D22 + , etc. (2.) Since Eq. (2) is to be true for all values of z, let z=n ; then (2) becomes (l4.n)«^^l4.Bn + Cn2 + D?iH, eta (3.) Subtracting (3) from (2), member from member, we have (l-}-2)m_(l4.^)m^I3(2-?0 + C(22-n2)-l-D(23-n3) + , CtC. (4) SERIES. 277 Dividing the first member of (4) by (1 + 2;) — (1 + ?i), and the second by its equal z—7i, we have (^+-)':-(|+"r=B+cg=^%Dg:^%, etc. (5.) (I + 2) — (1 + n) z—n z—71 ' But when z=n^ or l-{-z—l-\-7ij the first member of equation (5) becomes m{l-\- z)^-'^. 2;2 /y^2 Also, =:z-{-7i, when s=:n, becomes 2z. z — 71 Z — 71 :=z^-\-zn-\- 71^^ when z=zn^ becomes 82^, etc. I z—71 These values substituted in (5) give m (1 -f zf'-^ = B + 2C2; + 3D22 4. 4E23 + , etc. (6. ) Multiplying both members of Eq. (6) by 1 + 2, we have m(l + 2)^=B-f(2C + B>+(8D + 2C>2+(4E + 3D>3 + ,etc. (7.) If we multiply Eq. (2) by m, we have 7n{l-{-zy—m-{- 7nQz + mOz^ + 7iiJ)z^ + , etc. (8.) The first members of Eq. (7) and (8) are equal; hence their second members are also equal, and we have m -\- mB2 + 7nQz'^ + 7iiDz'^ -\- , etc. = B-f (2C+B)z+(3D + 2C)s2 + (4E-f8D)2;3+, etc. (9.) This equation is an identical equation ; that is, it is true for all values of z. Therefore the coefiicients of the like powers of 2; in the two members are equal each to each, and we have B^m. 2C+B=772B, whence Q = J"'^''^~~^^ • 2i .0 Substituting these values in (2), we have (l-^z)m^l^mz + -^~ 4^+ 2 8 '^' ^- '^ If in this equation we restore the value of e, which is ^, we have 278 ALGEBRA. and multiplying both members by x^^ we obtain ?n(m— l)(m— 2) {^■\-aY=x^-\-mx'^-^a^'^—^ ^a;'"-V + 2 3 ^ 3a3 + ,etc., (11.) which is the general formula for the development of any bino- mial (x-f-a)"*, whatever be the values of x and a, and whether m be positive or negative, integral or fractional ; and this for- mula is known as the Binomial Theorem of Sir Isaac Newton. 388. When the Series is Finite. — The preceding development is a series of an infinite number of terms ; but when m is a pos- itive integer, the series will terminate at the (??z + l)th term, and all the succeeding terms will become zero. For the second term of Eq. (11) contains the factor tw, the third term the factor m— 1, the fourth term the factor m — 2, and the (r/2 + 2)d term contains the factor m—m^ or 0, which reduces that term to 0; and since all the succeeding terms also contain the same factor, they also become 0. There will therefore remain only m-\-l terms. When m is not a positive integer, it is evident that no one of the factors m, m— 1, m— 2, m— 3, etc., can be equal to 0, so that in that case the development will be an infinite series. 389. Expansion of Binomials with negative integral Exponents This is effected by substitution in formula (11). Ex. 1. Expand r or (a + 6)-^ into an infinite series, ^ a + h ^ ' In (11) let m=--l, and we find the coefficient of the second term is —1, " " third " is "^^""^ ^-H, " " fourth " is±l-^^=-l, o " " fifth " is -1:1^^= 4-1, etc. SEKIES. 279 Hence we have {a + h)-^ = a-'^-a-%-{-a-%''-a-%''-{-, etc., 1 1 h h-" h' o^ — n:= ^+~^ — i+' ^^^-j a+0 a a^ a^ a^ which is an infinite series, and the law of the series is obvious. We might have obtained the same result by the ordinary meth od of division. Ex. 2. Expand - — —p- or (a +5)-^ into an infinite series. Ans. a-2_2a-25H-8a-462_4^-5/^3_|_ 5^-6^4 __^ etc.^ 1 2h 8^2 4^3 5^4 or -^ — gH — ^ 3-H — g--5 etc., a^ a"* a* a^ w" where the law of the series is obvious. Ex. 3. Expand j or (a—h)-'^ into an infinite series. Ans. a-^ -{■a-%-\-aT%'^-{-a-^h^ + , etc. Ex. 4. Expand -, r— or (a—h)-'^ into nn infinite series. Ex. 5. Expand (a4-^)~^ into an infinite series. Ans. a-^-Za-^h^Qa-^'^-lOa-^lP^lba-'^h^-, etc. Ex. 6. Expand {a—h)-^ into an infinite series. 7l7?s. a-^-{-4:a-^b-\-10a~%^-\-20a-'^b^ + S5a-^h^^, etc. Ex. 7. Expand {l-\-2x)~^ into an infinite series. Ans. 1 - 10a; + 60x2 -280a^3+, g^c. 390. Expansion of Binomials with positive Fractional Exponents, Ex. 1. Expand Va+6 or (a + Z))^ into an infinite series. Eepresent the coefficients of the different terms by A, B, Q D, etc. ; then A= +1, D=:CX E=Dx^--H^^f^, etc. 2 ~ 2.4' 7? -2 ■^2.4.6' 3 ~ n-3 1.3.5 4 2.4.6.8' 280 ALGEBRA. Hence we have + , etc. The factors which form the coefficients are kept distinct, in order to show more clearly the law of the series. The numer- ators of the coefficients contain the series of odd numbers, 1, 8, 5, 7, etc., while the denominators contain the even numbers, 2, 4, 6, 8, etc. Ex. 2. Expand {x—dy into an infinite series. Ex. 3. Expand {a^-\-xY into an infinite series. ^.,. ,+___. + ______ + , etc. ^^ ""^'la 2.4a^"^2.4.6a^~2.4.6.8a^"^'^^- Ex. 4. Expand (« + &)* into an infinite series. ^""^ i^-^3-a-0^^-^ 876:9^- 8.6.9. 12a- -^-^^^i Ex. 5. Expand {a^—h^y into an infinite series. . j-1 i^ 2^' 2.5/>^ , ) Ex. 6. Expand {a-^-xy into an infinite series. Ex. 7. Expand {a—hy into an infinite series. , ij-, ^^ 3/>2 8.7Z^3 3.7. IIM , ) Ans,a |l-4^-j:y-.-4:8n2^-4.8.12.16a*-''*"'f Ex. 8. Expand (1— x)^ into an infinite series. 6 6.10 6.10.15 6.10.15.20 ' 891. Expansion of Binomials with negative Fractional Eocpo nents. Ex. 1. Expand j or {a+h) ^ into an infinite series. SERIES. 281 The terms without the coefficients are a"* a~h, a'h'^, a~h^, a"V, etc. Kepresent the coefficients by A, B, C, D, etc. ; then A= +1, B= n =-i, T. r^ ^^-2 1.3.5 „ „ "-3 , 1.3.5.7 , Hence we obtain / i7\~2 ~2 -*- ~Y7 I 1-^ "Y/o 1-0.5 -5^70 , 1.0.0.7 --IT/ (a+i) =a -^a i + ^y^a ^^^-gTO" * +2X6:8'' ^ — , etc. _ 1 j & 8Z>^ 3.5Z;3 8.5.75^ ) ~ ^/7t\ 2a^2Aa^ 2.4.6a^~^2.4. 6.8a* ' ) Ex. 2. Expand (a^— a:)""^ into an infinite series. , 1 X l.Sx^ 1.8.5:r3 l,8.5.7x* a 2«^ 2.4a^ 2.4.6a^ 2.4.6.8fr Ex. 8. Expand , into an infinite series. Va;Ma* ^'''' xV 2x^'^2Ax' 2.4.6a;«^2.4.6.8x-«"'^*'''j Ex.4. Expand {a+x) ^ into an infinite series. -1 1-4 1.4-1^ 1.4.7 -1/- , 1.4.7.10 -ij3 Ex. 6. Expand (a^— a:^) ^ into an infinite series. . 1 {. x^ ^ 1.5.T* , 1.5.9X-6 , ^ ) ^^^- 7^ 1^+4^+0^+4:8:12^+' ^'^-j Ex. 6. Expand (1 + cc) ^ into an infinite series. , , X 6^2 6.11^3 6.11. lar^ Ans. 1— ^+;p— jT^— K -,,. -,^- 4- ^ -,,, T- .>.> —> etc. 5 5.10 5.10.15 ^.10.1o.2U ' 282 ALGEBRA. 392. Extraction of any Root of a Surd Number. — The approx- imate value of a surd root may be found by the binomial theorem by dividing the number into two parts, and consider- ing it as a binomial. Ex. 1. Find the square root of 10. Vl0=-v/9Tl=(9 + l)* If, in Ex. 3, Art. 890, we make a^ = ^ and x=l^ we shall have /r-TT q ■ ^ 1 . 3 3.5 3.5.7 ^ 2.3 2.4.3=' 2.4.6.3* 2.4.6.8.3' 2.4.6.8.10.3" ' ®'*^' The value of the first term is 8.0000000 " " second " + .1666667 " " third " - .0046296 " " fourth " + .0002572 " " fifth " - .0000179 " " sixth " 4- .0000014 " " seventh " - .0000001 Their sum is 3.1622777, which is the square root of 10 correct to seven decimal places. Ex. 2. Find the square root of 99. V99=Vl00-l=(100-lf. Substituting in Ex. 3, Art. 890, we have The value of the first term is 10.0000000 " " second " - .0500000 " " third " - .0001250 " '' fourth " - .0000006 Their sum is 9.9498744, which is the square root of 99 correct to seven decimal places. 393. The method here exemplified for finding the nth root of any number is expressed in the following RULE. Find^ hy trial, the nearest integral root (a), and divide the given number into two 2^arts, one ofzohich is (he 7ifh poiver of {a). Chn- SERIES. 283 sid^r these two parts as the terms of a binomial^ and develop it into a series by the binomial theorem. Ex. 8. Find the cube root of 9 to seven decimal places. , pi 2 2.5 2.5.8 "^8.22 3. 6. 2^ "^8. 6. 9. 2« 8.6.9.12.2^^"^'^^^*' = 2.0800838. Ex. 4. Find tlie cube root of 81 to seven decimal places. A qli_L^_ 2.4^ , 2.5.4^ 2.5.8.4^ , 1 1 "^3.27"8.6.272"^3.6.9.273~3.6.9.12.27*'^' ') = 8.1413806. Ex. 5. Find the fifth root of 30 to seven decimal places. . 2 2 2.4 2.4.9 6.16 5.10.16^ 5.10. 15. 16^ ' ' = 1.9748506. 284 ALGEBRA. CHAPTER XX. LOGARITHMS. 394. The logarithm of a number is the exponent of the power to which a constant number must be raised in order to be equal to the proposed number. The constant number is called the base of the system. Thus, if a denote any positive number except unity, and o?=m^ then 2 is the exponent of the power to which a must be raised to equal m; that is, 2 is the logarithm of m in the system whose base is a. If a^—m, then x is the logarithm of 9n in the system whose base is a. 395. If we suppose a to remain constant while m assumes in succession every value from zero to infinity, the corresponding values of X will constitute a system of logarithms. Since an indefinite number of different values may be attrib- uted to a, it follows that there may he an indefinite number of sys- tems of logarithms. Only two systems, however, have come into general use, viz., that system whose base is 10, called Briggs's system, or the common system of logarithms ; and that system whose base is 2.718 + , called the Naperian system, or hyperbolic system of logarithms. Properties of Logarithms in general, 396. TJie logarithm of the product of two or more numbers is equal to the sum of the logarithms of those nurnbei's. Let a denote the base of the system ; also, let m and n be any two numbers, and x and y tHeir logarithms. Then, by the definition of logarithms, we have a*=?7i, (1.) ay=^n, (2.) Multiplying together equations (1) and (2) member by mem- ber, we have a^'^y=.vin. LOGARITHMS. 285 Therefore, according to the definition of logarithms, a: -4-?/ is the logarithm of mn^ since it is the exponent of that power of the base which is equal to mn. For convenience, we will use log. to denote logarithm, and we have x-\-y — \og. mn=log. m+log. n. Hence we see that if it is required to multiply two or more numbers together, we have only to take their logarithms from a table and add them together; then find the number corre- sponding to the resulting logarithm, and it will be the product required. 397. The logarithm of the quotient of two numbers is equal to the logarithm of the dividend diminished by that of the divisor. If we divide Eq. (1) by Eq. (2), member by member, we shall have a^-y=z—. n Therefore, according to the definition, x—y is the logarithm of — , since it is the exponent of that power of the base a which an „, IS equal to — . That is, ^ n x—y=z\og. {—\=\og, m— log. n. Hence we see that if we wish to divide one number by an- other, we have only to take their logarithms from the table and subtract the logarithm of the divisor from that of the dividend ; then find the number corresponding to the resulting logarithm, and it will be the quotient required. 398. The logaritlim of any power of a number is equal to the logarithm of that number multiplied by the exponent of the power. If we raise both members of Eq. (1) to any power denoted byp, we have aP^ = mP. Therefore, according to the definition, ^ce is the logarithm of mP, since it is the exponent of that power of the base which ia equal to m^. That is, px=i\og. (mP)=p log. m. 286 ALGEBRA. Therefore, to involve a given number to any power, we multiply the logarithm of the number by the exponent of the power; the product is the logarithm of the required power. 399. The logarithm of any root of a number is equal to the log- arithm of that number divided by the index of the root. If we extract the rth root of both members of Eq. (1), we shall have a^=.\fm. X . Therefore, according to the definition, - is the logarithm of Vm, That is X , r/— log. m -=log. vm= ^ r r Therefore, to extract any root of a number, we divide the logarithm of the number by the index of the root ; the quotient is the logarithm of the required root. 400. The following examples will show the application of the preceding principles : Ex. 1. log. {abcd)—\o^. a+log. Z)+log. c+log. d. Ex. 2. log. f— -j=log. a-hlog. &4-log. c— log. c?— log. e, Ex. 8. log. {aH'^cP)=.m log. a-\-n log. b-\-p log. c. — ^j =m log. a-\-n log. b—p log. c. Ex. 5. log. Vob=^([og. a+log. b). Ex. 6. log.y '-^=i[log. a+2 log. 5+4 log. c-5 log. d\. Ex. 7. log. (a3 v/^)=log. {a^) =^ log. a. Ex. 8. \og.{a^—x'^)=\og,{{a-\-x){a—x)] =log. Ca+x)+log.Ca-x> Ex. 9. log. Va^-x''=i log. (a+x)+i log. {a-x). Ex. 10. log. (^|^)=i log. 3+i log. 4-i log. 6^:^ log. 2. LOGARITHMS. 287 401. In all systems of logarithms^ the logarithm of unity is zero. For in the equation a^—n^ if we make 92 = 1, the corresponding value of x will be 0, since a" = 1, Art. 75 ; that is, log. 1 := 0. 402. In all systems of logarithms ^ the logarithm of the base is unity. For a^=a; that is, log. a = l. Common Logarithms. 403. Since the base of the common system of logarithms is 10, all numbers in this system are to be regarded diS powers of 10. Thus, since 10" =1, we have log. 1 = 101 = 10, " log. 10 = 1; 102=100, " log. 100 = 2: 10^ = 1000, '' log. 1000 = 3, etc. From this it appears that in Briggs's system the logarithm of any number between 1 and 10 is some number between and 1 ; that is, it is a fraction less than unity, and is generally expressed as a decimal. The logarithm of any number between 10 and 100 is some number between 1 and 2 ; that is, it is equal to 1 plus a decimal. The logarithm of any number between 100 and 1000 is some number between 2 and 8 ; that is, it is equal to 2 plus a decimal ; and so on. 404. The same principle may be extended to fractions by means of negative exponents. Thus, since 10-i=tV or 0.1, we have log. 0.1 = -^1 10-2=^ or 0.01, " log. 0.01 = -2 10-3=.nroTr or 0.001, " log. 0.001 = -3 10-^=Toi(nr or 0.0001, " log. 0.0001 = -4, etc. Hence it appears that the logarithm of every number be- tween 1 and 0.1 is some number between and —1, or may be represented by —1 plus a decimal. The logarithm of every number between 0.1 and 0.01 is some number between —1 and 288 ALGEBRA. — 2, or may be represented by —2 jpliis a decimal. The loga- rithm of every number between 0.01 and 0.001 is some number between —2 and —3, or may be represented by —3 j^^t^ a decimal, and so on. 405. Hence we see that the logarithms of most numbers must consist of two parts, an integral part and a decimal part. The former part is called the characteristic or index of the logarithm. The characteristic may always be determined by the following RULE. The characteristic of the logarithm of any number is equal to the number of places hj which the first significant figure of that number is removed from the unit'' s place, and is positive when this figure is to the left, negative when it is to the right, and zei'o when it is in the unit's p)lax^e. Thus the characteristic of the logarithm of 397 is +2, and that of 5673 is +3, while the characteristic of the logarithm of 0.0046 is -3. 406. The same decimal part is common to the logarithms of all numbers composed of the same significant figures. For, since the logarithm of 10 is 1, it follows from Art. 397 that if a number be divided by 10, its logarithm will be dimin- ished by 1, the decimal part remaining unchanged. Thus, if we denote the decimal part of the logarithm of 3456 by m, we shall have log. 3456 =3 + ^^. log. 345.6 = 2 + 972. log. 34.56 = l-fm. log. 3.456 =0-fm. log. .3456= -1 + m. log. .03456= -2 + m. log. .003456= -3 + m. log. .0003456= --4+m. liable of Logarithms. 407. The table on pages 290, 291, contains the decimal part of the common logarithm of the series of natural numbers from 100 to 999, carried to four decimal places. Since these num- bers are all decimals, the decimal point is omitted, and the char- acteristic is to be supplied according to the rule in Art. 405. LOGAKITHMS. 289 408. To find the logarithm of any number consisting of not more thayi three figures. — Look on one of tbe pages of the ta- ble, along the left-hand column marked Ko., for the two left- hand figures, and the third figure at the head of one of the other columns. Opposite to the first two figures, and in the column under the third figure, will be found the decimal part cf its logarithm. To this must be prefixed the characteristic, according to the rule in Art. 405. Thus the logarithm of 847 is 2.5403 ; 871 is 2.9400. The logarithm of 63, or 63.0, is 1.7993 ; " 5, or 5.00, is 0.6990; " 0.235 is 1.3711. The minus sign is here placed over the characteristic, to show that that alone is negative, while the decimal part of the loga- rithm is positive. 409. To find the logarithm of any number containing more than three figures. — By inspecting the table, we shall find that writhin certain limits the differences of logarithms are propor- tional to the difi*erences of their corresponding numbers. Thus the logarithm of 216 is 2.8345 ; 217 is 2.3865; 218 is 2.3385. Here the difference between the successive logarithms, called the tabular difference^ is constantly 20, corresponding to a differ- ence of unity in the natural numbers. If, then, we suppose the logarithms to increase at the same rate as their corresponding numbers (as they do nearly), a difference of 0.1 in the numbers should correspond to a difference of 2 in the logarithms; a dif- ference of 0.2 in the numbers should correspond to a differ- ence of 4 in the logarithms, etc. Hence the logarithm of 216.1 must be 2.8347; 216.2 " 2.8849, etc. In order to facilitate the computation, there is given, on the right margin of each page, the proportional part for the fourth figure of the natural number, corresponding to tabular differ- N 29C '•'ABLE OF COMMON LOGARITHMS. 1 k 2 3 4 5 6 7 8 9 1 lO 0000 0043 0086 0128 0x70 0212 0253 0294 o334 0374 II o4i4 0453 0492 o53i 0569 0607 o645 0682 0719 0755 12 0792 0828 0864 0899 0934 0969 xoo4 io38 1072 X X06 i3 1139 1173 1206 1239 1271 i3o3 i335 1367 x399 i43o i4 i46i 1492 i523 i553 1 584 i6x4 1 644 1673 1703 1732 i5 I76I 1790 1818 1847 1875 1903 193X 1959 1987 20l4 i6 2o4l 2068 2095 2122 2i48 2175 2201 2227 2253 2279 I? 23o4 233o 2355 238o 24o5 243o 2455 2480 25o4 2529 i8 2553 2577 2601 2625 2648 2672 2695 27x8 2742 2765 ^9 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 20 3oio 3o32 3o54 3075 3096 3ix8 3x39 3i6o 3x8i 320X 21 3222 3243 3263 3284 33o4 3324 3345 3365 3385 34o4 22 3424 3444 3464 3483 35o2 3522 3541 356o 3579 3598 23 3617 3636 3655 3674 3692 37x1 3729 3747 3766 3784 24 38o2 3820 3838 3856 3874 3892 3909 3927 3945 3962 25 3979 3997 4oi4 4o3i 4o48 4o65 4082 4099 4ii6 4x33 26 4i5o 4i66 4i83 4200 42x6 4232 4249 4265 428X 4298 27 43i4 433o 4346 4362 4378 4393 4409 4425 4440 4456 28 4472 4487 45o2 45x8 4533 4548 4564 4579 4594 4609 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 3o 4771 4786 4800 48x4 4829 4843 4857 4871 4886 4900 3i 4914 4928 4942 4955 4969 4983 4997 5oxx 5o24 5o38 32 5o5i 5o65 5079 5092 5io5 5xx9 5x32 5x45 5x59 5x72 33 5i85 5198 52II 5224 5237J5250 5263 5276 5289 53o2 34 53i5 5328 5340 5353 5366 5378 5391 54o3 54x6 5428 35 5441 5453 5465 5478 5490 55o2 55x4 5527 5539 555x 36 5563 5575 5587. 5599 56xx 5623 5635 5647 5658 5670 37 5682 5694 5705 157x7 5729 5740 5752 5763 5775 5786 38 5798 5809 5821,5832 5843 5855 5866 5877 5888 5899 39 5911 5922 5933; 5944 5955 5966 5977 5988 5999 6010 4o 6021 6o3i 6042 ^6o53 6064 6075 6o85 6096 6x07 6x17 4i 6128 6i38 61496160 6x70 6x80 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 6284 6294 63o4 63x4 6325 43 6335 6345 6355 6365 6375 6385 6395 64o5 64x5 6425 44 6435 6444 6454 6464 6474 6484 6493 65o3 65x3 6522 45 6532 6542 655i 656i 6571 658o 6590 6599 6609 66x8 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 68o3 48 6812 6821 683o 68,^; 6848 6857 6866 6875 6884 6893 49 6902 691 1 6920 6928 6937 6946 6955 6964 6972 6981 5o 6990 6998 7007 7016 7024 7033 7042 7o5o 7059 7067 •5 1 7076 7084 7093 710X 71 10 7118 7126 7x35 7143 7i52 52 7160 7168 7177 7x85 7x93 7202 7210 72x8 7226 7235 53 7243 725 1 7259 7267 7275 7284 7292 7300 7308 7316 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 PROPOETlOrfAL 43 42 41 4o 4 4 4 4 9 8 8 8 i3 i3 X2 12 17 17 16 x6 22 2X 21 20 26 25 25 24 3o 29 29 28 34 34 33 32 39 38 37 36 38 37 36 35 34 4 4 4 4 3 8 7 7 7 7 IX XX X X X X xo x5 x5 i4 i4 i4 19 19 18 x8 17 23 22 22 21 20 27 26 25 25 24 3o 3o 29 28 f27 34 33 32 32 3i 33 32 3x 3o 29 3 3 3 3 3 7 6 6 6 6 10 10 9 9 9 i3 i3 12 12 12 17 16 16 i5 i5 20 19 19 18 17 23 22 22 21 20 26 26 25 24 23 So 29 28 27 26 28 27 26 25 24 3 3 3 3 2 6 5 5 5 5 8 8 8 8 7 X I IX 10 10 10 i4 i4 i3 i3 12 n 16 x6 i5 i4 20 '9 18 18. 17 22 22 21 20 19 a5 24 23 J 22 ja TABLE OF COMMON LOGARITHMS. %n No. 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 74o4 7482 7559 7634 7709 7782 7853 7924 7993 8062 8129 8195 8261 8325 8388 845 1 85i3 8573 8633 8692 8751 8808 8865 8921 8976 903 1 9085 9i38 9191 9243 9294 9345 9395 9445 9494 9542 9590 9638 9685 9731 9777 9823 9868 9912 9956 7412 7490 7566 7642 7716 7789 7860 7931 8000 8069 8i36 8202 8267 833i 8395 8457 85i9 8579 8639 8756 88 1 4 8871 8927 8982 9o36 9090 9143 9196 9248 9299 9350 9400 945o 9499 9547 9595 9643 9689 9736 9782 9827 9872 9917 9961 7419 7497 7574 7649 7723 7796 7868 7938 8007 8075 8142 8209 8274 8338 84oi 8463 8525 8585 8645 8704 8762 8820 8876 8932 8987 9042 9096 9149 9201 9253 93o4 9355 94o5 9455 95o4 q552 9600 9647 9694 9741 9786 9832 9877 9921 9965 7427 75o5 7582 7657 7731 7803 7875 7945 8oi4 8082 8149 8215 8280 8344 8407 8470 853i 8591 4 7435 75i3 7589 7664 7738 7810 7882 7952 8021 865i 8710 8768 8825 8882 8938 8993 9047 9101 9154 9206 9258 9309 9360 9410 9460 9509 9557 9605 9652 9699 9745 9791 9836 9881 9926 9969 8089 8i56 8222 8287 835i 84i4 8476 8537 8597 8657 8716 8774 883i 8887 8943 8998 9053 9106 9159 9212 9263 93i5 9365 94i5 9465 95i3 9562 9609 9657 9703 9750 9795 9841 9886 9930 9974 7443 7520 7597 7672 7745 7818 7889 7959 8028 8096 8162 8228 8293 8357 8420 8482 8543 86o3 8663 8722 8779 8837 8893 9004 9o58 91 12 745i 7528 7604 7679 7752 7825 7896 7966 8o35 7459 7536 7612 7686 7760 7832 7903 7973 8o4i 8169 8235 8299 8363 8426 8488 8549 8609 8669 8727 8785 8842 9165 9217 9269 9320 9370 9420 9469 95i8 9566 961/,' 9661 9708 9754 9800 9845 9890 9934 9978 8954 9009 9063 9117 9170 9222 9274 9325 9375 9425 9474 9523 9571 9619 9666 9713 9759 9805 9850 9894 993g 9983 8109 8176 8241 83o6 8370 8432 8494 8555 86i5 8675 8733 8791 8848 8904 8960 9015 9069 9175 9227 9279 9330 9380 9430 9479 9528 9576 9624 9671 9717 9763 9809 9854 9899 9943 9987 7466 7543 7619 7694 7767 7839 7910 7980 8o48 8116 8182 8248 83i2 8376 8439 85oo 856i 8621 8681 8739 8797 8854 8910 8965 9020 9074 9128 9180 9232 9284 9335 9385 9435 9484 9533 9581 9628 9675 9722 9768 9814 9859 9903 9948 9991 7474 755i 7627 7701 7774 7846 7917 7987 8o55 8122 8189 8254 83i9 8382 8445 85o6 8567 8627 8686 8745 8802 8859 8915 8971 9025 9079 9133 9186 9238 9289 9340 9390 9440 9489 9538 9686 9633 9680 9727 9773 9818 9863 9908 9952 99Q6 PROPORTIONAL PARTS 23 22 21 20 19 2 2 2 2 2 5 4 4 4 4 7 7 6 6 9 9 8 8 8 12 u 11 10 10 i4 i3 i3 12 II 16 i5 i5 i4 i3 18 18 17 16 i5 21 20 19 18 17 18 17 16 i5 i4 2 2 2 2 I 4 3 3 3 3 5 5 5 5 4 7 7 6 6 6 9 9 8 8 7 11 10 10 9 8 i3 12 II 11 10 i4 i4 i3 12 1 1 16 i5 i4 i4 i3 i3 12 II 10 9 I I I 1 I 3 2 2 2 2 4 4 3 3 3 5 5 4 4 4 7 6 6 5 5 8 7 7 6 5 9 8 8 7 6 10 10 9 b 7 12 II 10 9 8 8 7 6 5 4 I I I I 2 I I I I 2 2 2 2 I 3 3 2 2 2 4 4 ? 3 2 5 4 4 3 2 6 5 4 4 3 6 6 5 4 3 7 '6 5 S 4 292 ALGEBRA. ences from 43 to 4. Thus, on page 291, near the top, we see that when the tabular difference is 20, the corrections for .1, .2, .3, etc., are 2, 4, 6, etc. It is obvious that the correction for a figure in the fifth place of the natural number must be one tenth of the correction for the same figure if it stood in the fourth place. Such a correc- tion would, however, generally be inappreciable in logarithms which extend only to four decimal places. EXAMPLES. Find the logarithm of 4576. Ans. SM05. « " 18.78. Ans. 1.1892. , " " 1.682. ^725.0.2258. " " .08211. ^725.2^.5066. " " .4735. Ans. 1.6753. " " 15988. Ans. 4.2086. The logarithms here given are only approximate. We can obtain the exact logarithm of very few numbers ; but by taking a sufficient number of decimals we can approach as nearly as we please to the true logarithm. 410. To find the natural number corresponding to any loga- rithm. — Look in the table for the decimal part of the loga- rithm, neglecting the characteristic; and if the decimal is ex- actly found, the first two figures of the corresponding natural number will be found opposite to it in the column headed No., and the third figure will be found at the top of the page. This number must be made to correspond with the characteristic by pointing off decimals or annexing ciphers. Thus the natural number belonging to the logarithm 8.8692 is 2340; " " " '' 1.5878 is 84.5. If the decimal part of the logarithm is not exactly contained in the table, look for the neareM less logarithm, and take out the three figures of the corresponding natural number as be- fore. The additional figure or figures may be obtained by means of the proportional parts on the margin of the page. Find the number corresponding to the logarithm 8.8685. LOGARITHMS. 293 The next less logarithm in the table is .8674, and the three corresponding figures of the natural number are 288. Their logarithm is less than the one proposed by 11, and the tabular difference is 18. By referring to the margin of page 291, we find that, with a difference of 18, the figure corresponding to the proportional part 11 is 6. Hence, since the characteristic of the proposed logarithm is 8, the required natural number is '2836. EXAMPLES. 1. Find the number corresponding to the logarithm 2.5886. Ans. 845.6. 2. Find the number corresponding to the logarithm 0.2845. Ans. 1.716. 8. Find the number corresponding to the logarithm 1.9946. Ans. 98.76. 4. Find the number corresponding to the logarithm 1.6478. Ans. 0.4444. 411. Multiplication hy Logarithms. — According to Art. 896, to find the product of two numbers we have the following RULE. Add the logarithms of the factors ; the sum will he the logarithm of the product. The word sum is here to be understood in its algebraic sense. The decimal part of a logarithm is invariably positive ; but the characteristic may be either positive or negative. Ex. 1. Find the product of 57.98 by 8.12. The logarithm of 57.98 is 1.7688. " 8.12 is 0.4942. The log. of the product 180.9 is 2.2575. Ex. 2. Find the product of 0.00568 by 172.5. The logarithm of 0.00568 is '8.7505. 172.5 is 2.2868. The log. of the product 0.971 is L9878. Ex. 3. Find the product of 54.82 by 6.548. Ex. 4. Find the product of 8.854 by 0.5761. 294 ALGEBRA. 412. Division hy Logarithms. — According to Art. 397, to find the quotient of two numbers we have the following RULE. From the logarithm of the dividend subtract the logarithm of the divisor ; the difference will he the logarithm of the quotient. The word difference is here to be understood in its algebraic sense ; the decimal part of the logarithm being invariably pos- itive, while the characteristic may be either positive or nega- tive. Ex. 1. Find the quotient of 888.7 divided by 42.24. The logarithm of 888.7 is 2.9488. " 42.24 is 1.6257 . The quotient is 21.04, whose log. is 1.8231. Ex. 2. Find the quotient of 0.8692 divided by 42.32. The logarithm of 0.8692 is T.9891. 42.32 is 1.6265 . The quotient is 0.02054, whose log. is 2^3126. Ex. 8. Find the quotient of 880.7 divided by 18.75. Ex. 4. Find the quotient of 24.93 divided by .0785. 413. Involution hy Logarithms. — According to Art. 398, to involve a number to any power we have the following RULE. Multiply the logarithm of the number hy the exponent of the poiver required. It should be remembered that what is carried from the deci- mal part of the logarithm is positive, whether the characteris- tic be positive or negative. Ex. 1. Find the fifth power of 2.846. The logarithm of 2.846 is 0.4542. 5 The fifth power is 186.65, whose log. is 2.2710. Ex. 2. Find the cube of .07654. LOGARITHMS. 295 The logarithm of .07654 is 2^8839. 3 The cube is 0.0004484, whose log. is 46517. Ex. 3. Find the 20th power of 1.06. Ex. 4. Find the seventh power of 0.8952. 414. Evolution hy Logarithms. — According to Art. 399, to extract any root of a number we have the following RULE. Divide tlie logarithm of the number hy tlie index of the root re- quired. Ex. 1. Find the cube root of 482.4. The logarithm of 482.4 is 2.6834. Dividing by 3, we have 0.8945, which corresponds to 7.848, which is therefore the root required. Ex. 2. Find the 100th root of 365. Ans. 1.061. When the characteristic of the logarithm is negative, and is not divisible by the given divisor, we may increase the charac- teristic by any number which will make it exactly divisible, provided we prefix an equal positive number to the decimal part of the logarithm. Ex. 3. Find the seventh root of 0.005846. The logarithm of 0.005846 is 3".7669, which may be written 7+4.7669. Dividing by 7, we have 1.6810, which is the logarithm o^ .4797, which is therefore the root required. Ex. 4. Find the 10th root of 0.007815. 415. Proportion by Logarithms. — The fourth term of a pro- portion is found by multiplying together the second and third terms and dividing by the first. Hence, to find the fourth term of a proportion by logarithms, we have the following 296 ALGEBRA. RULE. Add the logarithms of the second and third terms, and from iheij sum subtract the logarithm of the first term,. Ex. 1. Find a fourth proportional to 72.34, 2.519, and 857.6. Ans. 12.45. Ex. 2. Find a fourth proportional to 43.17, 275, and 5.762. Ex. 8. Find a fourth proportional to 5.745, 781.2, and 54.27. Exponential Equations. 416. An exponential equation is one in which the unknown quantity occurs as an exponent. Thus, is an exponential equation, from which, when a and h are known, the value of x may be found. If a ==2 and 5 = 8, the equation becomes 2^=8, in which the value of a; is evidently 3, since 2^==8. If a =16 and h=2, the equation becomes 16^=2, in which the value of a? is evidently J, since 16* =2. 417. Solution hy Logarithms. — When h is not an exact power or root of a, the equation is most readily solved by means of logarithms. Taking the logarithm of each member of the equation a* =5, we have X log. a=log. h, whence x= , ^' . log. a Ex. 1. Solve the equation 3=^=20. log. 20 1.3010 ^^^^ Ex. 2. Solve the equation 5^=12. Ex. 3. Solve the equation f-j =|. Ex. 4. Solve the equation 10*= 7. Ex. 5. Solve the equation 12""= 3. 3 Ex. 6. Solve the equation 12^=7o LOGARITHMS. 297 Compound Interest. 418. Interest is money paid for the use of money. When the interest, as soon as it becomes due, is added to the principal, and interest is charged upon the whole, it is called comj)ound interest 419. To find the amount of a given sum in any time at com- pound interest. It is evident that $1.00 at 5 per cent, interest becomes at the end of the year a principal of $1.05 ; and, since the amount at the end of each year must be proportioned to the principal at the beginning of the year, the amount at the end of two years will be given by the proportion 1.00: 1.05:: 1.05: (1.05)2. The sum (1.05)^ must now be considered as the principal, and the amount at the end of three years will be given by the proportion 1.00: 1.05:: (1.05)2 :(1.05)^ In the same manner, we find that the amount of $1.00 for n years at 5 per cent, compound interest is (1.05)". For the same reason, the amount for n years at 6 per cent, is (1.06)^ It is also evident that the amount of P dollars for a given time must be P times the amount of one dollar. Hence, if we put P to represent the principal, r the interest of one dollar for one year, n the number of years for which interest is taken. A the amount of the given principal for n years, we shall have A = P (1 + r)". This equation contains four quantities. A, P, w, r, any three of which being given, the fourth may be found. The compu- tation is most readily performed by means of logarithms. Tak- ing the logarithms of both members of the preceding equation and reducing, we find log. A =: log. V-\~nx log. ( 1 + r), log. P =:log. A— nxlog. (1+r), , /-, , X log. A— log. P log. (1 + r) = -^ 2 — . n ^^ log. A-log. P log.(l + r) • N > 298 ALGEBRA. Ex. 1. How mucli would 500 dollars amount to in five years at 6 per cent, compound interest ? The log. of 1.06 is 0.0253 5 0.1265 The log. of 500 is 2.6990 The amount is $669.10, whose log. is 2.8255. Ex. 2. What principal at 6 per cent, compound interest will amount to 500 dollars in seven years? Ans. $332.60. Ex. 3. At what rate per cent, must 500 dollars be put out at compound interest so that it may amount to $680.30 in seven years ? Aiis. 4^ per cent. Ex. 4. In what time will 500 dollars amount to 900 dollars at 6 per cent, compound interest? Ans. lOyy years. Ex. 5. How much would 400 dollars amount to in nine years at 5 per cent, compound interest ? Ex. 6. What principal at 5 per cent, compound interest will amount to 400 dollars in eight years? Ex. 7. At what rate per cent, must 400 dollars be put out at compound interest so that it may amount to $620.70 in nine years ? Ex. 8. In what time will a sum of money double at 6 per cent, compound interest? Ex. 9. In what time will a sum of money double at 5 per cent, compound interest? Annuities. 420. An annuity is a sum of money stipulated to be paid an- nually, and to continue for a given number of years, for life, or forever. 421. To find the amount of an annuity left unpaid for any num.' her of years ^ allowing compound interest. Let a denote the annuity, n the number of years, r the in- terest of one dollar for one year, and A the required amount. The amount due at the end of the first year is a. At the end of the second year the amount of the first an- LOGARITHMS. 299 nuity is a(l-f-r), and a second payment becomes due; hence the whole sum due at the end of the second year is a-\-a{l-{-r). At the end of the third year a third payment a becomes due, together with the interest on a-^-a{l-\-r)] hence the whole sum due at the end of the third year is a+a(lH-r)4-a(l+r)2, or a{l + {l-i-r)-\-{l-\-ry}, and so on. Hence the amount due at the end of n years is G[l-f(l + r) + (l + r)H(l+r)^-h.... +(l + r)"-M. These terms form a geometrical progression in which the ratio is 1+7*. Hence, by Art. 332, the sum of the series is 422. To find the present value of an annuity ^ io continue for a certain number ofyears^ allowing compound interest. The present value of the annuity must be such a sum as, if put out to interest for n years at the rate r, would amount to the same as the amount of the annuity at the end of that period. If P denote the present value of the annuity, then the amount of the annuity will be P(l+r)", which must be equal to a,^ — ■ — <- . r Therefore p^a/l|r)"-l. r (1+7-)" Ex. 1. How much will an annuity of 500 dollars amount, to in 15 years at four per cent, compound interest? (1+r)^ =1.7987 (l+r)"-l= .7987, whose log. is 19024 the log. of .04 is 2".6021 L3003 the log, of 500 is 2.6990 The amount is $9983, whose log. is 3.9993. Ex. 2. What is the present value of an annuity of 500 dol- 800 ALGEBRA. lars to continue for 20 years, interest being allowed at the rate of four per cent, per annum ? (l + r)« =2.188 (l+r)^- 1 = 1.188, whose log. is 0.0748 the log. of (1+r)" is 0.8400 r.7348 -=12500, whose log. is 4.0969 The present value is $6787, whose log. is 3.8817. Ex. 8. How much will an annuity of 600 dollars amount to in 12 years at three per cent, compound interest ? Ex. 4. What is the present value of an annuity of 600 dollars to continue for 12 years at three per cent, compound interest ? Ex. 5. In what time will an annuity of 500 dollars amount to 5000 dollars at 4 per cent, compound interest? Ans. In 8| years. Increase of Population, 423. The natural increase of population in a country is some times computed in the same way as compound interest. Know- ing the population at two different dates, we compute the 7'ate of increase by Art. 419, and from this we may compute the pop- ulation at any future time on the supposition of a uniform rate of increase. Such computations, however, are not very reliable, for in some countries the population is stationary, and in others it is decreasing. Ex. 1. The number of the inhabitants of the United States in 1790 was 3,980,000, and in 1860 it was 31,445,000. What was the average increase for every ten years ? Alls. 344 per cent. Ex. 2. Suppose the rate of increase to remain the same for the next ten years, what would be the number of inhabitants in 1870? ^725.42,830,000. Ex. 3. At the same rate, in what time would the number in 1860 be doubled ? Aiis. 23J years. Ex. 4. At the same rate, in what time would the number in 1860 be tripled? LOGARITHMS. 301 To find the Logarithm of any given Number. 424. If m and n denote any two numbers, and x and y their logarithms, then "^ will be the logarithm of Vmn. For, ac- cording to Art. 396, a^^y=mn^ and, taking the square root of each member, we have a ^ —\mn. Therefore, ^ is the logarithm of -v/mn, since it is the exponent of that power of the base which is equal to ^/'mrl. Now, in Briggs's system, the logarithm of 10 is 1, of 100 is 2, etc. Hence the logarithm of VlOx 100 is ; that is, the logarithm of 31.6228 is 1.5. 1 + 1.5 So, also, the logarithm of VlO x 31.6228 is — ^r-^ ; that is, the logarithm of 17.7828 is 1.25, and so on for any number of logarithms. In this manner were the first logarithmic tables computed ; but more expeditious methods have since been discovered. It is found more convenient to express the logarithm of a number in the form of a series, 425. Logarithms computed hy Series. — The computation of log- arithms by series requires the solution of the equation a^=n^ in which a is the base of the system, n any number, and x is the logarithm of that number. In order that a and n may be expanded into a series by the binomial theorem, we will con- vert them into binomials, and assume a=l-\-h and n=l-\-m ; then we shall have where x is the logarithm of 1+m, to the base 1 + Z), or a. Involving each member to a power denoted by y, we have {X'\-h)^={l-\-m)y. Expanding both members by the binomial theorem, we have 302 ALGEBRA. i+^y6+ '^y("|-i) 6.+ ^y(^y-iK^y-2) fc3+, etc.= l+ym+^JS^,a'+ y(y-^)(V-^) ,n^ + , etc. Canceling unity from both members and dividing bj y, we have ,(i+^j3+M^)j3^_ etc.)= This equation is true for all values of y; it will therefore be true when y=0. Upon this supposition, the equation becomes ^(^-■2 + 3-' etc.)=m — -+- — , etc., m-— +— -,etc. whence x = log. {l-\-m)= ^2 — 13 • If we put M= — ^-j^, , ^-2 + 3-' ^^- the last equation becomes x=log. n=log. (l + m) = M(w— y+-p — , etc.). (1.) We have thus obtained an expression for the logarithm of the number 1+m or n. This expression consists of two fac- tors, viz., the quantity M, which is constant, since it depends simply upon the base of the system ; and the quantity within the parenthesis, which depends upon the proposed number. The constant factor M is called the modulus of the system, 426. To determine the Base of Napier'' s System. — In Napier's system of logarithms the modulus is assumed equal to unity. From this condition the base may be determined. Equation (1), Art. 425, in this case becomes m^ , m^ m* ic=m-y + -g— -^4-, etc. LOGARITHMS. 303 Eeverting this series, Art. 883, Ex. 3, we obtain r.4: 2 ' 2.3 ' 2.3.4 But, by hypothesis, a^=n=l-{-m ; therefore If X be taken equal to unity, we have "=2+1+0+2x4+' '*°- By taking nine terms of this series, we find a=2.718282, which is the base of Napier's system. 427. The logarithm of a number in any system is equal to the modulus of that system multiplied by the Naperian logarithm of the number. If we designate Naperian logarithms by Nap. log., and log- arithms in any other system by log., then, since the modulus of Napier's system is unity, we have log. (H-m)=:M(m-y4-g— , etc.), Nap.log.(l + 7n)=m— — +-g — , etc. Hence log. (1 + m) = M x Nap. log. (1 + m ), or M^ log.(l + ^) Nap. log. (1 + to)' where 1+m may designate any number whatever. 428. To render the Logarithmic Series converging. — The for- mula of Art. 425, log. (l + ,7i) = M(m^y H- — ^, etc.), (1.) can not be employed for the computation of logarithms when m is greater than unity, because the series does not converge. This series may, however, be transformed into a converging series in the following manner : 804 ALGEBRA. Substitute —m for m, and we shall have log. (l_m) = M(-m-|^-^'-, etc.). (2.) Subtracting Eq. (2) from Eq. (1), observing that log. (l-|-mj ~log. (1— m) = log. , we shall have ^ 1— m 1°& r:^^2M(m+-+_+, etc.). Now, since this is true for every value of ta^ put 1 , 1+m ^+1 m = - -, whence —^ , 22^ + 1 1— m p ' and the preceding series, by substitution, becomes log. P±}:=^oe.ip+^■)-^^g.p=m(^^^+J^^,+^^^,+, etc.): 429. This series converges rapidly, and may be employed for the computation of logarithms in the Naperian or the com- mon systems. It is only necessary to compute the logarithms of prime numbers directly, since the logarithm of any other number may be obtained by adding the logarithms of its sev- eral factors. Making ^ = 1, 2, 4, 6, etc., successively, we obtain the following Naperian or Hyperbolic Logarithms, log.3^1og.2 + 2(i4-3^+^+^+....) =1.098612 log. 4 = 2 log. 2 =1.386294 log.5=log.4 + 2(i+gl34-^4-y^+....) =1.609438 log. 6=log. 3+log. 2 =1.791759 log.7=log.6+2(J^+^+g^+^+....) = 1.945910 log. 8=3 log. 2 =2.079442 log. 9 = 2 log. 3 =2.197225 log. 10=log. 5+log. 2 =2.302585 etc., etc., etc. LOGARITHMS. 305 430. To construct a Table of Common Logaritlims. — In order to compute logarithms of the common system, we must first determine the value of the modulus. In Art. 427, we found 11^ log, (l + m) Nap. log. (l+m)* If l-|-m=a, the base of the system, then log. a=l, and we have ^=w-A — ; JNap. log. a that is, the modulus of any system is the reciprocal of the Naperian logarithm of the base of the system. The base of the common system is 10, whose ISTaperian log- arithm is 2.302585. Hence which is the modulus of the common system. We can now compute the common logarithms by multiply- ing the corresponding Naperian logarithms by .434294, Art. 427. In this manner was the table on pages 290-1 computed. 431. Results. — The base of Briggs's system is 10. Napier's '' 2.71828. The modulus of Briggs's system is 0.43429. " Napier's '' 1. Since, in Briggs's system, all numbers are to be regarded as powers of 10, we have 100.301^2, 100.602^4^ etc. In Napier's system, all numbers are to be regarded as powers Df 2.71828. Thus, 2.7180.693^2, 2.7181.098,^3^ 2.718^-3^=:4, etc. 306 ALGEBKA. CHAPTER XXI. GENERAL THEORY OF EQUATIONS. 432. A cubic equation with one unknown quantity is an equation in which the highest power of this quantity is of the third degree, as, for example, x^—Qx'^-{Sx—lb=0. All equa- tions of the third degree with one unknown quantity may be reduced to the form x^-\-ax'^-\-hx-\-c=0. A biquadratic equation with one unknown quantity is an equation in which the highest power of this quantity is of the fourth degree, as, for example, x"^ — 6x^ + 7x'^ + 5a:;— 4 =i 0. Every equation of the fourth degree with one unknown quantity may be reduced to the form x^-^ax^-^bx^-]-cx + d=0. The general form of an equation of the fifth degree with one unknown quantity is x^ + ax* + hx^-^cx^ + dx-{-ez=zO] and the general form of an equation of the ?2th degree with one unknown quantity is ci:" + Aa;~-i + Bx«-2+Ccc"-H .... +Tx+Yz±.0, (1.) This equation will be frequently referred to hereafter by the name of the general equation of the nth degree, or simply a3 Equation (1). An equation not given in this form may be reduced to it by transposing all the terms to the first member, arranging them according to the descending powers of the unknown quantity, and dividing by the coefiicient of the first term. In this equa- tion n is a positive whole number, but the coefficients A, B, C, etc., may be either positive or negative, entire or fractional, rational or irrational, real or imaginary. The term V may be regarded as the coefficient of a;°, and is called the absolute terra of the equation. It is obvious that if we could solve this equation we should GENERAL THEORY OF EQUATIONS. 307 have the solutioQ of every equation that could be proposed. Unfortunately, no general solution has ever been discovered ; yet many important properties are known which enable us to solve any numerical equation. 433. Any expression, either numerical or algebraic, real or imaginary, which, being substituted for x in Equation (1), will satisfy it, that is, make the two members equal, is called a root of the equation. It is assumed that Eq. (1) has at least one root ; for, since the first member is equal to zero, it will be so for some value of •T, either real or imaginary, and this value of x is by definition a root. 434. If a IS a root of the general equation of the nth degree, its first member can he exactly divided hy x—a. For we may divide the first member by x—a, according to the usual rule for division, and continue the operation until a remainder is found which does not contain x. Let Q denote the quotient, and E the remainder, if there be one. Then we shall have £c«+Acc"-i4-Bx"-24-.... +Tic+Y = Q(a:-a) + K. (2.) Now, if a is a root of the proposed equation, it will reducQ the first member of (2) to 0; it will also reduce Q(a?— a) to 0; hence K is also equal to 0. But, by hypothesis, E does nol contain x; it is therefore equal to 0, whatever value be attrib- uted to X, and, consequently, the first member is exactly di visible by x—a. 435. If the first member of the general equation of the nth de gree is exactly divisible by x — a, then a is a root of the equation. For suppose the division performed, and let Q denote the quotient; then we shall have a;"+Aa;«-i + Ba;"-2+.... -\'Tx-\-Y = Q,{x-a). If, in this equation, we make x=a, the second member re^ duces to 0; consequently the first member reduces to 0; and, therefore, a is a root of the equation. 808 ALGEBRA. EXAMPLES. 1. Prove that 1 is a root of the equation x3_6x2 + llx-6=:0. The first member is divisible hyx—l, and gives a:^— 5x+ 6 =0. 2. Prove that 2 is a root of the equation x^-x-Q=0. The first member is divisible by cc— 2, and gives ic2+2ic+ 3=0, 8. Prove that 2 is a root of the equation cc3_llcc2+86x-36=:0. 4. Prove that 4 is a root of the equation a;3 4-a?2_34:c+56=:0. 5. Prove that — 1 is a root of the equation x*-38cc3+210x2+538x+289=0. 6. Prove that —5 is a root of the equation a,5 + 6a:*-10a73-112a;2_207x-110 = 0. 7. Prove that 3 is a root of the equation 436. Every equation of the nth degree containing hut one un- known quantity has n roots and no more. Since the equation has at least one root, denote that root by a; then will the first member be divisible by cc— a, and the quo- tient will be of the form and the given equation may be written under the form (cc-a)(x«-i4- A'a;"-2+ .... ^T'x^Y')=0. (3.) Now equation (3) may be satisfied by supposing either of its factors equal to zero. If the second factor equals zero, we shall have £c"-i + A'a:«-2-f B'a;«-3 4- .... +T'cc+ Y'=0. (4.) Kow equation (4) has at least one root ; denote that root by b; then will the first member be divisible hy x—b, and equation (4) can be written under the form (x-Z))Gx^-2 4. A''^«-3+ .... +T'x+Y'')=0, which reduces Eq. (3) to the form of (:P_a)(cc-5)(a;"-H A^'x^-s^. .... T''a: + Y'0=O. GENERAL THEORY OF EQUATIONS. 809 Bj continuing this process, it may be shown that the first member will ultimately be resolved into n binomial factors of the form x—a^ x—b, x—c^ etc. Hence equation (1) may b^. written under the form (x-a){x-b){x-c){x-d).. . . {x-k){x-l) = 0. (5.) This equation may be satisfied by any one of the n values, x=za, x=b, x=c, etc., and, consequently, these values are the roots of the equation. The equation has no more than n roots, because if we ascribe to X a value which is not one of the n values a, J, c, etc., this value will not cause any one of the factors of Eq. (6) to be zero, and the product of several factors can not be zero when neither of the factors is zero. If both members of Eq. (5) be divided by either of the fac- tors x—a, x—b^ etc., it will be reduced to an equation of the next inferior degree ; and if we can depress any equation to a quadratic, its roots can be determined by methods already ex- plained. Ex. 1. One root of the equation x3+3a;2_16x 4-12 = is 1 ; what are the other roots? Ex. 2. Two roots of the equation a:*-10cc3-|-35a:2_50T+24=0 are 1 and 3 ; what are the other roots? Ex. 3. Two roots of the equation x*-12x^+4Sx^-6Sx-\-15=0 are 3 and 5 ; what are the other roots ? Ans. 2 ± V^. Ex. 4. Two roots of the equation 4x^ - 14^3 - 5xH 31cc + 6 = are 2 and 3 ; what are the other roots ? — 8 ± VE Ans. , 4 Ex. 5. Two roots of the equation x^-6x^+24:x-16=0 are 2 and —2 ; what are the other roots? Ans. 8± V6. 437. The n roots of an equation of the 72th degree are not necessarily all different from each other. Any number, and, in- 310 ALGEBRA. deed^ all ofthem^ may he equal. When we say that an equation of the nth degree has n roots, we simply mean that its first member can be resolved into n binomial factors, equal or un- equal, and each factor contains one root. Thus the equation x^—Qx^-\-12x—^=0 can be resolved into the factors (a;-2)(x-2)(x-2)=:0, or (x- 2)3 = 0; whence it appears that the three roots of this equation are 2, 2, 2. But, in general, the several roots of an equation differ from each other numerically. The equation x^=^ has apparently but one root, viz., 2, but by the method of the preceding article we can discover two other roots. Dividing x^— 8 by cc— 2, we obtain a:2-f-2x+4=0. Solving this equation, we find x= — 1± V— 3. Thus, the three roots of the equation x^=^ are 2; -i + vz:!; -l-V^s. The student should verify the last two values by actual mul- tiplication. Ex. 1. Find the four roots of the equation cc*— 81=0. Ex. 2. Find the six roots of the equation cc^— 64=0. 438. The coefficient of the second term in the equation of the nth degree is equal to the algebraic sum of the roots with their signs changed. The coefficient of the third term is equal to the algebraic sum of the products of all the roots, taken in sets of two. The coefficient of the fourth term is equal to the algebraic sum of the products of all the roots, taken in sets of three, with their signs changed. The last term is equal to the continued product of all the roots with their signs changed. Let a,b,c,d, I, represent the roots of an equation of the nth degree. This equation will accordingly contain the factors x—a, x^b, etc. ; that is, we shall have {x-a){x-b){x—c){x-d) (x-l)z^O. If we perform the multiplication as in Art. 351, we shall have GENERAL THEORY OF EQUATIONS. 311 c"— a a:" -i + aZ) ^n- -'^—abc -h ■\-ac — abd — c -{■ad — acd -d + bd -bed etc. etc. etc. X' n-3 4-.... ±:{abcd,.., l)=0', which results are seen to conform to the laws above stated. By the method employed in Art. 352 it may be proved that if these laws hold true for the product of ?i binomial factors, they will also hold true for the product of 72+1 binomial factors. But we have found by actual multiplication that these laws are true for the product of four factors, hence they are true for the product of five factors. Being true for five, they must be true for six, and so on for any number of factors. It will be perceived that these properties include those of quadratic equations mentioned on pages 203-5. If the roots are all negative, the signs of all the terms of the equation will be positive, because all the signs of the factors of which the equation is composed are positive. If the roots are all positive, the signs of the terms will bo alternately positive and negative. If the sum of the positive roots is numerically equal to the sum of the negative roots, their algebraic sum will be zero ; consequently the coefficient of the second term of the equation will be zero, and that term will disappear from the equation. Conversely, if the second term of the equation is wanting, the sum of the positive roots is numerically equal to the sum of the negative roots. Ex. 1. Form the equation whose roots are 1, 2, and 3. For this purpose we must multiply together the factors 'rr— 1, a;— 2, x—S^ and we obtain x^ — 6x^-{-llx—6 = 0. This example conforms to the rules above given for the co efficients. Thus the coefficient of the second term is equal to the sum of all the roots, 1 + 2 + 3, with their signs changed. The coefficient of the third term is the sum of the products of the roots taken two and two; tlins, Ix2 + lx3 + 2x3=zll. 312 ALGEBRA. The last term is the product of all the roots, 1x2x3, with their signs changed. Ex. 2. Form the equation whose roots are 2, 3, 5, and —6. Ans. x^-ix^-29x'+lbQx-lS0=0. Show how these coefficients conform to the laws above given. Ex. 3. Form the equation whose roots are 1, 1, 1, -1, and -2. Ex. 4. Form the equation whose roots are 1, 3, 5, -2, -4, and -6. Ans. x^ + 3x'-4.lx*-87x^-\-4.00x^+444:X-720=0. Ex. 6. Form the equation whose roots are l±V^ and 2±V~^. Ex. 6. Form the equation whose roots are l±-v/3i and2±'/3. 439. Since the last term is the continued product of all the roots of an equation, it must be exactly divisible by each of them. For example, take the equation x^—x—6 = 0. Its roots must all be divisors of the last term, 6; hence, if the equation has a rational root, it must be one of the numbers 1, 2, 3, or 6, either positive or negative ; and, by trial, we can easily ascertain whether either of these numbers will satisfy the equation. We thus find that +2 is one of the roots, and, by the method of Art. 436, we find the remaining roots to be — 1± V--2. If the last term of an equation vanishes, as in the example x*-\-2x^ + Sx'^ + 6x=0, the equation is divisible by x— 0, and consequently is one of its roots. If the last two terms van- ish, then two of its roots are equal to zero. 440. If the coefficients of an equation are whole numbers, and the coefficient of its first term unity, the equation can not have a root which is a rational fraction. Suppose, if possible, that y is a root of the general equation of the nth degree, where ^ represents a rational fraction ex- pressed in its lowest terms. Substitute this value for x in the given equation, and we have GENERAL THEORY OF EQUATIONS. 813 . i7.+ A^,Tn+Bj^,+ .... +T-+V=0. Multiplying each term by b''-\ and transposing, we obtain Now, by supposition, a, Z), A, B, C, etc., are whole numbers ; hence the right-hand member of the equation is a whole num- ber. But, by hypothesis, | is an irreducible fraction ; that is, a and b contain no common factor. Consequently, a" and b will contain no common factor; that is, ^ is a fraction in its low- est terms. Hence the supposition that the irreducible fraction T is a root of the equation leads to this absurdity, that an irre- ducible fraction is equal to a whole number. This proposition only asserts that every commensurahle root must be an integer. The roots can not be of the form of f , y, ■|, etc. The equation may have other roots which are incom- mensurable or imaginary, as 2ztVS, 1± V — 2. 441. Any equation having fractional coefficients can be trans-' formed into another which has all its coefficients integers^ and the coefficient of its first term unity. Eeduce the equation to the form Air" + Ba?"-HCcc"-2+ -(-Tcc-fY=0, in which A, B, C, etc., are all integers, either positive or neg- ative. Substitute for x the value x~~-. A' and the equation becomes r Br-^ Cyn-2 ,T^,y^.. which, multiplied by A*^-^, becomes 2/" + B?/^-^ + AC?/"-2+ + A"-2T?/+ A"-lVr:rO. O 314 ALGEBRA. in which the coefficients are all integers, and that of the first term unity. The substitution of ^ for x is not always the one which leads to the most simple result ; but when A contains two or more equal factors, each factor need scarcely ever be repeated more than once. Ex. 1. Transform the equation oc? — 9-+^^ — u~^ ^"^^ ^^" other whose coefficients, are integers, and that of the first term unity. Clearing of fractions, we have 86:c3-54cc2 4-45^^-8=0. Substituting ^ for cc, the transformed equation is or 2/^-92/2 4-45?/-48=0. Transform the following equations into others whose coeffi- cients are integers, and that of the first term unity. Ex.2. icH2x2+|-^=0. ^rz5. y3_^12?/2+97/-24=0. Ex.3. icH^'-^+2 = 0. ^715.2/3+23/2-93/4-432=0. Ex.4. x3 4-2ia;2 4-1-1=0. Ex.5. a;*-4ia:2 4-8aj4-2TV=0. 14^2 10 Ex.6. a;3_i^+7^_i!^^0. 442. If in any complete equation involving hut one unhnoum quantity the signs of the alternate terms he changed^ the signs of all the roots will he changed. Take the general equation of the nih. degree, ir"-f Ax"-HBcc"-3 4-Ca;"-8 4- =0. (1.) in which the signs may follow each other in any order whatever. GENERAL THEORY OF EQUATIONS. 815 If we change the signs of the alternate terms, we shall have x^ - Acc^-i 4- Bx^-2 _ Qx"-^ + =0. (2 .) or, changing the sign of every term of the last equation, -a?'^ + Aa:"-i-Bx"-2 + Ca?^'~^- =0. (3.) Now, substituting +a for x in equation (1) will give the same result as substituting —a in equation (2), if n be an even number; or substituting —a in equation (3), if n be an odd number. If, then, a is a root of equation (1), —a will be a root of equation (2), and, of course, a root of equation (3), which is identical with it. Hence we see that the positive roots may be changed into negative roots, and the reverse, by simply changing the signs of the alternate terms ; so that the finding the real roots of any equation is reduced to finding positive roots only. This rule assumes that the proposed equation is complete; that is, that it has all the terms which can occur in an equation of its degree. If the equation be incomplete, we must intro- duce any missing term with zero for its coefficient. Ex. 1. The roots of the equation x^ — 2x^—bx-\-Q—0 are 1, 3, and —2 ; what are the roots of the equation a;3-|-2a;2_5cc-6=0? Ex. 2. The roots of the equation cc^ — Ga^^-j-Hx— 6 = are 1, 2, and 3 ; what are the roots of the equation Ex.3. The roots of the equation x'^ — Qx^-^-bx^+lx—lO^O are —1, +5, 1 + V— 1, and 1— V — 1; what are the roots of the equation cc* + 60;^ + Sa;^ - 2ic - 10 = ? 443. If an equation whose coefficients are all real contains im- aginary roots^ the number of these roots must he even. If an equation whose coefficients are all real has a root of the form a+Z/V— 1, then will a—hV — 1 be also a root of the equation. For, let a-^-hV — 1 be substituted for x in the equa- tion, the result will consist of a series of terms, of which those involving only the powers of a and the even powers oihV —1 will be real, and those which involve the odd powers o^hV -^1 will be imaginary. 816 ALGEBRA. If we denote the sura of the real terms by P, and the sum of the imaginary terms by QV — 1, the equation becomes P + QV^^-O. But, according to Art. 243, this equation can only be true when we have separately P = and Q = 0. If we substitute a—bV — 1 for x in the proposed equation, the result will differ from the preceding only in the signs of the odd powers ofb\/—l^so that the result will be P— Q V — 1. But we have found that P = and Q=0; hence P— Q\/— 1=0. Therefore a—hV — 1, when substituted for x, satisfies the equa- tion, and, consequently, it is a root of the equation. It may be proved in a similar manner that if an equation "whose coefficients are all raiional,has a root of the form a+ Vb^ then will a— Vb be also a root of the equation. Ex.1. One root of the equation x^—2x-\-4: = is l-f--/— 1; what are the other roots? Ex. 2. One root of the equation a;^ — a;^ — 7a; + 15 = is 2+ V— 1 ; what are the other roots? Ex. 3. One root of the equation a:^ — a:^ -f 3a: -4- 6 = is 1 + 2-1/ — 1 ; what are the other roots? Ex. 4. One root of the equation x* — 4a;^ + 4x — 1 = is 2+ \/3 ; what are the other roots? Ex. 5. Two roots of the equation cc8-|-2a:6+4a;^+4a;*-8a;2-16a^-32=0 are — 1+ -/— 1 and 1— V— 3; what are the other six roots? 444. Any equation involving but one unknown quantity may be transformed into another whose roots differ from Hiose of Hie pro- posed equation by any given quantity. Let it be required to transform the general equation of the nth degree into another whose roots shall be less than those of the proposed equation by a constant difference h. Assume y=x—h^ whence x = y-\-h. Substituting y-^h for x in the proposed equation, we have (y-f /0"+ A(7/-f /0"-HB(z/-h/0"-H .... +V=0. GENERAL THEORY OF EQUATIONS. 317 Developing the different powers of y-[-h by the binomial formula, and arranging according to the powers of y, we have y-^ + ,etc.l J + A 4.(7i-l)AA +i(7,_l)(7i_2)A^2 + B +(n-2)B// + C which equation satisfies the proposed condition, since y is less than X by h. If we assume y=x-\-h, or x — y—h^ we shall ob- tain in the same manner an equation whose roots are greater than those of the given equation by h. Ex. 1. Find the equation whose roots are greater by 1 than those of the equation x^ -\-Zx'^ —4:X-\-l = 0. We must here substitute y—1 in place of a:. Ans. y^~-7y-\-7 = 0. Ex. 2. Find the equation whose roots are less by 1 than those of the equation x^ — 2x'^-\-Sx—4:=0. Ans. ?/^+?/2 + 2?/— 2 = 0. Ex. 3. Find the equation whose roots are greater by 3 than those of the equation x*-\-9x^-\-12x'^ — 14:X=0. Ans. 2/4_3^3-15?/2 + 49?/-12 = 0. Ex. 4. Find the equation whose roots are less by 2 than those of the equation 5a:;* — 1 2x^ + 3x^4- 4a:— 5 = 0. Ans. 5/+282/3+51?/2+32?/-l=0. Ex. 5. Find the equation whose roots are greater by 2 than those of the equation a;^ 1 Ox* + 42:cH 8 6a:2-f 70a: +12 = 0. A71S. y^ + 2y^^6y^-10y-{-8 = 0. 445. Any complete equation may he transformed into anothel wliose second term is wanting. Since h in the preceding article may be assumed of any value, we may put nh-\-A — 0^ which will cause the second term of the general development to disappear. Hence 7i— _— , and A ^^ x—y . Hence, to transform an equation into another which n wants the second term, substitute for the unknown quantity a new unknown quantity minus the coefficient of the second term divided hy the highest exponent of the unknown quantity. 318 ALGEBRA. Ex. 1. Transform the equation x^—6x'^ + 8x — 2 = into an- other whose second term is wanting. Futx=:7jJf-2. Ans. y^—4y— 2 = 0. Ex.2. Transform the equation x'^—16x^—6x-]-lb = into another whose second term is wanting. Put x=y+4:. Ans. y'-96y^-618y -777 = 0. Ex. 3. Transform the equation 0.-^ + locc* + 12a;' - 20ir2 4- 14a; - 25 =: into another whose second term is wanting. Ans. 2/^-78?/3+412?/2-7o7?/+401=0. Ex.4. Transform the equation x"^— 8x^-^6 = into another whose second term is wanting. According to Art. 438, when the second term of an equation is wanting, the sum of the positive roots is numerically equal to the sum of the negative roots. 446. I/two numbers^ substituted for tha unknown quantity in an equation, give results with contrary signs, there must be at least one real root included between those numbers. Let us denote the real roots of the general equation of the nth degree by a, b, c, etc., and suppose them arranged in the order of their magnitude, a being algebraically the smallest, that is, nearest to — a ; b the next smallest, and so on. The equation may be written under the following form, {x—a){x—b){x—c){x—d) =0. Now let us suppose x to increase from — a toward -f- a , assuming, in succession, every possible value. As long as x is less than a, every factor of the above expression will be nega- tive, and the entire product will be positive or negative accord- ing as the number of factors is even or odd. When x becomes equal to a, the whole product becomes equal to 0. But if a; be greater than a and less than 5, the factor x—a will be positive, while all the other factors will be negative. Hence, when x changes from a value less than a to a value greater than a and less than b, the sign of the whole product changes from -^ to — or from — to +. When x becomes equal to 6, the product again becomes zero; and as x increases from b to c, the factor GENERAL THEORY OF EQUATIONS. 319 jc— 5 becomes positive, and the sign of the product changes again from — to -f or from -f- to — ; and, in general, the prod- uct changes its sign as often as the value of x passes over a real root of the equation. Hence, if two numbers substituted for x in an equation give results with contrary signs, there must be some intermediate number which reduces the first member to 0, and this number is a root of the equation. If the two numbers which give results with contrary signs differ from each other only by unity, it is plain that we have found the integral 'pari of a root. If two numbers, substituted for x in an equation, give results with like signs, then between these numbers there will either be no root, or some even number of roots. The last case may include imaginary roots. For if a-hZ; V — 1 be a root of the equation, then willa— J-v/ — 1 be also a root. Now {x-a-h^^^){x-a-^l^^-i)^{x-ay■-^}P^, a result which is alwaj^s positive ; that is, the quadratic factor corresponding to a pair of imaginary roots of an equation whose coefficients are real, is always positive. Ex. 1. Find the first figure of one of the roots of the equa- tion x^-^-x^+x— 100=0. "When a: =4, the first member of the equation reduces to — 16; and when x=:b^ it reduces to -|-55. Hence there must be a root between 4 and 5 ; that is, 4 is the first figure of one of the roots. Ex. 2. Find the first figure of one of the roots of the equa- tion a;3-6x2+9a:-10=0. Ex. 3. Find the first figure of each of the roots of the equa- tion a;3— 4x2-6x+8:=0. 447. In a series of terms, two successive signs constitute a permanence when the signs are alike, and a variation when they are unlike. Thus, in the equation x^ — ^x^—bx+Q^O^ the signs of the first two terms constitute a variation, the signs of the second and third constitute a permanence, and those of the third and fourth also a variation. 820 ALGEBRA. Descartes' s Rule of Signs. 448. Every equation must have as many variations of sign as it has positive rooiSj and as many permanences of sign as it has negative roots. According to Art. 436, the first member of the general equa- tion of the nth degree may be regarded as the product of ?i bi- nomial factors of the form x—a,x—b, etc. The above theorem will then be demonstrated if we prove that the multiplication of a polynomial by a new factor, x—a^ corresponding to a pos- itive root, will introduce at least one variation^ and that the mul- tiplication by a factor, ic+a, will introduce at least one perma* nence. Suppose, for example, that the signs of the terms in the original polynomial are +H 1 1 f-, and we have to multiply the polynomial by a binomial in which the signs of the terms are -\ — . If we write down simply the signs which occur in the process and in the result, we have + + - + - + - - + - -- + + + - + - - + + + - - + - --f - ■ + + - ■ +±-±± + - + -±-1-- We perceive that the signs in the upper line of the partial products must all be the same as in the given polynomial ; but those in the lower line are all coritrary to those of the given polynomial, and advanced one term toward the right. When the corresponding terms of the two partial products have dif- ferent signs, the sign of that term in the result will depend upon the relative magnitude of the two terms, and may be either + or — . Such terms have been indicated by the double sign ± ; and it will be observed that the permanences in the given polynomial are changed into signs of ambiguity. Hence, take the ambiguous sign as you will, the permanences in the final product are not increased by the introduction of the posi- tive root 4- a, but the number of signs is increased hy oiie, and GENERAL THEORY OF EQUATIONS. 821 therefore the number of variations must be increased by one. Hence each factor corresponding to a positive root must intro- duce at least one new variation^ so that there must be as many variations as there are positive roots. lu the same manner we may prove that the multiplication by a factor, ic + a, corresponding to a negative root, must intro- duce at least one hqv^ permanence ; so that there must be as many permanences as there are negative roots. If all the roots of an equation are real^ the number of posi- tive roots is equal to the number of variations, and the number of negative roots is equal to the number of permanences. If the equation is incomplete, we must supply the place of any deficient term with ±0 before applying the preceding rule. Ex.1. The equation x^ — Zx^—bx^^lbx'^+4:X—12 = has five real roots ; how many of them are positive ? Ex.2. The equation a;* -Sa;^- 15x2+49^^-12 = has four real roots; how many of them are negative? Ex. 3. The equation a^6 _|„ 3x-5 _ 41x* - 87^3 + 400x2 + 444:r - 720 = has six real roots; how many of them are positive? Derived Polynomials. 449. If we take the general equation of the wth degree, and substitute y-\-h in place of a?, it becomes (?/4-A)«+A(?/+//)'^-i + B(7/+7z)«-24-.... +V = 0. Developing the powers of the binomial ?/ + A, and arranging in the order of the powers of A, we have yn _|_,^^n-l fB?/«-2+(72_2)B?/"-3 + T2/ +Y + T A + 72(n — 1)?/"-2 J^{n-V){n-2)k7f-^ + (n-2)(n^3)B?/^-4 .7/2 172+- The part of this development which is independent of h is of the same form as the original polynomial, and we will des O 2 822 ALGEBRA. ignate it bj X. We will denote the coefficient of h bj X^, the coefficient of — ^ bj Xg, etc. The preceding development may then be written X4-X,A+X,^+X3^+.... +/.n 450. The polynomials X^, Xg, etc., are called derived poly- nomials^ or simply derivatives. X^ is called the first derivative of X, Xg the second derivative^ and so on. X is called the prim- itive polynomial. Each derived polynomial is deduced from the preceding hy multiplying each term hy the exponent of the leading letter in that term^ and then diminishing the exponent of the lead- ing letter hy unity. Ex. 1. What are the successive derivatives of rlst. Sic^-Ux+a Ans. \ 2d. Qx-l^. { 8d. 6. Ex. 2. What are the successive derivatives of a;4_8a:34-14a?H4cc-8? Ex. 3. What are the successive derivatives of a;5+8a;*+2a:3_3x2-2a:-2? Ex. 4. What is the first derivative of ic'^+Aa^^-i + Bx^-H +Tx4-V? Equal Roots. 451. We have seen, Art. 436, that if a, 5, c, etc., are the roots of the general equation of the nih. degree, the equation may be written ^ = {x--.a){x-h){cc-c). . . . {x-lc){x-l)^0. When the equation has two roots equal to a, there will be two factors equal to x—a; that is, the first member will be di- visible \>y'{x—aY\ when there are three roots equal to «, the first member will be divisible by (cc— a)^; and if there are n roots equal to a, the first member will contain the factor (x—a)". The first derivative will contain the factor n{x^aY"^\ that is, GENERAL THEORY OF EQUATIONS. 328 x—a occurs (n — 1) times as a factor in tlie first derivative. The greatest common divisor of the primitive polynomial, and its first derivative, must therefore contain the factor x—a, re- peated once less than in the primitive polynomial. Hence, to determine whether an equation has equal roots, we have the following RULE. Find the greatest common divisor between the given polynomial and its first derivative. If there is no common divisor the equation has no equal roots. If there is a common divisor ^ place this equal to zero^ and solve the resulting equation. Ex. 1. Find the equal roots of the equation The first derivative is Sx'^—16x-\-21. The greatest common divisor between this and the given polynomial is a:;— 3. Hence the equation has two roots, each equal to 3. Ex. 2. Find the equal roots of the equation x^-lSx''-\-66x-76=0. Ans. Two roots equal to 6. Ex. 3. Find the equal roots of the equation a;3_7x2 + 16:r-12 = 0. A72S. Two roots equal to 2. Ex. 4. Find the equal roots of the equation x'-6x^—8x-S=^0. Ans. Three roots equal to — 1. Ex. 5. Find the equal roots of the equation x^-Sx^-^dx-\-27 = 0. Ex. 6. Find the equal roots of the equation x'-\-8x^ -{-20x4-16=0. Sturm^s Theorem. 452. The object of Sturm^s theorem is to determine the number of the real roots of an equation, and likewise the situation of these roots, or their initial figures when the roots are irrational According to Art. 446, if we suppose x to assume in succes- sion every possible value from — oc to + a, and determine the 824 ALGEBRA. number of times that the first member of the equation changes its sign, we shall have the number of real roots, and, conse« quently, the number of imaginary roots in the equation, since the real and imaginary roots are together equal in number to the degree of the equation. Sturm's theorem enables us easily to determine the number of such changes of sign. 453. Sturm's Functions. — Let the first member of the general equation of the Tith degree, after having been freed from its equal roots, be denoted by X, and let its first derivative be de- noted by Xj. We now apply to X and X^ the process of find- ing their greatest common divisor, with this modification, that we change the sign of each remainder before taking it as a di- visor; that is, divide X by Xj, and denote the remainder with its sign changed by E; also, divide Xj by E, and denote the remainder with its sign changed by E^, and so on to E„, which will be a numerical remainder independent of a;, since, by hy- pothesis, the equation X=0 has no equal roots. We thus obtain the series of quantities X, Xj, E, Ep Eg, .... E„, each of which is of a lower degree with respect to x than the preceding; and the last is altogether independent of a:, that is, does not contain x. We now substitute for x in the above functions any two numbers, p and q, of which p is less than q. The substitution ofp will give results either positive or negative. If we only take account of the signs of the results, we shall obtain a cer- tain number of variations and a certain number of perma- nences. The substitution o^ q for x will give a second series of signs, presenting a certain number of variations and permanences. The following, then, is the Theorem of Sturm. 454. If in the series of functions X, X^ E, E^, .... E,„ ive substitute in place of x any two numbers^ p and q, eithei' positive or negative, and note the signs of Hie residts, the difference beitvcen the number of variations of sign when x=p and when x=q is equal GENERAL THEORY OF EQUATIONS. 325 to the number of real roots of the equation X = comprised between p and q. Let Q, Qj, Q2, . . . . Qn, denote the quotients in the success- ive divisions. Now, since the dividend is equal to the product of the divisor and quotient plus the remainder, or ininus the remainder with its sign changed, we must have the following aquations : X=X,Q-R, (1.) X, = RQ, -R„ (2.) R:=R,Q,-R„ (3.) From these equations we deduce the following conclusions: 455. If in the series of functions X, Xj, R, etc., any number be substituted for x, two consecutive functions can not reduce to zero at the same time. For, if possible, suppose X^^O and R=:0; then, by Eq. (2), we shall have R^^O. x\lso, since R=:0 and R^ = 0, by Eq. (3) we must have R^^O; and from the next equation R3 — 0, and so on to the last equation, which will give R„=0, which is im- possible, since it was shown that this final remainder is inde- pendent of cc, and must therefore remain unchanged for every value of cc. 456. When^ by the substitution of any number for cr, any one of these functions becomes zero^ the tivo adjacent functions 'must have contrary signs for the same value of x. For, suppose R^ in Eq. (3) becomes equal to zero, then this equation will reduce to R=: — Rg; that is, R and Rg have con- trary signs. 457. If a is a root of the equation X = 0, the signs ofK and X, will constitute a variation for a value of x which is a little less than a, and a permanence for a value of x which is a little greater than a. Let h denote a positive quantity as small as we please, and let us substitute a-\-h for x in the equation X=:0. According 326 ALGEBRA. to Art. 449, the development will be of the form X+Xj/i-|-X2 7r+ other terms involving higher powers of A. Now, if a is a root of the proposed equation, it must reduce the polynomial X to zero, and the development becomes X^/i-l-Xg-^-f-other terms involving higher powers of /i, or A(X, + X,|+, etc.). Also, if we substitute a 4-^ for x in the first derived polyno- mial, the development will be of the form Xj + Xg/i + other terms involving higher powers of ^. Now a value may be assigned to h so small that the first term of each of these developments shall be greater than the sum of all the subsequent terms. For if A be made indefinitely small, then will Xg/t be indefinitely small in comparison with Xj, which is finite; and, since the following terms contain higher powers of h than the first, each will be indefinitely small in comparison with the preceding term ; and, since the number of terms is finite, the first term must be greater than the sum of the subsequent terms. Hence, when li is taken indefinitely small, the sum of the terms of the two developments must have the same sign as their first terms, Xj/i and Xj. "When h is positive, these terms must both have the same sign; and when li is negative, they must have contrary signs; that is, the signs of the two functions X and X^ constitute a variation when x = a—h^ and a permanence when x=a-\-h. 458. Demonstration of Sturm'' s Tlieorem. — Suppose all the real roots of the equations X=0, Xi=:0, R=0, R,=0, etc., to be arranged in a series in the order of magnitude, beginning with the least Let p be less than the least of these roots, and let it increase continually until it becomes equal to q^ which we suppose to be greater than the greatest of these roots. Now, so long as p is less than any of the roots, no change of sign will GENERAL THEORY OF EQUATIONS. 827 occur from the substitution of p for x in any of these functions, Art. 446. But suppose j9 to pass from a number a little small- er to a number a little greater than a root of the equation X=0, the sign of X will be changed from + to — or from — to +, Art. 446. The signs of X and X^ constitute a variation before the change, and a permanence after the change. Art. 457; that is, there is a variation lost or changed into a permanence. Again, while ^ increases from a number a little smaller to a number a little greater than another root of the equation X=0, a second variation will be changed into a permanence, and so on for the other roots of the given equation. But when jp arrives at a root of any of the other functions Xj, E, Rj, its substitution for x reduces that polynomial to zero, and neither the preceding nor succeeding functions can vanish for the same value of cc, Art. 455 ; and these two adja- cent functions have contrary signs, Art. 456. Hence the en- tire number of variations of sign is not aff* cted by the vanish- ing of any function intermediate between X and R^, for the three adjacent functions must reduce to +0— or — + . Here is one variation, and there will also be one variation if we supply the place of the with either + or — ; thus, 4-=fc— or — ±+. Thus we have proved that during all the changes of ^, Sturm's functions never lose a variation except when 'p passes through a root of the equation X = 0, and they never gain a variation. Hence the number of variations lost while x in- creases from ^ to g is equal to the number of the roots of the equation X==0, which lie between -p and q^. Now, since all the real roots must be comprised within the limits — a and + a , if we substitute these values for x in the series of functions X, X^, etc., the number of variations lost will indicate the whole number of real roots. A third suppo- sition that x — ^ will show how many of these roots are posi- tive and how many negative ; and if we wish to determine smaller limits of the roots, we must try other numbers. It is generally best in the first instance to make trial of such num- bers as are most convenient in computation, as 1, 2, 10, etc. 328 ALGEBRA. EXAMPLES. 1. Find the number and situation of the real roots of the equation x^ — 3cc^— 4x+18 = Here we have ^=x^—Sx'^—4:X+lS, and X^^Sx^— Ga?— 4. Dividing X bj Xj, we find for a remainder — 14x+35. Re- jecting the factor 7, and changing the sign of the result, we have R=2x— 5. Multiplying X^ by 4, and dividing by R, we find for a remainder — 1. Changing the sign, we have R^ = -|- 1. Hence we have X=x^-Sx''-4:X+lS, X, = Sx^-Qx-4:, R=2x-5, R,= + l. If we substitute — oc for cc in the polynomial X, the sign of the result is — ; if we substitute — a for cc in the polynomial Xj, the sign of the result is + ; if we substitute — oc for cc in the expression 2a:— 5, the sign of the result is — ; and R^, being independent of cc, will remain + for every value of ic, so that, by supposing x=— oc, we obtain the series of signs - + - +. Proceeding in the same manner for other assumed values of a?, we shall obtain the following results : ,s8iinied Values of ar. Resulting Signs. Variations. — OC - + - + givir ig 3 variations. -3 — + - + (( 3 -2 • + + — 4- (( 2 H- + ti 2 1 + + n 2 2 + + tl 2 2i + il 1 3 + + + + (( + 0C + + + + (( We perceive that no change of sign in either function occurs by the substitution for x of any number less than —8 ; but, in passing from —3 to —2, the function X changes its sign from — to +, by which one variation is lost. In passing from 2 GENEKAL THEORY OF EQUATIONS. 329 to 2^, the function X again changes its sign, and a second va- riation of sign is lost. Also, in passing from 2-J to 3, the func- tion X again changes its sign, and a third variation is lost; and there are no further changes of sign arising from the substitu- tion of any number between 3 and + a. Hence the given equation has 3 real roots ; one situated be- tween —2 and —3, one between 2 and 2^, and a third between 2-J and 3. The initial figures of the roots are therefore —2, + 2, and +2. There are three clmnges of sign of the primitive function, two of the first derived function, and one of the second derived function ; but no variation is lost by the change of sign of either of the derived functions; while every change of sign of the primitive function occasions a loss of one variation. 2. Find the number and situation of the real roots of the equation x^—6x^^8x—l = 0. Here we have X=x^-5x^+8x-l, X^ = Sx^-10x-\-8, E=2x-31, Ri=-2295. When x= — oc, the signs are — | , giving 2 variations, x=+ Ex.4. < .-5=61^ > Ex.5. ] 2-f^ 1 1(2. 6,+l)} 1 a:=42/ ) 861 ( llx—5y_Sx-\-y ) Ex.6. \ 22 ~~32"i (8x-6y=l ) Ex.7. Ex.8. Ex.9. Ex. 10. L8a"^66~ cc 3/ 1 1 a + 6 a— Z) a4-6 a; _^ y _ 1 ^a-{-b a—b a—b Soc—6y2x+y "~2~~+^-~5~ x—2y_x y ■"~F~~2"^8 =4 4aj ^ r7+8x 8cc-6?/ 10 2ic-8 " 5 1%+9 oi_L^.V+'^ 8.7/4-5^ L^^^*+"2 Tf^] Q A ns. y= 6. ^^^^- {.^12.' Ans. -j CC=:65. y::=66. X = 4:. Ans. { ^ Ans. Ans. Ans. x=7. y=lh x=Sa. iXz='6Q (2/=- 2k x= y= Ans. a a—b b a+b' x=12. 1^= 6. Ans. \ ' 362 ALGEBRA. Ex. 11. a h X z - + -: a c c {6x—6i/+4:z = 15\ Ex.12. hx-{-4:y-Sz=19l (2ic+ 2/+62=46) fx=.21-4y ] Ex.13, i 2 = 9-^ j^ U=64-7i.J ( 37+ ?/+ 2 = 5 Ex.14, -j Sx- 6y-\-7z=:75 (9;r-llz + 10=0 Ex.15. }7x+2i/-Sz=2 i (4:cH-3?/- 2 = 7 ) [ x—2y+Sz=6 \ Ex.16. hx + S2j-4:Z:=^20[ isx-2y + 6z=2Q) Ex. 17. ^ r 7x-3?/=i- 42-7?/ = 1 Ex. 18. Il2-7?^ = 1 [l9x-3w=l^ (Su-2y=2 ] 2x + 3?/=39 ! 5ir-7z=ll j 1 42/ +32 =41 J r2x-3y + 22=13 Fx IQ J 43/+22=14 E^-19-i4r.-2c.=30 [5y + 3u=32 (x=2a, Ans. to nnd the values of x and y. Ex.5. Given 2^ + 3?/=37, ^ns. a;=5 or 4j^^, 3/=: 9 or -y^^. a7+y =6, ) Arts. x=4: or 2, y=2 or 4. Ex.7. Given cc2 4- 2/2 = 10000, ) ^ ^ ^ .u i f a .^ , > to nnd the values of x and it X +y =z 124, ) ^ ^715. £c=96 or 28, 3/=28 or 96. Ex. 8. Given 12 : x : : 3/ : 3, ) ^ ^ , , , ^ , ^ j: M. to find the values of a: and ir. Va2 4-V?/=:5, ) J.775. a;=9 or 4, ?/=4 or 9. Ex.9. Given (3xH- 4?/) (7a: -2?/) + 3^+4?/ =^44, ) to find the (3x+4?/)(7x— 2?/)— 7aj+22/=30, ) values of x and y. Ans. cc=l or 1^^, y—2 ox — iV- Ex.10. Given —x'^-\-^xy—^y'^.-\-4^x~l2y= 4,) to find the a;2_2xy f 3?/2— 4x+5?/ =53, f values of x and y. Ans. ic = ll or — 7-|-, 2/=3 or — 3^. Ex. 11. Given 2 (a::^ + y"^) {x-\-y) = 15xy, ) to find the values 4:{x*—y'^)(x'^—y^)=4:6x'^y'^j j of a: and?/. Ans. x=2 or 1, y=l or 2. Ex.12. Given {x^—y^){x-^y) = 16a7y, | to find the values {x*-y*){x^—y^) = UOxYy ] of x and y. Ans. x=9 or 8, y=S or 9. Ex. 13. Given J? (cc+ 2/+ 2)= 27, ) ^ , , ^(^+y4.,)=:18, i to find the values of x, y, .(a: + ^ + .) = 36,) ^"^ ^- J.725. a: = 3, y=:2, z=4. Ex.14, Given a:y=z, "^ > to find the values of a:, ?/, z, and v. 2/2=v, yv = bXj^ Ans. x=—=, y=Vi, z=Va, v=ViiVi- yb EXAMPLES FOR PRACTICE. 375 Ex.15. Given x?/2 = 105,^ xzv=189, yzv=S16j to find the values of x, y, ;?, and V. A71S. x=S, y=6j z=7j v=9. to find the values of x and y. Ex.16. Given x2+-^+2/2=84, ^ +-+y =14, Ans. x—4:, y=2 or 8. Ex. 17. Given Vy— Va—x = Vy—x, 1 to find the val- 2Vy^-{-2V^^^=^6Va—x,j uesofxandy. Ans. x—^a, y — ^ Ex. 18. Given x^ ■\-x%P'=ay. ) ^ ^ -. ,i ^ n ^ -X 7 M- to find the values of ic and y. X y-\- y —ox^ ) Ans. cc = V Ti V = V r- Ex.19. (jiiYQn^j6Vx + 6Vy-\-Vx^-Vy= 10, f to find the \/^+ vy =275, ) values of x and y. Remark. Put z^=^x-\-Vy. Then, from Eq. 1, z = -y/l>', that is, -v/x + V^=^5. Next put '^x—^-\-v, and y^ — i~^- Substituting these values in Eq. 2, we find v^=\^ or v=±.^. . ^ ^ _13±5V;^ Ans. 0? — 9 or 4, or ^^ ~. y=4 or 9, or 13zp5V-5J Several of the following examples have imaginary or incom- mensurable roots "which are not here given. Ex.20. Given {x^^y-')xy = n090, ) to find the values of a! x-\-y — 18, f and y. Ans. x=7 or 11, ?/=ll or 7. Ex, 21. Given 5{x^+y^) + 4:xy =:S66, ) to find the values of x'^+y^-i-x-{ry=^ 62,) ic and y. Ans. x=4: or f>, ?/— 6 or 4. 376 ALGEBRA. Ex.22. Given {x'-^y^)xy=SOO,) to find the values of a? £c*+y* =337, ) and y. Ans. x= ±4, y= ±3. Ex. 23. Given {x^-{-y^){x^+y^)=^65, ) to find the values of x-\-y = 5, f X and y. Ans. x=:S or 2, y=2 or 3. Ex.24. Given ^!±^^±^^14 32 + 2/ = 18, to find the values of* x and y. x—y Ans. a;=:12, 2/ = 6. Ex.25. Given (a;2-a:?/+2/'0(^H2/^) = 91, ) to find the {x'^—xy-\-y'^){x'^-\-xy-{-y'^) = lZ2>, ) values of x and ?/. J.72S. x — ±^ or ±2, 3/= ±2 or ±3. Ex. 26. Given {x-\-y)xy = 30, ) to find the values of x (cc2-f 2/2)cc22/2=468, ) and y. Ans. x=2 or 3, y — S or 2. Ex.27. Given £c— 2/4- V^^^:——-, to find the values A71S. x= ±5, y= ±4. Ex.28. Given {x-^yY-{-x-{-y=SO, ) to find the values of ic x—y= 1, f and y. Remark. Multiply Eq. 1 by x-\-y, and we have (x+y)*+(x-ft/)» = 30(x+y). Add to each member 9(a:+y)=4-25, and the square root of each member of the equation may be extracted. Ans. x=2, y—1. Ex.29. Given (x -\-y){xy +1)= 18x?/ ) to find the val (x2-t-2/2)(x2?/2-f l)=208a;2?/2 ) ues of x and y. Remark. Divide Eq. 1 by xy, and Eq. 2 by x^y^ and we have ^ +y +\ +\ = '«• Put x-\--=z. and vH — =v. X y Then 2+v = 18, and 2» + u»=212. Whence 2=14 or 4, and w=4 or 14 ; and hence x and y are easily found. ^725. a; = 2±'/3, y=7±4\/3. EXAMPLES FOR PRACTICE. 377 PROBLEMS INVOLVING EQUATIONS OF THE SECOND DEGREE WITH SEVERAL UNKNOWN QUANTITIES. Prob. 1. If I increase the numerator of a certain fraction by 2, and diminish the denominator by 2, 1 obtain the reciprocal of the first fraction ; also, if I diminish the numerator by 2, and increase the denominator by 2, the resulting fraction, increased by 1^^, is equal to the reciprocal of the first fraction. What is the fraction ? Ans. ^. Prob. 2. It is required to divide the number 102 into three parts, such that the product of the first and third shall be equal to 102 times the second part, and the third part shall be IJ times the first. Ans. The first part is 34, the second 17, and the third 51. Prob. 3. A certain number consists of two digits. If I in- vert the digits, and multiply this new number by the first, I ob- tain for a product 5092 ; but if I divide the first digit by the second, I obtain 1 for a quotient with 1 for a remainder. What is the number? Ans. 76. Prob. 4. The fore wheel of a carriage makes 165 more rev- olutions than the hind wheel in going 5775 feet ; but if the circumference of each wheel be increased 2J feet, the fore wheel will make only 112 revolutions more than the hind wheel in the same space. Kequired the circumference of each wheel. Ans. The fore wheel 10 feet, the hind wheel 14 feet. Prob. 5. A piece of cloth, by being wet in water, shrinks one eighth in its length and one sixteenth in its breadth. If the perimeter of the piece is diminished 4J feet, and the surface 5J square feet, by wetting, what were the length and breadth of the piece? Ans. 16 feet long and 2 feet wide. Prob. 6. A certain number of laborers in 8 hours transport a pile of stones from one place to another. If there were 8 more laborers, and if each carried each time 5 pounds less, the pile would be removed in 7 hours; but if there were 8 less labor- ers, and if each carried each time 11 pounds more, it would re- 378 ALGEBRA. quire 9 hours to remove the pile. IIow many laborers were there employed, and how many pounds did each carry ? Arts. 28 laborers, and each carried 45 pounds ; or 36 laborers, and each carried 77 pounds. Prob. 7. A certain capital yields yearly $123J interest; a second capital, $700 larger, and loaned at i per cent, less, yields yearly $lli more interest than the first. How large was the first capital, and at what per cent, was it loaned? Ans. The capital was $3800, loaned at 3J per cent Prob. 8. A person bought a number of $20 railway shares when they were at a certain rate per cent, discount for $1500 ; and afterward, when they were at the same rate per cent, pre- mium, sold them all but 60 for $1000. How many did he buy, and what did he give for each of them ? Ans. 100 shares at $15 each. Prob. 9. A rectangular lot is 119 feet long and 19 feet broad. How much must be added to the breadth, and how much taken from the length, in order that the perimeter may be increased by 24 feet, and the contents of the lot remain the same? Ans. The length must be diminished 102 feet, and the breadth increased 114 feet. Prob. 10. There are two numbers such that their sum and product together amount to 47 ; also, the sum of their squares exceeds the sum of the numbers themselves by 62. What are the numbers? Ans. 5 and 7. Prob. 11. The sum of two numbers is a, and the sum of their reciprocals is h. Ecquired the numbers. Ans. 2 V 4 /;• Prob. 12. A and B engage to reap a field for $24 ; and as A alone could reap it in nine days, they promise to complete it in five days. They found, however, that they were obliged to call in C to assist them for the last two days, in consequence of which B received one dollar less than he otherwise would have done. In what time could B or C alone reap the field? Ans. B in 15 and C in 18 days. EXAMPLES FOR PRACTICE. 879 Prob. 13. The sum of the cubes ot two numbers is 85, and the sum of their ninth powers is 20,195. Kequired the numbera Ans. 2 and 3. Prob. 14. There are two numbers whose product is 300 ; and the difference of their cubes is thirty-seven times the cube oi their difference. What are the numbers? A71S. 20 and 15. Prob. 15. A merchant had $26,000, which he divided into two parts, and placed them at interest in such a manner that the in- comes from them were equal. If he had put out the first por- tion at the same rate as the second, he would have drawn for this part $720 interest ; and if he had placed the second out at the same rate as the first, he would have drawn for it $980 in- terest. What were the two rates of interest? Ans. 6 per cent, for the larger sum, and 7 for the smaller. Prob. 16. A miner bought two cubical masses of ore for $820. Each of them cost as many dollars per cubic foot as there were feet in a side of the other ; and the base of the greater contain- ed a square yard more than the base of the less. What was the price of each ? Aiis. 500 and 320 dollars. Prob. 17. A gentleman bought a rectangular lot of land at the rate of ten dollars for every foot in the perimeter. If the same quantity had been in a square form, and he had bought it at the same rate, it would have cost him $330 less; but if he had bought a square piece of the same perimeter, he w^ould have had 12J rods more. What were the dimensions of the lot ? Ans. 9 by 16 rods. Prob. 18. A and B put out at interest sums amounting to $2400. A's rate of interest was one per cent, more than B's; his yearly interest was five sixths of B's; and at the end often years his principal and simple interest amounted to five sev- enths of B's. What sum was put at interest by each, and at what rate? Ans. A $960 at 5 per cent, B $1440 at 4 per cent. Prob. 19. A person bought a quantity of cloth of two sorts for $63. For every yard of the best piece he gave as many dollars as he had yards in all ; and for every yard of the poor- 380 ALGEBRA. er, as many dollars as there were yards of the better piece more thau of the poorer. Also, the whole cost of the best piece was six times that of the poorer. How many yards had he of each ? Ans. 6 yards of the better and 3 of the poorer. Prob. 20. A commences a piece of work alone, and labors for two thirds of the time that B would have required to perr form the entire work. B then completes the job. Had both labored together, it would have been completed two days soon- er, and A would have performed only half what he left for B. Eequired the time in which they would have performed the work separately. Remark. Suppose A would have performed the work in z days, and B in y days. A labors -^ days, and performs -^ part of the work. o ox B performs 1 —^= — jr— ^ part of the work. ox OX Sx-2y . „ , , - — - — ^XM=time B labored. 6x 3xu-2y^- 2v , , . —^ — ^+-;^=wnole time consumed. ox O -+-=part both did in one day= -. X y ^ ' xy — 7^= the days of work if both labored together, x+y x+y 3x .3' Also, - X -±-=T, of —- 5-^- X x-\ry 2 3x Ans, A in 6 days and B in 3 days. EXAMPLES FOB PEACTICE. 381 PEOGRESSIONS. Ex. 1. What is the sum of the natural series of numbers 1,2, 8, etc., up to 1000? Ans. 600,500. Ex. 2. What is the sum of an arithmetical progression whose first term is 6, the last term 2833, and the number of terms 38? Ans. 53,941. Ex. 3. What is the first term and the sum of the terms of an arithmetical progression, when the last term is 24, the common difference -f, and the number of terms 22 ? A71S. First term 9, and sum of terms 368. Ex. 4. Kequired the number and the sum of the terms of an arithmetical progression, when the first term is — f , the com- mon difference — |, and the last term —2 If. Ans. Number of terms 25, and sum of terms —281 J. Ex. 5. The first term of an arithmetical progression is 5, the last term 23, and the sum of the terms 392. What is the com- mon difference and the number of terms ? Ans. Common difference |, and number of terms 28. Ex. 6. Between 7 and 13 it is required to interpolate 8. terras which shall form an arithmetical progression. Ans. 7f, 8i 9, 9|, 10^ 11, llf, 12^. Ex. 7. In an arithmetical progression, the sum of the 19th, the 43d, and the 57th terms is 827; the sum of the 27th, the 58th, the 69th, and the 73d terms is 1581. What is the first terra and the common difference ? Ans. The first term is 5, and the common difference 7 Ex. 8. In boring an artesian well 500 feet deep, $3.24 is paid for the first foot, and 5 cents more for each subsequent foot. How much was paid for the last foot, and how much for the whole well ? Ans. For the last foot $28.19, and for the entire well $7857J. Ex. 9. According to natural philosophy, a body falling in a vacuum describes in the first second of its fall 16tV feet, and 382 ALGEBRA. in each succeeding second 32 J- feet more than in the second immediately preceding. If a body has fallen 20 seconds, how- many feet will it fall in the last second, and how many in the whole time? Ans. 627J feet in the last second, and 6433J- feet in the whole time. Ex. 10. Divide unity into four parts in arithmetical progres- sion, of which the sum of the cubes shall be iV- Arxt 12 3 4 Ex. 11. A ship, with a crew of 175 men, set sail with a sup- ply of water sufficient to last to the end of the voyage ; but in 30 days the scurvy made its appearance, and carried off three men every' day ; and at the same time a storm arose which protracted the voyage three weeks. They were, however,, just enabled to arrive in port without any diminution in each man's daily allowance of water. Eequired the time of the passage, and the number of men alive when the vessel reached the harbor. Ans. The voyage lasted 79 days, and the number of men alive was 28. Remark. Put a:=:days the voyage was expected to last. a; + 21=days the voyage lasted. a:-l-21— 30=x— 9=the days after 30. On the 31st day the number of men was 172, etc. Last term =172-3(a:-10). X— 9 Sum of 8eries=(344-3(x-10))x-^. Then (344-3x-i-30)^ = 175(a:-30). Whence a:=58. Also, x-|-21=79 days the voyage lasted. Ex. 12. The number of deaths in a besieged garrison amount- ed to 6 daily ; and, allowing for this diminution, their stock of provisions was sufficient to last 8 days. But on the evening of the sixth day 100 men were killed in a sally, and afterward the mortality increased to 10 daily. Supposing the stock of provisions unconsumed at the end of the sixth day to support 6 men for 61 days, it is required to find how long it would sup- EXAMPLES FOR PKACTICE. 383 port the garrison, and the number of men alive when the pro- visions were exhausted. Ans. The provisions last 6 days, and 26 men survive. Remark. Put a:— number of men at first. a:— 42= number expected at end of 8 days. 2^-42 2 2a: -30 8=8j: — 168 = number of days' provisions. X6=6a:— 90=days' provisions exhausted at end of 6th day. 2 2x— 78 =366 =the remainder. Whence a:=222. 222 — 136 = 86= number of men after the sally. Put y= number of days the pi'ovisions lasted afterward. 172-10(^-1) 2 ^ Ex. 13. The first term of a geometrical progression is 1, the ratio 2, and the number of terms 13. What is the last term and the sum of the terms ? Ans. The last term is 4096, and the sum of the terms 8191. Ex. 14. The first term of a geometrical progression is 7, the ratio 3, and the number of terms 11. What is the last term and the sum of the terms ? Ans. The last term is 413,343, and the sum of the terms 620,011. Ex. 15. The sum of the terms of a geometrical progression is 411,771, the ratio 7, and the number of terms 7. Eequired the first and last terms. Ans. First term 3, and last terra 352,947. Ex. 16. Between 1 and |- it is required to interpolate 11 terms forming a continued geometrical progression. What are the terms? Ans. 0.9439, 0.8909, 0.8409, 0.7937, 0.7492, 0.7071, 0.6674, 0.6300, 0.5946, 0.5612, 0.5297. Ex. 17. What will $1200 amount to in 36 years at 4 per cent, compound interest? Ans. $4924.70. Ex. 18. A farmer sowed a peck of wheat, and used the whole crop for seed the following year; the produce of the second year he used for seed the third year, and so on. If in the 10th 884 ALGEBRA. year the crop was 1,048,576 pecks, by how many times must the seed have increased each harvest, supposing the increase to have been always the same? Ans. Four times. Ex. 19. There are three numbers in geometrical progression, the difference of whose differences is six, and their sum is forty- two. Kequired the numbers. Ans. 6, 12, and 24. Ex. 20. There are three numbers in geometrical progression^ the greatest of which exceeds the least by 24 ; and the differ- ence of the squares of the greatest and the least is to the sum of the squares of all the three numbers as 5 : 7. What are the numbers? Ans. 8, 16, and 32. Ex. 21. There are three numbers in geometrical progression whose continued product is 216, and the sum of their cubes is 1971. Kequired the numbers. Ans. 3, 6, and 12. Ex. 22. There are four numbers in geometrical progression whose sum is 350 ; and the difference between the extremes is to the difference of the means as 37 : 12. What are the num- bers? ^n5. 54, 72,96, 128. Ex. 23. There are four numbers, the first three of which are in geometrical progression, and the last three in arithmetical progression ; the sum of the first and last is 14, and that of the second and third 12 ; find the numbers. Ans. 2, 4, 8, 12 ; or — , -|, -, ^. Ex. 24. Three numbers whose sum is 15 arc in arithmetical progression ; if 1, 4, and 19 be added to them respectively, they are in geometrical progression. Determine the numbers. Ans. 2, 5, 8. THE END. ^4- iJ^)( ^€^C^\ ai^y-- 14 DAY USE RETURN TO DESK FROM WHICH BORROWED This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. 7 DAY USE DURING SUMMER SESSIONS Cu^w. IL K< V rvj 1 1 1 General Library LD 21-50m-8,'r)7 University of California (.C8481sl0)476 Berkeley I - '-' -t 6 -7 '•