# Full text of "Elementary Algebra: First[-second] Year Course"

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About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at |http: //books .google .com/I ELEMENTARY ALGEBRA THE MACMILLAN COMPANY NBW YORK • BOSTON • CHICAGO • DALLAS ATLANTA • SAN FRANCISCO MACMILLAN & CO., Limitbd LONDON • BOMBAY • CALCITITA MBLBOURNB THE MACMILLAN CO. OF CANADA, Ltd. TORONTO ^ ■J JOHN WALUS (1616-1703) Professor at the UniverBlQr of Oxfoi^ In 1686 he published it Treatise of Algebra which was the means of making the science of algebm more widely known in England. ELEMENTARY ALGEBRA ^econD i^ear Course BY FLORIAN CAJORI OOLORADO COLLJBOE AND LETITIA R. ODELL NOBTH 8IDB HIGH 80HOOL, DBNYBB THE MACMILLAN COMPANY 1916 • •• -•, • •••-•• • • ' • • •••••••• GOPTBIOHT, 1916, bt the magmillan company. Set up and electrotjrped. Published July, 19x6. CAJORI J. 8. Cashing Co. — Berwick A Smith Co. Norwood, Mass., U.S.A. PREFACE This book contains a brief review of the fundamental opera- tions of algebra followed by a thorough presentation of the topics usually included in the work of the third half year. The material is so arranged that the choice of topics for review or advance study may be easily made. The book contains not only a large number of practical problems but also practical applications of graphs. Material for drill in the manipulation of exponents and radicals will be found in Chapter VII. Great pains have been taken to make the subject of logarithms accessible to beginners. The most difficult part in learning to compute with logarithms is the process of " interpolation." Of the various arrangements of logarithmic tables which have been suggested, we have selected the one which renders interpolation easiest. The chapter on logarithms is introduced earlier than usual, on account of the great practical importance of the subject. If desired, it can be taken up later, after Chapter VI. The concept of disfunction does not receive isolated and abstract treatment; it is pre- sented as a fundamental idea in proportion, variation, and graphics. Its connection with problems of everyday life is firmly established. The aims of the First Year Book have been kept in view in the preparation of this text. Emphasis is thrown upon clear- ness of exposition and the use of expressions which recall the axiomatic processes involved. Continued stress has been laid upon oral exercises. 911236 vi PREFACE Several of the practical problems given in the text were suggested by the perusal of an English book, T. Percy Nunn's Exercises in Algebra (including Trigonometry), Part 1, 1913. The authors have received help from several teachers. Spe- cial mention should be made of Mr. E. A. Cummings of the North Side High School in Denver, who has worked most of the exercises and offered valuable suggestions. FLORIAN CAJORI. LETITIA R. ODELL. TABLE OF CONTENTS OHAPraB PAGB L Elembntabt Definitions and Operations .... 1 Fundamental Definitions 1 Addition and Subtraction 2 Order of Fundamental Operations 3 Multiplication and Division 6 Parentheses 7 Review Exercises 9 Equations 11 Simultaneous Linear Equations 13 Positive Integral Exponents .20 Fractional and Negative Exponents 21 Powers and Boots 25 Factoring 32 Fractions, their Multiplication and Division ... 37 Highest Common Factor and Least Common Multiple 43 Addition and Subtraction of Fractions .... 44 Quadratic Equations 48 Problems Involving Quadratics 52 n. More Advanced Theory and Operations .... 56 Fundamental Laws of Algebra 56 Remainder Theorem and Factor Theorem ... 59 Binomial Theorem (preliminary treatment) ... 65 Systems of Linear Equations. Determinants ... 67 Problems Involving Linear Equations .... 74 Division by Zero Impossible ...... 76 Equations with Fractions 77 Miscellaneous Practical Problems 80 vu Viii TABLE OF CONTENTS CEAVm PAGK III. Proportion, Variation, Function 86 Proportion 86 Functions 87 Different Modes of Variation 88 Variation Shown in Graphs 90 Problems in Proportion and Variation . ... 02 Graphs Exhibiting Empirical Data .... 101 IV. Logarithms 102 Logarithmic Curve 103 Fundamental Theorem 106 Finding Logarithms 108 Finding Antilogarithms Ill I*roblems 115 Exponential Equations 116 V. Quadratic Equations and their Propbrtils . . 119 Equations Quadratic in Foi^m 119 Relations between Roots and Coefficients . . . 120 Nature of the Roots 122 Graph of the Quadratic Equation ax^ -\-bz-\-c = y . 128 VI. Systems of Equations Solvable bt Quadratics . . 127 A System of Two Equations, One Linear . . 127 A System of Two Equations, Both Quadratic . . 134 Possibility of Solution by Quadratics .... 135 Problems 137 VII. Exponents, Radicals, Imaoinaries 139 Meanings of Different Kinds of Exponents . . . 139 Different Kinds of Numbrrs 140 Simplifying Radicals 141 Operations with Radicals 144 Square Root of a ±2v7) 148 Irrational Equations 149 Irrational Equations Quadratic in Form . . .152 Graphic Representation of Complex Numbers . . 154 VIII. Series and Limits 168 Arithmetical Series 168 Geometrical Series 163 Infinite Geometrical Series 170 Theory of Limits 173 TABLE OF CONTENTS ix PA6B SUPPLBMENT 177 The Highest Common Factor by the Method of Division . . 177 Solutions of Quadratic Equations 180 Mathematical Induction and Proof of the Binomial Theorem . 181 Applications of the Binomial Theorem 185 Table of Squares and Cubes, Square Roots and Cube Roots . 188 Review Exercises Selected from College Entrance Examinations 190 * ^ , J »* > ELEMENTARY ALGEBRA SECOND YEAR COURSE CHAPTER I ELSMSNTARY DEFINITIONS AND OPERATIONS FUNDAMENTAL DEFINITIONS 1. In arithmetic, numbers are commonly represented by Hindu-Arabic numerals. In algebra, numbers are represented also by letters. Any combination of numerals, letters and symbols of operation, which stands for a number, is called an algebraic expression. An algebraic expression may consist of parts which are separated by the -f- or — signs ; these parts, with the signs immediately preceding them, are called terms. Thus, in + a^ - 2 a& + 3 &3, there are three terms, + a^ — 2 a&, +3 b^. Each of the numbers which multiplied together form a product is called a factor of the product. A factor consisting of one or more letters is called a literal factor. Terms which have the same literal factors are called similar; terms which do not have the same literal factors are called dis- similar. In 4 a3& - 3 ab^ - 10 a% the terms + 4 a^b and > 10 a^b are siinilar ; the terms —Sdb^ and - 10 a^b are dissimilar. B 1 2 ' ELE^BmARY ALGEBRA . • • .• ••: .••••:: : •• : • All algebraic expression of one term is called a monomial, of two terms a binomial, of three terms a trinomial, and of several terms a polynomial. When a number is the product of two factors, either factor may be called the coefficient of the other factor. Often the word coefficient is applied only to the factor which is expressed in numerals. Thus, in 6 xyz^, the numeral 6 is called the coefBcient of xyz^. An exponent is a number placed at the right and a little above another number, called the base. When the exponent is a positive integer, it indicates how many times the base is taken as a factor. The exponent expresses the power to which the number is raised. Thus, the 4 in a* expresses the fourth power of a. Find a* when o = 2. Has o* the same value as 4 a, when a = 2 ? The absolute value of a number is its value regardless of its sign. Thus + 5 and — 5 have the same absolute value, 6. ADDITION AND SUBTRACTION 2. Addition : If similar terms have like signs, find the sum of the absolute values and prefix the common sign. If similar terms have unlike signs, find the difference of the absolute values and prefix the sign of the one which has the greater absolute value. If the terms are dissimilar, the addition is indicated in the usual way. Thus, the sum of 3 a and 5 & is 3 a + 6 &. To check an example in addition : Substitute numerals for the letters and find the sum. This sum must equal the result of substituting the numerals in the answer. ELEMENTARY DEFINITIONS AND OPERATIONS 3 3. Subtraction: Conceive the sign of the subtrahend to be changed, and then proceed as in addition. Check by substituting numerals for the letters. In what other way may subtraction be checked ? WBITTBN BXEB0ISB8 4. Find the sum of : 1. 7a-36-f 6c; 4a + 56-12c; -6a + 2b + Sc. 2. 4aj* — 3aj*y — 4a^*; 3a^ — 8a?y« — Saj*; 3aj* — 6a5«y. 3. 5a» — 6a"6" — 7c»; 2a" 4-3a»6» + 8c»; — 5a"6» — 2c". 4. 3(a - b)+4:{a + b) ; 7(a - 6)- 6(a + 6) ; - 6(a - b) -hll(a + 6). 6. a4-3a«4-9 a*-4a4-6; 7 a^ -}• 3a^ -^2 a^ + 10a + 7 -, -4a*-8a»-5. 6. From 2 m — 3 m« + 4 n take m — 3 mw — 9 n. 7. From 2a^'-2ab + Sb^ take a^ — ab — b^. 8. From 4 4- 3c -8d-9e take 7c 4- 5c -10 -2d. 9. From a* 4- 2 a*^*' — 6b^ take 2 a' — 5 a'ft*' + Sfe*'. 10. From 3(aj + y)—7(x — y) take 4(a: — y) — 6(a? 4- 3^). 11. From (a? — y)a — (« 4- y)b take (a? 4- y)a + (a — y)6. 12. What number added to 7 aj^ — 5 a?y* — 6 a?y will give 2a^2 — 3aj*y4-4ajy? ORDER OF FUNDAMENTAL OPERATIONS 5. The values of many algebraic expressions depend upon the order in which the operations of addition, subtraction, mul- tiplication, and division are performed. To avoid confusion it has been found necessary to adopt certain rules, so that an ex- pression shall always be interpreted in the same way and 4 ELEMENTARY ALGEBRA ^ represent the same value. The following rules are agreed upon: I. A succession of multiplications and divisions shall be performed in the order in which they occur. Thus 6. 8 + 4. 8 = 48 +4. 3 = 12. 3 = 36. II. A succession of additions and subtractions shall be performed in the order in which they occur. III. A succession of multiplications, divisions, additions, and subtractions shall be performed in accordance with I and II ; the multiplications and divisions being performed before any additions and subtractions. Thus, 20 + 6. 7-4-3. 6 = 20 + 42 -4- 16 = 43; 17 + 15 + 8-8. 2-*.4 = 17 4-6- 16 + 4 = 17 + 6-4 = 18. 6. If there are parentheses, apply the preceding rules to the expressions within the parentheses first ; then to the resulting expression as a whole. The forms ( ), [ ], { } go by the general name of " paren- theses" but they are designated by special names when it is necessary to distinguish one from the other. Thus, [ ] is called a " bracket," { } a " brace," but ( ) is always called a " parenthesis." The " vinculum " is also used to denote aggregation, thus a + &-*■ fit — ^« ORAL BXEB0I8B8 7. Simplify: 1. 16-2 4-14. 7. 28-5-7.2 4-5. 2. 3a?-42^4-8aj4-3y. 8. ^x-iy-^-^x + ^y. 3. 2a-7a-3a4-8a. 9. 15 • 3 -4-5 4- 10 -4. 4. 10 -5m -15 4- 20m. 10. 24 4- 5.5-10 - 7-(15-^2). 6. 5-(7_2)4-[8-i-3]. 11. 48 4-3(18-^-9.3)^9. 6. 25 -^ 5 4- 15 4- 40. 12. (12a-7a)-(13a-20a): ELEMENTARY DEFINITIONS AND OPERATIONS 5 MULTIPLICATION 8. To muUiply one term by another : Multiply the numerical coefficients ; annex the letters, giving to each letter in the product an exponent equal to the sum of its exponents in the two factors. Law of signs: Like signs give plus, unlike signs give minus. The degree of a term is the sum of the exponents of the literal factors. Thus 4 Qcy^afi is of the sixth degree. An expression is homogeneottSy if all its terms are of the same degree. In a^ + 2 a^ — 6 x^ every term is of the fourth degree ; the expression is homogeneous. To multiply a polynomial by a polynomial: Arrange the terms in both polynomials according to the Ascending or descending powers of some letter; multiply each term of the multiplicand by each term of the multiplier and add the partial products. Multiplication may be performed by detached coefficients. For example, (2ic*-7aj»-5aj*4-6aj- 3)(6 a^ - 4 a? 4- 6). 2-7-5+6-3 6-4+5 12-42-30 + 36-18 _ 8+28 + 20-24 + 12 10 - 35 - 25 +30-15 12-50+ 8 + 21-67 + 42-15 The product is 12 aj«- 50 a* + 8 aj* + 21 aj» - 67 a^ + 42 a? - 15. If any powers of x are lacking in either polynomial, the terms in question must be represented by zeros. Thus, in multiplying ac" + 2 05 — 4 by 05* + 05* + 1, we must write 1+0 + 2- 4and 1+0+1 + + 1. Check an example in multiplication by substituting numerals for the letters. 6 ELEMENTARY ALGEBRA JDIVISIOH . * ' • 9. To divide oiie term by another : Divide the numerical coefficients ; annex the letters, giving to each letter in the quotient an exponent equal to its exponent in the dividend minus its exponent in the divisor. Observe the law of signs : Like signs give plus, unlike signs give minus. To divide one polynomial by anothsr : I. Arrange the terms. II. Divide the first term of the dividend by the first term of the divisor and obtain the first term in the quotient. III. Multiply the divisor by this term in the quotient. IV. Subtract the product from the dividend. V. Treat the remainder as a new dividend and proceed as before. YI. Keep each new dividend arranged in the same order as the first dividend. Divide 14 a:* - 27 ax* + 21 aW - 32 a* + 12a8a; by 2x8 - 8 a« + 4 a«. 2a;2-3ax + 4a« 14jr* - 27 oa* + 21 a'^a + 12 a»x- 32(1* 14 re* - 21 ax» + 28 a%c2 7a:2-3ax-8a« - 60*8- 7a2a;2 + i2a»a5-32(i* - 6aofi+ 9 qagg -. 12 flgg - 16 a2a;2 -I- 24 a'te - 32 a* -16a2aja + 24a8'x-32a* By detached coefficients this division may be performed as follows : 14 _ 27 + 21 + 12 - 32 14 - 21 + 28 2-3+4 7-3-8 - 6- - 6 + 7 + 12- 9-12 -32 ^_ 16 + 24- 16 + 24 - -32 -32 The quotient is 7 a;^ - 3 ox — 8 a^. Division may be checked by substituting numerals for the letters, care being taken to avoid a zero divisor. ELEMENTARY DEFINITIONS AND OPERATIONS 7 ORAL BXBBOI8B8 iricaJ 10. 1. In what other way may examples in multiplication the and division be checked ? dend 2. If there is a remainder in a division, what must be done gns:' with it in checking? 3. If the multiplicand and the multiplier are arranged according to the descending powers of some letter, how will the product be arranged ? 11 of 4. If the multiplicand and the multiplier are homogeneous, what will be true of the product ? Of what degree will it be ? I eed as 2 I • I I I WRITTBN BXBRCISB8 11. Expand and check : 1. (7a-h2a2-44-2a»)(3a-3 + 4a«). 2. (aj*« — 2 aj«3/* -f y»)(af — ^). Divide and check : 3. a^-Sa^b + Sab^-b^hya-b. 4. ixfi — ]^ hy X — y, and by a? 4- y. 6. 12 aj-+< + 20 af+» - 107 ic»+2 ^ 37 0^+1 -f 33 af by 6 aj» - 11 a^. 6. m* -f w' 4- i>' — 3 mnp hy m-^n+p. 7. a»-&' + c» + 3a6cby a^+V + c^-f aft-ac + ^w. 8. 2,*-|2^-h||2/»-||2/«H.5y_^byy-|. PARENTHESES 12. I. A parenthesis preceded by a -j- sign may be removed without changing the signs of the terms within the parenthesis. II. A parenthesis preceded by a — sign may be removed, le provided that the sign of each term within the parenthesis be changed. L 8 ELEMENTARY ALGEBRA WRITTEN BXEBOI8BS 13. Simplify: 1. 18 -(7-2) -3. 2. 5a4-(6a-4a)-(9a-2a). 3. 8 a: - [2 aj 4- (3 aj - y) -f 7] + [5 « - 2 y + 3]. 4. -7m4-[2n- j4 m- (3 m-5n)- 8 ii{ -f 9m]. 6. 10c- {4d-(9c-5d)| - {(2c -d)- (4c-f 7d)}. 6. (aj-f 2)(aj-3)-5(a^-3« + 2)-10(aj + 3). 7. {5a«-3[-5+(a-f 2)(a-5)-7]| -a(a-l)\ 9. (a + 6)(a - 6) - (a - 6)2 4- (a 4- 2^)'. 10. 2{ -4-(7 -3) -f (-8- 2) -5(9-3 + 5)1 4-11. 11. m-{-[— (1— m)— l]-m|— {m— (5-4m)-(44-m)}. 14. Terms may be inclosed in a parenthesis by reversing the rules for removing parentheses. When terms are inclosed in a parenthesis preceded by a 4- sign, the signs of the terms are not changed. When terms are inclosed in a parenthesis preceded by a — sigriy tjiersigns of all the terms are changed, WRITTEN EXERCISES 15. Collect the coefficients of a? in a parenthesis preceded by a 4- sign, and the coefficients of ^ in a parenthesis pre- ceded by a — sign : 1. a>x — ay '\- bx — by, 4. 2x ^3y —- dx -{■ dy — fx. 2. 7nx + na; 4- f^/y — ^y. 5- py — 9^ —px-{- qy. 3. abx + cdy — cdx— aby, 6.-7 aaj+3 6y 4-4 bx— 5 ay. ■ ELEMENTARY DEFINITIONS AND OPERATIONS 9 REVIEW EXERCISES 16. liA = 2a^-^ab+7b', 5 = - 5a*+ 2a6-46«, C == S a" + 10 ab-- 6 b% D = 4a«-9 6« + 3a5, find the values of : 1. ^ + 5 + C4-A 4. B- A + G-D. 2. A-B-^O-D. 6, G-B + A-D. 3. ^A+B-C + D, 6. D-^A-G-B. If a = 1, 6 = 3, c = — 2, d = 4, n = 2, find the numerical values ' of : 2a-3c + 4d^ 3a^-2a--26-. 7c- 2d 8. ?^ + ^-. 13. (a+6)«-(c-d)». Q a^ + &^ 14. (c + d)(c-cr>+5^- • c2-hcP' ^"^ , ^ a" — 6* X6. 2 d -^ 3 c • dn — &c ^ d. 10. — • 11. 5a-^(6-c)H-d-n. 16. Vfe^+VS^. Find the sum of : 17. (a 4- b)x - (a - 6)2/ and (a - 6)aj + (a + ^)y. 18. |a-|6 + ic; ia + i6-^c;andfa + i6-Hc 19. |aj-fy + i2;; |a? + |y-i2;; and|aj-|y + |2. 20. [(a + c)aj 4- (^ + c)y] - [(a - c)a; - (& - c)y]. 21. [5(.'r4-2/)-7(aj-y)]-[-2(aj + y)-3(«-y)]. Simplify : 22. aj2-(?/2-2^)-[y^-(»''-«')] + [25'- (y^-aj«)]. 23. aj-22;-f3y-M?-[3a?-(5«-y-7w?)]-5« + 4a?{ , 24. (iB« + 2iC-^- 3af-2-l)(aj-l). 10 ELEMENTARY ALGEBRA 26. (a"+* — 4 a« + 5 a-"* + a'^^a + 1). 26. (a"4-&")(a"— 6»). 27. (|a'-|a6-hi6»)(|a-i6). 28. (a«6'»+^ — 6 a'6"* + 4 a'»6"»-i) -«- 2 a-ft*"*"*. 29. (aj»2r — af +y+^ 4- aj*--y*-*) -5- af -y-*. 30. (a** 4- a** + 1) -5- (a** - a* -f 1). 31. (9 a*» - 26 6^) -s- (3 a« + 6 fe'-j 32. (iaj»-|xV + M^-ty')-^(iaJ-iy). TYPE FORMS IN MULTIPLICATION 17. 1. (a ± 6)« = a* ± 2 a6 + &». Take the upper signs together, and the lower signs together. 2. (a-f 6--c)«=a*4-&*4-c2 + 2a6-2ac-26c. 3. (a + 6)(a — 6) = a2 - 6». 4. (aj + a)(aj 4- 6) = a.** 4- (a + ^)aj 4- a5. 6. (ax 4- 6)(ca5 4- cf) = ocaj* 4- {ad -\- bc)x + hd. 6. (a± 6)« = a«±3a*64-3a6»±6«. ORAL BXBROISB8 18. Write, by inspection, the va4ue8 of : 1. (a4-2)«. 6. (a?-^h)\ 9. (a»-6»)«. 2. {b-Sy. 6. (c'-d')*. 10. (aj»4-6')*. 3. (2 a: -1)2. 7. (c2 4-7(P)«. 11. (m»-2n2)«. 4 (3 a -4)*. 8. (5«-3y2)«. 12. (8a-6a?«)2. 13. (aj — y4-«)'. 18. (aj4-4)(aj4-9). 14. {2x- y- z)\ 19. (a? 4- 20)(aj - 15). 16. (5aj — 2 2/4- 25)*. 20. (aj - 8)(aj - 6). 16. (a? -33/ -4)*. 21. (a4-4c)(a — 3c). 17. (a?-7)(aj + 3). 22. (2a4- c)(3a - d). ^ ELEMENTARY DEFINITIONS AND OPERATIONS 11 23. (3aj4-4)(2a:-7). 31. (a - 2c)(a- 7 c). 24. (56-6c)(46+3c). 32. (a«4- 26)(a«4-36). 26. (x^y)(a — b). 33. (a» — 9 c)(a» + 8 c). 26. (3y-4)(3y4-8). 34. (6a-4 6)«. 27. (a« + c)(a« — c). 36. (a? + yy. 28. (2a + 6)(2a-6). 36. (3aj-4)«. 29. (5aj« — y»)(3/»-f 5aj«). 37. (x-yf. 30. («• 4- 3/*)(»' - 3/*). 38. (2 a 4- 1)'. Complete the squares : 39. 4aj*±12aj + ? 42. 4m«±?-f9w*. 40. 9a«±24a4-? 43. a*'±?+366»». 41. ?±18 6c + 81c». 44. 25aj*-20 4-? EQUATIONS 19. An equation expresses an equality. In other words^ an equation is a statement that two expres- sions stand for the same number. 2»-|-6 = 7a? — 4, and (a + 6)* = a* -|- 2 a6 -f 6* are equations. The first is an equation of condition^ because it is true only under the condition that a? = 2. The second is an equation of identity^ because it is true whatever values a and b may have. In this chapter we deal with equations of condition. The root of an equation, containing one unknown, is a number which, when substituted for the unknown, "satisfies" the equation by reducing both sides to identical numbers. An equation of the form ooj* + 6aj»-^ 4- ... 4-^ = 0, where n is a positive integer, is said to be of the nth degree. 12 ELEMENTARY ALGEBRA The degree of any equation is indicated by the exponent of the highest power of the unknown. Thus, a; -> 7 = 2, is an equation of the 1st degree, called linear. a;3 — 3 X = — 2, is an equation of the 2d degree, called quadratic. x8.62(^ + lla; = 6, is an equation of the 3d degree, called cubic. x« — 1 = 0, is an equation of the 4th degree, called quartic. etc. 20. The equilibrium of a balance is not disturbed so long as like changes in the weights are made simultaneously on bpth sides. So in equations, we may add the same number to both sides, or subtract the same number from both sides, or we may multiply or divide both sides by the same number (except divi- sion by zero). The equality is maintained during all these changes. Solve »(« + 4)=a^ — 3a? + 5. Remove the parenthesis, x^ -\-Ax = x^ — Sx-\- 6, Subtract x^ from both sides, 4x = — Sx + b. Add 3 x to both sides, 7 a; = 5. Divide both sides by 7, x = f . Check : In the given equation, substitute ^ for x, WBITTBN BXEBOI8E8 21. Solve and check : 1. (aj-2)(a?-3)=a:(« + 4). 3. (m — 4)*4-(m4-4:)2 = m(2m-f-l). 4. (r 4- 1)' - r(r — 1)= r2(r + 2)-f 5 r - 1. 6. y(t/ + l)+(y-hl)(2^ + 2)=(t/ + 2)(t/+3)+y(t/-f-4)-9. 6. «-.[4 + {4-(4 + 01]=0. 7. 6«-(4«-8)-{5-3«~(7aj-4)| = ll. ELEMENTARY DEFINITIONS AND OPERATIONS 13 8. s-|-3-[s-9-3{9-4(6-«)-sn = 3. 9. 20(1 - X)- 3(x - 6)- S[x + 8 -458 - 3(1 - «)}]==- 6. 10. (x 4- 8)(aj - 5)(x 4- 6) = (» + 2)(x 4- S)(x + 4). Solve each of the following equations for each letter in terms of the others : 11. A = iab. j3 ;S = -(a + Z). ^** ^i^i=^^2- 12. S = ia(b-^by ' ^ 16. = 1(27' -32). I SIMULTANEOUS LINEAR EQUATIONS 22. If a liQear equation contains but one unknown, one value may be found for that unknown. If a linear equation contains two unknowns, an unlimited number of simultaneous values may be found for them. For that reason the equation is called indeterminate. In « + y = 3, we have x = l, y = 2; a; = 2, y = l;aj=— 2, y = 5j a: = 0, y = 3 ; a; = 3, y = ; etc. If these pairs of values are plotted, the points thus obtained will be seen to lie on a straight line, as in Fig. 1 ; hence the name linear equation. Two points determine the posi- tion of a straight line. The graph of a linear equation in two un- knowns is most easily made by locating the points at which the line crosses the coordinate axes. ^^' ^• It was seen above that in the equation aj-|-y = 8, 05 = 8 when y = 0, and x = when y = 3. In Fig. 1, x = S, y = locates the point A. 35 = 0, y = 3 locates the point B, The two points A and B determine the line. I 1 t I I III l^*T^T— T— ' IIIIIIIIIII zz_sszzzzzz Q ^ ZX 14 ELEMENTARY ALGEBRA BXEBOI8B8 23. Make graphs of the following equations : 1. aj4-y=7. 6. 4a;~9y = 36. 11. y = 0. 2. aj — y = 2. 7. 5a? = 10 — 2y. 12. a: = 0. 3. re — 15 = 3 y. 8. aj = y. 13. x = 3. 4. 3aJ— 4y=*12. 9. aj — 3y = 0. 14. y = — 4. 6. 6-2y — a;=0. 10. 3aj — y = 6. 16.a; = — 2. 24. If the graphs of two linear equations in x and y inter- sect, the values of x and y at the point of intersection satisfy both equations ; this value of x and of y, called the coordinates of the point, constitute one solution. Since two straight lines cannot intersect in more than one point, two linear equations in x and y cannot have more than one solution* Two linear equations in x and t/, representing lines which intersect in one point, are called independent. WHi'ri'BN BXEB0I8BS 25. Make graphs of the following equations and determine, from the graph, the coordinates of the point of intersection. Check, by substituting the coordinates in the equations. We are not able to draw figures that are absolutely accurate. Hence, solutions obtained from graphs are usually only approximations to the true values. 1. a; + y = 6, 4. a: = 3, 7. 3« = 2y, re — y = l. 2a:— 3y = 9. a?-f4 = 7y. 2. 3aj — y = 7, 6. y = — 2, 8. a? — 3^ = 0, 2a;-y = 5. 6a: — 3y=16. 3a:-f23^ = 6. 3. a? 4- 3 2/ = 8, 6. a? = y — 6, 9. a? -f 1 = 0, 2aj-f-y = l. 6y = a: + 10. y — a?=6. ELEMENTARY DEFINITIONS AND OPERATIONS 15 26. The two linear equations, 2x-h2y = 10, can be satisfied by the same values of x and y ; for example, jc = 1 and y = 4, or a? = 3 and y = 2, or x = 5 and y = 0, etc. Moreover, by dividing both sides of the second equation by 2, the second equation can be reduced to the first. The two equar tions are therefore not independent ; they are really different forms of one and the same equation and represent one and the same straight line. They are called equivalent equations. 27. The two linear equations, x — y=z5, x-y=l, are called inconsistent^ for the reason that no finite value of x and of y can be found, such that x— y is equal to 5 and also equal to 7 ; no pair of finite values of x and y satisfy both V equations. This is geometrically evident from the fact that the graphs of the two equations are parallel lines. Two linear equations in two variables x and y belong there- fore to one of three groups : 1. The two equations are independent and represent two lines which intersect in one point, or 2. The two equations are inconsistent and represent two lines which do not intersect (are parallel), or 3. The two equations are equivalent and represent one and the same straight line. WBITTBN BXBBGISBS 28. Make graphs of the following equations : 1. 3aj-6y=8, 3. 2yH-3aj = 0, oj— 2y = 5. 9aj-h6y = 15. 2. 5y-2x = 4:, 4. 2a:-4y = -10, 4a:-10y=:-8. x-2y = -5. 16 ELEMENTARY ALGEBRA 6. a— 3y = 2aj-7, 7. aj = 3, 9. a;-h4 = 0, 3ajH-2y=4aj + 5y-7. y = -2. y-l = 0. 6. 5aj — 7y = ll, 8. aj-hy = 0, 10. aj = 0, 10aj-14y = 20. aj-y=sO. 3^ = 0. 11. Can you tell, before making the graphs, whether two equations are equivalent, or not? Inconsistent, or not? 12. Can you tell whether the graphs will be parallel, or not? 13. Can you tell whether the graphs will go through the origin, or not ? 29. Systems of independent linear equations may be solved by a process called elimination. The elimination may be accomplished in one of three ways : I. Addition or subtraction. — Make the coeflB.cients of one variable alike in both equations and then either add or sub- tract the sides of the equations. II. Substitution. — Find the value of one variable in terms of the other in one equation, and substitute that value in the other equation. III. Comparison. — Find the value of one variable in terms of the other in both equations, then equate these values and solve the resulting equation. EXBBCISBS 30. 1. Solve by addition or subtraction : 5aj-h3y = 69, (1) 4aj-7y=-20. (2) Solution. Multiply (1) by 4, (2) by 5, 20 x + 12 y = 276, (8) 20a;-36y=-100. (4) Subtract (4) from (3) , 47 y = 376, y = 8. Substitute in (1), a; = 9. CTieck : Substitute in (1) and (2), 46 + 24 = 69, 36 -66 =-20, ELEMENTARY DEFINITIONS AND OPERATIONS 17 2. Solve by substitution, 5 + 6 = *' (1) 7 8 14' (2) Solution. From (1), Substitute in (2), 6 y 1 7 8 14' 160 201, 7y= 4. 111^=164. y = 12. Substitute in (1), f+2 = 4. a; = 10. 3. Solve by comparison, 3 a - 13 y = 1, (1) (2) Solution, From (1), ^ 13 (3) From (2), y = 16a;-9H. (4) Compare (3) and (4), 8^=15x-9H, 3x-l = 195x-129, 192 a; = 128, x = f 1 Substitute in (3), y = A- Solve a,nd check : 4. 12 « + 11^ = 12, •• !+"-¥' 42 a -h 22 y = 4(H^. X • % 18 ELEMENTARY ALGEBRA 3 9 3 ^2 4 20 PROBLEMS 31. 1. The sum of two numbers is 180 and their difference is 12. Find the numbers. 2. One number is 27 more than another. Their sum is 143. What are the numbers ? 3. Of two numbers, twice the greater exceeds three times the smaller by 4 ; but the sum of the greater and twice the smaller is 44. Find the numbers. 4. Two pounds of sugar and three pounds of flour cost 19^, and three pounds of sugar and five pounds of flour cost 30 ^. What is the price of each per pound ? 5. Three yards of velvet and five yards of silk cost me $ 15.75. If I had bought two yards more of velvet and two yards less of silk, my bill would have been $ 18.25. What is the cost of the velvet and silk per yard ? 6. In five hours A can walk 1 mi. more than B can walk in 6 hours ; in seven hours A can walk 3 mi. more than B can walk in 8 hours. How many miles an hour can each walk ? 7. A, B, C, D, together have $ 300. A has $ 10 more than C ; B has $ 5 less than half as much as D ; and A and B have together $ 5 less than twice as much as C. How much has each? 8. A number consists of two digits whose sum is 12. If three times the sum of the digits be subtracted from the number, the digits will be reversed. What is the number? 9. The digits of a three-digit number will be reversed, if 396 be added to the. number. The units' digit is 3 times the hundreds', and if double the tens' digit be increased by 6, the result will equal the units' digit. Find the number. ^ELEMENTARY DEFINITIONS AND OPERATIONS 19 10. In a purse there are 3 times as many quarters as nickels, and twice as many half dollars as nickels. The total value of the coins is $ 3.60. How many of each kind are there ? 11. Some boys bought a boat and found upon paying for it that if there had been 2 more of them, each would have paid a dollar less ; but if there had been 2 fewer, each would have had to pay $ 1^ more. How much did the boat cost ? 12. A rectangle is four times as long as it is wide. If it were 3 in. shorter and 2 in. wider, its area would be increased 15 sq. in. Find its dimensions. 13. It costs as much to sod a square piece of ground at 25 ff a square yard as it does to fence it at $ 1 a yard. How long is the side of the square ? 14. A walk is laid 3 ft. wide around a rectangular court, which is 15 ft. longer than it is wide. The area of the walk is 960 sq. ft. Find the dimensions of the court. 15. A man invests part of $ 5480 at 5 % and the remainder at 4 %. The total annual income is $244. How many dollars has he in each investment ? 16. The yearly income from a 5 % investment is $ 97.70 more than that from a 6 % investment. The sum of the two investments is $ 6420. How much is invested at 6 % ? 17. A boy weighing 100 lb. is 6 ft. from the fulcrum of a seesaw. He balances a boy who is 8 ft. from the fulcrum. What is the weight of the second boy ? By careful measurement it has been ascer- i—r tained that, to maintain a balance, the lengths of the arms of a lever and the corresponding weights must conform to the law, hwi = hv32 ; that is, the product of one weight and its distance from the fulcrum is equal to the product of the other weight and its distance from the fulcrum. 18. Two boys together weigh 170 lb. They balance when one is 8 ft. and the other 9 ft. from the fulcrum. How much does each weigh ? 20 ELEMENTARY ALGEBRA 19. If the lever is 8 ft. long, how far from the fulcrum will two weights, 30 lb. and 50 lb., have to be in order to balance? 20. A weight of 180 lb. is carried between two men by means of a pole. One man is 5 ft. from the weight, the other is 4 ft. How many lb, does each man lift ? POSITIVE INTEGRAL EXPONENTS 82. In the expression a* we say a is raised to the nth power. When n is a positive integer : I. The nth power of a is the product obtained by using a n times as a factor. a • a • a • • • to n factors = a*. II. The nth power of a* is equal to a* • a* • a* • • • to n factors = a*". Hence (a*")" = a"". III. The nth power of a^ft** is equal to (a* • a* • a* • • • to n factors) (6' • 6" . 6' • • • to n factors) = a"*" • h^ = a'"*6'*. Hence {oFbpy = af^'^hf*'*. IV. The nth power of — is equal to a* • a* • a* • • • to n factors a*** ftp . 5j» . 5p . . . to n factors 6** Hence, ( — ) = — • ' \bpj bP'* 33. From the law of the signs in multiplication : 1. An even power of any real number is positive. 2. An odd power of any real number has the same sign as the number itself. ELEMENTARY DEFINITIONS AND OPERATIONS 21 BXBBCISBS 34. Write the values of : 1. (2a'6)'. jj /_^V 2. i-lahf)*. ^ ^^ 4. /'3<«f\*. 13. (2a»6c»)'. ^'^'^'^ 14. -(_a*6'c«)». 7. (0^2'')". ^^ f^^_f\\ 8. (a^+'ft*"")'. A ^^^*^ / 19. (-|a*6")»». 20. (6a^yh)^. 10. (23^)' MEANING OF FRACTIONAL EXPONENTS 35. When an exponent n is a positive integer, we know that a* = a • a • a • • • to w factors. That is, the exponent n indicates that a is taken n times as a factor. What is the meaning of a* ? It would be absurd to say that a* signifies a taken ^ times Vs a factor. A number can be taken ^s a factor only a whole number of times. For the purpose of attaching a meaning to a* we stipulate that we shall be able to multiply a' by a* according to the same rule by which a* is multiplied by a^ The product of a* and a* is found by adding the exponents; that is, a^>a^ = a*. If we add the exponents in the multiplication of a' by a% we obtain, 1 1 1 1 1 / 22 ELEMENTARY ALGEBRA It is seen that a' is one of the two equal factors of a, or the square root of a. Hence a^ is another way of writing the square root of a. Likewise, the meaning of a* is found by taking a* four times as a factor, thus, i s s « a* • a* • a* • a* = a*. Hence a* means one of the four equal factors of a' or the fourth root of aK In general, to find a meaning for a«, where p and g are pos- itive integers, we take a« as a factor q times, and add the exponents. We obtain, - - - . j» J -+ — r-+ ••• to c terms — _ a« . a« • a* • • • to g factors = a« « « = a« = a'. It is seen that a« is one of the q equal factors of a'. Hence t a« means the gth root of a^ Thus, The numerator of a positive fra^ional exponent indicates the power of the base and the denominator indicates the root of that power. In finding the value of 27*, we may square 27, which gives 729, and then take the cube root of 729, which is 9. Or we may take the cube root of 27, which is 3, and square 3, getting 9 as the result, that is, 27* = a/272 = {V27y. In general, we have. All irrational numbers which are expressed by the use of radical signs can be expressed by the use of fractional ex- ponents. In fact, the simplification of expressions usually can be effected more easily by the latter. Thus, ■v/16 a^h^d^ = 16*a*6^c* = 2 a^l^c. X X ELEMENTARY DEFINITIONS AND OPERATIONS 23 MEANING OF A ZERO EXPONENT 36. We proceed to assign a meaning to aP, where a is not itself zero. For simplicity we stipulate, as before, that in the multiplication of two powers having equal bases, we find the product by adding the exponents. Accordingly, a"* >a^=z a""*"® = a"*. Divide both sides by a*, 2!!j_?L — ?5!!. a" a"* Simplify, a* = 1. Therefore, any number y except 0, vnth a zero exponent is equal tO'l, MEANING OF NEGATIVE EXPONENTS 37. Let m represent a positive integer or a positive fraction. What is the meaning of a"^, where a is not zero ? Working on the assumption (§ 35 and § 36) that in multiplication we are permitted to add the exponents^ we obtain, a"* • a~*" = a'*""' = a\ Since a° = 1, a* • a"~ = 1. Divide by a**, a~** = — • a"* Therefore, any number with a negative exponent is equal to 1 divided by that number with a positive exponent. Thus, a-»6* 1 .6* 6* c 6*. • c a6« a-^b-^c " 1 a* 1 ~ — • c 6« c Therefore, a factor may be moved from the numerator to the denominator of a fraction, or from the denomi- nator to the numerator, provided the sign of its exponent be changed. 24 ELEMENTARY ALGEBRA -2 + 6 _5 Care must be taken to transfer only /actors. In ^ — ^^, ar^ is not a c factor of the numerator. "*" is not equal to , • Since a~^ = ^ « i + 6 it follow that ?:^±^ = ?l_ = li^. c c a^ PRINCIPAL ROOTS 88. Since 6 • 6 = -h 36, and ( - 6)(- 6) = + 36, it follows that both -h 6 and — 6 are square roots of 36. The two square roots of 36 are usually written in the form ± 6. Similarly the two square roots of a^ are written ± a. The positive square root of a number is called the principal square root Unless otherwise stated, we shall consider only the principal square root, and write V36 = 6. When V has no sign before it, or has the + sign before it, the principal square root is always understood. When we write — V we mean the negative square root. Thus, \/36 or (86)^ stands for -h 6, - V86 or -(36)* stands for — 6, ±V86 = ±(36)1=±6. It can be shown that there are 3 different cube roots of a number, four different fourth roots of a number, and, in general, n different nth roots 'of a number. For example, there are three different cube roots of &' ; namely, 6, ^^ "*" &, "" ^ ""^^^ 6. The fractional coeflacients in the last two 2 2 roots involve the imaginary number V— 3. Only one of the three roots is real, namely, the root b. We call h ihe principal root. In elementary algebra the principal root is the only root usually considered. I. .If a is a positive number, then the principal nth root of a is its positive value ; we designate it by a" or ^a. When n is an odd number, then this principal root is the only real root. ELEMENTARY DEFINITIONS AND OPERATIONS 25 1 When n is even there is still another real root, namely, — a* or Thus, the principal cube root of 64 is (64) t or v^ = 4 ; this is the only cube root of 64 that is real. The principal fourth root of 81 is (81) i or \^ = 8. There is another real fourth root of 81 ; namely, —3, for we see that ( — 3) ( — 3) ( — 3) (— 3) = 81. II. If a is a negative number, and n is an odd integer, there is no positive root ; there is a negative root and that is taken as the principal nth root. Thus, the principal cube root of — 27 is C— 27)^ or -^—27 = - 3. III. If a is a negative number, and n is an even integer, then all the roots are imaginary. Imaginary numbers will be dis- cussed more fully later. Thus, v^— 64 or ( — 64) • represents imaginary roots. POWERS AND ROOTS 39. It is readily seen that (a^y^^c^.a^^a^'C^^a."^ This result may be obtained at once by multiplying the exponents 3 and 4. In this process algebraic expressions are raised to powers ; it is called involution. In general, (a*)* = a"*" ((aryy = a~»» 11 It is agreed that each of the symbols a% Va, 6*, -Vb, (a6)n, ^^/ab shall represent only one nth root, namely, the principal 26 ELEMENTARY ALGEBRA nth root. For the principal real roots the following formulas can be shown to be true : (a"")* = a"* or Va"*" = a"*. Ill a* • 6" = {ahy or Va • Vh = Va6. ? = r?Yor^ y/a _ */a ^ 5^ W V6 - Express each of these formulas in words. • The square root of 2 cannot be exactly expressed by the Hindu-Arabic numerals. One can approximate its value by extracting the square root to two, three, or more decimal places, thus: V2 = 1.41+, V2 = 1.414+, V2 = 1.4147+, and so on. The radical V2 or 2*, and other radicals of the same kind, V3 or 3i, ^5 or 5^^, etc., whose values can be found ap- proximately, but not exactly, represent numbers called irra- tional numbers. ORAL BXBBCISBS 40. Express with fractional exponents and simplify : 1. V3 win*. 5. 2V2a^. 9. \^5 . Vo^. 2. A/2a^h\ ^ 10. -v/5.^35. 7. 3a/^. 3. V-21a'c\ ^^'_ 11. ^.■^. , 8. -7\/-. 7, 4. ay/^M. ^y 12. 6V2^^*. Express with radical signs : 13. 2(7 y)l 15. Sh^yl 17. 5(x + 2/)*. 14. Sajij/*- ^®* "1* 18- 4(aj*-y2)*. 4.27*. ELEMENTARY DEFINITIONS AND OPERATIONS 27 Simplify : 19. 49-i. 26 2^ *'■ (-3)"'-(-^)'- 20. 81*. ■ 8* ' 82- (M)-*. 21. 50 . 16i. ''' ^f!„r- ^'' <^>"'-_, , , 27. -^-^- 84. a».(|)-*. 22. 16-i.26i. 4-'a-> 23- iis) • 29. a-»(a^ + 6). ^_,^_, 24. (^)*. 80. 36-^-343*. **• '^i;=r* 37. £l^. 38. «1±^. m~* + b~* Multiply numerator and denomi- ' mr* — 6~*' nator by tfibf. Find the principal roots of : 40. Vxy. .g 1 256 a^b* »/ ->^a" 41. V4^. >'289m"- >/««««• 8. 42. V9^. 60. </256^. 56. Vl25a!>Y. 43. V26^. ^sr^^ "• A^ 44. VIOO ai6. ^^' XUoO'riOwie' h^^ si- 512 a^ 400 a^y « 45. ^-8a36i2. ^ 58. \/— ^ 46. ■\/64.afifz'^. x^ 47. ^ -343a36i« . 53. V/^?^. ^^' ^^^^' 48. V^/1^. e,.^ 60. #^'. \ 729 2/3 54. ^J/729^y%24; \ 76r Find the ^loo reaZ fourth roots of : 61. 10,000 a2&3c*(a + ^)*- «3. j\ x-*y'^^(x - yy. 62. 625 7/i8n«p3(a2 - 52)2. e4. .OOSl(x - yyzhi}-^. 28 ELEMENTARY ALGEBRA Find the real fifth root of : 65. -32iB6(m+7i-j9)-«y»24. 66. 243 a»(6 - c)i«ar«2*. 67. What kind of roots of positive numbers have two real values ? Which of those two real roots is the principal root ? 68. What kind of roots of positive or negative numbers have only one real value ? 69. What kind of roots of negative numbers have no real value? SQUARE ROOT 41. Since (a -h 6)2 = a^ + 2 a6 + d«, Va2-h2adH-62 = a + 6. Thus, by inspection, we can extract the square root of a tri- nomial which is a perfect square. Or we may find the square root in this way : a2-h2a6-h6* |a-h^ a* 2a6-h&* Trial divisor, 2 a Complete divisor, 2 a + 6 Thus the square root is a + 6. This simple case will enable us to devise a rule which is applicable to more complicated cases. The procedure is as follows : I. Extract the square root of the first term and subtract its square from the polynomial, leaving 2 a6 H- bK II. In 2aby we see the factor b which we know is the second term of the root. This factor b may be obtained by dividing 2ab by 2 a. Thus we call 2 a the trial divisor. Since a is the part of the root already found, we see that the trial divisor is double the root already found. After b h^s been found, add it to the trial divisor and we have 2 aft + 6, the complete divisor. ELEMENTARY DEFINITIONS AND OPERATIONS 29 III. Multiply 2ab + bhj by and subtract. This process may be extended to finding the square root of any polynomial. Find the square root of x* + 21xa-86a:-f 86 — Oa*. Arrange the terms according to the ascending or descending powers of some letter, x*-6a« + 21ic2-36a; + 36 \x^-Sx + e — 6ar« + 21x3 — 36x + 86, 1st remainder -6x8+ 9x2 12 x^ - 86 X + 86, 2d remainder 12x2-86x + 86 Ist trial divisor, 2 (x^) = 2 x^ Ist complete divisor, 2 x^- 3 x 2d trial divisor, 2(xa - 3x)= 2x2- 6x 2d complete divisor, 2 x^ — 6 x + 6 * 8d remainder The procedure is as follows : I. The square root of x* is x'. Here a = x^ ; 2 a = 2x2, the first trial divisor. II. 2 a6 = — 6x8 ; divide — 6x8 by 2x2 and we obtain — 8x, the value of &. Thus the first complete divisor is 2 x2 — 3 x. III. Multiply 2x2- 8x by —8x and we obtain —6x8 + 9x2. Subtract this product from the 1st remainder and we obtain 12x2-86x+86. IV. Treat this remainder as we did the first remainder and proceed as before. Since the third remainder isO, x2 — 3x + 6is the required square root. BXBBCI8BS 42. Find the square root of : 1. 4 a« - 28 a»6« -h 49 6". 2. 4a«-12a6 + 4ac + 962-6&c-hc*. 3. ai2 — 2icy-h6a:H-y2 — 6y-h9. 4. 4ic*-20ic»-30aj + 9H-37a^. 6. 9aj«-2ic*-3aj* + l-h 10 aj'- 12 a^-2 a. 6. 9 a« -h 9 d« -h 24 a'^ft -h 24 ab^ - 8 a^ft* - 8 a«6* -50 a»6». 7. ^-5aj + 25. 8. ^'-2 + i. 4 9 a* 30 ELEMENTARY ALGEBRA ^ 4 4 , , 187 , 15 ^25 9. -m* — rnr -\- ——nr :-wi4--r' 9 48 4 4 tix A 4a , a* , 2a' 4a' , a* 11. a' -H a5, to three terms. 12. 1 — a, to four terms. SQUARE ROOT OF ARITHMETICAL NUMBERS 43. 1« = 1, 2* = 4, ..., 9* =81, 10« = 100, ..., 90« = 8100, 100* = 10000, 1000» = 1,000,000. This is sufficient to show that a perfect square consisting of one or two digits has one digit in the root ; one consisting of three or four digits has two digits in the root ; one consisting oijive or six digits has three digits in the root ; and so on. In other words, the square of a number has twice as many inte- gral digits, or one less than twice as many, as the number itself. Hence, to find the square root we must separate the number into periods of two digits each, towards the left and right from the decimal point The period farthest to the left may have one or two digits, whereas the one farthest to the right must have two, a cipher being annexed, if necessary, to complete it. The square root has therefore one digit corre9ponding to every period in its square. Separating the square number into periods enables one to find one digit in the root at a time. This is seen more clearly if we consider that the square of any number of tens, say 4 tens (40), ends in two ciphers (1600) ; hence the two digits on the right are not needed to find the tens' digit (4), and are set aside until the unit's digit is to be found. Likewise, the square of any number of hundreds (400), ends in four ciphers (160,000), and all of these may be set aside until the hundreds' digit is found and they are needed for the tens' and units' digits. ELEMENTARY DEFINITIONS AND OPERATIONS 31 After pointing off the number into periods, the method is similar to the one used for polynomials. The procedure is based on the formula d^-{-2ab -^b^ = (a-\- by. When dividing by the .trial divisor, exclude, for brevity, the right-hand digit in the remainder. The reason for this is seen in the example which follows : the trial divisor 4 is equal to 40 of the next lower units. Now 14 divided by 4 gives the same digit as 148 -s- 40. Point off one decimal place in the root for each period in the decimal. Find the square root of 648.643241. 648.643241 1 23.421 4 148 1st remainder. 129 1964 2d remainder. 1866 9832 3d remainder. 9364 46841 4th remainder. 46841 1st trial divisor, 2 (2) = 4 1st complete divisor, 43 2d trial divisor, 2(23) = 46 2d complete divisor, 464 3d trial divisor, 2(234) = 468 3d complete divisor, 4682 4th trial divisor, 2(2342)= 4684 4th complete divisor, 46841 6th remainder. Point off three decimal places in the root. Thos, the square root is 23.421. To find the square root of a fraction, which is not a perfect square : Reduce the fraction to a decimal and find the square root of the decimal. EXERCISES 44. Find the square root of : 1. 6889. 6. 567,009. 2. 2025. 7. 2777.29. 3. 161,604. 8. 88.454025. 4. 337,561. 9. .00698896. 6. 7.3984. 10. .059049. 32 ELEMENTARY ALGEBRA Find the square root, to 3 decimal places, of : 11. 2. 13. 5. 15. 3|. 12. 3. 14. 1.005. 16. f FACTORING 45. Factoring is the process of finding two or more expres- sions whose product is a given expression. An integral expression is one that contains no fractions. A rational integral algebraic expression is one containing no fractions, no negative exponents, and no indicated roots. 2 + 3 aaj + aj'* is a rational integral expression ; J x* + J x + 1, a;-* + 6, y/a + x + y, X* + 10, are not rational integral expressions. A prime factor of an integer is an integral factor which is exactly divisible only by itself or one. Thus, and 13 are prime factors of 65. A prime factor of a rational integral expression is a rational integral factor which cannot itself be resolved into rational integral factors other than itself ojid one. Thus, a — 6 and a^ + aft + 6* are prime factors of a^ — 6«. In the following exercises we shall confine ourselves to find- ing 2>rime factors of rational integral expressions. In § 83 irrational factors will be required. I. Type form : aft + ac — ad. ab + a4i —ad=: a(b 4- c — d). 46. Factor.: 1. y^-2y' + Sy*-y. 3. a(x + y)-b(x + y). 2. 80x»y*-160 ajy -240 o^, 4. (a? - y)c -(« + y)c. 6. m{c — d)— n{c — 6i)-\-p{c — (Z). 6. a{m — n) + b{m — n). 7. m(a — &) — w(6 — a). ELEMENTARY DEFINITIONS AND OPERATIONS 33 8. a(x-Sy)+b(Sy'-'X), ®- (a? — y)— 2a(y — aj)-H36(y — a). 10. c(a — & — c)+d(6+ c — a). 11. Type form: ax -\- bx -\- ay + by. ax -{- bx +ay + by == a{x -^ y)'\' b(x '\' y)=^ix+y)(a + b). 47. Factor: 1. Sx + Sy-{-ax + ay. 6. 2ac — 36c — 2ad4-36d. 2. 005 — ay — &aj + 6y. 6. a"c" — &"c" -f- a"d" — 6*d*. 3. 2aaj — 2 6a5 + & — a. 7. a'— a6-Hac— ac+6c— c*. 4. 4a— 46 + ac — &c. 8. aV-h6V — aV — ^V* III. Type form : tf ± 2a6-H 6* and fl^-hft* + c* + 2a6 - 2ac- 26c a*±2a6-h6«=(a±6)*. a* -h 6* + c* 4- 2 a6 - 2 oo - 2 6c = (a + 6 - c )*. 48. Factor: 1. 9r*-24r« + 16a». 3. a* + a + i. 2. a*»-14a»-h49. 4. (aj-y)*-2(aj-3^)+l. 6. 4(c - 3)2 - 12(c2 - 9) + 9(c + Sy. 6. sfi-\'f/^'{-z* — 2xy^23DZ + 2yz. 7. a* + 6*4-c'--2a6 + 2ac — 26a 8. l + a* + 6* + 2a-26-2a6. 9. 4a* + 6* + c^-4a6 + 4ac-26c. 10. m*4-n»-h9 — 2mw- 6m + 6w. IV. Type form: tf-6*. 49. a«-6* = (a + 6)(a-6). a*-2a6 + 6«-c*=(a-6)«-c* = (a-6-hc)(a-6-c). a* — c* — 2cd— cP = a*— (c + d)* = (a-hc + d)(a — c — d). a«-2a6 + 6*-c» + 2cd-(f» = (a-6)«-(c-d)« =(a — 6 + c — d)(a — 6 — c -h d). 34 ELEMENTARY ALGEBRA 50. Factor: 1, c8~64. 6. 100aW-a*-hl2a6-366». 2, 121c«-144d*. 6. 36y» + 84y + 49-92!«. 3. aj* — m* — 2mn — w*. 7. m« + 10m» + 25 — 81n*. 4. 49a«-64 + 166-&«. 8. (a - 3)« - (6 - 2)*. 9. 46» — a?* + 4a^-ha*4-4a& — 4y«. 10. 26(f*-l-9c*d«-10ad + a2 + 6cd. V. Type form : jfi + bx + c, 51. Factor: 1. aj*-h5a; + 6. 6. r» — 4 r« — 77 ««. 2. a;2-5a? + 6. 7. b^c^ -^ 7 bey -{- 10 yK 3. yj- 7^4-6. 8. c^d* H- 7 cc«« - 120 (f*. 4. 2/2 + 72/-98. 9. <«-23«*-60. 6. 2!2 + 102 + 9. 10. a*2/® ■- 16 ic'y^ + 39 2^ VI. Type form : or^ + 6x + c. 52. ao?* + &« -h c = (pa? + g)(ra? 4- «), wherein pr =: a, ps -{- qr = 6, g« = c. By this method p, g, r, « must be found by trial, so that the three conditions just named will be satisfied. Factor 6a^ — 5x—6. Here a = 6f 6 = — 6, c = ^6. If possible, we must find numbers p, q, r, 8, such that pr = 6^ gs = — 6, p« + gr = — 5. pr = 6 suggests the values jp = 3, r = 2, orjp = 1, r = 6, etc. g« = — 6 suggests the values g = 3, « = — 2, or g = 2, s = — 3, etc. Try the different sets of values of p and r, whose product is 6, with each pair of values of g and s, whose product is — 6, until a combination is found which satisfies ps -{- qr= —6. ELEMENTARY DEFINITIONS AND OPERATIONS 35 Try the sets of values p = 3, r = 2, g = 2, a = — 8. Examine (3x 4-2)(2x-8) I I -9a; We see that the sum of the two cross products + 4 a; and ~ 9 x is — 6 a;, which is the middle term of 6 a^ — 5 x — 6. Hence 6a^-6a?-6=(3x + 2)J2x — 3). 53. Factor: 1. 6aj* + 5a? — 6. 6. 15 a? + 14 oa? - 8 a*. 2. 5aj«4-14a?-3. 7. 17 a«aj* - 36 oa; + 4. 3. 5aj2-14a?-3. 8. 20r» + 19r« + 35*. 4. 8y*-14y-39. 9. 9 aj» - 36 a^ - 13 y*. 6. 18m* + 37m + 19. 10. 28m2-48m + 17. VII. Type form: a* + o^ft* + &*• a* + a»6« + 6* = a* -h 2a*b^ + 6' - a^b^ = (a« + 6')* - a«&» = (a« + a6 -f 6*)(a« - aft + b^)- 54. Factor: 1. a;*-H«* + l. 6. a?* — 14 aj*y* + y*. 2. 9m* + 8mV4-4w*. 7. m* — 38mV-Hn*. 3. 16a* + 20a«6« + 96*. 8. 4 a* - 13 a'6« + 9 6*. 4. r8 + r*5* + «*. 9. 25c* + 26c«cP + 9d*. 6. a* - 5 a«6* 4- 4 6*. 10. 49a* + 110a*62 + 81 6*. 55. In factoring, first take out monomial factors. Then inspect the resulting polynomial, ascertaining to which type form it belongs, and factor it accordingly. In the final form, all factors should be prime. 36 ELEMENTARY ALGEBRA ORAL BXBROI8BS 56. Factor: 1. rn^-a}. 9. 20-9j5i-«*. 17. a^ + yhi. 2. ^a^— y^. 10. m* — n*. 18. aj* — ^a?. 3. a« + 6a!> + 96*. 11. p^-j:^. 19. a5»-4ajy». 4. 9aj»- 6ajy + y*- 12. aj» + 2aj-35. 20. 4a»-a6*. 6. 4aj*-12a?5+9«*. 13. aj«-2a?-35. 21. a^6-9&». 6. oaj*— 16ay«. 14. m*-2m — 3. 22. 5a' + a*6. 7. ca^ + cy\ _ 15. p^-Qp-^- 5. 23. a* - 2 a*6» + &*. 8. y«-5y + 6. 16. 4c*±4cd + cP. 24. 9a*±12a6+4&*. WBITTBN BXBBOI8BS 67. Factor: 1. 4a* — 20a6. 6. a*hc — ah*c. 9. 16 + 8 c + c«. 2. 6»-d». 6. 25r*-36«*. 10. l~a«. 3. m* — 9n*. 1. f -^^--y — z. 11. a^^ — 6*. 4. 4a*-.8a«6. 8. mV— 2mn*4-l. 12. a* - 5 a6 + 6 6». 13. cflj -f cy — cte — dy. . 24. m* — 2 mn + n* + m — n. 14. 6*-|.66« + 5. 26. v^'{-7*'\-r^ + r. 16. 64-&». 26. aj* + 2aj« + l. 16. 7 a? — 12 — aj*. 27. c*-cP — c — d. 17. a*®— a*. 28. 6* — 26c + c» + 5& — 5c. 18. 3aj» + 3aj«-27aj-27. 29. 60* + 13a6 + 66*. 19. a?*" + 2af-15." 30. a?" - ajyi*. 20. aj»-6aj*4-12aj— 8. 31. r* -4r» + 6r» — 4r + 1. 21. a^^a^\, 32. (a? — y)* - 9(aj + y)*. 22. r»-«*-2««-<». 33. a* + 12 a6 + 36 6*. 23. 26-4A:* + 4^*^-^«. 34. (a + 6)* - 7(a + 6) + 12. 36. (a? - 13 aj -h 42)(aj»+ 3 a? - 10). 36. (6-c)*+36-12(6-c). ELEMENTARY DEFINITIONS AND OPERATIONS 37 37. (a + («)*+ (6 + c)* - 2(o + d)(6 + c). 38. 25 a* - 41 a*6* + 16 6*. 39. a*-2a6 + 6«-2ac + 26c+<^. 40. c" — (?•. 41. a?-49a? + 360. 42. i^-47r»+l. 43. ab—abc^. 44. m* + 64n*. 46. (m — n)* — 1 + a(m — n -f- 1). 46. 3a*-10a6-86«-h3ad-H2W-9ac-6&c. 47. a?*-3iB* + 4. 48. 4a* + 816*. 49. 4a*~37a26«-H96*. 60. 64y*+128.v»2!2 + 81»*. « FRACTIONS 58. KfrcLCtion is the indicated quotient obtained by divid- ing one number by another. The fundamental principle of operations with fractions is — Both numercUor and denomincUor of a fraction may be multiplied or divided by the saws number, vnthout changing the value of the fraction. Thus, I = li, 2^ = ?, 8(^ + y ) =-2_. 8 16 2a a 4(x+y)(x-y) x-y 38 ELEMENTARY ALGEBRA BXEROISES 59. Reduce to lowest terms : 1. — — -n-n' O. 6a»&c* 3aj* + 3a;-18 2 ?i^. 7 (a-m«-i-ft)v 2a-26 g 4-(m-ny * a«-6« ' * n*-(m--2)«' - 5 ggg — 10 gy g a* — 2 a — 1 aj* — 5ajy4-6y«' " a» — 20*4-1 g 8(a^-6*) j^ aa-^c«-9-2a5-6c+&V 24a? + 246' ' 62-.a«-9-26c-6o+c*' 60. From the laws of signs in multiplication and division it is evident that a — a a . — a a a 6 -6' 6 b ' b -b From these relations we see that 1. The signs in both numerator and denominator may be changed without changing the value of the fraction. . 2. The sign of the numerator and of the fraction may be changed without changing the value of the fraction. 3. The sign of the denominator and of the fraction may be changed without changing the value of the fraction. 61. The following principles may be used effectively in operations with fractions. 1. The signs of an even number of factors may be changed * without changing the sign of the product. Explain. Thus, a • 6 • c = ( — a) • (— 6) • c = abc. 2. The signs of an odd number of factors may be changed, provided the sign of the product is changed. Explain. Thus, G'b- c= — (— a) 'b ' c = abc ; a'b 'C=^ (—«)(— b)(—c). ELEMENTARY DEFINITIONS AND OPERATIONS 39 MULTIPLICATION AND DIVISION OF FRACTIONS 62. In multiplication and division of fractions in algebra we have the same rules of operation as in arithmetic; namely, 1. To find the product of two fractionsy mtdtiply the numerators together for a new numerator , and the denaHrainaJtora together for a new denominator. 2. To find the quotient y invert the divisor and then proceed as in the multiplication of a fraction by a fraction. It is easy to establish the truth of these rules by means of the equation. For example, the second rale may be proved as follows : Let a c then a = &£, c = dy, and ad = hdx^ he = hdy. Dividing equals by equals, ad bdx X be hdy y But x_a . c y h' d' Hence a , c __ad h d he This proves the rule for the division of one fraction by another. Proceeding in a similar manner, prove the rule for the multiplication of fractions. Multiply m±^ by ^^'-^y\ hx -{-hy ax-\- ay 40 ELEMENTARY ALGEBRA BXEROISBS 63. Perform the indicated operations : ^ a b d ^ r-'S r* — «* 1, -■ • — I — • 6. • • c c T^ + ra r* — rs ^ 2a Sbx 2h ^ x^-3x-{-2 x'^-Sx X a 3aa a^ — 6x ijfi-\-x-'42 • 26-2'^6— l' ' a + b'^(a + by . 16m*n 8 m' ^ a^ + ah a-\-b 45 fl^ ^ ^ <^* — y* {'^ — yy « jj a^~-4a; + 3 gg-12a?4-35 5ar-10 aj2-7aj + 10' aj2H-2a?-3 '«2__3a.* --. a^ + oft a'c + 2 a^c + aft^c a2 4- 62 a^c — b^c 13. m *— n* j6 (m' — m^n + mn^ — n*) 14. 3 m2 4-3 n2 m^ — 2 mn + n« 18ag-9a , 2a2-f5a-3 r2-l "^r^-llr + io' gg — c2 -t- 2 a& 4- ft^ ?f^lAzi£. a2— c2 — 2 6c — 62'a + 5_c* 2a2 4-a-10 3a2 4-4a-4 17. a2_4 2-3a ELEMENTARY DEFINITIONS AND OPERATIONS 41 05*— J/* x^ — xy a?^ — 2 a?y + y^ 18. 19. 20. (x-hyy x-^-y a^-\-2xy + y^ 9 gg - 16 feg , 6 gg - 8 a5 3g2-llg6-20&2 • g2-5g6 * &2-136 + 42 a2 + 6g-27 a5-3 6-7.g + 21 * g2-81 COMPLEX FRACTIONS 64. A complex fraction is one which contains one or more fractions in its numerator, or denominator, or both. Complex fractions may be simplified in two ways : 1. Simplify the numsrator and denominator, then divide the numerator by the denominator, 2. Multiply both numeraJbor and denominator by the I. c. d. of their fractional terms. The pupil is probably familiar with the first method from his study of arithmetic. In simplifying complex fractions arising in algebra, the second method is usually the easier. It should be thoroughly m^^stered. Simplify, i-Zljk. i ~i r The l.c.m. of 2, 8, 4, 9 is 86. Multiplying the nomerator of the complex fraction by 86 gives 18 — 12. Multiplying the denominator of the complex fraction by 36 gives 9 — 4. Hence, Izil = l^-Jl? = ? . J-J 9-4 6 BIXBBCISES 65. Perform the indicated operations and simplify : 1. L o + ft *J a»-6» 42 ELEMENTARY ALGEBRA 3 ^ 'f ^ M ■ 1 - a* \1 - a 1 4- ay 4. (2aj*-6a? + 3)-!-/'i-iy 5. fl + i + lY?+l?+^y?iy:£. 7. \n sj \m r) \» y ^J \y ^ 10. (m^ 11. i+i 12. a b 13. I a " + 6 11 a & m + 1 V(m + 1)« m + i; 14. 2 + 2 a -r 15. 1+^ s a + 2> q~ & 16. a — b a+ 6 a — 6 , a4- ^ a 6 a+b 'a—b ELEMENTARY DEFINITIONS AND OPERATIONS 43 a?-3 — 17. ^• a? + 4 a« + ab 19 a« -6* X V. a a + b a — b r 3- r r 20. S + r r +'- r 18. -3. x + y-{- — y r + 3 r HIGHEST COMMON FACTOR AND LEAST COMMON MULTIPLE 66. Only expressions which contain no fractions or radicals will be considered here. The highest common factor (h. c. f .) of two or more expres- sions is the product of the prime factors that occur in each expression, every factor being taken the least number of times it occurs in any one expression. This definition indicates the process of finding the h. c. f . The lowest common multiple (1. c. m.) of two or more expres- sions is the product of all the prime factors that occur in the expressions, every factor being taken the greatest number of times it occurs in any one expression. This definition indicates the process of finding the 1. c. m. Observe that the 1. c. m. is the least expression which is exactly divisible by each of the given expressions, while the h.c.f. is the expression of highest degree by which each of the given expressions is exactly divisible. EXERCISES 67. 1. Find the h. c . f . and 1. c. m. of 4aa:- 4a3^, Sa***- Say, 12a;»- 122^. Factoring, 4 ax — 4 ay = ^a(x — y) . 8 a%e2 - 8 a^j^ = 2^a^(x -\-y){x-y). 12 ac» - 12 y8 = 28 . ^x - y){V^ + «y + V^)- The h. c. f. = 22(a; - y) = 4(x - y). The 1. cm. = 2« . 3 aHx - y)(x + y)(x2 + xy + y2) = 24a2(x + y)(«»-y»). 44 ELEMENTARY ALGEBRA Find the h. c. f. and 1. c. m. of : 2. 9 a6c, 18 a«6c», 27 a»6*c. 3. 16 aj»y V, - 30 a^i^z, 45 «V»' 4. (a - 6)*, a* - 6«. 5. 3a«+6a6 + 36«,9a«-96«. 6. 2(a - 6)»(a 4- 6), 3(a -f 6)»(a - 6). 7. a^-5»+6,aj* + 2aj-8. 8. 3a»-3a6«, 3a(a+&)«. 9. 3aaj»-15aaj4-18a, 6aV + 24a*a;-126a«. 10. sfi — y\ X — y, ax ^ ay ^ ex -{- cy. 11. a(a - 1)«, a«(a* - 1), 2 a(a« + 2 a - 3). 12. 5(aj» - y»), 10(aj + 3^)«, 15(x - y)«(a; + y). 13. A«-6A-14, ^'-10^ + 21, A«-49. 14. 16(ab 4- b), 8 a(a« - 6«), 24 ab(a} + 2ab + 6*). 16. m\m — 1)*, m(m' — 1), m\fn}n — w). 16. (y + 4)(y2-16),y«-y-20,ajy + 4aj-ay-4a. ADDITION AND SUBTRACTION OF FRACTIONS 68. The process is the same as in arithmetic. If the frac- tions do not have the same denominator, they must be reduced to fractions which do have the same denominator. The lowest common denominator (1. c. d.) is obtained by finding the lowest common muLtiplk of the given denominators. 1. Perform the indicated addition and subtraction : 1 — X . a-{-x 5a — X ' "^ — • 2 ax 3 aa^ 6 a*x ELEMENTARY DEFINITIONS AND OPERATIONS 45 The 1. c. d. is 6 a^^. The redaction of the fractiomi to the 1. c. d. is effected as follows : 1 — « Sax 3 ax — 3 ax^ 6a^^-i-2ax =3 ox, 2 ax Sax 6 a^^ 6a%c«-^3ax»=p2a, a+| . 2a^ 2a^ + 2ax , Sax^ 2 a QaH^ 6 a^x^ -i- 6 a^x = «, 5a — X X _ 6 ox — x^ 6a*e * x"~ 6a2x2 We obtain ^~^ 4. <^^ + ^ _ 5a — x _ 3 ax— 3 ax^H-2 a'^H-2 ax— 5 ax-\-oi^ 2ax 3ax2 Ca^x 60^x2 ^2a«-3e^^ + x^ ^ns. In practice, much of the work can be done mentally and need not be written down. 2. Perform the indicated subtractions : ^ 1 J^ a-b 2a 46* The 1. c. d. is 4 ab(a — b) . Reduce the fractions to the 1. c. d. : 1 4ab 4ab 4 a6(a — 6) + (a — 6) = 4 ab, a — b 4a6 4a6(a— fr) 4a6(a-6)^2a =26(a-6), -1 . 2^(<»- ^) = 2a6 - 2 6^ ^ ^ '^ ^' 2a 26((r-6) 4a6(a-6) 4a6(a-6)-f.46 =a(a--b\ i- • <»(«-^) = <»" " «^ . ^ ^ "V« >'» 4^ a(a-6) 4a6(a-&) Subtracting and simplifying, ^^^t.^^^7,^\ -4n». 4 ao(a — 0) 3. Perform the indicated additions : a* — 1 aj(aj + 1) 1 — a? The lowest common denominator is x(x^ — 1). Before reducing the last fraction to one with the denominator x(x2 - 1), it is a convenience to be able to write x - 1 in place of 1 — x. As 1 — x differs from x — 1 only 46 ELEMENTARY ALGEBRA in algebraic sign, we can do this, provided we change also the sign of the fraction or else change the sign of the numerator. The former change is simpler. Thus we have, 3^2 1 a;2 - 1 x(x + l) x-1 The reduction to the lowest common denominator, x(x* — 1), is effected as follows : x(x2-l)+a;(a;+l)=a;- 1, «« - 1 z »(aJ« - 1) 2 x—l 2a; — 2 x{z+l) « — 1 a(a:« — 1) x(a«-l)-i.(x-l) =x(« + l), — ^ • x(x_±lX__3i^±x_^ ^ ^ ^ ^ ^ ^' x^l xix+1) «(a;2_i) Weobtain-i- + — ^ L- = 3xH- 2x- 2- x«- x x2-l^a;(a+l) x-1 x(a^-l) 4 X - 2 - x« .^. = x(x«-l) - ^'^^ 4. Perform the indicated addition and subtraction : 1 1 + 1 (a - 6)(c - a) (a - 6)(c - 6) (c - 6)(a - c) Here the factor (a — b) occurs twice, and both times with the same order of the letters. The same is true of (c^b). But in the first de- nominator occurs the factor (c — a), in the last denominator occurs {a—c). It is a convenience to have (c — a) in both denominators. Since (a — c) differs from (c — a) only in sign, we write in the last denominator (c — a) and at the same time change the sign of the fraction. We obtain (a-ft)(c_a) (o-&)(c-6) (c-6)(c-a) The 1. c. d. = (a — 6)(c — a)(c — 6). Reducing to a common denom- inator and adding, e — b i = 0, Ana, ELEMENTARY DEFINITIONS AND OPERATIONS 47 BXEBCISBS 69. Perform the indicated operations : , 2x.^x . 6,6 1- -TT^—r* 4. - + 3 4 a?-l aj + l 2. • 5. y X X — y X — 2/ 3. JL 8_. 6. ^^^ + 5i^. aj + 1 x + 1 a + b a—^b „ a — b , b — c , c — a 7. 1 -f- • ab be ac (y -»)(»-») («-»)(» -y) (» - 2/)(2/ - «) g __3 7 4- 20a ' l-2a l + 2a l-4a«* 10. -^+ 1 1 11. a + 6 (a + 6)» a« — 6« 11 .V 2(a) -y) 2(3! +y) as'-y' 12. -J- + ^ll ?-+ 1 m+w m + 3n m — n m — 3n j„ g 6 , a' + ft' — oft a* - 62 a* + 62 ■ ft4 _ ^4 (a _^ ft)(a2 _|. 52) 14. ?-t^ + 1±1 + ^+^ {x-yXx-z) (y-^z){y-x) (z'^x)(z-y) io« — — -f- (a-bXa-c) (6-.c)(6-.a) (a-cXc-b) 48 ELEMENTARY ALGEBRA ,^ a*4-2a-f4 a*-2a4-4,4-.a» iv. — — -f- — * a + 2 2-a a-2 17. l^+.,-l^- 1^+ 1 4a + 4 4-4a 12-12a«3-3a 18. -^-i—.-^-J— ^+ 1 a* — a; — 6 aj^+6a5 + 8 12 — as — a^ 19. a;-.y-g ^ y-^g-,a; ^ g-g-y (aj-y)(aj-2) (y - z){y - x) (z - x)(z -- y) 20. — PiP-gXp-r) q(q- p) (q - r) pqr 21. :^-f ^ -^^ 2-a a + 3 a» + a-6 22. ^ +-^ nlf. a? + l 1 — a^ 05—1 23. ^ ^ ,+ ^ 24. 2-a:-aj» aj*+3a;+2 1 - a?* 2 _3 4 (a-2)(a-3) (3-a)(a-l) (l-a)(2-'a) 1.1.1 25. -— -H-: — -H- (a — 6)(6 — c) (c — a){b — a) (a— c)(6 — c) QUADRATIC EQUATIONS 70. The equation ao* + 6a5 + c = is a complete gt^odrottc equation. It is a quadratic equation because the highest power of the unknown x is the second ; it is complete because it con- tains a term involving the unknown x to the first power and a term c (called the ahsolvie term) which is free from x. ELEMENTARY DEFINITIONS AND OPERATIONS 49 Quadratic equations of tlie forms are called incomplete quadratic equationSj because either the term involving x to the first power or the absolute term c is absent. 71. Incomplete quadratic equations are easily solved. Take the form »* = c. Extracting the square root of both sides of the equation, x = ± Vc. The form ic* + 6a; = is solved by factoring, but may be solved also by the method of " completing the square," to be explained later. Factoring, we obtain, x(x + 6) = 0, Make the first factor equal to zero, x = 0. Make the second factor equal to zero, x + & = 0. x=— h, 72. The solution of quadratic equations leads to two values of the unknown quantity. As both of these values are usually of interest and importance in the solution of problems, it is customary, in the extraction of square roots, to write down both results, the principal value and also the second value. This is indicated by the use of the symbol ± . Thus, in " «= ± Vc," + Vc is the principal root, — Vc is the second root. Since, in finding a? = ± Vc, the square root of both sides of the equation has been extracted, it might be claimed that the sign ± should be written on both sides, giving ± aj=:^ Vc. But this result is the same as when we write a? = ± Vc. For, the equation ± « = ± Vc means here (1) +a;= + \/c. (8) ■\-x= — y/c, (2) -x = ->/c. (4) — a;=H->/c. Of these four sets, the first two are the same, and the last two are the same. Hence, %-=i±y/c gives all the values of x. E 50 ELEMENTARY ALGEBRA 73. A complete quadratic equation may be solved in three ways : (1) Bj factoring, (2) By completing the square, (3) By substitution in a formula obtained by the method of completing the square. 1. With our present knowledge of factoring, the first method is applicable only when the roots of the quadratic equation are ra- tional. The method depends upon the principle that, if the prod- uct of two or more factors equals zero, one factor must equal zero. Solve 3iB« = 2 + 5a;. Transpose all the terms to the first side, Sar^ — 5a; — 2 = 0. Factor, (3 a; + 1) (x - 2) = 0. Place the first factor equal to zero, 8 x + 1 = 0. x = -J. Place the second factor equal to zero, a; ^ 2 = 0. a; = 2. The required roots are 2, — |. II. The second method depends on the type form of a tri- nomial which is a perfect square : a^ ±2ah-\- ¥. If the b^ is lacking, we may take the square root of the first term, double it, divide the middle term by this, and square the quotient. Take, for illustration, x^ ± 6x, To complete the square, we take the principal square root of x^, which is x ; double it, 2 x ; divide 5 x by 2 x, { ; square j, ^. Hence x^±6x + ^\aa, perfect square. This results in the following rule for the solution of a quad- ratic equation : > 1. Transpose all terms containing a^ and x to the left side of the equation; all others to the right side. 2. Dimde both sides by the coefficient of a?. 3. Add to both sides the square of half the coefficient of x. 4. Extras the square root of both sides and solve the resulting equations. ELEMENTARY DEFINITIONS AND OPERATIONS 51 Solve 3aj«-5aj-2 = 0. Transpose, Sx^ — bx = 2. Divide by 3, a;2 _ j j. - |. Add(J.})2, x2-4a; + jj = Take the square root, x — f =; d: J. Whence, « = 4 ± {. a = 2, - }. ^n<., 2, - J. III. The third method depends upon the formula derived from the solution of the type form of the complete quadratic. Solve aa^ + bx^ c=:0. Transpose c, ox^ H- 6a; = — c. ^' . , (^^ Divide by a, ^ '^^ *aiL^__c (2 'a) Addfl.^V » a - a o 4a^ 4a2 /', Take the square root, » + ;^ = ± ^^^ - 4 gc 2a 2a - »:fc yy - 4 oc • 2a In using the formula, transpose all terms of the given equa- tion to the left side. Why ? Solve 3 aj* - 5aj = 2. Transpose the 2, 3 a;8 -. 5 » — 2 = 0. Here a = 8, 6 = — 5, c = — 2. .^^^6±v^6TM^i±7^2or-i/ 6 6* BXBBCISES 74. Solve ; if a numerical equation has irrational roots, aiy- proximate their values, to three decimal places. Use the table of square roots in § 197. 1. x^ + 4a; = 6. 3. «« — 8a: = — 11. 2. 3iB«4.aj-14 = 0. 4. a?»-2a;-15 = 0. 52 ELEMENTARY ALGEBRA 5. aj2+6aj = -9. 10. «* — 2 ooj = 6» - a*. 6. aj* + aj4-l = 0. n. m»aj« - 2 ma? = n - m^. 7. x^ — ax=c. „u 12. aa? -f- -^^ = (6 + c)a;«. 8. 6a?» — caj=:d. 6 + c 9. 3aj»-h»4-7 = 0. 13. (3 a: - 5)» = 4 ic 14. 12«»4-13aj-35 = 0. 16. -i^+ 1 1 05 — 1 05 — 2 a?— 3 16. (a: + 2)(aj-2)=7(aj-f 2)-6. 17. (ic-f 3)2-2(a;4-3)H-l = 0. 18. 5aj(aj-3)-2(aj2-6) = (a;H-3)(a; + 4). 19. l«^^±2 ^?^5 ^, (<^'-f)(y;-^^) = 2y. 2a; + l a;-3 c* 4- cP ^ (a? 4- m)* (a; — 7i)* 2y — a+2c 23. (aj-l)(a:+2)(aj* + aj-l)=0. 24. (a?*-7a:+12)(ar^+3a;+2)=0. PROBLEMS Ascertain in each problem whether both roots of the quadratic equation are applicable to the problem. 75. 1. Find two consecutive numbers whose product is 992. 2. Find the length and breadth of a rectangle whose area is 375 sq. in., and whose length exceeds its breadth by 10 in. 3. Express 71 as the sum of two numbers whose product is 448. 4. If the length of a square be increased by 2, and the width be increased by 3, the area of the resulting rectangle is 40. Determine the length of a side of the square. ELEMENTARY DEFINITIONS AND OPERATIONS 53 5. The height of a triangle exceeds its base by 8 ; if the area of the triangle is 1209, what is its base ? 6. The diagonal of a square is 2 ft. longer than the side. Find the side. 7. A cylinder 12 ft. in height has a capacity 125 cu. ft. Determine the diameter of its base. 8. When the edges of a cube are each increased by 6 in., the volume is increased by 936 cu. in. Find the dimensions of the original cube. 9. The sum of the numerator and denominator of a fraction is 77. If the numerator is increased by 111 and the denomi- nator is increased by 40, the fraction is doubled. Find the fraction. 10. Find two numbers which differ by 2, the cubes of which differ by 296. 11. The radius of one circle is twice the radius of another. Find the radii of both, if the difference of their areas is 75. 12. The difference of the volumes of two spheres is 100 cu. in. ; the difference of their radii is 5 in. Find their radii, correct to two decimal places. 4irr8 The volume of a sphere is 3 13. A woman paid $ 64 for silk. If she had bought 4 yards less for the same money, she would have paid $ 1 J more per yard. How many yards did she buy ? 14. The longer leg of a right triangle, exceeds the shorter leg by 3 ft. The area of the triangle is 135 sq. ft. Find the length of each leg. 15. A bookdealer sells a number of algebras for $ 87. Had he reduced the price of each book by 12 ^, he would have sold 16 more books for the same sum of money. How many books would he have sold at the reduced rate ? 54 ELEMENTARY ALGEBRA 16. If a man's daily wage had been $ 1 less, it would have taken him 15 days longer to earn $ 180. How many days did he work to earn $ 180 ? 17. A picture 10" X 14" is placed in a frame of uniform width. If the area of the frame is equal to half the area of the picture, how wide is the frame ? Draw a figure. 18. Find two consecutive even numbers whose product is 528. 19. Find two consecutive odd numbers whose product is 8099. 20. If the difference between the parallel sides of a trape- zoid is 5 ft., and, the altitude of the trapezoid is equal to the longer of the parallel sides, find the lengths of the parallel sides when the area is 2375 sq. ft. 21. A flower bed is 15' x 20'. How wide a walk, must sur- round the bed, to increase the total area by 770 sq. ft. ? 22. A tinner makes a square box 3 in. deep, with a capacity of 1587 cu. in. From each comer of a square sheet of tin a 3-inch square is cut and the four rectangular parts of the tin are turned up. What are the dimensions of the square sheet of tin? Draw a figure. 23. If a square has its length reduced by 7 in. and its width by 10 in., what are the linear dimensions of the resulting rec- tangle, if its area is 8370 sq. in. ? 24. An oil tank can be filled by one pipe in 2 hours less time than by another pipe. If both pipes are open 1^ hours, the tank will be filled. In what time can the tank be filled by each pipe ? 25. A number of postage stamps can be arranged in a rec- tangle, each side containing 60 stamps. If the same number ELEMENTARY DEFINITIONS AND OPERATIONS 55 of stamps be arranged in two rectangles so that each side of one rectangle will contain 12 more stamps than each side of the other, how many stamps does a side of each of the latter rectangles contain ? 26. A boat's crew can row at the rate of 9 miles an hour. What is the speed of the current in the river if it takes them 2 hours and 15 minutes to row 9 miles up stream and back ? 27. Divide $ 1248 among three persons, so that the second shall have $ 3 more than the first, and the third shall have as many times the share of the second as there are dimes in the first person's share. 28. The population of a city increases from 20,000 to 20,808 in two years. What is the annual rate of increase per hundred ? 29. A sum of $ 2000 drawing interest that is compounded annually, amounts to $ 2142.45 in two years. Find the rate of interest. CHAPTER II MORS ADVANCED THEORY AND OPERATIONS FUNDAMENTAL LAWS OF ALGEBRA 76. The operations of algebra obey certain fundamental laws which we have not formulated thus far. Nevertheless we have so accustomed ourselves to follow them, that we find it difficult to see how a new algebra might be made, in which a different set of laws would prevail. We shall now explain the laws which underlie our algebra. If several positive and negative numbers are added or sub- tracted, it matters not in what order the operations are per- formed ; the numbers may be commuted at pleasure. For example, 4 + 7 — 6 = 6, or 7 + 4-6 = 6, or -6 + 4 + 7 = 6. This is called the commutative law for addition.* Using let- ters, the law may be stated thus, a + b =:b + a. * In our algebra, addition and subtraction may be represented geometrically by the addition and subtraction of distances along a straight line. Let a and b represent distances measured off toward the right, then a + 6 and 6 + a both repre- — ^ ,.- ^ ^ ^ sent the same distance OC in Fig. 3; the Q I i C commutative law is obeyed. \y a Suppose, now, that a and b are assigned Fia. 3. meanings entirely different from those given above; suppose a means a rotation about the line OA as an axis, through 90 degrees, and b means a rotation about OB as an axis, through 90 degrees. 56 MORE ADVANCED THEORY AND OPERATIONS 57 Again, 5 + — 2 = 12, or (5 + 9)- 2 = 12, or 6-f(9-2)=12. That is, the final result is the same, whether the 9 be asso- ciated with the 5 or with the — 2. This is called the associative law for addition. Using letters, it may be expressed thus, a + b + c =(a -\- b) + c = a+(6 4-c). If an expression contains two or more factors, it matters not in what order the multiplications are performed. For example, 2.7 = 7-2. This is called the commutative law for miUtiplication,* Using letters, the law may be expressed thus : a ' b = b • a. Let a rectangle be the figure rotated (Fig. 4). Then a-\-b (i.e. the rotation about OAy followed by a rotation about OB) brings the rectangle in a position where " Alg/' is horizontal, as in Fig. 5. On the other hand, b-\-a (i.e. a rotation about OB, foUowed by a rotation about OA) brings the rectangle in Fig. 6. a position in which ** Alg." is vertical (Fig. 6) . Since the final positions of the rectangle are different, it follows that in this case, a+b^b+a. That is, the commutative law is not obeyed. (The symbol ^ means " is not equal to.") Whether the commutative law for addition is obeyed or not depends there- fore upon the definitions given to a and 6, and to the processes of addition and subtraction. * There is an advanced algebra, called quaternions, in which (;' =^ji; that is, the commutative law for multiplication does not generally hold true. Quaternions are used in the study of mathematical physics. 68 ELEMENTARY ALGEBRA Again, it matters not how the factors are associated or grouped, for 6 > 6 . 3 = 90. 6. (6. 8)= 90, (6 . 6) . 8 = 90. This is called the associative law for mtUtiplication. Using letters, the law may be expressed thus : a 'b • c=r:a» {b • c)={a • 6) • c. Again, a factor placed before or after a parenthesis contain- ing two or more terms may be distributed among the various terms without any change in the final result. That is, 6(9 - 4 + 6) = 6 X 9 - 6 X 4 + 5 X 6 = 66. This is called the distributive law for multiplication. Using letters, the law may be expressed thus : a(6 + c)==a6 + ac. raSTORICAL NOTE 77. It is a carious fact that, in the development of algebra, the funda- mental laws were the last things to be explained. The beginnings of algebra can be traced back to ab6ut 2000 years before Christ, but not until the nineteenth century were the fundamental laws of algebra formu- lated. In the earlier treatment the laws were tacitly assumed to be true. For instance, it was assumed that ab = hat a{b -\- c) = ab + ac, (ab)c=:a(bc)f without special attention being called to this matter nor special names being given to the relations assumed. The need of an explicit statement of the fundamental laws came to be recognized when it was perceived that, besides the algebra which we are studying, in which, for example, ab is always equal to ba, there could be established other algebras in which ab is not always equal to ba. Among those who helped to perfect the science of algebra along these lines were the Englishmen, George Peacock, D. F. Gregory, Augustus £>e Morgan, and Sir William Rowan Hamilton ; the Frenchmen, F. J. Servois, and A. L. Cauchy ; the Germans, Martin Ohm, Hermann Grassmann, and Hermann Hankel; and the American, Benjamin Peirce. The names "commutative law," " distributive law," were first used by Servois in 1814. Among the first to use the name ** associative law" was Sir William Rowan Hamilton. AnODSTUS DB MOBOUI (1806-lti71) Was professor of matheinaticB In London and wrote papers on Che tounda- tioDS of algebra and on loglo. To as English Journal, the Aihenieum, he con- tributed amaxing articles on circle squarers. These articles were aCterwarde ooUected iu a book, entitled A Budget of Paradozet. « 41 W V • » • w ». ^ C » . V «. k >. V. i. MORE ADVANCED THEORY AND OPERATIONS 59 REMAINDER THEOREM AND FACTOR THEOREM 78. Divide aj* — 5a^ + 7aj — 2bya? — a. X — a x2 +(a - 6)x + (a-* - 6 a + 7) (a - 6)a;2 + 7 x - 2 (a~6)a;'^-(a2-6a)ag (a2 _ 5 a + 7)a; - 2 (q^ - 6a + 7)a; - q8 + Sgg - 7 q a8-6a2 + 7a-2 Observe that the remainder, a' — 5 a^ + 7 a — 2, is the same as the dividend, if in the dividend a is substituted for x. This illustrates the Remainder Theorem : If a rational integral expression in x is divided by x— a, the remainder is the same as the original ex- pression with a substituted for x. Now aj' — 5aj* + 7a? — 2 would be exactly divisible by a? — a, if a» - 5 a2 + 7 a - 2 = 0. Hence such a polynomial will be exactly divisible by a binomial of the form a: — a, if , when a is substituted for x, the polynomial vanishes, i,e. equals zero. This illustrates the Factor Theorem : If a rational integral expression in x becomes zero when a number a is substituted for x, then x — a is a factor of the expression. Thus, the expression a:P -\-2sfi —05—2, fora;=l, becomes 1 + 2—1—2=0, hence x — 1 is a factor of the expression. Again, for a; = — 2, this expression becomes 64 — 64 + 2 — 2 = 0, hence X — (— 2)ora; + 2isa factor of it. A third factor can be found by dividing x' + 2a^ — x — 2by the prod- uct of a; — 1 and a; + 2 ; that is, by x^ + x — 2. The quotient is a:* + x' + a;^ + x+l. Hence we have, a:6 + 2a*-x-2 = (x- l)(x + 2) (xl^ -\- ofi + x^ + x -^ 1). In the example above we took x = 1 and a; = — 2. Notice that 1 and — 2 are both integral factors of the absolute term 60 ELEMENTARY ALGEBRA — 2 in the expression aj'4-2a^ — aj — 2. There is a theorem bearing on this point which we shall use along with the Factor Theorem. It relates to rational integral expressions, a?" + oo;"-! + 6aj— 2 H \' kx + k. The theorem, which we give without proof, is as follows : When a rational integral eacpression has 1 as the coefficient of the highest power ofx, and the otJier coefficients a, 6, "•, h, k are aU integral numbers (either positive or negative) , then, in search- i'f^g for rational va^vss of x that will make the given escpression zero, only integral factors of the absolute term k need he tried. Factor sc* + 6 x« + 42 a; - 49. The coefficient of a:* is 1 ; the other coefficients, 6, 42, — 49 are integers. Hence both conditions are satisfied, and we need to try only integral factors of 49 ; namely, ± 1, ± 7, ± 49. When « = !, 1+6 + 42 — 49 = 0. Hence, by the Factor Theorem, a; — 1 is a factor of the expression. When* a: =— 1, 1 — 6 — 42 — 49 ^ 0; a; + lis noJ a factor. Whenx = 7, 2401 + 2068 + 294 - 49 z^fc 0, a: — 7 is noJ a factor. When X = — 7, 2401 — 2068 — 294 — 49 = 0, a; + 7 is a factor. In the same way, a; — 49 and a; + 49 are found not to be factors. Dividing the expression by (x — l)(x + 7) yields another factor, aj2+7. Hence, a^ + 6x» + 42x- 49= (x- l)(x + 7)(x2 + 7). BXBBOISES 79. 1. Divide ^a^ — 1 a^ -{• ba^ — x — l\yj x — 6, then com- pare the remainder with the dividend. How does this illus- trate the Kemainder Theorem ? 2. Divide 5a!^ — 4aj8 + 6a? + 2bya? — a and illustrate the Remainder Theorem. 3. Using the Factor Theorem, factor aj4 _ 5 aj3 + 9 aj2 - 15 a? + 18. 4. Factor oj^ — 17 oj^ + 3 a? +- 54. * The symbol ^ means * * is not equal to." See § 76, footnote. MORE ADVANCED THEORY AND OPERATIONS 61 6. What is the remainder when 2aj3 — 3a;2 + 5aj— 1 is divided byaj — c? By x — b? By a? — 1? 6. What is the remainder when 5a^-{'2x^-{-Sx + 6 is divided by x -f-w ? By a? + 1 ? 7. Is a? + 2 an exact divisor of aj* — » — 6 ? 8. If a rational integral expression, with 1 as the coefficient of the highest power of a?, and with all other coefficients integers, is exactly divisible by a? — a, where a is an integer, what relation is a to the last term ? 9. Find a factor ofa?*^ — aj*-f-aj3 — aj^-f-aj — 1. 10. Find a factor of aj3 — 9 aj2 + 17 a? — 6. 11. Find a factor of 2 aj* - a^ - 2 aj2 - 3 a? - 10. 12. Find a factor of a^ — 13 a; -f 12. 13. Factor a^ — a**. If a is substituted for x, then a* — a^ = 0. Hence, by the Factor Theorem, x — a is a factor of ar^ — a*. Division of a^ — a^ by x — o yields a second factor. Hence 05*^ — a* = (x — a) (x* + x'a 4- «^a^ 4- ««* + «*). The quotient may be obtained by the following special method of short division : x^-4-x = x*. x*-^x•a = x*a. x^a -^ x • a = x^a^. x^a^ -t- X • a = xa**. xa^ h- x • a = (i*. 14. Factor a;*^ 4- a^ If —a is substituted for x, then (— a)* + a^ = — a* + a^ = 0. Hence by the Factor Theorem, x-^(— a) orx + aisa factor of x^ + a*. By division, we find a second factor. Hence x^ 4- a^ = (x + a) (x* — x^a 4- sc^a^ — xa* 4- a*) • The special rule of division given in Ex. 13 applies to this case, except that the signs in the quotient are alternately 4- and — . 62 ELEMENTARY ALGEBRA 16. Factor a^ + a*. If ± a is substituted for x, we obtain ( i a)® + o* = o* + o* t^ 0. Hence, neither a; + a nor a; — a is a factor of x* + o^. But afi + <jfi may be considered as the sum of two cubes (x^)^ + (a^)' and can therefore be factored, as j^ + 0^ is factored. 16. Explain the following results pertaining to rational and real factors : , . , . i. t. u i. j a* -f- 52 cannot be factored. a* 4- &* cannot be factored. a« + 6« = (a2 H- 62) (a* - a^b^ 4- ^O- a' 4- 6' cannot be factored. Note. The sum of the same two even powers of two numbers may be factored, if the exponent is the product of an odd and an even factor. One factor is the sum of the numbers with exponents equal to the even factor ; the other factor may be found by inspection or by long division. Thus, in a^^ + 6^^, the exponent 12 = 8 • 4, where 3 is odd, 4 even. One factor of a^^ + h^^ is a* + 6*. The other factor is a* — a*6* + &». Hence, a^ + h^^= (a* + 6*)(a8 - a*6* + b^). If numerical cubic equations of the form ofi + dx^ + 6x + c = 0, a, 6, c being positive or negative integers, have at least one root rcUional, then the equation can be solved vnth the aid of the Factor Theorem. 0^ + 4 x^ — 2x — 5 becomes zero when — 1 is substituted for x. Hence X 4- 1 is a factor, the quotient x' 4- 3 x — 6 is another factor. Write the cubic equation in the form (x + 1) (x^ + 8 x— 5) = 0.^ Make the first factor equal to zero, x 4- 1 = 0, and X = — 1. Make the second factor equal to zero, os^ + Sx — 6 = 0, Q_i_ "v/29 and solve the quadratic equation, x = — . Hence the roots of the cubic are -1, --3 + \/29 -3---\/29 '22 17. Solve (a? - l)(a;2 4- 6 a; 4- 8) = 0. Solve the following cubic equations : 18. aj»4- 5aj24-7a;4-3 = 0. 20. aj» -6aj2 - 8a? 4- 7 = 0. 19. a;'4-5aj2-3aj-22 = 0. 21. a;» 4- 18 02^32 a? 4- 55=0. MORE ADVANCED THEORY AND OPERATIONS 63 \ 80. Factor: 1. aj2 — 16y2. 2. 25a2-6l 3. 4a^ — 9y2. 4. 05^ + 3^. 6. 7? — j^. 6. aj' - 1. 7. 14-2/*. 8. aa;2 — ayK 9. aj2 _ 100. 10. iB» — 1000. 11. m» 4- 8. 12. m'-8. ORAL REVIEW 13. 27- a'. 14. 27 + a». 16. m^ + n*. 16. m*^ — w^ 17. a* — 6*. 18. a* + &*. 19. 2^ — 1. 20. 2^-y». 21. 05*4-0*. 22. x*—a?. 23. 10 aj' - 10. 24. 10 01? + 10. 25. iB»4-3ar2 4.3aj4. 1. 26. 2/»-3y24-3y-l. 27. a* 4- 2 a^* 4- 3^. 28. 0^4- 82/«. 29. 7? — 21v?. 30. a?* — 162^. 31. 10a^4-20ajy4-102^^ 32. «« — 1. 33. a:«4-l. 34. a?— (2y + a;)'. 36. aj2 — (2y-2)2. 36. 2a* — 2a. WRITTEN REVIEW 81. Factor: 1.. 8 a« 4- hK 2. ci«-di2, 3. ci2 + c2i2. 4. 27 a^- 1252^. 5. 27a^-2^. 6. 8m«4-27n». 7. 27a^-64 2/». 8. m*^ — n^^ 9. a* - h\ 10. c«4-64d«. 11. aj^— (2a?— y)*. 12. aj»+(y + «)». 13. a*+4a^-2a;2-4a;4-l. 14. aj«-4a:*-7aj2 + 28. 16. aj*-3aj» + 7a?-13a?4-6. 16. a?— a^ — 5 a^x 4- 5 a\ 17. m'- 6m* 4- 12m— 8. 18. ajio— yo^ 19. a.** 4- 35*2^* + ^. Hint. X* + ojV 4 y* = x* + 2icV4y*-a:*y*. 20. a*4-a' + l. 21. 4a?*4-3a;y 4-92^. 22. a* — 7 a262 4- 64. 23. 49p*H-68p2g*4-369*. 24. 64aj* + 55a^.y2 4-252^. 64 ELEMENTARY ALGEBRA IRRATIONAL AND IMAGINARY FACTORS 82. Thus far all factoring has been confined to the discovery of rationed factors. But it is sometimes advantageous to re- solve expressions into factors which involve irrational or even imaginary terin?»>^ Expressions like a' — 2 6*, a* — 3 b*, a? + 6^, or a* + 6* can be factored, if the restriction is removed that the factors must be real and rational. BXBROISBS 83. 1. Factor a» — 3. This may be done by the type form a^— 62=:(a -f- 6)(a — 6), where 62 = 3 and h = VS. We obtain a2-3=(a + V3)(a-V8). 2. Factor aj2 -f y\ This, too, may be factored by the type fonn a* — 6^ =(a + 6)(a — 6). Write «2 f y2 in the form x^-{-y^). Since V- y2 := yy/ZTi = fy, where % = V— 1, we have, x^ + y^^(x + %y){x-'iy). Kesolve into irrational or imaginary factors : 3. m«-5. 7. a* + 62. 11. 9a« + 166*. 4. a*- 12. 8. 7^-bsK 12. a*-36^ 5. a2 4.4. 9. 2a2-6». 13. a2— 6. 6. m2-h9n2. 10. aj^ — 8y*. 14. a* — 46*. 16. By factoring, find all four roots of the equation a^-l = 0. 16. By factoring, find all four roots of the equation a^-16 = 0. MORE ADVANCED THEORY AND OPERATIONS 65 THE BINOMIAL THEOREM 84. By multiplication, we find that (a ± by= a±b. (a ± 6)»= a» ± 3a26 + 3a5* ± ¥. (a ± 6/= a^ ± 4a»6 4- ea^fe^ ± 4a6» 4- 6*. (a ± 6)«= a» ± 5 a*6 + 10 a.»62 ± lo a^b" 4- 5 a5* ± 6». (a ± 6)«=:a» ± 6a»6 4- 15 a*^^ ± 20a'&» 4- l^a^b* ± 6ab^ +b^, etc., wherein a and 6 represent the first and second terms, respectively, of any binomial. A careful study of these products will show that they follow certain laws, by which they may be written down without recourse to laborious multiplications. In the products we observe the following laws : I. The first term isTyraiaed to the same power as that of the binomial. In each succeeding term the easponent of a decreases byl. II. The factor b does not appear in the first term. The ex- ponent of b in the second term is 1 and increases by 1 in each succeeding term, III. The coefficient of the first term is 1. The coefficient of any term after the first is found by multiplying the coefficient of the preceding term by the exponent of a in that term^ and divid- ing by 1 more than the exponent ofb, ' IV. If the binomial is a -i- b, the signs in the product are. all plus; if the binomial is a — 6, the signs are alternately 4- and — . V. The number of terms is one more than the exponent of the binomial, VI. Each term is of the same degree as the binomial, p 66 ELEMENTARY ALGEBRA Using these laws, find without actual multiplication the product of (a 4- by. The first term is by I, a', 1x7 The second term has by III the coefficient or 7, by I the factor j cfi^ by II the factor &. Hence the term is +1 <fib. The third term haa by III the coefficient l^i^ or 21, by I the factor a^, by II the factor h\ Hence the term is + 21 a^h^, ' The fourth term has by III the coefficient ^^ ^ ^ or 35, by I the fac- o tor a^, by II the factor &*. Hence the term is + 36 a*&*, and so on. By y the final result has 8 terms. We obtain, (a + by- aT + 7 (i«6 + 21 <fib^ + 86a*6» + 86a«6* + 21 a^lfi + 7 a6« + b\ All terms are of the seventh degree, as required by VI. 85. If n is a positive integer, then the products found above may all be represented by one general formula, as follows : (a + by = a- 4- m-»6 + 5^?^^ a-^fe^ + n{n^l){n^2) ^,.3^, This formula is called the Binomial Theorem. By actual multiplication it was shown above to be true for all positive integral valueg of n, up to 6. We found the expansion of (a + by on the tacit assumption that the formula holds for n = 7. Really we had no right to take for granted, without proof, that the laws given above do hold true for n>6. Frequently certain relations hold true up to a certain point, but no further. For example, the first three integers 1, 2, 3 are all prime numbers. A careless reasoner might be tempted to juntp to the conclusion that all ; integers are prime numbers, which is, of course, not true. That the Binomial Theorem is true for any positive integral value of n will, be proved later in § 193. MORE ADVANCED THEORY AND OPERATIONS 67 Expand (2 a; — 3 yy. Here a = 2 x, b = Sy. For conyenience in expanding, place 2 x and 8^ in parentheses, thus (2 x) and (3y). By Law IV of § 84, the terms in the expansion are alternately +, — . We obtain, (2a;-3y)*=(2a;)*-4(2x)8(8y)+6(2a;)2(3y)2-4(2aj)(3y)»+(3y)*. Simplify each term in the expansion : (2x - 3y)*= 16 X* - 96x8y + 216 xV - 216xy» + 81 y*. EZEBOISES 86. Expand: 1. (x-hyY- 4. (a-Sby. 7. (ox — yY^. 2. (a -6)8. 6. (c2-d2)4^ 8. (3a2 + 2 6»)». 3. (2x + yf. . 6. (3a- 62)6^ 9. (m — 2ny. 10. (2r3-3s2y. 11. Write the first three terms of (a—b)^, 12. Write the first five terms of (x + y)^. 13. Compare the coefficients of the first and last terms in the given expansions in § 84 ; of the terms next to the first and last. 14. What is the third term of {mx -f- ny)^ ? 16. Write the 9th term of (2 a^ - 3 by. SYSTEMS OF LINEAR EQUATIONS. PETERMINANTS 87. Systems of n independent linear equations ir. n xm- knowns may be solved by eliminating tho same unknown from all the equations. The resulting system contains in general one less equation and one less unknown. From this resulting system eliminate a second unknown, and continue this process until one equation containing only one unknown is obtained. 68 ELEMENTARY ALGEBRA BXBBOISES 88. 1. Solve the system of equations « H- y + « = 9, (1) x-h2y^z = S, (2) 2aj-33^-f 42 = 7. (3) SoluHon. Subtract (2) from (1), - y + 2 « = 1. (4) Subtract (8) from 2 . (1), 6 y - 2 = 11. (6) Add (4) and (6), * 4 y = 12, y = 3. Substitute 8 for y, in (4), «; = 2. Substitute 2 for z, and 3 for y, in (1), x = 4. Check by substituting the answers in all three equations (1), (2), and (3). 2. aj-y-h« = 20, S. 5x — 6y-{'Tz = 105, 2aj-3^--« = -30, 3aj-6y + 82; = 103, 3a. — 43^4.22; = 10. 3y — 4« = -44. 4. « -I- y -f 2; 4- to = 10, « — yH-2« — 3«; = — 3, 3aj — 2y + 2— w = 5, 3aj-3y-f-22 — 4m?=— 3. 89. In previous parts, three methods of elimination have been explained : (1) by addition or subtraction, (2) by sub- stitution, (3) by comparison. A fourth method of solution is by means of a formula, called a determinant. In deriving this formula we use the old methods of solution. Solve : ax + by = c (1) dx-{-ey=f. (2) Multiply (1) by d, (2) by a, adx + bdy = cd (3) adx + aey = c^f .. (4) Subtract (3) from (4), (ae — bd)y = c^f— cd ae — bd Similarly, a;=£lzi_St. ae — bd MORE ADVANCED THEORY AND OPERATIONS 69 The numerator af—cd may be written a c d f The term af is the product of a and / which lie on the diagonal from the upper left comer to the lower right corner. The term cd is the product of c and d which lie on the diagonal from the upper right comer to the lower left comer. The second product is subtracted from the first. 3 4 Similarly, 5 6 = 3x6-4x6. The numerator ce — bf may be written The denominator ae — bd may be written Expressions written in this form are called determinants. c b f e kTI a b jXI. d e Hence, a? = c b f e a b d e (5) y = a c d f a b d e (6) The determinants in (5) and (6) can be used conveniently as formulas for solving two linear equations in two unknowns. The determinants can be written down at once, according to the following rules : I. In the two denominators the determinant is the same. It is fortned by writing the coefficients of x and y as they stand in the equations (1) and (2). II. The determinant in the numerator of (p) is formed from the denominator by replacing , the coefficients of x, by the abso- lute terms . / 70 ELEMENTARY ALGEBRA III. The determinant in the numerator of (6) ia formed by b c replacing , the coefficients ofy, by the absolute terms, Example. Solve 3a?-f 4y = 34, 5 a? — 3 y = — 11. The denominator is 3 4 5 -8 The numerator for the value of x is 34 4 -11 -3 3 34 6-11 The numerator for the value of ^ is Hence, 34 4 ■ ^ 84.(-8)-4.(-ll) _ -102-f-44 ^ 3(«3)-4.6 -9-20 flC = -11 -3 8 4 6 -8 -68 -29 = 2, y = 8 34 6 -11 3 4 6 -8 3(^ 11) - 84 . 6 -29 - 33 - 170 - 203 -29 - 29 = 7. EXERCISES 90. Find the values of the following determinants : 1. 2. 3. 9 6 8 7 .9 .2 .5 .1 4. b . 3 • 4 -2 10 6. 9 a b 6. la 86 46 -3 a 7. a» 62 6-1 8. a be ae -6c 9. a-2 6-' Solve, by determinants : 10. 2aj-h32/ =4, 5 oj -♦- 6 y = 7. 11. aia; + 6iy=Ci, a^ 4- 62^ = C2. MORE ADVANCED THEORY AND OPERATIONS 71 12. 2iB — 5y= — 13, 14. x + my = ^m*, 3 oj — 4 y = — 9. oj 4- ny = — w*. 13. aa?-f 6y = a — 6, 16. aa; + 62/ = 2a6, oa? — 6y = a -f 6. 6a? -|- ay = a* + 6*. Solve, by any method : 16. 3a;-f 4y = 7, 20. 3a?-f-4y4-5 = 0, 4aj4-y=s3. 6a? + 7y4-8 = 0. 17. aj-|-2y = 8, 21. aaj + 6y=l, 3 a; + y = 9. ca? -|- dy = 1. 18. 8 a: — y = 34, 22. cx + dy=sa, x + SyssoS. mx -i-ny^b. 19. 3a? = 4y, 23. 2(a?-2) = 3(y + 3), 5aj=6y-4. 3(a?- 2)-|-2 = 6(y +3). 24. (r-l)(a + 2)=(r-3)(«-l)4-8, 4(2r-l)-16(«-2)=20. 25. (6 — a)aj4-(a + % = a'4-ftS 6a?-2ay = 6«-2a*. 26. m — w 33 6 — a, am — ac = 6/1 — 6c. 27. (m 4- n)x -f- (m -h />).y = m -|- n, (?/i + p)a: +(m + n)y = m 4-i>. 91. The method of solution by determinants may be ex- tended to systems containing three or more equations. For example : aa? + 6y + caj = d, (1) ex -hfy + gz = hy (2) kx + ly + rnz = n, (3) 72 ELEMENTARY ALGEBRA Using the addition and subtraction method, we obtain after some prolonged effort, __ €(fm 4- hlc + ngb — cfn — gld — mhb afm + 6/c + kgh —cfk — fjria — meb (4) __ g/im -^^ enc -\- kgd — cftA; — gna — med cj/w* + elc 4- A;(76 — c/A; ^gla — W€6 _ fl^n + el d + ifc^fe — djk — /iZg ~ ne6 afm 4- eZc 4- A;(76 — cfk — gla — meb (5) (6) These numerators and denominators may be written down more con- veniently in the form of 1^ determinants. The six terms of the denomi- nator may be written a b c e f 9 k I m provided that the f ollow- ing rule of expansion is observed (Fig. 7) : (1) Take the positive of the products of the terms marked by the 4- arrows. The three products are 4- <0», 4- e^c, 4- kgb, (2) Take the negative of the products of the terms marked by the — arrows. The three products, with their signs changed, are — c/fc, — gla^ — meb* It follows then that a h c e f 9 k I m = afm -\- elc 4- kgb — cfk — gla — meb. MORE ADVANCED THEORY AND OPERATIONS 73 This process of expansion must be fixed thoroughly in the mind. The numerators of (4), (5), and (6) can be expressed by the determinants in the same way. Three independent linear equations in three unknown quantities can be solved by determinants according to the following rules : I. In the three denominators of the vahies of x, y, and z, the determinant is the same. It is formed by writing the coefficients ofx, y, z, as they stand in equations (1), (2), (3). II. The determinant representing the numerator of (4) is formed from the denominator by replacing the column containing a, e, k, the coefficients ofx, by the absolute terms d, h, n. III. The determinant representing the numerator of (5) is formed from the denominator by replacing the column containing b, f I, the coefficients of y, by the absolute terms d, h, n. « IV. TVie determinant representing the numerator of (6) is formed from the denominator by replacing the column containing c, g, m, the coefficients ofz, by the absolute terms d, h, n. Example, Solve : The denominator is 1 2 3 2x-y-\-z = b, 3x-4y4-22! = 3. 1 1 -1 -4 1 2 = lx(-l)x2+2(-4)xH-3xlxl-lx(-l)x3-lx(-4)xl -2x2xl=-4 The numerator for x is 6 1 5-1 3 -4 1 1 2 =6x(-l)x2+6x(-4)xl+3xlxl-l>^(-l)x3-lx(-4)x6 -2x6xl=-12. 16 1 2 5 1 3 3 2 The numerator for y is =lx6x2 + 2x3xl4-3xlx6-lx 5x3-1x3x1-2x2x6= -8. 74 ELEMENTARY ALGEBRA The numerator otz\& 1 1 6 2-16 8-4 8 = lx(-l)x3+2x(-4)x6+3x6xl-6x(-l)x8-6x(-4)xl -8x2xl = -4. Hence x =(- 12) + (- 4)= 8. y=(-8)+(-4)=2. *=(-4) + (-4)=l. BXBBOI8B8 92. Solve: 1- a + y + » = 9, « - y + 2 = 3, 05 4- y — 2 = 1. 2. 2aj-3y + 2;= — 1, 3a; + 2y — 2 = 4, ^x — y-\- 5 2 = 17. 3. a + 26-c = l, 8a4-464-3c = 29, 3a-56-4c = -ll. 4. wi + n = 3, j!> 4- m = 12, n 4"i> = 6. 6. r + 2 « = 10, 3r-|-4« = 12, 6 « + 6 « = 22. 6. aj + y + « = 2, 3aj — 4y-f-62=41, 9aj-13y=-57. 7. a.-y-» = -^, 2aj — 3y4-42 = l, 4aj4-6y + 2 = 4^. B. aj + y4-2 = a, aj - y + 25 = 6, ic + y — « = c. 9. ax + by = m, c^ -I- cf2 = n, eaj 4-/2 =1>. 10. r4-» + «4-w=sO, 2r + 3a4-4iJ4-5w= — 6, 3r-4a+5^-6u = 2, 47.4.55 — 2« — tt = 2. PROBLEMS 93. 1. A man has 20 coins, all dollars and quarters, amount- ing to % 13.25. How many coins of each denomination has he ? 2. Weights of 13 and 17 lb. are fastened to the ends of a 9-foot pole. Where must the pole be supported in order that the weights will balance ? MORE ADVANCED THEORY AND OPERATIONS 75 3. Two men carry a weight of 300 lb. by means of a pole from which the weight is suspended. One man holds the pole at a distance of 3.7 feet from the weight, the other at a dis- tance of 4.3 feet. How much of the weight does each man carry ? 4. Two unknown weights balance when suspended 11 and 12 inches respectively from the fulcrum. If the weights are interchanged, 23 ounces must be added to the lesser weight to restore the balance. Find the two weights. 6. A farm laborer received $ 2.75 and board for every day he worked, and paid $.75 for board every day he was idle. After GO^days he received $154.50. How many days was he idle? 6. A •man has $12,000 at interest in two investments. From one he receives 5 % interest, from the other 6 % . His income from these investments is $ 645 per year. How is the money divided ? 7. Find the capital and the receipts of a railway company, if its receipts are distributed as follows : (1) Payment of a guaranteed dividend of 4 % on ^ of the capital stock. (2) Payment of a dividend of 3 % on the remaining capital stock, this dividend amounting to $ 162,000. (3) To working expenses and a reserve fund, which absorb 55 % of the entire receipts. 8. The area of a trapezoid is the product of half the sum of the two bases and the altitude. Find the two bases, if the altitude is 12 in., the difference between the bases is 3 in., and the area of the trapezoid is 150 sq. in. 9. How many gallons of each of two liquids, one 25% alcohol, and the other 75 % alcohol, must be taken for a 17- gallon mixture that is 45 % alcohol ? 76 ELEMENTARY ALGEBRA 10. How many ounces of silver, 75 % pure and 85 % pure, must be mixed to give 20 ounces of silver, 83 % pure ? 11. Tin and lead occur in the following ratios : In plumber's solder 1:2; in soft solder 2 : 1 ; in common pewter 4 : 1. How many pounds of common pewter and plumber's solder must be taken to make 10 lb. of soft solder ? DIVISION BY ZERO IMPOSSIBLE 94. To avoid perplexing errors in algebraic transformations it should be made very clear that it is not permissible to divide by zero. Let a and h be two fixed numbers. Both a and b^ie finite numbers, because in elementary algebra there are no fixed num- bers that are infinite. , a CL * * Letting - = a;, we define the quotient - as the value of x in b b the equation bx = a, I. As long as a and b are fixed numbers different from zero, it is always possible to find a value for x. II. If a is not zero, but b is zero, we are asked to find a number x which, multiplied by zero, gives a as a product. But every fixed number, when multiplied by zero, gives the product zero. Hence no number x exists which satisfies the equation •x=a. If, therefore, we perform operations which amount to divi- sion by zero, we must not be surprised if we obtain absurd results. Point out the error in the following : Let x = c. Multiply both sides oix = chy x^ x^ = ex. Subtract c^ from both sides, x^ — c^ = cx— c^. Factor, (x + c) (ic — c) = c(x — c). . Divide by (x — c), x-\- c = c. Since x = c, 2 c= c. Divide both sides by c, 2 = 1. MORE ADVANCED THEORY AND OPERATIONS 77 EQUATIONS WITH FRACTIONS 95. When an equation involves fractions we usually begin its solution by clearing it of fractions. Tbis operation rests on the principle that when equals are multiplied by equals, the results are equal. 1. Solve . -^ n = -^?^. m + n m — n Observe that this equation is meaningless and absurd when m ^n , for in that case it involves a division by zero. For the same reason we cannot have m:=^ n^ for then m + n = 0. Let us assume that neither denominator is zero. Then the 1. c. d. = (m + n) {m — n). Multiply both sides by (m + n)(riii — n). We obtain, (m — n)x — n(m^ — n^) z= m (m + n). Transpose, (m — n)x = n(m^ ^ n^)-\- m(m -\'n). Divide both sides by (m - n) , x = n(m2~ n2)+ m (m -f n) ^ m — n 2. Solve _^ + .^ZLi=:^. (1) Observe that if jc = 1, or x^ = 1, this equation represents an absurdity, since it involves division by zero. On the supposition that x^ =^ 1, the 1. c. d. = 4{x^ — 1). Multiply both sides by 4(x2-l), 4x(x + l) + 4(x-3) = 6(x2-l). (2) Simplify, 4x« + 4x + 4x— 12=5x2-5. Transpose, x* — 8 x + 7 = 0. Factor, (x — 7) (x - 1) = 0. Hence, x = 7, x = 1. Substitute in the original equation (1) ; we find x = 7 satisfies it, but X = 1 does not satisfy it, since it gives rise to division by zero. We must reject x = 1 ; x = 7 is the only correct solution of (1). It is of interest to notice that, while x=l does not satisfy (1), it does satisfy (2). In other words, equation (2) is satisfied by two values of a, while equation (1) is satisfied by only one value of a?. It is clear that the new root was introduced in the act of multiplying both sides of (1) by 4:(x^ — 1). Such new roota 78 ELEMENTARY ALGEBRA which do not satisfy the original equation, but do satisfy an equation that is derived from it, are called extraiieoua roots. When we multiply both sides of an equation by an expres- sion involving x, we obtain a new equation some of whose roots may not satisfy the original equation and are not correct solu- tions of it ; they are extraneous roots. This may happen even though all operations are performed free of error. Hence, to make sure that our values of x are correct, we must mihstitute the values found for x in the original equaJtioUy to see whether it is satisfied or not. BXBBOI8B8 96. Solve and check : 2 1 .43 1. oj— 1 aj-3 7« + 3 6«4-2 2. 5+4 = ^ + 6. 6. ^ = ^. y- y x-\-2 x + b 8 3 ^ 1 ^ 3m-l _ 3m-5 2r + l r-1 * 3m + 4 3m-|-2 1 1 7. {x - l){x - 4) (oj - 2){x - 6) 8. 4-A__=7-.?i^. y4-l y-3 10. (m* — 4)y = m ~ 2. m + n x 1 16. ^4.5^=3. 11. - = - a a c -„ ex 1 16. — - — = — ^2- J="^' Ax-^S 6aj-l 18. ?-l=a. 17. -^+ ^ '"^ 0? — 2 a: — 3 a — 4 MORE ADVANCED THEORY AND OPERATIONS 79 5 4 1 18. 19. 05— 5 a; — 4 05 — 3 1111 20. a?— 1 x—3 x — 5 x — 7 _3 1 ^ 5r-H5 r4-3 3-r r*-9 Solve and tell, by substitution in the original equation, which of the roots, if any, are extraneous. ' 21 ^_+^_ = S. x-1 a!-2 2 05 + 6 « — 7 (a! + 6)(a! — 7) 23. .i'^r^ =a;+3. 6a!« — 9a;4-6 a!«-9 26. 3ir^^^y_ y 2y-l 26. 2« 8 -3 x + 1 2a!«-a!-3 2a!-3 27. 3 _ -a!» + 34a!-31 2a;-3 3a;4-4 6aj2-.aj-12 oo 1 7. 2 6 28. r — 6 = y — 5 ^2 _ 52 80 ELEMENTARY ALGEBRA m MISCELLANEOUS PRACTICAL PROBLEMS 97. Applications of algebra, such as occar in the problems which follow, are frequently made by builders and manufacturers. Certain formulas which have been established by engineers and mathematicians are used in computing the desired results. We shall assume the formulas as true without giving the mode of deriving them. Care must be taken to use in every problem the proper units of measure. 1. Find correctly, to two decimal places, the side of a square whose area is 35.06 sq. in. 2. Compute the altitude of an equilateral triangle whose sides are a inches. 3. If one side of a rectangle is a inches long, and one of the adjacent sides is three times as long, find the length of the diagonal. 4. A ladder of the length 9 a is placed against a wall, with its foot at a distance 4 a from the wall. How high above the ground is the top of the ladder ? 6. In placing blackboards in schoolrooms, architects deter- mine the height of the chalk rail above the floor by the formula A = 26-hf (^-4), ' where h = height of the chalk rail in inches, g = the number of the grade to which the pupils in the room belong. Find the height of the chalk rail for pupils in the 5th grade ; also for pupils in the 8th grade and in the 3d grade. 6. The Baldwin Locomotive Works in Philadelphia have derived a formula, i2 = 3-|-^, for finding the approximate relation between the resistance, B, offered by a railway train, and the velocity, the train traveling on a straight and level track. In this formula B is the resistance in pounds per ton weight of the train, v is the velocity of the train in miles per hour. What is the resistance per ton of a train running at the MORE ADVANCED THEORY AND OPERATIONS 81 rate of 40 mi. an hour ? What is the resistance of this train, if it weighs 100,000 T. ? 7. The load that may be safely applied to an iron chain is given by the formula, I = 7.11 d^, where I is the load in tons (2000 lb.), and d is the diameter of the iron chain in inches. Find the largest safe load that may be applied to a chain for which (f is f in. Find d when Z = 13 T. 8. The distance which one can see from an elevation at the sea has been found to be cZ = 1.23 V^, where d is the number of miles one can see, and h is the elevation of the observer in feet. How far is the horizon from a man standing at the sea- shore, whose eye is 6 ft. from the ground ? • 9. In Ex. 8, how far can one see from a cliff 27 ft. high ? 67 ft. high? How high must a cliff be to afford a view of 20 mi. ? 10. From a cliff I can just see the lights of a seaport 12 mi. across the sea. If these lights are 20 ft. above the sea, what is the height of my eye ? 11. According to a rule sometimes used by architects, the "rise" (r inches) in the steps of a stair- case (Fig. 8) is connected with the " tread " [| T (t inches), by the formula r = ^ (24 — *), I where r and t are also subject to the limita- ' l \ u '* tions, 7 > r > 5|, 12 > « > 10. Explain the fw 8 formula in words. What is the " rise " when the tread is 10^' ? Hf"? 12. Engineers have determined that the velocity (v feet per minute) of a stream at the bottom of a river is connected with the velocity ( V feet per minute) at the surface, by the formula V = F+ 1 — 2VF. Find the velocity at the bottom when the velocity at the surface is 6 ft. per minute, 8 ft. per minute, 95 ft. per minute. G 82 ELEMENTARY ALGEBRA 13. A rectangular beam (Fig. 9) is supported at the ends and loaded in the middle. The weight it will just bear, with- out breaking, is w; = , where w is the weight in pounds, b the breadth of the beam in inches, d -p Q is the depth of the beam in inches, I the length of the beam in feet, and A: a coefficient equal to 3470 for wrought iron, 2540 for cast iron, 450 for red pine. Find the greatest weight which can be hung in the middle of a wrought-iron bar 5 ft. long, 1 in, wide, and 2 in. deep. 14. In £x. 13, find w when the beam is cast iron, also when it is red pine. 16. In Ex. 13, how deep must a beam of red pine be, to sup- port 2 tons, if it is 6 ft. long and 3 in. wide ? 16. At a curve in a railway track the outer rail is raised above the inner rail by an amount indicated by the formula h = , where h is the number of inches that the outer rail 1.25 r' is elevated above the inner, w is the width between the rails in feet, V is the greatest speed of a train in miles per hour, r is the radius of the curve in feet. If r = 2000 ft., w = 56^ in., V = 60 mi. per hour, what is ^ ? 17. The safe load which can be applied to a rope is i = ^iC*, its breaking load is 6 = k^c^ where I and b are measured in tons, and c (the circumference of the rope) in inches. For common hemp ropes ki = .032, Ajj = .18 ; for best hemp rope, ki = .100, k2 = .60 ; for steel-wire rope, ki = .450, Aij = 2.8. Cal- culate the safe load (a) for a common hemp rope 3.5 in. in circumference, (6) for a steel-wire rope 2.5 in. in circumference, (c) for a best hemp rope 4 in. in circumference. Calculate the breaking load for each of these. MORE ADVANCED THEORY AND OPERATIONS 83 18. Draw a graph showing the safe load of steel-wire ropes for different circumferences, up to 4 inches. 19. The breaking load of the best hemp rope is given by the formula, 6 = .60 c*, where b is the breaking load in tons, c is the circumference of the rope in inches. Draw a graph showing the breaking load from c = .3 in. to c = 4 in. From the graph determine the breaking load for three values of c not used in constructing the graph. 20. In the flow of water through pipes, a certain head is necessary merely to overcome the friction of the water against the pipes. This head is given by the expression, 1/ 0043 \ s* hss-l .0036 4- )— - , where a is the speed of the water in A -y/s /32 feet per second, I is the length of the pipe in feet, d is the diameter of the pipe in inches, h is the head of the water in feet. Find the head required to overcome the friction in a 3-inch pipe, 3000 feet long, in which water is running at the rate of 3 feet per second. 21. A house is to rest on piles as a foundation. In the erection of it one must know how much each pile can support. This is computed from the performance of the " pile driver," a machine that lifts a heavy " ram " and drops it on the pile. The max- imum load Z, in tons, that a pile will bear is obtained from the formula, Z= ^ ^ — -, where w is the weight of the ram in cwt., h is the height in feet from which the ram falls, d is the dis- tance in inches that the pile was driven in by the last blow, p is the weight of the pile in cwt. If lo = 6, A = 4, p = 15, d = 1^, what is the maximum load the pile will safely bear ? 22. A steam engine is operated by steam which enters a cylinder C (Fig. 10) and presses against one side of the piston P and moves it ; the motion is transmitted by the piston rod B to the crank shaft i^i, which turns the wheel around. The rate 84 ELEMENTARY ALGEBRA at which the engine does mechanical work depends upon the size of the cylinder, the pressure of the steam, and the number of revolutions per minute. The rate of doing work is measured in horse powers, H. If d is the diameter of the cylinder in / V Fig. 10. inches, I the length of the piston stroke in feet, n the number of revolutions of the wheel per minute, and p the mean steam pressure per square inch, then ^^ ^JL^. ' If w= 22, d = 55", I = 61", p = 20 lb., find H. If n = 25, p = 15 lb., I = ^ ft., what must d be, that the horse power may be 30 ? 23. The horse power required to move a ship is given approx- imately by the expression H = .0088 s^ (.05 A -h .005 S), where s is the speed in knots, A is the immersed cross section of the ship in square feet, S is the wetted surface of the ship in square feet. A and S remaining the same, state the nature of the variation of ^ as a function of s. What horse power is needed to move a steamer at 12 knots, when A = 230 and fi' = 3100? CHAPTER III PROPORTION, VARIATION, FUNCTION PROPORTION 98. A proportion expresses the equality of two fractions. The fractions are sometimes called ratios. Thus, — = £ is a proportion. It is written also a\h — cid. b d In a/b = c/dy the terms a and d are called the extremes, the terms b and c the means. In the ratio a/b, a is sometimes called the antecedent, b the conseqiient, but ordinarily there is no need of these terms. A mean proportional between two numbers a and b is the number m which satisfies the equation — = ^ . m b Since a proportion is really an equation, it is treated like an equation. Thus, in the proportion, ? = ^. b d 1. Multiply both sides by bd, ad = be. This relation is expressed by the theorem : In a proportion, the product of the means is equal to the product of the extremes, 2. When only three terms of a proportion are known numbers, the fourth tenn may be found. Thus, if a, b, and c are known, we obtain from ^ = -, b d ad= be and d = -. a 3. Add 1 to both sides of the equation, - + 1 = - -h 1, b d Whence, the new proportion, 85 a-\-b c + d b d 86 ELEMENTARY ALGEBRA BXBBOI8BS 99. 1. Prove that if the product of two numbers, e/, is equal to the product of two other numbers, gh, one pair may be taken as the means, the other pair as the extremes, of a proportion. Prove that, if - = -, then b d 2. a — 6 c — d b d 7. c d 3. a — b c — d a + b c4-d 8. 6* d* A a c 9. a» c» 'M» a -\-b c4-d 6» (f 6. a c 10. a* c" a — b c — d 6" d" 6. a 4- b _c + d a^b c—d 11. Simplify the following ratios by writing each in the form of a fraction and reducing the fraction to its lowest terms : 12. 126:675. 16. (a - 6)* : a* - 6». 13. 69 a»c» : 46 a*6c. 16. (8m« + n») :16(2m4-w). 14. a*-6*:(a + 6)c. 17. 4(p^ - 8 g*) : (p - 2 g). 18. (a* -h a*6* + 6*) : (a« - a5 + &«). 19. (aj» — ay — 2i^=):3(a; — 2y). 22. (aJ«-2ary4-y*}:(«' + »y-l-3/*). 23. (a?* + 1) : (ic* - a? V2 -f 1). 1 PROPORTION, VARIATION, FUNCTION 87 24. Find the mean proportional between 31.2 and 0.96. 26. Find the mean proportional between 3 a^c^ and 7 c^d^. 26. Determine which is the greater of the ratios 19 : 25, or 66 : 74. 27. Two numbers are in the ratio of 3 : 4, and their difference is to their product as 1 is to 18. Find the numbers. 28. Find the last term of a proportion whose first three terms are 27, 6, and 18. 29. Divide $ 644 among A, B, and C, so that A's share is to B's in the ratio of 2 to 3, and B's share is to C's in the ratio of 6 to 7. 30. The income of an estate should be divided between A and B in the ratio of 6 ; 4 ; $ 1100, however, is paid to A, and $ 700 to B. Which has been unjierpaid, and by how much ? FUNCTIONS 100. The formula 8 = 16.1 1^ is used in finding the distance (space) through which a body- falls from rest, in time t, where the time is measured in seconds and the distance in feet. During the fall, the time t and the distance 8 both increase ; t and 8 are variable8y while the number 16.1, of course, is con- 8tant. For every value of t it is possible to compute », 8 being de- pendent for its value upon the value of ^ ; 8 is said to be a function of t A variable i8 a function of another variable when its value i8 determined for every value of the other variable. We have used variables x and y in the drawing of graphs. In the equation of the straight line, y = 3 a? -|- 1, a; and y are variables ; 3 and 1, of course, are constants ; y is a function of oj, since for every value of x there exists a definite value of y. 88 ELEMENTARY ALGEBRA ORAL EXERCISES 101. 1. The velocity of a body falling from rest is given by the formula, ^ ^ ^^.2 t, where v is the velocity in feet per second and t is the time of fall in seconds. Which are the variables ? Which is the constant ? Why is V a function of ^ ? 2. The area of a circle is given by the formula Why is u4 a function of r ? Is tt a constant or a variable ? 3. If a man's monthly salary is % 125, his earnings may be expressed by the equation, E = 125 w, where n is the number of months during which he draws his salary, and E the earnings expressed in dollars. In this formula explain " constant," " variable," and " func- tion." 4. The area of a rectangle is given by the formula A = hh, where h and h are the " base " and " height " in linear units, and A is the " area " in the corresponding square units. Here A varies when h varies and also when h varies. That is, A is a function of the two variablesy b and h. Find A, when 6=2, ^ = 3 ; also when 6 = 4, ^ = 2. DIFFERENT MODES OF VARIATION 102. When one variable is a function of another variable, the variation takes on different forms in different cases. In' working problems it is necessary to know the exact mode of variation, else absurd results will be obtained. PROPORTION, VARIATION, FUNCTION 89 In geometry it is proved that the length of a circle varies as its radius, that the area of a circle varies as the sgware of its radius, that the volume of a sphere varies as the cvbe of its radius. In symbols, C : Ci = r : rj, V: Fi = r^ : Vi^, Hence, if the second radius is twice the first (ri = 2>), then the length of the second circle is twice that of the first, the area of the second circle is four times that of the first, the volume of the second sphere is eight times that of the first. Of special interest and importance are the problems in which one variable varies directly as another, or inversely as another. The words " directly " and " inversely " have the same signifi- cance in algebra as in arithmetic. For example, at any moment, the length (L) of the shadow of a ver- tical post cast upon level ground depends upon the height {H) of the post ; the taller the post, the longer the shadow. Here L varies directly as H ; we wriiie. _ _ _ . _ . _ _ __ _ _ . L:Li = n'.nuOT L = kJS, or Lccff, where kis9, constant and where L x ^signifies ^' L varies asH,^^ These three notations denote the same relation between L and H; all three in- dicate that L is a function of H, The time {t)it takes a train to travel from New York to Chicago de- pends upon its speed (v) ; the greater the speed, the less the time. Here t varies inversely as « ; we write t-.ti = Vi'.Vy OTt=k-<,OTtcC-, V V From t = k*-we obtain vt = k; that is, if one variable varies inversely V as another, the product of the two variables is a constant. Sometimes there are three variables. When one variable de- pends upon two other variables in such a way as to vary as the product of the two variables, we say that the first variable varies jointly as the two other variables. 90 ELEMENTARY ALGEBRA Ex. 4 in § 101 was of this sort. If we doable the length of the base (6) of a triangle and at the same time doable its height (A), the area (^) is four times greater. In symbols, the variation is expressed thus, AiA\-=hhi h\hu or ji = ib • &A, or ji K hh. There are many other kinds of variation. A variable may vary directly as a second variable and inversely as a third, as in- dicated by the formula y = A;-. Or, a variable may depend z upon three or more other variables. Thus, the price of lace depends not only upon the number of yards, but also upon the quality of the article, the cost of labor, the rate of profit ex- acted vby the seller. Practical problems involving variables as functions of other variables and exhibiting many different forms of variation were given in § 97. VARIATION SHOWN IN GRAPHS 103. Consider a proportion in which two of the terms are variables. An automobile travels 29 miles in 2 hours ; at this rate, how far does it travel in h hours ? As the distances are proportional to the times of travel, we have the proportion, d:29 = A:2, where d indicates the number of miles and h the number of hours. Here d and h are variables ; when h increases, d increases also, the value of d being dependent upon that of h. The relation between h and d can be expressed equally well by changing the equation to the form ^ 29, 2 This shows that h and d vary in such manner that the number d is always — as large as the number h\ d is a function of h. *4 PROPORTION, VARIATION. FUNCTION 91 Thia variation is exhibited by tlie tollowing graph (Fig. 11). Hours are measured along the horizontal line OX; miles aie measured along the ver- tieal line Y. For cou- venieuce the divisions representing bours are taken ten times longer than the divisions rep- resenting miles. The line OA is constructed ae follows : When h = 0, then ds=0. This determines the point 0. When ft = 4, then d = 5S. This deter- mines the point A. Through and ^ draw the line OA, Using this graph we can tell by inspection the distance traveled during a time not ex- ceeding 5 hours. For instance, to tell the distance traveled in 3.2 hr., we find on OX the point C, then pro- ceed parallel to OT to the point B; and from B to the left, parallel to OX. The answer is 46*^ miles. Actual multiplication gives 46.4 miles. How accurately can the distance be measured by this graph ? How accurately can the time be measured by this graph ? How can a graph be drawn that will give distances correctly to ^ of a mile ? 92 ELEMENTARY ALGEBRA EXBBOISB 104. 1. From the graph, Fig. 11, tell the distance traveled in 2.5 hr., 3.4 hr., 4.1 hr., 4.9 hr. 2. From the same graph tell the time needed for traveling 24 mi., 50 mi., 35 mi., 65 mi., 48 mi. 3. Draw a graph that will show the railroad fare for trav- eling short distances at the rate of 2| cents a mile. 4. Draw a graph showing corresponding values of x and y in the proportion x:y = 1.9 : 2.3. 6. A clerk in a New York store converts prices in " marks- per-meter" into ^* dollar s-per-yard.'' He draws a graph and roughly " checks " his computations by it. If 1 m. = 1.1 yd., and 1 M. = $ .24, draw the graph. Hint. 1 M. per 1 m. = |l .24 per 1.1 yd. = $ ? per 1 yd. 6. One knot (nautical mile) is equal to 1.15 miles. Draw a graph for changing from one unit to the other and then tell from it the speeds, in miles per hour, of cruisers making 21, 22, 24, 27, 29 knots per hour. 7. In lifting weights with the aid of a small, well-lubricated screw jack, part of the power varies with the weights that are raised, and the rest of the power is a fixed amount to overcome friction. It was found by trial that 5 lb. was necessary to lift a weight of 100 lb., and that 2.8 lb. was necessary to lift 50 lb. Derive an equation and draw a graph showing the relation between the power P and the weight W. By the conditions of the problem, the relation between P and TTcan be expressed by a linear equation of the form P= aW-\- b, where P and W are the variables, and a and b are constants to be found from our data. Observe that aW is the part of the power which varies with the weight, and that b is the constant part that is expended to overcome friction. Notice that P is a function of W. When P = 6 lb., W = 100 lb., hence 6 = 100 a + 6. When P= 2.8 lb., Tr= 60 lb., hence 2.8 = 50 a + 6. PROPORTION, VARIATION, FUNCTION 93 Solving for a and b we obtain a = .044, 5 = .6. Substituting these values we obtain F = .044 Tf 4- .6, which expresses ** the law of the machine." The graph of this equation is shown in Fig. 12. 8. In Fig. 12, tell by inspection the power, P, necessary to raise a weight, TT, of 30 lb., of 40 lb., of 65 lb., of 75 lb. To what fraction of a pound can you esti- mate the power by this graph? 9. In Fig. 12, tell by inspection the weight Wy which can be raised when P is 2^ lb., 3 lb., 4| lb. By this graph, can W be determined as closely as 1 lb. ? How would you draw a graph yielding more accurate results ? 10. In a system of well-lubricated pulleys a weight of 50 lb. is raised by a power of 10.7 lb., and a weight of 80 lb. is raised by a power of 16.7 lb. If part of the power varies as the weight, and part of the power is fixed, find the equation or " law of the machine " which shows the relation between P and W, Draw a graph exhibiting simultaneous values of PandTT. Y 100 / 90 J f / 80 / J f |70 / / &60 A r / •;j6o J J f "M'a^ / % 40 > "^30 J f / 20 J J f 10 / y r 1 1 r^ J / 5 ~xl T '0' .; rr in -P 3U n( 1 lo~ Fig. 12. 94 ELEMENTARY ALGEBRA PROBLEMS 105. 1. If XQcy, and x^^ when y = 6, find x when y = 7. First Solution. Since xocy, the yariation is direct. Hence, y:yi^x:xi, Taking y,=6, a?i=5, y=7, 7:6=:aj:5. Hence, a5 = 5f. Second SoltUion. Since ««y, we have o^sA;^. To determine the constant k, use the given simultaneous values of the variables x and y, viz., x = 5 when y = 6. We ^^*a^ 5 = A;. 6. Hence, ^ = f . Substituting the value of /c, xss^y. When y = 7, we have a? = ^ x 7 = &J. The student should master both solutions. 2. If x varies inversdy as y, and a; = 5 when y s= 6, find a; when y = 7. i^ir«< Solution. Since a? oc -, we obtain the proportion 7 : 6 : 5 : a;. x==^. Second Solution. Since a; x - , a; = — • Taking a? = 5, y = 6, ^ "^ fi' A; = 30. 30 Substituting the value of A;, a; = — • y When y = 7, a; = -8^ = 4^. PROPORTION, VARIATION, FUNCTION 95 3. If z varies jointly as x and y, and 2 = 5, when xssS and y = 4, find z when a? =a 7 and y = 6. First Solution, By proportion, z:zi = xy: x^yi. Take ajj = 3, yi = 4, «i = 5, a; = 7, y = 6. Then, « : 5 = 42 : 12, 2; = 17f Second Solution, Since zccxy^ z = kxy. To determine A;, substitute the simultaneous values, 2 = 5, a? = 3, y = 4, 5 = A: X 3 X 4, I. — . 6 a;— ly. Hence, « = -ji^ ajy. When aj= 7, y = 6, 2 = ^ x 42 = 17f 4. If z varies directly as x and inversely as y, and 2 = 5, when a? = 3, y = 4, find 2; when a; = 7 and y = 6. Solution. Here 2 oc -, hence z = k>-. y Substitute the simultaneous values 2 = 5, a; = 3, y = 4, 5 = A;.f. Hence, ^ = ^, A 20 a? ana 2 = — • - • 3 y Whena? = 7,y = 6, 2; = ^.-J = 7f Let the student give the solution by proportion. 6. The distance described by a body falling from rest varies as the square of the time. If in 2 seconds it falls 64.4 feet, find the distance it falls in 6^ seconds. Solution* The distance (d) varies directly as the square of the time {t). Hence, d = kfi. 96 ELEMENTARY ALGEBRA When « = 2, d = 64.4, hence, 64.4 = ifc . 2^, A: = 16.1, and d = 16.1 <2. When t = 6i seconds, d = 680.2 + ft. 6. The brightness or intensity of lig:ht varies inversely as the square of the distance from the source of the light. How many times brighter is the page of a book illuminated at a distance of 3 ft. from the source than at a distance of 5 ft. ? Solution. The intensity (i) varies inversely as the square of the dis- tance (d). That is, t = — , or i.ii = di^.d^, d^ Since we do not have sufficient data to determine the numerical value of k, it is easier to work the example by proportion. We obtain i :ii = ^: 3«. Or 1 = 25 = 24. ii 9 At a distance of 3 ft. the page is 2} times brighter than at 5 ft. 7. If a: varies as y, and if x is 144 when y is 3, find the value of y, when x is 360. 8. If X varies as the square of y, and if x is 144 when y is 3, find the value of y when x is 360. 9. If x varies jointly as y and z, and op = 10 when y = 15, and z = 18, find x when y =5 and z = 24. 10. If oj varies inversely as the cube of y, and a; = 54 when y = 3, find x when y = .1. 11. The areas of similar triangles vary as the squares of homologous sides. If homologous sides of two such triangles are 7 and 10, and the area of the larger triangle is 13.5, what is the area of the smaller triangle ? 12. The areas of two similar triangles are 36 and 43 ; one side of the smaller triangle is 5. Find the homologous side of the larger triangle, correct to two decimal places. PROPORTION, VARIATION, FUNCTION 97 13. Two pieces of round timber (right cylinders) have the same altitude ; their girths are 3.4 ft. and 3.1 ft. Find the ratio of their volumes. 14. If a carriage wheel 4 ft. 2 in. in diameter makes 240 revolutions in going a certain distance, how many revolutions will a wheel 4 ft. 8 in. in diameter make in going the same distance ? 15. The weight of a sphere of given material varies as the cube of the radius. If a sphere having a diameter of 3 in. weighs 4 lb., find the weight of a sphere of the same material, but with a radius of 2 in. 16. The rents of an estate should be divided between A and B in the ratio of 4:5. However, A is paid $ 250, and B is paid $ 275. Which has been overpaid, and how much ? 17. If a train, whose speed is 55 miles per hour, makes a certain trip in 3^ hr., how long will it take a slower train to travel | of that distance, its speed being to that of the fast train, as 4 is to 9 ? 18. If the pressure p exerted upon the air in a bicycle pump and the volume of the air v obey Boyle's law, accordiug to which pv = constant, and if the pressure is 25 lb. per square LQch when the volume is 27 cubic inches, what will the pres- sure be when the volume is reduced to 20 cubic inches ? 19. Part of the expenses of a certain school was fixed ; part of them varied with the number of pupils. If the yearly ex- penses were $ 75,000 when the number of pupils was 710, and $ 85,000 when the number of pupils had increased to 900, find the yearly expense when the number of pupils was 800. 20. If the ratio ofSx + ytoAx — Sy equals the ratio of 21 to 2, what is the ratio oix to y? 21. At the end of 12 days, 15 men finish one fourth of a piece pf work. How many additional men must be engaged in order to complete the remaining work lq 18 days ? H 98 ELEMENTARY ALGEBRA 22. If a body falls from rest through a distance of 16.1 feet during one second, how far will it fall in 8.3 seconds ? 23. Compare the intensity of illumination of the page of a book at a distance of 6.3 feet from a source of light with the intensity at a distance of 4.9 feet. 24. If the surface area of a certain sphere is 35 sq. in., what is the surface area of a sphere having a radius 4.7 times longer ? 26. If the volume of a certain sphere is 500 cu. in., what is the volume of a sphere whose radius is 2.9 times longer ? 26. If the base of a triangle is increased 10-fold and its alti- tude is reduced to -j^ its original length, find the ratio between the initial and final areas. 27. Two casks of similar shape have their homologous linear dimensions in the ratio of 2 : 3.4. Find the ratio of their capacities. 28. The radii of the bases of two cylindrical tanks are 1 ft. 5 in., and 1 ft. 11 in., respectively ; the heights of the tanks are 6 ft. and 5 ft., respectively. Compare the capacities of the tanks. 29. The area of a circle is found by the formula A = wR*, where v = 3.1416. Draw a graph by which the areas can be determined by inspection. B A Point i .79 A 1 3.14 B li 4.91 C H 7.07 D U 9.62 E 2 12.56 F 3 28.27 Q 4 60.27 H Assuming values of i?, as shown in the table, compute the values of the area A, Explain the construction of Fig. 13. J PROPORTION, VARIATION, FUNCTION 99 30. In Fig. 13, find by inspection the areas of circles having radii of 2J, 2^, 2f , 3j, ^, ^. 31. If a garrison of 2000 men is pro- visioned for 40 days, how long will their provisions last if they are joined by 375 other men ? 32. The surface of a sphere varies directly as the square of the diam- eter. If the diam- eter is 2 inches, the surface is 12.666 square inches. Find the surface when the diameter is 7 inches. 33. The value of a piece of land has risen in ten years from $ 1200 to $ 3200. How would you calculate the rise in value of another similarly situated piece which was worth $1750 ten years ago ? 34. The rent of a house in a town has gone from $ 85 to $ 70 per month during the last 15 years. If the value of other houses in that neighborhood has decreased in the same ratio, what was the decrease in the rent of a house now rented at $ 42 ? 36. The diagonal of a cube varies directly as the length of an edge. When the edge measures 3 inches, the diagonal measures 5.196 inches. What does the diagonal measure when the edge is 3.8 inches ? 1, { lO I 1 r > /v / *r / / — )> Z' f? > ' — ■i-\ t\ .y^ / ^} G I J f, f ■S J I lA / ~{ sU J / / Fj / 1 E / J T )/ ' c / B>f A /^ ^ 2C _, 2 S Radius of Circle Fia. 13. 100 ELEMENTARY ALGEBRA 36. The weight of a copper coin varies directly as its thick- ness and also as the square of its diameter. A coin weighing 11 grams has a diameter of 3 cm. and a thickness of .2 cm. What is the weight when the diameter is 2.5 cm. and the thickness .3 cm. ? 37. A's rate of working is to B's as 3 : 4 ; B's rate of work- ing is to C's as 6 ; 5. How long will it take C to do work which A can do in 8 hours ? 38. The distance through which a body falls from rest varies as the square of the time of falling. If a body falls 257.6 ft. in 4 seconds, how far will it fall in 5^ seconds ? 39. If a stone weighing 5 pounds falls from rest through a distance of 144.9 ft. in 3 seconds, how far will a stone fall from rest in 3 seconds, if its weight is 500 pounds ? See Ex. 38. 40. The tractive force or pull necessary to move a vehicle at a uniform rate (say 3 miles an hour) varies directly as the pressure (weight of vehicle and load) and inversely as the square root of the average radius of the wheels. If on level, paved, or macadam roads a pull of 57 pounds per ton of pres- sure is necessary when the front and rear wheels average 50 inches in diameter, what pull is necessary to move 3500 pounds when the wheels of the vehicle average 38 in. in diameter ? 41. If a tractive force of 75 lb. is necessary to keep a load of 1 ton in motion in a vehicle whose wheels average 38 in. in diameter, on a dry and hard earth road, what is the tractive force for 5 tons when the wheels average only 26 in. in diameter ? See Ex. 40. 42. If on an earth road, in sticky mud \ in. deep, the trac- tive force per ton is 119 lb. when wheels 38 in. in diameter are used, what tractive force is required to move 4 tons when the diameters are 50 in. ? 43. The tractive force per ton over dry, cloddy plowed ground is 252 lb. for wheels 50 in. in diameter ; what is the tractive force for J of a ton when the wheels are 26 in. in diameter ? PROPORTION. VARIATIpkf rtkc'TtON^ '|' \ IQl GRAPHS EXHIBITINO EkplRICAL 'iJATA 106. 1. By a certain plan of life inaurance a single premium is paid in order that the insured may receive $100 when he ia 60 years old or that Ms beneficiaries may receive $100 if he dies before that age. The premium depends upon the age at which the i: Age next birthday : 15 20 25 30 36 40 45 60 Premium: $41.20 $46.25$49.30 953.60 $60.10 $65.10 $72.05 $80.20 Draw a graph and use it to find the pre- mium one would pay at the age of 18, 23, 32, 38, 43. 2 It U convenient to g mark oB along the z-axie J the ages above 16 ; along a the {(-axis the premiums -g aftoiie $40. By this <le- J vice a much smaller sheet ** of Bquare paper can be used. See Fig. 14. 2. In a city in the northern part of the United States the times of sunrise on certain days in May, June, and July Miv ;20 .04 Jf: i^es above 16 years Fig. 14. as follows ; 46 Draw a graph showing the hour of sunrise from May 10 to July 30. Mark oS on one axis the number of days after May 10, od the other axis the number of minuteB after 3 : 44 a.h. • • • * • . • • - « CHAPTER IV LOGARITHMS 107. In an equation 10^ = N the exponent L is called the logarithm of the number N, to the base 10. For example, 102 = 100, hence 2 is the logarithm of 100, to the base 10. We write, 2 = log 100. 10> = 1000, hence 3 is the logarithm of 1000, to the base 10. We write, 8 = log 1000. 10^ = 3.16227+ (verify this), hence .6 is the logarithm of 8.16227+, to the base 10. ^^ ^^ite, .6 = log 3.16227+. 10* = 10i(10)i = 81.6227+, hence 1.5 is the logarithm of 81.6227+, to the base 10. ^^ ^^te, 1.6 = log 31.6227+ As in this chapter all logarithms are taken to the same base 10, no con- fusion will arise from the omission, hereafter, of the phrase *' to the base 10." Verify the following statements : 10» =1000 10* = 10(10)* = 31.6227+ 10* =100 IQi =3.16227+ 10^ =10 1 lft-i = — =.316227+ 10^ =1 ^^ lOi 10-1 = .1 .1 10-. = .01 io-i = ^-i|=. 0316227^ 10-» = .001 1 lO-f = A = .00316227+ 10* = 100(10)* = 316.227+ 10* 102 LOGARITHMS 103 Hence, by the definition of logarithms, 3 = log 1000 ^.5 = log 316.227+ 2 = log 100 1.5 = log 31.6227+ 1 = log 10 .5= log 3.16227+ O = logl -.6= log ,316227+ - 1 =. log .1 - 1.5 = log ^0316227+ - 2 = log .01 - 2^5 = log .00316227+ - 3 = log .001 LOGARITHMIC CUBVB 108. On a piece of square paper as small as that in Fig. 15, it is not convenient to draw a curve that will exhihit the logarithms of numbers ranging from ,001, the smallest, to 1000, the largest number considered above. The curve shows the logarithms of positive numbers between .1 and 50. The numbers are laid off on the as-&xiB, their respective logarithms on the y-axis. Fig. 15. — 1 = log .1 looatea the point A = log 1 locates the point B .6 = log 3.16+ IncatPB the point C 1 = log 10 locates the point D 1.6 = log 31.6-^ locates the point E curve, some additional values were used. 104 ELEMENTARY ALGEBRA ORAL BXBBCISBS ON THB LOGARITHMIC CXJBVB 109. 1. Where does the curve cut the ataxia ? How much is log 1 ? 2. What is the algebraic sign of the logarithms of all num- bers larger than 1 ? 3. For what range of numbers are the logarithms negative ? 4. Does the logarithmic curve extend to the left of the y-axis ? 6. Do negative numbers have real logarithms ? 6. Does the logarithm increase as a variable number in- creases ? By inspection of Fig. 15, find approximately the logarithms of the following numbers : 7. 5. 10. 35. 13. 2. 8. 15. 11. 45. 14. 4. 9. 25. 12. 50. 16. 6. By inspection of Fig. 15, find approximately the numbers corresponding to the following logarithms : 16. 1.6. 20. 1.2. 24. .8. 17. 1.5. 21. 1.1. 26. .3. 18. 1.4. '22. 1. 26. 0. 19. 1.3. 23. .7. 27. —.5. Find by inspection, how many times greater 28. log 4 is than log 2. 33. log 27 is than log 3. 29. log 8 is than log 2. 34. log 16 is than log 4. 30. log 16 is than log 2. 36. log 25 is than log 5. 31. log 32 is than log 2. 36. log 36 is than log 6. 32. log 9 is than log 3. 1 LOGARITHMS 105 Show by Fig. 15 that 37. log 50 = log 5 4- log 10. 40. log 48 — log 8 = log 6. 38. log 40 = log 5 + log 8. 41. log 24 - log 4 = log 6. 39. log 42 = log 6 4- log 7. 42. log 32 — log 4 = log 8. ASSUMPTIONS • 110. In drawing the logarithmic curve in Fig. 15 we located a few points and then drew a smooth curve through those points. This process is based on the following assumptions which admit of proof : (1) All positive numbers, whether rational or irrational, have logarithms ; (2) In all cases, the logarithm of a number increases when the number itself increases. FUNDAMENTAL THEOREM 111. In studying the logarithmic curve we noticed that log 50 = log 5 4- log 10. This illustrates an important theorem. Let N and Ni be any two positive numbers. Also, let JVr=10% and Ni = 10^. Then their product is, N- Ni = lO^+^i. By definition of logarithms, L = logarithm of N, and Zi = logarithm of Ni, X + Zi = logarithm of N • ^i. Hence, the theorem. The logarithm of the product of two positive numbers is equal to the sum of the logarithms of the numbers. 112. The integral part of a logarithm is called its charac- teristiCy and the decimal part is called its mantissa. For example, log 31.6227+ = 1.5. The characteristic of log 31.6227+ is 1. The mantissa of log 31.6227+ is .6. 106 ELEMENTARY ALGEBRA The tables of logarithms give only the mantissa ; the charao- teristic can be supplied by two easy rules. It has been found convenient to take the mantissas of all logarithms positive. The characteristic is positive in the logarithms of numbers larger than 1, and negative in the logarithms of numbers smaller than 1. That is, the char- acteristic of log, 100 is positive; the characteristic of log .01 is negative. If the characteristic is negative, the negative sign is placed over the figure, as a reminder that the — does not apply to the mantissa. Thus the characteristic of log .06 is 2. For the purpose of deriving the rule for determining the characteristic of the logarithm of a number, we restate some of the relations in § 107 : 1000 = 10* 1 =10° 100 = 10^ .1 =10-^ 10 = 10^ .01 =10-2 1 = 10° .001 = io-» Since 569.5 lies between 100 and 1000, its logarithm lies between 2 and 8. That is, log 569.6 = 2 + a mantissa. Since 86.6 lies between 10 and 100, log 86.6 = 1 + a mantissa. Since 7.03 lies between 1 and 10, log 7.03 = + a mantissa. Since .673 lies between .1 and 1, log .673 = ~ 1 + a mantissa. Since .046 lies between .01 and .1, log .045 = ^ 2 + a mantissa. Since .0078 lies between .001 and .01, log .0078 = — 8 + a mantissa. By inspection of these relations we obtain the rule : 1. If the first Bigni&CAnt figure of a number is n places to the ' left 1 / ' of units' place, the characteristic of the logarithm is I vOflv ( ■ ^ I 4- w 1 — n LOGARITHMS 107 The following illustrations will make this rule plainer : J S 1^ . Characteristic 6 6 9.6 2 . 8 6.6 1 7.03. 0.673 -1 0.046 -2 0.0078 -3 It will be seen that when a number is 10 or larger than 10, the first significant figure is to the left of units ^ place ; if the number is less than 1, the first significant figure is to the rig?U of units' place. This is shown by the dots placed above the digits ; the number of dots placed over a number indicates the number of units in the characteilstic. BXBBCISBS 113. By inspection, tell the characteristics of the logarithms of the following numbers : 1. 123. 4. 17.754. 7. .091. 10. 2000. 2. 97. 6. .7894. 8. .00034. 11. .00005. 3. 2.345. 6. .005. 9. .0034. 12. 13.764. 114. The following is an important property possessed by logarithms constructed to the base 10 : If two numbers contain the same figures in the same order , hut differ in the position of the decimxxl point, their logarithms to the base 10 have the same mantissa. Suppose the two numbers are 7896.1 and 7.8961. We have 7896.1 = 7.8961 x 1000. By §111, log 7896.1 = log 7.8961 + log 1000. But log 1000 = 3. Hence, log 7896.1 = log 7.8961 + 3. 108 ELEMENTARY ALGEBRA Since the two logarithms differ by the integer 3, the decimal parts (the mantissas) of the logarithms must be the same. For example, log 7896.1 = 3.8974. log 7.8961 = 0.8974. How the mantissa is found from the tables will be explained next. FINDING LOGARITHMS 115. 1. Find the logarithm of 789. The characteristic is 2. The mantissa is taken from the table, p. 110. The column on thcr ex- treme left contains the first two significant figures of the number whose loga- rithm is sought. In this case we look for the figures 78. In the same row with 78, and in the column headed * * 9 ' ^ ( 9 being the third digit in 789) , is the number 8971 ; this is the mantissa sought, expressed to four decimal places. Hence we have log 789 = 2.8971. 2. Find the logarithm of .02738. The characteristic is 2. The significant figures are 2738. On p. 109 look in the column on the extreme left for 27 ; in the column headed ** 3 '* and in the same row with 27 is the mantissa .4362. Hence log .0273 = 2.4362. But we want log .02738. This is not found in this table and must be determined by a process called *^ interpolation.'^ The table is so constructed that interpolation is made easy. In the column on the extreme right, headed **789,*' we find, under the 8, and in the same row with 4362, the number 13. Add 13 to 4362. We obtain 4376. This is the mantissa, obtained by interpolation. Hence, log .02738 = 2.4376. 116. It may be added that logarithmic computation in gen- eral is only approximate. With a four-place table like the one in this book, logarithms of numbers are found to four places only. Absolute accuracy is, therefore, out of the ques- tion. This fact does not materially diminish the value of logarithms, for the reason that in practical problems, results correct to three, four, or five decimal places are satisfactory. For instance, in the computation of the interest on a given sum, a result correct to the nearest cent is all we want. LOGARITHMS 109 117. Logarithms No. 10 11 11 IS 14 16 16 17 18 19 1 S S 0128 0681 0S99 1239 1668 4 0170 0569 0984 1271 1684 1875 2148 2405 2648 2878 6 0212 0607 0969 1308 1614 6 T 0294 0682 1088 1867 1678 8 9 1 S S 4 6 6 T 8 9 0000 0414 0792 1189 1461 1761 2041 2804 2653 2788 0048 0458 0828 1178 1492 1790 2068 2880 2577 2810 8082 3248 8444 3686 8820 0086 0492 0864 1206 1528 1818 2096 2856 2601 2888 8064 8268 8464 8666 8888 4014 4188 4846 4502 4664 0268 0646 1004 1886 1644 0884 0719 1072 1899 1708 1987 2268 2604 2742 2967 0874 0765 1106 1480 1782 4 812 4 811 8 710 8 610 8 6 9 17 2126 1619 28 14 17 21 18 16 19 12 15 18 29 88 87 26 80 84 24 28 81 28 26 29 2124 27 1847 2122 2880 2625 2866 1908 2176 2480 2672 2900 1981 2201 2455 2695 2928 1950 2227 2480 2718 2945 2014 2279 2629 2766 2989 8201 8404 8698 8784 8962 8 6 8 8 5 8 2 6 7 2 6 7 2 4 7 11 14 17 11 18 16 10 12 15 9 12 14 9 1118 20 22 26 18 2124 17 20 22 1619 21 16 18 20 80 SI SS SS S4 8010 8222 8424 8617 8802 8075 8284 8488 8674 8856 8096 8804 8502 8692 8874 8118 8711 8892 4065 4282 4898 4548 4698 8189 8846 8641 8729 8909 8160 8865 8660 3747 8927 4099 4265 4426 4579 4728 4871 5011 6146 6276 6408 8181 8386 8679 8766 8946 2 4 6 2 4 6 2 4 6 2 4 6 2 4 5 81118 81012 81012 7 911 7 9 11 15 17 19 14 16 18 14 15 17 18 15 17 12 14 16 86 S6 ST SS S9 SO S4 S6 S6 ST S8 S9 40 41 4S 4S M 46 46 4T 48 49 60 61 6S 6S 64 8979 4160 4814 4472 4624 8997 4166 4880 4487 4689 4081 4200 4862 4518 4669 4048 4216 4878 4688 4688 4829 4969 6105 5287 6866 4082 4249 4409 4664 4718 4867 4997 5182 5268 6891 4116 4281 4440 4594 4742 4188 4298 4456 4609 4767 2 8 6 2 8 6 2 8 6 2 8 6 18 4 7 910 7 810 6 8 9 6 8 9 6 7 9 12 14 15 1118 16 11 18 14 11 12 14 10 12 18 4771 4914 5061 5186 6816 5441 5668 6682 6798 6911 6021 6128 6282 6386 6185 6582 6628 6721 6812 6902 4786 4928 5065 6198 6828 5468 6576 6694 6809 6922 6081 6188 6248 6845 6444 6642 6687 67aio 6821 6911 4800 4942 6079 5211 5840 4814 4955 6092 6224 5858 4848 4988 5119 6260 5878 4886 6024 6169 6289 6416 4900 5088 5172 6802 6428 18 4 1 8 4 18 4 18 4 18 4 6 7 9 6 7 8 6 7 8 6 6 8 6 6 8 10 11 18 10 11 12 9 1112 9 10 12 9 1011 6465 5587 6706 6821 6988 6042 6149 6268 6855 6464 6478 6599 5717 6882 6944 6068 6160 6268 6865 6464 6561 6666 6749 6889 6928 7016 7101 7186 7267 7848 5490 5611 5729 6848 5956 6602 6628 5740 6866 6966 6076 6180 6284 68S5 6484 6580 6676 6767 6857 6946 7088 7118 7202 7284 7864 6514 6685 6762 6866 6977 6086 6191 6294 6895 6493 6527 5647 5768 5877 5988 5589 5668 5776 6888 6999 6107 6212 6814 6416 6618 6609 6702 6794 6884 6972 5661 6670 6786 5899 6010 1 2 4 1 2 4 1 2 3 1 2 8 12 8 6 6 7 6 6 7 6 6 7 6 6 7 4 5 7 91011 81011 8 9 10 8 910 8 9 10 6064 6170 6274 6876 6474 6671 6665 6768 6848 6987 7024 7110 7198 7276 7866 6096 6201 6804 6405 6503 6599 6698 6T85 6875 6964 7050 7185 7218 7800 7880 6117 6222 6826 6425 6522 1 2 8 12 8 12 8 12 3 1 2 8 4 6 6 4 6 6 4 6 6 4 6 6 4 5 6 8 9 10 7 8 9 7 8 9 7 8 9 7 8 9 6661 6646 6789 6880 6820 7007 7098 7177 7269 7840 6690 6684 6776 6866 6955 7042 7126 7210 7292 7872 6618 6712 6808 6898 6981 7067 7162 7285 7816 7396 12 8 12 8 12 8 12 3 12 8 4 6 6 4 5 6 4 6 5 4 4 5 4 4 5 7 8 9 7 7 8 6 7 8 6 7 8 6 7 8 6990 7076 7160 7248 7824 6998 7084 7168 7251 7882 7060 7148 7226 7808 7388 12 8 1 2 8 12 2 12 2 12 2 8 4 6 3 4 6 8 4 6 8 4 5 8 4 5 7 8 6 7 8 6 7 7 6 6 7 6 6 7 110 ELEMENTARY ALGEBRA Logarithms No. 7404 1 6 7419 8 4 6 7448 6 7451 7 7469 8 7466 9 1 6 8 4 6 6 7 8 9 7412 7427 7485 7474 1 2 2 8 4 ft 5 6 7 M 7482 7490 7497 7505 7618 7520 7528 7686 7548 7661 1 2 2 8 4 5 5 6 7 67 7069 7566 7674 7682 7589 7697 7604 7612 7619 7667 1 2 2 8 4 5 5 6 7 M 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 2 8 4 5 6 7 M 7709 7716 7728 7781 7808 7788 7810 7746 7752 7760 7767 7889 7774 2 8 4 5 6 7 60 7782 7769 7796 7818 7825 7882 7846 2 8 4 5 6 6 61 78m 7860 7868 7876 7882 7889 7896 7908 7910 7917 2 8 4 5 6 6 6S 7924 7981 7988 7945 7962 7969 7966 7978 7980 7987 2 8 8 5 6 6 66 7998 8000 8007 8014 8021 8028 8065 8041 8048 8066 2 8 8 5 5 6 64 66 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 2 8 8 6 5 6 8129 8186 8142 8149 8166 8162 8169 8176 8182 8189 2 8 8 5 5 6 66 8195 8202 8209 8215 8222 8228 8285 8241 8248 8264 2 8 8 5 6 6 67 8261 8267 8274 8280 8287 8293 8299 8806 8812 8819 2 8 8 6 5 6 66 8825 8881 8888 8844 8861 8867 8868 8870 8876 8882 2 8 8 4 5 6 66 8888 8895 8401 8468 8407 8470 8414 8476 8420 8482 8426 8482 8494 8489 8446 8606 2 2 8 4 6 6 70 8451 8457 OvOO 8600 2 2 8 4 5 6 71 8618 8619 fmt> 8681 86B7 8648 8549 8666 8661 8667 2 2 8 4 5 5 76 8678 8679 8686 8691 8697 8606 8609 8616 8621 8627 2 2 8 4 5 5 76 8688 8689 8646 8651 8657 8668 8669 8676 8681 8686 2 2 8 4 5 5 74 8692 8696 8766 8704 8762 8710 8768 8716 8774 8722 8779 8727 8788 8789 8746 8802 2 2 8 4 5 6 76 8751 8785 8791 8797 2 2 8 8 4 5 5 76 8806 8814 882(r 1B825 8831 8887 8842 8848 8864 8869 2 2 8 8 4 5 5 77 8866 8871 8876 8882 8887 8898 8899 8904 8910 8916 2 2 8 8 4 4 5 78 8921 8927 8982 8988 8948 8949 8954 8960 8965 8971 2 2 8 8 4 4 6 76 80 8976 9081 8982 9086 8987 9042 8998 8998 9004 9009 9015 9020 9074 9026 9079 2 2 8 8 4 4 5 9047 9058 9058 9068 9069 2 2 8' 8 4 4 5 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9188 2 2 8 8 4 4 5 86 9188 9148 9149 9164 9160 9166 9170 9175 9180 9186 2 2 8 3 4 4 5 66 9191 9196 9201 92()6 9212 9217 9222 9227 9232 9288 2 2 8 8 4 4 5 64 9248 9248 9299 9258 9258 9268 9269 9820 9274 9279 9880 9284 9885 9289 9840 2 2 8 8 4 4 5 86 9294 9804 9809 9815 9825 2 2 8 8 4 4 5 66 9845 9850 9856 9860 9865 9870 9876 9380 9886 9890 2 2 8 3 4 4 5 87 9895 9400 9405 9410 9415 9420 9425 9480 9486 9440 1 2 2 3 3 4 4 86 9445 9450 9465 9460 9465 9469 9474 9479 9484 9489 1 2 2 8 8 4 4 66 9494 9499 9604 9652 9609 9518 9618 9623 9671 9528 9576 9688 9638 1 2 2 8 8 4 4 90 9542 9547 9567 9562 9566 9581 9586 1 2 2 8 8 4 4 91 9590 9596 9600 9605 9609 9614 9619 9624 9628 9683 1 2 2 3 8 4 4 96 9688 9648 9647 9662 9657 9661 9666 9671 9676 9680 1 2 2 3 8 4 4 98 9685 9689 9694 9699 9708 9708 9718 9717 9722 9727 1 2 2 8 8 4 4 94 9731 9786 9741 9745 9791 9760 9764 9769 9806 9768 9768 9814 9773 9818 1 2 2 8 3 4 4 96 9777 9782 9786 9796 9800 9809 1 2 2 8 3 4 4 96 9828 9827 9882 9886 9841 9846 9860 9854 9869 9868 1 2 2 3 3 4 4 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 1 2 2 3 8 4 4 96 9912 9917 9921 9926 9980 9984 9989 9948 9948 9962 1 2 2 3 8 4 4 99 9956 9961 9966 9969 9974 9978 9983 9987 9991 9996 1 2 2 3 3 8 4 LOGARITHMS 111 BXBBCISBS 118. Find the logarithms of : 1. 870. 4. .01235. 7. 13.8964. 2. 63. 6. .003986. 8. 1089. 3. 708. 6. .7943. 9. 1008. FINDING ANTILOGARITHMS 119. 1. Given log n = 1.2355, find the antilogarithm n. The number n, frequently called the arUilogarithm, may be found by steps which are the reverse of those taken in finding the logarithm. At first consider only the mantissa 2366. On page 109 find the mantissa 2366. It occurs in the same row as the number 17 on the extreme left, and in the column headed ** 2.^* ^ The order of the figures in the answer is 172. ' Where should the decimal point be ? That is determined with the aid of the characteristic. All we need to do is to apply the rule relating to characteristics. Place the decimal point so that the characteristic of the resulting number is 1. That number is 17.2. Thus, n = 17.2. 2. Given log n = 2.8764, find the antilogarithm n. Look in the table for the mantissa 8764. It cannot be found. Look then for the next smaller mantissa. It is 8762 ; the order of figures in the antilogarithm of 2.8762 is 762. But we want the antiloga- rithm of 2.8764. This is found approximately by the following ** inter- polation.*' Find the difference between 8762 and the given mantissa. This differ- ence is 2. Look in the same row with 8762, and in one of the right-hand columns for the number 2. You see it in the column headed ** 3 '* and also in the column headed "4." Take either 3 or 4 as the figure found by interpolation. Which of these is the more accurate, we cannot tell from these tables. Take 3. Write 3 after 752, and we have the required order of figures ; namely, 7623. Place the decimal point so that, by the rule, the characteristic of the resulting number is 2. We obtain n = .07623, nearly. 112 ELEMENTARY ALGEBRA EXEBCISBS 120. Find the antilogarithms of the following : 1. 1.7931. 4. 1.7605. 7. 3.4567. 10. 3.6709. 2. 2.3181. ' 6. 0.9064. 8. 0.7064. 11. 3.5374. 3. 3.9876. 6. 2.1907. 9. 2.3706. 12. 5.9860. 121. 1. Using logarithms, find the product n = 7893 x 879G. By § 111, the logarithm of a product is the sum of the logarithms c the factors. We find , „^^„ ., „^«„ log 7893 = 3.8973 log 8796 = 3.9448 log n = 7.8416 We find the antilogarithm n = 69430000. Observe that this answer is only approximate. The true answer is 69426828. Closer approximations may be obtained by logarithms, by using a table containing figures to 6, 6, 7 or more places. 2. Find the product x = .9062 x .007362. The characteristics of the factors are I and 3. When logarithms are added or subtracted, it is easier to write a negative characteristic as the difference of two positive integers, thus : 1 = 9 — 10, 3 = 7 — 10. Accordingly, we get log .9062 = 9.9572 — 10 log .007362 = 7.8670-10 Add log X = 17.8242 - 20 Or loga:= 3.8242 X = .006671, nearly. BXEBCISBS 122. Using logarithms, compute the following products : 1. 123 X 978. 4. 37.61 x 3.945. 7. 9999 x 7777. 2. 12.34 X 75.34. 6. 1479 x 6984. 8. 19283 X 87045. 3. .0009638 X 7894. 6. 2673 x 7654. 9 93069 x 80493. LOGARITHMS 113 FUNDAMENTAL THEOREMS 123. In § 111, we proved the important theorem that Tlie logarithm of a product is equal to the sum of the logarithms of the factors. We proceed to establish two other theorems that are no less fundamental. The logarithm of a quotient is equal to the logarithm of the dividend minus the logarithm of the divisor. The proof is similar to that of the first theorem. Let N and Ny^ be any two positive numbers. Let also JV^=10^ and ^1 = 10a. Divide N by iVi, — = lO^-^i. By definition of logarithms, . L = logarithm of Ny Li = logarithm of Ni, and L — Li= logarithm of — Hence the theorem is proved. The logarithm of a positive number with the eocponerU n is n times the logarithm of the number. In this theorem n may be a positive integer or a positive fraction. Let -^be the number, and let JV^=1(P. Raise both sides to the nth power, N^ = (1(F)». Simplify N^ = lO"^, 114 ELEMENTARY ALGEBRA By the definition of logarithms, L = the logarithm of N and nL = the logarithm of N\ Hence the theorem is proved. If w =— , then this theorem gives the logarithm of y/W. m 124. 1. Using logarithms, find the quotient a?= ||^. We find log 9876 = 3.9946 log 6987 = 3.8443 log a; = 0.1603 The autilogarithm x = 1.414, nearly. 2. Find a? = (8.786)*. We find log 8. 786 = 0. 9438 Multiply by 6, 6 loga; = 4.7190 X = 52360, approximately. 3. Compute aj =(.01237)*. The characteristic of .01237 is 2 ; we write it 8 — 10. Hence, log .01237= 8.0923-10 Multiply by 7, 7 66.6461 - 70 Divide by 3. Since 70 -»- 3 introduces a fraction, it is more convenient to write the characteristic, 66 — 70, in the form 16 — 30. The logarithm is then 16.6461 -- 30. Dividing by 3, log a; = 6. 5487 - 10. X = .00003537, nearly. BXBBCISES jl25. Evaluate the following by means of the four-place table of logarithms : 1. 6.823 X 2.315. 4. 2293 x 4489 -^ 7895. 2. 46420 X 27.49. 6. .943 x 9855 -^ .0896. 3. 32.25-^6.923. 6. (2.519)*. LOGARITHMS 115 7. (.007668)'. ^^ 98.6 x 7.639 8. a/4978000000. -^^ 9. V752 X V3450. .. (273)«(743)« 10. (1.63)*. ' (897)V 11. (3.197)1 ^^ V3:598 ' -^479 , 12. V;03276. * V933 PROBLEMS 126. In the following problems let w = 3.1416, B = radios. 1. Find the length of a circle (i.e. the circumference) whose radius is 3.17 ft. 2. Find the area of a circle whose radius is 9.795 in. (Area 3. Find the area of a circle whose radius is 7.891 cm. 4. Compute the curved surface of a right circular cylinder whose height is 7.9 in. and whose base has a diameter of 79.86 in. 6. Calculate the volume of a sphere whose radius is 2.97 yd. Volume of a sphere = . 3 6. What is the area of the surface of a sphere when R = 0.7 in. . Area of spherical surface = 4 wlfi, 7. Compute the leg of a right triangle when the hypotenuse is 25.97 in. and the other leg is 17.8 in. In Fig. 16, X = Vh^ - 63 = V(h + 6) (A - 6). 8. Compute the area of a triangle whose sides are, respec- tively, 17.89, 20.96, 19.78. Area of triangle =>/«(« — a) (« — 6) (« — c) , where a, &, e are the sides of the triangle and s is half the sum of the sides. 116 ELEMENTARY ALGEBRA 9. In a right triangle the hypotenuse is 9.675, one leg is 7.98 ; find the length of the other leg. 10. Find the area of the surface of a hemispherical dome, the diameter of which is 40 feet. 11. A triangular piece of ground measures along the edges 315 yd., 541 yd., 479 yd. Find the area. 12. The volume of a pyramid is one third the product of its base and altitude. What is the volume of a pyramid whose height is 7.82 ft. and the sides of whose triangular base are 6.92 ft., 5.83 ft., and 4.91 ft. ? EXPONENTIAL EQUATIONS 127. An exponential equation is one in which the unknown occurs in an exponent. Thus, 12* =15, a**+'=6 are exponential equations which can be solved with the aid of logarithms. Since log 12* = x log 12, we obtain from 12* = 15, ' x log 12 = log 16. Divide both sides by log 12, x = l^i? . log 12 Finding the logarithms, we obtain x = — . 1 i7fii is computed like any other quotient. 1.0792 y J ^ log 1.1761 = 0.0705 log 1.0792 = 0.0.382 Subtracting, log x = 0.0373 X = 1.090, nearly. Note that in computations that are only approximate the answer 1.090 receives a different interpretation than the answer 1.09. The difference is this : the answer 1.090 shows that an effort was made to ascertain the third decimal figure and that it was found to be 0. On the other hand, 1.09 means that no effort was made to ascertain the value of the third figure. Consequently 1.090 signifies an answer carried to 3 decimals, while 1.09 signifies an answer carried to but 2 decimals. LOGARITHMS 117 EXERCISES 128. Solve for a?; 1. 7* = 17. 3. 9»+^ = 87. 6. 25*"+» = 795. 2. 126* = 98. 4. IS*' = 25. 6. 9* = 13. PROBLEMS 129. 1. In what time will $ 1 double itself at 4 %, interest compounded ajinually ? At the end of the first year the interest is $ .04, the amount is $ 1.04. At the end of the second year the interest is $ (1.04)(.04), the amount is $1.04 + (1.04) (.04) = I (1.04)2. At the end of the third year the interest is $ (1.04)2 (.04), the amount is $(1.04)8. At the end of x years the amount will be $ (1.04)'. We have the equation, (1.04)* = 2. x= 17.7 nearly. 2. In what time will $ 1 treble itself at 4 %, interest com- pounded annually ? 3. A sum of $ 1000 bears 5 % interest compounded annually. What is the amount after 20 years ? 4. Compute the compound interest on $ 879 at 4 % for 13 years. HISTORICAL NOTE 130. Logarithms constitute one of the most useful inventions in mathematics. A great French astroaomer once said that logarithms ^* by shortening the labors, doubled the life of the astronomer.*' Logarithms were invented by John Napier, Baron of Merchiston, in Scotland, who in 1(314 published a table of logarithms and described their use. An inde- pendent inventor of logarithms was the Swiss watchmaker and astron- omer, Joost Burgi, who published a table in 1620, at a time when Napier's logarithms were already known and admired throughout Europe. Napier's and BtLrgi's logarithms differed somewhat from each other and from the logarithms described in this book. Our system of logarithms to the base 118 ELEMENTARY ALGEBRA 10 was designed by John Napier and his friend Henry Briggs, conjointly, before the year 1617. At one time Napier lived in a beautiful castle on the banks of the Endrick. Tet even in such surroundings he was not free from annoyances which hindered intellectual work. On the opposite side of the river was a lint mill, and its clack greatly disturbed Napier. He sometimes desired the miller to stop the mill so that the train of his ideas might not be Interrupted. Much of Napier^s time was taken up in the management of his estate. Joost BUrgi was not rich, like Napier ; the necessity of earning a livelihood greatly hampered his scientific work. But both Bttrgi and Napier were men who loved mathematics and were able to achieve great results in spite of many hindrances. y 7 iti CHAPTER V QUADRATIC EQUATIONS AND THBIK PKOPBKTIXS REVIEW 131. In solving a quadratic equation of the form (j^ ^hx -h c = by " completing the square," the first step in our ex- planation was to divide both sides of the equation by the coeffi- cient of aj*. There are other methods of procedure which possess certain advantages ; fractions having large denominators may be avoided, or, fractions may be avoided altogether. These methods will be explained in § 191. In the following exercises solve by any method. To find the square roots of numbers, use the tables in § 197. 1. 6aj2-f-20a; = 17. 6. 35 a?,- 3a^= 11. 2. aaj» -I- afta; -f- 4 aft = 0. 6. 3aa^-f- 6 a; — 5a* = 0. 3. 5aj«-h21a;=30. 7. 40a?-3a^ = 12. 4. oo* -f- o&c -t- a« = 6». 8. (a+6)«*-|-2(a— 6)aJ-f-a=0. EQUATIONS QUADRATIC IN FORM 132. Equations like o^ -f- osff" -f- 6 = are said to be quad- ratic in form, because the unknown x occurs only with the exponents n and 2n, where one exponent is twice the other. Equations of this form can be solved by the methods used in solving ordinary quadratic equations. 1. Solve a;*-7aj*-f-10 = 0. Factoring, (a;« - 6) {x^ - 2) = 0. a;3 - 6 = 0, «a - 2 = 0. From the tables in § 107 we obtain x = ± 2.236, x =± 1.414, approxi- mately. 119 120 ELEMENTARY ALGEBRA 2. Solve aJ»-f-3aj»4-l=0. Complete the square, a^ + S a^ + { ^^ i 4. | = |. x=-y_,±:^. As there are in general three cube roots to a number, and there are in this ease two different numbers under the radical sign, it is seen that there are six roots of the g^ven equation. Of these, four are imaginary. As we wrote the answer, only the two principal roots are indicated. BXBBOISBS 133. Solve: 1. ic* 4- 6 a^ - 16 = 0. 6. 6 m« - 23 m» 4- 20 = 0. 2. a«-2aj*4-l = 0. 7. y»-2y»-143 = 0. 3. aJ»-h3aj»-hl = 0. 8. 2;* = 266. 4. a?«-8aj*-20 = 0. 9. y«"-52r + 6 = 0. 6. 2a?«-7a^-30 = 0. RELATIONS BETWEEN ROOTS AND COEFFICIENTS 134. From the solution of aac* -|- to -|- c = 0, we obtain — 6 ± V6» — 4 ac X =! • 2a If we designate the two roots by x^ and ajj, we have 2a V6*- -4ac 2 a V62- -4ac and Am = — ^ 2a 2a QUADRATIC EQUATIONS 121 The sum of the roots is _6 a The product of the roots is ^^ /- 6 + V6» - 4 acY^ b - V62-4 ac\ ^^ =(,^^ — 2^ — ^;(, — Ui — ) 6* — (6* — 4ac) 4ac c 4 a* 4 a' a If we divide both sides of ax^ -f- 6aj + c = by a, we obtain a a in which the coefficient of a^ is unity. We see that, in the last equation, (1) The sum of the roots is the coefficient of x with the sign changed; and (2) The product of the roots is the absolute term. To illustrate: Solve ac^ + 6« + 6 = 0. The roots are — 2 and — 8. Their sum is — 6, the coefficient of x with its sign changed ; their product is + 0) the absolute term. From this relation between the coefficients and the roots, we may form an equation, if the roots are known. Form the equation whose roots are — 1 and 6. We assume that, in the required equation, the coefficient of x^ is 1. Then, xi + X2 = 6f the coefficient of x with its sign changed ; xix% = — 6, the absolute term. Hence, the equation isx^ — 6x— 6=0. Or, we might write the equation thus : xa-(-l + 6)x+(-l-6)=0. Whence, ac* — 6 a; — 6 = 0. 122 ELEMENTARY ALGEBRA 135. Form the equation whose roots are : 1. + 5, 4- 2. 6. 2^, 3|. 9. V2, - V3. 2. -3,-8. 6. -i, -f 10. I4-V2, 1-V2. 3. 4,-6. 7. —a, —6. 11. — V5, — VS. 4. 7, 0. 8. m2, - 3 m\ 12. |V2, - 1 V2. If you know one root of a given quadratic equation, how can you calculate the other: (1) from the coefficient of x, (2) from the absolute term, (3) by the Factor Theorem ? Use each of these methods in turn, to find the second root in each of the following equations : 13. aj2 — 8.7 05 -I- 17.6 = 0, one root 5.5. 14. aj2 — 2.1 a; + .9 = 0, one root 1.5. 16. a;2 — 1.7 a; — 4.8 = 0, one root 3.2. 16. a;2 4- 11.8 X + 34.17 = 0, one root — 6.1. 17. aj2 -f 3.6 a: - 25,92 = 0, one root - 7.2. 18. a;2 — .085 x + .00175 = 0, one root .05. NATURE OF THE ROOTS 136. We shall discuss the nature of the roots of the quad- ratic equation asc* + 6» -f c = 0, in which the letters a, 6, c are assumed to be real and rational numbers* When some of these letters are imaginary or irrational, the conclusions which we shall draw do not necessarily hold. The solution of ax2 -j- 6aj 4- c = is . — b± Vb^ — 4 ac x = • 2a The nature of the roots depends upon b^ — 4:acy which is called the discriminant. I. When b^—4:ac=0, the roots are real and equal. Explain. II. When 62 _ 4 etc > 0, the roots are real and uneqtuil. If 62 — 4ac is a perfect square, both roots are rational; if QUADRATIC EQUATIONS 123 6* — 4ac is not a perfect square, both roots are irrational. Why? III. When 6* — 4 ac < 0, both roots are imaginary. Explain. To iUustrate: 1. In as^- lOx + 26 = 0, 6^ _ 4 ^j^ = 100 - 100 = 0. Hence the roots are real, rational, and equal. The roots are 6, 6. 2. In flr2 - Sjc - 28 = 0, 62 «. 4 ^ - 9 _|. 112 = 121. Hence, the roots are real and unequal; since 121 is a perfect square, both roots are rational. 3. Inx«-2a: + 6 = 0, 63 — 4ac = 4 — 24 =- 20. Hence both roots are imaginary, EXEBOISBS 137. Determine, without solving, the nature of the roots of : 1. aj2-aj-12 = 0. 6. 7a;2- 12ajH-4 = 0. 2. iB24.8a;-16 = 0. 7. 4 «2 - 12 «'= - 9. 3. a.2__8ic-25=:0. 8. 4 aj2- 12 a? -55 = 0. 4. 5aj2-f.7a; + l = 0. 9. a?2_2a;H-3 = 0. 6. 2aj2 4-3a; = 8. 10. 3aj2-f.l3a:-f-l = 0. For what values of a are the roots of the following equa- tions equal ? Real and unequal ? Imaginary ? 11. aaj24.4a; = — 1. 13. 3a«-h3aj + a = 0. 12. aj^^ cue 4- 5 = 0. 14. ax'^ -{- ax -\- 2 = 0. GRAPH OF THE QUADRATIC EQUATION, ax^-^bx-^c = y 138. The value of the expression, ooj* -^ bx-\-c, for given values of a, 6, and c, depends upon the value of the variable x. For that reason the expression is called a function of x. If we write aoi^ -f- fta? + c = y, then for every value of x there is a corresponding value of y. "^ The values of x and the corresponding values of y may be taken as the coordinates of points on the graph of the equation aa? -h 6a5 4- c = y. 124 ELEMENTAHY ALGEBRA 1, Draw the graph of a!* — 4 a! — 5 = y. " V Point 16 A 7 B C -6 D -8 B -9 F -8 -6 S -1 I -2 7 J -8 16 . K ThuB it will be seen (Fig. 17) that the graph crones the x-azis at the points (5, 0) and (-1,0). Upon solving ib«-4i-6^0, we find tbat X = + 6 and — 1, and these an th« values of x which make ^ = 0. QUADRATIC EQUATIONS 2. Draw the graph of a? - - f Poinl 6 A 6 i B C 3 D 2 E 1 F In rig. 18 It is seen that tbe grapb touches the z-axis, but does not ctobs it. The graph tumB and proceede upward. Instead of two points of intersection with the x-asis, as in Fig. IT, there is la Fig. 18 one point o( contact; the two points have UDlted in a single point. Evidently there are two roots at that point. Upon solving the equation we find that z = 3, S ; 3 is called a doable root. BLBMBIfTARY ALOBBBA 3. Diaw the graph of a^ — 2x+ 3 = y. =« * Point 6 18 A 4 11 B 3 6 C 3 8 D 1 2 E 8 F _ 1 6 -2 11 H -3 18 I Fig. 19 UluHtratea the fact that not eyeiy gnph cati the z-axis. K we solve z> - a x 4- 8= 0, we find that K = 1 ± v'^. Now V^^ l8 an imaginary Dumber ; therefore, there U no real valae of z which will make x> - 2 x + 8 equal to 0. The grapb makes this fact evident at a glance. We see that the two niota of the eqoaUon ax* + bz -1- c = may be real and unequal, as in Fig. IT, leal and equal, as in Tig. 18, oi imaginary, as In Fig. 19, CHAPTER VI SYSTEMS OF EQUATIONS SOLVABLE BY QUADRATICS I. ONE EQUATION IN THE SYSTEM IS LINEAR 139. Solve aj« + 3/2 = 26. « - 3/ = 1, (1) (2) Let us solve this system, first graphically, then alge- braically. The equation a; — y = 1 is linear and is represented by a straight line. The graph of aj* -f- y* = 25 is a circle. X -y = 1 X y Point 1 -1 A D x^ + y2 = = 26 X y Point 6 C 4 ±3 D,E 3 ±4 F, G ±6 H,I -3 ±4 K,L -4 i3 M, N -6 Fig. 20 shows the graph of x^ + y'^ = 26, (1), and « - y = 1, (2). The line (2) intersects the circle in two points (—3, — 4) and (4, 3). The coordinates of these points satisfy both equations and are there- fore the values sought ; namely, 05 = 4 and y = 3 ; x = — 8 and y = — 4. 127 ELEMENTARY ALGEBRA 140. Observe that x and y, when subject to the one condi- tion a? + t^ = 25, are variables, because they are capable of taking on successively a never ending aeries of different values. For the same reason x and y are also variables when subject to the one condition x~y = l. But x and g can no longer aa- same successively a never ending series of different values, ■when they are subject at the same time to both conditions^ ('' ~ ( . Hence x and y are now no longer variables, x^y=\ f but constants. It is the purpose of the solution to find the values of these constants. In drawing graphs we are dealing with variables; in finding the points of intersection of these graphs, we are finding the values of constants which, at the outset, are unknown. SYSTEMS OF EQUATIONS 129 141. The algebraic solution of aj« 4. y« = 26 (1) x-y = l (2) can be effected easiest by the method of suhstitutixm. From (2), x = 1 + y. Substitute in (1), 1 + 2 y + y2 _|_ ^a = 26. Whence, y^ ^ y _ 12 = 0. (y + 4)(y-3)=0. y = — 4, or + 3. Substitute in (2), x = - 3, or + 4. The two sets of roots are, ac = — 3, y = — 4, and as = 4, y = + 3. In checking, substitute the values of x and y in both of the given equations ; the answers might be wrong, yet might satisfy one of the two equations. For instance, x = 3, y = 2 will satisfy (2), but not (1). If in Fig. 20 the line should move, parallel to its present position, until it were tangent to the cirele, we would have just one set of values satis- fying the pair of equations. Show that this is the case, if the equations are 052 ^ y2 ::= 26 and ac — y = 6 V2. If the line should not intersect the cirele, nor be tangent to it, we would have a set of imaginary values satisfying both equations. Show that this is the case, if the equations are x^ + y^ = 25, x — y = 10. 142. Instead of elimination by substitution, some special device is frequently used. 1. Solve ic«-y» = 5, (1) x-y = -b. (2) Divide (1) by (2), x + y = - 1 (3) x-y=-6 (2) Add (2) and (3), 2 x = - 6 X =-3 Subtract (2) from (3), 2 y = 4 y = 2. Ans. x=— 3, y = 2. ELEMENTARY ALGEBRA '- -v=- 6 j?_^ = S « V Po[nt - V Points -5 6 A B 3 -8 « -4 ±a ±2 ±3.3* ±8.3* A,Ai S,Bi C, C, E,E, Ezplaiu bow Fig. 21 exLibita the values of X and y which nitisEy eqnations (1) and (2). Square (1), (2)- 4, Add (8) and (4), Extract square root, Add (6) and (1), Snbtroct (1) from (6), The MlB of roots are iEys= — 5, -ixv + y^ = 36. x' + 2aK + v^ = 16. x + y = a) (2) (S) (<) (5) («) (1) SYSTEMS OF EQUATIONS 131 8. Solve a' + y» = 20, (1) xy = 8. (2) MulUpl? (2) by 2, 2 a^, = 16. (3) Add (1) and (8), i" + 2 37 + k^ = 36. (4) Subtract (1) and (S), a? - 2 xy + v* = 4. (6) Extract square root of (4) itnd (6), Whence, 2a:=±8, ±4, a: =±4, ±2. 2y=±i, ±8, If = ± 2, ±4. The sets of rootfl are, i =± 4, v = ± 2; a: =± 2, v = ±*- The graphic soIuUon is exhibited in Fig. 22. 132 ELEMENTARY ALGEBRA X^ ' + y« = 20 X y Point ±4.47 A,Ai ±4.47 B,Bi 2 ±4 C, Ci -2 ±4 A A 3 ±3.3+ ^, ^1 -3 ±3.3+ F,Fi 4 ±2 a, Gi -4 ±2 H,Hi jcy = 8 X y Point ±2 ±4 a, A ±3 ±2} /, /i ±4 ±2 ^,^1 ±5 ±^ A A 4. Solve aj«-y» = 91, a? — 3/ = 7. t Divide (1) by (2), x^^-xy-\-y^ = 13. Square (2), 052 - 2 ay ± y2 -- 49. Subtract (4) from (3), 3 xy = - 36, 3cy=-12. Add (5) and (3), a;2±2xy±ya = l. Extract square root a5 + y = ± 1- 05 — y = 7. Whence 2 X = 8, 6, X = 4, 3. And 2y=-6,. -8, y=-3, . -4. The sets of roots are aj = 4,y=-3; x = 3,y 6. Solve aj*-h 2^ = 97, a; 4- y = 5. =-4. Let x = t« + r; y = w — t>. Substitute in (1) and (2), «* + 4 m8» + 6 mV 4.4Mt>8 + v*±t«*-4M8v±6 m%2_ 4 u«« + «* Whence, 2 u* + 12 u2»2 + 2 1;* = 97, and u±t> + M — 1> = 6. Whence, 2 u = 6, (1) (2) (8) (4) (6) («) (7) (2) (1) (2) 97. (3) (4) (6) SYSTEMS OF EQUATIONS 133 Substitute u = J in (4), a|A + 75 »2 + 2 r* = 97. (6) Whence, 16 r* + 600 »2 = 151. Completing the square, 16 «* + 600 1?2 4 5625 = 5776, 4v« + 75=±76, 4 1?2 = 1, - 151, t^ = i -H^» Hence, « = 3, 2, J ± J V- 161, The sets of roots are y = 2, 3, J q: i V- 151. 05 = 3, y = 2; x = 2, y = 3; a; = J±J V- 151, y = J ^ } V - 151. Check each set of real roots. The check for the imaginary roots is as follows : (1) 3ff.^ i|i V- 161 - u^fiA T ^i^ V- 151 + ^¥<P + W :p ijA V-IST _ xi|2A ± ij4 V- 161 + »3^ = 97, i|Ji = 97, 97 = 97. (2) jijVirirH-f TiV^=T5l = 5, 5 = 5. 143. The equations aj* -|- ^ = 97 and x-\-y = 5 are called symmetrical equations, because they remain unaltered when x is written for y and y is written for x. An expression is symmetrical with respect to two or more letters, if it remains unaltered when the letters are inter- cttanged. Every system of two equations that are symmetrical, or sym- metrical concept for the signs, can be solved by assuming x^u-^-v and y = u — V, Instances of equations symmetrical except for the signs are a* — y* = a, a — y = 6. 134 ELEMENTARY ALGEBRA II. A SYSTEM OF TWO EQUATIONS IN X AND F, BOTH QUADRATIC 144. Any tioo equations of the form aa^ ■i'by^=c,or aa? + hxy = c, or oo* + bxy -\- cy^ == d, can be solved by assuming y = vx, and determining the constant v. It is seen that in these equations all the terms which contain x and y are of the second degree in z and y. Some authors call these equations homogeneous. We avoid this term for the reason that the word homoge- neous is more commonly used to designate equations like ax^'^hxy-\'Cy^=0, in which every term of the equation contains x and y^ and is of the same degree in x and y. See § 8. Solve a^-\-xy = — l, (1) y'-2xy = S. (2) Let y = vx. Then x^ + rx* = - 1, (3) and fAc^^2vx^ = 8. (4) From (3), (1 + v)x^=-h and «» =-^J-. (6) 1 + 1? From (4), («2 - 2 v)x^ = 8, and x^ = — ^ (6) •• i + t,-t^-2i? . ^-^ Whence, - «« + 2i? = 8 + 8», (8) and t^ + 6i? + 8 = 0, (9) and (v + 4)(i? + 2) = 0. Then v = - 4, - 2. Substitute in (6), x^=^ = l, and a;2 = — = 1, K=±jV3, ±1, y=zvx = Tiy/S, T2. The two sets of roots are a; = ± J VS, y = =F J\/3 ; x =±1, y = T2, To find the value of y care must be taken to use the value of v with that value of x which was obtained by substituting the value of v. When — 4 was substituted for v in (5), yielding x = ±}V3, the — 4 must be multiplied by ± J\/3 (since y = vx), to obtain the correspond- ing value of y. Hence the values of x and y must be carefully paired as shown above. When ± and =F are used in the answers, the upper signs go together and the lower signs go together. SYSTEMS OF EQUATIONS 135 145. The method just explained possesses the great advan- tage of always yielding results. Very often, however, much shorter solutions can be given by special devices. Frequently a third equation can be derived from the two given equations which is simpler than one or both of those given. The solu- tion is then obtained by the use of this simpler equation along with one of the original ones. For example, solve a* — 3 ajy = -143, (1) 2^ + ajy = 168. (2) By adding (1) and (2) the shnpler equation is obtained, x^ -2icy-hy2=26. Extract the square root of both sides, « — y = ± 6. (8) (3) is a linear equation. It gives X = ± 6 + y. Substitute in (2), y2±5y4-y2=168, 2y^±6y = l6S, Taking the upper sign in ±6y, _ 6 ± V1369 Taking the \ower sign in ± 6y, y = ^'r'' Hence, y = 8, -8,V. -¥• Substitute the values of y in (2), a: = 18, -13, Y, -^. The sets of roots are a; = ±13, y = ±8; x = ±^,y=±^. POSSIBILITY OF SOLUTION BY QUADRATICS 146. It is readily seen that when, in a system of two equa- tions, one equation is of the second degree and the other is of the first degree or linear, a solution may always be obtained by quadratics. If, however, both equations are quadratics, this is usually not the case. Only special types of quadratic equations, such as have been studied in this chapter, and others of similar 136 ELEMENTARY ALGEBRA character, admit of being solved by quadratics. The general case is far more complicated. Given '^ aal^ + bxy -\- q/^ -\- dx -\- ey -\-f= and Oiic* + b^xy + c^ + d^x + e^y +/i = 0, to find X and y. If y is eliminated by substitution, the result is a quartic equation, which is of the fourth degree and cannot be solved by quadratics, except in special cases. It is shown in more advanced algebras that the algebraic solution of a quartic equation depends in general upon the solution of a certain cubic equation or equation of the third degree. The solution of cubic and quartic equations is not explained in this book. The given equations of the second degree in x and y can be solved by quadratics whenever the auxiliary cubic equation here mentioned possesses a rational root. This rational root can be found by the factor theorem (see § 78) ; the other two roots of the cubic can then be found by quad- ratics, as can also the four sets of roots of the given equations. (See F. Cajori, Theory of Equations, New York, 1914, pp. 71-73.) BXBBCISBS 147. Solve 1. 2 ic* — 3 a?y = 5, 6. ic* + 3 a?^ = — 5, oj — y = 2. 2xy + y^ = -'l^, 2. a?-|-y = l, 7. a + 2/ = a, iK* + y2 = 85. 4:xy — a^=z — b^, X y 15' x^ — y^ = 4: mn. 1_^1^_34 9. aj*-y* = 2401, a^ f 225' 352^2/2^49. 4. aj» H- y» = 28, lo. ar' + 7 a?y = - 104, a + y = -2. 5xy-'y^=:-129. 6. xy = 45, 11. a5 + y* = 33, aj + y = 14. x + y = 3. SYSTEMS OF EQUATIONS 137 12. x-y = ly 15. aj3 -I- y3 «. 109^ a , y _. 2 J 05*^ + ojy* = — 36. 13. L-3y = 0, "• f-y'=10. 7a^ + 3a,-4.V = 43. 2a- + 6;.^ + 33^' = 14. 14. a:^-'y^ = m^y 17. 3aj2-32/2 = 9, 18. a^—7x + y^—7yy 3(a + 2/) = -2icy. PROBLEMS 148. 1. Prices of two kinds of bicycles are such that 7 of one kind and 12 of the other kind can be obtained for $640. Three more of the latter can be purchased for $180 than can be purchased of the former for $120. Find the price of each. 2. A person lends $ 2500 in two separate sums at the same rate of interest. At the end of one year the first sum with interest amounts to $997.50; at the end of two years the second sum with (simple) interest amounts to $1705. Find the two separate sums and the rate of interest. 3. Prove that if the sum of two real numbers is multiplied by the sum of their reciprocals, the product cannot be less than 4. Let X and y be the two real numbers, p the product considered ; solve for the unknown ratio -. y 4. A traveler starts from A toward B, another traveler starts at the same time from B toward A. In two hours they meet 20 miles from A. When the second traveler arrives at A, the first is still 13^ miles from B. Find the distance be- tween A and B. 138 ELEMENTARY ALGEBRA 6. If in aoc^ + to -|- c = 0, the coefficients are related to each other in such a way that a -h 6 = ^ and a = 2 c, what must be the value of a, so that 8 will be a root of the given quadratic equation? 6. The difference between two numbers is 12, and the difference between their cubes is 7488. What are the two numbers ? 7. A sum of money at simple interest for four years amounts to $2240. Had the rate of interest been 1 % higher, the sum would have amounted to $80 less than this in two years. Find the capital and the rate. 8. The sum of the areas of two circles is 694.2936 sq. in. ; the sum of their radii is 21. Find their radii. 9. A circular track is constructed so that the width of the track is ^ of the inside diameter. The area of the track is 2600 sq. yd. What are the inside and outside lengths of the track? 10. Find two numbers such that their sum is equal to their product and also to the difference of their squares. 11. Find two numbers such that the sum of the numbers is equal to the sum of their squares and also to twice their product. 12. Find two numbers whose sum is 8 and whose product is 26. 13. Find three numbers such that the sum of the squares of the first two is equal to three times the square of the first minus the square of the second, and is also equal to the first minus the second plus twice the third, and also to the sum of the three numbers. CHAPTER VII EXPONENTS, RADICALS, IMA6INASIES MEANINGS OF DIFFERENT KINDS OF EXPONENTS 149. The different kinds of exponents which have been stud- ied thus far have been interpreted in the following manner, m and n being taken to be positive iategers : a" = a • a • a ••• (to n factors), 1 -- 1 a~* = -^i or more generally, a " = — ;;;-, where a ^ 0, a •» a° = 1, where a =^ 0, While a number a has in general n different nth roots, it is agreed that 1 only one of them shall be represented by a** or Va, namely, the so-called principal root This restriction was made in order to avoid unnecessary complication and confusion in the interpretation of expressions and equa- tions involving radicals. Accordingly, Vi = + 2, — Vi = — 2, \^ = + 2, -\/8=-2, \/^^ = -2, -v^^=4-2, v^=+2, etc. The exponents considered above are all rational numbers. But other exponents have been brought to our attention. In the study of logarithms, mention was made of the fact that logarithms are often irrational numbers. Since logarithms are realjy exponents, it follows that exponents may be irrational. In more advanced books still another kind of exponent is considered, namely, the exponent that is an imaginary number.* * For a fuller treatment of the theory of exponents, consult H. B. Fine, College Algebra^ 1904, p. 376. 139 140 ELEMENTARY ALGEBRA OPERATIONS WITH EXPONENTS 150. Operations involving exponents are subject to the fol- lowing laws : 1. Law of multiplication : a"^ -a'* = a"*"*^. 2. Law of division : a"^ -^ a"" = a"*"". 3. Law of involution : (a*)* = a"»*. 1 m 4. Law of evolution : (a") ** = a" . DIFFERENT KINDS OF NUMBERS 151. In the study of arithmetic and algebra several different kinds of numbers have come to our notice. These may be classified as follows : Numbers Real (Classified according to signs) (Classified according to express- ibility in Hindu-Arabic nu- merals) Positive Negative Rational Irrational, as V6, ir = 3.14169... Imaginary Pure Imaginary, as hV— 1, where &^0 Complex Numbers, as a-\-hy/— 1, where 6^0 Arithmetic deals with real numbers, all positive, but some of them irrational. The irrational numbers of arithmetic arise in the process of finding roots. In algebra use is made of the numbers encountered in arith- metic, but convenience forces upon us the need of considering also negative numbers for the purpose of indicating relations of opposition, as temperatures above or below a certain fixed point, debts or assets, distances to the right or left, etc. In the solution of quadratic equations we meet a still different type. If we try to solve ic2 -|. 1 = 0, we are confronted with the symbol V— 1. EXPONENTS, RADICALS, IMAGINARIES 141 This result greatly embarrassed mathematicians of the eight- eenth and of earlier centuries ; no satisfactory explanation of it could be made at first. But now the symbols are recognized as constituting a new type of number, the so-called imaginary number, which deserves a legitimate place in algebra and is of great service in certain advanced developments of algebra, that are useful in the study of polyphasy electric currents and of other advanced topics in mathematical physics. Just as nega- tive numbers have been found truly useful in elementary algebra, so imaginary numbers have been found useful in ad- vanced algebra. The symbol V— 1, or this multiplied by any real number 6, such as 6V— 1, is called a pure imagi- nary number. Expressions of the type a ± 6V— 1, which are the sum or difference of a number a and of a pure imaginary b^/— 1 (a and b being any real numbers, but b ^ 0), are called GfmSplex numbers. The term " complex " was intro- duced because the parts of the expression are partly real and partly imaginary. SIMPLIFYING RADICALS 152. A radical is said to be in its simplest form : (a) When the index of the root is as small as possible, (b) When the expression under the radical sign, called the radicandy is integral, (c) When the radicand contains no factor with a negative ex- ponent, or raised to a power equal to or greater than the index of the root. -\/25 is not in its simplest form, because the index 4 of the root is not as small as possible ; we have y/2b = 6* = 5* = V5. ^\ is not ill its simplest form, because the radicand, \, is fractional. y/c^ is not in its simplest form, because the exponent 4 is greater than 8, the index of the root ; we have ■\/a*6 = ay/ab. \/3 ab^ is in its simplest form, since it fulfills all three conditions. 142 ELEMENTARY ALGEBRA 1. Simplify J^^. 3 gi gj 8 3 * Notice that the denominator was rationalized by multiplying numer- ator and denominator by 8^. 2. Simplify </3 a*b-^c. 6* ft ft 3. Other illustrations of simplifying are : '^7a« ^ 3^ g* _ 3^ ■ q ■ aM _ a j/Wab^ V72 - 36 VS = 36^(2 - v^)* = 6 V2 - V3. BXBBOISBS 153. Remove all negative exponents, compute the values, and simplify the expressions : 1. 27*. 4. 16"*. 7. 64"*. ^^' (A)"'- 2. 16* 5. 64-i. 8. (!)-». 11. (.16)*. 3. 27-*. 6. 64^. 9. 8"*. 12. (.125)*. 13. (_| + 6)0. 16. (100«+10)-^ ^g gm-n ^ g-mft 5-«-25-\ 14. 70x(.26)-*. 17- — 1 20. ^""^ 16. (.49/-'. 18. g'-^-g'". 21. «:l±«^+l. EXPONENTS, RADICALS, IMAGINARIES 143 22. V60. 25. V2000. 28. VST. 31. V- 800. 23. Vm. 26. 7V20. 29. \/48. 32. v^729. 24. V98. 27. 12 V8. 30. ^"08. 33. \/2450. 34. 2V6x5>/3. 38. 3Vlbx24V6-f-Vl5. 36. 5V2x7V6. 39. 3 ViO X 2 Vl5 -4- 3 V2. 36. 2Vl6 X 5V2 X 2V3. 40. 5Vl6 X 3V6 -4- 2V5. 37. 2Vl5-j-5V3x2V2. 41. (6.25)-*. 42. (2.25)1 46. (.216)1 60. * (^)*. 43. (.25)"*. 47. (.0625)-*. 61. ~ (6.25)i 44. (.125)"*. 48. (-.125)*. 62. (a*6*)«. 46. (-.008)-*. ^49. (fj)*. 63. (a26»)-i 64. (a-«6-»)"'^. S6. c^-^c^xcK 66. a^-^a^y. a*. 67. 2* x 64* -4- 729*. Eationalize the denominators of the following fractions and then find their values to three figures : 58. -i-. Hint. J-=-i..^ = ^. 69. ^- V2 V2 V2 V2 2 2V3 60. -4=- 61. -4=r- 62. V|. 63. ^^^. Vl4 V2.3 V:22 Calculate the values of the following expressions when m = V2, n = V3, p = V5, using the tables in § 197 : 64. 7 m* — n*. 67. phi^ — nha\ 70. (n+J>)w. 66. 1©* — wiV. 68. — -^ 71. ^ ^^ 2p pn « • . « s «A wi — w »»« ^* — ^** 66. mH^-\-rM. 69. ^ ^ - 72. 5jp* — mnp 144 ELEMENTARY ALGEBRA 78. Which of the expressions is the larger, V5 or v/lT ? V6 = 5* = 5* = 126^ \^ = 11*=11* = 121*. But 126^ > 12li Hence V5 is the larger. Arrange in the order of magnitude in each example : 74. \/ii, </2l. 77. 2, Vily v^. 76. V2, -v^^. 78. •^, ^/Vf, VO. 76. VO, \/9. 79. \/3, -5^, vT7. 80. Using logarithms, compute the value of y/7943. Log 7943 = 3.9000. Log \/7943 = 1.8000. v'mS = 19.96, nearly. 81. Using logarithms, compute the value of x =(59.45)"^. (59.46)"* = — ^ (69.46) I Log 1 = 0.0000 Log 69.46 = 1.7742. Log (59.46)' = j(1.7742) = 1.1828 Log X = 2.8172 X = .00664, nearly. Compute the values of the following radicals : 82. ^/96S, 84. -^790. 86. (154)"*. 83. VTOSS. 85. (12)A 87. (376)J. ADDITION AND SUBTRACTION OF RADICALS 164. The indicated additions and subtractions of radicals can be reduced to one term, only wh^n the terms in the ex- pression are similar after they are simplified. Thus, \/l8 + V60 = SV^ + 6 V2 = 8 V2; EXPONENTS, RADICALS, IMAGINARIES 145 EXBBCISBS 155. Perform the indicated operations : 1. V63+\/784. 3. A/24 + </8i+\/375. 2. V294+-v^576. 4. 50\/^^+<^7000-^56. 6. Vr^ - V9a^ - V16 a^dP. 6. 6V3+8Vf-|-V^. 7. 4V^-|-15v^-60V|. 10. w3\/--«S^+wV\/-. MULTIPLICATION OF RADICALS \ 156. E.eal radicals of the same index are multiplied to- gether according to the formula a -y/b X c -y/d = ocVftd. When the real radicals have different indices^ it is easiest to change to the exponential notation, then perform the multipli- cation and, after reducing fractional exponents to a common denominator, return to the radical notation. This may be shown as follows : m n Special care must be exercised when the expressions to be multiplied involve imaginary numbers. It is customary, for brevity, to represent the pure imaginary V— 1 by the single 146 ELEMENTARY ALGEBRA letter. t. Hence the complex number a+V— 16 may be written a -f ib. The following has been adopted as a definition of V— 1 or f : 7%e number V— 1 is one whose square is — 1. Accordingly, (V— 1)* = i* = — 1. All multiplications involving imaginary numbers must be made to conform with this fundamental definition. Particular care must be exercised to avoid the following procedure: V^n. times \/^n[=V(-l)(-l) = >/TT= 1. This procedure is in violation of the definition of V— 1 given above, and must therefore be avoided. Always take V— 1 x V— 1 = — 1, or The multiplication of expressions involving imaginaries can be performed easiest^ if we take the precaution to vrrite a radical V— a in the form V— 1 Va or i Va. For example, V— a X V— 6 = i Va X I* V^ = i^ Voft = — VoS. 1. Multiply 2 Va - 3 V26 + 4 V3 a6 by 5 Va6. 2 V5-3\/26 + 4 V3a6 10aV6-166 V2a + 20a6V3 2. Multiply Va + 2 V- 6 - 3 c by V-a6. Va + 2f>/6-3c tVa6 ai iVb^2hy/a-Sciy/ab EXPONENTS, RADICALS, IMAGINARIES 147 BXBBOISBS 157. Perform the indicated operations and simplify the result : 1. V5.V125. 18. V2^^-V2T«. 2. \/3-\/9. 14. ^^r^.vsTft. 8. SVa.SVoft. 15. (V2a+V36)». 4. -^cC^-i-y/a^xVaK le. (a+V6)». 6. -^ -!- Va X \/a». 17. (a + V^^)*. 6. aV^x26V^=n:. 18. Vab-^-Va. 7. 3i.4iV5. 19. </^^Va6. 8. 4iVc"j-2t. 20. SVx^^'^V23r^. 9. 5V^V3x6V^^. 21. V^^^a6-4-V^=^ 10. - M2xV|. 22. iaXi6V2xic. 11. (^a-bVcy. 23. V^Ta . v"=^ . V^=^. 12. — V2a' X V2a&. 24. t V« • t V^ • * Vy2. 26. ^J^^-g X \/^^ X ^v^^=^. 26. (2Vi — 3Vy+V2«)-5Vc^. 27. 6^^(2V«+Vy-«). 28. (2V2-3V^^H-^v/"=^-2</2)\/2. 29. (2V5-</6-2V^2. 80. (Vs-sV^^)*. 81. Change to an equivalent fraction having a rational V3+V2. denominator, V3-V2 If >/3 ->/2 is multiplied by VS + V2, the product is (>/3)2-(V2)2 or 3 — 2. Hence, multiply both numerator and denominator of the given fraction by VS + V2. We obtain (V3+V2)2 ^ 8 + 2V6-h2 ^ 6 + 2v^ ^g ^ g^/g (V3)2_(v^)2 3-2 1 148 ELEMENTARY ALGEBRA Change to equivalent expressions having a rational denomi- nator: 82. ' ^ . 34. ^-^-A 86. t±^. V3-VO Va-Vft 2-hV3 33. — 1 36. ^■^^. 37. ^ V64.V5 V2-V6 2V5-3V3 38. Is — 1 -h V5 a root of the equation aj2 4-2aj — 4 = 0? Write - 1 + V6 for X, (_ 1 4. V6)a + 2(- H-V6)- 4 = 1 - 2V6 + 6 - 2 + 2V6 - 4 = 0. Hence, — 1 + V6 is a root of the equation. Is the binomial written after each equation a root of that equation ? 39. aj»-4aj-l = 0, 2-hV3. 40. aj2-a;-3 = 0, |(l-Vi3). 41. a;2-h3a;-hl = 0, 3-hV5. 42. 2aj2-aj-2 = 0, i(l-Vl7). THE SQUARE ROOT OF a±2y/b 158. Since ( V3 ± V2)«= 3 ± 2 V6 + 2 = 5 ± 2 V6, it follows that V6±2V6 = V3± V2. We notice that 5 is the sum of 2 and 3 ; 6 their product. This suggests the following rule for finding the square root of a-h2V6: Find two numbers whose sum is a and whose product is b. Write down the square root of the larger number^ plus (or minus) the square root of the smaller. This rule is of practical use only when the two numbers sought are rational. In other cases the resulting expression is more complicated than the given expression; hence to be rejected. EXPONENTS, RADICALS, IMAGINARIES 149 1. Find the square root of 8 + V^O. Writing the expression in the form a ± 2Vbj 8 + 2\/i6. The two numbers whose sum is 8 and product 16 are 6 and 8. Hence, V8+V60 =Vb + VS. 2. Find Vg - 4 V2. Changing to the form a ± 2 V6, 6 — 2v^. The two numbers whose sum is 6 and product 8 are 4 and 2. Hence, Vo- 4V2 = Vi - v^ = 2 - V2. BXBBCISBS 159. Find the square root of : 1. 3-h2V2. 7. 9-h6v^. 2. 7-4V3. 8. I-V2. 3. 9-h2Vi4. 9. t^-VJ. 4. 9 4-2V20. 10. X — Vaj» - 1. 6. 16 4-6V6. 11. (a2-h4 6)-4aV6. 6. 8-h2V7. 12. x-^2y + 2V2xy. IRRATIONAL EQUATIONS 160. An irrational or radical equation is an equation in which the unknown number x and expressions containing it occur under radical signs or with fractional exponents. For example, 2 — V« = Va? — 1, aj^ -|- 5 «» = 6. It is agreed among mathematicians that in equations of this sort, only the principal roots ot^/x, Va; — 1, x^ are to be taken. The solution of irrational equations may be explained by examples. 150 ELEMENTARY ALGEBRA 1. Solve 2 - Vic - Va? - 1 = 0. Solution. Transpose one of the radicals, 2 ^ Vx = y/x— 1. Square both sides, 4 — 4 V« + x = x — 1. Simplify and transpose to the right all terms except 4 Vie, 4>/x = 5. Sqoare both sides again and solve for x, ^ = il' Substitute in the given equation, 2 — Vf | = V{J — 1, Hence x = f | is the value sought. »=t 2. Solve 1 -h Va; -h 3 = V«. Square both sides, 1 + 2Vx + 3 + x + 3 = x. Simplify and isolate the radical, Vx + 8 = ^ 2. Square both sides again, x + 3 = 4« x=l._ Substitute x in the given equation, 1 + \/4 = VI, 1 + 2 = 1. This is absurd ; x = 1 is not a root of the given equation. This is an example of a radical equation which cannot be satisfied by any value of x. The equation is impossible. It is therefore not true of radical equations that values of x satisfying them always exist. It will be instructive to examine just how the false value x = 1 was obtained. When both sides of an equation involving x are squared, new values of x may be introduced, giving rise to extraneous roots. By sub- stitution we can ascertain in which place the extraneous root came in. It is found that x = 1 does not satisfy equations (1) and (2), but it does satisfy 3. Hence, the extraneous root x = 1 was introduced when both sides of the equation Vx + 3 = — 2 were squared. Extraneous roots arise here in the same way as in the solution of frac- tional equations. One simple way of determining whether a value of x obtained by squaring both sides is an extraneous root or not, is by sub- stitution in the original equation ; if it is not satisfied, then the root is extraneous. From the explanation given above it is evident that, though all opera- tions in finding x may be performed without error, the answer may never- theless be false. It is therefore necessary to substitute the value of x in the original irratiohdl equation, to ascertain whether that value is true or false. EXPONENTS, RADICALS, IMAGINARIES 151 While x = l IB not a root of 1 + Vx + 3 = Vx, it will be noticed that it 18 a root of 1 — Vx + 3 = — Vx. The solution of the latter equation gives rise to equation (3), and to x = 1. 3. Solve -Vx'^ -h V3a;-h3 « 3. (1) Transpose so that the most complicated radical stands alone on one side of the equation, , , ' Va;-2-3=-V3a; + 3. (2) Square both sides, a; — 2 — 6y/x — 2+9 = 3a; + 3. (3) Simplify, keep only the radical on left side, — 6 Vx — 2 = 2 x — 4. (4) Divide both sides by 2 — 3 Vx-2 = x - 2. (6) Square both sides, 9x— 18=x3— 4x+4. (6) Solve the quadratic x^ - 13 x + 22 = 0. (x-2)(x-ll) X = 2, X = 11. Substitute x = 2 in (1), y/2^^ + V9 = 3. 3 = 3. Hence X = 2 is a root of (1). Substitute x = 11 in (1), Vll-2 + V33 + 3=3. 3 + 6 = 3. Hence x = 11 is not a root of (1). Nor is it a root of (2), (3), (4), and (6) . But it is a root of (6). It follows that x = 11 is an extraneous root, obtained in squaring (6). 4. Solve V6 — a; -h V« + 7 = 6. Place one radical on one side by itself, V6-x = 6— Vx+7. Square both sides, 6 — x = 26 — 10 Vx + 7 + x + 7. Simplify and keep only the radical on left side. 10Vx+7 = 2x + 26. Divide both sides by 2, 6 Vx+7 = x + 13. Square both sides, 26 x + 176 = x^ + 26 x + 169. Solve the quadratic x* + x — 6 = 0. (x + 3)(x-2) =0. X = - 3, X = 2. 152 ELEMENTARY ALGEBRA Substitute x = ^ 3 in the given equation, >/9 + \/4 = 6, 3 + 2 = 6. Hence z = — 3 is a root of the given equation. Substitute x = 2 in the given equation, V4+\/9 = 6, 2 + 3 = 6. Hence x = 2 is also a root of the given equation. BXBROISBS 161. Solve the following irrational equations, testing care- fully each value of x that may arise : 1. Va? + 4 = 4— Vof — 4. 2. Va?-2+Vaj-|-3 = 6. 3. Va;— l4-Va-h6 = V4aj-|-9. 4. Vic+Vx+-l = V2a;-|- 1, 6. Vy^^ = Vy-hV2. 6. vV-hl-Vy+l = 0. 7. 4V5aj*- 5a: -1-9 + 3 = 31. 8. Vaj* + 21+a=21. IRRATIONAL EQUATIONS QUADRATIC IN FORM 162. The equation a;« + 5 «» + 6 = is said to be quadratic in fomiy because it contains only two different powers of x, one exponent of x being double the other exponent. If we put x^ = y, then %' = y^ and we obtain the quadratic y^ + 5y + Q = 0. Solving this we obtain y = — 2 and 2( = — 3. Hence, x ■ = — 2 and a;» = — 3. Cubing both members, x=—S and a; = — 27. EXPONENTS, RADICALS, IMAGINARIES 153 Substituting x = — 8 in the given equation, (_8)*+5(-8)* + 6 = 0. Simplify, (-2)2 + 6(-2)+6=0, 4 - 10 + 6 = 0. Hence, x = — 8 is a root of the given equation. Substituting x = — 27 in the given equation^ (-27)*H-6(-27)t + 6 = 0, (-3)2+6(-3)+6 = 0, 9-16 + 6 = 0. Hence, x = — 27 is a root of the given equation. Solve a:* - 7 «* = 8. The exponent | is double the exponent } ; hence the equation is quad- ratic in form. Let X* = y, then x* = ^, Solving, y2 _ 7 y _ 8^ y = 8, y = -l. We obtain x' = 8 and x^ = — 1. (1) The exponent f in x* signifies the third power of the fourth root of x ; we wish to find x itself. To find the value of x^, we must extract the cube root of both sides of (1). We obtain x* = + 2 and x^ = — 1. (2) To find the value of x, we must raise both sides of (2) to the fourth power. We obtain x = 16 and x = 1. Substitute x = 16 in the given equation, (16)^ — 7(16)^ = 8, (4)3_7(2)« = 8, 64 - 66= 8. Hence, x = 16 is a root of the given equation. Substitute x = 1 in the given equation, (1)^ — 7(1)* = 8, 1-7 = 8. Hence, x = 1 is not a root of a given equation. Where did this false value enter ? We see that x = 1 satisfies neither x* = — 1 nor x* = — 1 ; hence this extraneous root appeared when the sides of x^ = — 1 were raised to the fourth power. 154 ELEMENTARY ALGEBRA In solving (1), we might have reversed the two operations, by first rais- ing both sides to the fourth power and then extracting the cube root. But the operations would have involved larger numbers iu the case of %* = 8. Be careful to avoid mistakes in solving equations like x* = 8. A frequent error is to conclude from x* = S that X = 8*. As a matter of fact, x = 8>. The safer way is to proceed by two steps as is done above : The first step (extraction of the cube root) changes the exponent of x from i to }. The second step (raising to the fourth power) changes the exponent of x from ^ to 1. * BXEBOI8B8 163. Solve: 1. aj*-7xl4-10 = 0. 2. 8ajl-66ajl4-8 = 0. 3. 2* -32* + 2 = 0. 4. (a? -h 1)* - ll(a; + l)i 4- 30 = 0. 6. a;*-a;i + 132 = 0. 6. y* -h 19 yA- 216 = 0. 7. (a; -h 1)» - 60(a; + l)i = 256. 8. a;*-h6a?*— 55 = 0. GRAPHIC REPRESENTATION OF COMPLEX NUMBERS 164* The graphic representation of ordinary positive and negative numbers is familiar to all students of algebra. On a straight line, usually drawn horizontally, a point is chosen as the starting point or origin ; positive numbers are shown by distances to the right, negative numbers by distances to the left. The graphic representation of complex numbers like a + 15, where i = \/— 1, is not so obvious. In the seventeenth century John Wallis of EXPONENTS, RADICALS, IMAGINARIES 155 the University of Oxford made attempts to find a geometrical interpre- tation of such numbers, but was not able to devise a general and con- sistent scheme. Not till the close of the eighteenth century did a satisfactory plan suggest itself to mathematicians. In 1797 a Norwegian surveyor, by the name of Caspar Wessel, found a graphic representation which agrees with that now adopted. But his writings failed to attract the notice of mathematicians. A little later a Frenchman by the name of Argand had a similar experience. More successful in reaching the ear of the mathematical public on this matter was the German mathematician, Carl Friedrich Gauss. He placed the theory of complex numbers on a firm basis and, by his reputation, induced others to enter upon a more careful study of this subject and to adopt the geometrical interpretation that is given now in works on algebra. 4, -3,-2, -i, ■I i 1 H +1, +2, +3, +4, +6, ■H 1 1 1 H -•-2/1 Fio. 23. 165. This interpretation of imaginary or complex numbers is really quite simple. Two axes are drawn perpendicular to each other, one axis horizontal, the other vertical, as in Fig. 23. Along the horizontal axis real numbers are marked off, positive 156 ELEMENTARY ALGEBRA numbers to the right and negative numbers to the left of the origin. The pure imaginary numbers -|-V— 1, -|-2V— 1, -f 3V— 1, ••• are marked off along the perpendicular axis, from the origin up. The pure imaginary numbers — V— 1, — 2V~ 1, — 3 V— 1, ••• are marked off along this same perpendicular, but from the origin douni. Complex numbers are pictured in this diagram by points in the plane. In case of 3 -f 1 4, measure off three unit distances on the horizontal axis to the right, and 4 unit dis- tances up. This locates the point A in the plane (Fig. 24). The point ^ in the plane is accordingly the graphic representation of 3 + *4. To find the geometric representation of 3 — 1 4, pass from the origin 3 units to the right, as before, and 4 units down; the point B in Fig. 24 represents 3— i 4. Similarly, C represents — 3 — i 4, and D represents -3 4-i4. The representation is similar for other complex numbers. It is seen that every complex number can be represented in this way, and that every point in the plane stands for some number. All real numbers are confined to the horizontal axis ; all pure imaginary numbers are confined to the vertical axis ; all other numbers a ± ih lie in one or another of the four quadrants. The axis for the pure imaginaries is used in much the same way as the y-axis when the graph of an equation containing the variables x and y is drawn. Only now the vertical axis carries imaginary numbers, while before the y-axis carried real values of the variable y. " "" 1 ■\ , A J J ^ ^ A f^ 1 , T J< ' >u Fig. 24. EXPONENTS, RADICALS, IMAGINARIES 157 BXBBCISBS 166. Locate the' points whicli represent the following com- plex numbers : 1. 5 4-* 4. 4. 5 — 16. 7. 3_4V31. 2. _5_i6. 6. -1-h*. 8. 44-5y^^. 3. -5 4-* 6. 6. 6 4-t6. 9. V3 + V^. 10. (2-ht3)-h(l4-i2). IHint. (2 + i 3) + (1 + i 2) = 3 + <6 ; locate 3 + » 5.] 11. (4 + t)-(2 + i2). 12. (5-i2)4-(3-t4). 13. (6-t5)-'(6-i7), CHAPTER VIII SBRISS AND LIMITS ARITHMETICAL SERIES 167. The numbers 6, 9, 13, 17, 21 appear upon examination to have been selected according to some law and arranged in a definite order. Each number after the first is greater than the one immediately preceding by 4. Such a regulated succession of numbers is called series, , When, as here, the increase is the same throughout, the series receives the special name of arithmetical series or arithmetical progression. An arithmetical series is a succession of numbers in which each number after the first minus the preceding one always gives the same difference. This difference is called the common difference. Instead of increasing, the numbers in the series may decrease, so that the first number is the largest and the common difference is negative. Arithmetical series are frequently encountered in the study of mathematics, hence it is desirable to develop certain formulas relating to such series. BXBBCISBS 168. State which of the following series are arithmetical series : 1. 10,8,6,4,2. 4. a, a-hd, a-|-2d, a4-3d. ^ 9 7 5 3 1 2. i, 1, H, 2, 2^, 3. 5. _,_,_,__. 3. 10, 9, 7, 6, 4. 6,x,x — y,x—2y,x^S y, 158 SERIES AND LIMITS 159 THE LAST TERM 169. If a stands for the first number or term of an arith- metical series, and d for its common difference^ then the series may be written in general terms thus, a, a -h d, a -h 2 d, a + 3 d, a -h 4 d, etc. Let I stand for the last term of the arithmetical series. The value of the last term evidently depends upon three things ; namely, the value of a, the value of d, and also the number of terms in the series. Denote the number of terms by n. It is to be observed that the coefficient of d in the second term is 1, in the third term is 2, in the fourth term is 3, in the fifth term is 4. If n denotes the number of terms, what must be the coefficient of d in the last term ? From what we have observed it must be one less than the number of the term, that is, n — 1. Hence we have the formula for the nth term, I = a +(n - l)d. (A) BXBBCISBS 170. 1. Find the 10th term in the arithmetical series, 2, 7, 12, 17, . . . Here a = 2, d = 6, w = 10. Substitute these values in (A), Z = 2 +(10 - 1)6, « = 47. Hence the 10th term is 47, as may be verified by writing down all the terms to the 10th term. 2. Find the 16th term of the series —2, —4, —6, — 8, • • •. 3. Find the 12th term of the series 1, 1^, 2, 2^, 3, • • .. 4. Find the 20th term of the series 5, 5 -\-2x, 5 4- 4 «, 6. Find the 24th term of the series V2, V2 4. 1, V2 -h 2, V2+3, .... 160 . ELEMENTARY ALGEBRA 6. Find the (n — l)th term of the series 3, 6, 9, 12, 15, • • •. Find the expression for any given term (the nth term) of the following series of numbers ; . 7. 3, 6, 7, 9, . . .. 10. I, V^, IVV, • • •. 8. 7, 12, 17, 22, . . .. 11. 66, 59, 62, • • .. 9. 13, 17|, 22, 26^, . . .. 12. 2.4, 2.1, 1.8, • • .. 13. Find the (n — 2)th term of the series 5, 1, — 3, — 7, • • -. 14. A bullet is fired vertically upward so that at the end of the first second it has a velocity of 200 ft. per sec, at the end of the second second a velocity of 168 ft. per sec, at the end of the third second a velocity of 136 ft. per. sec, and so on. Compute the velocity at the end of the sixth second, at the end of the tenth second. Interpret the second answer. 15. A body falling from rest falls 16 ft. during the first second, 48 ft. during the second, 80 ft. during the third, 112 ft. during the fourth, and so on. How far will it fall during the ninth second ? ARITHMETICAL MEANS 171. The arithmetical means between two numbers are num- bers which, together with the two given numbers as first and last terms, form an arithmetical series. If the two given numbers are 5 and 50, then 14, 23, 32, 41 are four arithmetical means, because 5, 14, 23, 32, 41, 50 is an arithmetical series. EXBSBCI8BS 172. 1. Insert six arithmetical means between 7 and 63. The six terms to be found, and the given numbers 7 and 63, will make 8 terms. We have n = 8, a = 7, Z = 63. Substitute in (-4), 63 = 7+(8-l)d, 63=7 + 7d, d=8. Hence the required series is 7, 15, 23, 31, 39, 47, 55, 63. SERIES AND LIMITS 161 2. Insert 5 arithmetical means between 6 and 72. 3. Insert 4 arithmetical means between 7 and — 23. 4.' Insert 8 arithmetical means between 7 and S^, 5. Insert 3 arithmetical means between x and y, 6. Insert 7 arithmetical means between V7 and 10V7. 7. Insert an arithmetic mean between 100 and 133. * SUM OF AN ARITHMETICAL SERIES 173. The sum of the terms of an arithmetical series can always be found by writing down all the terms, and adding them. But, if the number of terms is great, this operation is quite laborious. We proceed to derive a formula by which the sum of a large number of terms may be computed with less labor. Observing that, if the last term is I, the term immediately preceding may be written I — d, the term before this I — 2dy and so on, we may indicate the sum of the series thus, ^ = a+(a + d) + (a-h2d)+ ... +(Z-2d)-h(Z- d)+L (1) Reverse the order of the terms in the right side of (1), ^=:Z+(Z-d)4.(?-2d)-h ... -h(a + 2d)+(a + (f)+a. (2) Adding (1) and (2), In (3) there are as many parentheses (a + 1) as there are terms in the series ; hence. 162 ELEMENTARY ALGEBRA BXBBCI8BS 174. 1. Find the sum of 20 terms of the series 99, 103, 107, 111, .... Hero n = 20, a = 09, d = 4. Henoe, I = a + (fi - 1)<2 = 99 + 19 x 4 = 175, and ^ = |(a+Z) = y(99+176) = 10x274=2740. The roqaired sum is 2740. 2. Find the sum of 25 terms of the series — 60, — 46, - 40, • . . . 3. Find the sum of 18 terms of the series 11, 8, 6, 2, • • • . 4. Find the sum of the first 200 integers, 1, 2, 3, • • • . 6. Find the sum of all the integers between --60 and 76, excluding —50 and 75. 6. Find the sum of 15 terms in 1^, If, 2, 2f, • • •. 7. In formula (J3) substitute for I its value as given in formula {A) and derive a formula for S which does not con- tain L 8. Find the sum of the first one hundred odd numbers. 9. Find the sum of the first one hundred even numbers. 10. How many of the integers 1, 2, 3, • • • must be added to yield the sum 55 ? Use the formula ^9 = - (2 a +(n - l)(l), derived in Ex. 7. Do both answers satisfy the conditions of the problem ? 11. How many terms of the series 5, 4, 3, • • . are necessary to yield a sum 9 ? Do both answers satisfy the conditions of the problem ? 12. The second term of an arithmetical series is 11, the fifth term is 20 ; find the 14th term. 13. The first week a store was opened the expenses ex- ceeded the income by % 52.25. The second week the loss was SERIES AND LIMITS 163 $41.75. If the improvement in the trade continued at the same rate, how much profit was made in 24 weeks ? 14. A man enters an office at a salary of $ 1200, which is increased annually $ 75. How much will the firm pay him during 18 years ? 15. The three formulas found below give the salaries offered by three companies to men entering their employ. S is the monthly salary in dollars earned after a given number of years (n). Calculate which company after 20 years' employ will give the highest salary. How much does a man earn in 20 years in each case ? (a) 5'=100 4-|w, (b) /S' = 95 4-4n, (c) S=SS+^n. GEOMETRICAL SERIES 176. The series 3, 6, 12, 24, is not arithmetical ; the dif- ference between successive terms is not the same. The successive terms are formed in accordance with a diflerent law. It is readily seen that any term after the first is derived from the preceding one by multiplying by 2. Such a series is called a geometHcal series or a geometrical progression. A geometric series is a succession of numbers in which each number after the first, when divided by the preceding number, always gives the same quotient. The quotient is called the common ratio. EXERCISES 176. Which of the following series are geometrical and which arithmetical? 1. 7, 35, 175, .... 5. 2, - 4, 8, - 16, • . .. 2. 6, 12, 18, .... 6. 2, 1, 0, - 1, • • •. 3. 2, 8, 32, .... 7. V2, 1, V|, . • -. 4. 2, 8, 32, 40, .... 8. 2i, 2i, 2f , . . .. 164 ELEMENTARY ALGEBRA LAST TERM OF A GEOMETRICAL SERIES 177. If the first term of a geometrical series is a, the com- mon ratio is r, the number of terms n, then an expression for /, the last term, may be obtained by inspecting the terms in the general geometrical series, a, ar, ar^, ar*, ar^^ • • •• We observe that in the second term, the exponent of r is 1, in the third term it is 2, in the fourth term it is 3, in the fifth term it is 4. Evidently, in the nth term, the exponent of r is n— 1. Hence we have the formula for the last term of a geomet- rical progression, ^ ^ ^^-i ^q^ BXBBOISBS 178. 1. Find the seventh term in the geometrical series, 16, 32, 64, . . .. Here a = 16, r = 2, n = 7. Hence, I = ar^^ = 16(2)^-1 = 16 x 64 = 1024. The seventh term is 1024. This result may be verified easily by writ- ing down the first seven terms of the series. 2. Find the 8th term of the series 6, 2^, 1 J, • • •. 3. Find the 6th term of the series 130, 390, 1170, • • .. 4. Find the 9th term of 3, - 6, 12, - 24, . • .. 5. Find the 8th term of 3, — 1, i, — |, • • •. 6. Indicate the 15th term of the series — -y 1, -— -, •••. V2 6 GEOMETRICAL MEANS 179. The geometrical means between two numbers are numbers which, together with the two given numbers as first and last terms, form a geometrical series. If the two given numbers are 8 and ^, then 4, 2, 1 are three geometrical means, because 8, 4, 2, 1, ^ is a geometric series. SERIES AND LIMITS 165 BXBBCISBS 180. 1. Insert four geometrical means between 5 and 160. The two given numbers and the four means make together 6 terms. We have a = 6, 1 = 160, n = 6. We must find r. By(C), l = ar^\ 160 = 6 »*. r = 2. Hence the required geometrical series is 5, 10, 20, 40, 80, 160. 2. Insert six geometrical means between 10 and — 1280. Solve r^ = - 128 by trial. 3. Insert four geometrical means between 3 a and 96 a*. 4. Insert one geometrical mean between 133 and 1197. What is the geometrical mean between two numbers ? What is the arithmetical mean between two numbers ? 6. Two numbers differ by 6, and their arithmetical mean exceeds their geometric mean by 1. Find the numbers. SUM OF A GEOMETRICAL SERIES 181. The sum of the tirst n terms of a geometrical series may be indicated thus, S = a-^ar-^ar^-\ f- aV"-^ -f ar^"^ + ar^-K (1) Multiply both sides of (1) by r, r/S = ar 4- ar* 4- a^^ 4- • • • 4- a^*"* 4- ar""* 4- ar*'. (2) Subtract (1) from (2) and observe that all the terms dis- appear in subtraction, except a and ai**. We obtain, rS — S = ar^ — a S(r - 1) = a(r-- 1) r — 1 166 ELEMENTARY ALGEBRA BXBBOISB8 182. 1. Find the sum of six terms of the geometrical series 11, 22, 44, ... . Here a = 11, n = 6, r = 2. Hence. /? = -<'!^rL!l = yC^.rJl = ll(?lrill= n x 63 = 603. r-1 2-1 1 2. Find the sum of seven terms of 1, ^, ^, • . • . 3. Find the sum of six terms of 1, — V2, 2, — 2 V2, • • • . 4. Find the sum of five terms of a?, — ay', ajy*, .... 6. Show that formula ((7) may be written 8 = ^\ "~ — ^ • 1 — r For what values of r is this form more convenient than (C) ? CL ^— 1*1 6. Show that (0) may be written S = 1 — r 7. What will $1000 amount to in four years, interest 3 %, compounded annually ? 8. What will $ 600 amount to in two years at 4 ^ annual interest, compounded semiannually ? 9. What will $ 700 amount to in eight years at 4 % annual interest, compounded semiannually ? We simplify thia computation by the use of logarithms. The amount is given by the expression x = ^700 (1.02)". log 1.02 = 0.0086, log (1.02)w = 16 X 0.0086 = 0.1376 log 700 = 2.8451 logaj = 2.9827 flc = f 961. This answer is only approximate. Had we used a six-place table of logarithms, instead of the four-place table, the answer would have come out 9 960.06, correct to the nearest cent. SERIES AND LIMITS 167 10. What sum of money, at 3 % interest, compounded annually, will amount to $ 1000 in 20 years ? Let Jjjl X be the sum, in 20 years this amounts to $x(1.03)*>. Hence the equation, x(l.OS)^ = 1000 1000 (1.03)20 Log 1000 = 3.0000 Log 1.03 = 0.0128. Log (1.03)«> = 20 X 0.0128 = 0.2660 loga; = 2.7440, x = $664.60. The answer, correct to the nearest cent, is $ 663.68. 11. Wliat sum of money, at 4 % interest, compounded annually, will amount to $ 695 in 10 years ? 12. A sum of $ 497 draws compound interest at 3^ % for 7 years. Find its amount. 13. If $ 1 could have drawn 6 % compound interest during the past 500 years, what would be its amount now ? 14. Owing to the introduction of automobiles the number of horses in a town decreased in five years from 560 to 336. Assuming a constant annual rate of decrease, find the number after four more years. 15. Determine the present value of $ 1000 due in 5 years, on the supposition that money can be invested so as to yield 3 % interest when compounded annually. 16. A man undertook to pay $ 100 to a charity one year, $ 90 the next year, ^ of the latter sum the third year, and so on, until his death. He died after making 21 payments. What was the total of his gifts ? Use logarithms in the computation. 17. Another man pledged $ 175 and promised to increase the amount by ^ every year. What is the total of his dona- tions after 15 years ? 168 ELEMENTARY ALGEBRA 18. A man saved $ 250 every year and invested it in a busi- ness that brought him 4 % per annum compound interest. He made these investments f6r six consecutive years. What was the total sum standing to his credit immediately after he paid in his last $250? 19. A man's business is increasing at a uniform rate, and so rapidly that his income is doubled in two years. What is the rate of increase or growth-factor f If at starting his business was 9 a, it was 9 or at the end of the first year, and 9 ar^ at the end of the second. We have then aH = 2 a, or r = \/2. The growth-factor = >/2. 20. What is the annual rate of increase, or annual growth- factor, if income increased uniformly and doubled in the course of four years ? 21. In 4 years the population of a town increased from 2000 to 2662. Determine the growth-factor on the supposition that the rate of increase was uniform. 22. A young tree grew in 6 years from a height of 32 in. to a height of 20 ft. 3 in. What constant annual growth-factor would account for this increase ? 23. The population of a mining town, now 10,000, is falling off at the rate of 1 % per annum. What will the population be 2 years from now ? 24. Between the ages 9^ and 14^ the average height of boys increases from 60 to 60 inches. Find how tall a boy would be at 10|, 11|, 12^, and 13| years of age, on the supposition that the growth-factor was the same for these years. Compare your results with the statistical averages which give the heights as 51.9, 53.6, 55.4, 57.5 inches, respectively. Use logarithms. SERIES AND LIMITS 169 INFINITE GEOMETRICAL SERIES 183. An important kind of series arises when the number of terms in it is no longer restricted to a fixed finite number, but is permitted to increase without end. Consider the series 1, ^, J, \, ^. It has five terms. Imagine now that the number of terms is much larger, and that this number is steadily growing. We obtain a series which may be written, 1» h h h^y^f-hy-^ infinity. (1) The number of terms is taken larger than any large finite number that we may name. We say that the series is infinite. The characteristic prop- erty of it is that it has no last term. Whenever we try to seize upon any one term, say ■— -, as being the last one, we can always go one step farther and write down the term — ^ which comes after it. Considerations of this sort compel us to accept the conclusion that there is no last term in such a series. Another noticeable feature about the series (1) is that its terms decrease toward the right; that is, the terms farther from the starting point are always the smaller. The nth term may be written By taking n sufficiently large, we may make this nth term smaller than any small number which we may name. For instance, the nth. term will become smaller than -j-J^, if we 1 11 take n = 8. In this case — - becomes — = — -, which is less 2»-i 27 128' thauy^. If you name some smaller number, say y^T^, then again a value of n can be chosen, so that the nth term becomes smaller than it. Take, n = 11, then h. = hxh = :^y^l= ^ 2w 2' 2» 128 8 1024' 170 ELEMENTARY ALGEBRA which is less than y^Vir* ^^^ ^^"^7 name a still smaller fraction, and again a value can be assigned to n in the expression for the nth term, which makes that nth term less than that small fraction. No matter how small a fraction is written down, we can always find some term in series (1) which is smaller than that fraction. In such a case we say that the term — -^ approojohea zero as a limitf as n increases wUhoiU end. SUM OF AN INFINITE GEOMETRICAL SERIES 184. We proceed now to find the sum of the infinite series IH 1 1 [-•••H : -h • • • to infinity. Examine formula (C) ; we see that o __ a(r* — 1) _ a(l — r *) __ a — ar^ __ a^ ar* r — 1 1 — r 1 — r 1 — r 1 — r We know that this series gives the sum correctly for any finite and fixed number of terms. What is the sum when the series becomes infinite ? In the infinite series 1, -J., J, i., . . . we have r = J^, a = 1. Other geometrical series may be written, but in all of them that we shall consider now, we shall assume that the ratio r is numerically less than 1. In that case the power r* becomes smaller for larger values ofn and approaches the limit zero as n increases without end. That this is so, may become clearer by a second illustration. Let r = I, then r^ = ^, r" = ^^y, r* = \^, r* = ^^^, and so on. The diminution in the value of the power is not as rapid now as in the case, r = ^ ; nevertheless, a value of the integer n can always be found, such that r" becomes less than any small frac- tion previously named. To make r" less than y^itj ^® need only take n = 12. Since -jj 64 .f 12 «8 -^ 64x64 4096 r = , we get r^^ = r^ *t^ = 729' ° 729x729 531441 SERIES AND LIMITS 171 In this fraction the numerator is seen to be less than the one-hundredth part of the denominator. 185. What happens to the fraction , when the integer 1 — r n LQcreases without end, and r is numerically less than 1 ? Bear in mind that a, the first term of the series, is a fixed number ; r is also a fixed number, hence the denominator 1 — r is a fixed number. Only the factor r" changes, as n increases without end. And we have seen that, under these circum- stances, r* approaches the limit zero. Hence the numerator ar** also approaches the limit zero. Hence, the entire fraction, , approaches the limit zero as n increases without end. 1-r The series, S = a -\' ar -\- aj^ -{- ...4- ar"""^ + . . . which for a fixed, finite number of terms n has the sum a ar^ aS = 1 — r 1 —r has the sum, 8 = - , (D) 1 — r when the series becomes infinite, so that n increases without end, provided that r is numerically less than 1. From what we have said it appears that by the " sum " of an infinite series we mean the value which the sum of the first n terms of the series approaches as a limit, when n increases without end. When the common ratio r of the geometrical series is 7iot numerically less than 1, the above reasoning does not apply. In that case r" does not approach the limit as w increases without end. In the series 1, 2, 4, 8, • • • r = 2, and ?•* = 2" ; 2" becomes greater and greater as n increases. Thus 2* = 4, 2* = 8, 2* = 16, etc. It is seen that 2» increases without limit 172 ELEMENTARY ALGEBRA when n increases. Hence the fraction does not ap- 1 — r proach the limit zero. When in an infinite geometrical series the ratio r is numeri- cally less than 1, so that the sum of the first n terms ap- proaches or converges to as a limit, the series is said to 1 — r be convergent. Convergent aeries play a very important part, both in theo- retical and in applied mathematics. BXBROISBS 186. 1. Find the sum of the infinite geometrical series, ^2 4 8 ^2"-^ Here a = 1, r = J. Hence ,8^ = _?_ = -J_ = 2. 1-r 1-J The Bum of the series is 2. To convince himself of the correctness of this result the student may find the sum of 8, 4, 6, or more terms and see that the smn gets very close to 2, the closer when more terms are added. Let 8n mean the sum of the first n terms, then we see that 82 = 1.5, 8z = 1.76, 84 = 1.876, 8s = 1.9376, 8^ = 1.96875. By actual addition we can approach to the limit 2 closer and closer, but we cannot reach it, because we can perform this addition for only a finite number of terms. 2. Find the sum of the infinite geometrical series 1, |^, |^, ift^? • • •• 3. Find the sum of the infinite geometrical series 1, ^, 4. Find the sum of the infinite geometrical series 2, — J, h — T^7) • • •• 6. Find the sum of the infinite geometrical series 1, — , -, 11 2* 2 ^'2-^'-- SERIES AND LIMITS 173 6. Find a common fraction which expresses the exact value of the repeating decimal, .252525 • • •. The repeating fraction may be written as an infinite geometrical series thus, .2625 ... = tW + tAW + Tirt^inj + -• Here o = fW* »• = nflftnr ■*• ^ = rir- Hence ^ = _«_ = .JL= 25,^^ ^26 ^^ l_r 1-tJ^ 100 100 99' Find the common fraction which is the exact value of each of the following repeating decimals : 7. .333.... 8. .151515.... 9. 7.363636.... 10. .555... 11. .325325. ... 12. 7.8212121.... 13. A man undertook to pay $500 to a charity one year, $ 450 the next year, -^ of the latter sum the third year, and so on, until his death. What was the outside limit of the expectation of this charity? If the man died after making 15 donations, how much did his total payments fall short of that sum ? Use logarithms in performing the computation. THEORY OF LIMITS 187. In deriving the sum of an infinite converging geomet- rical series we touched upon the theory of limits which plays a fundamental r51e in higher mathematics. The ideas which we brought out suggest the following definition of the limit of a varying number. 2%e limit of a variable is a fixed number which the variable approaches in siich a way that the difference between the variable and its limit becomes and remains numericaUy less than any num- ber, however smallj but never becomes zero. Thus, in fi; = 1+1 + 1+...+ ^ 2 4 ^2*»-i' the variable is 8n^ which increases when the integer n increases. 174 ELEMENTARY ALGEBRA The limit of 8^ is the number 2. As n increases without end, the variable Sn approaches 2 and may be made to approach it so closely that 2 — i9» is less than any previously assigned small number different from zero, such as ^ithm ^^ looiooc A variable of some importance is the fraction -, when n n takes in succession the values 1, 2, 3, 4, • • •. It is seen that the variable - becomes smaller and smaller and approaches n the limit zero. Under the same conditions the fraction -, n where a is any fixed number, will likewise approach the limit as the integer n increases without end. Quite different is the behavior of -, when a is a fixed num- X ber different from zero, and x takes in succession the values 1, ^, i, i, • • •. For simplicity, let a = 1. Then, for a; = 1, - = 1, for a; =i, - = 8, X S X fora: = i, ^ = 2, for a: = 1, ^= 16, 2 a? 16 a? for a? = - , - = 4, for a; = — - , - = 32, 4' a? ' 32' a? ' and so on. The smaUer the denominator, the larger the value of the fraction. We let the variable x approach the limit zero in such a way that it does not reach zero. We stipulate that x shall not be- come zero for the reason that we cannot divide by zero, as was explained in a previous chapter. If, then, x is permitted to approach as a limit, without reaching its limit, then the fraction - increases in value without limit. That is, we can x 1 select a value for a;, such that - will be greater than any pre- viously assigned large number. SERIES AND LIMITS 175 For instaoice, select the number 1000 ; we can take x so small that - is greater than 1000. Let « = — - = X ° 210 2*^.2^ 32.32 = , then - = 1024, which is more than 1000. In the same 1024' X ' way, X can be so chosen, that - exceeds 10,000, or any other X fixed number. Hence the theorem, As the variable x is made to approach the limit zero, without rea>ching zero, - increases so as to exceed any large number that X may be previously assigned. This theorem is sometimes condensed in this manner : As X approaches zero, - approaches infinity. X A still further compression of the theorem consists in the use of the symbolism This symbolism is objectionable, because it may convey the idea that division by is permissible. The reader must re- member that these syjnbols are intended merely as a con- venient abbreviation for the theorem given above. EXEBCISBS a I 35 188. 1. What value does the fraction — -^— - approach, when b — X a and b are fixed numbers (b ^ 0) and cc is a variable approach- ing the limit zero ? Under these conditions the numerator a + ^ approaches a, the denomi- nator h — x approaches 6 . Hence the fraction approaches -• h 176 ELEMENTARY ALGEBRA 2. What value does the fraction — approach, when x in- creases without limit ? a + x The reasoning is easier, if we first divide numerator and denominator by X, so that x occurs in the denominators of the minor fractions. «-l We obtain, a — x^x a-^x ^ A.I X Since a is a fixed number, and x increases without limit, the fraction - approaclies the limit zero, hence, the numerator - ~ 1 approaches — 1 X X as a limit ; the denominator - -f 1 approaches + 1 as a limit ; the com- ae plex fraction approaches -^^— or — 1 as a limit. What is the limiting value of the following fractions when X approaches zero as a limit ? ' 4 — 6aj' ' 1— «* * 2 — aj*' *• ~ • o. — -. o. aj4-6 S-haj* aj*-f2a:*-i-3 What is the limiting value of the following fractions when the variable x increases without limit ? X l-|-2aj a — ar 10. _?_. 12. «±^. 14. ^+£±«. x + 6 c + dx x^ — x-i-b SUPPLEMENT 177 SUPPLEMSNT THE HIGHEST COMMON FACTOR BY THE METHOD OF DIVISION 189. The method of finding the h. c. f. of two polynomials not easily factored is similar to the method for finding the h. e. f . of two numbers in arithmetic, neither of which can be easily factored. The method was first suggested by Euclid (300 B.C.). BXAMPIiB Find the h. c. f . of 1343 and 3002. Id43)3002[2 268.6 316) 1348 [4 1264 79)316 [4 316 Process : Divide 3002 by 1343. Take the remainder 316 as a new divisor, and 1348 as a new dividend. Take the next remainder 79 as a new divisor, and 316 as a new dividend. The next remainder is zero. The last divisor 79 is the h. c. f . of 1343 and 3002. Proof: Let ^ = 1343, i? = 316. B = 3002, Q = 2. We have, as always. Dividend = Divisor x Quotient + Remainder. B = AxQ +B, or B = B-AQ From the last equation we see that any factor of both A and B must be also a factor of B — AQ^ and therefore a factor of the re- mainder B. Hence, all common factors of A and B are common factors of A and B, Instead of finding the h. c. f. of A and B, we may therefore find the h. c. f . of A and B ; the latter process involves smaller numbers. Repeating the above process, divide A by B. Let Q' be the new quo- tient 4, and B' be the new remainder 79. As before, all common factors of A and B are factors of B' ; for, A = Q'B + B' and B' = A— Q'B. 178 ELEMENTARY ALGEBRA Hence, the problem reduces itself farther, to finding the h.c. f. of B and B' ; that is, to finding the h. c. f. of 816 and 79. As 816 = 79 x 4, it follows that 79 is the h. c. f . required. This process may be used in finding the h. c.f. of poly- nomials which cannot be easily factored. Numerical factors common to the polynomials are easily detected by inspec- tion. Such numerical common factors, as well as other common factors that can be found by inspection, should be removed at once. BXAMPLB Find the h.c.f. of 2iB»-22aj-12 and 4iB»-|-16aj«-|-16rc-|-12. Process : 2 x« - 22 « - 12 = 2(a* - 11 a:- 6). 4 x» + 16 a;a + 16 a + 12 = 4(x» + 4 a;2 + 4 x + 8). The numerical factor 2 is common to the polynomials. x»-llx-6)a* + 4a;2+ 4x + 8Ll z« -11 a; -6 4a:«-|-16x + 9)4x»-44a;-24 |a;-16 -16x2- 63x- 24 4 -60x2« 212 X- 96 -60x2- 225 X- 136 18 13x-h39 X + 3)4 x2 + 16 X + 9 |4x4-8 4x2+ 12x 3x + 9 3x + 9 The h.c.f. =2(x + 8). To avoid fractions any expression may be multiplied or divided by any number which is not a factor of the other. In the example above, when 4 x2 in the first remainder will not be contained in x^ a whole num- ber of times, the expression x'— llx — 6 is multiplied by 4. Again when 4 x2 is not contained in — 16 x2 a whole number of times, the SUPPLEMENT 179 expression — 15 x^ — 63 2c — 24 is multiplied by 4. In the remainder 13 X + 30, the factor 13 may be discarded because the expression 4 a;2 _|_ 15 x -j- does not have a factor 13. It would be wrong to multiply 4 x^ + 16 X + 9 by 13, because 13 is a factor of 13 x + 30. Continue to divide until the remainder is of a lower degree than the divisor. Then take the remainder for a new divisor and the previous divisor for a new dividend and proceed as before. The 1. c. m. of two expressions not easily factored may be found by first finding their h. c. f . The 1. c. m. required will then be the product of either expression and the quotient obtained by dividing the other ex- pression by the h.c. f. It is seldom that resort to this long process be- comes necessary. EXERCISES 190. Find h. c. f. of : 1. 2a2 + a — 3and4a'-|-8a2 — a — 6. 2. 2a^ — 5a;2 — 2aj + 2and2aj» — 7a;2 + 9aj — 3. 4. a* — a' + 2 a2 — a 4- 1 and a* + a' -|- 2 a2 + a + 1. 5. 8 am«-|-24 am^-}- 22 am-\-6 a and 6 m'+13 m2 + 8 m + 1. 6. 2a:3-10aj2-14aj + 70anda^-3aj2-7a;-15. 7. S xlh/—10 a^y-\-7 xh/—2 xy and 6 a^y --11 oiih/ -^8 ah/— 2 xh^. 8. 4a» - 4a2 - 5a + 3 and 10a2- 19a + 6. The h. c. f . of three or more expressions may be found as follows : Find the h. c. f. of two of them ; then find the h. c. f. of this result and another of the given expressions, and so on. The last h. c. f . is the one required. 9. 2a^-2aj2_2aj-4, 3iB»-6a;2+9aj-18,and8aj»-12a;2 -4aj-8. 10. a^x — 6 a2aj -f- 11 ox — 6 aj, cj^x — 9 a2a; -|- 26 ox — 24 aj, and a«aj2 — 8 a2aj2 4. 19 aaj2 — 12 x^. 180 ELEMENTARY ALGEBRA SOLUTIONS OF QUADRATIC EQUATIONS 191. In previous solutions of the quadratic equation ax^ 4- 6a; 4- c = 0, obtained by completing the square, the first step was to divide both sides by a. There are other methods of procedure which possess certain advantages. If a, 6, c are integers and it is desired to avoid fractions in the process of completing the square, the Hindu method of completing the square may be used. It is as follows : Given ox* + ftx = — c. Multiply both sides by /our times the coefficient ofx^^ or 4 a. We obtain 4 a^c^ -\- 4 abx = - 4 ac. Divide 4 abx by twice the square root of 4 a'x' ; this gives 5. Add the square of 6 to both sides, 4a^x^ + 4abx + fe^ = 52 _ 4 <|c. Take the square root of both sides, 2ax + b = ± y/h^ - 4 oc. Transpose h and divide by 2 a, x ^ - & J:>/"&^ - 4ac ^ 2a Sometimes fractions may be avoided by the use of a smaller factor than 4 a. Consider the equation, 3a? -h 4a; = 8. Make the first term a perfect square by multiplying by 3, 9a;2+12x = 24. Divide 12 x by twice the square root of 9 x^ ; this gives 2. Add the square of 2 to both sides, 9«3 + i2a;-|-4 = 28. _ Then 8a; + 2 =±^28, _ -2 J:>/28 ""- 3 ' -2±2V7. 3 Solve by completing the square without introducing frac- tions : 1. 2a«- 13a; 4-10 = 0. 3. 4a;* + 6aj- 41 =0. 2. 5a;2-h7a;-13 = 0. 4. 7a;2 - 14a; - 3 = 0. SUPPLEMENT 181 MATHEMATICAL INDUCTION AND PROOF OF THE BINOMIAL THEOREM 192. There is an important method of reasoning in mathe- matics, called mathematieal induction, which we shall use in proving the binomial formula. The method will be grasped more readily, if we give a non-mathematical illustration of it. Non 'mathematical Illustration, For the sake of argument, suppose it established as true (1) That there was once an Indian named Hiawatha* (2) That every Indian named Hiawatha (if ever such an Indian ex- isted) had a grown son named Hiawatha^ From these two propositions certain conclusions can be drawn. By (1) we know that an Indian named Hiawatha once existed. By (2) we know that he had a grown son named Hiawatha. Desig- nate him Hiawatha II. By (2) we know that Hiawatha II had himself a grown son named Hiawatha. Designate him Hiawatha III. By (2) we know that Hiawatha III had himself a grown son named Hiawatha. And so on. Thus the inference is drawn from propositions (1) and (2) that there existed an unbroken, never-ending line of descent of Indians named Hiawatha. Mathematical Illustration It is to be proved that tf^e sum of the first n positive integers i« ^ (n -h 1). I. We see by trial that 1 + 2 = J(2 + 1), 1 -h 2 + 3 = i(3 -h 1). This shows that the theorem is correct for the particular cases n = 2 and n = 8. II. We establish the conditional proposition, that if the theorem is true for some value of n, say n = m, then the theorem must be true for n = m + l. If the theorem is true for » = m, we have l-|-2 + 3 + .-.+m = ^(w+l). Add (m -I- 1) to both sides, 1 + 2 + 3-1- ... +m+(w+l) = ^(m+l)+(w + l). 182 ELEMENTARY ALGEBRA Combine terms on the right side, 1 + 2 + 3 + . . . + m + (m + 1)= 5L±i([w, + 1]+ 1). As this result is the original theorem for n = m + 1, we have shown that the theorem is true forn = m + 1, provided it is true for n = m. Next we apply the reasoning called mathematical induction : By I, the formula 1 + 2 + 3 + . .. + n = |(n + l) is true for n = 3. By II it follows that, being true for n=3, it must be true also for n=4. By II it follows that, being true for n = 4, it must be true also for n = 6. And so on for n = 6, 7, 8, • • •. Consequently, the theorem is true for any positive integral value of n, no matter how great it may be. Thus, the theorem is established. Note. Very often a beginner fails to i)erceive the need of such an involved argument. If a theorem is true for certain special cases, he may conclude at once, without further argument, that it is true for all cases. But such jumping at conclusions is dangerous. The relation a^ = 2 a holds for a = 0, also for a = 2 ; is it true generally ? Is it true for any other value of a ? PROOF OF THE BINOMIAL THEOREM WHEN THE EXPO- NENT OF THE BINOMIAL IS A POSITIVE INTEGER 193. By actual multiplication it is found that (a ± 6)*= a?±2db-\-h\ {a ± by= a« ± 3a«6 -+ 3a6» ± b\ (a ± by= a* ± 4 a'6 4- 6 a'fe* ± 4a&» +- b\ By inspection we find that these products follow the follow- ing laws : I. The first term is a raised to the same power as that of the binomial. In each succeeding term the exponent of a decreases by 1. SUPPLEMENT 183 II. The factor b does not appear in the first term ; the ex- ponent of 6 in the second term is 1 and increases by 1 in each succeeding term. III. The coefficient of any term after the first is found by multiplying the coefficient of the preceding term by the ex- ponent of a in that term, and divided by one more than the exponent of 6. , IV. If the binomial is a + b, the signs of the product are all plus ; if the binomial is a — &, the signs are alternately -f- and — . V. The number of terms is one more than the exponent of the binomial. VI. Each term is of the same degree as the binomial. If we use the exponent n, these six laws are embodied in the following formula : * (a + by = a» -h na^'^b + ^l^^^lDa^-^^^ _^ n(«-l)(«-2) ^^y + . • . + naly-^+ b". (1) Thus far, this formula has been shown to be true only for the particular cases n = 2, 3, or 4. (Substitute 4 for n and show that the product of (a + by is obtained.) Is formula (1) true for all positive integral values of n ? Were we to attempt to prove its generality by actual multiplication, we would soon weary ; the task would be impossible. By mathematical in- duction the proof is short. Let us establish the conditional proposition that if formula (1) holds for, say n = m, it must hold also for m +1. In (1) write m in place of n. Then multiply both sides of the re- sulting equation by (a + b). We obtain, * By the notation 2 1 we mean 2 x 1 » by 3 ! we mean 3x2x1; generally, nl=»n(n-l)(n — 2) ...3.2.1. WecaUn! "factorial n." 184 ELEMENTARY ALGEBRA (a + 6)« = a* + ma^'^b+ 5*l5Lziiia«»-a6a + • • • + mab"^^ + ft*, (2) (a + 6)= a-}- h (a + 6)"+i 2J ^1 « I + (m+l)a6« + &«+i. (8) The operation given above should be studied very carefully. Several steps require explanation. We have multiplied both sides of equation (2) by a + 2). Examine every step carefully. In the right sides we first multiplied by a, then by b. The dots stand for terms that could not all be written down, because the number of terms is m + 1, and m is not given in Hindu-Arabic numerals. The term *»(t»— l) q,>-25> which is 2! obtained when we multiply by 5, is not written down and is among the terms represented by dots. The term ^(^~ ^/ qaftm-i ig obtained by multiplying by b the term ^(^~ ^^ q2fc«>-a which in (2) lies among the terms represented by dots. Be sure to verify the addition of the partial products. We observe that (3) is the same in form as (1) ; that is, if in (1) we write m -f- 1 for w, we obtain (3). By the above multiplication we have proved that the for- mula (1) is true for n = m -f- 1, provided it is true for n = m. Now we proceed to the argument by mathematical induction : Formula (1) is known from actual multiplication to be true for n = 4. But we have shown that if (1) is true for w = 4, it must be true for n = 5. Again, if (1) is true for n = 5, it must be true for n:=6. And so on, for n = 6, 7, 8, • • • . Hence formula (1) is true for any positive integral value of the exponent. SUPPLEMENT 185 BXEBCISBS 194. 1. Expand (a + 3 y)\ In formula (1) substitute a = a;, 6 = (3y), n = 6. We obtain, Simplifying, («+3y)«= a^ + 18a^y+186a:*yH540a:»y8+1215a;2y*+1468icye+729 y«. 2. Expand (2p — qy. In formula (1) substitute a = 2p, 6 =(— g), n = 6. (2i>- g)6 =(2p)6 + 5(2p)*(- g) + ^ (2p)»(-g)« Simplifying, (2p - g)6 = 32 j)6 - SOp^g + 80i)«g« - 40p2ga + I0|)gf4 - g6. Expand : 3. (a-26)». 7. (2i> 4-3^)5. 11. (a + 6)'. *• («-iy)*- 8. (V2a-6y. 12. (a + V^6)». 6. (3aj-|2/. _ / , /I Y 9. (V2a;+^/2y)«. 13. (a -V- 16)*. \m "" **/ ' 10. (a - h)\ 14. ( V^ «+ V^ y)*. 195. It can be proved that, under certain limitations, tlie Binomial Formula is applicable to cases in which the exponent w of (a + 6)** is not a positive integer, but is a negative integer, or a positive or negative fraction. In such cases the binomial expansion becomes an infinite series. Such an infinite series can be used for purposes of computation only when it is con- vergent; that is, only when the sum of the first r terms of the series approaches a finite constant as a limit, as r increases without end. When a > 6, the expansion, expressed in ascend- ing powers of 6, is always a convergent series. Omitting all proofs, we proceed to applications. 186 ELEMENTARY ALGEBRA 1. Find approximatiiely the cube root of 27^. This can be obtained by the regular process of extracting the cube root, or by the use of logarithms. A third method is by means of the binomial theorem. We notice that (27.2)> is a little over 8. Take 27-2 = 27 + .2, where 27 is a perfect cube. Then by the binomial formula, taking a = 27, 6 = .2, n = 1, (27 + .2)* =(27)1 + l.(27)-*(.2) + Ki.Ilil(27)**(.2)2 + . . . Here .2 is small when compared with 27. Hence the series is convergent. In the expansion, .2 occurs to the first power in the second term, to the second power in the third term, and to higher powers in the terms foUowing which are not put down. The simplified value of the second term is + ^\^ = .007407, of the third term is — ^^^ = — .000018. If we take only the first two terms in the expansion, and neglect all the rest, we obtain an approximate value for the cube root, 8.007407. If we take the first three terms, the more accurate value, 3.007889, is obtained. If in a calculation, account is taken of a number h which is small com- pared with the other numbers involved, but the square and higher powers of h are discarded, then the calculation is said to be carried to the fir^t approximation. If both the first and second powers of b are used, but no higher powers of &, then the calculation is carried to the second approximation. And so on. In the above example, where h = .2, the first approximation to the cube root of 27} is 8.007407 ; the second approximation is 8.007889. EXBBCISBS 196. Find to a first approximation the values of the follow- ing expressions in which the second number is much smaller than the first : 1. (32^)*. 4. (a* 4- 6)*. 2. (63)* =(64-1)*. 6. Va« + 2 c. 3. V99.=V100-1. 6. -^/IpTq, 7. (a«-36)*. 8. (9m«-n)^. 9. ■\/a^—-\/a. SUPPLEMENT 187 10. A brass cube 1 inch each way is heated until its edges are 1.003 inches long. Compute to the first approximation the area of each face and the volume of the cube. 11. A metal cube 2 inches each way is cooled so that its volume is reduced to 7.998 cubic inches. What is now the length of an edge ? 12. Compute to a second approximation the amount after 10 years, of $ 575 at 2 %, interest compounded annually. Evaluate 676 (1 -h «02)^<> by retaining the first and second powers of .02 in the binomial expansion, but discarding higher powers. Does the first approximation give results differing from those due to simple interest ? 13. In measuring the diameter of a sphere whose diameter is actually 5", an error of 1 % is probable. What is the probable error, computed to a first approximation, of the volume derived from the inaccurate diameter ? The volume of a sphere = | irt*. Take r» = \(6 + .06)». 14. How great an error is made by assuming 1-a^ 1 + a = 1 — a + a* — a' + a*. when a = ^ ? What is the ratio of the error to the whole of the fraction ? 16. Proceed on the assumption that the earth is a sphere, the radius of which is 4000 miles. On the supposition that its radius was at one time 10 miles greater than at present, calcu- late to the first approximation, and also to the second approxi- mation, the amount that it has lost during the contraction, (1) in superficial area, (2) in volume. [The superficial area = 4 irr^.] 188 ELEMENTARY ALGEBRA 197. A Table of Squares and Cubes, Square Boots and Cube Roots of Numbers from 1 to 200, Square • Cubs Sqvars Cubs 8QVAXB8 CVBSt No. — * — 1 K0OT8 Roots ft SQVAKn CVBM No. 61 Roots Roots 1 1 1.000 1.000 2,601 132,651 7.141 3.708 4 8 2 1.414 1.260 2,704 140,608 62 7.211 3.733 9 27 8 ,M^.?32 1.442 2,809 148,877 68 7.280 3.756 16 64 4 2.000 1.587 2,916 167,464 64 7.348 3.780 26 125 6 2.236 1.710 3,026 166,376 66 7.416 3.803 36 216 6 2.449 1.817 3,136 176,616 66 7.483 3.826 49 343 7 2.646 1.913 3,249 186,193 67 7.660 3.849 64 612 8 2.828 2.000 3,364 195,112 68 7.616 3.871 81 729 9 3.000 2.080 3,481 206,379 69 7.681 3.893 100 1,000 10 3.162 2.164 3,600 216,000 60 7.746 3.915 121 1,331 11 3.317 2.224 3,721 226,981 61 7.810 3.936 144 1,728 12 3.464 2.289 f 3,844 238,328 62 7.874 3.958 169 2,197 18 3.606 < 3.351 • 3,969 260,047 68 7.937 3.979 196 2,744 14 3.742 2.410 4,096 262,144 64 8.000 4.000 225 3,376 16 3.873 2.466 4,226 274,625 66 8.062 4.021 266 4,096 16 4.000 2.620 4,356 287,496 66 8.124 4.041 289 4,913 17 4.123 2.671 4,489 300,763 67 8.185 4.061 324 6,832 18 4.243 2.621 4,624 314,432 68 8.246 4.062 361 6,859 19 4.359 2.668 4,761 328,609 69 8.307 4.102 400 8,000 20 4.472 2.714 4,900 343,000 70 8.367 4.121 441 9,261 ' 21 4.583 2.769 6,041 357,911 71 8.426 4.141 484 10,648 22 4.690 2.802 6,184 373,248 72 8.486 4.160 629 12,167 28 4.796 2.844 5,329 389,017 78 8.644 4.179 676 13,824 24 4.899 2.884 6,476 405,224 74 8.602 4.198 625 15,625 26 6.000 2.924 5,626 421,876 76 8.660 4217 676 17,676 26 6.099 2.962 5,776 438,976 76 8.718 4.236 729 19,683 27 6.196 3.000 6,929 466,533 77 8.776 4.254 784 21,952 28 6.292 3.037 6.084 474,562 78 8.832 4.273 841 24,389 ; 29 6.385 3.072 6,241 493,039 79 8.888 4.291 900 27,000 80 6.477 3.107 6,400 612,000 80 0.9S% 4.309 961 29,791 81 5.568 3.141 6,561 631,441 81 9.000 4.327 1,024 32,768 82 5.657 3.175 6,724 651,368 82 9.056 4.344 1,089 35,937 88 6.746 3.208 6,889 671,787 88 9.110 4.362 1,156 39,304 84 6.831 3.240 7,056 692,704 84 9.165 4.380 1,225 42,875 86 6.916 3.271 7,226 614,125 86 9.219 4.397 1,296 46,656 861 6.000 3.302 7,396 636,056 86 9.274 4.414 1,369 50,653 87 6.083 3.332 7,669 658,503 87 9.327 4.431 1,444 54,872 88 6.164 3.362 7,744 681,472 88 9.381 4.448 1,621 59,319 89 6.245 3.391 7,921 704,969 89 9.434 4.465 1,600 64,000 40 6.325 3.420 8,100 729,000 90 9.487 4.481 1,681 68,921 41 6.403 3.448 8,281 753,571 91 9.539 4.498 1,764 74,088 42 6.481 3.476 8,464 778,688 92 9.691 4.614 1,849 79,507 48 6.657 3.503 8,649 804,367 98 8.644 4.631 1,936 85,184 44 6.633 3.530 8,836 830,584 94 9.695 4.647 2,025 91,125 46 6.708 3.557 9,025 857,375 96 9.747 4.563 2,116 97,336 , 46 6.782 3.683 9,216 884,736 96 9.798 4.679 2,209 103,823 47 6.856 3.609 9,409 912,673 97 9.849 4.595 2,304 110,692 48 6.928 3.634 9,604 941,192 98 9.900 4.610 2,401 117,649 49 7.000 3.659 9,801 970,299 90 9.960 4.626 2,500 125,000 60 7.071 3.684 10,000 1,000,000 100 10.000 4.642 SUPPLEMENT S4IM1B Cdbh Ko. Root* Etwm IKIFIBES ClTBia No. 8OT.« Roots Roan 10,201 1 " 01 10,050 *.667 1 1 12.288 O.320 10,<HH 1 OS 10.100 4.672 1 2 12.329 6.337 10.609 1 27 10.149 4.688 S 12.369 0.348 10,816 1 64 10.198 4.703 4 12.410 6.360 11,025 1 25 10.247 4.718 ISS 12.500 5.372 11236 1 16 10.296 4.733 1 B 12.490 6.383 lt.449 1 43 10.344 4.747 1 7 12.530 5.395 i: a 1 12 10.392 4.763 1 8 12.670 6.406 I )i I 39 ' 10.440 4.77V I 8 12.610 5.418 i: » 1 00 10 10.488 4.791 160 12.649 6.429 i: a 1, , 31 1 10.536 4.806 1 1 . 12.689 6.440 11 H 1,404,928 1 10.683 4.820 2 12.728 5.461 i: 19 1,442,897 1 :o.&to 4.830 3 12.767 5.483 l,4Ht,544 1 10.677 4.849 84 a 5.474 i: » 1,020,870 1 10,724 es 5.486 i; 16 1,660,896 1 10.770 4^877 66 12.884 5.496 i: » 1,601,613 1 10.817 4.891 7 12.923 6J107 i; 14 1,643,032 1 10.863 8 12.961 5.018 H 11 1.685,169 1 10.909 4^919 eg 13.000 6.029 u « 1,728,000 £0 10.954 4.932 13.038 5.040 U 11 1,771,061 2 IIJWO 4.M6 71 13.077 6.660 1B1B848 11.IM0 4.960 7S 13.116 5.661 U 19 1,W«B«T 8 11.091 4.973 73 13.103 6.572 11 '6 1,! 4 24 11.136 4.987 4 13.191 5.583 U » i; s U 11.180 6.000 7fi 13229 5.593 11 '6 2,1 6 s 11.220 0.013 6 13.266 5.604 If !9 2/ 3 2 11.269 6.027 7 13.304 5.615 1« H 2,< a 8 11.314 6.040 78 13,342 0.820 It H 2, 8 29 11.368 0.053 79 13.379 6.636 1( 10 2 S 11.402 6.066 80 13,416 5.646 1^ 11 2 1 S 11.446 81 13454 5.607 n » 2.1 8 s 11.489 6:092 82 13.491 5.667 n 19 2r 7 3 11.533 0.104 183 6.677 n i6 2,' 4 8 11.576 6.117 184 13!s6fl 0.688 If !9 Z 5 8 11.619 5.130 186 13.601 0.698 It e 2J 6 S 11.662 6.143 186 6.708 It 19 3.. 8 3 11.705 0.155 187 13,870 0.71S 11 4 2,1 2 S 11.747 6.168 188 13.711 6.729 1! 11 2) g 3 11.71« 0.180 89 13.748 0.739 11 10 2.' 4 11,832 5.192 ™...0 190 13.784 5.749 li a 2.: ;l 4 11.874 0.206 36,481 191 13,820 5.769 a M 2; e 4 11.916 6.217 36,864 1 2 13.866 6.769 20,419 2,' 1 4 11.968 0.229 37.249 193 13.892 5.779 20,736 2; 4 4 12.000 5.241 37,636 194 13.928 5.789 21,026 3.1 4 12.042 5.2H 38,025 9S 13.964 21,316 3 4i 4 12.083 5.266 38.416 96 14.000 5.809 21800 3; 3 4 12.124 0.278 38,809 97 14.036 5.819 21,904 3,; S 4 12.1H6 5.290 39,204 98 14.071 0.828 22.201 3,. 9 4 12.207 5.301 30,601 9 14.107 5.838 22,600 3;. S 12.247 5.313 40;000 800 14.1*2 6.848 REVIEW EXERCISES (Sblbotbd from Collbob Entbanob Examinations) FACTORING 198. 1. Factor 6 a* + 10a6 — 4&«, aj» + 125, 1 — a? — a^ + a^, a^ + a?* + 1. Univ, of State of N.Y. 2. Find the prime factors of (a) (a?- ««)» + (« - !)• + (1 - x)K (6) (2aj + a — ft)* - (aj - a + by. Mass. InstUute of Tech. 8. Factor a? — 2 mas -|- wi* — n', a^-^^ — 11 aj»+^"» -h 18 ajy»* Univ. of Pa. 4. (a) Resolve the following into their prime factors : (1) (aj«-y«)«-3^. (2) 10aj»-7aj-6. (b) Find the h. c. f . and the 1. c. m. of aj»-3aj*4-a5-3, aj' — Sa^ — a5-f-3. Colunibia. 6. Find the h. c. f . : »* — y*, a?* -h 2«y — 3 y*. Cornell. 6. Factor the following expressions : (a) ai — 6i, (5) aj*yV — aJ*2 — yH; -h 1, (c) 16(a; -f- y)* — (2 a; — y)*- Jfoun^ Holyoke. 190 REVIEW EXERCISES 191 FRACTIONS a 199. 1. Simplify tj- a fe' + cb b Harvard. 2, Simplify the expression ic-hy — aj-hy- xy x-\-yl a^ — y^ Cornell. 3. Divide /^g^ziJ^^ ^4-y Wa^ + y^^^±y.\ \a?-y^ Q^ — xyJ Xx — y xy — yy Sheffield. 4. Simplify the expression A y\A ab-¥ \ a* ^ a--b \ ay\ a* Ja^ + b^ ' a^-^b^' Mass. Inst. Tech, 6. Find the product of Vasaar. 12 3- 6. If m = T » n = -, p = — ;—p^f what is the value of m + a + l' a + 2'^ a4-3 + 1 — m 1 — w 1 —p Univ. of Pa. EQUATIONS 200. 1. Solve ^ + ^±^ = ?. aj 4- X -{- a 2 Tale. 2, Solve for x : — z=. H — ^=3 = Va — b. Univ. of Pa. Va — X -y/x — b 192 ELEMENTARY ALGEBRA 3. Solve for a; : ?i^-?-6=s Cornell x-1 a?-|-2aj-3 4. (a) Solve for a? : V2 a? - 3 a + V3 a? — 2 a = 3Va. (b) Solve for m: 1-- = — —.4- ^" . Univ. of Pa. 6. Solve H = 0. Princeton. a? 4- 1 05 — 1 a? — 3 a? — 5 6. Solve the equation V2 x -}- 5 ^ V6 — as = 1. Uwiv. of Pa. 7. Solve 3 aJ' - 11 aj = 70. CTniv. o/ /Seoie o/ N. T. 8. Solve the quadratic equation aj» — 1.6 a? 4- 0.3 = 0. Harvard. 9. If d is one of the roots of the equation aa? + 6aj -|- c = 0, find the other root. Univ. of Pa. 10. Solve — ^ 4- - = 3. Univ. of State ofN. T. x-\- 1 X 11. ■ Solve Vaj-f- a -f- Vi 4- -y/x — a = 0. Uhry. o/ State of N. Y. 12. (i) Solve for t: {t- 2)(t 4- 3) = (3 « 4- 4)(2 - t). (ii) Solve for x : ■^ H -^ = . -. Univ. of Pa. V^T2 V3aj-2 V3a:»4-4a;-4* REVIEW EXERCISES 193 RADICALS 201. 1. Find the square root of af^^2ix^ — x^-\-Sa^-\-2x + l. Univ, of State of N.T. 2. 3. Univ, of State of N, Y. Solve for x and y the equations Vx-^-y = a H- 6, * — y = (« — ft) Va? + y. C/niv. of Pa, 4. Simplify (a? -f 1)' — x-y/x H- 1, V— m'n • V— mn', i 1 6*^ C/niv. o/ Pa, 5. Simplify the product of (ayx~^)^y (bxy~^y, and (y*a"'6~*)*. Princeton. 6. Solve the simultaneous equations x-i-h2y-i^i, 7. Simplify (a) Ve - V20. (6) i^v^Tl + o:^ 8. Find the value of a^6^, if a = a?^y » and 6 = ^ «"^y*, FoZe. Cornell, and reduce the result to a form having only positive exponents. Haroard. 9. Simplify — — ^- — ^— , and compute the value of the fraction to two decimal places. Yale. o 194 ELEMENTARY ALGEBRA SYSTEMS OF LINEAR EQUATIONS . 1. Solve = a, - + - = 0, = c. X y y ^ z X Univ. of State of N.T. 2. Solve the equations, 2 a + 5 y = 85, 2^4-52 = 103, 2 2 -I- 5 a? = 57. Vassar. 3. Solve the following set of equations : « -H y = - 1. aj-h3y + 22 = -4. 05 — y + 42 = 5. Cornell. 4. Solve the equations : 7a?-h6 ig^ 5g-13 8y-x 11 "^^ 2 5 ' 3(3iB + 4) = 10y-15. Mass, Inst. Tech. 6. Draw the lines represented by the equations 3 a— 2y= 13 and 2a;-|-5y= — 4, and find by algebra the coordinates of the point where they in- tersect. Univ. of Col. 6. Solve the simultaneous equations : a + a and verify your results. Harvard. SYSTEM OF EQUATIONS, ONE OR BOTH QUADRATIC 203. 1. Solve the system : aJ* + y' + «y = 54^, 2. Solve a? — y = 2, a^ — a^ = 30. Univ. of Pa. Univ. of State of N. T. REVIEW EXERCISES 195 3. Solve for t and u: - H — = 74, = 2. Univ. of Pa. 4. Solve the equations 2 a? — 3 y = 0. Columbia. 5. Plot the following two equations, and find from the graphs the approximate value of their common solutions : a^ -f y2 = 26. 4 a;* -f- 9 y* = 144. ColunMa. 6. Solve the following pair of equations for x and y : a' + y' = 4, a? = (1 H- V2)y - 2. Oomc/Z. 7. Solve a; — y = 2, and sketch the graphs. Mass. Inst. Tech. BINOMIAL THEOREM 204. 1. Write in the simplest form the last three terms of the expansion of (4 a' — o^x^y. Tale. 2. Expand (1 — V2)*^ by the Binomial Theorem. Univ. of Pa. 3. Write out by the binomial theorem the first ^ve terms of /^2 a - — y . Univ. of State of N. T. 4. Find the seventh term of f a + - ) • Columbia. \ ay 6. In the expansion of ( 2a?-f-^--] the ratio of the fourth term to the fifth is 2 : 1. Find a?, Princeton, 196 ELEMENTARY ALGEBRA PROBLEMS 205. 1. Two cars of equal speed leave A and B, 20 mi. apart, at different times. Just as the cars pass each other an accident reduces the power and their speed is decreased 10 mi. per hour. One car makes the journey from A to B in 56 min., and the other from B to A in 72 min. What is their common speed ? Tale. 2. The sum of two numbers is 13, and the sum of their cubes is 910. Find the smaller number, correct to the second decimal place. Vassar. 3. A man arranges to pay a debt of $ 3600 in 40 monthly payments which form an A. P. After paying 30 of them he still owes ^ of his debt. What was his first payment ? Princeton. 4. A page is to have a margin of 1 inch, and is to contain 35 square inches of printing. How large must the page be, if the length is to exceed the width by 2 inches ? Mount Holyoke. 5. Insert 10 arithmetical means between 10 and 61, and find the sum of the entire series. Univ. of Pa. 6. A man bought a number of cattle which cost him in all $672. If each head had cost him $4 less, he would have been able to buy 3 more. How many did he buy and at what price? Univ. of Pa. 7. A boy is 5 years older than his sister and ^ as old as his father ; the sum of the ages of all three is 51. Find the age of the father. Univ, of State of N. T. 8. Find two consecutive numbers whose product is 306. Univ. of State of N. T. 9. A man engaged to work a days on these conditions : for each day he worked he was to receive b cents, and for each day he was idle he was to forfeit c cents. At the end of a days he received d cents. How many days was he idle ? Univ. of Pa. REVIEW EXERCISES 197 10. For what values of m will the roots of the equation (m -h |)a?2 — 2(m + l)aj -f 2 = be equal ? Form the equation whose roots are | and ^. Univ. of Pa. 11. If a number of two digits be divided by the product of its digits, the quotient will be 6. If 9 be added to the number, the sum will be equal to the number obtained by interchanging the digits. What is the number ? Univ. of Pa. 12. A and B each shoot thirty arrows at a target. B makes twice as many hits as A, and A makes three times as many misses as B. Find the number of hits and misses of each. Univ. of Cat. 13. The sides of a triangle are a, h, c. Calculate the radii of the three circles having the vertices as centers, each being tangent externally to the other two. Harvard. 14. The force P necessary to lift a weight IT by means of a certain machine is given by the formula P=a-h6Tr, where a and h are constants depending on the amount of fric- tion in the machine. If a force of 7 pounds will raise a weight of 20 pounds, and a force of 13 pounds will raise a weight of 60 pounds, what force is necessary to raise a weight of 40 pounds ? (First determine the constants a and h.) Harvard. 16. Find the sum of 8 terms of the progression 6 + 3^ -h 2| -f .... Halyard. 16. How many terms must be taken in the series 2, 5, 8, 11, • • ., so that the sum shall be 345 ? Mass. Inst. Tech. INDEX Numbers refer to sections. Absolute, term, 70. value of a number, 1. Addition, 2. checking, 2. fractions, 68. radicals, 154. Algebraic expression, 1. Antecedent, 98. Antilogarithm, 119. Approximations, 195. Arithmetical mean, 171. Arithmetical operations, order of, Arithmetical series, 167. Associative laws,' 76. Base, 107. Binomial, 1. surds, 158. theorem, 84, 85, 192, 193. Braces, 6. Brackets, 6. Characteristic, 112. Checking, addition, 2. division, 9. equation, 20. multiplication, 8. subtraction, 3. Coefficient, 1. detached, 8. Commutative laws, 76. Complex numbers, 151, 164. Consequent, 98. Constants, 140. Contact, point of, 138. Convergent series, 195. Coordinates, 24. Degree, of equation, 19. of term, 8. Determinants, 87. Discriminant, 136. Dissimilar terms, 1. Division, 9. by zero, 94. checking, 9. of fractions, 62. of radicals, 157. Double root, 138. Elimination, by addition, 29. by comparison, 29. 5. by substitution, 29. Equation, 19. axioms used in, 20. complete quadratic, 70. containing fractions, 95. cubic, 19. degree of, 19. double root, 138. exponential, 127. general quadratic, 70. graph of linear, 22. graph of quadratic, 138, impossible, 160. incomplete quadratic, 70. indeterminate, 22. involving parentheses, 21. irrational, 160. linear, 19, 22. of condition, 19. of identity, 19. quadratic, 19, 70, 131. quadratic form, 132, 162. quartic, 19. radical, 160, 162. root of, 19, 146. Equations, equivalent, 26, 27. homogeneous, 144. inconsistent, 27. independent, 24, 27. linear, 87. 199 200 INDEX Equations — continued one linear, one quadratic, 139. simultaneous linear, 22. simultaneous quadratic, 139. solution by quadratics, 146. symmetrical, 143. Exponents, 1, 149, 150. fractional, 35. negative, 37. positive integral, 32. sero, 36. Extraneous roots, 95, 160. Extremes, 98. Factor, 1. H. C. F., 66, 189. imaginary, 82. irrational, 82. literal, 1. prime, 45. rational, 45. rationaUxing, 152, 157. theorem, 78. Factoring, 45, 79. Formulas, 31, 75, 97. 100, 101, 102, 105, 126. Fractions, 58. addition, 68. complex, 64. division, 62. multiplication, 62. reduction of, 58. subtraction, 68. value of, 60. Fu{iction, 100, 101, 103, 138. Geometrical mean, 179. Geometrical series, 175. infinite, 183, 184, 185, 195. Graphs, 103, 106. identical, 26. of complex numbers, 164. of linear equation, 22. of quadratic equation, 138. of simultaneous linear equations, 24. parallel, 27. variation shown by, 103, 105. H. C. F.. 66, 189. Hindu method, 191. Historical Notes, 77, 130, 164. Homogeneous expressions, 8, 144. Imaginary, factors, 82. numbers, 138, 151, 152, 156. Inconsistent equations, 27. Integral expression, 45. Interpolation, 119. Intersection, 138. Involution, 39. Is not equal to, 76, note. Laws of Algebra, 76. L. C D., 68. L. C. M., 66. Limits. 167, 185, 187. Logarithmic, curve, 108, 110. table, 117. Logarithms, 107, 117, 123. Mantissa, 112. Mathematical Induction, 192. Mean proportional, 98. Means, 98. Monomial, 1. Multiplication, 8. by detached coefficients, 8. checking, 8. fractions, 62. radicals, 156. type forms, 17. Number, absolute value, 1. Numbers, complex, 151, li4. imaginary, 138. 151, 166. irrational, 39, 151. negative,. 151. positive, 151. rational, 45, 151. real, 151, 165. Order of fundamental operations, 5. Origin, 164. Parentheses, 6, 12. equations, 21. insertion of, 14. removal of, 12. Polynomial, 1. INDEX 201 Portraits, de Morgan, 77. Wallis, frontispiece. Power, 1, 39. law of signs, 33. Principal roots, 38, 39, 160. Problems, practical, 31, 93, 97, 104, 105, 126. 129. Products, special, 17. Proportion, 98. Quadratic^ equation, 19, 70. complete, 70, 73. general, 70. graph of, 138. incomplete, 70. solved by completing square, 73, 191. solved by factoring, 73. solved by formula, 73. nature of roots, 136. relation of roots and coefficients, 134. Quaternions, 76. Radicals, 152. addition, 154. division, 152^ multiplication, 156. rationalizing, 152, 153. simplifjring, 152. subtraction, 154. Radicand, 152. Ratio, 98. Rational integral expression, 45, 78. Remainder theorem, 78. Review exercises, 198. Root, double, 138. of equation, 19. square, 41, 43, 158. Roots, 39. extraneous, 95, 160. nature of, 136. principal, 38, 39, 160. relation between roots and coeffi- cient, 134. Series, arithmetical, 167. convergent, 185, 195. geometrical, 195. infinite, 183, 195. Similar terms, 1. Simultaneous, equations, 22, 139. homogeneous, 144. sjrmmetrical, 143. Square root,_41. of a+2V6, 158. of fractions, 43. of numbers, 43. tables of, 197. Subtraction, 3. checking, 38. fractions, 68. radicals, 154. Tables of square roots, 197. Terms, dissimilar, 1. similar, 1. Trinomial, 1. Variables, 101, 138, 140. Variation, 102, 103. Vinculum, 6. Zero, division by, 94. exponent, 36, Printed in the United States of America. 'TpHE following pages contain advertisements of a few of the Macmillan books on kindred subjects. Elementary Algebra — A First Year Course By FLORIAN CAJORI Colorado College, and LETITIA R. ODELL North Side High School, Denver Cloth, i2fno, illustrated^ vii and 206 pages, 65 cents The joint product of a great scholar who is also a great teacher, and of an exceptionally able high school teacher whose success is based upon thorough scholarship as well as upon a comprehensive knowledge of school conditions and the abilities of boys and girls. 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