OURSE i ALGEBRA HAWKES -lUBY-TOUTON J GIFT OF MATHEMATICAL TEXTS FOR SCHOOLS EDITED BY PERCEY F. SMITH, Ph.D. PROFESSOR OF MATHEMATICS IN THE SHEFFIELD SCIENTIFIC SCHOOL OF YALE UNIVERSITY SECOND COURSE IN ALGEBRA BY HERBERT E. HAWKES, Ph.D. PROFESSOR OF MATHEMATICS IX COLUMBIA UNIVERSITY WILLIAM A. LUBY, A.B. HEAD OF THE DEPARTMENT OF MATHEMATICS KAKSAS CITY POLYTECHNIC INSTITUTE AND FRANK C. TOUTON, A.M. FORMERLY PRINCIPAL OF CENTRAL HIGH SCHOOL ST. JOSEPH, MISSOURI REVISED EDITION GINN AND COMPANY BOSTON • NEW YORK • CHICAGO • LONDON ATLANTA • DALLAS • COLUMBUS • SAN FRANCISCO tt33 ENTERED AT STATIONERS' HALL COPYRIGHT, 1911, 1918, BY HERBERT E. HAWKES, WILLIAM A. LUBY AND FRANK C. TOUTON ALL BIGHTS RESERVED 921.3 gbc satftengum 3pre<< GINN AND COMPANY • PRO- PRIETORS • BOSTON • U.S.A. x^ PREFACE This revision of the " Second Course in Algebra " has been carried out in the same spirit as was the recent revision of the " First Course in Algebra " by the same authors. The exercises and problems, mainly new, have been carefully graded, and the exposition has been wholly rewritten. Some advantageous changes have been made in the order of topics, and some chapters for which no well-grounded demand exists have been omitted. Such simplifications of exercises and exposition have been made as are consistent with a standard course in third- semester algebra. In the chapters devoted to a review of first-year algebra the fact was borne constantly in mind that the material would be handled by students who had not pursued the study of algebra during the preceding year. It was consequently thought desira^ ble to have work in equations come much earlier than before. The subject of fractions, the topic usually most in need of review, has received full and careful treatment. Linear sys- tems have been presented without the use of determinants. Instead of treating square root, radicals, and exponents in one chapter, the work under these topics has been made more accessible by giving a separate chapter to each. The needs of classes, even under almost identical conditions, differ widely, one class needing more review on a certain topic than does another. Consequently the review material has been expanded so as to afford ample work for any class. It is not intended, however, that all the exercises and problems should be solved by any one student. V 459908 SECOND COURSE IN ALGEBRA CHAPTER I REVIEW OF FUNDAMENTAL OPERATIONS 1. Order of fundamental Operations. The numerical value of an arithmetical or an algebraic expression involving signs of addition, subtraction, multiplication, and division depends on the order in which the indicated operations are performed. It is understood that In a series of operations involving addition^ subtraction^ multiplication^ and division, first the midtiplications and' divi- sions shall be performed in the order in which they occur. Then the additions and subtractions shall be performed in the order in which they occur or in any other order. Within any parenthesis the preceding rule applies. If the multiplication of two numbers or number symbols is indi- cated by juxtaposition without any symbol of multiplication, it is customary to think of the multiplication as already having been 9 a performed. Thus 9 a ^ 7 a^ = EXERCISES Simplify : 1. 6 - 2 + 8 - 7. 5. 36 -- 4 . 3 - 8 + 2. 2. 8 + 12-3-5. 6. (19-3.5)(8-5)-(14-f-7). 3. 8-6-4. 7. 50 - 4(20 - 4 • 2)- 3 + 2 -7. 4. 24 - 2 -r- 3. 8. 12 - 8 -f- 2 + 10 . 3 - 6 + 8 . 2. 9. (28 - 14 . 24 - 8 - 3 + 10) (30 - 3 - 5 -^ 2). 10. (16 - 32 X 48 H- 8 - 8 -h 3) -f- (42 - 6 . 7 - 42 - 6) . 6. 1 9 2 • SEGO-JSTB COURSE IN. ALGEBRA ' ' Fmd'tiie name^ieai v.alue of : 11. 9ic-7 itx = 3. 12. 2x^-5x + 3iix = 4.. 13. t^-St''-^St-lift = 2. • 14. Does 4(2;:c-5) + 15 = 3(£t;4-10) if cc = 7?. 15. Does (t -\-4:)(t-{-3)-(t-\- 1) (z^ + 2)= 42 if ^ = ic T^ 3 • 2x 3 + 5^-2.^2. „ 16. Does — - = — iix = 2? X — 1 x -{-1 x^ — 1 2. Similar terms. Terms which are ahke iii every respect except their coefficients are called similar. 3. Addition. In algebra, addition involves the uniting of similar terms which have the same or opposite signs into one term. For this we have the following rules: /. To add two or more positive numbers^ find the arith- metical sum of their absolute values and prefix to "this sum the plus sign, II. To add two or more negative numbers^ find the arith- metical sum of their absolute values arid prefix to this sum the minus sign. III. To add a positive and a negative number, find the dif- ference of their absolute values and prefix to this difference the sign of the number which has the greater absolute value. Obviously 2 + 4+7=2+7+4 = 7+24-4, etc. Even if we have a series of positive and negative numbers, the order in which they occur does not affect their sum. This principle of addition is called the Commutative Law for Addition. For the addition of polynomials we have the Rule. Write sitnilar terms iri the same column. Find the algebraic sum of the terms in each column and write the results ift succession with their proper signs. EEVIEW OF FUNDAMENTAL OPERATIONS 3 4. Subtraction. For the subtraction of polynomials we have the Rule, Write the subtrahend under the minuend so that similar terms are in the same column. Then change mentally/ the sign of each term of the sub- trahend and apply the rule for addition to each column. EXERCISES Add: 1. 12, - 8, 4- 4, - 3, and 6. 2. 4 a, 3 a, — 7a, 6 a, and — 2 a. 3. 5 a - 3 c 4- 6, 2 a - 6 c 4- 11, and 4 a — c - 9. 4. 3s-4i^ + 6, -7!f + 6s-8, and t-\-s^l^. b. X - 2 f - ?> z, ^ f - ^x -{- 2 z, d.n& 4cz - f. 6. ^2 - 3 a -f 1, - 2 «2 _ 7f^ _|- 6, and 3 cv" - 4 + 5 a. Write so that x, y, or t shall have a polynomial coefficient; 1. ax — 2x. 11. at — aH — 2 st. Solution. ax — 2x = {a — 2) x. 12. ay -\- by -{- y. 8. ace + bx -\- ex. IZ. ^ ax — A:bx -{- ^ x. 9. 5 at — Abt — 2t. 14. 4 a; — abx — x. 10. 4 2^ — 3 a?^ — S2^. 15. 7x — 3ax~ 4:a^x. Write the following so that the binomial will have a binomial coefficient : 16. (a-3)x-c(a-3). 18. S(a -{- b)- G(a -^ b). 17. A(a-x)-^c(-x-{-a). 19. 6a(^c - 2 c)- S(x - 2 r). 20. 3a(x-l)-2b(x-l). 21. 4:b(3x-2)~Sc(3x-2). 22. 4 7w,(5 a — 3 c) — 6 /t(— 3 c + 5 «). Subtract the second expression from the first : 23. 6a, 4a. 25. 4x4-3, 8.T4-6. 24. 8 a\ 15 a^ 26. Tor^ - 10, 5x^-h 20. SECOND COURSE IN ALGEBRA 27. 5x-6, 20" + 8. 28. x^-5x-{-6, 2cc2_|_3^_io 29. a'-4:ac-3c',4:a'-^10ac-e'. 30. Sa--2b-6c,4:a-h6b-7c-2. 31. 3a^-2c''-6ac,5a' + 4:c'-Sac. 32. a^-Sa^c + 6 ac\ 7c^-h4:a^c-2 a\ S3. X — Sf -\- z — 4:ac -\- 7 ax, 4:X — f + 8 — 5ax -h 9 ac. 34. a^—c-\-Sx — a^m — 8 ac, 4 a* + m — 8 x — 10 ac + 4 Find the expression which added to the first will give the second: 35. Sx^ - 5x -\- 2, 6x^ -llx -^ S. 36. 4:X^^Scx + c%10x^-j-Scx-9 g\ Find the expression which subtracted from the second will give the first : .37. 4.a'-2ab-\-b^, 7 a^ -lOab -h 6b\ 38. c'-ecx -i-Sx% 9x^ - lOcx + 4 + c^. 39. From the sum of 2^^ — 4 # — 9 and 3 ^^^ _ g ^^ _|_ ^1^ ^^i^^, ^.^^g sum of 3 2^ - 6 + 4 ?J2 ^nd 5 - 8 2^2 + 4 1. 40. From the sum oi ax — ac — ^ ^ and 4 c^ — 3 ac take the sum of 4 c^ — 8 ace -|- a^ and Aac -\- Sax — 5 c\ ORAL EXERCISES 1. What name is given to each 3 in a^ -\-Sa? Define both. 2. Distinguish between an exponent and a power. What is the meaning of 4 in cc* ? of 2 in 3^ ? of a in x^ ? 3. What is the coefficient of x in 3 a'^bx ? ofa^? ofb? 4. What is the coefficient of x in the expressions ax -\-x? 4:X — ax? ex — ax — X? What is the meaning of 3 in 3 ic ? of 10 in lOic? of a in OK? KEVIEW OF FUNDAMENTAL OPERATIONS 5 5. What is meant by the absolute vahie of a number? Illustrate. 6. What is a literal exponent ? Illustrate. 8. x^^x'^? x« -^ a;' = ? a^^^+i -^x^ = ? 9. How can the correctness of the result of addition be checked ? of subtraction ? 10. What is meant by arrangement of an algebraic expres- sion with respect to a certain letter ? 11. Arrange a^ -\- b* — 4: a^b — 6 a%^ + 4 ab^ according to the descending powers of b. 12. Arrange t^ — Sf^ — 5 -\-t — 2t^ according to the ascend- ing powers of t ; according to the descending powers of t. 13. Is arrangement of divisor and dividend in the same order, desirable ? Why ? 5. Multiplication. In multiplying one term by another the sign of the product, the coefficient of the product, and the exponent of any letter in the product are obtained as follows : 7. 7^e sign of the product is plus if the multiplier and the multiplicand have like signs, and minus if they have unlike II, The coefficient of the product is the product of the coeffi- cients of the factors. III, The exponent of each letter in the product is determined hy the general law n°xnf>= n«+ K For the multiplication of polynomials we have the Rule. Multiply the midtiplicand by each term of the multi- plier in tu/m, and add the partial products. SECOND COURSE m ALGEBEA EXERCISES Perform, the indicated multiplication-: 1. (6x^-3x-h4.)Sx. 5. Qix^ - 2x + 6)(7 - 3x^ - x). 2. (3x-5)(4:X + 3). 6. (2x^ - 5x-i-3)(x^- 5x-{- 6). 3. (2s-St)(4:S + 5t). 7. (a*-}-2a''-4)(a^-2a-3). ^. (x^-.x-^2)(3x-4:). 8. (t^-2t-\-6)(t*-3t''-ty 9. (2s^-3st-\-f-)(s^-{-5st-4:t^. 10. (2(4 _|_ 8 4- 4 2^2) (f^ + 8 - 4 1^). 11. (6^2 - ttc + c') («' + c" + ac). 12. (a'* -^2ab-[- h^) {a} + b- ab). 13. {t^ — t^ + t) (at^ + a-\- at). 14. (4 A2 _j_ 6 Jc^ + 9 AA)) (4 A^ 4. 6 A;^ - 9 M). 15. (3 c/.z^2 _ 2 a?^^ 4. 5 ^) (6 a?^' - 2 at - 4. af). 16. (2 ^2 _ 3 c« 4. 4 ac8) (2 a^ - 3 c« - 4 ac«). ,^ /a 2a« «^/» 2^2 a^ ^®* V2~T""4A2~ 3 ~ 4.J' 19. (?^^ + 1 + f) {t'-\-l- f) (f' + 1-0- 20. (x^ -\^ 9f -{- l(y -h Sxij - 4:x -\- 12y)(x - 3y + 4). Find the value of : 21. 3x^-^2x-\-r) if ic=:5. 22. 3t^-it-\-Siit=-3. 23. 9 - 8 jJ + 5 ?{' - 3 ^» if ^ =5 2. 24. 2 «« - 3 a^c 4- 4 ac2 - 3 c2 if a = 3 and c =^ -- 2. 25. Does 15(x - a)- 6(x -i- a)= 3(5 a - 3x) itx = 2a? ^6. Does ax(a -\- 3) -\- a(10 - a^)^ x + 3 if x =^ a - 3? ^7. Does T -T = 5 — ; ifa; = 0? ifx = -2? REVIEW OF FUNDAMENTAL OPERATIONS 7 6. Division. In dividing one term by another the sign of tlie quotient, the coefficient of the quotient, and the expo- nent of each letter in the quotient are obtained as follows : /. The sign of the quotient is plus when the dividend mid the divisor have like signs, and minus when they have unlike signs. II. The coefficient of the quotient is obtained hy dividing the coefficient of the dividend hy that of the divisor. III. The exponent of each letter in the quotient is determined hy the law tv^ -i- n^ = n°- *. The method of dividing one polynomial by another is stated in the Rule. Arrange the dividend and the divisor according to the descending powers of some common letter, called the letter of arrangement. Divide the first term of the dividend hy the first term of the divisor and ivrite the result for the first term of the quotient. Multiply the entire divisor hy the first term of the quotient, write the result under the dividend, and suhtract, heing careful to write the terms of the remainder in the same order as those of the divisor. Divide the first term of the remainder hy the first term of the divisor to get the second term of the quotient, and proceed as before until there is no remainder, or until the remainder is of lower degree in the letter of arrangement than the divisor. EXERCISES Perform the indicated division : 1. (2cc2-5a;-f 3)--(2x-3). 2. (6a^2-13£c + 6)-(3-2£c). 3. {^x^ -[-Bxy-2y'')^(x-\-2if). 4. (6a2 4-23o-!^-55?^2)-f-(3ci-5)5). 5. (6a« + 6a2-28-26a)-(2c^ + 4). .' 8 SECOND COURSE IN ALGEBRA 6. (6x^ -5x' + 25cc» - 17x^)-^(5x^ - 2a^«). 7. (6-«-7s^-8)^(6'2_^4 + 2s). 8. (x^-5 a'x + 2 a«) -^ (o? -\- 2 ax - a").' 9. (2 ^* - 12 jJ^ _ 2 _|. 11 25 _ 7 jj8^ ^ (^j^ _ 3 ^ _ 2 ^2>)^ 10. (23s2-13s^4-2s''-60-s)--(5 4-35-6-2). 11. (32a;*- 60- 2£c- 104x^ + 92 ic2)--(5 + 6x-4a;2). 12. (a?-2ah-\-h^-^x')^{^x-)rh-a). 13. (4^»c4-l-4c2-^»2)^(2c-^»-l). 14. (a;^ + i»' + 8ic2+8)H-(£c2-2x + 4). 15. (27 a - 18 a^'^ _ 3 ^9 ^ 3 ^J4^) ^ (3 _ i^ + ^4>^. 16. (3£c^ + 9cc2 + 2 - 5aj - 8a;« - cc*)h- (x - 1)1 18. (4.0" -{- ^ - ^ah + h^ - Qb + 12 a)^(h - ?> - 2 a). 19. (£c« - 4icy + lQy^)^(x' + 2;rV + 2x^/4- 4ic/ + 4/). 20. (x8 + 8/ + 12o-30«2/)^(a; + 2?/ + 5). 21. (a?« + 2/' + ^' - ^xyz)^{x + 2/ + ^)- 7. Parentheses. The removal of a parenthesis preceded by a plus sign is governed by the Principle, A parenthesis and the sign before it, if plus, may be removed from an expression without changing the signs of the terms which were inclosed by the parenthesis. The removal of a parenthesis preceded by a minus sign is governed by the Principle. A parenthesis and the sign before it, if minus, may be removed from an expression, provided the sign of each term which was inclosed by the parenthesis be changed. The principle just stated is equivalent to the principle which holds in the subtraction of one polynomial from another. EEVIEW OF FUNDAMEJSTTAL OPERATIONS 9 When one parenthesis incloses another, either the outer or the mner parenthesis may be removed first. Usually it is best to use the Rule. Rewrite the expression^ omitting the innermost paren- thesis, changing the signs of the terms which it inclosed if the sign preceding it he minus aiid leaving them unchanged if it he plus. Comhine like terms that may occur within the new inner- most parenthesis. Repeat these processes until all the parentheses are removed. EXERCISES E/emove parentheses and simplify : 1. a + {a-h)-{a-Zh). 3. ^b -\_{a - h)-{c - a)']. 2. ^_(a-6') + (2c-3a). 4. 2c - 2(a - c)-f 3(c - a). b. x-2j^-^(x-y)-A.(2x-y). 6. 4ic-a-f[-(3c-a')-(2a-3cc)]. 7. a_[_(a_3) + (3c-2a)-5«]+6c. 8. X - 3 - (ci - 2 ic) - [2 (a - X - 5) - 3 (6 - 2 a)]. 9. 2t^-^t-2t{^ + t). 10. x'-^-(x-l){x-2). 11. ^x" - ^ a^ -{a - 2x)(2>x - a). 12. 6cc+(3c- 8x + 2)-(c-a;-2). ' . 13. 6a;-[-(a-c)4-(3c-4a)]. 14. 7c-[(3c-4)-6-(4a^-3a-c)]. 15. 4.T. -2(£c-3)-3[x-3(4-2a;)+ 8]. 16. 6cc-4(3-5cc)-4[2(x-4)+3(2ic-l)-(ic-7)]. 17. 3x - 2[1 - 3(2 £c - 3 - a)- 5{a -(3cc - 2 a) - 4}]. 18. 2t''-lt-(2t-l)(t-{-l). 19. (x - 4>(£c - 3) - (x - 3) (;r + 2). RE 10 SECOND COUESE IK ALGEBRA 20. (a + b)a -(a — b)b-\-3 ab. 21. 3a(a - b)-(a + b)(a - b). 22. (x -3)(x-4:)-{x- 5) (x + 3). 8. Important special products. Certain products are of frequent occurrence. These forms should be memorized so that one can write or state the result without the labor of actual multiplication. I. For the square of the sum of two terms we have the formula . (a+bf = a" + 2ab+Ir'. This expressed verbally is : The square of the sum of two terms is the square of the first term plus twice the product of the two terms plus the square of the second terin. II. For the square of the difference of two terms we have the formula , ,.„ 9 « i. , 1.0 (a —hy = a^ — 2ah-\- b". This expressed verbally is: The square of the difference of two terms is the square of the first term, minus twice the product of the two terms, plus the square of the second term. III. For the product of the sum and tlie difference of two terms we have the formula , ' {a + b)(a-b) = a'-bi'. This expressed verbally is: T^e product of the sum and the difference of two terms equals the difference of their squares taken in the same order as the difference of the terms. IV. For the product of two binomials having a common term we have the formula {x+a){x^-b) = jc^ -f {a^b)x-\-ab. REVIEW OF FUNDAMENTAL OPERATIONS 11 This expressed verbally is: - The product of two binomials having a common term equals the square of the common term^ plus the algebraic sum of the unlike terms multiplied by the common term^ plus the algebraic product of the unlike terms, V. The square of the polynomial a-\-b — c gives the formula ^^^^_^y =: ^2 ^ ^ _^ c'+2ab~2ac-2hc. This expressed verbally is: TJie square of any polynomial is equal to the sum of the squares of each of the terms plus twice the algebraic product of each term by every term that follows it in the polynomial, VI. The cube of the binomial a + b gives the formula {a +by = (^ + Sa'b+3ab^ + b". This expressed verbally is: The cube of the sum of two numbers equals the cube of the first, plus three times the square of the fivst times the second, plus three times the first times the square of the second, plus the cube of the second. VII. Similarly, {a). 21. (a^ + 3)(.^ + 4). 22. {x + 2){x + b). 23. {x + l){x + l). 24. (x-3)(x-5). 25. (cc - 2) (ic - 7). 26. {x - 3) (a; + 4). 27. (a; H- 6) (ic - 7). 28. (£c H- 2) (£c - 9). 29. (x - 5) (£c + 10). 30. (o^cc — 3) {ax -\- 5). 31. (2a;-l)(2^ + 3). 32. (3cc + l)(3a; + 4). 33. (4x-2)(4ajH-3). 34. {a^-3a){a''^4.a). 35. (a + ^>4-c)2 36. {a^-G + xf 37. (c* + c - cc)*'^. 38. (a-c + ic/ 39. {a + e-\- Vf 40. (a-c + l)2 41. (a + c + 2)2 42. (a + c -h 6)-^. 43. (2 a -I- c + 1)2. 44. (a + 20- -3)2. 45. (a -3 a; 4- 2)2. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. (,, _ 3 ^ - cf. a:^ + b^ + c' + ) (a - &) = a^ _ j2 ^re identities, for in each, if the indicated operations be performed, the two mem- bers become precisely alike. An identity involving letters is true for any set of numerical values of the letters in it. Thus the identity a(h — c) = ah — ac becomes 2 (9 — 5) = 18 — 10, or 8 = 8, when, for example, a = 2, & = 9, and c = 5. 11. Equation of condition. An equation which is true only for certain values of a letter in it, or for certain sets of related values of two or more of its letters, is an equation of condition, or simply an equation. Thus 2 a; + 5 = 17 is true for a: = 6 only; and a: + 2 ?/ = 10 is true for a: = 8 and y = 1 and for many other pairs of values for x and y, but it is not true for x = ^ and y = 2 and for many other (though not for all) pairs of values for x and y. 15 16 SECOND COUKSE IN ALGEBRA J.2. Satisfying an equation. A number or literal expres- sion which, being substituted for the unknown letter in an equation, changes it to an identity, is said to satisfy the equation. Thus X = Q a satisfies the equation 2x + 5 a = 17 a, for on sub- stituting 6 a for X we have 2-Qa + 5a = 17a, or 17 a = 17 a, which is an identity. After the substitution is made it is usually necessary to simplify each member before the identity becomes apparent. 13. Root of an equation. A root of an equation is any number or number symbol which satisfies the equation. Thus 8 is the root of the equation 3 x + 2 = 26, for it satisfies the equation. 14. Axioms. An axiom is a statement the truth of which is accepted without proof. Some of the axioms most fre- quently used are Axiom I. If the same number is added to each member of an equation^ the result is an equation. Axiom II. If the same number is subtracted from each member of an equation, the residt is an equation. Axiom III. If each member of an equation is multiplied by the same number, the result is an equation. Axiom IV. If each member of an equation is divided by the same number (iiot zero), the result is an equation. Each of the foregoing axioms is used in the solution of the EXAMPLE Solve^-i = x + 7. Solution. -=z X + 7. (1) Multiplying (1) by 2, ox-l = 2x-\-U. (Ax. ITT) (2) LINEAR EQUATIONS IN ONE UNKNOWN 17 Adding 1 to each member of (2), (Ax. I) 5x = 2x + 15. (3) Subtracting 2x from each member of (3), (Ax. II) 3 a; = 15. (4) Dividing (4) by 3, (Ax. IV) X = 5. (5) Check. Substituting 5 for x in (1), we have 2^5 _ i := 5 _,_ 7^ or 12 = 12. Since substituting 5. for x satisfies (1), 5 is the root of (1). 15. Transposition. In the solution of the foregoing sec- tion by the appKcation of Axiom I to (2), the term — 1 is omitted from the first member and -f- 1 is combined with the second member. Again, by applying Axiom II to (3), the term + 2 :?:; is omitted from the second member and — 2 2^ is combined with the first member. It thus appears that a term may he omitted from one member of an equation., provided the same term with its sign changed from -{- to — or from — to -\- is written 4n or combined with the other member. This process is called transposition. Hereafter, in order to simplify an equation, instead of subtracting a number from each member or adding a number to each member, as illustrated in the foregoing example, the student should use transposition, since it is usually more rapid and convenient. He should, however, always remember that the transposition of a term is really the subtraction of that term from each member of the equatio7i. 16. Equivalent equations. Two or more equations in one unknown, even if of very different form, are equivalent if all are satisfied by every value of the unknown which satisfies any one of them. 18 SECOND COURSE IN ALGEBRA Equations (2), (3), (4), and (5) of section 14 are each equivalent to equation (1) and to each other, for all are satisfied by the same value of the unknown. Of the four axioms or assumptions of section 14 we shall make constant use. If the '' same number " referred to in each is expressed arithmetically, the result is always an equation equivalent to the original one. Further, if identical expressions involving the unknown be added to or subtracted from each member of an equation, the resulting equation is equivalent to the first. If, however, both members of an equation be multiplied by identical expressions con- taining the unknown, the resulting equation mag not be equivalent to the original one. Multiplying each member of the equation x — 2 = 3 by a; — 1, we get x"^ — Q X -\- 2 = 'd X — '3, or x^ — Q X + = 0. Now this last equa- tion has the roots 1 and 5, whereas the given equation has the root 5 only. Here the root 1 was introduced by multiplying the given equation by a: — 1. Results obtained from the use of Axiom III with multipliers which contain an unknown should always be carefully checked. When a root is obtained which does not satisfy the original equation, this root should be rejected. The use of Axiom IV when the divisor contains the unknown may result in the loss of a root which the process of checking will not discover. If an equation is divided hy a factor containing an unknown^ this factor should be set equal to zero. The root thus obtained is a root of the given equation. For example, if each member of a:^ — 4 = 3 a; -I- 6 is divided by a: + 2, the result is a; — 2 = 3, whence x- = 5. But a; = — 2 satisfies a;2 — 4 = 3 a; + 6. This root was lost by dividing by a; -H 2. With these and with certain other rare exceptions which will be noted later, the application of the axioms will produce an equation equivalent to the given one. LINEAR EQUATIONS IN ONE UNKNOWN 19 For solving equations in one unknown which do not involve fractions we have the Rule. Free the equation of any parentheses it may contain. Transpose and solve for the unknown involved. Reject all values for the unknown which do not satisfy the original equation. Checking the solution of an equation is often called testing or verifying the result. For this we have the Rule. Substitute the value of the unknown obtained from the solution in place of the letter which represents the unknown in the original equation. Then simplify each member of the resulting identity until the two members are seen to be identical. If the correct substitution of the root for the unknown does not transform the equation into an identity, an error has been made in the solution. EXERCISES Solve, and check the results as directed by the teacher: 1. hx-\-\ = 2x-\-l. 3. l + £c + 5£c + 17 = 0. 2. 6cc + 10 = 10x4-2. 4. 15cc = 3(4£c-5). 5. 2(£c + 2)-(a; + 5)=0. 6. 3(2£c-l)-(5a^-l)=0. 7. 3(2a;-7)-2(5-2^)-f-l = 0. 8. 3(£c4-2)-5<2x-3)-:0. 9. 6(cc + 4)-4(a; + 2)- 0. 10. 4(6j^4-2)-3(7 25 + 3)=0. 11. 6(4z^-5)-ll(2^-3)=0. 12. 6(2a^-l)+4(3-4a;)=0. 13. 4(4ic-l)+3-2(34-^)=0. 14. 5 7i-9(2 7i-f 4)-2(7z,-9)=0. 15. 4(x-2)+3(2-x)-3* = 6(ir + l). • 20 SECOND COUESE IN ALGEBRA 16. Sn - 5(4: - 71)= 5 - S(l-\- n). 17. (x-\-l)(x-2)=x^-[-3. 18. (x -{-5)(x + l) = (x- 3) (x-2)-^ 10. 19. (4cc-3)(2x-5)-(4a;-7)(2x-l)=0. 20. (x + 2)2 -f- 48 = (x - 4)2. 21. (a^ + 3)2 + 40=(cc + 5)2. 22. (2x-\-Sy = 4.(l-xy. 23. (x + 4)2 - (2 - £c)2 = 84. 24. (x-^ 3) (6 £c 4- 5) - (2 a: + 4) (3 £c - 8) = 38. 25. (1 - x) (x + 2) + (x + 3) (x + 4) = 0. 26. (2/_4)(6-2/) + (2/ + 2)(2/-4)=0. 27. (x + 4)(x4-3) = (x + 2)(x + l)+42. 28. (2v-S)(3v + 2)-(4.-6v)(l-v)=0. 29. (5x-3)(4-6x) + (3x + 4)(10x-21)-9 = 0. 30. 3(x + 2)(x-4)+5(cc-l)(x + 3) = 4(2x-l)(.T-2)+l. 31. X — 2a = 4:a — X. 32. c — ic = X — c. 33. 8s — x = X — 4r. 34. ax — 2 ah = A: ah — ax. 35. 2t ax — 5 ac = 2f ac — ax. 36. 3c(ic- 2a)=2c(a-ic). 37. m (x — 5 tt) — 3 m (^i — ic) = 0. 38. x(h-\-a)= ah-{-a^. • 39. x(a — g)= a 2 6'2 4c2. 41. 2ax-4a2-f 4a = l4-a^. 42. ax-a2 + 5a = e-j-3x. LINEAE EQUATIONS IN ONE UNKNOWN 21 43. ax + 1 — a^ = x. 44. 2mx-\-5m-3 = 3x-\-2m^. 45. (x -j- a)(x -{- b)= x" -\- 2a^ -^ Sab, , 46. (a — c){x — m) = (m — c) (x — a). 47. J ax - Slab = l^a? + lbh^ - bhx. 48. a^x — a^-\-Sax = S — 10a-{-x. 49. c(l + a:)+m(ic + l)-ic(m4-c + l)=0. 50. m (m — 2 x) + 2 am = a (2 £c — a). 17. Solution of problems. In the solution of problems lead- ing to simple equations the following steps are necessary : /. Read the problem carefully aiid find the facts which will later be expressed by the equation. II. Represent the unknown number by a letter and express any other unknown involved in terms of this letter. III. Express the conditions stated in the problem as an equation involving this letter. IV. Solve the equation. V. Check by substituting in the problem the value found for the unknown. In the preceding sentence the words " in the problem " are of importance, for substituting the value found m the equation would not detect any errors made in translating the words of the problem into the equation. PROBLEMS 1. The sum of two consecutive numbers is 975. Find the numbers. 2. One number is five times another, and their sum is 102. What are the numbers ? 3. The sum of two numbers is 54. Twice one of them equals ten times the other. What are the numbers ? 22 SECOND COURSE IN ALGEBRA 4. The sum of three consecutive even numbers is 1044 Find the numbers. 5. The product of two consecutive even numbers is 1416 less than the product of the next two consecutive even num- bers. Find the numbers. 6. Of four consecutive numbers the product of the sec- ond and fourth exceeds the product of the first and third by 201. Find the numbers. 7. One pupil is four years older than another. Eight years ago the first was twice as old as the second. Find their ages now. 8. Two men are 48 and 18 years of age respectively. How many years hence will the older be twice as old as the younger ? 9. One man is three times as old as another. Fifteen years ago the first was six times as old as the second. Find their ages now. 10. A's age is double B's. Twelve years ago B's age was one fourth of A's. How old is each ? 11. A is twice as old as B, and C is three times as old as D. B is 8 years older than D. In ten years the sum of their ages will be 113. How old is each ? 12. If each side of a square is increased 7 feet, its area will be increased 329 square feet. Find the side of the square. 13. A certain rectangle is 8 feet longer than it is broad. If it were 2 feet shorter and 5 feet broader, its area would be 60 square feet greater. What are its length and breadth? 14. A certain, rectangular plot of ground is 15 yards longer than it is wide. If it were 20 yards shorter, it would have to be 40 yards wider in order to have the same area. What are its dimensions ? 15. A certain square plot is surrounded by a border 6 feet wide. The area of this border is 816 square feet. What is the side of the square ? LINEAR EQUATIONS IN ONE UNKNOWN 23 16. A certain picture is 4 inches longer than it is wide, and the frame is 2 inches wide. The area of the framed picture is 192 square inches greater than that of the picture alone. What are the dimensions of the picture ? 17. Can a rectangle of perimeter 112 inches be drawn which has a length 5 inches greater than twice the width? If so, give its dimensions. 18. A sum of |15.75 consists of dollars, quarters, and dimes. If there are 6 more quarters than dollars, and twice as many dimes as quarters, find the number of coins of each kind. 19. A certain sum consisting of quarters, dimes, and pennies amounts to $8.62. The number of dimes equals twice the num- ber of quarters, while the number of dimes and quarters together is 2 greater than the number of pennies. Find the number of coins of each kind. 20. A collection of 109 coins is made up of quarters, dimes, and nickels. There are 7 fewer dimes than quarters, and 3 less than five times as many nickels as dimes. Find the amount of the collection. 21. The sum of the digits of a certain two-digit number is 14. If the order of the digits is reversed, the number is decreased by 36. Find the number. 22. Change the word "decreased" to "increased" in Prob- lem 21 and solve. 23. The digits of a certain three-digit number are consecutive odd numbers. If the sum of the digits is 15, find the number. Facts from Geometry.* The area of a circle is the square of the radius multiplied by 7r(7r = ^Y^- approximately). This is expressed by the formula A — irR^. The circumference of a circle equals the diameter times ir. The usual formula is C = 2 ttR. 24. If the radius of a given circle is increased 14 inches, the area is increased 1232 square inches. Find the first radius. 24 .SECOND COURSE IN ALGEBRA 25. By adding 2 inches to the radius of a circle whose radius is 8 inches, how much is the circumference increased? the area ? 26. Substitute R inches for 8 inches in Problem 25 and solve. Interpret your results. 27. Imagine that a circular hoop 1 foot longer than the cir- cumference of the earth is placed about the earth so that it is everywhere equidistant from the equator and lies in its plane. How far from the equator will the hoop be ? 28. Compare the result of Problem 21 with the one obtained when a similar process is carried out with a sphere 18 inches in diameter, instead of with the earth. 29. If the height of a square is increased 4 feet and its length is increased twice that amount, the area of the figure will be increased 248 square feet. Find the side of the square and the area of the rectangle. 30. The rope attached to the top of a flagpole is 5 feet longer than the pole. The lower end of the rope just reaches the ground when taken to a point 25 feet from the base of the pole. Find the height of the pole. 31. The length of a given rectangle is three times its width. A second rectangle is 9 inches shorter and 1 inch wider than the first, and has a perimeter one half as great. Find the dimensions of each rectangle. CHAPTER III FACTORING 18. Definition of factoring. Factoring is the process of finding the two or more algebraic expressions whose product is equal to a given expression. In multiplication we have two factors given and are required to find their product. In division we have the product and one factor given and are required to find the other factor. In factoring, how- ever, the problem is a little more diJfficult, for we have only the product given, and our experience in multiplication and division is called upon to enable us to determine the factors. 19. Rational expressions. A rational algebraic expression is one which can be written without the use of indicated roots of the letters involved. Thus 2, ox, 3?/— v2, and a^ are rational expressions. In this chapter factors which involve radicals will not be sought. 20. Integral expressions. If a rational expression can be written so as not to mvolve an mdicated division in which an unknown letter occurs in a denominator, it is said to be integral. Thus 3, 7 a, -> and 4 a; — 3 are integral expressions. In this chap- o 1 ter factors which involve fractions will not be sought. 21. Prime factors. An integral expression is prime when it is the product of no two rational integral expressions except itself and 1. «^ 25 26 SECOND COURSE IN ALGEBEA It must be remembered that to factor an integral expres- sion means to resolve it into its prime factors. The methods of this chapter enable one to factor integral rational expressions in one letter which are not prime, as well as some of the simpler expressions in two letters. No attempt is made even to define what is meant by prime factors of expressions which are not rational and integral. There is no simple operation the performance of wliich makes us sure that we have found the prime factors of a given expression. Only insight and experience enable us to find prime factors with certainty. A partial check that may be applied to all the exer- cises in factoring consists in actually multiplying together the factors that have been found. If the result is the original expression, correct factors have been found, though they may not be prime factors. 22. Polynomials with a common monomial factor. The type form is ab+ac-ad. Factoring, ab -\- ac — ad = a(lb + c — tl), ORAL EXERCISES Factor : 1. 3a + 6. 10. bax — 2ax^. 2. 5£c + 15. 11. 2c + 4c--2cc?. 3. a^+a. 12. 4.a-10a^-2a\ 4. 2c-Qc\ 13. 6ac-3^c4-3c. 5. 9x^-3cc. 14. 10cc+15a;«-5«?/. 6. cd^c^d\ , 15. 14a*-7«« + Ta*-7rt'. 7. ax" - A. 16. 3 c^ - 6 c^ -h 9 c» - 15 c\ 8. 4fx-8c2. 17. 6rV-3r8«2 + 3/'V-3rV. 9. 14 A - 21 A^A;. 18. 10 xY - 2 a-/ -f 2 a^/ - 2 xy. FACTORING 27 23. Polynomials which may be factored by grouping terms and taking out a common binomial factor. The type form is ax+ay+bx^hy. Factoring, ax -{- ay -{- hx -\- hy = (ax + ay') + (hx + hy) = a(x-\-y')-]-b(x-^y) = (_x + y)(a-hh'). EXERCISES Separate into polynomial factors : 1. 2(a-\-b)-{-x(a-['b). 5. a(c - d) -b(c - d). 2. S(b-^5)-{-a(b-j-5). 6. 2(x - y)- x{x - y). 3. 5x(a — cy-^y(a — c). 7. h(m +Sn) — 2k(m -{-3n). ^.2a(x-2y)-\-b(x-2y). 8. 2r(5x-4:2j)-9s(5x-4.tjy 9. -2a(Sh-k)—b(Sh-k). 10. x(r — s)-\- y(s — r). Hint. Write in the form x{r — s)— y{r — s). 11. 2a(c-Sd)-\-b'(3d-c). . 12. 5r(Sm-2n)-2s(2n-Sm). 13. ac -{- ad -}- be -\- bd. 14. ac -\- 2 ex -\- 3 ay -{- ^ xy. 15. mx — my -\- nx — ny. 16. cd-3cf-\-2d-^f. 17. «./z,r -|- aki' — ahs — (7A;s. 18. r^s + 2rs- 37^t-^ rt. 19. 4 /^m + 8 Ati - 6 A^m - 12 A;/i. 20. 2 a^ — 2 ax — ac + ca?. 21. x^-^xy^-^x^y + 'dy^. 22. aa? + ay -\- bx -{- by -\- ex -\- cy. 23. mr — 2 r + ms — 2 s + m^ — 2 ^. 28 SECOND COURSE IN ALGEBRA 24. Trinomials which are perfect squares. The type ^^^^ ^^ a^±2ab+lf. Factoring, a^± 2ab -{-b'^=(a ± by. ORAL EXERCISES Separate into binomial factors : 1. a^-\-2ax-^x\ 7. r^ - 10 r*- + 25 si 2. x^-2xt/-\-f. 8. 9 + 6a-\-a\ 3. m^ - 2 7nn + n^ 9. 16 - 8 ao; + aV. 4. a;2_|_4^_^4 IQ, 9(r2-12x?/4-4/. 5. 4_4a;H-ic2 11^ 1 4- 10 a^> + 25 a^^A 6. a^-4: ah + 4 Z-'l 12. {a + 5)^.- 2(a + 6) + 1. 13. (r-.s)2-6(r-.9)+9. 14. 4 (a + 2)2 - 12 c {a + 2) + 9 ) (c 4- 2) + 4 (6' + 2)1 18." {a + Z;)2 - 6(a + ^) (c - ^) 4- 9(c - 6/)l 19. a2n_i2a^^ + 36. 20. x?"" - 14 x^/* + 49 ?/^ 25. A binomial the difference of two squares. The type form is cfi — W Factoring, cfi—l)^={a-\- b) (a — b). More generally, «2^2aft4-^2_^4-2cc?-c?2 = a^+2ah + b^-((^-2cd+ d^) = (a-hby-(c-dY = (a-}-b-\'C-d){a-^b-c-\-d). FACTORING 29 ORAL EXERCISES Factor : 1. a^-o'^ 10. 25^2 -36 5V. 18. x^-1. 2. m^ - 'n?. 11. 36a2^»2_49c2c^l 19. x'-l. ' 3. a^-4. 12. a^-25 6^ 20. x'' - 81. 4. x" - 9. 13. xy-9. 21. 16 - a\ 5. 16 -xl 14. xy-64^^ 22. 625 - x\ 6. a^-16h\ 15. 81a2_l00^»V. 23. 81 - c\ 7. l-25cl 16. a* -16. 24. ^2m yin 8. 9a;y-l. Hint. Find three factors. 25. ^2m _ j2n^ 9. 160^2-25 2/^. 17. a^'-h'. 26. C''* - d""^. EXERCISES Factor : 1. a'-x\ 7. a?-{b + cf. 2. x'if-z\ 8. 4r'^-(r-s)*. 3. (tt-2f-c2. 9. 9m2- 4(71 + 3)1 4. 4 (a^ + 3)^ - 2/1 10. (a - c)^ - (cZ + ef. b. 16{x-yf-z\ II. a (/• + 2 s)^ - e^ (a^ - 2/)^ 6. 9(x-^yf-16z\ 12. x2(2^-A:)2-cc2(77^-27^)*. 13. 4 a{a - cc)^ - 9 a(c - 2 d)\ 14. a2cc(cR-2/)2-Z'2£c(3c-^)2. 15. m2 + 2m7i + 7i2-(£c2-2x?/ + ^). 16. x'-2xij + 2f 2 ah- y. 17. cc^ + 6a3 + 9 - ^2 4. 2 ^/^ - z^. 18. a^ - 22 a 4- 121 - &2 + 20 &c - 100 cl 19. 2-8£c2 + 8.x^-22/'-8?/^- 8.-;l 20. «2^<2 + 2a& -f 1 - c^ + lOccZ - 25 (Z'^. 21. ^c^-a^-2ab-l}'. 22. 50 6^%2 _ 8 ^,2 _^ 8 Z>c - 2 c^. 80 SECOND COURSE IN ALGEBEA 23. 49 x'- 49 2/2_ 14 ^_ 1. 28. r" + rs - {i^ - s^. 24. 121 ic»- 1-18 1/^-81 2/^. 29. nv" - n- - m - n. 25. (^2 - ^2 - (a - 6). ' 30. m-{-n- m^ + nl *26. cc^ - / 4- («^a; + a^/). 31. x^-4:f-\-x-2 y. 27. r'-rs-(7^-s'). 32. 7^ - r - S s - 9 s\ 26. The quadratic trinomial. The type form is x^ + bx+c. Since (x -{- h) {x -\- k^ = x'^ + (li + ¥)x-\- hk, it follows that a^ -\-hx-{- c can be factored into the bino- mial factors (x -\-}i)(x-\- k^ if two numbers h and k can be found which have the sum h and the product c. The method of determining these factors is illustrated in the EXAMPLE Factor x^ -2x-15. Solution. Here — 15 = 1 • — 15 or — 1 • 15 or + 5 • — 3 or + 3 • — 5. Of these pairs of factors of — 15 only the pair — 5 and + 3 give the sum — 2. Hence z^ - 2x -15 = (x - d)(x + 3). For factoring expressions of the type x^ -{-hx + e we have the Rule. Find two numbers whose algebraic product is + c and whose algebraic sum is + b. Write for the factors two binomials both of which have x for their first terms and these numbers for the second terms. EXERCISES Separate into binomial factors : 1. x^-h5x-\- 6. 4. a'' + 8 ^ + 12. 7. d^-Sd-10. 2. x' + 7x + 12. 5. d"" + 11 r/ + 18. 8. ^2_2^_35. 3. x' + 7a; + 6. 6. a^ -\- 10 a -\- 25 9. 0^-4:0-12. . FACTOEING 31 10. r" -r- 90. 18. a" -^ 9 a - 10. 11. m^ -3m- 18. 19. (a -{- bf - 2 (a -{- h) - 8. 12. a^2_2ic-24. 20. (x-yy + 4.(x-y)-{-3. 13. 9-lOic + a^l 21. a^n _|_ 12 ct^ + 35. 14. 7-2 - 4rs + 35^. 22. c^^ - 7 c« - 18. 15. m^ - 7 mn + 10 Til 23. m'"' - 11 m^ - 12. 16. 1 - 5x + 6x\ 24. a^^ - 5 a^^ - 6. 17. 1 _|_ 2 Tfc - 24 Til 25. yy -2h^y - 35. 27. The general quadratic trinomial. The type form is For many trinomials of this type two binomial factors of -the form Qix + It) (rnx + n) may be found. The method of factormg such trinomials is illustrated in the EXAMPLE Factor 2 x^ + 7 a:; — 15. ? a; + ? (\\ Solution. 2a:2+7x-15 = (?a: + ?)(?a: + ?). ?a: + ? (2) To find the proper factors we must supply ^x ■{■ , x such numbers for the interrogation points in + ?a: —15 (1) and (2) as will give 2 2'^ + 7 a: - 15 (3) 2 3^ for the. product of the first two terms of the binomials, — 15 for the product of the last two terms of the binomials, -\- 1 X for the sum of the cross products. Now 2x^ = 2x-x, (4) and - 15 = - 1 • + 15 ; 1 • - 15 ; + 3 . - 5 ; - 3 • + 5. (5) The factors of 2 and — 15 from (4) and (5) may be substituted for the interrogation points in (1) and (2) to form the following pairs of binomials, each having a product containing the first and last terms of the trinomial : 2x-l 2x+15 2i;+l 2x-15 2a: + 3 2ar-5 2x-Z 2a: + 5 a:+15 x—1 a; — 15 x-\-l x~5 x + 3 x + 6 x — d 32 SECOND COUESE IN ALGEBRA By trial we find that only the seventh pair has +7 x for the sum of its cross products, which gives the middle term of the trinomial. Therefore 2 a;^ + 7 a; - 15 = (2 a; - 3) (a; + 5). After a little practice it will usually be found unnecessary to write down all of the pairs of binomials that do not produce the required product. If none of the pairs gives the required product, the given trinomial is prime. If an expression of the form ax^ + bx+c is not prime, it can be factored by applying the Rule. Find two binomials, such that I. The product of the first terms is ax^ ; //. The product of the last terms is + c\ * ///. The sum of the cross products is + bx. EXERCISES Factor : 1. 2oc^ + ^x + 2. 4. 4^2 4- 7a + 3. 7. 4;r" + 8.T + 3. 2. 2a;2 + 7a^ + 6. 5. 3 .t^ + 13 .r + 12. 8. 6 ^-^ + 7 r -}- 2. 3. 2a2 + 9a + 10. 6. ^x' + llx + W. 9. 2b^-5h^2. 10. ?>x?-^x-\-^. 21. 9r7.2 + 3a-2. ir. 6(-2^7c4-2. 22. 12 7'2 + 10r-12. 12. 3 .r^ - 11 a: + 6. 23. 10 /-^^ - 19 r.s- - 15 s\ 13. 3.x' - 11 .T 4- 8. 24. iSa' -lla'^h-ld^bK 14. 4 .r' - 13 a; + 10. 25. 6 x» + 10 xhj - 4 xf. 15.10x^-29x4-10. • 26. 2x2_5^y_3y2 16. 12x''-llxy + 2f. 21.-2x^^Bxy -12y\ 17. 2a'^ + ^a-2. 28. ^a^-2a%-% ah\ 18. S7^ + r-2. 29. 6 a^"" - 7 a"" + 2. 19. 2a^-a-15. 30. 3 a1»» - 10 a" - 8. 20. 8 .s-2 -6s -9. 31. 10 a»" - a,'' - 3. FACTORING 33 28. Expressions reducible to the difference of two squares. The type form is ^4 _|_ ^^^2^^ ^ ^^ If k has siach a value that the trinomial is not a perfect square, a trinomial of this type can often be written as the difference of two squares. Thus, if ^ = 1, the addition and subtraction of aW accomplishes this result. EXAMPLES 1. Factor a" + (1%'' -\-h\ Solution. «4 + 0.%"- + &* = a^ + 2 a^U^ + h^ - a^l)^ = (a2 + V^y - (aby = (a2 + &2 4. ab) (a2 + &2 _ ab). 2. Factor 49 h"" + 34 h^k^ + 25 k\ Solution. If 36 A^p is added, the expression becomes a perfect trinomial square. Adding and subtracting 36 h^ki^, we have 49 h^ + 34 AU-2 + 25 ^4 = 49 ¥ + 70 /j^F + 25 yl'* - 36 ^2^^2 = (7h'' + Dky -(6hky = (7A2 + 5 ^'2 4. 6 7<^-) (7A2 + 5 ^2 _ g J^J.y EXERCISES Factor : 1. x^ 4- xy -f ?/. 11. 25 X* - 19 £c2 4- 9. 2. c* + c2(^2 + d\ 12. 9 ax« - 28 axY + 4 r/7/l 3. a' -\- a:'b' -\- b\ 13. 4 A + 3 c^^^^x + 9 ^>«ic. 4. a'' + 3a%^-i-4.b\ 14. 4a* + l. 5. m* + m^ + 1. Hint. 4a* + 1 = 6. x^^5x'^9. 4a* + 4a2 + l_4a2. 7. c*H- 4c2 + J6. 15. c^ + 4i7^ 8. 25a*-19a2 + i. le. 64aV + icl 9. 25cc*-lla^2 4-l. 17. 4a*^ + ^^^ 10. 47'*-44r^52_j_49^.4 ^^ x""^ -\- 4.y^'\ 34 SECOND COUESE IN ALGEBKA 29. A binomial the sum or the difference of two cubes. The type form is c^zkW a^ + h^ divided hj a-\-h gives the quotient a?' — ah -{- b% and a^ — h^ divided hj a— b gives the quotient a? -\- ah-\- b\ Therefore a^ + 5^ = (a + 5) (^2 - ab -^ b^), (1) and a^-b^ = (a-b}(a^-i-ab-hl^'). (2) Formulas (1) and (2) above may be apphed as in the EXAMPLES 1. Factor (^^ + 27. Solution, a^ + 27 = a^ + S^ = (a + 3) (a- - a • 8 + B^) ' = (a + 3) (a2 - 3 a + 0). Solution. 8 - 2:8 = : 23 - a:8 = (2 - a-) (2^ f 2x + a:2) = (2 - a;) (4 + 2x4-2:2). EXERCISES Factor : 1. x' + 7,A 9. x' - 2^ 17. ««+(/>2/. 2. 6t^4-^^'. 10. a^-27. 18. c« + ^«. 3. a^4-2«. 11. m^ - 64. 19. m« - nl 4. c' + 51 12. m8-(2 7i)». 20. (2ay-(3by. 5. c^'^ + S. 13. 8;r«-7/«. 21. 8a:«-27v/«. 6. 6/« + 27. 14. 7^-27s\ 22. 125x« + 8//. 7. a-« + 125. 15. 64-a;».- 23. (a -{- by + c^ 8. x' - y\ 16. 125 -a;«. 24. ^?8 4- />» + or. -h ^>. ^5. x^-y''-\-x -?/• 28. ,3_8.s.8 4.r-25. 26. m^ — n^ — m -f- w. 29. ^8» ^ /,8n^ 27. ic» - 8 / + ^; — 2 ?/. 30. ^8m _ pn^ FACTOEIKG 35 30. The Remainder Theorem. If any rational integral expression in x be divided by x — n, the remainder is the same as the original expression with n substituted for x. This fact is illustrated in the EXAMPLE Divide x^ — 5x -\- 6 hj x — n. • Solution. x^ — 5 X + 6 X + (n — 5^ (n-5)x + 6 (n — o)x ~ n^ -{- 5 n n^ — 5 n + 6 = Remainder Here the remainder n^—3n + 6is the same as a:^ — 5 .r + 6, the given expression, when n is substituted for x. EXERCISES 1. Divide x^ -\- bx -\- c hj x — n and show that the remain- der is n^ -{- bn -{- c. Z. Divide x'^ -{- bx -{- c hj x — a and find the remainder. 3. Divide x^-j- ax^-\- bx + chj x — n and find the, remainder. 4. Ill (x* -{- x'^ — 5 x + S) -i- (x — 2) find the remainder (a) by division, (b) by the Remainder Theorem. Solution (b). 2^ + 2^ - 5 • 2 + 3 = 5. 5. In (x^ — X -\- 5) -^ (^x — 3) find the remainder (a) by divi- sion, (b) by the Remainder Theorem. By use of the Remainder Theorem find the remainders in the following : 6. (x^ -\- x^ - 5x -\- S)-ir(x - 3). . 7. (x^-3x-15)-^(x-^4.). 8. (x^-2x''-100)^(x-5). 9. (x^ - 2x^ -2x- 3)^(x - 3). 10. (x^-2x^-^x- 2)^(x - 2). 36 SECOND COURSE IN ALGEBEA 31 . Factor Theorem. By substituting 2ioTxmx^—5x-\-6 we obtain 4 — 10 + 6, or 0. Hence a; — 2 is an exact divisor (or factor) ot a^ — 3 x -\- 6. Again, if 3 is substituted for x m x^ — 5 X -\- 6, the expression equals zero. Hence a; — 3 is a factor oi x^ — 3 x -\- 6. These examples illustrate the Theorem. If any rational integral expression in x becomes zero when a number n is substituted for x^ then x—n is a factor of the expression. The Factor Theorem may be used to factor some of the preceding exercises and, in addition, many others which are very difficult to factor by previous methods. Note. By means of the Factor Theorem we are able to solve cubic and higher equations when the roots are integers. The solu- tion of the general cubic equation is one of the famous problems of mathematics and one which is accompanied by many interesting applications. This problem was first solved by the Italian, Tartaglia, about 1530, but was published by, Cardan, to whom Tartaglia ex- plained his solution on the pledge that he would not divulge it. For many years the credit for the discovery was given to Cardan, and to this day it is usually called Cardan's Solution. When searching for the values of x which will make an expression zero, only integral divisors of the last term of the expression (arranged according to the descending powers of x') need be tried, for the last term of the factor must be an integral divisor of the last term of the expression. EXAMPLE Factor cc» + 2cc-3. Solution. If X — n is a factor oi x^ ■\- 2x — 3, then n must be an integral divisor of 3. Now the factors of — 3 are 1, — 1, 3, and — 3. If 1 is put for X, then x^ -I- 2 x — 3 equals zero, hence a: — 1 is a factor of a:^ + 2 a: — 3. Dividing a,*^ -f- 2 a; — 3 by a; — 1, we obtain the quotient a;^ + a: + 3. Since a;^ -f- a: + 8 is prime, the factors of x* -H 2 a; — 3 are a: — 1 and ar^ + a: + 3. FACTORING 37 ORAL EXERCISES 1. Is X — 1 a factor of ic^ + 3 cc — 4 ? 2. Is X - 2 a factor of 2x^-^x' - 20? 3. Is a - 2 a factor of a^ - 3 (^ + 2 ? 4. Is X — 1 a factor of cc^ + 3 x^ — 4 ? 5. Is r 4- 1 a factor of r'^ — 4 r^ — 4 r + 1 ? 6. Is r — 3 a factor of 2 r^ — 7^ + 5 ? 7. Is 5 +1 a factor of 3s^ - 5^^ + 8 ? 8. Is A: - 3 a factor of 2 A:^ - 5 ^b^ - 9 ? EXERCISES Factor : 1. x^-{-x-2. 8. y^-^f -^y-{- 9. 2. x^+-2xH-3. ' 9. .T'^-Tx'-ex. 3. tt«4.«2_36. 10. x^ -7x^4- 4x4-12. 4. ^3 ^ .^ _io. 11. 2x^ - 2x2 - X - 6. 5. ^^ + ^"-12. 12. x3-x2-4. 6. x^-2x2-5x + 6. 13. 3x^-2x2 + 2x- 3. 7. ^8_x24-4x-4. 14. a^-Q>a^ +lla'-^a. 32. The sum or difference of two like powers. The type form is a" zfc 6". The cases m which a^ ± If' is divisible by a 4- ^ or a — h can be determined by the Factor Theorem. Thus in a^ — If^ n being either an odd or an even in-" teger, substitute h for a. Then a^ — h^ becomes b'^ — h^ = 0. Therefore a— b is always a factor of a^ — h^. In a" — 6^, n being even, put — b for a. Then d!^ — b^ becomes b^ — b'^ = 0, since (— 5)^ is positive when n is even. Therefore when n is even a 4- ^ as well as a — 6 is an exact divisor of a^ — b^. 38 SECOND COURSE IN ALGEBRA In a'* + h''\ n being even, put either -\-h or —h for a. Then a^ + V^ becomes If^ + h^^ which is not zero. Therefore ^i _|_ yti ^g i^eyer divisible hy a-\-b oy a — h when n is even. In a** + h''\ n being odd, put — b for a. Then a" + 6" becomes (~ by^ + ^'^ = 0, since (— 5)" is negative when ?i is 06?(i. Therefore when n is odd a-\-b is a divisor of a^. + 5". Summing up: I. a" — b" is always divisible by a — 6. II. a" — &", when n is e^^e/i, is divisible both by a + & and by a — &. III. a" + &" is never divisible by a—b. IV. a" + &", when n is oc7c?, is divisible by c + 6. ORAL. EXERCISES I For each of the following, state a binomial factor : 1. x'-f. 6. x''-2f. 11. 7?^'^+2l 16. 1-7-1 2. x^-5\ 7. o:^ - 2^. 12. a' + 8. 17. l-r» 3. 27 - a\ 8. 2^cc^-7/l 13. iK^+7/. 18. 1 + 7-^ 4. cc^-2^ 9. a'^ - c-". 14. r^ + 2'. 19. l + /« 5. 32^^-/. 10. a«-^A 15. a^+32. 20. 5« + l. 21. 8^-1. 23. Is 10' -f 1 divisible by 11? 22. 10^-1. 24. Is 10^-1 divisible EXAMPLE by9? Factor x^ + y^ Solution. By division, £! -i-^ = x*- xh + xhr - xy^ + v^. Hence a;'^ + y* = (a: + ?y) (.r^ — ar*?/ + a-V — 2:?/' + y^). Note that the signs of the second factor are alternately plus and minus. Also note the order in which the exponents occur. FACTORING 39 EXERCISES Factor : 1. x^-^z\ 8. a^-32af. 18. 1 - r^. 2. x'' + 1. 9. (2x)^-2432/^ Hint. Write only the first five terms and the 3. a^-^2\ 10. a} - x\ last term of the poly- 4. x' + 32. 11. 1 - r^. nomial factor. 5. (ay + (hy '. 19. 1 + r^ 6. cc^-«^ 12. x^ - 128. 20. x^ — if. Hint. Find the , see- 13. x^' + y'K Hint. Factor first ond factor by division and observe the signs 14. o}'^Z2x^K as the difference of two squares. of the terms and order in which the the ex- 15. x' 4- a\ 21. x« - y\ ponents occur. 16. \^t\ 2^2. a}''-}?-\ 7. a' - 2K 17. 128:r^-M. 23. a"-"--}?^.. 33. General directions for factoring. The following sug- gestions will prove helpful in factoring : /. First look for a common monomial factor^ and if there is one (other than i), separate the expression into its greatest monomial factor and the corresponding polynomial factor. II. Then from the form of the polynomial factor determine with which of the following types it should he classed^ and use the methods of factoring applicable to that type. 1. ax+ay+bx+by. 5. ax'^ + bx+c. 2. a'±2ab-\-1y', 6. a^ + kaW + b\ 3. a'-b'. . 7. a^zkb^. 4. x' + bx+c. 8. fl"±6". ///. 'Proceed again as in II with each polynomial factor obtained^ until the original expression has been separated into its prime factors. IV. If the preceding steps fail, try the Factor Theorem. 40 SECOND COURSE IN ALGEBRA REVIEW EXERCISES Factor : 1. 6x^ + 2x^-{-2x\ 29. x^ - Sx^-4.x-^12. 2. 5a^ + 2a^-15a- 6. 30. x - x^ - x^ -^ x\ 3. a^-\-4.ab-\-4: h\ 31. (a + xf + 10(a + x)-\- 25. 4. 2cH-%cd\ 32. {x + rf-^{x + r)-l^. 5. 3m«-3m2-18m. 33. x^ - Sx"" - x -[- ^. 6. 2^2 + 3 aa; + al 34. a^ + 27 6i"'. 7. x'-lxhf-^-^y^' 35. 64c^8-|-2^«. 8. a^c - ac^. 36. a^'^ - a^»V + a^^c - a6c». 9. 2x\j-2xy\ 37. (ic + 2/)2- 6(a; 4-2^)^ + 9;5;2. 10. cc^ - 2 cc^ - 9 a;2. 38. (a - xf - 16(m - ti)^ 11. ac^2hG-ad-2hd. 39. 2 ic^ - 2 ic^ - 12 cc. 12. 18 r^ - 24 r^s + 8 7-/. 40. 10 a?c - 15 aV - 70 ac». 13. 45xV-20x/. 41. a* -11^2 + 1. 14. 2 A% 4- 4 «2 _ 30 ;j,3 42. 6t^ + aVK 15. 3a2-10a/> + 3^;^. 43. a^ + ct'i^ 16. ic^ + 7 .T^ + 16. 44. x^{x - 1)2 - .T(a; - 1). 17. aH-%ad\ 45. 4(a-^')2-12(a-^>)^ + 9<'2. 18. 2 6i^ + 64 dK 46. a^ - 10 aZ» + 25 ^'^ _ c^. 19. m^n-\-mn\ 47. cc2_2ic(«_^,)_35(a-^)l 20. x^ -\-^x- 5. 48. 6^2 - 13a: 4- 6. 21. ax'-^a^Zx^-Vl. 49. ««- 8 ar» + 17 a; - 10. 22. a%'' - 4 a2^»2 _|_ 4 a6. 50. ic* - ^if + a;» - a^V- 23. aV-c2-a2^_i. 51. ^2_i2,r + 36-a*. 24. xy - 13 x\f - 14 a;y. 52. a"" - b^ -\-(a- bf. 25. 27» -b't^s-Srs". 53. c2-2c(Z + rf''-2(c-r/)-85. 54. 2{a-\-by-^^e{a + b)-2c'. 55. r??/* - 7 wV + 71*. 56. a8 + ^<8 + 3a'''^ + 3a^'^. 26. x'- -l^x^ + 36. 27. a^- - a2/>8 - - a»^»'-^ ^^'^ 28. x"^ -x\ FACTORING 41 57. ^2x^-xh/\ 69. 9ic'-4i/2-3x-2y. 58. b{a-bf-a + b. 70. x" -b(2x- 5). 59. c2 + 4^2_^2_4^^ rj^ a^_S-7a^ + Ua. 60. (a-2x}c'-\-(2x-a)d:\ 72. 3a2-14a(^>-c)4-8(*-c)l 61. x^-20-\-x\ 73. aV4-4. 62. 6a^-\-Sa^-3 al 74. ccV + x'z^ 63. 16a* + 7a^ + l. 75. cc^ - lOx - 3. 64. ccy - 64. 76. x' - xS/ - a-* (cc^ - ij"). 6b. x' -lSx^-21x\ 77. a«-5a* + 4. 66. a' - a^Z*^ + ^2^ - a^ 78. ti* - Qa^ - « + 3. 67. a* + 8 fi^^ + a' + 2 «^>. 79. x - 1 + a-^ - x\ 68. a2 + 2a + l-Z'' + 2^»c-cl 80. x^ + x - y - f. %l.,x' + 2xij + f-2^ a^ -lOa-1. 82. a''-b^-2ah{a'-b''). 83. 2 ir^ + 3 x^ + a;. 92. 7^'"^ - 2 A"^A:" + A:^^ 84. m^ - 8 m - 7 m*. 93. a^"^ - ^'^^^ 85. 3 (^^ - 9 a (2 a - 3). 94. x^'^ + (r + s)^^'' 4- ^'s. 86. 2x^-10x^-\-4:X. 95. /^^-x'*^ + hrx'' + Ascc^ + As. 87. 10 a -7 a" -6 a\ 96. a*"* + a^'^b^'' + &*'\ 88. a^-^-l-^-Sa' + Sa. 97. a«"^ + ^'^ 89. 12x^-Sx-^x^-6x\ 98. a«"* - b^"". 90. a^x^ + 2 (^»ic + a« 99. a^"* + ^>s^ 91. mr — ms — m- + ws. 100. a^^ — ^^^ 101. Solve for x, ax -\-bx — 3a = 3b. Solution. Rewriting, ax + bx = ^ a -^ ^b.^ Factoring in each member, a:(a + &) = 3 (a + 6). Dividing each member by a + &, a; = 3. 102. Solve for x, ex -\- Ux = (^ -\- ed. 103. Solve for m, wa — ?w^ -f Jc = ai. 42 SECOND COURSE IN ALGEBRA 104. Solve for y, 5 ay — S bi/ — 5 a^ -\- S ab = 0. 105. Solve for z, az — 3 ad = bz — S bd. " 106. Solve for ic, cic — 2 dx = c^ — 4 c?l 107. Solve for m, am — m -\- 1 — a^ = 0. 108. Solve for y, by -^S dy - b^ + 9d^ = 0. 109. Solve for r, r(2a -7c)= 4.a^ - 28 ac + 49 c^. 110. Solve for m, m(Sa — c)— 9 ad = 2 ec — 6ae — Scd. 111. Solve for a;, — ^- 6. -7i z r^J a; — 2' x + 2 3 2c^ a^ — 9' 'a-3 3 a 2 a' 7. 8. x^ — 5ic + 6 — x^ + 4 2 a ' 5 «. 2a2 + 3a + l'2«^-hl' 2a ^4-3 2^2 + 5 (^-2)2^-2 a + 2 3a + 13a2 + 7a-f2 a 2 a 5a 9. 10. 2^ 3x x'-\-2x 2x4-1 5 a;2-2ic + 4 a;^ + 8 x -^ 2 2x-9 -2 11. -^ ^» 12 ^5_32 _a;8 + 2x2 x'-{-2x^ + 4.x^+Sx'-\-16x ax — hx — ar -\- br x' — i^ o^ — IP" ax -\- hx -\- ar -\- hv a? — ly^ x? — r^ 2x 5x 4 3ic + 2 13. -;; -y- J x^-S 2-x 4:-x' x^ + 2x' + 4.x 40. Addition and subtraction of fractions. To find the algebraic sum of two or more fractions in their lowest terms we proceed as in the EXAMPLE Find the algebraic sum of — H — r r- ° a a'- ah Solution. TheL.C.D. isa% Rewriting with common denominators, and adding numerators, ^® ^^^® 2^ . ?i! _ 5l^ = 2 «&' + 3 ?>2 - 5 g 64 SECOND COUKSE IN ALGEBRA Check. Substituting 2 for a and 3 for b, we have 6 9 5 _ 36 + 27-10 2 4 6" 12 53^53 12" 12' For finding the algebraic sum of two or more fractions we have the Rule. Reduce the fractions to respectively equivalent frac- tions having the lowest common denominator. Write in succes- sion over the lowest common denominator the numerators of the equivalent fractions^ inclosing each polynomial numerator vn a parenthesis preceded hy the sign of the corresponding fraction. Rewrite the fraction just obtained^ removing the parentheses in the numerator. Then combine like terms in the numerator and^ if neces- sary^ reduce the resulting fraction to its lowest terms. Check. Set the original expression equal to the final result and substitute in each member numerical values for the letters involved. The equation should be an identity. EXERCISES Find the algebraic sum of : 2x 3^_2£r . c 2c 510 15 • "c-4c-f3 2c5c^c a 2a 1 ^ a-b a-Zb Sa 3a;-l 3 ab a" y x'-2x^-Z x-2 2r8^-\-r r-Qs _ s^ 2g + l 4 r's ~ 3r^ ~2r x^-^ x-Z 5a 2 ,. «-3 a- 3 5 10. 3 3 a-^-4 2-« ' a + 2 FEACTIOKS 55 6^ + 2 c-5 ,, ^ x^ ^''- c(c-2) 8-c^ ^-2 13. ^^ _ 1 _ ^m_-2^ 17. a' + ab + P- -^ m — 1 m m^ — 1 a — o 14.2 + -^- . ''■''-f^S-^- X — 7 20. -i- ^^ ^^^ a; + 2 ' X ^x' + 2x-2 c-^2d 1 • 2 22. 23. c^^2Gd + d'' d-G c + 2^ a_3 2-a 1 a,2_2o^-3 a2_3^_^2 l-a^ ^^* 4-^2 a-2 "^ a + 2 a + 2 a-\-2 25. a2_7^_,_12 6 + ^2- 5a 6a -a^- 8 2S.*±_«_2(*-^). — a \a a — 01 28. ,. '.+ ' . + \*'. . , + " + • (^ — c) (c — a) ac — a?" — be -\- ab (a — b)(b — c) x^ — (m — ny m^ — {x — nY t? —{x — TYif (x + ny — Tf? {x -\- KYif — n^ (rti + rtf — x^ i •1 = = M- a c ac b d -u' 5.1f = -i-h'- = -V-. n a n a na h * 56 SECOND COUESE IN ALGEBRA 41. Multiplication of fractions. In algebra, as in arith- metic, the product of two or more fractions is the prod- uct of their numerators divided by the product of their denominators. Thus Similarly, and In like manner Integral and mixed expressions are reduced to fractional form before the multiplication is performed. Factors com- mon to any numerator and any denominator are canceled the same number of times from each. EXAMPLE Multiply 20«^(l-i)(^^^j^). Solution. Writing the above in fractional form, we have 20 g^ g^ - 4 6 a 1 ' a2 '5a3-10a2* Factoring and canceling, 4 1 ^ -^rtJ^XjOr — -^ a a To find the product of two or more fractions or mixed expressions we have the Rule. If there are integral or mixed expressions, reduce them to fractional form. Separate each numerator and each denominator into its prime factors. FRACTIONS 5T Cancel the factors (^factor for factor^ common to any numerator and any denominator. Write the product of the factors remaining in the numerator over the product of the factors remaining in the denominator. Check, Set the given expression equal to the final result and substitute in each member numerical values for the let- ters. Simplify each member. The result should be an identity. 42. Division of fractions. For division of fractions we have the Rule. Reduce all integral or mixed expressions to fractional form. Then invert the divisor or divisors and proceed as in mul- tiplication of fractions. Check as usual. EXERCISER Perform the indicated operations : 6 a 10 xip' ^ a^ — 4 4 a 25 xV 4 A 2^2 ^ ^ 3(^ 4c ^ 3^2 2. 5c. TT^^-TT-,- 9. lOc^^ ^d x + 3 ^x^ laV' ^^cU jQ a}-\ ^a? 5a-5 ^cd"- Ua% 2a a'-2a-{-l 3(l-a) ^ Sah 21 c^ 20 a%^ ,, (^ , 1\/ a'' \( 1\ 4 ^ 2a^ _^ 12. A-tV 2 ' 3x'' l^a 13. If H-lf: 6. lOa.I^.^. 14. 1^^ ^^- 21 (T^ "• 15 ax" 4:m\ 6ns 2mV 120 e^d 100 cd 9 7iV 4.m^s 2mn^ *"' 42c^x * 147 c^ic^ RE 58 SECOISTD COUESE IN ALGEBRA 24 a;* _8^ ^ _48^ 22 mV . 52 m'n , 275 n-x ^<^xif' 22x'y '' \2\xY 35^3 ' 125 ar^ ' 39m-^x _ 15 rs^ 300ri(« 1605^* ,^ 5x^ ^{x-yf 1 14^2^ 98^2^ 28rV x-y 2^^ x{x-y) x + S a;^-4a;"H-4 , x'-2x x^~4:' 2x-t-6^ ' x^ + 2x' x^^2x + 4l ^ a;»-8 . x-^ ~ 2a; a;'-2x + 4 ' ic2 + 2x * x^ + 8 * „ 2a;^ + 9a; + 9 ^ x^ - 9 a;'-6x + 9 2a;2-9cc + 9 * 4x^-9x' 2x2- 3« x^ — y^ . i^^ + i^y + ^z"* ^'^ — ^2/ + 3/^ *a7^ + y^' x + y a; — 2/ a^ + 32 , a« + 2a2 af" -^ 2a^ + Aa"" -^ Sa + 16 a" -32 ' a^-2a' a^-2a^ + 4.a^-Sa + 16' 25. 26, x^-Qx' + llx-S , a;2-5a; + 6 a;'' + 1 a;*-l ' x^-\-2x-^l' x-\-l' 2x + 5 ^ 25-4a;2 ^ 2:^2 + a; - 15 ^3_6^2^_18.^_27 • a;» + 27 ' x-3 29. 30 -©)"-Ki5)'©)«(if.)e)' -•(— D(^)fei)• 1 4. «2 • \^-^ a« + 1/ V «« + 1 / 33 34 FEACTIONS 59 9 • '^;rri^\Sa'-{-la + 2}v''~l~V 35 — : — 12" X- 5 + a? -|- 3 Hint. An expression of this form is called a complex fraction. It is simply another way of writing \ x + 3/ \ x + 3/ X — 2 \^ — y / \x — y / , 3 X-1-- 2 11 X X + 1 + 38. — b a-\-b a — b x__ 45_ X 39. a -{- b a — b a , b a-b~a+b b a b ^ + 1 + ^^ 46.'^^ + ^^ « + ^ a . a 1 — a a -\- 1 a 41. ^ a 1 — a 48^ -^2 X a — - a T f 1 1 ^^- a;" •s^ 2«/ a — i " 4- 2 ^>^ A.__2 "M— -- i_- ^j'.^ — />^ 2 a \ c a c I b^cll - ' 4a6 a 60 SECOND COUESE IN ALGEBRA 50. If ic = ■; > y = — ; — ) and z = -> find the value 6 + c a -\- c a -{- n , ,.; when a = x— - and b = x -\- - a/ -^b^ X X 51. What value has ^ , ,., ; when a = x— - and b —x-V-'i 52. 1 ^-. 53. 1 ?: — 54. 1 + 1 + ^ 1- 1 + ^ 1-| 1 + i Z 6 a 55. Brouncker (1620-1684) proved that ir (the circumference of a circle divided by its diameter) is four times the following fraction : 1 1 + 2 + 2 + ?^ 24- '^ 2 + etc. 225 (cC) Rewrite the fraction, continuing it to 2 + etc. (&) Stopping with 2 -f- -^^- , find the difference in value between four times the value of this fraction and 3.1416, the approxi- mate value of TT. Note. William Brouncker, one of the brilliant mathematicians of his time, was an intimate friend of John Wallis (see " First Course in Algebra," Revised Edition, p. 48). These two scientists were among the pioneers in the study of expressions with a countless number of terms. The complex fraction in the exercise, if continued indefinitely according to the law which its form suggests, is called an infinite continued fraction. Brouncker was the first to study the properties of such expressions. FRACTIOKS 61 43. Equations involving fractions. Equations involving fractions are solved as in the EXAMPLE Solve ' 5-« = ^(^. (1) X X +1 Solution. The L. CM. of the denominators is x(x + 1). (1) . x(x + 1), 5a:(a; + 1) - 8(x + 1) = 5^2- 5a:. (2) From (2), 5a;2 + 5a: - 8 a: - 8 = 5a;2 - 5a;, (3) or . 2a: = 8. Whence a: = 4. Check. Substituting 4 for x in (1), 5 - 2 = 3, or 3 = 3. For solving equations in one unknown which may or may not ipivolve simple fractions we have the Rule. Where polynomial denominators occur^ factor them if possible. Find the L.Q.M. of the denominators of the fractions and multiply each fraction and each integral term of the equation by it, using cancellation wherever possible. Transpose and solve as usual. Reject all values for the unknown which do not satisfy the original equation. EXERCISES Solve the following for x and check as directed by the teacher : 1. 2 2 2^- 2. ¥-l = f + 3. 3. 2x , , 7 4. llx 2 9 3 = "- 3x 16 2 3 25 = x--^ ^x 21 6 5 ^x-2 10 2(^-4) 3 3 2(0.-2) 2 + 5a; 9 62 SECOND COURSE IN ALGEBRA ^ 2x + l lzL?^_Qi 1^ a: + 5_10 ^' ~4 3~~ ~ ^^- ^^=^ ~ 3 ' 3 7 4' 14. ^ = — ^ 4 a? 16 3 a; 1, 2x-7 3 • 3a; 5x 45 3x-2 3 12. 3(a; + 2)-4(2a;-3) + 2 = 0. a; + 7 "^ 5 17. (a; + 5)(a; + l)-(ir-3)(x-2) = 10. 1ft _A_ . __J___o 99 ^ + l _ x + 4 ^** 2a;-5^2a;-l"^' ^'^* a; + 3~a; + 2* 2a;-5 ^ 3a;-14 a;+3 3a; + 4 * 2a;4-7~ 3a;-2* 2aj -f 1 ~ 6a; - 2* „^ 5a; + 3 5a; + 7 ^^ 17-3a; 13-3a; 20. = • 24. = 2a;-3 2a; 3»-13 3a;-8 «^11 — ^/v «,.^ — 3h 5 21. 7i 5 = 0. 25. — ; = 1 a?6 — a;* a;4-7 x -\- 1 26. 3(1- 8a;) 2(1+ 8a;) _ 1- 34a; 5x 8a; —1 5x 27._1- + ^^±1==1_ ^^ 28. 29. 2x+l l-2a; 4a;2-l c -\- X _ x(c —1) ~ x-\-l " c (a; + 1) ' l+2a; _ 3 a; +1 _ 8 + 3a; -4a;'' 2a;-l x+1 ~ 2a;2 + a;-l 30. a^- — h. a x^a x-\rh x -- x ^ 31. :; — == — a. 33. tH —:r=^Za, a b ' a —1 a +1 a X ^ <( -\- 2 o« 1 , 1 1 oA ^ a; +16 . 32. - + - = -• 34. — —rr ?r = 4a. 36. FRACTIONS 63 2a;+l 3x-7 _ 9-Sx-a^ 'x-{-32~x~x'^x-6^ X -\- cd X — cd _2 chl — 2 cd'^ c -\~ d c — d c^'— d^ 37. f^-|^^ = 0. Lx — a ox — c 5x-\- A 1.3 a; -.05 30.35 -8 a; ^^* '~~^~ + 4 - 2A • 5 — 23- r 4-2 39. 11.3-^-— ^ = 2.3 -(5- 7a;) + --^. ,' 2x-.3 .4x + 5 275 ^, 41. ;r = — 9.4. Jx ' X .25 X PROBLEMS 1. Separate the number 286 into two parts such that the greater will be 2^ times the less. 2. Separate the number 1010 into three parts such that the second will be -^- of the first and the third will be -^-^- of the first. 3. By what number must 352 be divided so as to give a partial quotient 15 and the remainder 7 ? 4. What number must be subtracted from both terms of the fraction |-| to give a fraction equivalent to J- ? 5. Separate 133 into two parts such that their quotient is 2^. 6. Separate 96 into two parts such that 56 exceeds two thirds of the one by as much as the other exceeds 16. 7. A boy is 12 years old and his sister 8 years old. In how many years will the boy be f as old as his sister ? 8. Two thirds a man's age now equals |- his age 30 yearg ago What is his age ? 64 SECOND COUKSE IN ALGEBRA 9. The square of a certain number is 4 greater than two thirds of the product of the next two consecutive numbers. Find the number. 10. The length of a certain rectangle is 2|- times the width.. If it were 10 yards shorter and Ij yards wider, its area would be 1260 square feet less. Find the dimensions of the rectangle. 11. A square court has f the area of a rectangular court whose length is 4 yards greater and whose width is 3 yards less. Find the dimensions of the square court. 12. A can do a piece of work in 10 days and B in 12 days. How many days will they both require working together ? Hint. Let x = number of days required by A and B together. Then - = fractional part of the piece of work that they can do in 1 day. 13. A can do a piece of work in 10 days and B in 15 days. After they have worked together 5 days, how many days will A require to finish the work ? 14. A tank has a supply pipe which fills it in 4 hours and a waste pipe which empties it in 6 hours. If the tank is empty and both pipes are opened, how much time must elapse before the tank is filled ? 15. A tank has a supply pipe which fills it in 4 hours and two waste pipes which empty it in 6 and 8 hours respectively. If the tank is full and all three pipes are open, how much time will be required to empty the tank ? 16. If all three pipes of Problem 15 were outlet pipes, how long would be required to empty the tank ? 17. If in Problem 15 the supply pipe had been closed after 4 hours, how much more time would have been required to empty the tank? 18. The diameter of the earth is 3| times that of the moon, and the difference of the two diameters is 57 GO miles. Find each diameter in miles. FEACTIONS 65 19. The diameter of the sun is 3220 miles greater than 109 times the diameter of the earth, and the sum of the two diame- ters is 874,420 miles. Find each diameter in miles. 20. The diameter of Jupiter is lOj-J times the diameter of the earth, and the sum of their diameters is 94,320 miles. Find each diameter in miles. 21. A man who can row^ miles per hour in still water rows up a stream the rate of whose current is Ij- miles per hour. After rowing back he finds that the entire journey re- quired 10 hours. Find the time required for the trip upstream. 22. A man who can row 4 J miles per hour in still water finds that it requires 6i hours to row upstream a distance which it requires 2|- hours to row down. Find the rate of the current. 23. A passenger train whose rate is 40 miles per hour leaves a certain station 2 hours and 45 minutes after a freight train. The passenger train overtakes the freight in 5 hours and 15 minutes. Find the rate of the freight train in miles per hour. 24. A man invests a part of $8000 at 5% and the remainder at 4%. If the yearly interest on the whole investment is $345, how much was invested at each rate ? 25. A man invests $6800 in two parts : the first part at 5%, and the second part at 4%. If the average rate of interest is 4f %, find the amount of each investment. 26. Two thousand dollars of Mr. A's income is not taxed. All of his income over that amount is taxed 2%, and all above $10,000 is taxed 2% in addition. He pays a tax of $180. What is his income? 27. How much water must be added to a gallon of alcohol 95% pure so as to make a mixture 10% pure ? Hint. Let w = the number of gallons of water to be added. Then nkll = i?-, etc. 1 + w 100 66 SECOND COURSE IN ALGEBRA 28. How many ounces of alloy must be added to 45 ounces of silver to make a composition 60% silver ? 29. It is desired to mix coffee which sells for twenty-five cents per pound with coffee which sells for thirty-five cents per pound so as to obtain a 10-pound mixture which may be sold at the same profit for thirty-two cents per pound. In what ratio must the parts be taken f r^m each grade ? 30. Milk of a certain grade is known to test 20% cream. How much water must be added to 25 gallons of this milk to make a mixture 18% cream ? 31. Grun metal of a certain grade is composed of 16% tin and 84% copper. How much tin must be added to 410 pounds* of this gun metal to make a composition 18% tin ? Hints. Since the composition is 16% tin, ^W • 410 = the number of pounds of tin in the first composition. Let X = the number of pounds of tin to be added. . Then \- x = the number of pounds of tin in the sec- ond composition, and 410 + X = the number of pounds of both metals in the second composition. 16.410 + X Therefore ■' = — , etc. 410 + X 100 32. How many "gallons of alcohol 90% pure must be mixed with 12 gallons of alcohol 96% pure to make a mixture 93% pure ? 33. A certain lot of pig iron contains 93% pure iron. How much pure iron must be melted with 10 tons of pig iron to make iron 98% pure ? 34. The arms of a lever are 5 feet and 6 feet in length respectively. Excluding the weight of the beam, what weight on the shorter arm will balance 70 pounds on the longer ? Hint. The products of the weights by their respective arms are equal. FKACTIONS 67 35. The arms of a balanced lever are 7 feet and 12 feet respectively, the shorter arm carrying a load of 36 pounds. Find the load on the longer arm. 36. If the load on the longer arm in Problem 35 be reduced 7 pounds, how many feet from the fulcrum must a 12-pound weight be placed on the longer arm to restore the balance ? 37. A beam 12 feet long supported at each end carries a load of 2 tons at a point 4 feet from one end. Find the load in tons on each support. 38. At what time between 3 and 4 o'clock will the hands of a clock be together ? Solution. The minute hand moves twelve times as fast as the hour hand. While the minute hand travels x spaces, the hour hand travels — spaces. Hence a; — — equals the number of spaces gained by the minute hand in any given time x. In the time from 3 o'clock until the hands are together, the minute hand must gain 15 minute spaces to overtake the hour hand. X Therefore x — — = 15. Whence x = 16 y\. Hence the hands are together 16 y\ minutes after 3 o'clock. 39. At what time between 9 and 10 o'clock are the hands of a clock together ? 40. At what times between 3 and 4 o'clock are the hands of a clock in a straight line ? 41. At what time between 5 and 6 o'clock is the minute hand 10 minute spaces ahead of the hour hand ? 10 minute spaces behind the hour hand ? 42. At what times between 4 and 5 o'clock are the hands of a clock at right angles ? 43. If the average height of n boys is x inches, what is the sum of their heights in yards ? 68 SECOND COURSE IK ALGEBKA 44. If 3 m 4- 5 packages weigh j9 pounds, wliat is the weight of n of them ? 45. If it takes a man t hours to do a piece of work, what portion of the work can he do in 1 hour ? What portion of the work would n men do in 1 hour ? What portion would k men do in h hours ? 46. If it takes n men h hours to do a piece of work, how long will it take x men to do it ? 47. A man buys bananas at d cents a dozen and sells them for h cents each. What does he gain on n dozen ? 48. A man bought m articles for c cents per hundred. He sold them all for |10. How many dollars did he lose ? 49. Itn yards of ribbon cost c cents, find the cost of x yards. 50. If «/ yards of ribbon cost x cents, how many yards can be bought for d dollars ? 51. A man buys goods for b dollars and sells them for s dollars. What is his per cent of gain ? 52. One man can do a piece of work in d days, another can do the same work in n days. How many days will it take both, working together ? 53. If it takes h hours to mow m acres, how many days of 10 hours each will it take to mow n acres ? 54. A train goes y yards in t seconds. If this equals m miles per hour, write an equation involving y, #, and m. 55. If 71 men can do a piece of work in d days, how many men would it be necessary to hire if the work had to be done in t days ? 56. A transport plying between two ports is under fire for / feet of the way. If she steams k knots per hour, for how many minutes is she under fire ? Hint. 1 knot = 6080 feet. CHAPTER V LINEAR SYSTEMS 44. Graphical solution of a linear system. The construc- tion of the graph of a single linear equation in two unknowns or of a linear system in two unknowns depends on several assumptions and definitions. It is agreed: I. To have at right angles to each other two lines, X^OX^ called the jr-axis, and F'OY, called the y-axis. II. To have a line of definite length for a unit of distance. Thus the number 2 will correspond to a distance of twice the unit, the number 4| to a distance 4^ times the unit, etc. III. That the distance (measured parallel to the a:-axis) from the ?/-axis to any point in the paper be the j:-distance (or abscissa) of the point, and the distance (measured parallel to the ?/-axis) from the :r-axis to the point be the y-distance (or ordinate) of the point. IV. That the a;-distance of a point to the riglit of the ?/-axis be represented by a positive number, and the x- distance of a point to the left by a negative number ; also that the ^-distance of a point above the a:-axis be repre- sented by a positive number, and the y-distance of a point helow the a:-axis by a negative number. Briefly, distances measured from the axis to the right or upward are positive-, to the left or downward are negative. V. That every point in the surface of the paper corre- sponds to a pair of numbers, one or both of which may be positive, negative, integral, or fractional. 69 70 SECOND COUKSE IN ALGEBRA • VI. That of a given pair of numbers the first be the measure of the a^-distanee and the second the measure of the ^-distance. Thus the point (2, 3) is the point whose x-distance is 2 and whose ^-distance is 3. The point of intersection of the axes is called the origin. The values of the a;-distance and the ^-distance are often called the coordinates of the point. The relation between an equation and its graph may be stated as follows : The equation of a line is satisfied hy the values of the x-distances and the y-distance of any point on that line. Any pointy the values of whose x-distance and whose y- distance satisfy the equation, is on the graph of the equation. The graph of a linear equation in two unknowns is a straight line. Therefore it is necessary in constructing the graph of such an equation to locate only two points whose coordinates satisfy the equation and then to draw through the two points a straight line. It is usually most convenient to locate the two points where the line cuts the axes. If these two points are very close together, however, the direction of the line will not be accurately determined. This error can be avoided by selectmg two points at a greater distance apart. The graphical solution of a linear system in two unknowns consists in plotting the two equations to the same scale and on the same axes and obtaining from the graph the values of x and y at the point of intersection of the lines. Through the graphical study of equations we unite the two subjects of geometry and algebra, which have hitherto seemed quite separate, and learn to interpret problems of the one in the language of tke other. LINEAR SYSTEMS Tl EXAMPLE Solve graphically the system 3^ _ 4?/ + 20 = 0, Solution. Substituting zero tor x in S x — 4: y + 20 = 0, we obtain y = b. Substituting zero for y, we obtain a:=— 6|. This may be expressed in tabular form : li x = -6| then y = 5 Similarly, for the equation 2x + 2^4-6 = we obtain the following table : If x = -3 then y = -6 Then, constructing the graph of each equation as indicated in the adjacent figure, we obtain for the coordinates of the point of intersection of the two lines a; = — 4 and y = 2. Y y 5 ^^ V ^y 5 -..^4 V ^^ * \,yr \T\ X ^ 3 X /'-« -♦ A'' " ' j^ 3r 2 4 =^ IS -4 ^ A; ^ 3 r r Solve graphically : 2. aj - 2/ = 6, 5x + 4t/ = — 15. 3 aj 4-2^4-11 = 0, ' y-x = l. EXERCISES iK + 2y=4, 6y4-2x = 20. ^ + 4 = 10x, 2 - a; = 0. 2x4-42/ = 20, 22/-2 = x. 72 SECOND COURSE IN ALGEBRA x + y = ^, x-y = b, * 2/ + 2 = 0. 3a^-f-2t/ = 6. 0^ + 5 = -32/, 3a^-42/ = 33, • 62/ + 2£c-ll = 0. 4« + 32/=-6. «H-?/ = 4, 2x-4ty = % ' x + 2y = 7. x-2y=S. 13. In each of the first three exercises will the values of the X- and ^/-distances of the point of intersection of the two lines, as obtained from the graph, satisfy the equation obtained by- adding the two given equations ? 14. Graph the equation x — 2 y = S. Then multiply both members by 3 and graph the resulting equation. Compare the two graphs. Th^n use — 2 as a multiplier and graph the result- ing equation. Compare the three graphs. What conclusion seems warranted? 15. What are the coordinates of the origin ? 16. Is a graphical solution of a linear system ever impossible? Give an example. 17. What is the form of the equation of a line parallel to the £c-axis ? the ^/-axis ? 18. The boiling point of water on a centigrade thermometer is marked 100°, and on a Fahrenheit 212°. The freezing point on the centigrade is zero and on the Fahrenheit 32°. Con- sequently a degree on one is not equal to a degree on the other, nor does a temperature of 60° Fahrenheit mean 60° centigrade. Show that the correct relation is expressed by the equation C = f (F — 32), where C represents the number of degrees centigrade and F the number of degrees Fahrenheit. 19. Construct a graph of the equation in Exercise 18, using C and F as abscissa and ordinate. Can you, by means of this graph, express a centigrade reading in degrees Fahrenheit and vice versa? LINEAR SYSTEMS 73 20. By means of the graph drawn in Exercise 19 express the following centigrade readings in Fahrenheit readings and vice versa: (a)60°C.; (^) 150°F. ; (c)-20°C.; (d)-SO°¥. 21. From the graph determine what reading means the same temperature on both scales. 45. Elimination. In order to find values of x and 2/ which satisfy the equation 32:+2y=20, (1) when we know that ^ = 22: + 3, (2) we may substitute for 7/ in the first equation the value of y from the second, obtainmg the single equation in x, Sx-h2(^2x-{-S)=20, or 72^ = 14. The process by which we have obtained one equation containing one unknown from the two equations (1) and (2) each of which contains two unknowns illustrates one method of elimination. In general, the process of deriving from a system of n equations a system of n — 1 equations, containing one variable less than the original system, is called elimination. For example, when n = 2, if we have a system of two equations in two unknowns, the process of elimination leads to one equation in one unknown. Since we can always solve such an equation, it appears that we can solve a system of two equations in two unknowns whenever it is possible to eliminate one of the unknowns. We shall see that only in certain exceptional cases is elimination impossible. This is either because more than one unknown is removed when we try it or because the result of attempted elimination is not an equation. Only two methods of solution will be considered — that involving elimination by substitution and that involving elimination by addition or subtraction. 74 SECOND COUKSE IN ALGEBEA 46. Solution by substitution. The method of solving a system of two linear equations by substitution is illustrated in the EXAMPLE Solve the system/^ ~^^,^^^^;, W ^ 18 0- + 11 2/ = 18. (2) Solution. From (1), a; = 13 ?/ + 81. (3) Substituting this value for x in (2), 8 (13 2^ + 31) + 11^ = 18. Simplifying, 104 z/ + 248 + 11 ?/ = 18. Combining terms, 115 y = — 230. Whence y =—2. Substituting — 2 tor y in (3), and solving, x = 6. Check. Substituting 5 for x and — 2 for y in (1) and (2), 5-13(-2) = 31, or 31 = 81, and 8-5 +11 (-2) = 18, or 18=18. The method of the preceding solution is stated in the following Rule, Solve either equation for the value of one unJmown in terms of the other. Substitute this value for the unknown which it represents^ in the equation from which it was not obtained^ and solve the resulting equation. In the simplest of the preceding equations which contains both unknowns, substitute the definite value just founds and solve, thus obtaining a definite value for the other unknown. Check. Substitute for each unknown in both original equa- tions its value as found. If the resulting equations are not obvious identities, simplify them until they become so. 3x -22/: = 18, 42/ + Sx. = 0. 8m -Sn + 6/ 2 4 m -1 = -Sn. LINEAR SYSTEMS 75 EXERCISES Solve by substitution : Sx-Sy = 20, 5. X — 61/ = 0. 2x + 52/ = 8, ^ 2(x + y)+Stj = 4:, 4.t-2n = lS, ' 5==x-hy- ^- 202^ = 7/1 + 63. ^ 16x + 7 = 152/, 6a; + 38 = 12y, • 4a:4-52/ = 0. ^' 4:X-Sy = 0. 47. Solution by addition or subtraction. The method of solving a system of two linear equations by addition or subtraction is illustrated in the EXAMPLE c 1 .1. . fl3x + 3y = U, (1) Solve the system |^^_^^^^^^; (2) Solution. First eliminate y, as follows : (1) • 2, 26 a; + 6 ?/ = 28 (3) (2). 3, 21a:-6.y:^66 (4) (3) + (4), 47 a: =94 (5) (5) ^47, a; = 2. (6) Substituting 2 for x in (2), we obtain y = — 4. Check. Substituting 2 for x and — 4 for y in (1) and (2), 26 - 12 = 14, or 14 = 14, and 14 + 8 = 22, or 22 = 22. Before starting to eliminate, the system should always be inspected carefully to determine which unknown can be removed most conveniently. In this case the elimina- tion of y is the simpler, because it involves multiplication •by smaller numbers than does the elimination of x. 76 SECOND COUESE IN ALGEBRA The method of the preceding solution is stated in the following Rule. If necessary^ multiply each member of the first equa- tion hy a number^ and each member of the second equation by another number, such that the coefficients of the same unhnoum in the resulting equations will be numerically equal. If these coefficients have like signs, subtract one equation from the other ; if they have unlike signs, add ; then solve the equation thus obtained. In the simplest of the preceding equations which contairis both unknowns, substitute the value just found and solve for the other u7iknown. Check. As on page 74. ORAL EXERCISES Solve the following systems : ' x-y = 2, ' 2x + y = ^. *2ic-3?/ = 9. g a^ + 27/ = 3, ^ 2aj+32/ = 8, ^hx-6y = 7, ' X -\- y = 2. ' x — y = 1. ' Ax — Sy = 2. 48. Special cases. The equation x + y = 10 has as roots any set of two numbers whose sum is 10. li x-\- y = 5 is taken as the other equation of a system, one can see im- mediately that tlie two equations have no set of roots in common, since the sum of two numbers cannot be 10 and 5 at the same time. A system of equations which has a common set of roots is called a simultaneous system. A system of equations which does not have a common set of roots is called inconsistent or incompatible. LINEAE SYSTEMS 7T The attempt to solve an incompatible system results in getting rid not only of one but of both unknowns and leads to a statement in the form of an equation which is false. Consider x -\- y = 10, (1) x + y = 5. (2) (1) - (2), = 5, which is false. If, on the other hand, the equation x-\-i/ = 10 is taken for one equation of a system and 2a: + 2?/=20 for the other, it appears that any set of numbers which satisfies one equation satisfies also the other, since if the sum of two numbers is 10, the sum of twice those numbers is 20, and any one of the countless sets of roots of one equa- tion is a set of roots of the other. In fact, the second equation may be obtained from the first by multiplying each member by 2. If one equation of a system can be obtained from one or more of the other equations of the system by applica- tion of one or more of the axioms, it is called a derived or dependent equation. If it cannot be so obtained, it is called independent. Thus equations (1) and (2) in the example on page 75 are independent, while equation (5) is derived from them. An attempt to eliminate one unknown from a system of two equations in two unknowns which are not inde- pendent results in gettmg rid not only of both unknowns but of the constant terms as well, so that only the identity = remains. Thus x + y = 10, (1) 2x + 2y = 20. (2) (1).2, 2a; + 2^ = 20. (8) (2) -(3), = 0. 78 SECOND COURSE IN ALGEBRA ORAL EXERCISES Which of the following systems are incompatible, which are simultaneous, and which are dependent ? x + y = 2, ^x-22j=4:, 'x-\-y = l. *5ic — 102/ = 4. 3x + 32/ = 6, ?>x-y = 2, ^' x + y = 2. • 9^^-32/ = 6. 2 a; + 2/ = 4, 3^ -2/ = 2, "*• 4a^ + 23/ = l. * 9a: + 32/ = 6. 0^ + 2/ = 3, 2t/-^ + 6 = 0, *• x-2/ = l. • 3a; -67/ -21 = 0. 2a;4-32/ = 4, 2a3 = 72/-5, 4a; + 62/ = 8. * 8a; = 282/ 4- 3. EXERCISES Solve the following systems : 2a; , _ 11 1 3 +^= ^' 6. "^ ^ 6 2 3_4 5a; -3^/ = 105. 3r 7 s a; 2/~3* 4 2 12' 1 1 r+8=-2s. Hint : Solve first for - and - X y 32/ + 1 ^ + 22 = 3 4 12 > 2r-h4s 38 «-2y = l. 2r-s ~ 3 ' ^ ^ - .5x + .73, = |, 7* S .8a;-.22/ = 3f. w -h 2 71 2m — 71 5 2m + Sn-2 4 __ ;:::: — , 5 10 2 m+^+6 3 w, + 71 m — n 71 4 7 '2* LINEAR SYSTEMS 79 x-^ y ' 3 -11 ' X — ^y _2 5 5 5 '~n' 2x + y = l. 2.^h-\-l-.lk ^ X y 6j/ 2x k-l(^ + h ' 5 "^2 2 "^ 5 .8A-2.2"^35-5.5A: ' ^^- 4 2 2h-Sk = 3, X 1 4-2y 4 n. ^-^^ 5^±^4.2= ^^ ^— = 39. ^^-. + 7 17 + . _Q 3 « 5-« Solve for ic and y : jg 5a; + 4?/ = 10a + 4, jc — 2 ay = 0. 7x + 5y = 21c, 2 acc — 3^*^ = 7, * 5aic + 7Z>y = 3. Zx y 2-2 = 6, 19. " " ^ + 1_ = 13 2a 3c Sax-6by = 5c. 20. a Sa X y Za a 1 ^ y~2' 21. 4a / a 22. X y a -\-h ah ab a^-Jy" X — y = ; ^ ah oo 2 ace ~ 4 6?/ = 7 c, ^0 SECOND COURSE IN ALGEBRA ^^ ax + hy = c, ^^ ax + hy = c, kax + khy — ck. ' dx -\-fy = ff. 26. Show that if af—hd = 0, the equations in Exercise 25 are inconsistent, unless they form a dependent system. 27. Solve the system of Exercise 17 for a and h in terms of X and y. 28. Solve the system of Exercise 23 for a and b in terms of Xy y, and c. 49. Equations in several unknowns. We have already seen that the equation x-{-y = 10 is satisfied by an un- limited number of sets of roots, since there is an infinity of pairs of numbers whose sum is 10. An equation or a system of equations which is satisfied by an infinite number of sets of roots is said to be indeterminate. If a simultaneous system is satisfied by only a limited number of set.* of roots, it is said to be determinate. The system x + ?/ = 10, a; — y = 2, is determinate and has the set of roots (6, 4). The system 2 x + 2 y = 20, x + y = 10, is indeterminate. When we consider systems of equations in three un- knowns, the question arises whether two such equations form a determinate system. For example, the equation x + y-^z=\0 (1) is satisfied by an infinite number of sets of roots. If we consider a system consisting of (1) and ^-(y + ^)=2, (2) it appears from inspection that the system is satisfied if x = Q and ?/ -f 2 = 4. But the equation y -\-z = ^ is satis- fied by an infinite number of sets of roots. Hence equar tions (1) and (2) form an indeterminate system. LINEAR SYSTEMS 81 If, however, we adjoin a third equation to the system, as y-z=% (3) it becomes determinate, since ?/ + ^ = 4 and y — z=2 are satisfied only by the set of numbers 3, 1. It is usually true that three equations in three unknowns form a deter- minate system. In general, when the number of unknowns in a system of linear equations exceeds the number of equations, the system is indeterminate. If the number of equations equals the number of unknowns, the system is usually determinate and simultaneous. If the number of equations exceeds the number of unknowns, the system is usually inconsistent. There are many special cases wliich arise in the study of hnear systems in n unknowns, corresponding to those mentioned for two unknowns in section 48, but they become very complicated for larger values of n^ and. a thorough study of them is quite beyond the scope of this text. Note. It is not a little remarkable that the writings of the first great algebraist, Diophantos of Alexandria (about a.d. 275), are de- voted almost entirely to the solution of indeterminate equations ; that is, to finding the sets of related values which satisfy an equa- tion in two unknowns, or, perhaps, two equations in three unknowns. We know practically nothing of Diophantos himself, except the information contained in his epitaph, which reads as follows : ".Diophantos passed one sixth of his life in childhood, one twelfth in youth, one seventh more as a bachelor; five years after his mar- riage a son was born who died four years before his father, at half his father's age." From this statement the reader was supposed to be able to find at what age Diophantos died. As a mathemati- cian Diophantos stood a;lone, without any prominent forerunner or disciple, so far as we know. His solutions of the indetermi- nate equations were exceedingly skillful, but his methods were so obscure that his work had comparatively little influence upon later mathematicians. 82 SECOND COURSE IN ALGEBRA 50. Determinate systems. The method of obtaining the set of roots of a determinate system in three unknowns is illustrated in the EXAMPLE (x-\-62j-5z = 21, (1) Solve the system iSx — Sy-\-z=— 5, (2) [5x-7y-^2z = 4:. (3) Solution. First eUminate one unknown, say z, between (1) and (2) : X + 6 3/ - 5 2 = 21. (1) (2). 5, 15x-^0y + 5z=-2o. (4) (l) + (4), lQx-Uij=-4:. (5) Now eUminate z between (2) and (3) : (2) -2, Qx-lQy + 2z=-10. (6) 5x-7y -\-2z = ^. (3) (6)-(3), x-9y=-U. (7) The equations (5) and (7) contain the same two unknowns x and y. lQx-Uy=-^. (5) (7) -16, 16 a: -144?/ =-224. (8) (5) -(8), 110 y = 220. y = 2. Substituting in (7), a; = 4. Substituting both these values in (1), 4 + 12 - 5 ^ = 21. Whence z= — l. Check. Substituting 4 for x, 2 for y, and — 1 for 2 in (1), (2), and (3) respectively, 4 + 6-2-5(-l) = 21, or 21 = 21. 3.4-8.2 + (-l)=-5, or -5=-5. 5.4-7.2 + 2(-l) = 4, or 4 = 4. LINEAR SYSTEMS 83 For the solution of a simultaneous system of equations in three unknowns we have the Rule. From an inspection of the coefficients decide which unknown is most easily eliminated. Using any two equations^ eliminate that unknown. With one of the equations just u^ed and the third equa- tion again eliminate the same unknown. The last two operations give two equations in the same two unknowns. Solve these equations. Substitute in the simplest of the original equations the two values found, and solve for the third unknown. Check. Substitute the values found in the original equations and simplify results. . A system of four independent equations in four un- knowns may be solved as follows : Use the first and second equation, then the first and third, and lastly the first and fourth, and eliminate the same unknown each time. This gives a system of three equations in the same three unknowns, which can be solved by the rule given above. EXERCISES Solve for x, y, and z and check the results : x + ^y — ^z = 2, x + y + z = l, I. 2x — y — z = 1, ^. X -{- y — z = 2, Sx-{- 5y-7z=-10. x-?/ + ^ = 3. 2x+Sy-i-4.-z=-14:, . x + 2y -\- z = 1, 2.x — y + 3z = 0, 5.2x-^y — z = 0, 5x-\-2y + z = 14:. x-\-2y -{-z = 0. x+'2y-\-z = -l, 2x-y-^5z=0, 3. 2x-y-}-z=-20, 6. Sx -^7 y -\- z = SS, — X — y — 5 z = IS. X — 6y — z = 7. 84 SECOND COUESE IN ALGEBEA ' X y z X 1^ y x-\-y £_2. 7/ 4- ^ X ^ ""4 2 "12' Sx + 2y=12-3z. x-6y-\-Sz = ^ 4x-32/ = «, 8. s = £c -4- 2/, 2x = 32/ + l. 2x + 2/ = S + «, 9. 05-2^ = 6, 32/ + 2« = aj. ^_2y = 10, 10. 3^4-4^=-!, 5 X — ^ = 18. ^ = 3^ + 2, 11. y = .x-7i, ^ = 6i/-l. 4x-22/ = 0, 12 . 6^-82/=-2, x^z = \\. 18. A« J[-hy — lz = 2 hk 19. A^i/ — /^ic + ^^ = 2 A;^, hx — hy + 1^ = 2 hi. 20. 2a; + 2/ + ^ + ^ = ^» jC _ 1/ _ ;2 + 2 ?^^ = 4, X + 2 7/ — ^ — 'i^ = 0, oj —'2/ + 2 ^ — i^; = 1. i. ^3-- + ^ = t"' ^^•4x-22/ + ^ — =-9 ^ + " = ^^' 2x-32/-2. + ^^ = 35c4.2i/ = 6a-2^, 14. x-5« + 62/ = 2a-ll^ x + 2/-« = ^j 6a; -82/ = 12 a + 8^. ^ _|- 2/ - w; = 2, ic-?/ + « + w; = S) ax + hy = 0, 2sc-3y-« + i^ = 7. 15. cx-6« = 2 6c, bx -\- az — cy = b\ ^ ^y -\- z = l, .3x + .22/ + .4. = 1.9, 23. ^"^"""^"i' 16. .02x = .l-.01//-.02^, x-z-w:^-5, a; + 2/ + « = 6. z — « + !(; = 0. LINEAR SYSTEMS . 85 PROBLEMS Express the conditions of the following problems by means of simultaneous systems, solve, and check the results : 1. The sum of two numbers is 109 and their difference is 49. Find the numbers. 2. A workman is hired for 30 days. He is paid $3.50 per day and board, but is credited with 80 cents for each day that he does not receive board. At the end of the 30 days he re- ceives $113. How many days did he receive board ? 3. Thirty -nine tons of material are to be moved by motor trucks and drays. It is found that the work can be done in a given time either by 10 trucks and 6 drays, or by 8 trucks and 10 drays. What is the capacity of a truck and of a dray ? 4. Two men travel from New York to the same station by rail. It costs one of them three times as much for excess baggage as it costs the other. One pays $7.40 in all, the other $10.60. How much does each pay for his ticket? 5. A man and a boy can do in 18 days a piece of work which 5 men and 9 boys can do in 3 days. In how many days can 1 man do the work ? 1 boy ? 6. A and B together can do a piece of work in 5 days. If they work together 3 days and A can then finish the job alone in 4 days more, how many days does each require alone ? 7. Two sums are put at interest at 5% and 6% respectively. The annual income from both together is $100. If the first sum had yielded 1% more and the second 1% less, the annual income would have decreased by $2. Find each sum. 8. If ax -{-by = 2 is satisfied by x = 2 and y = S and also hj X = 6 and y = 5, what values must a and b have ? 9. li 2x-\-b't/ = c is satisfied when x = l and y = — l and also when x = 5 and y = 4=, what values must b and c have ? 86 SECOND COUESE IN ALGEBRA 10. A sum of $4000 is invested, a part in 5% bonds at 90, and the remainder in 6% bonds at 110. If the total annual income is |220, find the sum invested at each rate. 11. The purity of gold is measured in carats, 18 carats meaning that 18 parts out of 24 are pure gold. A goldsmith has 20 ounces of pure gold which he wishes to use in making 16-carat and 10-carat alloys. How much pure gold can he use for each alloy if he makes 39 ounces in all ? 12. A bag weighing 18 ounces contains two sizes of steel balls — ounce balls and |^-ounce balls. There are 23 balls in all. Find the number of balls of each size. 13. If the length of a rectangle is decreased by 7 feet and the breadth is increased by 8 feet, the area is unchanged. If the length is increased by 14 feet and the breadth is decreased by 4 feet, the area is also unchanged. Find the dimensions of the rectangle. 14. A man has |4.50 in dimes and quarters. If he has 36 coins in all, how many has he of each ? 15. A man has $6.00 in quarters, dimes, and nickels. He has as many quarters as he has dimes, and three times as many nickels as dimes. How many of each has he ? 16. A is half as old as B. Seven years ago A was one third as old as B. How old is each now ? 17. A man can walk 4 miles per hour. He reaches a point 20 miles from his starting point in three hours, having been taken part of the way by a stage traveling 12 miles per hour. How far did the stage carry him ? 18. A man and his two sons can do a piece of work in t of a day. The two boys together can do it in 1|- days and one of them can do it in 1 day less than the other. What portion of the work does each do when they work together? LINEAR SYSTEMS 87 19. Two automobiles 25 miles apart travel toward each other and meet in 1 hour. If they had both traveled in the same direction, the faster would have overtaken the slower in 5 hours. Find the rate of each. 20. An aeroplane travels a certain distance in 3 hours. If the distance had been half again as great, the aeroplane would have been forced to travel 50 miles per hour faster in order to cover it in the same time. Find the distance and the speed of the aeroplane. 21. One angle of a triangle is twice another, and their sum equals the third. Find the number of degrees in each angle of the triangle. 22. The sum of three numbers is 108. The sum of one third the first, one fourth the second, and one sixth the third is 25. Three times the first added to four times the second and six times the third is 504. Find the numbers. . 23. The sum of three numbers is 217. The quotient of the first by the second is 5, which is also the quotient of the second by the third. Find the numbers. 24. If the tens' and units' digits of a three-digit number be interchanged, the resulting number is 27 less than the given number. If the same interchange is made with the tens' and hundreds' digits, the resulting number is 180 less than the given number. The sum of the digits is 14. Find the number. 25. In 1 hour a tank which has three intake pipes is filled seven-eighths full by all three together. The tank is filled in 1^ hours if the first and second pipes are open, and in 2 hours and 40 minutes if the second and third pipes are open. Find the time in hours required by each pipe to fill the tank. 26. The sum of two sides which meet at one of the vertices of a quadrilateral is 20 feet. The sum of the two which meet at the next vertex is 27 feet. The sums of the two pairs of 88 SECOND COUESE IN ALGEBRA opposite, sides are 23 feet and 29 feet respectively. Find each side. (Two solutions.) 27. Two chairs cost h dollars. The first cost m cents more than the second. Find the cost of each in cents. 28. Find two numbers whose sum is a and whose difference is h. 29. Find two numbers whose sum \s> a-\-h and whose differ- ence \s, a — h. ' 30. Two relays of messengers carry a message k miles. The first relay travels c miles further than the second. How far does each go? 31. A man has a dollars and b cents in dimes and quarters. If he has c coins in all, how many of each kind has he ? ^ 32. A man has a dollars in quarters and nickels, with b more quarters than nickels. How many of each has he ? 33. A and B together can do a piece of work in m days. B works G times as fast as A. How many days does each require to do the work alone ? 34. A man rows m miles downstream in t hours and returns in a houi's. Find his rate in still water and the rate of the river. 35. Two contestants run over a 440-yard course. The first wins by 4 seconds when given a start of 200 feet. They finish together when the first is given a handicap of 40 yards. Find the rate of each in feet per second. 36. A train leaves M two hours late and runs from M to P at 60 % more than its usual rate, arriving on time. If it had run from M to P at 25 miles per hour, it would have been 48 min- utes late. Find the usual rate and the distance from M to P. 37. A train leaves M 30 minutes late. It then runs to N at a rate 20% greater than its usual rate, arriving 6 minutes late. Had it run 15 miles of the distance from ilf to iV at the usual LINEAE SYSTEMS 89 rate and the rest of the trip at the increased rate, it would have been 12 minutes late. Find the distance from M to N and the usual rate of the train. 38. It is desired to have a 10-gallon mixture of 45% alcohol. Two mixtures, one of 95% alcohol and another of 15% alcohol, are to be used. How many gallons of each will be required to make the desired mixture ? Hint. Let x and y = the number of gallons of 95% and 15% alcohol respectively. Then —. '- — - = — , and x 4- y = 10. ^ 10 100 ^ 39. A chemist has the same acid in two strengths. Eight liters of one mixed with 12 liters of the other gives a mixture 84% pure, and 3 liters of the first mixed with 2 liters of the second gives a mixture 86% pure. Find the per cent of purity of each acid. 40. When weighed in water the crown of Hiero of Syracuse, which was part gold and part silver, and which weighed 20 pounds in air, lost 1^ pounds. How much gold and how much silver did it contain? Hint. When weighed in water 19^ pounds of gold and 10^ pounds of silver each lose 1 pound. 41. Find the positive integers which satisfy the equation 5x + 22/ = 42. 4.2 — 9 y Solution. X = -• 5 42 — 2 y If X is to be a positive integer, -^ — - must be integral ; that o is, 42 — 2 y must be a positive integral multiple of 5. Hence y can only have the values 1, 6, 11, and 16. The corresponding values of X are 8, 6, 4, and 2. The various related sets of integTal values of the unknowns which satisfy an equation may be effectively represented to the eye by the graph of the equation. Since the equation 5 re + 2 y = 42 has the integral sets of roots (8, 1), (6, 6), etc., the line of which this is the equation passes through the points whose coordinates are these RE 90 SECOND COURSE IN ALGEBEA sets of integers. If the line does not enter the first quadrant, we can see at a glance that the corresponding equation has no positive integral sets of roots. 42. Solve in positive integers 7x -|- 2y = 36, and illustrate the result graphically. 43. In hov^ many ways can a debt of ^73 be paid with 5-dollar and 2-dollar bills ? Illustrate the result graphically. 44. A man buys calves at |6 each and pigs at |4 each, spending |72. How many of each did he* buy? 45. In liow many ways can |1.75 be paid in quarters and nickels ? 46. A farmer sells some calves at |6 each, pigs at $3 each, and lambs at |4 each, receiving for all |126. In how many ways, could he haye sold 32 animals at these prices for the same sum ? Determine the number of animals in the various groups. Hint. Eorm two equations in three unknowns. Then eliminate one of the unknowns. 47. In how many ways can a sum of $2.40 be made up with nickels, dimes, and quarters, on the condition that the number of nickels used shall equal the number of quarters and dimes together ? Determine the various groups. CHAPTER VI- EXPONENTS 51. Fundamental laws of exponents. The four laws of exponents used in the preceding chapters are: I. Law of Multiplication, If a and b are positive integers, we have x'^ = X ' X ' X ' X ' ' • to a factors, and x^ = X- ' X ' X • ' ' to h factors. Hence x^ * x^ =(x • x * x - - -to a factors) x (^x - x > x - - » to b factors) = X'X'X'-' to a-^b factors _ ^a+6 \yj ^i^Q definition of an exponent. II. Law of Division, Xa-^ X^ = X°-^. Again, if a and b are positive integers, we have . x^ X' X ' X • ' ' to a factors X^^X^=—: — = —T x'^ X' X ' X ' • ' to factors If b is less than a, the b factors of* the denominator may be canceled with b factors of the numerator, leaving a — b factors in the numerator. x^ Hence — = x^~^. x^ • 91 92 SECOND COUESE IN ALGEBRA If h is greater than a, III. Law of Involution, or raising to a power, As before, if a and h are positive integers, we have (x^y = x^ ' x^ ' x^ ' ' 'to b factors _ ^a+a + a+ ' ' ' . . . to 6 terms of the exponent IV. Law of Evolution, or extraction of roots, Law I may be stated more completely thus : x"" 'X^ 'X' " . = 2:^+^ + «+--'. Law Ill includes the more general forms (a) Qx^yhy — ^ac^ftc^ (^) ((xf'yy ' . . = x^^''-'. When Laws I, II, and III were used in previous work in multiplication and division, we always assumed that a and b were positive integers and, in Law II, that a was greater than b. In the work on radicals (see "First Course in Algebra," Revised Edition, pp. 251-252) the meaning of an exponent was extended so as to include fractional exponents, as defined by Law IV. Though Laws I-IV have thus far been restricted to positive integers and frac- tions, they hold, nevertheless, for any rational values of a and b. This fact will be assumed without proof. We shall now explain the meaning which, according to these laws, must be given to a zero or to a negative exponent. EXPONENTS 93 52. Meaning of zero as an exponent. From Law II, But a^^a^ = - = l. Therefore ;c** = 1. More generally, 2f^ -h sf' = af-'^ = x^, and, as before, aP =1. That is, any number (except zero) whose exponent is zero is equal to 1. Hence 4^ = (f )^ = (— 6)^, for each equals 1. Again, if x is not zero, (5 xy = 1, and if a; is not 1, (x^ — 2 x + 1)*^ = 1. 53. Meaning of a negative exponent. From Law II, a' d> Obviously, a^ _j_ ^5 _ Therefore ar'^ is another way of writing — • Then 4-3^1_ ^ .' 43 64 Also a" t z= — z= — ^=^ . 1 In general terms, x~^ = — Consequently = — = Jf°. Similarly, we obtain the more general results hx~ " = — and = hxf^. 94 SECOND COURSE IN ALGEBRA Therefore, Any factor of the numerator of a fraction may he omitted from the numerator and written as a factor of the denominator, and vice versa, if the sign of the exponent of the factor he changed. It follows that any expression involving negative exponents may be written as an expression involving only positive exponents. That is to say, negative exponents are not a mathematical necessity but merely a convenience. The extension of the laws of exponents which brings with it the zero exponent and the negative exponent is an illustration of what is called the Law of Permanence of Form. It is to be understood that the part of the rule for muItipHcation (p. 5) and of the rule for division (p. 7), which determines the exponents in the product or m the quotient, applies to all numbers, whether positive or nega- tive, integral, fractional, or literal. Hence those rules need not be restated here. ORAL EXERCISES Fractional, Negative, and Zero Exponents Simplify : 1. xKcci 10. x^-^x-\ 19. {a^-a^).aK 2. cc^^l^ 11. x^ -x-^. 20. {al^a^j^l)a\. 3. x^.xl 12. x^ 'X-^. 21. ^8a-l ^ x'^-'2a 4. x^.xi. 13. x:'-^x-\ 22. x« - x\ 5. X" -T- X^. 14. x^-i-x-\ 23. x2«-i.xi 6. x' . x\ 15. {bxy-nx'. 24. Vx 'X. x^.x. 16. lx''x\ 25. ^X -h x\ 7. 8. x" . x°. 17. x^ 'X^. 26. vs.^i. 9. aj« 'X. 18. ai . a^ ' a^. 27. Vi*-- Vi. EXPONENTS 95 28. ^x . Vx\ 33. {x'^y . x\ 37^ ^2 _^ J_ 29. V^. V^. 34. (x^a-x-''-. X 1 30. a^^-^-x^-^. 35. a^^^^- 38. ^ ^ : ^_, x~^ 31. x2«-^--cc^-2«. ^ • 39. ici-^*-T-x-2«. 32. (x2)3 . x\ ^^' ^^" ^ ^ ' 40. ax-^ -^ a-^x\ Kead the following with positive exponents and simplify the results : 41. 42. 2a-\ 49. 4c« xy-^ 54. 10- ^a bc^ 43. 44. Sab--\ lxhj-\ 50. y-' 55. 45. 46. x~'^y~^z. 51. 5-\aby 10- •■^^»'^ 56. 4-V-V- 47. 3 52. ■3a^b-^c 57. 2 s-' 48. 4:X 53. Vlxhj-' 2yx-^ 58. 2e-: ^ Arrange terms so that the exponents of one letter occur in the descending order : 59. a^-\-a-'-ea-{-Sa''. 60. a^-]-a~^^~-a^ -i-a + Sa'. 61. a + a^ + 2 + a-^ i-a-\ 62. a^ + a~^ + a~^ -\- a -j- S. 63. a^ + -^ + « + - + 6^'. 67. Arrange the polynomials in Exercises 17 and 21 on page 97 in descending order with respect to the letter a, and Exercises 22 (p. 97) and 28 (p. 98) with respect to x. 64. a^ a 65. ••4.+^& 66. --S-?-¥-^ 96 SECOND COURSE IN ALGEBRA EXERCISES IN MULTIPLICATION Perform the indicated multiplication and simplify : 1. {x + x^-^x-^)2xi. 13. (e^ + e-y. 2. \a^x-[-ax^ — a^x ^)ax^. ^ ^ 3. (x^ — a^)x^a^. ^ /I 1\ 4. (x^-5ax + 6a^)Jx-k ^^' ^^'" +.^"H^ " ^)* 5. (x^ + y^jx^yK ^ ^ ^ /I IX / 1 tx 18. (3a-^ + 2a*)'. 19. (x-i-3x-2ir-2)2. 7. (cc^ + y'^) (cc^ — 2/^). " / 1 ^ 1 ^ 3\2 ^ ^^ -^ ^ 20. U-^4-2a;^-3x^). 8. (a-2 + 3)(a-2-5). 9. (.--3.«)(.^-2.). ^1- (-* + -^+l)U^--^+l)- 10. (a-i - a)l ^ o , 11. (6i«-2«-y. 23. {x^-(^){x^-\-a^-J^aS). 12. (a- 1 _ 2 a + 3 a- 2)2. 24. (a^ - a^cc + 4 ic^) (a^ + 2 ic). 25. (x^ + 2 ?/^) (x3- - 2 x^?/^ + 4 2/^). 26. (a + a^^>^ + ^)U + ^^ - f^^^'^). 27. (a* - /^^-5V^). Hint. This is best written (ai — 6 a^) (a^ — 6 a^). 30. (5 V^ + -v^^ - a -v/^'f. 31. W-^^n^-v;^). 5. EXPONENTS 97 EXERCISES m DIVISION Perform the indicated division and simplify : 1. x*^x\ 6. (x«- 2x2^-1 + 3 ic^«-2)-f-x2«-i. 2.x^^x^. 7. (6 67.3+*^-9a^™-2 + 12a2-n^)_^3^»-2^ 3, x^ ^ x^. 8. (x-y)-^ {x^ — y^). 4. axi-i-Jx^. 9. (x + y)^{x^ + yi). ax-a^x\ 10. (a^-8 2/)-(aj*-2 2/^). a^a:^ 11. (16a^^-A096f)-i-{2x^-^Syi), 12. (at + Z,)-^(V^-^^). 13. (a^ + c»6-i + ^*-2)^(^^ _ Jf^-h + ^-i). 14. (e2^4-e-2^ + 2)--(e-^ + e^). 16. (6 + e-^^ + e^o; _ 4 ^-20; _^ 4 e^^^)-- (e^ - e"^). 17. (a + 2 Z^ + 2 ah^ + a*^»*) - («^* + 2 5^). 18. (m' - 7m-2 + m"* + 7 m^ + 8) - (5 - 7/^-^ + m^). 19. (9a;^*^-^-.T3«-2_^2cc2«-i-)_j_^2ic'^-i + 3ic2"-2). 20. (I6 £c - 8 y~^ -\-x~^y-2x~ V) ^ (^~ ^y~^ -S xhj- 1). 21. (40 ah - 2^a-h^ - 16 aH^) - (- 4 a'^V^ + 5 a^h-^) 22. (2:r-2« - 28.T-« + 33 + lla^^^ + 38a:« + cc^^) -^(4 + a;«-2cc-«). Divide : 23. a^ - ah^ + a^^ - h^ by a^ - 5^ 24. 3£c-i° + x«-4a^-«by 2ir-2 + a^2_^3a^-« 98 SECOND COUESE IN ALGEBRA 25. x^ — y^ by \_{x^ — y^)-i-{x^^ + 2/^)]. 26. 9m + 4m-i-13 by 3w^-5 + 2m~i 27. ^2a ^ 4^-2a _ 29 by a!« _ 2cc-« - 5. 28. 9x2« + 25£c-*« - 19x-« by 5a:-2« + Sic*^ - Ix 29 (^ + ?I^ ?9- ' 8 ^ 64 ^ [/^^ -f ^"j (64 a; - 96 x^y^ + 144 y)l. Note. To us, who use the notation of exponents every day, it seems so simple and natural a method of expressing the product of several factors that it is difficult to understand why such a long time was necessary to develop it. But here, as in many other in- stances, it required a great man to discover what to us seems the most obvious relation. The man who brought the notation of exponents to its modern form was John Wallis (1616-1703), an Englishman. Though the idea of using negative and fractional exponents had occurred to writers before Wallis, it was he who showed their natu- ralness, and who introduced them permanently. He also was the first to use the ordinary sign oo to denote infinity. MISCELLANEOUS EXERCISES ON EXPONENTS Find the numerical value of the following : 1. 3-^ 10. 1- 18. 16"^. 2. 4-^ 3° 19. 8-1 3. 2-* • 3^ 20. 16-1 4. 2--. 3-". 12. 5.2°- .(5 . 2)^ 21. 25 H 5. 7 . T'' . 0. ,o 4-^3- -2 22. (- 64)-*. 6. a)-^ 13. ,_. ~* 23. (- 32)i 7. (|)-^.4o. 14. 32-1 24. (32)-*. 8. (i)-^ •(¥)-'• 15. 0^ . h\ 25. (-125)- i 2 16. 4"i 26. ■v^27-^ *-3-- 17. 8-^. 27. -^8-^. EXPONENTS 99 ay- 22 23 28. {^/^sy. 30. (I)" I ^^^'^- ^r-7. — ::r-.= . ' . ^ etc. 31. (.04)1 32. (.027)" i 33. (.064)- 1 34. (.00032)1 Write with positive exponents and simplify the results 36. 2-1 2- -2 2-3 TTttx rT. 2-1 2-^-2-3 37. 3- -2 2-2 3- ■1-2-1 38. 2- ■1 + 3-1 2- -3_^3-8 QO 3^ -3 9-3 40. Hint. 44. 2 5-2 J. _ 1_ -2 46. etc. 41. 42. 43. + ^'- 45. a-%- 7.-4 -hb- 47. a-^-b-' a-^ + b- -1 a a-'~ - b- -2 Bse" s-- + e- -2 18 '~' + ^»-l *®- a-3 + Z»-3 -27-1 -3-^ "V^rite without a denominator and simplify the results : 50. 51. 52. 53. 2xy b' ' 4 6-c-^ 2-is-' 54. 55. 56. 57. 12 a%^ 4.xf ' Ix-hf 2-y * c{x-yf 58 59. ^(^-?/)-^ ^6'iC (iC — y) 42m-«?i2"* 60. 61. r~V(s — r)* 100 SECOND COURSE IN ALGEBRA Simplify : 62. (28/. 71. (x-^yK 80. (a8)2^ • (a2)8^. 63. (28)-2. 72. (Sx-'^y. 81. (a^+i)^ • (ai-^)^ 64. (2-')-\ 73. (5(^^)-^ 82. (««)^ + ^-(a'')^-^ 65. [(f)-^^ 74. (0-^)^. 83. (a%y • (W)^ 66. W- 75. (5o.2«.38)i 8^. (.^-.->3. 67. U^j. 76. (25a^^«)-^ 68. {x^)\ 77. (2a?y . 8-4^. ^2^8\2a? yn C2 71 88. ^^-;:2_ = 2 86. 2'^. 42 = 2?. 87. 2" .4^ = 2?. 69. (xy\ 78. (6i2^^)«(a + Z.). 2 88 — 70. (x-«)-2. 79. (a^)^'^ . a^^. * 43^^ 89. 2'' .4^+^^ 2'' = 2?. 4.TC + 1 J^^+1 Oft : _J_ 5 — 9 2"('4'''~^V* * /4» + iyi-i "^ Solve for w: ^^ -95. 81.27^^=(9")2. 92. 3«.9^ = 812. g,„ 93. 9'* . 3« = 27^\ ^^- (^^'')'' == (l25y^' 94. 2^'^"^^ • 4" + '^ = 8^". 97. 2^'"''^^ • 4^^+^ = ('8"')". Solve for x : 98. a-^ = 8. 103. x-^ = 4. 108. ^x~^ = 2. 99. x-^ = 5. 104. x^ = 2. 109. (cc"*)"* = 49. 100. x~^=25e. 105. £c^ = 16. 110. (aici)"* = 27. 101. x"i = 2. 106. ic"^ = 32. V^ _ \/25 102. x^ = 4. 107. x^ = 343. * ^x^ -^ CHAPTER VII SQUARE ROOT 54. Square root. The square root of any number is one of the two equal factors whose product is the number. From the law of signs m multiplication it follows that Every positive number or algebraic expression has two square roots which have the same absolute value but opposite signs, 55. Square root of a monomial. For extracting the square roots of any monomial we have the Rule. Write the square root of the numerical coefficient preceded by the sign ± and followed by all the letters of the monomial^ giving to each letter an exponent equal to one half its exponent in the monomial. A rule similar to the preceding one holds, for the fourth root, the sixth root, and other even roots. 56. Cube root. The cube root of any number is one of the three equa,l factors whose product is the number. For extracting the cube root of a monomial we have the Rule. Write the cube root of the numerical coefficient fol- lowed by all the letters of the monomial^ giving to each letter an exponent equal to one third of its exponent in the monomial. A rule similar to the preceding one holds for the fifth root, the seventh root, and other odd roots. 57. Principal root. For a given index the principal root of a number is its one real root if it has but one, or its positive real root if it has two real roots of that index. 101 102 sircc»:^rD course in algebra Then the principal square root of 9 is + 3 ; the principal fourth root of 16 is + 2, not — 2. The square root of — 4 or — 9 is not real ; such numbers have no principal square root. The principal cube root of 8 is 2, of — 27 is — 3. The principal fifth root of 32 is +2, of - 32 is - 2. Every number has more than one root of given odd index. The number 8, for example, has two other cube roots besides its principal cube root 2. What they are and how they are obtained will be made clear in the chapter on Imaginaries, where the consideration of the square roots of negative numbers will also be taken up. Unless otherwise specified, only the principal odd root of a number will be considered. ORAL EXERCISES Find the principal square root of the following : 1. 16. 2. 25. 3. 4.a\ 5. 36 a«. 6. 49 t\ 7. 64 1\^ 9. 10. h xJ" ^' 13. x-\ 15. 9a;V*. 16. 4x\ 17. x^. 4. 9 a*. 8. 81a?«. 11. 18. x^. Find the principal cube root of the following 19. 8. 24. 27 rr*. 29. -125a«. 34. 343 a«. 20. 27. 25. 64 x\ 30. - 27 a\ 35. -512a«. 21. 64. 26. -8. 31. -64a\ 36. - 27 a^. 22. 8 a". 27. -27. 32. -%a\ 37. -^x-\ 23. 8a«. 28. - 64. 33. 216 a\ 38. -27ic-«. Find the principal fourth root of the following : 39. 16. 42. a\ 45. 16 x\ 48. x^, 40. 81. 43. x\ 46. 625 a^ 49. 16 a;-*. 41. 266. 44. x"^. 47. 16x-\ 50. 81 x"^. SQUARE BOOT 103 Give the principal root and one other root for the following : 51. The fourth root of 81; of w" \ of x~^. 52. The sixth root of 64; of a«; of x-\ 53. What is the sign of the principal odd-Yoo^, of a positive number ? The principal odd root of a negative number ? 54. What is the sign of the principal even root of a positive number ? 55. State the rule for extracting the fourth root of a mono- mial. 56. State the rule for extracting the fifth root of a monomial. 57. Can one obtain the fifth root of a monomial by extract- ing the square root of its cube root ? by extracting the cube root of its square root ? Explain. 58. Square root of polynomials. Extractmg the square root of a number is essentially an undoing of the work of multiplication. The square of any polynomial* may be represented by Qi^t^uy' = h^ + 2]it + t'^^-2hu + 2 tu 4- w2 A little study of this last form and a comparison with the example which follows will make clear the reason for each step of the process. EXAMPLE J^ First trial divisor, 2 h + 2]it-\-t^ =(2A+ 0^ First complete divisor, 2 A + ^ Second trial divisor, 2^ + 2^ |+2AM + 2m+u2 = (2^ + 2«+w)u Second complete divisor, 2 ^ + 2 « + m \+2Tiu-\-2tu-\-u'^ = {2h-\-2t-\- u)u Therefore th^e required square roots are ± (h -{■ t ■\- u). 104 SECOND COURSE IN ALGEBRA EXERCISES Extract the square root of the following : 1. x^-{-4.x^-2x^-12x-\-9. 2. a^-10d^-4.a^-\-25a^-{-20a-\-4.. 3. x^ + 4lx^ + 16-4.x^-{-Sx^-16x. 4. t^ ^ 4:t^ -{- 9 + 4.t' - 6 1^ - 12 t\ 5. 4 a* + 9 aH^ + t^ +*12 aH - 4 aH"" - 6 a^^- 6. 4 a« + 12 c^* - 7 - 24 c*"* + 16 «"«. 7. 49 c-« - 28 c-4 + 74 c-^ -20 + 25 e". 8. 9x2 + 4a^ + l + 12ir*4-6x + 4£A 9. 9i»*-6a^^ + a;^-66x^ + 22ic2_^121:r. 10. 25x^-10x''-\-90xi + x-lSxi-j-Slxk 11. 16 772.-^- 12. -^ + 7 + /-7, + -* + 104 m - 26 m* + 169 m« + m-\ 3 2/ iC Solution. Arranging terms in descending powers of x and apply- ing the method of page 103, we obtain the following : (t) 4x2 12 y^ y 4a:2 X 4:X^ _ + 3-^ r ff-)= 4x y i^ + 3 4a: 12 X y y "(t-)^ + 6 i£ + „_JL V 2 a; 2_^ + JC a: 4ar2 -2 a: 4 a:2 \?/ ^ 2a:/\ 2x1 Therefore the square roots are ±(— + 3 — ^j- SQUARE ROOT 105 13. x« + 6a;'^-^ + 2x + -• 15. ^ ^g -2a;+ 3 +^4- 4 9 o 17. ^a' - 2a' -\-7^a' - -^^a -{- ^K 18. 9c*-12c^ + 4c2-- + i + 6. 19. ^'-18 + -, + 9(^2-2 + 2a2. 9 a^ 20. ^ 1 + -^ + 4i. 21- I^ + ^ + ^'^' + 2^2 + 2 + 2^2. x^ a^ X a g^ 2a2 117 40 16 **'4c*^ ^25^* c^^Sac 5^2 ^^- 9c2"^ a^ "*"3c"^ g "^3' J[__3£ 9^ _a^_6^ J^ 4a* a« ■*" a"- "^25:^2 5 "^5a;' Find the first four terms in the square root of the following : 27. l + 2a;. 28. ^'^-a?. KE 106 SECOND COUKSE IN ALGEBRA 59. Square root of arithmetical numbers. The abbrevi- ated process of extracting the square roots of an arithmetical number is as follows : 7^32^67^89 1 2706.8 + 4 47 332 329 5406 54128 36789 32436 435300 433024 2276 Therefore the square roots of 7326789 are ±2706.8+. It follows from the preceding example that the work of extracting the positive square root of a number may be a never-ending process. The number 7,326,789 has no exact square root, and no matter how far the work is carried, there is no final digit. As the work stands we know that the required root lies between 2706.8 and 2706.9. The method just illustrated for extracting the positive square root of a number is the one eommonly used. For it we have the Rule. Begin at the decimal point and point off as many periods of two digits each as possible : to the left if the num- ber is an integer^ to the right if it is a decimal^' to both the left and the right if the number is part integral and part decimal. Find the greatest integer whose square is equal to or less than the left-hand period and write this integer for the first digit of the root. Square the first digit of the root^ subtract its square from the first period and annex the second period to the remainder. SQUARE KOOT 107 Double the part of the root already founds for a trial divisor, divide it into the remainder (omitting from the latter the right- hand digit^, and write the integral part of the quotient as the next digit of the root. Annex the root digit just found to the trial divisor to make the complete divisor, multiply the complete divisor hy this root digit, subtract the result from the dividend, and annex to the remainder the next period, thus making a new dividend. Double the part of the root already found, for a new trial divisor, and proceed as before until the desired number of digits of the root have been found. After extracting the square root of a number involving decimals, point off one decimal place in the root for every decimal period in the number. Check. If the root is exact, square it. The result should be the original number. If the root is inexact, square it and add to this result the remainder. The sum should be the original number. EXERCISES Find the positive square root of the following : 1. 6241. ' 5. 53.29. 9. 2,932,900. 2. 9216. 6. 1.4641. 10. 7,049,025. 3. 15,129. 7. 216.09. 11. 3.9601. 4. 56,169. 8. 988,036. 12. .0061504. Find to three decimal places the square root of the following: 13. 7. 15. .01235. 17. f. 19. ^K 21. 23^\. 14. .63. 16. .96384. 18. 4|. 20. |. 22. 89i. 23. Find the hypotenuse of a right triangle whose legs are 136 and 273 respectively. 108 SECOND COUESE IN ALGEBRA 24. A baseball diamond is a square 90 feet on each side. Find the distance from the home plate to second base, correct to .01 of a foot. 25. The hypotenuse of a right ttiangle is 207 feet, and one leg is 83 feet. Find the other leg, correct to .01 of a foot. 26. The hypotenuse and one leg of a right triangle are respectively 5471 and 4059. Find the other leg. 27. The side of an equilateral triangle is 17 inches. Find its altitude, correct to .1 of an inch. 28. Find the side of an equilateral triangle whose altitude is 15 inches, correct to .001 of an inch. Fact from Geometry. If a, ^, and c represent the sides of a triangle and s equals one half of a + & + c, the area of the triangle equals Vs(s — a){s — b) (s — c). 29. Find the area of a triangle whose sides are 12, 27, and 35 inches respectively, correct to .001 of a square foot. 30. By the method of Exercise 29 find to .01 of a square inch the area of a triangle each side of which is 22 inches. 31. Find to two decimals the sum of all of the diagonal lines that can be drawn on the faces of a cube whose edge is 11 inches. 32. Find to two decimals the radius of a circle whose area is 70 square feet. 33. Find to two decimals the diagonal of a room whose dimensions in feet are 15, 22, and "28. 34. Find to two decimals the diagonal of a cube whose edge is 8 feet. 35. A room is* 24 feet by 40 feet by 14 feet. What is the length of the shortest broken line from one lower corner to the diagonally opposite upper corner, the line to be on the walls or the floor, but not through the air ? CHAPTER VIII RADICALS 60. Radical. A radical is an indicated root of an algebraic or aritnmetical expression. Thus V9, V5, V2x, and Va;^ — a; — 12 are radicals. 61. Index. The small figure like the 3 in v^7 is called the index of the radical. The index determines the order of the radical and indicatefcj the root to be extracted. For square root the index is usually omitted. Thus V 2 and ViS mean v2 and vl8 respectively. 62. Radicand. The radicand is the number or expression under the radical sign. In V5 and \^2ax the respective radicands are 5 and 2 ax. 63. Fractional exponents. Radical expressions may be written in either of two ways : with radical signs or with fractional exponents. Thus v/5 and 5^ have the same meaning, and Va^ equals as:, etc. 64. Rational numbers. A rational number is a positive or a negative integer or any number which can be expressed as the quotient of two such iutegers. Thus 7, — 6, |, or 2.871 are rational numbers. 65. Irrational numbers. Any real number which is not rational is irrational. (See section 67.) 109 110 SECOKB COURSE IN ALGEBRA If a number under a radical sign is such that the root indicated cannot be exactly obtained, the radical represents an irrational number. For example, V? and Vi are irrational. Approximate values for these are given on page 274. A repeating decimal, though endless, is not an irrational number, ' for any repeating decimal can be expressed as a common fraction, and is therefore rational. Thus the repeating decimal .272727 • • • is not irrational, as it exactly equals ^^. Similarly, .2857142857142 • • • exactly equals f , etc. Note. The number f reduced to a decimal repeats the digits in groups of six each, and the mere fact that a decimal does so repeat is proof that it is a rational number. On the other hand, the num- ber TT is known to be irrational, and its value has been computed to 707 decimals, showing, of course, no repetition. The fact that it does not repeat in 700 digits is, however, no proof that tt is irrational, for decimals with even more than that many digits do repeat. For example, the fraction WgV^ equals the decimal 1.29 + , which repeats in groups of 7698 digits each. 66. Imaginary. An indicated square root of a negative number is called an imaginary number. Thus V— 4, V— 7, and V— 12 are imaginary numbers. And 3 + V^^ is also imaginary, though, as will be seen later (Chapter XII), such numbers are better called complex numbers. 67. Classification of numbers. All the numbers of alge- bra then may be placed in one or the other of two classes : real numbers and imaginary numbers. Real numbers, as we have seen, are of two kinds, rational numbers and irrational numbers. 68. Surd. A surd is an irrational number in which the radicand is rational. Thus Va, V^, etc., are surds. But v 2 + Vi} and Vtt are not surds. EADICALS 111 69. The algebraic sign of a radical. The square root of 4 is both + 2 and — 2. The symbol Vl, however, sig- nifies only + 2, the principal root (section 57). Similarly, ^81 is + 3 and V64 is + 2^ But -V9 is - 3, and -^l6 is — 2. The symbol ± V25 denotes both + 5 and — 5. Further, +^27- +3, -^27 = - 3, and -^327== +3. The foregoing remarks apply also to fractional expo- nents. Thus 4^ = + 2 only, and 81^ = + 3 only, etc. It should be noted that these statements really define the meaning of such symbols as V~? ^? v^ » etc. Such an un- derstanding as this avoids all the ambiguity which would arise if Vl6 meant both + 4 and — 4. The distinctions here made are especially needed in radical equations. ORAL EXERCISES Find the numerical value of the following : 1. -Vi. 7. -^81. 13. 8* 19. 16^ 2. -V9. / — 8. v^32. 14. 36t 20. (i)^. 3. Vl6. 9. -^-125. 15. 49^. 21. (f)*. 4. Vs. 10. --v^-32. 16. 8^. 22. (- 64)*- 5. ^-8. 11. ^64. 17. 16*. 23. (49)1 6. -^/i6. 12. -^625. 18. 27^. 24. (121)1 Bead in radical form : 25. x\ 30. Srx^. 35. a(a-l)^. 26. x^. 31. 4.ax^. 36. 2c(2x-3)i. 27. (at)\ 32. 2aV. 37. x^'k 28. (3t)i 33. 5ux^. 38. 2ixk 29. St^. 34. 4Ai 39. a 2 71 xH'^. 112 SECOND COURSE IN ALGEBRA Read with fractional exponents : 40. V^^ 43. V'^K 46. -^{a + x)\ 41. V^«. 44. 2V^. 47. -s/^^x-ay. 42. Va\ 45. ^v^. 48. -^x'^y - 1). 49. What are the two square roots of 36 ? 50. What are two fourth roots of 16 ? of 81 ? of 625 ? 51. What is the value of a/16 ? of "v^ ? of -v/625 ? 52. What are two sixth roots of 64 ? 53. What is the distinction between a rational number and an irrational one ? 54. Which of the numbers 8, |, 343, VI, Vs, and ir (it = 3.14159 + ) are rational ? Which are irrational ? 55. Give a geometrical illustration of an irrational number by means of a right triangle. 56. Is a radical always a surd ? Illustrate. 57. Is a surd always a radical ? Illustrate. 58. Distinguish between a surd and a radical. 59. Which of the numbers Vs, VI, ^27, V^, \/2 + ^^3, and V2 tt are surds ? Which are radicals ? 60. What is the principal root of ± Vi, Vs, and V— 8 ? 61. Name the order of V6 ; of a^; of VS; of c*; of Vm^. 62. Give an example of (a) a real number ; (b) an imaginary number; (c) a rational number; (^Z) an irrational number; (e) a radical; (/) a surd; («7) an index; (h) a radicand; (i) the principal odd root of a positive number ; (j) the prin- cipal even root of a positive number; (k) the principal odd root of a negative number. 70. Simplification of radicals. The form of a radical expression may be changed without altering its numerical value. It is often desirable to change the form of a radical RADICALS 113 so that its numerical value can be computed with the least possible labor. The simplification of a radical is based on the general A radical is in its simplest form when the radicand /. Is integral. n. Contains no rational factor raised to a power which is equal to, or greater than, the order of the radical. m. Is not raised to a power, unless the exponent of the power and the index of the root are prime to each other. For the meaning of I, II, and III study carefully the EXAMPLES Examples of I : - 1. Vf = Vf = vT^ = Vi^ = i^- 2. 6\4 = 6Vf = 6vTr3 = 6.iV3 = 2V3. Examples of II : 2. 5 ■\/24.x' =5^Sx^-Sx^ = 5 V(2x)^ 'Sx^ = 10a; VS^. 3. Vl6 - 8 V2 = \/4(4 - 2 V2) = 2 V4 - 2 Vi Examples of III : 1. ■^ = -v^ = 2* = 2^ = V2. 2. ^ = ^2_3* = 3^ = -v^3. 3.' -\/'^^ = Jb^ = ah--=b\^. 114 SECOND COURSE IN ALGEBEA A radical of the second order is simplified by the use of the Rule. Separate the radicand into two factors one of which is the greatest perfect square which it contains. Tiien take the square root of this factor and write it as the coefficient of a radical which has the other factor as radicand. If the original radical has a coefficient other than the num- ber i, multiply the result obtained above by this co&^cient. A similar rule holds for simplifying radicals involving the cube root and roots of higher orders. EXERCISES Simplify : 1. Vl8. 6. V52. 11. Vl92. 16. S Vu. 2. V20. 7. V63. 12. 2V45. 17. V^. 3. V28. 8. V68. . 13. Vi6. IS. ax". 4. Vii. 9. V75. 14. 4V54. 19. VaV 5. V5O. 10. VT08. 15. -v^. 20. nVT^\ 21. V|. Solution. V^ = V5 = V^~^ = i V3. 22. x/f. 23. Vf 24. V\. 25. V|. • 26. ^. ya _ Solution. ^ ft = ^/£ = ^/i . « = - V^. \a \a2 \a2 a 34. 6v 27. \ — 28. «^ -35. VWl?. IT 3|2^ 32. Vi_ 36. v.^^£(|y. \^«' ^^- NT* 33. V-|. 37. x/r+(|y. RADICALS 115 38. 44. \/a^ + aW3. 45. V16-8V3. 39. 4'+gJ- 46. V54-9Vi8. 42. Vrr^V!. 49. ^/J»-|V3. Hint. V4-8V3 = V4 (1 - 2 ^ 43. V36 + I8V5. 51. -^. v/3). 54. 55. 56. 50. Soli ition. 21 = 2^= V2. 52. -\/4. 53. -Va^t «l4a« Express entirely under the radical sign : 59. 2V7. 63. a-Va. 68. e^Ve^ + e"^. Solution. 2V7= 64. 2c^. ^^ , . in^O_ 60. 3V5. ee.x^^\ 70. (2^ + 1)^)^^ 61. 4 Vs. ^„ a 3f9 „^ .x-3a 3| 125 62.2^8. S^a^' 5 \(a5-3a)2 Express in simplest form with one radical sign : 72. VV2. 73. VV^. 76. VV^» Solution. VV2=V2^ 74. VVa. 77. V Vo-. = 2i:=^. 75. vWl 78. V-?/^. 116 SECOND COUESE IN ALGEBRA 79. 82. VsVsVS. 85. VVV^. 80. V-\/Sd^x. 81. V3V3. 83. Vs. 84. 2\/2V2. Find by the formula of Exercise 28, page 108, the areas of the triangles whose sides are 88. 6, 8, and 10. 90. 33, 56, and 65. 89. 7, 24, and 25. 91. 104, 153, and 185. 71. Addition and subtraction of radicals. Similar radicals are radicals of the same order, with radicands which are identical or which can be made so by simplification. The sum or the difference of similar radicals can be expressed as one term, while the sum or difference of dissimilar radicals can only be mdicated. Simplify and collect : 1. V8 + VI8. Solution. Vs + V18 = EXERCISES >V^. 2 V2 + 3 V2 2. VI -f 3 V2. 3. V5O + V98-V32. 4. V12 + 5V75-2V27. 5. 3VI8-V98 + V128. 6. V75 + 3V147- V12. 7. 2V54 + V24-V96. 8. V45 - V20 + 5 V245. 9. 3V275 + 2V99-5V44. 10. -v/ie + "v/si - 3 -v^. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. + V375. V192 - 4 -V24 -, V54 + -v/ie - VT28. ^625 + ^40 + Vi35. loVf-V^ + V^- 3x/f + 3Vi-2V^. a Vac^— -VA— 5 Va^. \Sa . 13 X lax [a fa |5x* EADICALS 117 21. v^32 x^ 4- a/1250 x - 4^512 x - V2^. 22. V(a ^cf-c ^{a + c)2 + 2 c V {a + cf. 23. -^(ct _ cf + c ^^2 - 2 ac + c^ + (» + c) -v/^ 24. \^--aJ-+aJ— -^— + 2--^J- ■'±^'_9 ac 3y 25. ^^ _|- ^(3 c^ + 9) (ct + 3)^ - V8i + c, Vg - 4 -V3, 26. 2 V9 a^ - 9 a^^* - 3 V9 aV" - 96* + V(a--^ - V') (a + &). 36a^^-36^^^ a-hh 27. (a-^»)-J^ + V25a^-25&^ + ^A| ^ ^ ya—h a—h > 72. Multiplication of real radicals. Real radicals of the same order are multiplied as follows : Example 1. Multiply 2 -\/x — 3 V« — 4 ^ax by 2 Vox. Solution. 2 Vx_— 3 V« — 4 Vaa; 2 Vaa: 4 a: V« — 6 a Vx — 8 aa: Real radicals of different order are multiplied as follows : Example 2. Multiply Vn by -^x. Solution. Vn = n^ = n5 = "V^n^ "v/x = x^ = xt = "v/^^. Then Vn • -^a; = -v^ • V^ = xn^x^. The method of multiplying real radicals is stated in the Rule, If necessary, reduce the radicals to the same order. Find the products of the coefficients of the radicals for the CQefficient of the radical part of the result. Multiple/ together the radicands and write the product under the common radical sign. Reduce the result to its simplest form. The preceding rule does not hold for the multiplication of imaginary numbers. This case is discussed in Chapter XII. 118 SECOND COURSE IN ALGEBRA EXERCISES Perform the indicated multiplication and simplify the products : 1. V3V27. 5. VI- Vi- VI- 2. Vi2Vi8. 6. (V3-Va)V2. 3. Vf . Vf . 7. ( V2 - 3 VS) V5. 4. V|->/t- Vf- 8. (V3-2V2)(V2-V3). 9. (V5-3V2)(2V5-V3). 10. (Va — Vax)(Va + 2 Va^). 11. (V2 + V3)(V2- V3). 16. (V3^-V2^)^ 12. (3V5- V2)(3V5 4-V2). 17. {-^x - 3)1 13. (V7-V5)(V7 + V5). 18. (2V3a:-l)'. 14. (2V3- V3)(2V3 + V3). 19. (V^ - V.X- - 2)'. 15. (4V5 + 2V7)(4V5-2V7). 20. 3Vaj-3V4x-8, 21. (V5 - V3 - V2)(V5 4- V3 - V2). 22. (3 V2 + 2 V3 + V30)(2 V2 + 2 V3 - 2 V5). 2J.(;,-fV3)(2« + ^V5). Square : 26. V2 - Va: - 3. 29. Va- - 3 - Va: + 3. 27. -v^ _ Va^4-4. 30. 2 Vx — 3 V2 .r + 1. 28. Vrr - 3 + Vic + 5. 31. 3 V;r - 1 -f- 2 vT a;. RADICALS 119 Perform the indicated multiplication : 32. (a + Va + ^)(a - Va -t- b). 33. (Va — b — ^a){-y/a — b + Va). 34. (V2x - 3 - V3^)(V2x - 3 + Vs^). Express as radicals of same order : 35. V2 and V^3. 37. Va^ and VS. 36. VS and V^. 38. V^ and VS. Multiply the following : 39. V3, -v^. 43. ^, V2. 47. V^, v^. 40. VS, VS. 44. Vl2, x4. 48. xl^ , ^P • 41. V^, Vs. 45. Vc, VS. 49. ^y27^ V3^. . 42. VS^, VS. 46. VS^, VS^. 50. VST^; VS+^. 73. Division of radicals. Direct division of radicals co- efficient by coefficient and radicand by radicand is often possible. Thus 6V5-^3V3 = 2V| = §V15, and 3 Vox -t- 2 VS = | Va. Direct division of radicals when the divisor is a radical expression with more than one term is usually very diffi- cult. In sucji cases a rationalizing factor of the denomi- nator is used. We then carry out the operation of division indirectly by resorting to multiplication. 74. Rationalizing factor. One radical expression is a rationalizing factor for another if the product of the two is rational. 120 SECOND COURSE IN ALGEBRA A rationalizing factor for Vz is Vt, since Vz • V? = 7. For "v/5 a rationalizing factor is ■v^25, since "Vb • ■v^25 = 5. Similarly, Vs — V3 is a rationalizing factor of V5 + Vs, as their product, (Vs - V3) (V5 + V3), is equal to 5 - 3, or 2. In like manner (3 Vt + 2 Vs) (3 V? - 2 V5 ) = 63 - 20 = 43. Two important radical expressions are V^+Vi and Va — Vb. Two such binomials are called conjugate radicals, and either is a rationalizing factor for the other. Rationalizing factors are used in division of radicals as follows : , , /;: /r V6V5 V30 Example 1. V 6 -^ V5 = \_ ,_ = -—- • V5V5 5 Example 2. (6 V2 - 15 Vs) . 3 Vs = ^^^"l^^j'^ 3V5.V5 _ 2VlO-5\^ 5 Examples. (Vs +^)^(V5- Vii) = M±^^1M±2^ (V5-V3)(V5 + V3) _ Vi5 + ViQ + 3 + Ve 5-3 = K^^ + ^^ + 3 + Ve)- Therefore when direct division of radicals is impossible use the Rule. Write the dividend over the divisor in the form of a fraction. Then multiply the numerator and denominator of the fraction by a rationalizing factor for the denominator and simplify the resulting fraction. This rule applies in all cases, while the rule for direct division fails when dividing a real radical by an imaginary number. RADICALS 121 EXERCISES Find a simple rationalizing factor for 1. Vt. 2. 3 Vs. 3. Vs. 4. Vi. 5. Vs. 6. Vs. 7. Vs-Vr. 8. V3 - 2. . 9. 3V6-2VII. 10. V3a-V2x. Perform the indicated division : 16. V8^V2. 17. eVio^Vs. 18. Vl2^V3. 19. V8--V24. 20. Va^H-Va^. 21. V2a^-v-V3a^. 11. Vx-3-V3. 12. Vic - 3 - 2 V3^. 13. Va — ^ + Va -}- 6 14. V8+V2-V5. 15. V3H-Vi8-V^. 22. (Vi4-Vi0)--V2. 23. (2yiO-3V5)--2V5. 24. ( Va^ — Va^) -^ VS. 25. 8 --4 Vs. 26. 8 --2 Vs. 27. 24--3V3. 28. V3^V2. „ , ^. V8 V3.\/4 V3.^4 ^27 • 16 ^^^"'"" vi = v^:^ ^ -^ ^ -^— 29. V5-V3. 35. Vi-VJ. 30. V^--V2. 31. V8--V2. 32. Va-f-Va. 33. V2^^ 4 -^ V 2 O. -f- C ViC. 34. V32-V2. R£ 37. V3^(V3-2). 38. V5--(V5+V2). 39. (2V3+V5)--(V3-V5) 40. (V7+V3)-(V7-V2). 122 SECOND CGHTESE m ALGEBRA Change to respectively oqiaiYail^nt fractions having rational denominators: ., , , , V5---^ ^. 2V5+"3V7 . ,„ Va!-3+V3 ■'42: —^ ■%. y^. 48. ^^^^ V5;-V2, Vcc+Vc V2-V2 Perform the indicated division : 50. ( ViO - V5)-( VlO ^i'W):^^^ i>i^toiL/ii 61LI nnol-i..i/,,!, y>',miLy6'iq iloiiiw anoift89iqz9 \imm Solution. 9-V66=J,^-^^A(^^ dfl-gjiodi od won v/ifn smiiq We must now find two numbers whose sum is 9 and w;hose product is 14. These are 7 and 2.' ' - " ^ ~ "' '" rwiiy Then 9 - 2 Vii = 7 - i^^' ^f'^^^ -^ . ^'^^ Heiice th^ sqtmi^ rdot^^f ^i^^^^^i^ "tg (H^H)^. ^^^^^ «I •• -iL^-j.Jjjn iO -- • ' iijrl) hoojaiohnn yIibsIo .8ioio.Bl oipXERCiSES-t Qflj fii ginsioifiooo odd ei Find the positive squafe roots in Exereise^^d[*-:l^iri^'89'i oT ^'\B^k V^- ^.11-3 Vs. ,; ' ^3 tMte /n-^Lim 4. 7-V40. 8. Il-Vi20. 4 .Soj^fd 12. 2 cc + 2 Vx2 - 49. 14. V^ + 3 VS = V? 4-^. 13. a + Va^ - 1. 15.' V15 - 5 VS = ? '* 16. V« + V^^^^^^^ = ? 17. V ??i"^ + 7?^ H- 2, ?i -f- 2 ??i V m + 2 71 = ? Note. In the writings of one of the later Hindu mathematicians (about A. D. 1150-)'we find a method of extracting the square root of suTd.s which is poetically the.ipame as that given in^the te?:t.§In fact, the formula for the operation is given, apart froin the modern symbols, as follows : Va + Vh = V « + Z> + 2 Vah. The study of ex- pressitdTS of the t^i^" V Va i ^6 had been carried to a i^tto^tTeii^rk- able degree of acQuracy by the Greek, Euclid. His researches on this subject, if original with him, place him among the keenest mathe- maticians of all time, but his work and all of his results are ex- pressed in geometrical language which is very far reindved f^oirPthe algebraijc aymboli^iXJi,of tp-d^ _ g _ - ^^ ^ .01 124 SECOND COURSE IN ALGEBRA . 76. Factors involving radicals. In the chapter on Factor- ing it was definitely stated that factors involving radicals would not then be considered. This limitation on the character of a factor is no longer necessary. Consequently many expressions which previously have been regarded as prime may now be thought of as factorable. Thus 3 x2 - 1 = (a: V3 + l)(a: VS - 1), and 4.x^-5 = (2x + V6)(2x-^^'). In this extension of our notion of a factor it must be clearly understood that the use of radicals is limited to the coefficients in the terms of the factors. To restrict the use of radicals in the way just indicated is necessary for the sake of definiteness. Otherwise it would be impossible to obey a direction to factor even so simple an expression as x^—y"^; for if the unknown is allowed under a radical sign in a factor, x^ — y^ has countless factors. Thus x^ — y^ = (x ■\- y^(x -^ y) = (x + y) (Vx + Vy) ( V.r — Vy) = (x + y) (Vx + Vy) (Vx + V^) (Vx — Vy), and so on indefinitely. EXERCISES Factor : 1. ar^-11. 3. a;« + 3. 5.3x^-27, 2. 3a^2-8. 4.0^^-12. 6. 6x«4-125. Find the algebraic sum of the following : 2-Vb 2 x + c x^ -^c^ ' a-b V«-f V^^* * V^-Vc ^-c Solve by factoring, and check the results : 9. x'-b^Q. 11. (T* + 144 = 26x1 10. 2x2-3 = 0. 12. 4.x'-{-c = x^+A:Cx\ RADICALS 125 PROBLEMS (Obtain answers in simplest radical form.) 1. The side of an equilateral triangle is 8. Find the altitude. 2. The side of an equilateral triangle is s. Find the altitude and the area. 3. The altitude of an equilateral triangle is 24. Find one side and the area. 4. Find the side of an equilateral triangle whose altitude is a. 5. Find the altitude on the longest side of the triangle whose sides are 11, 13, and 20. Find the area of the triangle. Hint. Let the altitude on the side 20 be x, and the two parts into which the altitude divides side 20 be y and 20— y ; then set up two equations involving x and y, and solve. 6. Find the altitude on the longest side of the triangle whose sides are 10, 12, and 16. ' ' E n Fact froTYi Geometry. A regu- / \ lar hexagon may be divided into / \ six equal equilateral triangles by / \ lines from its center to the vertices. fI 2 \fy In the adjacent regular hexa- \ gon,AB = BC = CD, etc. is the \ 7 center and OK perpendicular to \/ ^^ is the apothem of the hexagon. A ' K B 7. Find the apothem and the area of a regular hexagon (a) whose side is 18; (h) whose side is s. 8. Find the side and the area of a regular hexagon (a) whose apothem is 30; (h) whose apothem is h. Facts from Geometry. The volume of a pyramid or cone is — > where a is the altitude and h is the area of the base. The altitudes of an equilateral triangle intersect at a point which divides each altitude into two parts whose ratio is 2 to 1. 126 SECOND Ce^ESE' IN ALGEBRA 9. The base of a pyramid is a square, each side of which is 10 feet. The other four edges are each 20 feet. Find the altitude and the volume of the pyramid. ,3 [ , jjp^ r jTJ;ie: side pf an equilateral triangle ift ^^ , j|!in^ the two parts into which each altitude is divided by the othey altitudes. '>i 'A' regular- tetrahedron is '^^i^P'^ 'ifJ^ to oLiJ^jjUu siiT .£ pyramid whose four faces are .Bs^kaif.^ hua mFjih eq^ual equilateral trianglesJ •rr.^tiiliirpo iir> io oMh dkM)[ii . sl-The altitude of a regulajrf.j no obufi/B }\\\^1 .a t%%ahedron , (>Z>^ in the adja-,os; bar, ,^/,tl ojias^H sao/lw §^^t iigui-e) ^eets,the,base,^t ,,i, ,,0 9/jijiB ..iJ.li .\rH o rsihedT^. If each, edge is 24,- =''''^N<^' ''■'<\^^)}^^''^>:yC'ff find C/!^'VK'^a,ndt;-^ lastly, tlm^^yi A av^''^J?*v^'^^<>'\V. "^^'^ altituae Z)^. oJiii bobivib.;:Jt:l v(^i;iii>8<{AaKyiI ud J%. Find the altitudeandvai ;:''^""":' '^"''""f" *" ^" . ' ,' ui^e of a regdlar tetrahedron each of whose ed^esis ^6 iiiclies. ik Show that the /ititude and the voluthe^of a»tfegulartetra^ hedron\j^hose-^4^is e are respectively % "V^ and tt: ^(2. MISCELLANEOUS EXERCISES f r* 1 iRediice! fc6 respectively equivalent fractions haviingf tatibnal denominaltoaB i^i'MiJoqij o^.uii ^/ (A), -(jii rii tiirj(l,jo + V60,^^ + a^^>^ . , . ,, 14. Show that x = ig^^ root of the equation 3 ax^ + abx = 5 b. ~ ~ Simplify : 15. 2V18H-3V5- V8-*V32. 17, V3 - V6. — : , — -. ^ ., .lo 16. sVf + 2\/^-^Vf H-'^^^^- 18- V5H-3V2. 20. (5 V7 - ^)i;^( v^ + V7)r ^^ 21. sVi + Ss/S+f V96-V66|. 22. ; ^^ - ^, ^ ^^ <,-) 25. 03 . i (a^ - xYH- 2 ^) - («' - ^')^. 26. (x^)^ his) x'(S x^) - (x^ 4- 5) 4 x^ x^ + 20 (x*)2 • x^ 128 SECOND COURSE IN ALGEBRA (x^ - If 28. 29. 30. X x^ x\ax^- x'-l -\-l)2x {xy x^^'nx^- x' -1 - x'' . 2ax^^-^ (x^y . ,2 a 31. 32. x-'(- 2x-^)-(x^-\-S)(- 5x-*) (x-y 2 e"'^ + 1 33. ^^ ^pr-^ ■\x •^ + 1 CHAPTER IX FUNCTIONS AND THEIR GRAPHS 77. Functions. One of the most important concepts of mathematics is the notion of function. As the term is used in mathematics the basic idea is the dependence of one quantity upon another or upon several others. Countless functional relations exist in everyday- affairs. The velocity of a falling body is a function of the time since it started to fall, the interest on a definite sum of money is a function of the time and the rate, and the quantity of water transported yearly by the Mississippi is a function of the rainfall (and the snowfall) in its basin. Relations such as these can often be expressed either exactly or approximately by equations. For a body falling from rest near the earth's surface, s = 16 t\ Here the distance in feet, s, is expressed as a function of the time, ^, in seconds. In y = 7? -\- 2> X -\-l the value of y is a function of x. Such a relation is often represented more clearly and strikingly by a graph of the equation than by the equation itself. 78. Names of functions. A function is called linear, quad- ratic, or cubic according as its degree with respect to the unknown or unknowns is first, second, or thirS respectively. Thus 4 a: — 7 is a linear function of a: ; 3 x^ — 8 a; + 1 is a quadratic function of x ; and a:^ — 3 a:^ + a: — 10 is a cubic function of x. In the study of functions the unknown is often called the variable, since from this point of view the problem is not so much the finding of an unknown as it is the study of the changes of a variable quantity. 129 130 SECOND COUESE IN ALGEBRA 79. Notations for a function. After a function of any variable x has once been given it is usual to refer to it later in the same discussion by the symbol /(.r), which is read the function of Jr, or, more briefly, / of x. 80. Linear functions^ THd dxpresfeion 3 a; 4- 2 is a func- tion of x^ and the value of this binomial varies with x. The following table gives a partial view of the relative ^change of val,u^s,,J?ptw^l^|}5 a::,,an4, tl;^^ fung|4^^7§^ f i oij.mtr ^ t If n,i. i? theii/(i) = 8x + 2=i^l0! tt^ v4i4i im^' -iQ'! i'5 • 'i8-'Iiili(j)I ,•>., .?, u. This relation can, be xepo^^^anted graphically l;>yiTisuigtih,e same ir-axis as beforQj (section 44) and using the.j/-axis as the func- tion axis; that is, laying ; off values of x horizour. , tally aad corresponding .j values of the function^ \ ^x + 2 vertically. Thev graph resulting from, ^ the above t^J^le iqf y^U^k ues is shown in the, ^CQmpanying figu^p,,,. J^ic^ bja show^ th^^t, M the graph ofr a Imea^ , , , function is , always „ a straight Un^-,,j.,„^,^ ^l^^^f,y ^> ^j or . i • , j; - •'■•, I.jik ;x lu aolUnn^ Mift i)'>rii;') n'ttlii «f iiwoii /({EXERCISES 'i'fojiult ^o \Imi\<'. 'mH nl il'UMU OK Jtur yi n!''iifo'iif -Mit /■'•'i • ',"■""! HfiM, rnn-fj -yiiiiij ^'tUlnhivf 1. Construct , the graph of the function 3 a: ~L., .. ,, 2. Construct the graph of the function/^ w,rln .3. .Mjiirijv ru3srcTiONS axd their graphs 131 81. Quadratic functions. The function x^-{-x—6 may be represented. graphically by proceeding a^ :^ollows: Qd! a 1;h^ii'J^(i9±i:?ar*'{P'dt-t^6t=it.q(i6f \>i^ : u.4 ' i'iM k,M.[ l' ;! l! ^S rhrrrrr a^-l'. ^e «' -6 -4 ^ s ■6!' -^f' Plotting the\ *^'6ints corre- sponding to the numbers in thj^^'r ■-'..: T-rr •In IT, "\ I , 1 1 j up-: ;t8y I if i , GRAPH OF ii -lo iiq: 1 / table we obtain ^- i'l'J i'l'J \i/ m- ^^f. ;aj- B ; tn ' ; ,( if • r i /' the accompany-' - - "" ■\. jil A ::■: :n ' .1 !!''! " li / ' ing graph* ^-^oUoi 8S •o,.i \ ' r 1 I '/ (i V !];;■ j~-Th;e - g df a quae ttmvtK 1 1 tip 11 .ratio r I, _ V- - 1-^ / i \ / fmnctionii ijone 1 \ / . . _ . 1 '• \ / . 1 y' variable is a 3 c . ^-' J -i i /2 3 ,,:■■ [fiurve called K . \ / liurabola. Itmayl / ;.> \ ~ ) 1 ' ~^ / :ib l!<.>' f'.' ' . •■■■'- ■>■; - ' ; '» \ b^ sha^ r ierj --, \ \y 'flatter thai a the; hV ; ^^ .■ : : ! ■]'■ i ' ' ' " j accompanyijig^ 1\ -h '-J !!) p* ' graphs jbd + ^ ' i ' ^n..^ ' .1 V'-i- ,- ^ y.L- : I \i the same general shapfi^anfl-the opening may be upward or downwfi^d^ .J'.i:...ii i ^no :ji[} nl f^- j T j EXERCISES »i^^ 8088010 OYlir'o OflT ^1 ri^^T L „, « .:.. ^ •„ .:;^!ffiii 6^-nI} aixB--^; Uonstruct tna graph of the lollowmg : ^ r 7. A body falling from rest near the earth's surface obeys the law s = 16t^, where s is in feet and ^ is in seconds. Con- stwt(&t the gi?^h 4ii fit) = 16 1^. ii ^ .xioirniiebxii alliil Ji 1 . 132 SECOND COUKSE IN ALGEBRA - Note. In the study of analytical geometry one takes up system- atically the curves which represent equations of the various degrees beginning with the simplest. It turns out, as we have already seen, that the linear equation is represented by a straight line.. Equations of the second degree in x and y lead to the so-called "conic section." One of the most interesting and important aspects of the graphi- cal method is the fact that the simplest equations correspond to the most useful curves both in pure science and in nature. The com- monest curves in nature are the circle and the parabola. Their equations are the. very simplest equations of the second degree. 82. Graph of a cubic' function. The graph of a cubic function is obtained in the same general way as that of a quadratic function. The function a^— 5x+^ may be represented graphically by proceeding as follows: If x = - 4 -3 -2 -1 1 2 n 3 then/(a;)=a:«-5a: + 3 = -41 -9 5 7 3 -1 1 6i 15 Plotting the points corresponding to the numbers in the table (except the first and last), we obtain the points ^(-3, -9), B, C, D, E, G, and II, in the order named. The curve crosses the a:-axis three times : once between 1 and 2, again between 0. and 1, and a third time between — 2 and — 3. Above If the curve rises indefinitely, and below A it falls indefinitely. In each case it becomes more F GR WoF 1 f(3 f) = x'- 5£C +3 / P / / fN. Ih- Yn / / / \ / \ ^ X X f ^ V I i -1 6 \ \ y 1 \ E 1 1 1 1 1 f o \ A F' FUNCTIONS AND THEIR GRAPHS 133 and more nearly straight as it recedes from the a:-axis, never crossing either axis again. In forming a table of values two pairs are sufficient for a linear function, but more are needed for a quadratic function and still more for a cubic function. Usually the higher the degree of the function, the more points are needed in constructing its graph. It should be noted that in making a good graph the number of points is not so important as is their distribution, which should be such as faithfully to outline the entire curve. Where the graph curves rapidly or makes sharp turns the points should be close together. Such places are difficult to locate before the graph is constructed ; hence one should make a table of values which appears to be suffi- cient and plot them. Then inspection of the plotted points will usually show where sharp turns or rapid curvature exists. The table of values should then be properly extended and the additional points located. Repetition of this last step will enable one to draw a graph which accurately pictures the variation of the function. It should be observed here that scales on the two axes need not be the same. Some experience is required to choose for the two axes the scales which are best suited to bring out clearly the shape of the curve. In general the graph should be drawn to as large a scale, in both directions, as the size of the paper permits. What that will be for each axis can be decided by inspecting the table of values. For example, when the dimensions of the preceding graph are once deter- mined one can see from the table that all values of x are easily repre- sented but that it is undesirable to try t.o represent the values of the function, — 41 and 15. EXERCISES Construct the graph of the following : 1. x^-Sx-^l. 3. x»-4cc-2. 2. x^-Sx + 2. 4. £c*-ll£c2 + 24. Note. The notion of a function is one of the three or four most fundamental ideas in modem mathematics. Only the simplest ex- amples are given in this book, but many others involving expressions of the utmost complexity have been studied by mathematicians for many years. An important reason for the-study of functions is found 134 SECOND COURSE IN ALGEBRA i|Q, the fact that all kinds of facts and principles which we meet in the study of nature can be expressed symbolically, by means of , functions, and the discovery of the properties of such functions helps us to understand the meaning of the facts. A cohipl6te understaiidiiig of the laws of falling bodies, light, electriditj^, o'l* 'teoiind ' could tLevei^%(e reached Without the study of .i( 83. Graphical solution of equations in one, unknown. Jn of the preceding sections is their use in solving equations^ in one unknown. The ideas involved can be made clear by questions? il:l 10 'i^Or'- e»ni m ^ofno?. r> ^v;h>\ ?,o o* a^OItAL EXERCISfifi^'' ^' ^'^* iBiJ-^no^ xiT .svtjjo ndJIi;i,.ij;n:t JxMiV/' -^iuniKf v,(ns'f ^^. i k, '•^■l^'niitJ) .^.^\vyV^vuV AaoA X. ¥rom the graph m sectibn 80 id^termine,the value fOi tna function ^ 3 x -f- 2 at the 'ppmt yrhev^e it^ ,g;^'f!'P,^ iC?^^^s^^.^ i^h^ ,«?7^fi9f> 3. Does this value of aa satisfy the equation 3 a;-}- 2. *i)0t2n! 8. Solve 3a5 + 2 = without reference tb th^' gtaphl'^ f''^"»« 4. What point on the. gi;^ph represents" the root of 3a; + 2 = 0? ^'^^^'' 5. Irom the graph in section 81 determine the vulues ui the function x^ ^ x — 6 at tHe points where its graph crosWs the a;-axis. .*^- 4- '^ - + ''-^ " ' ' -^ 6. Do these values of « satisfy the equation'^4^'5'— 6ii/0? 7. Solve x^-{- x—6 =="(y without referving'to' tie graphl '. ^ ^ , ,i8,„TO^,,piQip^,pp,,tia^ ngxaph repre^Qftii tbft .^ootBiipf, F^KCTIONS .iLasrE> THEIR GRAPHS 135 9. From >tbe giapiiiin action 82 determine tlie values of the functioixiqff!^ 5.;?^ rb 3 at the points where its graph crosses It). Do these values of oj mak^-the functioa i^-^&As^f B equal zero ? .3ini gi eifia 11. What method could beHsi&d' to solve the ^qiuatloii 84. The process of graphical solution. From what pre- cedes, it is apparent tlmt tfee steps in. the graphical solution of an equation ip one unknown are: , m^ ^,, Transpose the terms so that the right member 4$ ^^(%idBi}pe ^iilGrmjpJp ^HJkM^^9V( in the, left {m^mf^er, , : , : ; , , j r , ,,, . r... ; t The values of xfor the points wlipretfi^ gi^ph- crp,^s&»,t^, X-axis are the real roof$ ^f the eqpation.\Q imyi dirjji8i!00 erij The algebraic solutions of a linear equation and a quad- ratic equation in one unknown are so simple that except for the purpose of illustration their graphical solution is comparatively unimportant. The algebraic solution of cubic and higher equations, except in simple eases, is much more difficult than the solution of the quadratic and is never iresented\ij^s^an elementary course. For this i»^asOil and ijoT the insiglit it gives into equations in general, the Graphical solution of the cubic and higher equations is i|mportant Ti!q[d: - fflnminating. For: subfe' equations if the method of factoring fails, the graphical method is the only method open to the student at this point in bis. progl^ss. I, x^-Tx + S, f^jO.xo' a = 't 8ort4.[^? -8^ = 0. , )(c. - 5.) 136 SECOND COURSE IN ALGEBRA 85. Imaginary roots. An equation of the second or a higher degree often has imaginary roots. Such roots can- not be obtained by the graphical methods so far considered. A study of the graphs which follow will make clear why this is true. Consider the following equations: a^-4:x-5 = 0, (1 ) 2^_4:^; + 4 = 0, (2) ^_42; + 13 = 0. (3) The graphs of the functions in the left members of equations (1), (2), and (3) are given in the accompanying figure. The three functions differ only in their constant terms, for 9 added to the constant term of (1) gives the constant term of (2), and 9 added to the constant term of (2) gives the constant term of (3). Apparently, as the constant term is in- creased the graph rises without change of shape and without motion to the left or to the right. From the graph the roots of a? — 4 x — 5 = are seen to be 5 and — 1. These results are obtained from factoring ; a^ — 4a: — 5 = 0, or (^x — 5)(a; + 1) = 0. Whence x=5 or — 1. If we imagine curve (1) to move upward, the two roots change in value and become the single root of, curve (2), F V ~! \ Vf- t1 Iv 71 \ i \ / ' / \* ' '7 3aS ttt Cv^ - U SjuA. 9n Z-J-i- \5^v 6 -/^Q Xx^^.'X yJ-l- yyj^i-^'^ztz \ "^ 4 •'' / X \ N (2) ..' / X -2 -\ 1 2 3 A ^ y4 1 y' ' -8*^4-^ F 1 FUNCTIONS AND THEIR GRAPHS 137 which touches the a;-axis at a point where x equals 2. Solv- ing 2j2 — 42^ + 4 = by factoring gives (x — 2) (a; — 2) = 0. Whence x=2. If we now imagine curve (1) to move still farther upward from its position (2), it will no longer cut the a^-axis. Further, when the curve reaches the position of (3) it does not cut the a;-axis at all, and hence cannot show the values of the roots of the equation a:^ _ 4^_j_13 _ 0^ as, in fact, it does not. The graph does show, however, that the value of a:^ — 4a; + 13 at the lowest point of the curve is 9. This means that for every real value of a?, positive or negative, x^ — 4:x+l'^ is never less than 9. The graph of (3) makes clear that no real number if substituted for x will make a:^ — 4 a; + 13 equal zero. It can be shown by the method of section 87 that the roots of a^-4ic + 13 = are 2 + 3V^ and 2-3^/^. Note. It required the genius of no less a man than Sir Isaac Newton first to observe from, the graph of a function that two of its roots become imaginary simultaneously. He also saw that an equation with two of its roots equal to each other is, in a certain sense, the lim- iting case between equations in which the corresponding roots appear as two real and distinct roots and those in which they appear as imaginary roots. 86. Imaginary roots for a cubic equation. If we attempt to solve 2;^ — 2^; — 4 = graphically, we obtain the graph of the accompanying figure. The curve crosses the 2;-axis at x=2. This is the only real root the equation has ; the other RE F J ] X ' X _9 3 1 / / / ^ 1 / ^ J / -5 / GRA PH OF ^ n xS= X'- -Zx ^4 "r _ F 138 SECOND COURSE IN ALGEBRA two are imaginary. The roots can here be obtained by fac- toring a^-2x-4: = 0; thus (x-2)(x^'{-2x + 2} = 0. The roots of 2:24- 2 a:;— 2 = are the two imaginary roots of the euhic equation. EXERCISES As far as possible solve graphically, finding results to one decimal place : 1. x^-5x-9 = 0. 6. x^-4:X-^5 = 0. 2. x^ = 4.x-5. 7. X* - 7 ic + 4 = 0. 3. a:^ — 3 X + 4 = 0. 8. ic*-4x» + 12 = 0. 4. a;3 + ic - 4 = 0. 9. a;* = 10x^-9. 6. x^-lx'-2x + S = = 0. 10. x^- 2x^ = 0. CHAPTER X QUADRATIC EQUATIONS 87. Solution by completing the square. An equation of the form a7? -{- 62; -f- c = 0, where a, 5, and c denote num- bers or known literal expressions, is called a quadratic equa- tion. Any such equation can be solved by the method of completing the square. This method gets its name from the fact that in the course of the solution there is added to each member of the equation a number making one member a perfect square. ORAL EXERCISES What terms should be added in order to make the following expressions perfect squares ? 1. x2 + 2ic + ? 5. ir2+a^ + ? 9. cc2 + |x + ? 2. x^-2x^l 6. a;2_3^_^9 10. x^-\x+l 3. a;2_6^4-? 7. £c2_^fx + ? 11. cc^ + |x + ? 4. ic2 + 8x + ? 8. a^2_4^_|.? EXAMPLE 12. x'-\-ax-\-'i Solve5ar^-3x — 2 = and check the result. Solution. 5 :r2 - 3 a: - 2 « Transposing, 5 a;^ — 3 a: Dividing by the coefficient of a:^, x^ — \x Adding (— ^-^y to each member, = 0. (1) = 2. or 139 140 SECOND COURSE IN ALGEBRA Extracting the square root of each member, X h ±/zT- Whence ^ = i% ± tV = 1 ^^^ ~ §• Check. Substituting 1 for x in (1), 5- 12 -3- 1-2 = 0, 5-3-2 = 0, = 0. Substituting — f for x in (1), 5(-f)^-3(-f)-2 = 0, 1 + 1-2-0, = 0. The method of solving a quadratic equation illustrated in the preceding example' is stated in the Rule, Transpose so that the terms containing x are in the first member and those which do not contain x are in the second. Divide each member of the equation by the coefficient of x^ unless that coefficient is +1. In the equation just obtained^ add to each member the square of one half the coefficient of jc, thus mahing the first member a perfect trinomial square. Rewrite the equation^ expressing the first member as the square of a binomial and the second member in its simplest form. Extract the square root of both members of the equation^ and write the sign ± before the square root of the second member^ thus obtaining two linear equations. Solve the equation in which the second member is taken with the sign +, and then solve the equation in which the second member is taken with the sign — . The results are the roots of the quadratic. QUADEATIO EQUATIONS 141 Check. Substitute each result separately in place of x in the original equation. If the resulting equations are not obvious identities^ simplify each until it becomes one, EXERCISES Solve by completing the square and check real results as directed by the teacher : 1. £c2 _^ 4£c -h 3 = 0. 9. 6ic2 + ic - 35 = 0. 2. 0^-2^-8 = 0. 10. 12x2 -25a; + 12 = 0. 3. ^2 - 5 - 2 = 0. 11. 3 ^2 _^ 8 ^ + 4 = 0. 4. x2-h2=-3a:. 12. x-{-2 = ^x\ 5. 2x2 + 5^2 + 3 = 0. 13. x-4 + x2 = 6-2x2 + 8x 6. 3x2 + 7x-6 = 0. ,^^+14 7. 2^2- 3x- 5 = 0. 4 ' 8. 5x2-7a!-6 = 0. 15. (3x - 2)^ + (x - 1)^ = 1. In Exercises 16-25 obtain results to three decimal places : 16. £c2-12x + 31 = 0. Hints. By applying the rule we get X = 6 + VS, and X = 6 — Vs. , From the table on page 274, V'5 = 2.236. Hence we get the result, to three decimals, a; = 8.236 and 3.764. 4~ 3 17. x^-x-^l^^. 22. Sx' 18. n^ + 2 = 5n. 23. x'- 19. 3x2_l2x + 9 = 2. Q 20. 5a;2 + 8x + 2 = 0. 24. 21. 10-2x' = x^-Tx. 25. (y-\ 25. (2/+l)(2/ + 2) = 32/(y-4). 142 SECOND COUESE IN ALGEBRA 26. £c* - 3 cc^ + 2 = 0. Note. This is not a quadratic equation, but many equations of this form can be solved by the methods applicable to quadratics. i- Solution. x4 - 3 a:2 + 2 = 0. xi-Sx^=-2. x^-3a:2+|=-2 + | = x^-^=±h a:2 = 2 or 1. Whence x = ±V2, ±1. Check as usual. Note. It should be particularly observed that the equation of Exercise 26 has four roots instead of two. In general an equation has a number of roots equal to its degree. Thus the equations in Exercises 29 and 31 have six and eight roots respectively, although some of them are imaginary and the student at present should not be required to find them at all. 27. ic*-5a;2 + 4 = 0. 32. 6 cc^ - 11 ic^ -j- 3 = 0. 28. x' - ISoc^ + 36 = 0. , 33. 4m« - 15 = 7m«. 29. x^-j-S = 9x\ 34.. x^-4.x^-\-S = 0. 30. x^ - 7x^ = 8. 35. 2/' + 7/ = 8. 31. a;«- 17 0-^ + 16 = 0. 36. m*-4 V2m2 + 7 = 0. 37. (x^ - 2xy- 7 (x'' - 2x) = -12. Solution. Let x^ — 2 x = y. Substituting y for x^ — 2xy we obtain Solving, .y = 3 and 4. Then a;^ - 2 a: = 3. Whence a: = 3 and — 1. Also a;'* — 2 .r = 4. Whence • x = l± V5. QUADRATIC EQUATIONS 143 In Exercises 38-41 do not expand, but solve as in Exercise 37 : 38. (x-iy-\-4.(x-l)=5. 39. (x^-4.xy~5(x'-4.x)-24: = 0. 40. S(2f-^Syy-7(f + S2j)=20. 41. (x - ^y+ Jx-^-5 = 0. 42. aa^ + 6x -f c = 0. Solution, ax^ + bx + c = 0. x^ ■{ — X = — -' a a .^b , / bV J2 ^ j2_ a:2 + -a;+ — - =—--- = — - a \2 a/ 4 a^ rt 4 — iac , b Vft^ — 4 ac 2a 2 ___6_ \^b^ - '^ ac _ - b ±Vb^ - 4: ac ^~ 2a'^ 2 a ~ 2 a 43. x^-\-Sax-h2a^ = 0. 46. *cc2-a(J+l)a? + ^t2=0. 44. 2f + b2j=6b\ ,. h ^ b , 45. bx^ + x = l+bx. 2a 2a 48. cx^ -{(^-itd)x + cd = 0. 49. a^ic^ — aj (a + ^) Va^ + aZ> = 0. 50. 3a.*-aV-2a^ = 0. „ 3 , ^ 1 . 9a^ 53. 7cc^ — 2aa:-h-a2 = -- — 51. 6ay-7ar2/ + 2?-2 = 0. 4 2 4 52. (3ax + 4&)2 = (2ax-&)l 54.. aV4-3aicV^-10Z» = 0. 55. 9A:2-cc2 = 2;ta;V3. 56. (aa; + fey^4-3(r^r + i)+2 = 0. 57. («V - 3 ax)2 = 14(aV - 3 aa^) - 40. 144 SECOND COUESE IN ALGEBRA 88. Solution by formula. In Exercise 42, above, the general quadratic ax^ -{-hx-\- c = Q has been solved and the roots found to be x = (F) 2a ^ ^ The expression (i^) is a general result and may be used as a formula to solve any quadratic equation in the standard form aa^ -\- bx -\- c = 0, in which a, b, and e may represent numbers, single letters, binomials, or any other form of algebraic expression not involving x. If the numbers a, b, and c are such that the expression 6^ — 4 ac is negative, the formula contains the square root of a negative num- ber, which is a kind of number not yet fully considered in this text. In the exercises that follow it will be assumed that only such numer- ical values of the literal coefficients are involved as will not make 6^ — 4 ac negative. A discussion of the case here ruled out will be found in Chapter XII. EXERCISES Solve for x by formula and check the results as directed by the teacher : 1. 4:X^-^SX = S. Solution. Writing in standard form, 4ar2 + 8a:-3=0. Comparing with ax^ + bx + c = 0, evidently 4 corresponds to a, 8 to 6, and — 3 to c. Substituting the values in (F) gives 8±V64-4«4-(-3) 2-4 8±V64 -H48 -8±Vll2 Check as usual. -8db4V7 8 8 8 QUADKATIC EQUATIONS 145 2. x"- ex -16 = 0. 9. 3£c + 4 = x\ 3. x^-x-l = 0. 10. Ua^ -Wx = 26. 2 -^ 2" 5. 4a;2 + 5a^-3. 12. x + f = fa;2. e.2x-\-4. = x\ 13. f -i£c2z=2x. • 7. X = 1 - xi 14. 6x == 1 + 2£c2 8. 4^2 - lliT - 60 = 0. 15. .^2 + 2cc = 8 - a-^ 16. 3j9V^j9x + 4. Solution. Writing in standard form, Here a = 3 j?^, b= — p, and c=— 4. Substituting these values in the formula (F), ^ - (-p)±^(-py - 4 • Sp^(- 4) 2.3jo2 _ p±^p^ + 48 jp^ _ JP =fc 7jp _ 4 -1 Check as usual. 17. x''-2kx-Sk'' = 0. „^ .^,, 13A;i 871- 25. 6 h^ H 18. 3a^ = 7ax + 6x^. 19. 4ic2 ^ A;a! - 14/^2 =0. «« ,2 . . . 1 26. ic^ _^ ^>£c + - = 0. 20. bx = 12x^-b\ .2^2 c 21. a'x^-\-4.abx-{-Sb^ = 0. 27. - + a^ + 3^ = 0. 22. 12pV-4j9ra;-?- = 0. 23. Sa^x^-\-Sabx-h4.b^=0. 28. — - - _ i = 0, 24. X - 3 VJ - — = 0. 2 ic 29. 13 a%^x^ = 9 ^/ -f 4 aV. 146 SECOND COURSE IN ALGEBRA 30. 2x^ + 5x = cx^-{-Scx-\-S. Solution. (2 - c) a;2 + (5 - 3 c) a; - 3 = 0. Here a = 2 — c, 6 = 5 — 3 c, and c = — 3. Hpi ace 3 c - 5 ± V(5 - 3 c)2 + 12 (2 - c) XlCJ 2(2-c) 3c-5±V49-42c + 9c2 4-2c _3c-5±(7-3c) 4-2c 2 1 ^ 6c-'l2 4-2 c' ^^ 2-c' "^ 4-2c Check as usual. 31. P' -\-x'' = 22)x-j-2x-2p. 32. (X -lf = a(x-x^). 33. x' + 2x = kx^-\-kx-l. 34. Sx + ax^ = 2(x^ -{-ax- 1). 35. cc^ — ex — ^(ax — ac). 36. 4aV + 4c2 = 16aV + cV. 37. x« -a^x^=Sb'x^-Sa%\ 38. a^ -i-x'' = 2ax-\-b\ 39. 12jpq = x^-}-4:qx-Spx. 40. x^ - aV + a%' = ^>*a;l or -3. 89. Comparison of the various methods. Four methods have been given for the solution of tlie quacbatic equation : (a) Solution by factoring. (J) Solution by graphing, (c) Solution by completing the square, (c?) Solution by formula. In practice, the first and the last of these methods are most convenient. QUADRATIC EQUATIONS 147 When an equation with integral coefficients can be solved by factoring, as explained in Chapter III, the roots are always rational numbers. For example, 3x2 + 2a;-8 = factors into (3 a; - 4) (a: + 2) = 0. Hence the roots are x = ^ and — 2. An inspection of the formula -h± Vb^ -4:ac 2a shows that, if a, b, and c are integers, one always obtains roots involving radicals unless the expression under the radical sign, 5^ — 4 ac, is a perfect square. In this case, however, the values of the roots are rational numbers. For example, in the equation 3a;^ + 2a: — 8 = the value of the expression 6^ — 4 ac is (2)^ — 4 • 3 (— 8) = 100, which is a perfect square. Hence the roots of 3 a:^ + 2 a: — 8 = are rational num- bers, since the radical term, in the roots can be expressed as a rational number. Hence to determine whether a quadratic equation of the form aa^ -j-bx-^ c = can be solved by factoring into ratienal factors, we have the Rule. Compute the value of b^ — 4: ac for the equation. If the result is the square of an integer, the left member of the equation can be factored and the roots are rational. ORAL EXERCISES Determine which of the following can be solved by factoring : 1. x" + lOcc + 25 = 0. 6. x^ - 4x - 4 = 0. 2. x'-^x- 4.00 = 0. 7. 3 a;2 _ ic + 4 = 0. 3. x2_7^_3 = 0. ' 8. 3a^2-5ic + 8 = 0. 4. 2x2_f_3^_^-^^Q 9^ 2£c2-10£c-25 = 0. 5. 2cc2 + 3x + 2 = 0. 10. 3x^-h2ax-a^==0. 148 SECOND COUESE IN ALGEBRA REVIEW EXERCISES Solve the following by the method best adapted to each 1. ^x" -7x - 10 = 0. 6. x' - 7x2 + 12 = q 2. 5x2 + 14.r + 8 = 0. 7. 6x«-20 = 2a;l 3. 4x2 4-3x-2 = 0. S.x^-x = 0. 4. x^ -1.1 X- .84 = 0. 9. x^ = x. 5. 8x2-f-2V5x-15 = 0. 10. x^ + 64x = 0. 11. (a^-3x)'^-2(x2-3x)=8. 12. 5^2- 9^ + 3-0. ,_ x^-l x^ + 1 _ x' + l 14. x2 + xV2-V6 = xV3. 19 14 X - 1 ^ X + 1 a; + 4x-2 x + 2 13. .03x' + .01x = .l. x2 + xV2-V X — 1 X -\-l ^^' X - 6 ' ^^ 2ic -1 1-x x + 1 X — 1 cc — 2ic + 2 X — 6 _ X — 4: xHi ~ 2a; + 6* ^1. pqx^ - rqx -\- psx = rs. x±l x + 3^22 22.x'-2ax-\-a'~b' = 0. 24. (x2 + 5x + 2)2-6(«2-f5x + 2)=16. 3x-l x + 1 48 25. 5.^ + 1 x-1 (5x + l)(l-a:) a;»-10a^ + l 27. 15^2 -1.95x4- .054 = 0. 28. 10-9x = 7x''. 29. x^ + 961 a^ = 62 ax. 30. (x8 + 4)^ = 4 + a:»: 31. (2x + 3)(x-4) = (3x-8)(4x-l). 32. (3x - 2)(x - 5) = (4x - 3)(x + 1). QUADRATIC EQUATIONS 149 33. 34. 35. 36. X -j- b X + o 38. 17a; = 6ic2-10. 39. 7x*-4t2x'-\-25 = 13x''-5x^-\-4.00. 40. 2k^x^ -h Skx = 5(3kx -1). 1 2 1 x-^7 5 3-a: ^ a: + 2 ar^ + Scc- 4"^x + 4 £C + 3 x-1 x-2 x^-3x-\-2 x-2 x-1 5x-ll l-3a^ 2x-l x-1 ' cc + 2 0^ + 1 x + 12 ^ , x + 7 41. 37j9^a: = 210^2-pV. 42. .x + 1 ic-1 2ic-f-l x^-1 x^ + 1 (x^^x + l)(x^- -0^ + 1) 43. 3x-2 2 5(^ + 1) 0,-1 '3~^6-4^_i 44. cc*-a;2^4(9 + ^'). 45. ^ , 4:k k-2x e — Za3 c 46. 47 4-30^ 2x 2x-\-5 X 1-x 3-2ic 6cc2^x-5 12x=^H-2x-4 4ic2-£c-2 8a^2-2£c + 4 48. 8.4x2 4- .005 a: -.15 = 0. 49. 324 a;^ + 1936 = 1665x1 50. 3x2 + .7x=.2. 51. 3 aV _ 10 abx - 25b^ = 2(2abx - aV). 150 SECOND COURSE IN ALGEBRA 1 1 1 52. ^ -7^ + 53 54. x'-Zx + 2 x^ + x-^ ?,-2x-x^ x + 2 2 3(2-0;) (x + 4.)(x- 1) (x + 4)(3 -x)~ {x-'d){l-x)' 2(ar^-2)-l ^ x'' + 2 x^-1 ~(x'-2)-4.' 55. (3x - 7)(2x -^1) - (5x -^ 2)(2x - S) = 0. . 56. (x - 2)2- (1 - 2x)(Sx-\- 5) = 5 - (1 - 2x)(3a3 + 2). 57. (a32-2cc-2)2+2(cc2-2x-2)-3 = 0. 58. (£c2 - 4ic + 1)2 _ 4(cc2 - 4x) - 16 = 0. 59. (2a;-^>)2 = ci(2£c-Z>) + 2a2. 60. « + - = w H x m PROBLEMS 1. Separate 42 into two parts such that the first shall be the square of the second. 2. Find two consecutive odd numbers whose product is 143. 3. The sum of the reciprocals of two consecutive numbers is \^. Find the numbers. 4. The time, in hours, required for a trip of 216 miles waa 6 greater than the rate, in miles per hour. Find the time and the rate. 5. The altitude of a triangle is 6 feet less than the base. The area is bQ> square feet. Find the base and altitude. 6. One leg of a right triangle is 9 feet shorter than the other, and the area is 45 square feet. Find the three sides. 7. The area of a trapezoid is 180 square feet. One base is 4 feet greater than the altitude, the other is three times the altitude. Find the two bases and the altitude. 8. A rectangle whose area is 27 square inches has a perim- eter of 21 inches. Find the length of the shorter side. QUADRATIC EQUATIONS 151 9. A polygon of n sides always has ^n(n — S) diagonals. How many sides has a polygon with 119 diagonals ? 10. If AB, in the accompany- ing figure, is a tangent to the circle and BD is any secant, then (ABf = BC . BD. li AB = 6 and CD = 10, find BC. 11. A requires 4 more days than B to do a piece of work. Work- ing together they require 2|^ days. How many days will each require alone ? 12. In selling an article at $20, a merchant's per cent of profit is 9 greater than the original cost of the article in dollars. Find the original cost and the per cent of profit. 13. A 3-inch square is cut from each corner of a square piece of tin. The sides are then turned up to form an open box of volume 12 cubic inches. What was the side of the original square ? 14. The number of lines on a certain printed page is 20 less than the average number of letters per line. If the number of letters per line is decreased by 16, the number of lines must be increased by 15 in order that the new page may contain as much matter as the old. How many lines and how many letters were there on the original page ? 15. For what positive value or values of x will the product of 0? — 5 and a? -f- 5 be 1 greater than their difference ? 16. What values of x will make the product of 4ic — 5 and X — 2 equal in value to sc -f 4 ? 17. The length of a rectangular room exceeds its width by 2 feet. A rug placed in the middle of the floor leaves a mar- gin of 1 foot all around. If the area of the rug is eight times that of the margin, find the dimensions of the room. 152 SECOND COURSE IN ALGEBRA 18. A sum of |4000 is invested, and at the end of .each year the year's interest, plus |400, is added to the investment. At the beginning of the third year the investment afnounts to $5230. What is the rate of interest? 19. The cost of an outing was $60. If there had been two more in the party, the share of each would have been a dollar less. How many were there ? 20. The dimensions of a rectangular box are expressed by three consecutive numbers. Ics surface is 214 square inches. Find its dimensions. 21. The radius of a circle is 28 inches. How much must this be shortened in order to decrease the area of the circle by 1078 square inches ? 22. Two bodies, A and B, move on the sides of a right triangle. A is now 5 feet from the vertex, and moving from it at the rate of 10 feet per second. B is 35 feet from the ver- tex, and moving toward it at 5 feet per second. At what time (past or future) are they 75 feet apart ? 23. How high is a mountain which can just be seen from a point on the surface of the sea, 40 miles distant ? Hint. See Exercise 10. 24. The distance in feet, s, through which a body fallg from rest in t seconds, is given by the formula s = I gf/^, where <7 = 32, approximately. A bomb dropped from an aeroplane struck the ground below 8 seconds later. How high was the aeroplane at the time ? 25. If a stone be thrown vertically upward, with a velocity of 100 feet per second, it is known that, neglecting the resistance Of the air, its height after t seconds will be 100^ — 16^^ feet. After how many seconds will it be 136 feet high? Explain the double answer. 26. After how many seconds will the stone of Exercise 25 return to its starting point ? Explain the zero root. CHAPTER XI IRRATIONAL EQUATIONS 90. Definitions and discussion. An irrational equation in one unknown is an equation in which tlie unknown occurs under a radical, or is affected by a fractional exponent. Thus 3x -\-5Vx = l, x-x^ + 1 = 0, and Vx^ -3x4-2 = 6 are irrational equations. The chief difficulty involved in the solution of sucK equations arises from the fact that sometimes results are obtained which do not satisfy the given equation and hence are not roots of that equation. A result of this kind is called extraneous. EXAMPLE (a) Solve Vic — 6 — 4 = :0. (^) Solve -Vcc-6-4: = 0. Solution. Transposing, Vx - 6 = 4. (1) -Va:-6 =4. (1) Squaring, a; — 6 = 16. (2) a: - 6 = 16. (2) Solving, X = 22. X = 22. Check. V22 -6-4 = 0. - V22 -6-4 = 0. Vl6 -4 = 0. -Vi6-4 = 0. 4-4 = 0, -4-4 = 0, lich is true. which is not true. It appears from a study of these solutions that state- ments (1) differ only in the signs preceding their left members. Consequently this distinction disappears after »K 163 154 SECOND COUESE IN ALGEBRA squaring, and equations (2) are identical. Since the re- mainder of the work in both («) and (5) consists in the solution of (2), the result obtained is really the root of this equation. Whether the root obtained satisfies both (a) and (5), or only one of them, can be determined only by substitution. Hence it appears that (a) is an equation and that (5) is not, but is merely a false statement in the form of an equation. In any case, all of the roots of the original equation are sure to be among the results found, provided no factor containing the unknown has been divided out. But no result should be called a root unless it satisfies the origiual equation. This means that all results must be checked. In irrational equations, as m all the work up to the present, it is understood that unless a radical or an ex- pression affected by a fractional exponent is preceded by the double sign ± it has only the one value, just like any other number symbol. Thus Vl6 means + 4, and not — 4. Also 4^ means + 2, while — ^^ means — Vi, or — 2. . If this fact is kept in mind, it is clear from an inspec- tion of (5), above, that it could have no root, since the sum of two negative numbers could not possibly be zero. The method of solving equations containing radicals is stated in the Rule. Transpose the terms so that one radical expression (the least simple one if there are two or more} is the only term in one member of the equation. Next raise both members of the resulting equation to the same power as the index of this radical. Combine like terms in each member and^ if radical ex- pressions still remain^ repeat the two preceding operations lERATIONAL EQUATIONS 155 until an equation is obtained which is free from radicals ; then solve, this equation. Check. Substitute in the original equation the values found and reduce the resulting numerical equation to its simplest form by extracting roots, but not by raising both members of the equation to any power. Finally, reject all extraneous roots. EXERCISES Solve the following for real roots and check results : 1. Vcc + 1 = 5. 7. Vic + l=V3x~5. 2.. -s/?>x-5 = 7. • 8. 2V8x = xVi. 3. ■\/2a^-h3 = 3. 9. 3 V2a: + 6 = VGx^ - 6. 4. 2 Vx + 4 = 4. 10. -\/4x + 3--v/4-3a^ = a 5. 3V2x-8-7 = 17. 11. V2x + 2=V2x-2. 6. 7 + 2a/2cc-1=:13. 12. ^iTl=V^^^. 13. 1+Vcc-2=V^. Solution, Transposing, Squaring, Transposing and collecting, Squaring, Check. Substituting | for x in the original equation, l + V|^=V|, 14. V9ic- 4 = Vcc. 15. V3ic-2 + 5 = Vcc4- 35. 16. V3 .'^5 + 4 — Vcc + 5 = Vd — X. 17. 3Va; + 4 + Vl-.T==3Vl-ic + 2Vic + 4. 18. Va; + 2 = Vi + V2. 19. a; -f 1 = V^ic^ + S; 156 SECOND COUKSE IN ALGEBKA 20. ir -h 4 + V^ = V^M^e. 21. V2x-4 4-Va; + 6 = VT^ 22. -Wx -\- 2 -\- -Vx -1- ^3 X -\- 3 = 0. 23. V3cc+1- Vic+1= Vx-4. 24. x^- 9x^ + 20 = 0. Note. The equation here given is not a quadratic equation, but it is of the general type ax^^ + 6a;'» + c = 0. Here x occurs in but two terms, and its exponent in one term is twice that in the other term. Many equations in this form can be solved by completing the square (compare Exercise 42, p. 143). Solution. x^ - 9 xt + -\L = _ 20 + -S|^. art = 5 or 4. Whence a; = ± 5* or ±8. Check. Substituting ± 5^ for x in the original equation, (±5f)|_9(± 51)1 + 20 = 0, or 52 - 9 . 5 + 20 = 0, or = 0. Substituting ± 8 for a: in the original equation, (± 8)i-9(± 8)^ + 20 = 0. 16 - 36 + 20 = 0, or = 0. 25. x^ + 5a;' -14 = 0. 30. 2a;^ - 7 -^sc + 6 = 0. 26. .'-10x1=11. 31...-! = 4x-i + 32. , 32. aji - a;i - 6 = 0. 27. 6 = x^-{-x. OK 1 33 r-i — -4- i — 28. Sxi -{-5x^-\-2 = 0. SQxi 9~^' 29. x-''-17x-^-\-52 = 0. 34. 7ic^-6=2a;i 35. x^-^5x + aVa^^ + Sx - 54 = 0. Hint. Let y = x^ ■}■ 6x. lEEATIOKAL EQUATIONS 157 36. 3x2-4x-llV3ic^- 4a; +28=0. 37. 2x^-'^x-^--\/2x'- -3a^-l+l = 0. 38. x^-lx-h-^x"- -2x- -4 + 2 = 0. 39. V2x-4 + 5 = l. 40. =^^\x-X 41. 12£c^-27ic'^ = 20£c3 _ -45. 42. 15-2Vir2_^9 = x^ + 9. 43. 27(8 + 27a;*-ic^) = 8xi V2a^-V8 44. 4Vic + 2 = 3Vic + 4. 45.- ^x + 4 = Va^ - 8. 2V3 \/2a: + 12 V2x-3 3V3 48. Vx + 1 = Va; - 4 + 1. 49. 4(cc - 5) - 2(aj _ 5)* - 2 = 0. 50. ^x^'x = {x -\- 3) V9^. ^, Va; + 2- ^8-^ 8 51. , — , = o * V8 - cc Vx + 2 3 52. 5x^ + «7x^ - 200 = 16 - 3icl 53. 4 V^- 21 ^^ + 27 = 0. 54. ix" - 5a; + 2)* - h{x? - 5cc + 2)^ + 6 = 0. 55. V 17 + 2 V3 + s + V7+?- 5 = 0. 56. 3cc2-a;-22 = 6V3ic'-cc- 6. 57. If # = TT-v -J solve for I and cr. >^ 58 . VtT = \x V3 and ^ = 2 7rr^, express A in terms of x. 59. If ^^•= 2 7'^ and C = 2 vrr, express C* in terms of a. 60. If ^ = — - V 3 and a; = -|- r V3, express A in terms of x. 61 . If i!L = 3 ?-^ and a = J ?■ v 3 + V2, express i^ in terms of a. CHAPTER XII IMAGINARIES 91. Definitions. When the square root of a negative number arose in our previous work, it was called an imaginary, but no attempt was then made to use it or to explain its meaning. The treatment of imaginaries was deferred because there were so jnany topics of more im- portance to the beginner. It must not be supposed, however, that imaginaries are not of great value in mathe- matics. They are frequently used in certain branches of applied science ; and it is unfortunate that symbols which can be employed in numerical computations to obtain prac- tical results should ever have been called imagmary. By such a name something unreal and fanciful is suggested, to obviate which it has been proposed to call imaginary num- bers orthotomic numbers^ but this name has been httle used. The equation a:^ + 1 = 0, or x^ = — l^ states that a:; is a number whose square is —1. By defining a new number, V— 1, as one whose square is —1, we obtain one root for the equation x^ + l = 0. Similarly, V— 5 is a number whose square is — 5. And, in general, V— rv is a number whose square is — n. Obvi- ously, V— 6 means something very different from V5, and V— n from Vn. The positive numbers are all multiples of the unit -f 1, and the negative numbers are all multiples of the unit — 1. Similarly, pure imaginary numbers are real multiples of the imaginary unit V— 1, as 2V— 1, — 5V— 1," and 5V— 1. 168 IMAGINARIES 159 Furthermore v — 4 = Sv— 1 ; -y — a^ = a^/— 1 ; and V^=V5.V^. __ The imaginary unit V— 1 is often denoted by the letter I ; that is, 3 V— 1 = 3 z. If a real number be united to a pure imaginary by a plus sign or a minus sign, the expression thus obtained is called a complex number. Thus — 2 + V— 1 and 3 — 2V— 4 are complex numbers. The general form of a complex number is a + hi, in which a and h may be any real numbers. Note. Up to the time of Gauss (1777-1855) complex numbers were not clearly understood and were usually thought of as absurd. The situation reminds one of the time when negative numbers were similarly regarded, and the veil was removed from both in about the same way. It was found that negative numbers rfeally had a sig- nificance — that they could be used in problems that involve debt, opposite directions, and many other everyday relations. The inter- pretation of imaginary numbers is not quite so obvious, and is not considered in this text. But as soon as it was seen that an interpre- tation was possible the ice was broken, and it needed only the insight and authority of a man like Gauss to give complex numbers their proper place in mathematics. ORAL EXERCISES Express as multiples of V— 1, or i : 1. V-16. 5. 3 V- 6. 2. V-25. 6. 2^-c. 3. V-6^2_ 7. V3.V-3. 4. 2V-3. 8. V2.V-5. 11. ^-x'-2x-l. 12. V-?/ + 62/-9. 92. Addition and subtraction of imaginaries. The funda- mental operations of addition and subtraction are performed on imaginary and complex numbers as they are performed on rational numbers and ordinary radicals of the same form. 160 SECOND COURSE IN ALGEBKA Thus and 5V^-3V^=2V^. Also (8 + 5V^)+(4-2V^) = 7+3V^. Similarly, (a + bi) + (c + di) = a -{■ c ■\- (b + d) i Simplify : 1. 3V^+2Viri. 2. 4V^4-V^. EXERCISES 7. V-18 + V^. 3. V- 25 - V- 16. 4. V^ + V^. 5. V-4 4-V~16. 6. (_ 8)^ + (- 32)^. 8. 4V-25cc2_2V-36ar». 9. 24-3V^ + 6-5V^. 10. 7VI^2-5a + 4V^ir;^. 11. (Sa- 6ib)-]-(a-\-{h). 12. 4-8i + 16- 3V^. 13. 5 4-3V-49cc2_6V^6x2H-4. 14.^ 18 - 3(- 1)* + 6(- 25)* + 4. 15. 5 V^ + 3 V^ _ V- 27 + 2 V^s. 16. (8-5 V^Il6)-(7 + 3 V- 25). 17. 4 V- 9 a* - 6 a^ V- 16 + 3 18. (5x-6iy)-(Sx-\-2iy). 6 + 5 V- 54. 93. Multiplication of imaginaries. By the definition of square root, the square of (— ny is — n. Therefore {y/'^f = -l, (V3i)3 = (V3i )VZi = _ V3T. (v::i )^ = (vzi )2 ( vzT)2 = (_ 1 ) (_ 1) = 1. IMAGIKARIES 161 '^) To multiply V— 2 by V— 3, we first write v32'^V2.viri, and V^ = V3.V3l. Then V32 . V^g = ( V2 . V:^) ( Vs • VI3) == V6 . V3i . V^=-V6. Similarly, (2 Vi:5)(- 3 Vr2)= (2V5 . vC3) . (- 3V2 • a = _6VlO(-l) = 6VlO. In general, if V^^ and V— 5 are two imaginaries whose product is desired, they should first be written in the form Va . V— 1 and y/b • V— 1 and the multiplication should only then be performed. This method will prevent many errors. In this connection it must be clearly understood that one rule followed in the multiplication of real radicals (see page 117) does not apply to imaginary numbers. In the case of ordinary radicals we have But the product of two imaginaries like V— 2 • V— 3 does not equal V(— 2)(— 3), for this equals Vc. We have seen above that In multiplying two complex numbers, write each ex- pression in the form a ± hi and proceed as in the following EXAMPLE Multiply 4 + V^ by 2 - V^. Solution. 4 + V^ = 4 + V5 . V^, 2 _ VITe = 2 - Ve . V^. (1) 12) Multiplying (1) by (2), 8 + 2 Vs • V^ - 4 V6 • V^- V30 (- 1). Rewriting, 8 + 2 V^ - 4 V^ + V30. 162 SECOND COUKSE IN ALGEBRA EXERCISES Perform the following Indicated multiplications and simplify results : 17. Vm + n ' V— 7n — n. 18. (3+v:ri)(3-V:^). 19. (5+V^)(5-V^). 20, (3-4V2^)(3 + 2V2^). 21. (4+v:^)(5-V:r3).. 22. (5-3i)(6-5V2i). 23. {a + ib){c + id). 32. (2 + 2 aAITs)' - (2 - 2 V^ 25. {a-\-ib)(a-ib). 26. (-i + iVirs)^ 27.(-i-iV33;. 28. (-2 4-2V^)'. 29. (-2-2V^)'. 30. (x — iy)\ 31. {a + ibf-{a-ibf. Sf. 33. (a + ^ Vl - b^) {a - I Vl - h'^). 34. Determine wh ether the su m an d the product of the numbers 5 + 6 V— 2 and 5 — 6 V— 2 are real. 35. Determine whether the sum and the product of the numbers 2 + V— 3 and 2 — V— 3 are real. 94. Division of imaginaries. One complex number is the conjugate of another if their product and their sum are real. Thus a + hi and a — hi are conjugates. Conjugate IMAGINAEIES 163 complex numbers are used in division of imaginary ex- pressions as conjugate radicals are used in division of real radicals. In case either the numerator or the denominator of a fraction is imaginary or complex, the division may be performed as in the following 1. V^-v-V2. EXAMPLES V-6.V2 2V-3 OVIUUIUU. V2 (V2)^ 2 2. V8--V^ ^. Solution. V8 V8.V-2 4ti Vi:2 (V^)' - 2 -21 3. V-6--^ 7:^2. Solution. V- () V- 6 . V- 2 V6 • '•^^•'_,/3 v:^ C^/^)' -2 ^^• 4. 3-(2+- s/33). Ortll-l<-Trt« 3 _ 3 (2 - V- 3) 6-3V-3 2 + V-3 (2 + V-3)(2-V- 6 - 3 V- 3 -3) 4 + 3 The method of the above examples is stated in the Rule. Write the dividend over the divisor in the form of a fraction. Then multiply both numerator and denominator of this fraction hy the simplest expression which will make the new denominator real and rational. Reduce the result to its simplest form. 164 SECOND COUESE IN ALGEBRA 1. V- 12 . -V2. 2. Ve^v ^-2. 3. 2VE^ 4V-1. 4. V-9^ -V-1. 5. i^Vi: ~3. 6. 4-^V- "5. 7. Vs^v '-2. 8. (- 49)* ^(-64)*. 9. ^by^^ ^-y. 10. V— m - f-V— 7i. 21. . (3 + 2i)(l EXERCISES Perform the indicated operations : 11. (- 6/>a;)*-f-(-5a;)t 12. [(-a^6■)i_(-_^2^^i]^(^_^)i 13. 3-(l+V^). 14. 2-(l-V^). 15. 2V^^(V^H-6). 16. 2V^-f-(3V^ + 3). 17. (_i_v^)^(-i+v:r3). 18. (l+2^)--(3-4^). 19. X -r-(£C + ^^/). 20. {(i-^ih)^{c-^id). 22. Is 4- 1 - V- 3 a cube root of - 8 ? 23. Does a;2_6ic4-12 = 0ific = 3± V^^s ? 24. Does X = f(V-lO), y=- |-(V-10), satisfy the sys- tem x'-xy- 12 2/-2 = 8, x" + xy-10y^ = 20 ? 95. Equations with imaginary roots. The student should now be able to check the solution of an equation which has imaginary roots. EXERCISES Solve the equations which follow, and check the results : 1. x' + 4a; + 8 = 0. 6. cc^ + a; + 1 = 0. 2. 3,2 _ 8a; + 24 = 0. 7. oi^ - a; + 1 = 0. 3. ic2 + 3a; + 9 = 0. 8. 5a;=^ - 6a; + 14 = 0. 4. a;2-6a; +16 = 0. 9. 6 a;^ + 10 a; + 21 = 0. 5. 3a;2 4-2a; + 4 = 0. 10. 3ar^ + 16a; + 21 = 0. IMAGINARIES 165 11. x^ = l. Hints. If x^ = 1, x^ - 1 = 0. Hence (x - 1) (x^ + x + 1) = 0. Then . x - 1 = 0, and x2 + X + 1 = 0. 12. ic^ + 1 = 0. 15. x^ = l. 18. x^ = 64. 13. x^ = 8. 16. x^ = 9. 19. x^ = 64. U.,x^=-27. 17. £c' = l. 20. x^ = -125. 21. How many square roots has any real number? cube roots ? fourth, roots ? sixth roots ? 22. What do the preceding exercises suggest regarding the number of Tith roots which any real number has ? 23. 27 ic^- 8 = 0. 27. a^« + 7a;^-8 = 0. 24. 64a;«f 125 = 0. * 28. 3ic* + 16cc2 + 21 = 0. 25. x'' - 2ic2 -.8 = 0. 29. 27 o;^ -I2x''- 64. = 0. 26. x^-\-x^-2x-2 = 0. 30. 6x^ + 21x^ + 9 = 0. 31. 25£c^ + 40a^2_^64 = 0. 32. (x^ 4- 4)(.x2 4- 3 a; + 7) = 0. 33. (^2 + 20^)2 + 15 (x^ + 2a:) + 54 = 0. 34. (x^ + 5xy + 9(x^ + 5x)- 112 = 0. 35. Solve a: + ?/ = 4, x^ — 3 xi/ — y^ = — 39, and check the results. 36. Solve X -\- 2 y = 4z, 'if — X = 0, and check the results. 37. Solve x^ -\- if = 4:, X — 1/ = 6, and check the results. Note. Long before the time of Gauss, mathematicians had performed the operations of multiplication and division on com- plex numbers by the same rules that they used for real numbers. As early as 1545 Cardan showed that the product of 5 -f V— 15 and 5 — V— 15 was the real number 40. However, he was not always equally fortunate in obtaining correct results, for in another place he sets -( — \ — -) = — — = = -• 4\ >J 4/ V64 8 166 SECOND COURSE IN ALGEBRA Even the rather complicated formula for extracting any root of a complex number was discovered in the early part of the eighteenth century. But all these operations were purely formal, and seemed to most mathematicians a mere juggling with symbols until Gauss showed clearly the place and usefulness of such numbers. 96. Factors involving imaginaries. After studying radi- cals we enlarged our previous notion of a factor and, with certain limitations, employed radicals among the terms of a factor. Now in a similar manner, with like restrictions, we extend our notion of a factor still farther and use imaginary numbers as coefficients or as terms in a factor. For example, x^ + 1 may hereafter be regarded as factorable, for a^-\-l=2^-(-l) = {x+V^){x-^'^). Similarly, 4 :^^ + 9 = 4 2^2 _ (_ 9) _ (2 ^^ + 3 V^) {2x - 3 V^) and a:2 4- 6 = a:2 - (- 6) = {x + V^^)(2: - V^). Further, a^ -1= (^x -1} (^x^ -{- x -hi) . Hitherto the trino mial a:^-{-x-\-l has been regarded as prime ; but the stu- dent can easily prove that a^ + x+lis equal to the product (2; + l4.jV33)(a: + | -^V^). Therefore a^-1 has three factors, these two and x — 1. Note on the use of imaginaries. We have explained the laws of addition, subtraction, multiplication, and division for imaginary (and complex) numbers and have made some use of them. It is largely because imaginaries obey these laws that we call them numbers, for it must be admitted that we cannot count objects with imaginary numbers. Nor can we state by means of them our age, our weight, or the area of the earth's surface. It should be remembered, however, that we can do none of these things with negative numbers. We may have a group of objects — books, for example — whose number is 5 ; but no group of objects exists whose number is — 5, or — 8, or any negative number -wliatever. If it is asked, How, then, can negative numbers and imaginary numbers IMAGINARIES 167 have any practical use ? the answer is this : They have a practical use because when they enter into our calculations and we have performed the necessary operations upon them and obtained our final result, that result can frequently be interpreted as a concrete number like those dealt with in ordinary arithmetic. Moreover, if the result cannot be so interpreted, it is, in applied mathematics at least, finally rejected. In that part of electrical engineering where the theory and meas- urement of alternating currents of electricity are treated, complex numbers have had extensive use. Their employment in the difficult problems which there arise has given a briefer, a more direct, and a more general treatment than the earlier ones where such numbers are not used. In theoretical mathematics complex numbers have been of great value in many ways. For example, numerous important theorems aboat functions are more easily proved under the assumption that the variable is complex. Then, by letting the imaginary part of the complex number become zero, we obtain the proof of the theorem for real values of the variable. Indeed, the student need not go very far beyond this point in his mathematical work to learn that, if e is 2.7182 + (see page 253), e ""^l^ + e-"^/-i is equal to the real number 1.082 + • At the same time he will learn also how such a form arises, and something of its importance. In a way which we cannot now- explain, even so involved an expression as (a + 16)'' + *'' has in higher work a meaning and a use. If the student pursues his mathematical studies far enough, that meaning and use and a multitude of other uses for complex numbers will become familiar to him. But the numbers which we have learned in this book to use, namely fractions, negative numbers, irrational numbers, and compl-ex numbers, com- plete the number system of ordinary algebra, for it can be proved that from the fundamental operations no other forms of number can arise. CHAPTER XIII THEORY OF QUADRATIC EQUATIONS 97. Formation of equations with given roots. According to section 34, the equation (x—2)(x—S)=0 has the roots 2 and 3. In general, the equation (a; — r^) (a; — rg) = has the roots r^ and r^, because either of these numbers, when substituted for x, satisfies the equation. Hence we can always find an equation whose roots are two given numbers r^ and r^ by setting the product of the binomials x — Ti and x — r2 equal to zero. For example, an equation whose roots are 3 and 4 is seen in (a; - 3) (x - 4) = 0, or a;^ - 7a; + 12 = 0. EXERCISES Form an equation whose roots are the following : 1. 2,3. 5. 2,|. 9. 1 + V3, 1-V3 2. 3, 7. 6. 5, f 10. 3 ± V7. 3. 1, - 3. 7. - I, 1 . 11. 2 + V^, 2 - V^. 4.-2,-5. is. -|,-|. 12. -7+V:r5, -7-V^. 13. i+Vf,i-\/|. 17. 1,-1,2. 14. i+vr|,i_vri. i8..i,i,.i. _ 15. 2, 3, 4. 19. Vir2, - V- 2, 2. Hint, (x - 2) (x - 3) (x- 4) = 0. 20. 1 + V2, 1 - V2, 3. 16. 1, 3, 5. 21.1,- 1, V^, ~ VI^i. 22. V2 - V3, V2 + V3, - V2 - V3, - V2 + Vs. 168 THEORY OF QUADRATIC EQUATIONS 169 98. Relations between roots and coefficients. By direct multiplication we obtain from (x-r^)(x-r,)=0 (1) the equation x^ — (r-^ + r^) x + r^r^ = 0. (2) Since r-^ and r^ are the roots of (1), it appears from an inspection of (2) that the quadratic equation x^-\-hx-\-c=0 has the roots r-^ and r^^ provided h = — (r-^-\- r^) and c = r^r^. For example, we may form at once the equation whose roots are 4 and 9, as follows : x-2-(4 + 9)x + 4 • 9 = 0, or a:2 _ 13 a; + 36 = 0. Similarly for the cubic equation x^ + bx^ + ex + d = 0, whose roots are r^, r^, and r^, we have (x — r^)(x — r^) (x — r^) = 0. Then b =-(r^ -{- r.-, -\- r^), c = r^r^ + r^r^ + r^r^, and d=— r^r^r^. ORAL EXERCISES Form equations whose roots are the following : 1.2,9. 4.-3,-5. 7. 2V2, -2V2. • 2. 4,5. 5. -7,2. 8. 3 + Vt, 3 - Vz. 3. -1,6. 6. V3, -V3. 9. I+V2, I-V2. We will now show the relations which exist between the roots and the coefficients of the general quadratic equation a2^+ hx-\- c=Q. By section 88 the roots of ax^ -\-hx+ c = are and r« = . 2a 2 2a 170 SECOND COUKSE IN ALGEBEA Adding r^ and rg, we have - b + Vja _ 4 ac - J - Vj2 _ 4 ae -26 2a h '■i ' '^- 2« a Therefore — (r^ + ^2) ~ " * Multiplying r^ by r^^ we have -4ac 4 a2 4 ^2 a Therefore r^^ = - • These results may be expressed verbally as follows: For the equation ax^ + ftjr + c = 0, /. The sum of the roots with its sign changed is -• c ^ II. The product of the roots is - • These relations frequently afford the simplest means of checking the result of solvmg a quadratic equation, as illustrated in Exercises 18-31, below. EXERCISES Form the equation whose roots are the following : solution. -:(l-i\=-L^^., (?\/i\=_5 = £. \2 5/ 10 a \2/\ ;5/ 6 a Hence the required equation is a;2 - — a: - 5 = 0, or 10 a;2 - 7 a: - 12 = 0. 10 5 THEORY OF QUADRATIC EQUATIONS ITl 2. hh 3. h-h 4. H, - H- 5. 4.41, 1.59. 6. 2 + 3V3, 2- -3 Vs. 7. 4 + V- 2, 4 -V-2. 8. f + Vt, 1— V7. 9. HiV-3, i-iV: 11. ±V-1 12. 1 a, —-' a 13. S a 5 a 2' 2 1 + 1. 14. 1 + a, 1 a. 10. If ± V3. a -{-b a — h ^^' T' — TT' a — a -\- 16. 6, 8, f 17. - 4, 4, f Solve the following equations by the use of the formula and check the result by the use of I and II, above r 18. cc2_5ic + 6 = 0. 19. x'-x-^^O. 20. x2-2x-4 = 0. 21. x^- 9a; -10 = 0. 22. a;2 + 2a; + l = 0. 23. ^2 + 8a: + 16 = 0. 24. £c2 + 5a; + 5 = 0. 25. 2r»2_|_3x- 6 = 0. 26. 3cc2+3£c-5 = 0. 27. 5a;2-6a; + 10=0. 28. x^ 4- £c + 1 = 0. 1 29 - + - + - 3 ^4^5 0. 30. - + 4x-7 = 0. 31. Qx'-^.x- 0. Find the value of the literal coefficient in the following : 32. a;^ + 2 i^ — c = 0, if one root is 2. 33. 0?^ — a? — c = 0, if one root is 6. 34. a^ — ca- — 70 = 0, if one root is 7. 35. ar^ + 2 ^>x + 20 = 0, if one root is - 4. 36. a?^ — 8 a; + c = 0,if one root is twice the other. 37. a;^ + 7 a: + c = 0, if one root exceeds the other by 2. 38. x^ + 11 X + ^ = 0, if the difference between the roots is 10. _^,+V62- - 4 ac 2^ -h-^h^- - 4 ac 172 SECOND COURSE IN ALOEBEA 99. Character of the roots of a quadratic equation. It is often desirable to determine whether the roots of a given quadratic equation are real or imaginary, rational or irra- tional, equal or unequal, without solvmg the equation. This can be accomplished by use of the formulas for the roots of the quadratic ax^ + 5a; + c = : (1) r,= ^^^^~^-^. (2) These expressions are seen to differ from each other only in the sign preceding the radical. The expression ft2 - 4 ac, which appears under the radical sign, is called the dis- criminant of the quadratic. The only way in which r^ or ^2 can be a complex number is for the discriminant to be negative. If all of the coefficients a, 5, and c a re rational, r^ or r^ can be rational only when Vj2 — ^ac is rational ; that is, when the discriminant is a perfect square. Hence if a, 6, and c are rational, an inspection of (1) and (2) shows that the following statements are true: /. If y^— 4iac is positive and not a perfect square, the roots are real, unequal, and irrational. For example, ina:^ — 8a; + 2 = the discriminant h^ — 4:ac equals (— 8)* — 4 • 1 • 2 = 56, which is not a perfect square. The roots of the equation are the real, unequal, and irrational numbers 4 ± vl4. //. If y^ — ^ac is positive and a perfect square, the roots are real, unequal, and rational. For example, in the equation 2 a;^ — 3a; — 9 = the discriminant 62 — 4 ar equals (— 3)^ — 4-2-(— 9)= 81, wfiich is a perfect square. The roots are the real, unequal, rational numbers — J and + 3. THEORY OF QUADRATIC EQUATIONS 173 ///. i/ ^2 _ 4 ^^ {g 2^ro, the roots are equal. For example, in the equation 4:x^ — 12x + 9 = 0, the discrimi- nant P - 4 ac equals (- 12)2 _ 4 • 4 • 9 =^ 144 - 144 = 0. The only number which satisfies this equation is f , so in one sense the equa- tion has only one root. But since the left member has two identical factors each of which affords the same root of the equation, it is customary to say that the equation has equal roots. IV. If IP' — 4: ac is negative^ the roots are imaginary. For example, in the equation 2 x^ — b x -\- ^ = the discrimi- nant 62 _ 4 ac equals (- 5)2 - 4 • 2 • 4 = 25 - 32 = - 7. The roots of the equation are the conjugate imaginaries and + 5-V~7 ^ 4 ORAL EXERCISES Determine the character of the roots of the following equations without solving: 1. ic'-f 5aj + 6 = 0. Q. x'-{-x+l=0. 2. 3a;2-4ic+l=0. 7. x^- 8a; +16 = 0. 3. 2x2- 6a;- 3 = 0. 8. 2x2h-3x + 5 = 0. 4. 2x'-3x-2 = Q. 9. 4x2-4x+l=0. 5. 5x2-5x + 4 = 0. 10. 3^2 -2x- 2 = 0. EXERCISES Determine the value of k which will make the roots of the following equations equal : 1. x2-A;x-M6 = 0. Solution. a = l,h=-k,c = lQ. Hence ' h'^-^ac = k^- 64. In order for the roots to be equal, U^ — ^ac must equal zero. Therefore P - 64 = 0. Whence 1- = ± 8. 174 SECOND COURSE IN ALGEBRA Check. Substituting 8 for k in the original equation, a;2 - 8 X + 16 = 0. Whence x = 4, only. Similarly, substituting — 8 for k, x^ + 8 X + IQ = 0. Whence x = — 4, only. 2. x^-^kx-j-16 = 0. 7. 9cc2 + 30£c + A; 4- 9 = 0. 3. x^ - lOcc -\-k = 0. S. 4.kx^- 60ic + 25 = 0. 4. 2x^-hSx + k = 0. 9. 9k^x^-S4:X-\-4:9 = 0. 5. x^-3kx + S6 = 0. 10. 49x2 - (A: + 3)a; + 4 = 0. 6. 3£c2 + 4A;x + 12 = 0. 11. (k"" -\- 5) x^ - SO x -{- 25 = 0. Determine the values of a for which the following systems will have two sets of equal roots : 12. ^' ^ ^ "^^ 13. ^' + ^' = "^'^ 14. ^2/ = «^' '2/ = ic + a. *y = x + l. *a;H-2/ = l. 100. Number of roots of a quadratic. In section 87 we found that every quadratic equation has two roots. Up to the present we have assumed that a quadratic equation can have no other root than the ones found by the method of completing the square. This fact can be proved as follows: Proof. If we writ/e the equation ax^ + 62; + c = in the form x2 ^^, ^ ^ E - and substitute therein from (I) and (II) on a a page 170, we get x^ — (r^ + r^) x + r^r^ = 0. This can be factored and written as (x — r{) (x — r^) = 0. Now if any value of x different from rj and rg, say r, be a root of this equation, such a value when substituted for x must satisfy the equation (x — r^) (x — r^) = 0. Hence (r — r^)(r — r^) must equal zero. By assumption, however, r is different from r^ and r^. Consequently neither the factor r — r^ THEORY OF QUADRATIC EQUATIONS 175 nor r — r^ can equal zero, and therefore their product cannot equal zero. This proves that no additional value r can satisfy the equation x^ — (r^ + rg) a: + r^^ — 0. As this equation is but another form of 03?" + 62; + c = 0, the latter has only two roots. 101. Factors of quadratic expressions. Let r^ and r^ be the roots of a^ + 5a: + c = ; then I c or a7^-\- hx -\- c = a(x — r-[) (x — r^. Therefore the three factors a, x — r-^^ and x — r^ of any quadratic expression can be found if we first set the ex- pression equal to zero (see section 34) and solve the equa- tion thus formed. Obviously the character of the roots so obtained will determine the character of the factors. Hence by the use of the discriminant l^— 4:ac we can decide whether the factors of a quadratic expression are real or imaginary, rational or irrational, without factoring it. EXERCISES Determine which of the following expressions have rational factors : 1. a;2-3cc-40. 3. Tic^-Qx + lS. 5. 120? -11x^1. 2. 2a:2 + 5aj-7. 4. ^^.x'-x-lO. 6. 5ic2 + 3cc-20. 7. 3^2 - 9a- + 28. 9. x" - 2«cc + {a? - b^. 8. 33 A^ - 233 A, - 6. 10. ahx" - {W- + a')x + ah. Separate into rational, irrational, or imaginary factors : 11. 2x2 + 5x-8. Solution. Let 2 a:2 + 5 a: - 8 = 0. Solving by formula, x = — ^^ ^ = 176 SECOND COURSE IN ALGEBRA rr. -5+V89 ^ -5-V89 1 hen r, = and r„ = 4 Therefore 2x^-{-5x = 2\x -S+Vsoir _5-V89' |(4x + 5-V89)(4a; + 5 +V89). 12. x'-lx-T. 21. £c2 + 7£f + 8. 13. x'-Ax- 1. 22. x^-\-x + l. 14. x^-}-2x-^2. 23. x^ + 1. 15. a^2 4.4^_9. 24. £c2 + 9. 16. 4a;2_l2£c-9. 25. cc^ - 2 ao: + a'^ - ^». 17. 25x2 + 20ic + 4. 26. a:^ + 6acc + 9'a2_4^ 18. 6;z^2^14£c-40. 27. 4£c2 + 4aic + a^ _ 4c. 19. 10-9a;-9a;2. 28. x"" - ^ ax + 4: a^ -\- c. 20. 10x2 4-12-26cc. 29. a^t-^ _,_ ^^ _|_ ^ 30. x^ — xy-j- 5x — 2y + 6. Solution. Let x^ — xy •{■ 5 x — 2i/ +6 = 0. Then x^ + (6 - y)x - 2 y + Q = 0. Solving for x in terms of y by the formula, (5 - y) ± V (5 - ?/)2 - 4 (- 2 y + 6) _ -5 + y±-y/y^-2y + l ~ 2 ' ' Whence x =— 2 and y — 3. Therefore x^ - xy -\- 5 x - 2 y -\- Q = (x -\- 2)(x - ?/ + 3). 31. Sx^ - 6xy -\-14:X - 4:y + S. 32. x^-xy-2f-\-Sx-6y. 33. ar^ - 4xy - ?/ + Sy- - 2 - cc. CHAPTER XIV GRAPHS OF QUADRATIC EQUATIONS IN TWO VARIABLES 102. Graph of a quadratic equation in two variables. Before solving graphically a quadratic system, the method of graphing one quadratic equation m two variables must be clearly understood. EXAMPLES 1. Construct the graph of ic^ = 3 y. Solution. Solving the equation for y, y = — . o We now assign values to x and then compute the approximate corresponding values of y. Tabulating the results gives : If x = 6 4 3 2 1 - 1 — 2 -4 -6 then y = 12 ^/ 3"" 1 i h t ¥- 12 Using an a;-axis and a y-SLxis as in graphing linear equations, plotting the points correspond- ing to the real numbers in the table, and drawing the curve determined by these points, we obtain the graph of the adja- cent figure. The curve is called a parabola. The graph of any equation of the form x^ = ay is a parabola. Y \ p AR AB )Li ^ 1 ^ c mi IPHQ] 7 ^ X •i— 3y Y J \) q f V J A / X r\ s 1 / X -2 j- 1 > I \ Y' 1.77 178 SECOND COUESE IN ALGEBRA 2. Graph, the equation xy -{- S = 0. 8 Solution. Solving for y, y = X Assigning values to x as indicated in the following table, we then compute the corresponding values of 2^ : If x = -6 -5 -4 -3 -2 -1 -i \ 1 2 3 4 5 6 8 then y — 1 f 2 1 4 8 16 -16 -8 -4 -! -2 -1 -i -1 Proceeding as before with the numbers in the table, we obtain the two-branched curve of the figure below, which does not touch either axis. The curve is called a hyperbola. Y y Q HI fPE RB OL A A Y H X X -7 -^ -. l-X ] . I 3 ^ t yA H U^ GRAPH OF -2 X ^:2/H ■8- =0 f r _J The graph of any equation of the form xy = K is a hyperbola. The curve for xy — K (Jr=any constant) is always in the same general position ; that is, if K is posi- tive, one branch of the curve lies in the first quadrant and the other branch in the third. If A' is negative, one branch lies in the second quadrant and the other in the fourth. GRAPHS OF QUADRATIC EQUATIONS 179 3. Graph the equation x- -\- y^ — X^. Solution. Solving for y, y = ± Vl6 — x^. Assigning values to x as indicated, in the following table, we obtain from page 274 the corresponding values of y : If x = -5 -4 -3 -2 -1 1 2 3 4 5 theni/ = ±3V^ ±2.64 ±3.46 ±3.87 ±4 ±3.87 ±3.46 ±2.64 ±3V^ For values of x numerically greater than 4 it appears that y is imaginary. The points corresponding to the pairs of real numbers in the table lie on the circle in the accompanying figure. The center of the circle is at the origin, and the radius is 4. The graph of any equa- tion of the form oP- -\- tf^^ r^ is a circle whose radius is r. This can be proved from the right triangle FKO, If P represents any point on the circle, OK equals the 2;-distance of P, KF equals the ^/-distance, and OF equals the radius. Now ~0K^ ■\-KF'^ =0F'^ \ that is, o?'-\-y^ — r^. It follows, then, that the graphs of 2^-|-?/2=9 and a^ + «/2 = 8 are circles whose centers are at the origin and whose radii are 3 and V8 respectively. Hereafter, when it is required to graph an equation of the form 2^ 4- ?/2 = 7^, the student may use compasses and, with the origin as the center and with the proper radius (the square root of the constant term), describe the circle at once. In all the graphical work which follows, the student will save time by obtaining from the table on page 274 the square roots or cube roots which he may need. Y ^A M ' Q H k. / r,s •■ V 2 > / \ C IRC LE X' \ X K - ] , I > J i V / \ k -2 J / N K ^ l> Y GRAPH OJ i' ^■'f2/i = 16 r 180 SECOND COURSE IN ALGEBRA 4. Graph the equation I6x^ -\-9y^ = 144. Solution. Solving, y = ± ^ V9 — x^. Proceed as in Example 3 If x = -4 -3 -2 -1 + 1 + 2 + 3 + 4 then y = ±^v^ ±2.98 ±3.77 '±4 ±3.77 ±2.98 ±^V^ For any value of x numerically greater than 3, y is imaginary. The points corresponding to the real numbers in the table lie on the graph of the adjacent figure. The curve is called an ellipse. The graph of any equation of the form of ax^ -\-by^ = c in which a and b are unequal and of the same sign as c is an ellipse. Note. These three curves — the ellipse, the hyperbola, and the parabola — were first studied by ihe Greeks, who proved that they are the sections which one obtains by cutting a cone by a plane. Not for hundreds of years afterwards did anyone imagine that these curves actually appear in nature, for the Greeks regarded them merely as geometrical figures, and not at all as curves that have anything to do with our everyday life. One of the most important discoveries of astronomy was made by Kepler (1571-1630), who showed that the earth revolves around the sun in an ellipse, and stated the laws which govern the motion. Those comets that return to our field of vision periodically also have elliptic orbits, while those that appear once, never to be seen again, describe parabolic or hyperbolic paths. The path of a ball thrown through the air in any direction, except vertically upward or downward, is a parabola. The approxi- mate parabola which a projectile actually describes depends on the elevation of the gun (the angle with the horizontal), the quality of the powder, the amount of the charge, the direction of the wind, and various other conditions. This makes gunnery a complex subject. Y y^' H^ Y- 3- "Si ^ 3 I -: «> e 2 ^ t V \ t tL Lx -2 - 2 i.± m , "^ ELLIPSE / ^ 2 t \ -2 f- \ J- S. z GRAPH OF >jv^ ^ 16xH 91/^14' III Y GRAPHS OF QUADRATIC EQUATIONS 181 EXERCISES Construct the graphs of the following equations and state the name of each curve obtained: 1. x' = 4:y. 3, x'+i/ = 49. 5. x^ - f = 16. 2. y^-{-2x = 0. 4. x^-{-t/ = 18. 6. xij = 12. 7. xy = -6. 9. 1603^-9?/^ = 144. 8. 9x''-\-4.y^ = S6. 10. 25 x^ -\- 9 if = 225. 103. Graphical solution of a quadratic system in two variables. That we may solve a system of two quadratic equations by a method similar to that employed in section 44 for linear equations appears from the EXAMPLES 1. Solve graphically [^^^_^^^ _ {j (1) (2) Solution. Constructing the graphs of (1) and (2), we obtain the straight line and the parabola shown in the adjacent figure. There are two sets of roots correspond- ing to two points of intersection, which are Note. If the straight line in the adjacent figure were moved to the right in such a way that it always remained parallel to its present position, the points A and B would approach each other and finally coincide. The line would then be tangent to the parabola at the point x = 4, y = 1. Were the straight line moved still farther, it would neither touch nor intersect the parabola and there would be no graphical solution (see page 182). Y \ ^^ \ A ^ v^ \ Q N \ o N .<2) \ N X ^ L K -^ ~ 1 > o\ L \ > 5 r > b\ > ^ y 1) ^ ^ \ ^ F' 182 SECOND COUESE IN ALGEBEA 2. Solve grapMcally i 20^ + 2/ = 12, (1) y + 4:x = 19. (2) Solution. The graphs of (1) and (2) are the straight line and the parabola of the adjacent figure. These curves have no real points of intersec- tion. There are, however, two pairs of imaginary roots. Solving (1) and (2) by substitution, X = JJ- +V^ or ^7^- — v — 1, and y = l-2V^ or 1 + 2V-1. The essential point to be em- phasized here is that real roots of a simultane- ous system cor- respond to real intersections, and imaginary roots correspond to no intersections of real graphs. ex 3. Solve graphically | x^-f = 4, £c - 6 = 0. (1) (2) Solution. Constructing the graphs of (1) and (2), we obtain the hyperbola and the parabola of the figure on the opposite page. There are four sets of roots corresponding to the four points of intersection, which are approximately = -2.7. 1^ = 3.1. ^1^ = 1.8. ^ a: =-2.7, y=-1.8. 3.7, -3.1. If the two curves had been so chosen as to intersect only twice, their equations would have had only two sets of real roots. GRAPHS OF QUADRATIC EQUATIOKS 188 Examples 1, 2, and 3 partially illustrate the truth of the following statement: If in a system of two equations in two variables one equation is of the mth degree and one .of the nth, there are usually mn sets of roots (real or imaginary) and never more than mn such sets. r \ /^ \ ^^'^ A >■ / \ s. ^^ Y_ m — . \ ^£ / l/ ^ \ / X ' y' f X V -^ ] \ ? ^ J \ / ^ * — -2 \ V ) i!l / / -\ ^ — ■— / -4 \( 1) y Y' EXERCISES If possible, solve graphically each of the following systems 2/^ = 4 cc, x" 4- tf = 25, 2. -^ 2/ 3. 2x = 10. 4. 5. x2/= 6, 10. 11. x^2y = 17, 9. ^2^4y = 17, x-}-22/ = 12. «2 + 7/2 = 25, x^ + 2/^ = 9, a; 4- ?/ = 10. 0^2 + ^2 = 36, x^-y' = 25. 12 13 a^2 + y^=9, ^- X^ - 7/2 = 16. x?/ = 12, ^' 2a; + 2/ = 10. 0^2 = 4 7/, x'-\-y = h, if + x = Z. ic_2/V8 = 0, . x^ = 7/8 - 9 7/. 9. 184 SECOND COURSE IN ALGEBRA 104. Graphical presentation of numerical data. A great variety of statistics can be presented graphically in a very striking manner. Business and commercial houses have during the past few years used the method extensively not only to present facts but also to aid in interpreting them and in indicating their tendencies. The following exercises illustrate some of the possibili- ties in the graphical presentation of numerical data. EXAMPLE For the year 1912 the total income and expenses .per mile of line of all the railroads in the United States having a yearly revenue of one million dollars or more was as follows : Jan. Feb. Mar. Apr. May June July Aug. Sept Oct. Nov. Dec. Income in dollars . Expenses in dollars 930 730 970 710 1050 750 980 720 1040 740 1080 750 1120 760 1220 780 1210 780 1320 830 1220 810 1170 800 Delia rs > y y' V. B Nl« ^ "»^ ^ V ^ uu ^ — - ^ -^ _ -1 _ i-i PE NS E- __ _. =» — — — - — — - — — — — — — — — OO RC 2 T - - =- •— V ^ "^ ^ •ja n. F< b. M ar. A Dr. M ay Ju n. Ji 1. Ai Iff. s« p. k. N w. D ^c. - GRAPHS OF QUADRATIC EQUATIONS 185 The foregoing graph represents the given data on the same axes. The lowest curve on the diagram shows the profits of the railroads. It was obtained by plotting the differences of the numbers in the table. EXERCISES 1. The average incomes of 155 members of a certain college class for the first ten years after their graduation is given in the following table : Years after graduation 1 2 3 4 5 6 7 8 9 10 Income in dollars . . . 706 902 1199 1651 2039 2408 2382 2709 3222 3804 Graph the data. What tendencies do you note '! 2. The number of inches of rainfall during the month of July and the number of bushels of corn yielded per acre for a term of years in a certain locality is given in the following table : Year '89 '90 '91 '92 • '93 '94 '95 '96 '97 '98 '99 '00 '01 '02 Rainfall Corn yield .... 5.4 32 2.6 23 5.1 27 3.7 27 3.4 24 1.9 18 4.8 5.4 30 38 3.7 25 4.3 26 4.6 28 4.7 30 1.2 19 6.0 32 Plot these data on the same axes.. How do you account for the similarity of the curves ? 3. The numbers of hundreds of telephone calls in certain business and residential sections of New York City are given in the following table for various hours of the day. Plot both sets of data on the same axes and explain the reason for differ- ences in the shapes of the graphs. Time of day 7 8 ■9 10 11 Noon 1 2 3 4 5 6 7 8 9 10 11 12 Business district . Residence district 1 .25 2 4 35 55 73 108 76 101 65 78 52 75 48 93 47 87 40 67 38 35 38 8 36 2 32 1.5 25 1 15 1 4 1 3 RE 186 SECOND COURSE IN ALGEBRA 4. Measurements of the breadth of the heads of a thousand students in a certain school were as follows : Head breadth in inches Number of students . . ,5.5 3 5.6 12 5.7 43 6S 80 5.9 131 6.0 236 6.1 185 6,2 142 6.3 99 6.4 37 6.5 15 6.6 12 6.7 3 6.8 2 Construct a graph of the above data. 5. The chest measurements of 10,000 soldiers were tabulated as follows : Chest measure- -» ment in inches / Number of soldiers 322 38 1305 39 18G7 40 1882 41 1628 42 1148 38 48 Construct a graph of the above data. It may appear accidental that the foregoing measure- ments should group themselves with any regularity. But if the number of measurements of this type is large and each is made with care, they obey a law called the law of proba- bility. In fact the graph of the data in Exercises 4 and 5 is a close approximation to what is called the probability curve. The equation of this curve is y = e"^* when eJ = 2.7 approximately. 6. Construct the graph of the equation y = e'^"^ between the values cc = — 2 and x = 1. Hint. Let x = — 2, — |, \, 0, etc. Note. It is well established that physical characteristics, such as those illustrated by the graphs of Exercises 4 and 5, obey the law of probability. If the graph of Exercise 4 is carefully consid- ered, it may raise the question, Do mental characteristics also obey the law? An interesting aspect of this is given by the fact that an increasing number of high schools and colleges assume that such is the case, and grade, or mark, their students by a system based GEAPHS OF QUADRATIC EQUATIONS 187 on the law of probability. Such a system assumes that if one hun- dred or more students in any subject are examined, the number of students and their degree of mastery of the subject arrange them- selves according to the probability "curve shown below. Obviously between very many students the differences in grades attained will be small, between many others they will be moderate, and between only a few will they be great. In the statistical study of problems which have their origin quite remote from each other, this curve frequently occurs, and occupies a central 'position in the mathematical theory of statistics. CHAPTER XV SYSTEMS SOLVABLE BY QUADRATICS 105. Introduction. The general equation of the second degree in two variables is ax^ + hy^ + cxy -\-dx + ey -i-f = 0. To solve a pair of such equations requires the solution of an equation of the fourth degree. Even the solution of x^-{- y = 3 and ^2 _j_ ^. _ 3 requires the solution of such an equation. In fact only a limited number of systems of the second degree in two variables is solvable by quadratics. By the graphical methods of Chapter XIV the student can solve graphically for real roots any system of quadratic equations, provided the terms have numerical coefficients. The algebraic solution of such systems will in many cases be possible for him only after further study of algebra. 106. Linear and quadratic systems. Every system of equations in two variables in which one equation is linear and the other quadratic can be solved by the method of substitution. EXAMPLE Solve the svstem i , * * ^ „ ' -^ la; + 2 7/ = 13. (1) (2) Solution. Solving (2) for y in terms of x, 1^-x y 2 ' Substituting — - — for y in (1), (8) .^-..f«r)=r. (4) From (4), 2 x" _ 13 x - 7 = 0. 188 (5) SYSTEMS SOLVABLE BY QUADRATICS 189 Solving (5), a: = 7 or — ^. Substituting 7 for a: in (3), 13-7 = 3. Substituting — ^ iov x in (3), y-- 13 + Jj 2 = 6i The two sets of roots ^vQx = l,y-- = 3, and ; r =- -i,y = 6|. Check. Substituting 7 for X and 3 for y in (1) and (2), 49 - 42 = = 7, « 7+6 = = 13. Substituting - i for X and 6f for 3/ in (1) = 7, = 13. and (2), EXERCISES Solve the following systems, pair results, and check each set of roots : 1. x-\-y = 5, 2. 2x-^y = l. 3. x^ -\-xy = 15, x + y = ^. 4. x'' + 2xy = 21, x + 2y = ^. 5. x' + 2y = ll, 2x-y = 2. A ^_10:r = 6. x4-3y = 13. 8 2 ar2-a;y = 70, o» ic+-4y = 23. Q ^2 + 25?^ = 108, V • 4s-3^=6. 10. 5^-352^ = 22, 4s + 2^ = 2. 11. 5^ +. ^2 ^ 169, 2s + ?J = 22. 12. s^ + 2 ^=^ = 27, 4^-25 = 18. SECOND COURSE IN ALGEBRA 13. s^ + Sts-\-t" = 44, 2s-t = 0. 22.i + | = a, 14. s^-\-ts + t^ = 12, s-\-t = 2. 1 X 5 • 2/ 2~4* 15. 16. 2s^-ts -{-t^ = 16, 2s-t = 5. 6 4_2 s t~ S' 4:X -\- y = 6. o. ^ + 13 4.^ + ^^=28, 6- + 1 2 • 2a:2 + 3a^2/-98. ^ + 4_1 -4-^ = 2, 2/ .T 6 19. -4-^ = 2, 32/ + 2x = 2. ?/ a? 3^5 + 22/ = 5. 26 ^' + '^' + 2^^ = 40, * s-\-t + 2 = 0. on ^ 1 1 Sx-y = 2. y — X = V2, X- 2, 98 ^' + y' + 4x + 6./ = 40, 10_10^3 29^ + ^^^ = ^^ y X 2 • ^Z^ + x^zzzieo^. 107. Homogeneous equations. If both the equations of a system are quadratic, an attempt to solve it by substi- tution usually gives an equation of the fourth degree. In most cases such an equation could not be solved by fac- toring, and at the present time its solution by any other method is beyond the student. With certain types of SYSTEMS SOLVABLE BY QUADRATICS 191 systems, however, which occur more or less frequently, we can employ special devices and avoid the solution of any equation of higher degree than a quadratic. Among these systems are the so-called " homogeneous " systems. An equation is homogeneous if, on being written so that one member is zero, the terms m the other member are of tHe same degree with respect to the variables. Thus x^ + y^ = xy and x^ — oxy-{-y^ = are homogeneous equa- tions of the second degree. 2x^ -{• y^ = x'^y — 3 xy"^ is a homogeneous equation of the third degree. The system \ { ^ „ ' )► is a homogeneous system. Systems like i ^ „ „ , ' ^ are often called homogeneous 1,2 a:^ — 2/2 4- 4 a;y = 6 J systems, but strictly speaking they are not. As will be seen, the method of solving such systems is about the same as the method of solving a homogeneous system. Hence they are classed with homogeneous systems. 108. Systems having both equations quadratic. Occasion- ally when both equations are quadratic the terms which occur in the two are so related that the elimination of terms involving both variables or of the constant terms can be performed. The following system is of such type: EXAMPLES x" 1. Solve the system r; , ^2/ " 2, (1) Solution. First eliminate xy by addition (§ 47), (1) -4, 4 a:2 - 4 a:y = 8. (3) (2) + (3), 7x2 = 28. Whence j: = ± 2. 192 SECOND COURSE IN ALGEBRA Substituting + 2 for x in (1), 4 — 2 ?/ = 2, whence y = \. Substituting — 2 for x in (1), 4 + 2 ?/ = 2, whence ^/ = — 1. Therefore a: = 2, - 2, and ' ?/ = 1, — 1. These values may be checked as usual. The method of solving a system in which every term in each equation except the constant terms is of the second degree is as follows : -xy-\-'iif = ^, (1) x^-\-y' = 10. (2) 2. Solve Solution. First we combine the two equations to obtain a homo- geneous equation : (1) -5, 5 xy + 15 2/2 :== 30. (2) -3, 3 a:2 + 3 2/2 = 30. (3) - (4), - 3 x2 + 5 xy + 12 y2 = 0. Solving (5) for x in terms of y, and Substituting ^y iox x in (2), From (8), Substituting from (9) in (6), 4 x^^y, 4« 9 ?/2 + ^,2 ^ 10.- ar=±3. 16 7/2 Substituting - —^ for a; in (2), — -^ + y'^ From (10), Substituting from (11) in (7), 10. y=±^Vi0. (3) (4) (5) (6) (7) (8) (9) (10) (11) When X = 3 -3 + tVio -^VIo then 2/ = 1 - 1 -^ VlO + '^Vio" Each pair of values can be checked as usual. SYSTEMS SOLVABLE BY QUADRATICS 193 A quadratic system in which one equation is homoge- neous is easier to solve than the system of Example 2, as can be seen from what follows: 3. Solve the system |^. _^ ^ _ 5 ^ ^ 3^ ^2) Hints. Solving (1) for x, x = 2 . 2x^-^xy + if=(). '^3.^ + 2.^ = 7, 2a^-.^ + 2^ = 12, 2x -^xy = -^. 8. 2x''+xy-{-2f=^. ^if-xy = 2, ^* 2 2/=^ + 3 0^1/ = 38. 9 20^2 + 2/2 = 3a;y, x^ + ^xy = lQ. y"^ — xy = 2 x^^ ' 2x^-{-xy = 16. _ 52_3^^^4 5 x''-Sy-{-S = 0, ''']^-,y=,r y^-\-x'-25. s^ + 2st-\-4.t^ = lS, 6. Solve Example 3 completely. " t^ + S — — 2st 109. Symmetric systems. A system of equations in X and 1/ is symmetric if the system is not altered by sub- stituting X for y and 1/ for x. Thus x^ + ^2 _ g and x + y = 11 is a symmetric system, but x^ — y^ = Q and z -{- y ~ 11 is not. 194 SECOND COURSE IN ALGEBRA Certain symmetric systems or systems which are nearly symmetric can be easily solved by the method of substi- tution. Of such the following are types: 1x1/ = 4:. \x±y=2. l2a;±«/ = ll. A few other systems which are symmetric or nearly so are more easily solved by certain special methods. The following list contains typical systems, and the methods applicable are given in Exercises 1, 10, and 12. EXERCISES 1. Solve <^ ^^^ )^{ \x2j = 6. (2) Hints. These equations can be combined in such a way as to obtain definite values ior x + y and x — ?/ as follows : (2). 2, 2xy = 12. (3) {l) + (3), a;2 + 2x^ + 2/2 = 49. (4) From (4), x-\- y= ±7. (5) (l)-(3), x^-2xy + y^ = 26. (6) From (6), x-y = ±6. . (7) (5) and (7) combined give four systems of equations : rx + 2/ = 7, (8) fx-{.y=-7, (11) -^\x-y = 6. (9) \x-y = 5. (9) rx+7/ = 7, (8) fx^y=-7, (11) ^\x-2/=-6. (10) \x-?/=-5. (10) The solution of systems A^ B, C, and D is left to the student. The pairs of roots found for the four systems A, B, C, and D will check in the original system. .r*^ + ;//=:100, x'' + xy + 9f==19, ^ x^-\-4y^ = 20, xy = 48. * xy = 6. ' o-y = 4. SYSTEMS SOLVABLE BY QUADRATICS 195 9x' + t/ = ei, xy = 10. 4.x''-{-xy + f = So, xy = 12. 9. 6 x = -■ y 7 "3 4x^ + 252/^ = 41, '• xy = 2. 10. x' y' 1 = 6. xy x'-xy-\-y' = S, Hint. To clear of fractions in Exercise 10 would merely increase the difl&culty of solution. Instead we solve in a manner similar to that of Exercise 1 (see Exercise 6, p. 78). 1 = 12. xy Then 1 X2 + ^ + i = xy j/2 :25. Whence 1 X + ' = y :±5. In like manner 1 X 1_ ±1. — = 12. . xy 13. 11 13 ?>x^ 2y~ 6' 11 97 9ic2 ' 4y^"36* 11 5 ~1 + ~5 = 7' ^^ x^ ^r 4 12. 1 1 _ 1 x y~ 2 14. x^ x^^2/^ '' i = 6. ^2/ XI 12 1 Hints. — 1 = x2 xy ?/2 Then — = xy 1 "4" = 1, etc 15. x^ xy f l + i = 7. 196 SECOND COUKSE IK ALGEBRA 110. Equivalent systems. Equivalent systems of equations are systems which have tlie same set or sets of roots. If the two systems r^ ,^ = 12, (1) ^^^ r. , = 2, (8) lx + y = ^. (2) Vx-\-y = Q. (1) are solved, only one pair of roots, x — 4: and y = 2, is obtained for A and the same pair for B. Systems A and B are equivalent, though usually a system which consists of a linear and a quadratic has two sets of roots and hence cannot be equivalent to a linear system (see page 183, second paragraph). Sometimes an equation simpler than either of those given in a system can be derived from this system by dividing the left and right members of the first equation by the corre- sponding members of the second. In systems like those of the following Ust the equation so obtained taken with one of the first two gives an equivalent system more easily solved than the original one. EXERCISES Solve, using division where possible, pair results, and check each set of real roots : "" x + 2y^2. 4. 4a^^-y = 16, 2x + y = S. Hint. Division gives the equiv- 5. R^h - 75 = 0, alent system Rh = 15. x-22/ = 10, X — y = 1. 6. X y ^- 2y-x=-\l. 7. x'-¥y' = 2^, a; + 2/ = 4. SYSTEMS SOLVABLE BY QUADRATICS 197 In the following figure, (1), (2), and (3) are the graphs of x^ -{- y^ = 28, x^ — xy •\- i/ = 7, and x -\- y = ^ respectively. These equations are all used in the solution of Exercise 7. The graph makes clear in a striking way that the system (1) and (3) is equivalent to the system (2) and (3). JF 1 ^ 3)^^ S^ \ X 4^ ^ 1^ ^ "^^^ =^ S ~^ "^--^^^ ^'^.> l^^S -7 ^ J-^-A / 1 "^ / >\ X' ,: t ^^ :^ /-2 -1 U 1 2y 3 \ ^ 4- 1 i^ ^ -yr -1 ^1 ^ ^T ^ \ '3:^ 4^^ y" \ % ^---^ ^ (1^. J5 \ Y a^«-8y^ = 35, x — 2y = h. X y 1, y"^ x^ X y 152, MISCELLANEOUS EXERCISES Solve by any method and pair results. If any system can- not be solved algebraically by the methods previously given, solve it graphically. a! 4- 2/ = 4. • x-'dy = \, ' £C + «/ = 7. 198 SECOIS'D COURSE IN ALGEBRA ^^o A i« 0^^ + 2^2/ + 22/^ = 10, • 5. y "^ ' 3x'-xi/-f = 51. Q (x + yf = 9, ' x' + y = 11. • (^_3/). = 49. ^^ ^. + ^,^ + ^. = 7, ^•.^ 2aj2_^2/2^33, x2 + 2/' = 10. -^ 072 4-0-?/ + 2/ = 0, 3 ^2 _ 8 A)2 = 40, * £c2 + ici/ + a; = 0. 5 ^2 + ^^2 ^ 81 — 4- — = 13 4i2f + 3 = 9i2|, - 0)-^^/ 12i2f + i2|-^9^. 1 1 = 1. ^^' a;^ + y = 20. -^8-^=^ 0^2 + 2/2 = 169, 23. ^ ^'^ a? ?/~ * xy -\- X = 18, ajy = 60, 24^ 40.2 + 2/2 = 289, £C2/ = 60. 25 0^^ + 2/^ = 9, ' a; V = 20. • x + 2/ = 5- 27. 9x -^2/=lS = ^2/- 28.4. + J = 46 = ?|^-i. 5 Tr,2- 6.8 >F| = 99.55, TF2-Tr| = 20. _i_ + — !— = -, 20 ^-2 2/-2 4 0^2^ + 22/^ = 2, l_l = i. ^ ' 3iC2^ + 62/^ = 2, X y 12 12. x^ = y, xy = S. 13 x — xy = 5, AO. 2y-\-xy = 6. 14. x'-y' = 19, x'-\-xy-\-7/ = 19. 4 7^2 + 7 ?/i2 = 9, 15. 2 7^2 _ 9 ^ ^,,2^ SYSTEMS SOLVABLE BY QUADEATICS 199 88, x-1 34. 30. X — 2/ = 6. 31. xy = c, X -\- y = a. 32. x-^ + y-^ = 2. 33. 3, 35. 36. 37. PROBLEMS 2/-1 / + y+l _13 x^-x+1 43' 4a;^-13a;y+9V=9, ^¥ — y^ = ^' x^-\-2xy=16, 2>x'-4.xy-\-2f= 6. a-« = ^ + 37, x^y z^xy^ -\- 12. (Reject all results which do not satisfy the conditions of the problems.) 1. Find two numbers whose difference is 6 and the differ- ence of whose squares is 120. 2. The sum of two numbers is 20 and the sum of their squares is 202. Find the numbers. 3. Find two numbers whose sum plus their product is 132 and whose quotient is 3. 4. It takes k)^ rods of fence to inclose a rectangular lot whose area is one acre. Find the dimensions of the lot. 5. The area of a right triangle is 180 square feet and its hypotenuse is 41 feet. Find the legs. 6. The area of a pasture containing 15 acres is doubled by increasing its length and its breadth by 20 rods. What were the dimensions at first ? 7. The difference of the areas of two squares is 495 square feet and the difference of their perimeters is 60 feet. Find a side of each square. 8. The area of a rectangular field is 43^ acres and one diagonal is 120 rods. Find the perimeter of the field. 200 SECOND COUESE IN ALGEBRA 9. The value of a certain fraction is |-. If the fraction is squared and 44 is subtracted from both the numerator and the denominator of this result, the value of the fraction thus formed is -^^. Find the original fraction. 10. Two men together can do a piece of work in 4|- days. One man requires 4 days less than the other to do the work alone. Find the number of days each requires alone. 11. The perimeter of a rectangle is 250 feet and its area is 214 square yards. Find the length and the width. 12. The base of a triangle is 8 inches longer than its alti- tude and the area is 1^ square feet. Find the base and altitude of the triangle. 13. The volumes of two cubes differ by 316 cubic inches and their edges differ by 4 inches. Find the edge of each. 14. The perimeter of a right triangle is 80 feet and its area is 240 square feet. Find the legs and the hypotenuse. 15. The perimeter of a rectangle is 7 a and its area is a^. Find its dimensions. 16. A man travels from A to B, 30 miles, by boat and from B to C, 120 miles, by rail. The trip required 6 hours. He returned from C to B by a train running 10 miles per hour faster, and from B to A by the same boat. The return trip took 5 hours. Find the rate of the boat and of each train. 17. There were 1400 fewer reserved seats at a certain sale than of unreserved seats, and the price of the latter was 15 cents less than the price of the former. The total proceeds were |490, of which |250 came from the reserved seats. Find the number of each kind of seats and the price of each. 18. If a two-digit number be multiplied by the sum of its digits, the product is 324; and if three times the sum of its digits be added to the number, the result is expressed by the digits in reverse order. Find the number. SYSTEMS SOLVABLE BY QUADEATICS 201 19. The sum of the radii of two circles is 31 inches and the difference of their areas is 155 tt square inches. Find the radii. 20. The yearly interest on a certain sum of money is $42. If the sum were |200 more and the interest 1% less, the annual income would be |6 more. Find the principal and the rate. 21. A vheelman leaves A and travels north. At the same time a second wheelman leaves a point 3 miles east of A and travels east. One and one-third hours after starting, the short- est distance between them is 17 miles, and 3|- hours later the distance is 53 miles. Find the rate of each. 22. A starts out from P to Q at the same time B leaves Q for P. When they meet, A has gone 40 miles more than B. A then finishes the journey to Q in 2 hours and B the journey to P in 8 hours. Find the rates of A and B and the distance from P to Q. 23. A leaves P going to Q at the same time that B leaves Q on his way to P. From the time the two meet, it requires 6| hours for A to reach Q, and 15 hours for B to reach P. Find the rate of each, if the distance from P to Q is 300 miles. GEOMETRICAL PROBLEMS 1. The sides of a triangle are 6, 8, and 10. Find the altitude on the side 10. Hint. From the accompanying fig- ure we easily obtain the system (x^ + y^ = S6, \ (10 - x)2 + ?/2 = 64. 2. The sides of a triangle are 8, 15, and 17. Find the altitude of the triangle on the side 17 and the area of the triangle. 3. The sides of a triangle are 11, 13, and 20. Find the altitude on the side 20 and the area of the triangle. UE 202 SECOND COURSE IK ALGEBRA 4. The sides of a triangle are 13, 14, and 15. Find the altitude on the side 14 and the area of the triangle. 5. The sides of a triangle are 12, 17, and 25. Find the altitude on the side 12 and the area of the triangle. 6. Find correct to two decimals the altitude on the side 16 of a triangle whose sides are 16, 20, and 24 respectively. 7. The parallel sides of a trapezoid are 14 and 26 respec- tively, and the two nonparallel sides are 10 each. Find the altitude of the trapezoid. Hint. Let ABCD be a trapezoid. Draw CE parallel to DA and CF perpendicular to AB. Then EC = 10, and AE = 14, and EB = 26- 14, or 12. If we let EF = x, FB must equal 12 — x ; then we can obtain the system of equations rx2 + ?/ :::: IQO, \ (12 - X)2 + 2/2 = 100. 8. The two nonparallel sides of a trapezoid are 10 and 17 respectively, and the two bases are 9 and 30 respectively. Find the altitude of the trapezoid. 9. The bases of a trapezoid are 15 and 20 respectively, and the two nonparallel sides are 29 and 30. Find the alti- tude of the trapezoid and the area. 10. The sides of a trapezoid are 7, 9, 20, and 24. The sides 24 and 9 are the bases. Find the altitude and the area. 11. The sides of a trapezoid are 21, 27, 40, and 30. The sides 21 and 40 are parallel. Find the altitude and the area of the trapezoid. SYSTEMS SOLVABLE BY QUADEATICS 203 12. The sides of a trapezoid are 23, 85, 100, and x. The sides 23 and 100 are the bases, and each is perpendicular to the side x. Find x and the area of the trapezoid. 13. The area of a triangle is 1 square foot. The altitude on the first side is 16 inches. The sec- ond side is 14 inches longer than the third. Find the three sides. L B 14. A rectangular tank is 8 feet 6 inches long .and 6 feet 8 inches wide. A board 10 inches wide is laid diagonally on the floor. What two equa- tions must be solved to determine the length of the longest board that can be thus laid ? HiKTS. Let BR the triangle AKL. X and DE = y. The triangle DKE is similar to CHAPTER XVI PROGRESSIONS 111. A sequence of numbers. In all fields of mathe- matics we frequently encounter groups of three or more numbers, selected according to some law and arranged in a definite order, whose relations to each other and to other numbers we wish to study. There is an unlimited variety of such groups, or suc- cessions, of numbers. Only two simple types will be considered here. 112. Arithmetical progression. An arithmetical progression is a succession of terms in which each term after the first is formed by adding the same number to the preceding one. Thus, if a denotes the first term and d the common number added, any arithmetical progression is represented by a, a -h d, a -{• 2 d, a + ^ d, a + 4: d, ' • . This common number d is called the common difference and may be any number, positive or 7ieijative, It may be found for any given arithmetical progression by subtract- ing any term from the term which follows it. The numbers 3, 7, 11, 15, • • • form an arithmetical progression, Muce any term, after the first, minus the preceding one gives 4. Simi- larly, 12, 6, 0, — 6, — 12, • • • is an arithmetical progression, since any term, after the first, minus the preceding one gives the com- mon difference — 6. In like manner 5, 5, 6^, • • • is an arithmetical progression whose common difference is 1^. 204 PKOGEESSIONS 205 ORAL EXERCISES State the first four terms of the arithmetical progression if : 1. a = 2,d = 3. 6. a = 100, d = -10. 2. a = 5, d = 4.. 7. a = 20x, d = — 2x. 3. a = 10, d= 6. S. a =x, d = 2x. ^. a = 20, d = 5. 9. a = 17 m, d = — 2 m. 5. a = 18, ^--3. 10. a=V5,d = l-\-V5. From the following select the arithmetical progressions, and in each arithmetical progression find the common difference : 11.1,10,19,.... 16. 8,9f,10|,.... 12. 4, 12,36,.... 17. 15,3, -12,.-.. 13. 19, 11, 3, . . .. 18. 6a, 10a, 14«, • • .. 14. 9, 12, 1, 16, . . ..' 19. 18 a, 14 a, 12 a, • . .. 15. 3, ^, -V-, .... 20. 5 a, 5 a + 3, 5 a + 6, . . .. 21. Sx, X — 2, —X — 4:, • . .. 22. 2x-l, X, 1, 2-cc, ...."^ ^' 23. — 5 Va, — 2 Va, Va, 4 Va, .... 24. 5V^-1, 4V^-2, 3V^-3, .... 113. The last or Trth term of an arithmetical progression. In the arithmetical progression a, a + d, a + 2 d, a + 3 (f, a + 4 ff, • . . one observes that the coefficient of d in each term is 1 less than the number of the term. Hence the ?ith, or general, term is a -\- (n — 1) d. li I denotes the nth term, we have l=a+(n-i)d. 206 SECOND COURSE IN ALGEBRA EXERCISES Eind the required terms of the following arithmetical progressions : 1. The fifteenth term of 2, 7, 12, • • •. Solution. Here a = 2, c? = 5, and n = 15. Hence I = a -{■ (n —lyd becomes Z = 2 + 14 • 5 = 72. 2. The eighth term of 1, 4, 7, • • •. 3. The eleventh term of 15, 9, 3, - 3, • • .. 4. The twenty -first term of a, 4 a, 7 <7, • • •. 5. The sixteenth term of Qcc, jr, — 7ic, • • •. 6. The sixth term of ^^-, f , |, • • •• 7. The eleventh term of — y, |-, 1, • • • . 8. The sixth term of V3, 4 V3, 7 Vs, • • •. 9. The eighth term of 3 Vs, V5, - V5, - 3 Vs, .... 10. The eighth term of 5 + a, 7+ 3 a, 9 + 5 «, . . -. 11. The twelfth and the twentieth terms of 0, — 5x + 2, 4 -10a;, .... 12. The tenth term of V^ + 3, 5 Va +7, 9 Va + 11, . . .. 13. The tenth and the twentieth terms of 9 Va — 3, 5 Va, Va 4- 3, .... ■yj^ 5 "v £C 9 "v £C 14. The fourteenth term of -—> — - — > — r— 5 ' ' '• z z z 15. The ninth and the twelfth terms of V3, — 7=? 3 Vs, . . ..' V3 V iC 3 V cc 16. The eighth and the sixteenth terms of -jr- 4- 1, —75 h 2, 5 Vcc + 3, 17. The fifteenth term of ^^ - 12, - 5, ^^^ ^ ""^ PROGRESSIONS 207 18. The twenty-ninth term of a, a -\- d, a -\- 2 d, - - -. 19. The mth term of a, a -^ d, a -\- 2 d, • - • . 20. The (m — l)th term ot a, a -\- d, a -{- 2d, • • •. 21. The (n - 2)th term oi a, a -\- 5, a -{-10, ". o _ 22. The (n - 5)th term of —^, 3 V3, 5 Vs, • • -. 23. rindthe(7i-3)thtermofV5-l,2V5-2,3(V5-l), .... 24. Find the nth term of -> ? 2 — -? • • •• a a a 25. The first and third terms of an arithmetical progression are 2 and 22. Find the seventh term ; the nth term. 26. The first and second terms of an arithmetical progres- sion are r and s respectively. Find the third term and the nth. term. 27. The edges of a box are consecutive even integers with n the least. Express in terms of n, (a) the sum of the edges ; (b) the area of the faces ; (c) the volume. 28. A body falls 16 feet the first second, 48 feet the next, 80 feet the next, and so on. How far does it fall during the twelfth second ? during the nth second ? 29. The digits of a certain three-digit number are in arith- metical progression. If their sum is 24 and the sum of their squares is 194, find the number. 114. Arithmetical means. The arithmetical means between two numbers are numbers which form, with the two given ones as the first and the last term, an arithmetical progression. The insertion of one or more arithmetical means between two numbers is performed as in the following examples: 208 SECOND COURSE IN ALGEBEA EXABIPLES 1. Insert four arithmetical means between 7 and 72. Solution. 1= a + (n — l)d. There will be six terms in all. Therefore 72 = 7 + (6 ~ 1) d. Solving, ^ = 13. The required arithmetical progression is 7, 20, 33, 46, 59, 72. 2. Insert one arithmetical mean between h and k. Solution. I = a -\- (n — 1) d. There will be three terms in all. Therefore k = h + 2d. Solving, d = —^' Therefore the progression is h, h -\ — , k, or 7i, — -— , k. It follows from the above that the arithmetical mean between two numbers is one half of their sum. EXERCISES Insert one arithmetical mean between : 1. 3 and 15. 2. 3 and 27. 3. h and 5 h. 4. m and n. Insert two arithmetical means between : 5. 2 and 8. 7. 49 and 217. 9. x and 3 //. 6. 5 and 29. 8. x and x -{- 6 y. 10. h and k. Insert three arithmetical means between : 11. 10 and 34. 13. - 12a; and Ux. 12.-5 and 31. 14. h and k. 15. Insert seven arithmetical means between —13 and 131. 16. Insert five arithmetical means between — | and ^^. PROGRESSIONS 209 17. Insert four arithmetical means between — 2 Vs and 18 V5. 18. Insert five arithmetical means between 7 a — Sb and 13 a + 9&. /- 27 19. Insert six arithmetical means between VS and p=- 2V3 3 20. Insert two arithmetical means between 7= and 4V3-6. 1-^ 21. In going a distance of 1 mile an engine increased its speed uniformly from 15 miles per hour to 25 miles per hour. What was the average velocity in miles per hour during that time ? How long did it require to run the mile ? 22. At the beginning of the third second the velocity of a falling body is 64 feet per second, and at the end of the third second it is 96 feet per second. What is its average velocity in feet per second during the third second ? How many feet does it fall during the third second ? 23. The velocity of a body falling from rest increases uni- formly and is 32 feet per second at the end of the first second. What is the average velocity in feet per second during the first second ? How many feet does the body fall during the first second ? the second second ? 24. Find the average length of twenty lines whose lengths in inches are the first twenty odd numbers. 25. Find the average length of fifteen lines whose lengths in inches are given by the consecutive even numbers beginning with 58. 26. With the conditions of Problem 23 determine the average velocity per second of a body which has fallen for 12 seconds. 27. A certain distance is separated into ten intervals, the lengths of which are in arithmetical progression. If the shortest interval is 1 inch and the longest 37 inches, find the others. 210 SECOND COUKSE IN ALGEBEA 115. Sum of a series. The indicated sum of several terms of an arithmetical progression is called an arith- metical series. The formula for the sum of n terms of an arithmetical series may be obtained as follows: S = a+{a + d^+(a + ^d)+ . . . J^(l-2d)-^(l-d)+l. (1) Reversing the order of the terms in the second member of (1), S = l+(l-d)+(l-2d^+ . . .+(a + 2c?)4-(a + (^)-h«. (2) Adding (1) and (2), 2^ = (a + Z) + (a + + (^ + 0+---+(^ + + (« + + (a + Z) = n(6)^ + Z). Therefore S= -(« + /). EXAMPLE Required the sum of the integers from 7 to 92, inclusive. Solution, n = 86, a = 7, Z = 92. Substituting in 5 = - (a + Z), Therefore the sum of the integers from 7 to 92 is 4257. ' ORAL EXERCISES Using the formula S = -(a-\-l)j find the sum of the following : 1. The six-term series in which a = 2 and I =17. 2. The ten-term series in which a =1 and I = 46. 3. The twelve-term series in which a = — 12 and I = 21. 4. The seven-term series in which a = S and I = 63. 5. The twenty-term series in which a = 5 and I = 86. 6. The twelve-term series in which a = — 175 and I =125. progeessio:n^s 211 Yt The formula S = -(a + l) enables one to find the sum of a series if the first term, the last term, and the number of the terms are given. If, however, the last term is not given, but instead the common difference is given or evident, we can use a formula obtained by substituting m. S = -(a -\- V) the value of I from the formula l = a-\-{n — 1) c?, on page 205. VI Then s = j.\a -^ a -^ (n -V)d\ or S = -[2a + (n-l)tf]. JL EXAMPLE Required the sum of the first fifty-nine terms of the progression 2,9, 16,-.. . Solution, n = 59, a = 2, and d = l. Substituting in 5 = ^ [2 a + (/i - 1) rf], 5- = ^ (4 + 58 • 7) = ^ (410) = 12,095. EXERCISES Find the sum of the following : 1. The first eight terms of the series in which a = 2 and c? = 4. 2. The first ten terms of the series in which a. = 8 and d = 5. 3. The first nine terms of the series in which a = 1 and d = ll. 4. The first -fifteen terms of the series in which a = 17 and d = S. 5. The first twenty terms of the series in which a = 100 and d=-5. 212 SECOND COURSE IN ALGEBRA 6. The first eight terms of the series 2 + 4 4- 6 + • • • . 7. The first ten terms of the series 1 + 9 -h 17 + • • •. 8. The first ten terms of the series — 8 + (— 4) + + • • • . 9. The first eighteen terms of the series 1 + 5 + 9 + • • •. 10. The first twenty terms of the series 10 + 8 + 6 -f • • -. 11. The first twelve terms of 1 + | + 2, • • .. 12. The first twelve terms of 15 + 12i + 10 H . 13. The first one hundred integers. 14. The first one hundred even numbers. 15. The first one hundred odd numbers. 16. Show that the sum of the first n even numbers is n{n + 1). 17. Show that the sum of the first n odd numbers is n^. 18. Find the sum of the even numbers between 247 and 539. 19. How many of the positive integers beginning with 1 are required to make their sum 861 ? Hint. Substitute in the formula *S = - [2 a + (n — 1) d] and solve for n. 20. How many terms must constitute the series 7 + 10+13 + ... in order that its sum may be 242 ? 21. Beginning with 90 in the progression 78, 80, 82, • • ., how many terms are required to give a sum of 372 ? 22. The second term of an arithmetical progression is — 2 and the eighth term is 22. Find the eleventh term. 23. Find the sum of the first t terras of -> > -^ • • •• a a a 24. If Z = 25, a = 1, and <^ = 4, find n and s. 25. If a = - 20, 6^ = 11, and s = 216, find n and I. 26. It d = — 9, n = 15, and s = 0, find a and I. PEOGKESSIOiS^S 213 27. The first and second terms of an arithmetical progres- sion are h and k respectively. Find the sum of n terms. 28. If 5 = 9 A, a = 12 - 10 h, and ti = 9, find I and d. 29. If s = 66 V3, a = - 4 V3, and ^Z = 2 Vs, find n and I. 30. A clock strikes the hours but not the half hours. How many times does it strike in a day ? 31. A car running 15 miles an hour is started up an incline, which decreases its velocity ^ of a foot per second. («) In how many seconds will it stop ? (J)) How far will it go up the incline ? 32. A car starts down a grade and moves 3 inches the first second, 11 inches the second second, 19 inches the third second, and so on. (a) How fast does it move in feet per second at the end of the thirtieth second ? (h) How far has it moved in the thirty seconds ? 33. An elastic ball falls from a height of 24 inches. On each rebound it comes to a point \ inch below the height reached the time before. How often will it drop before coming to rest ? Find the total distance through which it has moved. 34. The digits of a three-digit nmnber are in arithmetical progression. The first digit is 3 and the number is 20^ times the sum of its digits. Find the number. 35. A clerk received $60 a month for the first year and a yearly increase of |75 for the next nine years. Find his salary for the tenth year and the total amount received. 36. If a man saves |200 a year and at the end of each year places this sum at simple interest at 6^, what will be the amount of his savings at the time of the sixth annual deposit ? 37. Assuming that a ball is not retarded by the air, deter- mine the number of seconds it will take to reach the ground if 214 SECOND COURSE IN ALGEBEA dropped from the top of the Washington Monument, which is 555 feet high. With what velocity will it strike the ground ? Hint. See Exercise 28, p. 207. 38. A ball thrown vertically upward rose to a height of 256 feet. In how many seconds did it begin to fall ? With what velocity was it thrown ? 39. By Exercise 28, p. 207, it is seen that a falling body obeys the law of an arithmetical progression. Show from the data of that exercise that the general formula S = -(2 a -\- (n — 1) d) at'' becomes the special one S = ^y which is used in physics for all such problems. 40. A ball thrown vertically upward returned to the ground 6 seconds later. How high did it rise ? With what velocity was it thrown ? 41. A and B start from the same place at the same time and travel in the same direction. A travels 20 miles per hour. B goes 30 miles the first hour, 26 miles the second, 22 miles the third, and so on. When are they together ? Note. In the earliest mathematical work known a problem is found which involves the idea of an arithmetical progression. In the papyrus of the Egyptian priest Ahmes, who lived nearly two thousand years before Christ, we read in essence, " Divide 40 loaves among 5 persons so that the numbers of loaves that they receive shall form an arithmetical progression, and so that the two who receive the least bread shall together have one seventh as much as the others." From that time to this, the subject has been a favorite one with mathematical writers, and has been extended so widely that it would require many volumes to record all of the discoveries regard- ing the various kinds of series. 116. Geometrical progression. A geometrical progression is a succession of terms in which each term after the first is formed by multiplying the preceding one always by the same number. PEOGRESSIONS 215 Thus, if a denotes the first term and r the common multiplier, then any geometrical progression is represented by a, ar, ar^, ar^, ar\ • • • . The common multiplier is called the ratio. It is evident from the above that the ratio r in a geometrical progression is found by dividing any term by the preceding one. The numbers 2, 10, 50, 250, • • • form a geometrical progression, since any term, after the first, divided by the preceding one gives the same number 5. Similarly, the numbers 3,-3 V2, 6,-6 V2, • • • form a geometrical progression, since any term, after the first, divided by the preceding one gives the common ratio — V2. ORAL EXERCISES Determine which of the following are geometrical progres- sions and state the ratio in each case : 1. 1, 3, 9, 27, .... 7. i Ve, VO, V54, .... 2. 2,4,16,.... 8. Vf, -V|,2,.... 3 2, 6, 18,.... 9. 7a,S5a, 175a,.... 4. 5, 1, i, •••• 10. 8V5, -2V5, V5, .... 5. 18, - 3, i . . .. U. 5x% lOa^^ 20a:^ • • .. 6. 2, 1 V2, i V2, . . .. 12. 3y\ 12y, ^Sy\ . . .. 13. Find the condition under which a, b, and c form a geometrical progression. State in order the first four terms of the geometrical progression in which the first term is 14. 1 and the ratio is 4. 17. 64 and the ratio is J. 15. 3 and the ratio is 10. 18. — 243 and the ratio is ^. 16. — 3 and the ratio is 2. 19. 2 and the ratio is V3. 216 SECOND COURSE IN ALGEBRA 117. The nth term of a geometrical progression. Since by definition any geometrical progression is represented by a, ar, af", ar\ - - -, it is evident that the exponent of r in any term is 1 Iqss than the number of the term. Therefore, if t„ denotes the nth, or general, term of any geometrical progression, EXERCISES 1. Find the fifth term of 4, 12, 36, • • -. Solution. Here a = 4, r = 3, n — 1 = 4. Substituting these values in the formula ty^ = ar^^-'^, <^ = 4.3* = 324. 2. Find the eighth term of 1, 2, 4, • • .. 3. Find the tenth term of 3, 6, 12, • • -. 4. Find the eighth term of 3, 2, |-, • • •. 5. Find the twelfth term of 7, - 14, 28, • • -. 6. Find ^^ of 15, -5, +|, .... 27 a^ 7. Find t^ of 12 a", 9 a^ -^, ••-. — 2 c* — 5 8. Find t^ of — g— , - 1, Y^» . . •• 9. Find fg of 4 V2, 4, 2 V2, . • -. 1 1 V2 10. Find t„ of :> t:' ^' 2V2 2 -2 11. Write the twentieth term of $4.12(1.01), $4.12(1.01)^, H.12(1.01)«, .... PROGRESSIONS 217 12. Since the nth. term of a geometrical progression is ar'^~^, what is the (n - l)th term ? the (n - 2)th ? the (n - 3)th ? the (n + l)th ? the (n + 2)th ? 13. The first and second terms of a geometrical progression are m and ti respectively. Find the next two terms. 118. Geometrical means. Geometrical means between two numbers are numbers which form, with the two given ones as the first and the last term, a geometrical progression. Thus ar and ar^ are the geometrical means in a, ar, ar^, a?-^. EXAMPLES 1. Insert two real geometrical means between 9 and 72. Solution. There are four terms in the geometrical progression, and a = 9, n = 4, and tn = t^= 72. Substituting these values in t^ = ar*^-^, 72 = 9 r3. Whence r = 2, and — 1 ± V— 3. The required geometrical progression is 9, 18, 36, 72. 2. Insert one geometrical mean between h and k. Solution. There are three terms in the progression, and a = //, w = 3, and t„ = k. Substituting these values in tn = ar"-i, we have h — Ji-r^. Solving, r=±A/-- Hence the progression is ^, ± ^-y/-, ^, or A, ± VA^, k. It follows from the above that the geometrical mean between two numbers is the square root of their product. K£ 218 SECOND COUESE IN ALGEBKA ORAL EXERCISES Insert one geometrical mean between : 1. 1 and a^. 3. 1 and 4ic^ 5. a and a*. 2. 1 and x^. 4.3 and 75. 6. a^ and d^. Insert two geometrical means between : 7. 1 and icl 10. 1 and 125. 13. x"- and x^. 8. 1 and 2«. * 11. a and a\ 14. 2 and 16. 9. 1 and 27. 12. x and x\ 15. 5 and 40. EXERCISES Obtain progressions involving real terms only : 1. Insert two geometrical means between 21 and 168. 2. Insert two geometrical means between 15 and 405. 3. Insert three geometrical means between 3 and 243. 4. Insert one geometrical mean between 9 and 81. 5. Insert one geometrical mean between aP' and a*. 6. Insert three geometrical means between — 9 and — 144. 7. The fourth term of a geometrical progression is 16, the eighth term is 256. Find the tenth term. 8. The second term of a geometrical progression is 4 Vs, the fifth term is ^. Find the first term and the ratio. 9. Show that the geometrical means between a and«c are ± Vac. 10. The first and fourth terms of a geometrical progression are a and c respectively. Find the second and thii-d terms. 11. Insert three geometrical means between h and k. 12. The sum of the first and fourth terms of a geometrical progression is 56. The second term is 6. Find the four terms. PROGRESSIONS 219 B D 10, find BD and DC. 13. In the accompanying figure, ABC is a right triangle and AD i?, perpendicular to tl^e hypotenuse BC. Under these conditions the length oi AD is always a geometrical mean between the lengths of BD and DC. {a) If BD = 4: and DC = 9, find AD. (b) If BD = S and BC = 21, find AD. (c) If BC = 25 and AD 14. In the accompanying fig- ure, AB touches and AD cuts the circle. Under such conditions the length of AB is always a geometrical mean between the lengths of AC and AD. (a) If AD = 36 and ^C = 4, -find AB. (b) If DC = 90 and AB = 24, find AC and AD. 119. Geometrical series. The indicated sum of n terms of a geometrical progression is called a geometrical series. The process of obtaining in its simplest form the. expression for this sum is often called finding the sum of the series. The expression for the sum is derived as follows: Let Sji denote the indicated sum. S = a-\- ar :{- ar^ -h • ar' -\-ar"-'-\-ar" = a(l-\-r + r^ + 1 _^^«-3_^^r^-2^^n-l^^^ S„ = a d (1) (2) 1-r since the polynomial in (1) is equal to the fraction in the parenthesis of (2) (see Exercise 18, p. 39). a — ar" Hence s„ = 1-r 220 SECOND COURSE IN ALGEBRA EXERCISES 1. Find the suni of the first nine terms of 3, — 6, 12, • • •• „ , , . ,^ a — ar^ Solution. Sn = 1 — r By the conditions, a = 3, r =— 2, and n = 9. Substituting, __3-_H-^ l-(-2) 2. Find the sum of 1, 5, 25, • • • to nine terms. 3. Find S^for 2,-4, 8, • • •. 4. Find S^ for 40, 20, 10, ••.. 5. Find ^g for - 180, 90, - 45, . • •. 6. Find 5^ for I, 1,1-,.... 7. Find S^^ for a% a\ a\.". 8. Fina S^ for 2 Vs, 6, 6 Vs, • • -. 9. Find >S^ for 4, 12, 36, • • .. 10. Find 5„ for 125, - 25 Vs, 25, .... 11. Find >s;,_i for 3, 12, 48, . . •. 12. Find S,, for 3, - 15, 75, .... 13. Find »S„,_2 for x, 4:x\ 16x',- - -. 14. Show that for a geometrical progression S,i = — ;-• Hint. Substitute I for the factor rtr"-i in the last term of the luimerator in the formula of Exercise 1. 15. A rubber ball falls from a height of 60 inches and on each rebound rises 60% of the previous height. How far does it fall on its sixth descent? Through what distance has it moved at the end of the sixth descent ? 120. Infinite geometrical series. If the number of terms of a geometrical series is unlimited, it is called an infinite geometrical series. PEOGEESSIONS 221 In the progression 2, 4, 8, • • • the ratio is positive and greater than 1, and each term is greater than the term preceding it. Such a progression is said to be increasing. Obviously the sum of an unlimited number of terms in such an increasing geometrical progression is unlimited. If r >1 (read "r is greater than 1"), the sum can be made as large as we please by taking enough terms. In the progression 2, 1, i, i, i, Jg, • • . the ratio is positive and less than 1, and each term is less than the term preceding it. Such a progression is said to be decreasing. Though the number of terms of such a geometrical pro- gression be unlimited, the sum of as many terms as we choose to take is always less than some definite number. The sum of the first three terms of the series 2 + 1+^+:^+^ + • • • is 3^ ; of four terms, is 3| ; of five terms, is 3| ; of six terms, is 3^f ; of seven terms, is 3|^, etc. Here for any number of terms the sum is always less than 4. The limit of the sum of the series 2 + l4-i+i+i+t's"+*" can be seen by reference to the following diagram : D E^FGX i— -r-i Here the terms of the above series are represented by intervals on the line AXj which is four units in length; that is, AB = 2, BC — 1, CD = I, DE= 1, EF= I, FG = ^V etc. A study of the several intervals reveals the fact that as we pass from A toward X each new interval is just one half of the then unused portion of AX. It follows, then, that the distance laid off gradually approaches, though it never equals, AX, for at no time is an interval adjoined which is more than one half of that which then remained. Therefore there will always be an unused interval between the extremity of the last interval used and the point X. Since the sum of the intervals laid off approaches but never quite equals the interval AX, we have in the above diagram an illustration of the fact that the sum of the series 2+l+i + ;|^ + ^ + TV+ ••• approaches, yet always remains less than, 4. 222 SECOND COURSE IN ALGEBRA a — ar^ The formula for the sum of a geometric series S^ — may be written as the difference of two fractions, 1 — r 1 — r For the series 2 + 1 + J + | + • • • it follows that Now (1)2=1, (1)8=1, (-1-)".= ^^, .(J/ = 3:V Conse- quently (|)^ becomes very small if n is taken very great. Therefore 2(|)^, the numerator of the last fraction in (3), decreases and approaches zero as n increases without limit. Since the denominator of the fraction remains ^ while the numerator approaches zero, the value of the fraction decreases and approaches zero as n increases. Then if >S^oo denotes S^ where n has increased without limit, we may write 9 Soo approaches » or 4. 2 In general, if r < !('' if r is numerically less than 1"), the numerical value of fraction approaches zero as n 1 — r increases without limit. Under such conditions the formula S,, = -z becomes aSoo = -z • 1— rl— r \ — r This means that for r numerically less than 1, S^ approaches •> but for any definite value of n it is always numerically less than this number. Hence whenever we speak of the sum of such a series we mean the limit which the sum approaches as n increases indefinitely. PROGRESSIONS 223 EXERCISES Find the sum of the following series : 8. 5 + V54-I4- 1. 4 + 1 + 1 + . . Solution. So Substituting, So i-i 9- l + ; + 3 + (x>l) 2. l + ^ + i + .... 3. 3 + (-l) + i + ---. 4. 5-|-(-2)+A + .... 5. 2 + V2 + H . _ 5a 5a 6. 5a + — + — +.... 7. l + x + a;2+ .... (x] (1 + r) - ;. } (1 + r) - p. (3) The sum due at beginning of fifth year is [{['^(1 + -i^] (1 + ^) -i>} a + -i>] a + r) -p. (4) By the conditions of the problem, (4) = 0, for all the debt has then been paid. Setting (4) equal to zero and simpUfying, S(l + ry-p{l + ry-p(l-{-ry-p{l+r)-p = 0. (5) Solving (5) ior p, S(l + ry (1 + ry + (1 + ry + (1 + r) + 1 r But the denominator in (6) is a geometrical series whose sum by the formula *S„, = is -'^ ^ ■- 1— r . r Substituting this last value for the denominator of (6), ^ (1 + ry-l ^^ In the general case, if we have n annual payments, the exponent Sr(l + r^'^ 4 in (7) would be replaced by n, and then p = -——^ ^ — p= .. ,„.s,.rv:;^ , - w 226 SECOND COUESE IN ALGEBRA 9. A loan of flOOO is to be repaid in three equal annual payments, interest at 6%. Find the payment required. 10. A loan of |2000 bearing interest at 5% is to be repaid in five equal annual payments. Find the payment required. 11. The machinery of a certain mill cost |10,000. The owner figures that the machinery depreciates 10% in value each succes- sive year. What was the estimate on the value of the machinery at the end of the sixth year ? 12. A vessel containing wine was emptied of one third of its contents and then filled with water. This was done four times. What portion of the original contents was then in the vessel ? (J 13. In the adjacent figure the tri- angle DEF is formed by joining the mid-points of AB, BC, and CA respec- tively. Triangles GHI and KLAI are formed in like manner. If ABC is equilateral, prove that the successive perimeters of the triangles form a geometrical progression, li AB = 5, find the sum of the perimeters of all the triangles which may be so formed. 14. Each square in the adjacent fig- ure except the first is formed by join- ing the mid-points of the ftquare next larger. It AB = 4:, show that the perim- eters of the squares form a decreasing geometrical progression. Find the sum of the perimeters of all the squares which may be so drawn? CHAPTER XVII THE BINOMIAL THEOREM 121. Powers of binomials. The following identities are easily obtained by actual multiplication : (^ j^hy = d^-\-2ah + h\ (1) (a Jrhy = a^ + Z a% + Zal^ + lA (2) (a 4- 6)4 = ^4 + 4 a% + 6 a.262 + 4 aS^ _^ h\ (3) (a + 6)5 = ^5 _^ 5 ^46 +10 ^352 + 10 ^263 _^ 5 a64 + 65. (4) If a -f- 6 is replaced by a — h, the even-numbered terms in each of the preceding expressions will then be negative and the odd-numbered terms will be positive. 122. The expansion of (a+by. The form of the expan- sion for the general case will now be mdicated : The first term is a" and the last is 6". The second term is -h na"~^b. The exponents of a decrease hy 1 in each term after the first. The exponents of b increase hy 1 in each term after the second. The product of the coefficient in any term and the 'exponent of a in that term., divided hy the exponent of b increased hy i, gives the coefficient of the next term. The sign of each term- is -\- if a and b are positive ; the sign of each even-numhered term is — if b alone is negative, 227 228 SECOND COURSE IN ALGEBRA According to the above statement we have (a + b)" = a" + 7 a"-^b+ "^f ~ ^ a"-^bi' 1 • 1*2 n(n-l)(n-2) ■^ 1.2.3 ''""^ + . . . + &^. This expresses in symbols the law known as the binomial theorem. The theorem holds for all positive values of n and also, with certain limitations, for negative values. Note. The coefficients of the various terms in the binomial expansion are displayed in a most elegant form as follows : 1 1 1 12 1 13 3 1 14 6 4 1 In this arrangement each row may be derived from the one above it by observing that each number is equal to the sum of the two num- bers, one to the right and the other to the left of it, in the line above. Thus 4=1 + 3, 6 = 3 + 3, etc. The next line is 1 5 10 10 5 1. The successive lines of this table give the coefficients for the expan- sions of (a + &)"■ for the various values of n. The numbers in the last line of the triangle are seen to be the coefficients when n = 4 ; the next line would give those for n = 5. This array is known as Pascal's triangle, and was published in 1665. Tt was probably known to Tartaglia nearly a hundred years before its discovery by Pascal. ORAL EXERCISES What is the second term in the expansions of Exercises 1-4 ? 1. (a + by\ 2. (a-b)"". 3. (a-\-by\ 4. (a - bfK Assuming that the terms in Exercises 5-10 occm- in an ex- pansion of the binomial a -\- b, find (a) the exponents, (b) the coefficient in the next following term of the expansion. 5. 10a%\ 7. 15a*b\ 9. 252 a^b^. 6. Sab\ 8. 11 . 5 . 9a'b\ 10. 20 a^b\ THE BINOMIAL THEOKEM 229 EXERCISES Expand by the rule : 1. (a + b)\ 3. (a + 1)'. 5. {a + 3)«. 2. {a - If. 4. (a + 2)^ 6. (2 - af. Obtain the first four terms of the following : 7. (a + by\ 8. (a + b)"". 9. (a + 1)^. 10. (a - 2)2^. Expand : 11. (a^ + 2by. Hints. To avoid confusion of exponents first write (a2)& + (a2)4 (2 6)1 + (^2)3 (2 6)2 + (a2)2 (2 6)3 + (a2)i (2 6)* + (2 6)5. Then in the spaces left for them put in the coefficients according to the rule of this section. Finally, expand and simplify each term. 12. («.^ + 2)«. 13. (a^-24)'. 14- (^ J- 15. (^)'. Obtain in simplest form the first four terms of the following ; 16. (a^ + 3 b)"". /a^ 2^Y 17. (a'^-Sby. ' ^^' ^*/ 19. (t + -^) • 23. [^ + ^J. 20. (^-^)- 24.(1 + -J ■■&-sf' -('-;: 25. Write the first six terms of the expansion ol (a + by\ and evaluate it for n = 1, n = 2, n = S, n = 4:. How does the number of terms compare with n ? What is the value of each coefficient after the (t^ -h l)th ? Why does not the expansion extend to more than five terms when n = 4^? 230 SECOND COURSE IN ALGEBRA Compute the following to two decimal places. (In each case carry the computation far enough to be certain that the terms neglected do not affect the second decimal place.) 26. (l.l)^*'. 29. (.98)". Hint. (1.1)1° z= (1 + .l)io etc. Hint. (.98)" = (1 - .02)" etc. 27. (5.2)«. 30. (4.9)^ 28. (1.06)«. 31. (2.9)1 123. Extraction of roots by use of the binomial expansion. The expansion in section 122 may be verified for any par- ticular integral value of n without difficulty by direct mul- tiplication, as in section 121. But if n has a negative or fractional value, a laborious proof is required to show that the expansion is still valid when a is numerically greater than h. Since none of the factors of the coefficients, as n, n — 1, n — 2, vanish for fractional or negative values of n^ it appears that for such exponents the expansion is an infinite series. For example, (a + h)^ = a^ + \arh -\'a~h'^ + -^^ . a" V+ • • .. By giving n the values |- or J, one can compute the square root or the cube root of a number to any required degree of accuracy. In such computations it is desirable to let the number which corresponds to a in the binomial exceed the one corresponding to h by as much as possible and at the same time to have a** an integer. Note. 'The process of extracting the square root and even the cube root by means of the binomial expansion was familiar to the Hindus more than a thousand years ago. The German Stifel (1486- 1567) stated the binomial theorem for all powers up to the seven- teenth, and also extracted roots of numbers by this method. THE BINOMIAL THEOEEM 231 EXERCISES Eind the first four terms of the following : 1. (l + x)\ 3. (S-x)i. 5. (2 + x)^. 2. (2-\-x)K 4. (l + x)k 6. (S-x)^. Eind to at least three decimals by the binomial theorem : 7. (27)i Solution. (27)^ = (25 + 2)^ = 25^+ i.25-2. 2-^-25-t.22 + V^.25-^.23 = 5 + .2 - .004 + .00016 . • • = 5.196 + . It is proved in more advanced books that when the terms of an infinite series are alternately plus and minus, and each term is numerically less than the preceding one, the value of the entire sum from a given term on cannot exceed that term. This fact renders these so-called " alternating series " especially convenient for com- putation, since a definite limit of error is known at each stage of the computation. In this example the error cannot exceed .00016. 8. (lT)k 9. (28)1 10. (38)i 11. (78)^ 12. (125)i 13. (61)i Solution. (61)^ = (64 - 3)i = 64i - J- • 64- 1 . 3 - ^ . 64- ^ . 3^ -/r-64-t.33-f,-. = ^ ~ T^?r — tttV? — t-^At^w + . . . = 4 - .0625 - .00097 - .00000259 • • • = 3.93653 - . Here three terms give the result to five figures. 14. (79)i Hint. (79)^ =: (81 - 2)i = 81^- -| • 81"^ • 2+ . • .. Here (81 — 2)^ yields more accurate results with fewer terms than does (64 + 15)i. 15. (28)i 16. (66)i 17. (30)i 18. (700)i 232 SECOND COUESE IN ALGEBRA 124. The factorial notation. The notation 5 ! or [5 sig- nifies 1 . 2 . 3 . 4 . 5, or 120. Also 4! =1 . 2 . 3 • 4 = 24. In general, t^ ! = 1 • 2 • 3 • 4 . . • (w — 2) (/i — 1) n. The symbol nl or [w is read "factorial ti." In the factorial notation the denominators of the fourth and fifth terms of the expansion of (a + by become 3 1 and 4 I respec- tively (see formula, p. 228). EXERCISES Evaluate : 1. 6!. 3. 5!. 2!. 5. 4! -3! • 2!. 2. 4!. 3!. 4. 6!--2!. 6. ti! ^(/i -1)!- ^ , ^ n(n-l)(n-2) -.- (n-r-^2) ^ Evaluate -^ ^ — , \^, — ^' when: (r-l)l 7. n=7,r = 5. 10. n = 20, r =15. 8. 71=15, r = S. 11. n=lS,r=17. 9. 71 = 21, r = 12. 12. n = 10, r = 11. 125. The rth term of (a + &)". According to the bmomial theorem the fifth term of the expansion on page 228 is n(n-l)(n- 2) (n - S)a''-^^ . "^ 4! • If we note carefully this term and the directions on page 227, we can write down, from the considerations that follow, any required term without writing other terms of the expansion. The denominator of the coefficient of the fifth term is 4 I From the law of formation the denominator of the sixth term would be 5 !, of the seventh term 6 1, etc. Conse- quently in the rth term the denominator of the coefficient would be (r —1)1. THE BINOMIAL THEOEEM 233 The numerator of the coefficient of the fifth term contains the product of the four factors ?^(7^— l)(/i— 2)(?i— 3). The numerator of the sixth term would contain these four and the additional factor 7i — 4. Similarly, the last factor in the numerator of the seventh term would be n — 5, etc. Hence the last factor in the rth term would be 7i — (r — 2), and the numerator of the coefficient of the rth term is 7i(n-l)(^-2)(^-3) ... (w-r + 2). The exponent of a in the fifth term is n — 4, and in the sixth term it would be n — 5, etc. Therefore in the rth term the exponent of a is n — (r — 1), or n — r + 1. The exponent of b in the fifth term is 4, in the sixth term is 5, etc. Therefore in the rth term the exponent of h is r — 1. The sign of any term of the expansion (if n is a posi- tive integer) is plus if the binomial is a + b. If the binomial is a — b, the terms containmg the odd powers of b will be negative. In other words, the sign in such cases depends upon whether the exponent r — 1 is odd or even. Hence the rth term (r not equal to 1) of (« + b^^^ equals n(n-l)(/2-2)(n-3). • .(n-r+2) (r-1)! .a"-'+^lf-K (1) If we wanted the twelfth term, we would in using (1) substitute 12 for r. EXERCISES Write the indicated terms : 1. Fifth term of (a + bf^. 2. Sixth term of (a + bf. Solution. Substituting 10 for n ,-. • .^ f 1 /ix • 3. Fourth term of (a + ^)^. and o for r m the formula (1) gives ^ ' ^ lQ-9-8-7 ^ 10-9 -8 -7 4. Seventh term of (a - bf 4! 4-3.2 = 210 a«&*. 5. Eighth term of (a — bf^ 234 SECOKB COURSE IN ALGEBRA 6. Fourth term of (a + -) • 8. Sixth term of ( ] • 7. Fifth term of (a^ - hf". 9. Middle term of {x" - xf. 10. Seventh term of (-; ) • (v-.-^r 11. Fifth term of Find the coefficient of : 12. x^ in (1 + xy. 14. x^^ in (x^ -f- 1)'^ 13. x^ in (1 + xY' 15- x^' in (^^ - »"')" Note. The binomial theorem occupies a remarkable place in the history of mathematics. By means of it Napier was led to the dis- covery of logarithms, amd its use was of the greatest assistance to Newton in making his most wonderful mathematical discoveries. But to-day the results of Newton and of Napier are explained with- out even so much as a mention of the binomial theorem, for simpler methods of obtaining these results have been discovered. It was Newton who first recognized the truth of the theorem, not only for the case where n is a positive integer, which had long been familiar, but for fractional and negative values as well. He did not give a demonstration of the general validity of the binomial develop- ment, and none even passably satisfactory was given until that of Euler (1707-1783). The first entirely satisfactory proof of this difl&cult theorem was given by the brilliant young Norwegian Abel (1802-1829). CHAPTER XVIII RATIO, PROPORTION, AND VARIATION 126. Ratio. The ratio of one number a to o, second number b is the quotient obtained by dividing the first by the second, or -. The ratio of ^ to 5 is also written J. ^ a: 0. It follows from the above that all ratios of two numbers are fractions and all fractions may be regarded as ratios. Thus -' — ' — ' and — — : are ratios as well as fractions. 5 ^d c-d V3 Since ratios like the above are fractions, operations which may be performed on fractions may be performed on these ratios. Hence the value of a ratio is not changed by multiplying or dividing both numerator (antecedent) and denominator (consequent) by the same number. a Thus - = and - = t- h b' X ^2. y EXERCISES Simplify the following ratios by considering them as frac- tions and reducing the fractions to lowest terms : ,2 -i^-^^^-i^*^^^ 2. 1 kilometer : 1 mile. (1 kilometer = .62 miles.) 3. 1 liter : 1 quart. (1 liter = .001 cubic meter ; 1 quart = ^^ cubic inches ; and 1 meter = 39.4 inches.) 235 236 SECOND COURSE IN ALGEBRA 4. A city lot 100 x 160 feet : 1 acre. (1 acre = 43,560 square feet.) 5. Area of printed portion of this page : total area of the page. 6. If |24,000 is divided between two men so that the portions received are to each other as 5:7, how much does each receive? Hint. Let 5 x and 7x be the required parts. 7. Separate 690 into four parts which are to each other as 2 : 5 : 7 : 9. X X -\- S 8. Show that -< ;: if x is positive. X -\- 2 X -^ 5 Hint. Reduce the given fractions to respectively equivalent fractions having a common denominator, then compare the numerators of the fractions so obtained. 9. Arrange the ratios 3 : 4 and 7 : 9 in decreasing order of magnitude. 127. Proportion. A proportion is a statement of equality between two ratios. Four numbers, a, h, show that the following equalities are true : di ^^ 5a 5c ,« 2a 2g 7b Id 18. 19. P ac bd 5a -\-b _5g -\- d 5a — b 5g — d 20. 21. 22. 23. b^ d^ b' a''-7P d^ c'-ld^ 2 -(^2 2ab 2Gd 7a2_3^2 7c2_3^2 5ab 5Gd 24. a'-^-ab-Jrb^ _a'- ah -f- b'' g' + cd -\-d''~ c'-Gd-\-d'' 240 SECOND COUKSE IN ALGEBRA 25. In the proof which follows give the reason for each step and state the result as a theorem : o_c_e a-\-c-\-e _a _c _e l^d^f ^^' b + d+f~b~d~f Proof. Setting each of the given ratios above equal to r, - = r, - = r, and - = r. (1) Then from (1), a = hr, c = d?^ e =fr. (2) Adding in (2), a + c + e = br -]- dr +fr. (3) Factoring in (3), a + c + e = (h + d +fy. (4) Therefore 1^7^ = '' ^^^ Hence by (1) and (5), a-\-c + e_a_c_e bTJTf ~b~d~f 26. If T = 3 = - 5 show that = - or -• Solve, using theorems of proportion : 27. 2a^:(ic + 8) = 10:3. 29. 5 : 4 = (.t - 3):(£r - 4). 28. 25:x = x:169. 30. (15 + x):(15 - «) = 13 :17. 31. (10 + £c):(20 + 3cc) = (10-ic):-3ir. 32. V^ + 2 ^^+1 ■y/x-2 x-7 33. Show that the mean proportional between two numbers is the geometric mean between these numbers. PROBLEMS 1. The surface of a sphere is 4 7r72^ If S represents the surface of a sphere, R its radius, and D its diameter, show 2. Find the ratio of the surfaces of two spheres whose radii are in the ratio 1:10, KATIO, PEOPORTION, AND VAKIATION 241 3. If the diameter of the earth is 7920 miles and that of Mars is 4230 miles, find the ratio of their surfaces. 4. If the diameter of the moon is 2160 miles, find the ratio of its surface to that of the earth. 5. The volume of a sphere is — - — • If V represents the o volume of a sphere, R its radius, and D its diameter, show that for any two spheres, y j^s j)s 6. The diameter of the sun is approximately one hundred and nine times the diameter of the earth. Find the ratio of their volumes. 7. From Exercises 3 and 5 find the ratio of the volumes of Mars and the earth. 8. Find the ratio of the volumes of the earth and the moon. 9. The areas of two similar triangles are to each other as the squares of any two corresponding lines. If the corre- sponding sides of two similar triangles are 12 and 20 and the area of the first is 90 square inches, find the area of the second. 10. The areas of two similar triangles are 147 and 300 respectively. If the base of the first is 10, find the correspond- ing base of the second. 11. If ^jBC is any triahgle and KR is a line parallel to BC, meeting AB at ^-and AC at R, then area ABC _ AB^ _ AC^ _ BC^ area AKR ~ AK^ ~ AR^ ~ KR^' If in the accompanying figure area ABC = 225 square inches, area AKR = 81 square inches, and AB = W inches, find AK, 242 SECOND COUESE IN ALGEBRA 12. In the figure of Exercise 11, if ABC = 845 square inches, BC = 13 inc*hes, and KR = 7 inches, find the area AKR. 13. If in the figure of Exercise 11 triangle AKR equals ^j of the trapezoid KBCR and AC = 15 inches, find AR and RC. 14. In Exercise 13 substitute ^ for 2^ and solve for AR to two decimals. 15. If in the figure of Exercise 11 the triangle is equivalent to the trapezoid and AK = 10, find KB to two decimals. 16. A certain flagpole casts a shadow 45 feet long at the same time that a near-by post 8 feet high casts a shadow 4: J feet long. Find the height of the pole. 17. The bisector of an angle of a triangle divides the oppo- site side into segments which are proportional to the adjacent sides. In triangle ABC ii AB = 15, BC = 24:, and CA = 25, find the segments of i^C made by the bisector of angle A. \S. If a plane be passed par- allel to the base of a pyramid (or cone), as in the accompany- ing figure, cutting it in KRL, then pyramid D — ABC : pyra- mid D - KRL = DH^ : DS^, etc. If in the adjacent figure the volumes of the pyramids are 8 and 27 cubic inches respectively, and the altitude DH equals 15 inches, find DS. 19. If DH in the accompany- ing figure is 18 inches and the volume of one pyramid is one third the volume of the other, find DS to two decimals. 20. In the accompanying figure the frustum is seven eighths of the whole pyramid, (a) If DH equals 16, find DS-, (b) if DH = X, find DS. RATIO, PROPORTION, AND VARIATION 243 21. If a plane parallel to the base divides the whole pyra- mid into two parts having equal volumes and DH = 75, find to two decimals the parts into which the plane divides DH. 22. The volumes of two similar figures are to each other as the cubes of any two corresponding edges. Compare the vol- umes of two similar solids, the edge of one of which is 60% greater than the corresponding edge of the other. 23. Compare the radii of two spheres whose volumes are to each other as 125 : 27. 133. Variation. The word quantity denotes anything which is measurable, such as distance, rate, time, and area. Many operations and problems in mathematics deal with numeri- cal measures of quantities, some of which are fixed and others con- stantly changing. Such problems as deal with the relation of the numerical measures of at least two changing quantities are called problems in variation. The theory of variation is really involved in proportion. This fact will become evident after a study of the illustra- tions of the different kinds of variation here given. The equation x='^y may refer to no physical quantities whatever, yet it is possible to imagine y as taking on in succession every possible numerical value, and the value of X as changing with every change of ?/, and consequently always being three times as great as the corresponding v^lue of y. In this sense, which is strictly mathematical, X and y are variables. The symbol for variation is oc, and x cc y is read " x varies directly as ?/ " or '' x varies as ?/." 134. Direct variation. One hundred feet of copper wire of a certain size weighs 32 pounds. Obviously a piece of the same kind 200 feet long would weigh 64 pounds, a piece 300 feet long would weigh 96 pounds, and so on. 244 SECOND COURSE IN ALGEBRA Here we have two variables, W^ (weight) and L (length), so related that the value of ^depends on the value of X, and in such a way that W increases proportionately as L increases. That is, W is directly proportional, or merely pro- portional, to L, Hence, if W^ and W^ are any two weights corresponding to the lengths L^ and L<^ respectively, W,: W, = L,:L,. (1) The fact expressed by (1) can be stated in the form of a variation, thus: WccL. In general, if x cc ^, and x and «/ denote ani/ two corre- sponding values of the variables, and x-^ and ?/j a particular pair of corresponding values of these variables, then £ = 1. (2) From (2), x = (^\^. (3) X But -^ is a constant, being the quotient of two definite numbers. Call this constant K, and (3) may be written x = Ky. That is, if one variable varies as a second, the first always equals the second multiplied hy some co7istant. Thus for the copper wire just mentioned, W= j'VV ^» ^^ ^ ^• Here, though W varies as L varies, W is always equal to L multi- plied by the constant 5%. EXERCISES 1.. If X QC y, and x = S when ^ = 8, find x when y =12. Solution. The variation is direct. ■" Therefore - = — X 12 Solving, a; = 4^. RATIO, PEOPORTION, AJ^D VARIATION 245 2. li x^y, and ic = 8 when y = 15, find y when x = 10. 3. If. xcc 2/, and x = h when y = k, find y when x = r. 4. If a? oc 2/, and a? = 2 when y = 5, find iT. 135. Inverse variation. If a tank full of water is emptied in 24 minutes through a smooth outlet in which the area of the openmg A is, 1 square inch, an outlet in which A is 2 square inches would empty the tank in one half the time, or in 12 minutes; and an outlet in which ^ is 3 square inches would empty the tank in 8 minutes. Suppose it possible to increase or decrease A at will. When A is doubled t is halved ; when A is trebled t is divided by 3 ; and so on. We then have in t (the time required to empty the tank) and in A (the area of the opening) two related variables such that if A increases, t will decrease proportionally, while if A decreases, t will increase proportionally. Now let t^ and ^2 be any two times corresponding to the areas A^ and A^ respectively; then t^:% = A^:A^. (1) The letters and the subscripts in (1) say : The first time is to the second time as the second area is to the first area. The proportion (1) may be written ^^ : ^2 = — - — A^ A^ where the subscripts on the ^'s and those on the ^'s come in the same order. First, from (1), t^. A^ = t^. A^. • (2) Dividing (2) by ^1^2^ ^ = ^' (3) Whence '<:];) = 21. V 10 3 34. 8 ■1. •/ 22. ? 10 »> 35. 4 8 V 23. ? 1( 3 1 36. .1 10 •> liead in tlic notation of lojjiiritlinis : 37. 300 =10-^*^ 42. 1730 = 103-*^. 38. 65:::=10^«^ 43. 173 = W--^«^ 39. 4=:10"". 44. 1.73 = 10-'«. 40. 1 = 10^ 45. .173 = 10-1 + •-'^\ 41. .10 = 10-^ • 46. .0173 = 10- '- + •-««. Read Exercises 47-49 and 54-56 as powers of 10 : 47. log 3 = .48. 48. log 20 = 1.301. 49. log 4.9 = .69. 50. log^^lOO + log^^lOOO 4- log,olO,000 = ? 51. log^,10 + log^^.Ol - logj^l = ? 54. log 490 = 2.69. 52. log^S + log327 4- log,l = ? 55. log .0049 = - 3 + .69. . 53. log39 + log^64 = ? 56. log 381 = 2.58. LOGARITHMS 255 Biographical Note. John Napier. Although many scientists have been honored with titles on account of their discoveries, very few of the titled aristocracy have become distinguished for their mathematical achievements. A notable exception to this rule is found in John Napier, Lord of Merchiston (1550-1617), who devoted most of his life to the problem of simplifying arithmetical operations. Napier was a man of wide intellectual interests and great activity. In connection with the management of his estate he applied himself most seriously to the study of agriculture, and experimented with various kinds of fertilizers in a somewhat scientific manner, in order to find the most effective means of reclaiming soil. He spent several years in theological writing. When the danger of an invasion by Philip of Spain was imminent he invented several devices of war. Among these were powerful burning mirrors, and a sort of round musket-proof chariot, the motion of which was con- trolled by those within, and from which guns could be discharged through little portholes. " But by far. the most serious activity of Napier's life was the effort to shorten the more tedious arithmetical processes. He invented the first approximation to a computing machine, and also devised a set of rods, often called Napier's bones, which were of assistance in multiplication. His crowning achievement, however, was the invention of logarithms, to which he devoted fully twenty years of his life. 140. Steps preceding computation. Before computation by means of the table can be taken up, two processes requiring considerable explanation and practice must be mastered. /. To find from the table the logarithm of .a given iiumher. II. To find from the table the number corresponding to a given logarithm, 141. Characteristic and mantissa. Unless a number is an exact power of 10, its logarithm consists of an integer and a decimal. This fact is illustrated in Exercises 37-46, p. 254. The integral part of a logarithm is called its characteristic. The decimal part of a logarithm is called its mantissa. Log 200 = 2.301. Here 2 is the characteristic and .301 is the mantissa. 256 SECOND COURSE IN ALGEBRA The characteristic of any number is obtained not from a table of logarithms but by an inspection of the number itself, according to rules which will now be derived. 104 = 10,000; that is, the log 10,000 = 4. 103 = 1000; that is, the log 1000 = 8. 102 = 100; that is, tlie log 100 = 2. 101=10; that is, the log 10 = 1. 100 = 1; that is, the log 1 = 0. 10-i=.l; that is, the log .1 = -1. 10-2 = .01; that is, the log .01 = —2. 10-3 =.001; that is, the log .001 = -3. The preceding table indicates between what two integers the logarithm of a number less than 10,000 lies. This determines the characteristic. Since 542 lies between 100 and 1000 (that is, between 102 and 103), log 542 must lie between 2 and 3 and must equal 2 (characteristic) plus a decimal (mantissa). And since .0045 lies between .001 and .01 (that is, between 10"^ and 10"'^), log .0045 = — 3 plus a positive decimal or — 2 plus a negative decimal. For the determination of the characteristic of a positive number we have the following rules: /. The characteristic of a number greater than 1 is one less than the number of digits to ike left of the decimal point, II. The characteristic of a number less than 1 is negative and numerically one greater than the number of zeros between the decimal point and the first significant jigwre. Accordingly the characteristic of 2536 is 3; of 6 is 0; of .4 is -1; of .032 is - 2 ; of .00086 is -4. JOHN NAPIER LOGARITHMS 25 ORAL EXERCISES What is the characteristic of the following : 1. 347. 5. 35. 9. 97.2. 13. .00972. 2. 9864. 6. 972. 10. 9.72. 14. 30.467. 3. 95. 7. 9720. 11. .972. 15. .5000. 4. 7. 8. 97200. 12. .0972. 16. .000375. The table on pages 2G6-267 gives the mantissas of numbers from 10 to 999. Before each mantissa a decimal point is understood. The numbers 5420, 542, 5.42, .0542, and .000542 are spoken of as composed of the same significant digits in the same order. They differ only m the position of the deci- mal pohit, and consequently their logarithms to the base 10 will have different characteristics, but they Avill have the same mantissa. The last two points are easily illustrated b}^ any two numbers Avhich have the same significant digits in the same order. log 5.42 = .734, or 5.42 = 10-'34; 5.42 . 102 _ 542 = 10-1 . 102 _ 102.734. Therefore log 542 = 2. 734. The property just explamed does not belong to a system of logarithms in which the base is any number other than 10. Thus, if the base is 100, the most convenient number after 10, the logarithms of 5420, 542, 54.2, and 5.42 are respec- tively 1.8670, 1.3670, .8670, and .3670. While a certain regularity in characteristic and mantissa can be seen here, it is obvious that the rules for obtaining them would not be so simple as they are for the base' 10. Moreover, it can be seen that tables of a given accuracy are far shorter when the base is 10 than they would be with any other base. 258 SECOND COURSE IN ALGEBRA 142. Use of the table. To obtain the logaritlim of a imiiiber of three or fewer significant figures from tlie table, we have the Rule. Determine the characteristic hy inspection. Find in colmnn N the first two sif/nificant figures of the given 7iimiber. In the roiv with these and iii the column headed hy the third figure of the number, find the required mantissa. ORAL EXERCISES Find tlie logarithm of the following : 1. 263. 4. 56. 7. 3.7. 10. 7. 2. 375. 5. 560. 8. 3700. 11. 932. 3. 729. 6. 37. 9. 5. 12. .932. Solution. The characteristic of .932 is — 1 and the mantissa is .9694. Hence log .932 = — 1 + .9694. This is usually written in the abhreviated form, 1.9694. The mantissa is always kept positive in order to avoid the addition and subtraction of both positive and negative decimals, which in ordinary practice contain from three to five figures. Negative characteristics, being integers, are compara- tively easy to take care of. (The student should note that log .932 is really negative, being —1-1- .9694, or — .0306.) 13. .563. 15. .00376. 17. .0202. 19. 3.86. 14. .0637. 16. .00468. 18. 725000. 20. .987. 143. Interpolation. The process of finding the logarithm of a number not found in the table, from the logarithms of two numbers which are found there, or the reverse of this process, is called interpolation. If we desire the logarithm of a number not in the table, say 7635, we know that it lies between the logarithms of 7630 and 7640, which are given in the table. Since 7635 is halfway between 7630 and 7640, we assume, though it is not strictly true, that the required logarithm is halfway LOGAllITHMS 259 between their logarithms, 3.8825 and 3.8831. In order to find log 7635 we first look up log 7630 and log 7640 and then take half (or .5) their difference (this difference may usually be taken from the column headed D) and add it to log 7630. This gives log 7635 = 3.8825 +.5 x.0006 = 3.8828. . Were, we finding log 7638, we should take .8 of the difference between log 7630 and log 7640 and add it to log 7630 as follows : log 7638 = 3.8825 +.8 X .0006 = 3.8825 +.00048 = 3.8825 +.0005 = 3.8830. Observe that in using four-place tables one should not carry results to five figures. If the fifth figure is 5, 6, 7, 8, or 9, omit it and increase the fourth figure by 1 ; that is, obtain results to the nearest figure in the fourth place. For finding the logarithm of a number we have the Rule. Prefix the proper characteristic to the mantissa of the first three significant figures. Then multiply the difference hettveen this mantissa and the next greater mantissa in the table (^called the tabular differ- ence') by the remaining figures of the number preceded by a decimal point. Add the product to the logarithm of the first three figures^ taking the nearest decimal in the fourth place. In this method of interpolation we have assumed that the increase in the logarithm is directly proportional to the increase in the number. As has been said, this is not strictly true, yet the results here obtained are nearly always correct to the fourth decimal place. 260 SECOND COURSE IN ALGEBRA EXERCISES Find the logaritliin of the following : 1. 3625. 5. 646.8. 9. 705.50. 13. 30.07. ' 2. 464.7. 6. 82.543. 10. 3.0075. • 14. 3.1416. 3. 52.73. 7. 10.101. 11. .00286. 15. 2.71828. 4. 42.75. 8. 500.35. 12. .0007777. 16. .0023456. 144. Antilogarithms. An antilogarithm is the number cor- responding to a given logarithm. Thus antilog 2 equals 100. If we desire the antilogarithm of a given logarithm, say 4.7308, we proceed as follows: The mantissa .7808 is found in the row which has 53 in column N and in the column which has 8 at the top. Hence the first three sig- nificant figures of the antilogarithm are 538. Since the characteristic is 4, the number must have five digits to the left of the decimal pohit. Thus antilog 4.7308 = 53,800. Therefore, if the mantissa of a given logarithm is found in tlie table, its antilogarithm is obtained by tlie Rule. Find the roiv and the colmnn in which the given mantissa lies. In the row found take the two figures which are in column N for the first two significant figures of the antilogarithm and for the third figure the number at the top of the column in tvhich the mantissa stands. Place the decimal point as ind/'cafcd li/ ihc rltaractenstic. ORAL EXERCISES Find the antilogarithm of the following : 1. 3.8768. 6. 7.5866-10. 10. 4.1335. 2. 1.8035. Hint. 7. 58«0 - 10 = 8.58(5(5. 11. 6.9154. 3. .5763. 7.9.2455-10. 12.8.3464. 4. 1.3747. 8. 4.1335 - 10. 13. 0.8882. 5. 2.7649. 9. 5.7875 - (>. 14. 5.9689. LOGARITHMS 261 If the mantissa of a given logarithm, as 1.5271, is not in the table, the antilogarithm is obtained by interpolation as follows: The mantissa 5271 lies between .5263, the mantissa of the sequence 336, and .5276, the mantissa of the sequence 337. Therefore the antilogarithm of 1.5271 lies between 33.6 and 33.7. Since the tabular difference is 13 and the dif- ference between .5263 and .5271 is 8, the mantissa .5271 lies -^^ of the way from .5263 to .5276. Therefore the re- quired antilogarithm lies ^^ of the way from 33.6 to 33.7. Then antilog 1.5271 = 33.6 + ^83 x .1, • and 33.6 +.061= 33.66. Therefore when the mantissa is not found in the table we have the Rule. Write the number of three figures corresponding to the lesser of two mantissas hetiveen which the given mantissa lies. Subtract the less mantissa, from the given one and divide the remainder by the tabular difference to two decimal places. If the second digit is 5 or more., increase the first digit by 1; if less than 5., omit it. Annex the resulting digit to the three already found and place the decimal point where indicated by the characteristic. EXERCISES Find the antilogarithms of the following : 1. 1.5523. 5. 1.2566. 9. 9.2664 - 10. 2. 2.3821. 6. 7.3572 - 10. 10. .7729. 3. 0.6790. 7. 9.8327 - 10. 11. 7.1060 - 10. 4. 2.5720. 8. 5.9613 - 8. 12. 6.2318 -:- 10. 262 SECOND COURSE IN ALGEBRA 145. Multiplication. Multiplication by logarithms de- pends on the Theorem. The logarithm of the product of tivo numbers is the sum of the logarithms of the numhers. . That is, for the numbers a and x \og,,{a . X) = log^a -h log^a-. Proof. Let log^a = l^, (1) and log/^x = h. (2) From (i), a = bh. (3) From (2), x = hk. (4) (3) X (4), ax = ¥1 + 12. (5) Therefore lt)g^«.r = li-^ U = logf,a + \ogf,x. Solution. EXERCISES Perform the indicated operation by logarithms 1. 18 X 25. log 18 =1.2553 log25 = 1.3979 log (18 X 25) = 2.G532 (adding) antilog 2.6532 = 450. 6. 386 X 27. 7. 432 X 263. 8. 589 X 375. 9. 4326 X 497. 2. 37 X 28. 3. 29 X 9. 4. 9.8 X 6. 5. 42 X 3.3. 10. 2870 X 3754. 11. 286.7 X 2.391. 12. 3.412 X 2.526. 13. 432 X .574. Solution. log 432 = 2.6355 = 2.6355 log .574 = 1.7589 = 9.7589 10 log (432 X .574) = 2.3944 = 12.3944 - 10 (adding) autilog 2.3944 = 247.9. LOGARITHMS 263 Since the mantissa is always positive, any number carried over from the tenths' cohimn to the units' column is positive. This occurs in the preceding solution, where .6 + -7 = 1.3, giving + 1 to be added to the sum of the characteristics + 2 and — 1, in the units' column. Mistakes in such cases will be few if the logarithms with negative characteristics be written as in the 9—10 notation on the right. In the preceding example and in others which follow, two methods are given for writing the logarithms which have negative character- istics. This is done to illustrate those cases in which the second of th<5^two ways is preferable. It should be understood that in practice one, but not necessarily both, of these methods is to be used. 14. 385 X .647. 19. .6381 x -.01897. 15. 571 X 073 Hint. Determine by inspection the sign of the product. Then operate as 16. 37.6 X .00865. if all signs were positive. 17. .0476 X 673. 20. 675 x - .0286. 18. .07325 X 6.354. 21. -.437 x -.0046. 146. Division. Division by logarithms depends on the Theorem. The logarithm of the quotient of two numbers is the logarithm of the dividend minus the logarithm of the divisor. That is, for the numbers a and x logfc a ' X = logf^a - log^.r. Proof. Let i log^a = /j, (1) (2) From (1), a = Uu (•^) From (2), X = hh. (4) (3) -^(4), X Therefore = logta - logftA-. 264 SECOND COURSE IN ALGEBRA EXERCISES Divide, iising logarithms : 1. 891^27. Solution. log 891 = 2.9499 log 27 =1.4314 log (891 -^ 27) = 1.5185 (subtracting) antilog 1.5185 = 33. 2. 96-5-32. 5. 439-- 27.1. 8. 9896 -- 52.78. 3. 888 -T- 47. 6. 3860 -^ 4.32. 9. 6732 -f- 7.81. 4. 976^361. 7. 4627^281. 10. 3.26 -- .0482. Solution, log 3.26 = 0.5132 = 10.5132 - 10 log. 0482 = 2.6830=8^6830-10 log (3.26 -i- .0482) = 1.8302 = 1.8302 - antilog 1.8302 = 67.64. 11. 2.35-^.0683. _ 347 x (- 625) 12. 4.86 --.751. * 346 13. .0635-^.277. 473.2 x 4.78 14. .2674 --3.66. * -68.3 15. .07882 H- 68.72. 9.63 x -0892 16. 356 X 392-- 128. ^^' .00635 147. Involution. Involution by logarithms depends on the Theorem. The logarithm of the rth power of a number is r times the logarithm of the number. That is, for the numbers r and x, log^a^ = r log^a;. Proof. Let logftx = I. (1) Then x = ¥. (2) Raising both members of (2) to the rth power, xr = bri. Therefore logta:^ = rl = r log&a:. LOGARITHMS 265 EXERCISES Compute, using logarithms : 1. (2.73)«. Solution. log 2.73 = .4362. log (2.73)3 = 1.3086 (multiplying by 3). antilog 1.3086 = 20.33. 2. (6.32)*. 3. (34.26)1 4. (6.715)1 5. (.425)1 Solution. log .425 = 1.0284 = 9.0284 - 10. log (.425)3 = 2.8852 = 28.8852 - 30. antilog 2.8852 = .07677. 6. (.362)1 ' 9. (486.2)2 . (3.85)3. 7. (.0972)1 10. (.375)^ • (62.5)1 8. (.003597)^ 11. (^.25)^ - (1.232)3. 148. Evolution. Evolution by means of logarithms de- pends on the Theorem. The logarithm of the real rth root of a number is the logarithm of the number divided hy r. That is, for the real numbers r and w, logf^^n= — logjn. Proof. Let log,/* = f- (1) Then n = bK (2) Extracting the rth root of both members of (2), 1 11 (ny = (h^Y = hr, (3) 1 7 log, n Therefore log. (nV = - = -^^^^ • (4) ^ r r RE 266 SECOND COURSE IN ALGEBRA N 1 2 3 4 5 6 7 8 9 D 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 42 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 38 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 35 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 32 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 30 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 28 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 20 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 25 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 24 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 22 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 20 22 3424 3444 3464 3483 3502 3522 3641 3560 3579 3598 19 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 18 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 18 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 17 26 4150 4166 4183 4200 4216 4232 4249 4266 4281 4298 16 27 4314 4330 4346 4362 4378 4393 4409 4'425 4440 4456 16 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 16 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 15 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 14 31 4914 4928 4942 4955 4969 4983 4997 5011 6024 5038 14 32 5051 5065 6079 6092 5105 5119 5132 5145 6159 6172 13 33 5185 5198 5211 6224 5237 5250 5263 5276 6289 6302 13 34 5315 5328 6340 5353 6366 6378 6391 5403 5416 5428 13 35 5441 5453 6465 6478 6490 5502 6514 5527 6639 5551 12 36 5563 5575 5587 6599 5611 5623 5635 6647 5658 5670 12 37 5682 5694 5705 6717 6729 5740 5752 5763 6775 5786 12 38 5798 5809 5821 6832 6843 6855 6866 6877 6888 5899 11 39 5911 5922 5933 6944 6955 6966 6977 6988 6999 6010 11 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 11 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 10 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 10 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 10 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 10 45 6532 6542 6551 6561 6571 6680 6590 6599 6609 6618 10 46 6(J28 6637 6646 6666 666.5 6676 6684 6693 6702 6712 9 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 9 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 9 49 6902 6911 6920 6928 6937 6946 6965 6964 6972 6981 9 60 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 9 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 8 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 8 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 8 64 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 8 LOGARITHMS 267 N 1 2 3 4 5 6 7 8 9 D 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 8 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 8 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 8 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 6 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 6 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 6 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 6 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 6 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 6 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 6 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 6 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 6 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 6 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 6 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 6 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 5 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 5 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 5 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 5 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 5 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 5 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 5 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 5 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 5 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 5 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 5 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 5 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 5 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 5 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 5 94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 5 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 5 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 5 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 4 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 4 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 4 268 SECOND COURSE IN ALGEBKA EXERCISES Compute, using logarithms : 1. -y/sje. Solution. log 376 = 2.5752. log V 376 = .8584 (dividing by 3). Then antilog .8584 = 7.218. 2. ^783 3. S/1435. 4. ^3421 5. A/.000639. Solution. log .000639 = 4.8055. If one divided 4.8055 as it stands by 3, he would be likely to confuse the negative characteristic and the positive mantissa. This and other difficulties may easily be avoided by adding to the char- acteristic and subtracting from the resulting logarithm any integral multiple of the index of the root which will make the characteristic positive. Thus log .000639 = 2.8055 - 6. iog\/.000639 = .9:^52-2 (dividing by 3). Then antilog 2.9352 = .08614. 6. VX)756. 11. (- 6.387)^. 14. a/269. 7.^.0007624 ,^^3-^^ 15. v/i74^. t: 12. - 8. V.005679. M (3.423)« ^- ^^^-^^ • 1 (43.56)^.7.984 10. (4.925)'. N (7.623)« ' 17. ^^V? ^-07241. 18. Determine the logarithms of 5732, 573.2, 57.32, and 5.732 to the base 10 and to the base 100. Compare the results. What fact about logarithms do these results emphasize ? LOGARITHMS 269 Note. The preceding four-place table will usually give results correct to one half of one per cent. Five-place tables give the man- tissa to five decimal places of the numbers from 1 to 9999 and, by interpolation, the mantissa of numbers from 1 to 99,999. Such tables give results correct to one twentieth of one j)er cent, a degree of accuracy which is sufficient for most engineering work. Six-place tables give the mantissa to six decimals for the same range of numbers as a five-place table, but the labor of using a six- place table is much greater than that of using a five-place one. Seven-place tables contain the mantissas of the numbers from 1 to 99,999. Such tables are needed in certain kinds of engineering work and are of constant use in astronomy. In place of a table of logarithms engineers often use an instru- ment called a slide rule. This is really a mechanical table of logarithms arranged ingeniously for rapid practical use. Results can be obtained with such an instrument far more quickly than with an ordinary table of logarithms, and that without recording or even thinking of a single logarithuL. A slide rule ten inches long usually gives results correct to three figures. In work requiring greater accuracy a larger and more elaborate instrument which gives a five- figure accuracy is used. 149. Exponential equations. An exponential equation is an equation in which the unknown occurs in an exponent. Many exponential equations are readily solved by means of logarithms, since log a^ =x log a. Thus let ci^=c. Then xloga = logc. Whence a: = logc-f- loga. EXAMPLE Solve for x 8"^ = 324. Solution. log 8-^ = log 324, or x log 8 = log 324. Whence . = l2£^ = 1^ = 2.78+. logs .9031 The student must overcome his hesitation actually to divide one logarithm by another if, as here, it is necessary. 270 SECOND COURSE IN ALGEBRA MISCELLANEOUS EXERCISES 1. Can you find the logarithm of a negative number to a positive base ? Explain. Find, without reference to the table, the numerical values of; 2. log^O. 6. 51og^9. 3. log^8. 7. log^8H-31og3 4. 4. log32. 8. 21og^81-41og3,27. 5. 4 log^27. 9. 3 log^l25 + 2 log,25 - 2 log^o. 10. 4 log,(i) - 5 log/Jy) + 2 log^T^- Simplify : 11. log I + log if 13. log -V- + log U - log i- 12. log ^V - log f |. 14. 2 log 3 + 3 log 2. 15. 31og4 + 41og3-21og6. Show that : 16. log la — ^J= log (a + x) + log (a — .r) — log a. 17. log V^'-x^ = J [log(« + x)-\- log(«. - x)]. 18. log Vr(.s' - a) = I [log s + log (.s -a)y > s — a lllogs + log(.s - h) + log(.v - c) - log{s - u)^ Solve, using logarithms (obtain results to four figures) : 20. The circumference of a circle is 2 wR. (tt = 3.1416, li = radius.) (a) Eind the circumference of a tdrcle whose radius is 85 . inches. (b) Eind the radius of a circle whose circumference is 3281 centimeters. LOGARITHMS 271 21. The area of a circle is ttR^. (a) Find the area of a circle whose radius is 5.672 feet. (b) Find the radius of a circle whose area is 67.37 square feet. 22. The area of the surface of a sphere is 4 irR^. (a) The radius of the earth is 3958.79 miles. Find its surface. (b) Find the length of the equator. 23. The volume of a sphere is — o (a) Find the radius of a sphere whose volume is 86 cubic feet. (/>») Find the diameter of a sphere whose volume is 47 cubic inches. 24. If the hypotenuse and one leg of a right triangle are given, the other leg can always be computed by logarithms. In the adjacent figure let a and c be given and x required. Then X = Vc^ - a' = V{r-\-a)(c-a). Whence logic = i log(c + «) 4- 1^ logO; - a). (a) The hypotenuse of a right triangle is 377 and one leg is 288. Find the other leg. (b) The hypotenuse of a right triangle is 1285 and one leg is 924. Find the other leg. 25. The area of an equilateral triangle whose side is s is — V3. Find in square feet the area of an equilateral triangle whose side is 34.23 inches. Solve for x : 26. 3^ = 25. 30. 3 = (1.04)-^. 34. 10*^ "^ = 3. 27. 64^ = 4. 31. 2^ = 64. 35. 8-^ + '^ = 6. 28. 16-^ = 1024. 32. 42*^+1 = 84. 36. (.3)"^ = 5. 29. (- 2y = 64. 33. 3^ + ^ = 6561. 37. (.07)-^ = 9. 272 SECOND COURSE IN ALGEBRA Find the number of digits in : 38. (a) 3^2; (b) 2^-, (c) 2« • 3« . 5'. 39. In how many years will $1 double itself at 3% interest compounded annually ? * Solution. At the end of one year the amount of $1 at 3% is $1.03 ; at the end of two years it is |(1.03) (1.03), or $(1.03)2 ; at the end of three years it is $(1.03)^, and at the end of x years it is $(1.03)^. If X is the number of years required, (1.03)^ = 2. Taking the logarithms of both members of the equation, X log 1.03 = log 2. o 1 • log 2 .3010 ^^^ . t^ Solvmg, x = - — ^ = = 23.5 + . *& ^ log 1.03 .0128 V^ 40. In how many years will $1 double itself at 6% interest compounded annually? 41. In how many years will any sum of money treble itself at 4^ interest compounded annually ? 42. In how many years will $450 double itself at 3-|-% interest compounded annually ? 43. In how many years will $4000 amount to $7360.80 at 5% interest compounded annually ? 44. About 300 years ago the Dutch paid $24 for the island of Manhattan. At 4*^ compound interest, what would this payment amount to at the present time ? 45. In how many years will $12 double itself at 3% interest compounded semiannually ? 46. Show that the amount of P dollars in t years at ?•% interest compounded annually is P (1 + ry ; compounded semi- annually is P(H--) ; compounded quarterly is pfl + jj ; and compounded monthly is P f 1 + — * In making computations of this nature by the aid of logarithms, care must be exercised not to retain more significant figures in the jesult than are given with accuracy by the process. LOGARITHMS 273 47. Find the amount of |5000 at the end of four years, interest at 4% compounded {(i) annually; (b) semiannually; (c) quarterly. 48. Find the amount of |4.12 at the end of five years, interest at 4^^, compounded quarterly. 49. Set up and solve the equation used to determine the amount which should be paid for a |5 certificate to be paid in five years, interest at 4'^, compounded quarterly. Note. It is not a little remarkable that just at the time when Galileo and Kepler were turning their attention to the laborious computation of the orbits of planets, Napier should be devising a method which simpUties these processes. It was said a hundred years ago, before astronomical computations became so complex as they now are, that the invention of logarithms, by shortening the labors, doubled the effective life of the astronomer. To-day the remark is well inside the truth. In the presentation of the subject in modern textbooks a loga- rithm is defined as an exponent. But it was not from this point of view that they were first considered by Napier. In fact it was not till long after his time that the theory of exponents was understood clearly enough to admit of such application. This relation was noticed by the mathematician Euler, about one hundred and fifty years after logarithms were invented. It was by a comparison of the terms of certain arithmetical and geometrical progressions that Napier derived his logarithms. They were not exactly like those used commonly to-day, for the base which Napier used was not 10. Soon after the publication (1614) of Napier's work, Henry Briggs, an English professor, was so much impressed with its importance that he journeyed to Scotland to con- fer with Napier about the discovery. It is probable that they both saw the necessity of constructing a table for the base 1 0, and to this enormous task Briggs applied himself. With the exception of one gap, which was filled in by another computer, Briggs's tables form the basis for all the common logarithms which have appeared from that day to this. The square roots and the cube roots on the following page are corrected to the nearest digit in the third decimal place. 274 SECOND COURSE IN ALGEBRA No. Squares Cubes Square Roots Cube Roots No. Squares Cubes Square Roots Cube Roots 1 1 1 1.000 1.000 51 2,601 132,651 7.141 3.708 2 4 8 1.414 1.260 52 2,704 140,608 7.211 3.733 3 9 27 1.7.32 1.442 53 2,809 148,877 7.280 3.756 4 16 64 2.000 1.587 54 2,916 157,464 7.348 3.780 5 25 125 2.236 1.710 55 3,025 166,375 7.416 3.803 6 36 216 2.449 1.817 56 3,136 175,616 7.483 3.826 7 49 343 2.646 1.913 57 3,249 185,193 7.550 3.849 8 64 512 2.828 2.000 58 3,364 195,112 7.616 3.871 9 81 729 3.000 2.080 59 3,481 205,379 7.681 3.893 10 100 1,000 3.162 2.154 60 3,600 216,000 7.746 3.915 11 121 1,331 3.317 2.224 61 3,721 226,981 7.810 3.936 12 144 1,728 3.464 2.289 62 3,844 238,328 7.874 3.958 13 169 2,197 3.606 2.351 63 3,969 250,047 7.937 3.979 14 196 2,744 3.742 2.410 64 4,096 262,144 8.000 4.000 15 225 3,375 3.873 2.466 65 4,225 274,625 8.062 4.021 16 256 4,096 4.000 2.520 66 4,356 287,496 8.124 4.041 17 289 4,913 4.123 2.571 67 4,489 300,763 8.185 4.062 18 324 5,832 4.243 2.621 68 4,624 314,4.32 8.246 4.082 19 361 6,859 4.359 2.668 69 4,761 328,509 8.307 4.102 20 400 8,000 4.472 2.714 70 4,900 ^43,000 8.367 4.121 21 441 9,261 4.583 2.759 71 5,041 357,911 8.426 4.141. 22 484 10,648 4.690 2.802 72 5,184 373,248 8.485 4.160 23 529 12,167 4.796 2.844 73 5,329 389,017 8.544 4.179 24 576 13,824 4.899 2.884 74 5,476 405,224 8.602 4.198 25 625 15,625 6.000 2.924 75 5,625 421,875 8.660 4.217 26 676 17,576 5.099 2.962 76 5,776 438,976 8.718 4.236 27 729 19,683 5.196 3.000 77 5,929 . 456,533 8.775 4.254 28 784 21,952 5.292 3.037 78 6,084 474,552 8.832 4.273 29 841 24,389 5.385 3.072 79 6,241 493,039 8.888 4.291 30 900 27,000 5.477 3.107 80 6,400 512,000 8.944 4.309 31 961 29,791 5.568 3.141 81 6,561 531,441 9.000 4.327 32 1,024 32,768 5.657 3.175 8'.i 6,724 551,1368 9.055 4.344 33 1,089 35,937 5.745 3.208 83 6,889 571,787 9.110 4.362 34 1,156 39,304 5.831 3.240 84 7,056 592,704 9.165 4.380 35 1,225 42,875 5.916 3.271 85 7,225 614,125 9.220 4.397 36 1,296 46,656 6.000 3.W2 86 7,396 636,a56 9.274 4.414 37 1,369 50,653 6.083 3.3;52 87 7,569 6.58,503 9.327 4.431 38 1,444 54,872 6.164 3.:362 88 7,744 681,472 9.^381 4.448 39 1,521 59,319 6.245 3.391 89 7,921 704,969 9.4M 4.465 40 1,600 64,000 6.325 3.420 90 8,100 729,000 9.487 4.481 41 1,681 68,921 6.403 3.448 91 8,281 753,571 9.539 4.498 42 1,764 74,088 6.481 3.476 92 8,464 778,688 9.592 4.514 43 1,849 79,.507 6.557 3.503 93 8,649 804,357 9.644 4.531 44 1,936 &5,184 6.6;^3 3.530 M 8,836 830,584 9.695 4.547 45 2,025 91,12,5 6.708 3.557 95 9,025 857,375 9.747 4.563 46 2,116 97,336 6.782 3.583 96 9,216 884,7.% 9.798 4.579 47 2,209 103,823 6.856 3.609 97 9,409 912,673 9.849 4.595 48 2,304 110,592 6.928 3.6;m 98 9,604 941,192 9.899 4.610 49 2,401 117,649 7.000 3.659 99 9,801 970,299 9.950 4.626 50 2,500 125,000 7.071 3.684 100 10,000 1,000,000 10.000 4.642 INDEX Abel, 224, 234 Addition, algebraic, 2 ; commuta- tive law of, 2 ; of fractions, 53 Ahmes, 214 Alternation, 238 Antecedent, 235 Antilogarithm, 260 Apothem, 125 Axes, X- and y-, 69 Axiom, 16 Axioms, 16 Base, 252, 257 Binomial theorem, 227 ; extraction of roots by, 230 ; rth term of (a + 5)", 232-233 Binomials, difference of two squares, 28; powers of, 227; sum or difference of two cubes, 34 Briggs, 253, 273 Brouncker, William, 60 Cardan, 36, 165 Cauchy, 224 Characteristic, 255 Checking, rule for, 19 Circle, 179 Complex number, 159 Conjugate iniaginaries, 162 Conjugate real radicals, 120 Consequent, 235 Constant, 244, 246, 248 Coordinates of a point, 70 Cubes of numbers, 274 Decimals, repeating, 110 Diophantos of Alexandria, 81 Discriminant of a quadratic equa- tion, 172 Distance, x~ and y-. 69 Division, by logarithms, 263 ; rule for, 7 Elimination, 73 Ellipse, 180 Equation, definition of an, 15 ; of condition, 15 ; derived, 77 ; inde- pendent, 77; root of an, 16 ; satis- fying an, 16 Equations, definition and typical solution of irrational, 153 ; equiv- alent, 17 ; exponential, 269 ; for- mation of, with given roots, 168 ; homogeneous, 190 ; with imagi- nary roots, 136, 164 ; indetermi- nate, 80 ; involving fractions, 61 ; linear, solution by addition and subtraction, 75 ; in one un- known, rule for solving, 19 ; rule for the solution of irrational, 154 ; in several unknowns, 80 ; solution of, by factoring, 42 ; special cases, 76 ; use of division in, 196 Euclid, 45, 123 Euler, 234, 273 Evolution, by logarithms, 265 ; law of, 92 Exponent, logarithm as an, 253 ; meaning of a fractional, 109 ; meaning of a negative, 93 ; mean- ing of a zero, 93 Exponents, fundamental laws of, 91 Expressions, integral, 25; reducible to difference of two squares, 33 Extremes, 237 Factor, highest common, 44 Factor Theorem, 36 Factorial notation, 232 Factoring, definition of the process, 25 ; general directions for, 39 ; solution of equations by, 42 Factors, prime, 25 Fraction, changes of sign in a, 49 275 276 SECOND COUESE IN ALGEBRA Fractions, addition and subtrac- tion of, 53 ; complex, 59 ; division of, 57 ; equivalent, 52 ; multi- plication of, 56 ; operations on, 47 Function, definition of a, 129; graph of a cubic, 132 ; graph of a linear, 130 ; graph of a quadratic, 131 ; notations for a, 130 Functions, 129 ; names of, 129 Galileo, 273 Gauss, 134, 159, 165, 224 Graph, of a cubic function, 132 ; of a linear function, 130 ; of a quadratic equation in two vari- ables, 177 Graphical representation of nu- merical data, 184 Graphical solution of an equation in one unknown, 134; the process of, 135 Graphical solution of a linear sys- tem, 69, 70 Graphical solution of a quadratic system in two variables, 181 Hindu mathematicians, 123 Hyperbola, 178 Identity, 15 Imaginaries, addition and subtrac- tion of, 159 ; conjugate, 162 ; defi- nitions, 158 ; division of, 162 ; equations with imaginary roots, 164 ; factors involving, 166 ; mul- tiplication of, 160 ; note on the use of, 166 Imaginary, 110 Imaginary roots, 136 Index, 109 Infinite geometrical series, 220 Integral expressions, 25 Interpolation, 258 Inversion, 238 Involution, by logarithms, 264 ; law of, 92 Irrational numbers, 109 Kepler, 180, 273 Klein, Professor Felix, 134 La Place, 224 Leibnitz, Gottfried Wilhelm, 237 Line, straight, 130 Logarithms, definition of, 252 ; antilogarithm, 260 ; interpola- tion, 258 ; systems of, 253 ; table of, 266-267 Mantissa, 255, 263 Means, 237; arithmetical, 207; geo- metrical, 217 Multiple, lowest common, 50 Multiplication, law of, 5 ; rule for, by logarithms, 262 Napier, John (Lord of Merchiston), 234, 255, 273 Newton, 137, 223, 234 Numbers, classification of, 110; complex, 159; imaginary, 110; irrational, 109; orthotomic, 158; pure imaginary, 158 ; rational, 109; real, 110 Operations, order of fundamental, 1 Origin, 70 Oughtred, 237 Parabola, 177 Parentheses, removal of, 8 Pascal's triangle, 228 Polynomials, with a common monomial factor, 26 ; factored by grouping tenns and taking out a conmion binomial factor, 27 Powers, sum or difference of two like, 37 Probability curve, 187 Problems, solution of, 21 Products, important special, 10 Progressions, connnon difference of arithmetical, 204 ; definition of arithmetical, 204; definition of geometrical, 214 ; nth term of arithmetical, 205 ; nth term of geometrical, 216 ; ratio of geometrical, 215 Proportion, 236 ; test of a, 237 ]*roportional, fourth, 237; mean, 237; third, 237 INDEX 277 Proportions, derived by addition, by subtraction, and by addition and subtraction, 239 ; from equal products, 237 Quadratic equation, character of the roots of, 172 ; comparison of the various methods of solution of the, 146 ; formation of, with given roots, 168 ; number of roots of, 1 74 ; relation between the roots and the coefficients of, 169 ; solu- tion of, by completing the square, 139 ; solution of, by formula, 144 Quadratic expressions, factors of, 175 Quadratic trinomial, 30 ; rule for solving, 30 ; the general, 31; rule for solving the general, 82 Quantity, 243 lladicals, 109 ; addition and sub- traction of, 116 ; algebraic sign of. 111; conjugate, 120 ; division of, 119 ; factors involving, 124 ; multiplication of real, 117; simi- lar, 116; simplification of , 112 Radicand, 109 Ratio, 235 Rational expressions, 25 Rational numbers, 109 Rationalizing factor, 119 Remainder Theorem, 35 Root, cube, of arithmetical numbers, 274 ; cube, of monomials, 101 ; defi- nition of cube, 101 ; definition of square, 101; principal, 101; rule for extracting square, 106; square, of arithmetical numbers, 106, 274 ; square, of a monomial, 101 ; square, of polynomials, 103 ; square, of surd expressions, 122 Roots, imaginary, 136 ; imaginary, for a cubic equation, 137; table of square and cube, 274 Series, arithmetical, 210 ; geomet- rical, 219 ; infinite geometrical, 220 ; sum of, 210 Slide rule, 269 • Squares of numbers, 274 Stifel, 230 Subtraction, of fractions, 53 ; of polynomials, 3 Surd, 110 Systems, determinate, in three variables, 82 ; equivalent, 196 ; incompatible or inconsistent, 76 : simultaneous, 76 ; solution by addition and subtraction, 75 ; so- lution by substitution, 74; solu- tion of, when one equation is linear and the other quadratic, 188; solution of, when both are quadratic, 191 ; solution of a lin- ear, in two variables, by graphs, 69 ; special methods for solution of, 194 ; symmetric, 193 Table, of cubes and squares, 274 ; of logarithms, 266-267 Tartaglia, 36, 228 Terms, similar, 2 Transposition, 17 Trinomials, perfect squares, 28 Variable, 130, 244 Variation, 243; direct, 243; inverse, 245 ; joint, 247 Wallis, John, 60, 98, 237 Zero as an exponent, meaning of, 93 VB 35926 459908 UNIVERSITY OF CAUFORNIA LIBRARY