SECOND COURSE IN ALGEBRA BY WEBSTER WELLS, S.B. AUTHOR OF A SERIES OP TEXTS ON MATHEMATICS AND WALTER W. HART, A.B. ASSISTANT PROFESSOR OK MATHEMATICS, UNIVERSITY OF WISCONSIN COURSE FOR TH3 TRAINING OF TEACHERS D. C. HEATH & CO., PUBLISHERS BOSTON NEW YORK CHICAGO iv PREFACE * students a rather complete treatment of these important top- ics, even though they may not care to cover all of the examples and problems of the text. Special attention is directed to the emphasis placed upon getting roots of quadratic equations in decimal form. (See § 78.) (c) Chapter IX is an elaboration of Chapter XVI of the First Year Algebra. The advantage of postponing the topics in this chapter until well into the third semester of algebra is apparent. Isolating these topics as in this text makes it pos- sible for schools that desire a brief course to omit the chapter altogether. In this chapter, topics such as those in §§89 to 95 find a logical place. (d) The remaining chapters contain the topics which appear among the various college entrance requirements in algebra. In the main, all colleges enumerate the topics considered in Chapters I to XIV. The other topics are enumerated by one or more institutions. Obviously, teachers will need to select the chapters that meet the needs of their classes. (e) Special attention is directed to Chapters XIII and XIV, on Exponents and Radicals. These chapters, while of con- siderable mathematical interest, probably are retained in sec- ondary courses largely because they appear among college entrance requirements. Usually they are taught in the first course. They are so difficult that students acquire there little knowledge and very little skill in dealing with them. Where preparation for an examination is a special aim, the authors are confident that instruction on these topics toward the latter part of the third semester of algebra will be found not only more pedagogical but more timely. The examples selected for this text are as simple as possible. More difficult examples on Exponents occur in the supplementary set F. Attention is directed to the practical turn given to radicals in § 126. (/) Chapter XXIV contains a number of topics that will have interest for some teachers CONTENTS CHAPTER PAGE I. The Fundamental Operations . . . 1 II. Special Products and Factoring • . . . 14 III. Fractions . .24 IV. Simple Equations 33 V. Graphical Representation . . . . . 46 VI. Simultaneous Linear Equations . . . 50 VII. Square Root and Quadratic Surds . . .63 VIII. Quadratic Equations 74 IX. Special Products and Factoring .... 100 X. Quadratic Equations having Two Variables . 117 XI. Simultaneous Equations 124 XII. The Theory of Quadratic Equations . . 135 XIII. Exponents 140 XIV". Radicals 150 XV. Logarithms . . 170 XVI. Progressions 187 XVII. The Binominal Theorem 202 XVIII. Ratio, Proportion, and Variation . . . 206 XIX. Graphical Representation of Complex Numbers 220 XX. Equations in the Quadratic Form . . . 224 XXI. The Binomial Theorem (Concluded) . . .227 XXII. Permutations and Combinations .... 231 XXIII. Determinants ........ 289 XXTV. Supplementary Topics ...... 256 INDEX 283 v ALGEBRA I. THE FUNDAMENTAL OPERATIONS UPON ARITHMETICAL NUMBERS 1. The Number System of Arithmetic consists of the integers and the common fractions. The following facts about arith- metical numbers are collected here for reference : (a) The sum, the product, and the quotient of two arithmeti- cal numbers is always an arithmetical number ; the difference between two such numbers, however, is an arithmetical num- ber only when the minuend is greater than the subtrahend. Division by zero is not allowed. (b) The Associative Law of Addition. The sum of three or more numbers, addends, is the same in whatever manner the addends are grouped. Thus, a + b + c =(a + b) + c = a + (b + c). (c) The Commutative Law of Addition. The sum of two or more addends is the same in whatever manner the addends are arranged. Thus, a + b + c = a-\-c-\-b = c+b-\-a. (d) The Associative Law of Multiplication. The product of three or more numbers, factors, is the same in whatever man- ner the factors are grouped. Thus, a • b • c = (a • b) • c = a • (6 • c). 1 2 ALGEBRA (e) The Commutative Law of Multiplication. The product of two or more factors is the same in whatever manner the factors are arranged. Thus, a,'b-c = b-a-c = C'b'a. (/) The Distributive Law of Multiplication. If the sum or the difference of two (or more) numbers is multiplied by a third number, the product may be found by multiplying each of the numbers separately by the multiplier and connecting the results by the proper signs. Thus, a (6 + c — d)= ab -f ac — ad. 2. Positive and Negative Numbers. To avoid the difficulty that subtraction is sometimes impossible in the system of arithmetical numbers, the negative integers and fractions are added to the number system. The combined system of posi- tive and negative integers and fractions is the System of^ Rational Numbers. The following facts are learned in the first course in algebra : (a) To every arithmetical number, there corresponds a posi- tive and a negative number. The arithmetical number is the Absolute Value of each of the corresponding signed numbers. Thus, 3 is the absolute value of + 3 and of — 3. (b) The rational scale is : -5 -4 -3 -2 -1 +1 +2 -1-3 +4 45 1 t— 1 1 1 1 1 H- 1 1 The fractions correspond to points between the points marked by the integers. Any negative number precedes all of the positive numbers, and may, therefore, be regarded as being less than any positive number ; of two negative numbers, the one having the greater absolute value is the less. (c) The sum of any positive number and the corresponding negative number is zero. Thus, (+3) + (-3)=0. THE FUNDAMENTAL OPERATIONS 3 (d) To add two signed numbers having the same sign, add their absolute values and prefix their common sign. Thus, (_10) + (-2)=-12. (e) To add two numbers having unlike signs, find the differ- ence between their absolute values and prefix the sign of the one having the greater absolute value. Thus, (+2) + (-9)=-7. . (/) To subtract one signed number from another, change the sign of the subtrahend and add the result to the minuend. Thus, (_3)-(-8) = (-3) + (+8)= + 5. (g) To multiply one signed number by another, find the product of their absolute values, and make it positive if the numbers have the same sign, but negative if they have unlike signs. Thus, (- 3) • (+ 9)= - 27, and (- 5) . (- 16)= + 80. (h) To divide one signed number by another, find the quo- tient of their absolute values, and make it positive if the numbers have the same sign, but negative if they have unlike signs. Thus, (+39)-*-(-3)=-13, and (- 45)-s-(+5) = - 9. 3. The Fundamental Operations are ' addition, subtraction, multiplication, division, involution, and evolution. Involution is the process of finding the product when a given number is used as a factor two or more times. The number itself is called the Base; the result is called a Power of the •base ; the Exponent, written at the right of and above the base, indicates the number of times the base is used as a factor. Thus, 3* = 3 • 3 • 3 • 3 = 81, and (2 6)8 = (2 6) • (2 6) • (2 6) = 8 6 3 . 4 ALGEBRA Evolution is the process of finding what number must' be used as a factor a specified number of times to produce a given number as product. The given number is called the Radicand ; the result is called a Root of the radicand ; the Radical Sign, V , with the proper Index denotes the desired root. Thus, VS indicates the cube root of 8 ; 8 is the radicand, 3 is the index of the root, and the root itself is 2, for 2 3 = 8. Evolution is not always possible in the system of rational numbers ; thus, there is no rational number which exactly ex- presses the square root of 2. Order of Operations. In a sequence of the fundamental opera- tions on numbers, it is agreed that operations under radical signs or within symbols of grouping shall be performed before all others ; that, otherwise, all multiplications and divisions shall be performed first, proceeding from left to right, and afterwards all additions and subtractions, proceeding again from left to right. EXERCISE 1 1. Illustrate by arithmetical examples the five fundamental laws in paragraph 1. 2. Give an arithmetical example in which subtraction is impossible. Give the result when dealing with signed numbers. 3. Subtraction is said to be the inverse of addition, and division of multiplication. What is meant by these statements ? 4. What are the two signed numbers corresponding to J ? Give their sum, their difference, their product, and their quotient. 5. Perform the following indicated operations : («) (+6) + (+10). (e) (-4) + (-3). (6) (_4) + (+9). (/) (+6) + (-3). (c) (_8) + (+6). (fir) (+6)-(-5). (d) ( + 2) + (-3). (A) (_4)-( + 2). THE FUNDAMENTAL OPERATIONS (9 ("5)-(-3). ($ (-5). (-9). 0) (-4)-(-8). (o) (+12). (-3). (*) (+3)-(+9). (p) (+25). (-5). (0 (+6). (-5). (?) (_28) + (+4). (m) (-7). (+8). (r) (-22) + (-11). 6. What is the value of 3 4 ? of 2 5 ? of 6 3 ? of (-2) 2 ? of (_3)3? of (-4) 2 ? of (-o) 3 ? 7. What is the sign of the product of an even number of negative factors ? of an odd number ? Give an example of each. 8. What is the sign of an even power of a negative num- ber ? of an odd power ? Give an example of each. 9. Read the following product: btfytfw. Give the expo- nent of each of the literal numbers. What is the value of the product when x = 2, y = — 3, z = — 2, and w = 5. 10. Find the values of the following for the values of x, y, z, and w given in Example 9 : (a) Sx 2 . (c) ^ + 2/3. (e) x 2 -3x + 4. (6) x 2 y. (d) w 2 -z 2 . (f) z 2 -4z-4:. (g) z 2 -2zw + w 2 . (j) a^+y 4 . (h) **""**• (k) !° + l_I (t) x A -2x* + x 2 -x + l. h J z 2 x 2 y z Write the following in symbols and find their values : (I) the sum of the squares of x and y. (m) the difference between the cubes of w and x. (n) twice the product of y and z, diminished by three times the quotient of x and iv. 11. The following formulae occur in applications of algebra. Those marked with an asterisk, *, occur in geometry ; try to recall the theorems they express. Express the others in words. 6 ALGEBRA (a) * A = | a&. Find J. when a = 25 and 6 = 40. (&)* .4 = £ft(& + 6'). Find .4 when/* = 10, 6 = 30, and &'=20. (c) ^4 = pfl + ^-V Find A when P = 5000, r = 5, and t = 4 * ' (if) F= 4 tt# 3 . Find F when tt = 3±- and R = 3. (e) £ = 7t a/- • Find * when I = 8 and g = 32. (/) s = a£-fi^ 2 . Find s when a =50, t = 3, and # = 32. (0) * 7i 2 = a 2 + & 2 . Find h when a = 6 and b = 8. (A) S = ttI(R +r). Find # when I = 10, 72 = 5, and r = 3. (f) p = a 2 -fft 2 - c 2 Findi> when a = 8? b = 9j and c = 10 ( j) a = ^+- ch. Find .4 when h = 5 and c = 30. THE FUNDAMENTAL OPERATIONS UPON NUMBER EXPRESSIONS 4. (a) An Expression is a symbol for a number, consist- ing of numerals and literal numbers connected by some or all of the signs denoting mathematical operations; as, ^2 ajty + 4 ^ - iv\ (b) A Monomial or Term is an expression whose parts are not connected by the signs + or — ; as, 6 rV£. (c) A Binomial is an expression having two terms. (d) A Trinomial -is an expression having three terms. (e) A Polynomial is an expression having more than one term. (/) A polynomial is said to be arranged in descending powers of one of its letters if the term containing the highest power of that letter is placed first ; if the next lower power is placed second ; and so on. (g) Any factor of a product is the Coefficient of the product of the remaining factors. If one factor is expressed in nu- THE FUNDAMENTAL OPERATIONS 7 merals and the remaining factor in letters, the former is called the Numerical Coefficient of the latter. (h) A Common Factor of two or more terms is a factor of each of them. (?') Like or Similar Terms are terms that are alike in their literal parts ; Unlike or Dissimilar Terms are terms which are not alike in their literal parts. Terms are like with respect to one or more factors if they have these factors as common factors ; thus, 3 a(x — y) and 4 b(x — y) are like w T ith respect to (x — y). 5. Addition and Subtraction of Expressions. Rule. — To add two or more like terms : 1. Multiply their common factor by the sum of its coefficients. Thus, 2 a(x - y) + 3 b(x - y) = (2 a + 3 b) (x-y). This rule follows from the distributive law of multiplication, §1,/ Rule. — To add polynomials ; 1. Write the polynomials with like terms in vertical columns. 2. Add the columns of like terms, and connect the results by their signs. This rule follows from the commutative and associative laws of addition, § 1, b and c. Rule. — To subtract one term from a like term or one polynomial from another : 1. Write like terms in vertical columns. 2. Imagine the signs of the terms of the subtrahend changed, and add the resulting terms to those of the minuend. 6. Parentheses, ( ), Brackets, [ ], Braces, j J, and the Vin- culum, ~, are symbols of grouping, used to indicate terms which are to be treated as parts of a single number expression. Thus, 3a -(2x + y - z) means that 2 x + y - z is to be subtracted from 3 a. 8 ALGEBRA Rule. — To remove parentheses preceded by a plus sign : Rewrite all terms which are within the parentheses without changing their signs. Rule. — To remove parentheses preceded by a minus sign : Rewrite all terms which are within the parentheses but change their signs from + to - , or from - to -f . Sometimes terms must be inclosed within parentheses. Rule. — 1. To inclose terms within parentheses preceded by a plus sign, rewrite the terms without changing their signs. . 2. To inclose terms within parentheses preceded by a minus sign, rewrite the terms, changing their signs from + to - , or from - to + . Thus, r + s-t-r + (s-t) = r — (— s + t). EXERCISE 2 1. Consider the monomial 5 ab 2 &{x — ?/). (a) What are its factors ? (b) What is its numerical coeffi- cient ? (c) What are the exponents of a, b, and c respectively ? (d) What is the coefficient of (x — y) ? of b 2 ) + 8(c - d). THE FUNDAMENTAL OPERATIONS 9 8. Add 14(* + y) — 1"0 + z), KV + z ) - 9 ( 2 + »)i and — 3(x + y)—7(z + x). 9. Add 2 ax + 3 &x — 4 ex. 10. Add 5 mx 2 -j- 2 nx?/ +J9?/ 2 and tx 2 — rxy + qy 2 . 11. From 8 a; + 2// — 7 z subtract 8 x — 2y + 7 z. 12. Subtract 5 n 3 - 9 - 14 n 2 + 16 n from 7n 2 + 20n 3 -5 + 13 ». 13. Take 49 x 2 -f 16 m 2 - 56 mx from 25 m 2 + 36 x 2 - 60 mx. 14. Subtract — 5{a + b) + 9(c - d) from 7(a + 6) - 6 (c — a"). 15. From 3(x + ?/) 2 -2(x + 2/)+5 take (a?+y) 2 + 3(a? H- y) — 7. 16. What expression must be added to 3 x 2 — x -f 5 to give 0? 17. By how much does 2 m — 4 m 2 — 15 -f 17 m 3 exceed — 9 + 6 m 3 — 11 m — 14 ra 2 ? 18. From the sum of 2x 2 — 5xy — y 2 and 7 x 2 — 3 xy + 4 ?/ 2 , subtract 4 x 2 — 6 xy + 8 y 2 . (Do it all in one operation, if possible.) 19. From 7 x— 5z — 3y subtract the sum of 8 ?/ + 2 x — 11 z and 6 z — 12 y + 4 x. 20. From the sum of 7 x 3 — 4 x 2 + 6 x and 3 x 2 — 10 x — 5, take the sum of - 5 x 3 + 4 x + 12 and 8 x 3 - 11 x 2 - 2. Remove parentheses and combine the terms: 21. 2x-3y + (5x-y)-(-$x + 7y). 22. 5 a -(7 a -[9 a + 4]). 23. 2x-(8?/ + 5x-J5x-?/D-(-92/ + 3x). 24 . 8 a 2 - 9 - \ 5 a 2 - ( 3 a + 2) J + j 6 a 2 - (4 a - 7) j . 25. 5m-[7m-S-3m-(4w + 9)|-56m-8|]. 26. 25 -(-8 -[-34 -16 -47]). 10 ALGEBRA Inclose the last three terms of the following in parentheses preceded by a minus sign : 27. a 2 -4?> 2 + 12 6 — 9. 29. a 2 + b 2 - c 2 + d 2 . 28. 4 x 2 — y 2 — 2 yz — z 2 . 30. n 4 — 8n 2 -f 6 n , + 7. 7. Multiplication. The Law of Signs for Multiplication is stated in paragraph 2, (g). (a) The Law of Exponents. The exponent of any number in a product is equal to its exponent in the multiplicand plus its exponent in the multiplier. This law is proved for certain exponents in paragraph 115. Rule. — To find the product of two monomials : 1. Find the product of their numerical coefficients, using the Law of Signs for Multiplication. (§ 2, g.) 2. Multiply this result by the product of the literal factors, using the Law of Exponents for Multiplication. This rule is a consequence of the commutative and associa- tive laws of multiplication. (§ 1, d and e.) Rule. — To find the product of a polynomial and a monomial : 1. Multiply each term of the polynomial by the monomial. 2. Unite the results with their signs. Rule. — To find the product of a polynomial and a polynomial : 1. Multiply the multiplicand by each term of the multiplier. 2. Add the partial products. These last two rules are consequences of the distributive law of multiplication. (§ 1, /.) It is desirable to arrange both multiplier and multiplicand according to the same order of powers of a common letter. THE FUNDAMENTAL OPERATIONS 11 Example. Multiply x 2 - y 2 4- 2 xy by y 2 + x 2 - 2 a#. t Solution : x 2 + 2 xy — J 8 x 2 - 2 xy + y 2 x 4 + 2 x 3 y — x 2 2/ 2 - 2 x 3 ^ - 4 x 2 y 2 + 2 xy 3 x 2 y 2 + 2 xy 3 - y 4 x 4 — 4 x 2 y 2 + 4 xy 3 — y 4 Note. Multiplication by detached coefficients is considered in § 256, and may be studied at this time, if desired. 8. One of the useful forms of multiplication is illustrated by the following example. Example. Find the product of 2 x — 3 y and a? — 3xy — 5 y 2 . Solution: 1. (2 x — 3 y) (x 2 - 3 xy - 5 y 2 ) 2. = 2 x(x 2 - 3 xy - 5 y 2 ) - 3 y(x 2 -3xy-5y 2 ) a = 2 x 3 - 6 x 2 y - 10 xy 2 - 3 x 2 y + 9 xy 2 4 15 y 3 4. = 2 x 3 - 9 x 2 y - xy 2 + 15 */ 3 . Note. The second step is often omitted. EXERCISE 3 Multiply : 1. - 8 a 4 by 7 ab\ 3. 9(a + 6) 2 by 3(a + 6) 3 . 2. -5a^c by - 12 afc* 4. 13(x-y) by -2(z-2/) 2 . 5. 10 a 3 6 + 7 «6 4 by - 6 ab\ 6. 3 a 2 - 2 a& - 4 b 2 by - 4 a 3 6 3 . 7. m 2 — m — 3 by— 2 m. 9. a 2 — 2 a& 4- b 2 by a — 6. 8. x 2 — 2 «?/ + 4 ?/ 2 by - 3 ay. 10. c 2 4- 2 ed 4- d 2 by c — eZ. 11. a; 2 — 2 xy + y 2 hy x 2 + 2 xy 4- ?/ 2 . 12. - 6 x 4- 2 a 2 + 8 by - 4 + x 2 4- 3 x. 13. a 3 + 53 + 2 a6 2 4- 2 a 2 6 by b 2 4- a 2 - 2 a6. 14. (a 4- 6 -2 c) 2 . 16. (a-2 6) 4 . 18. (^-m-^n* h: { «-z y y. 17. g*W; 19. |*-|y v 12 ALGEBRA Find the following products as in § 8 : 20. (2w 2 -4n-f7)(> + 2). 21. (4 m 2 + 9 n 2 - 6 mn)(2 m + 3 n). 22. (3a 2 -2a + 4)(2a-l). 23. (5m 2 + 3m-4)(6m-5). 24. (a 2 +4 xy -+ 16 ?/ 2 )(a — 4 y). 25. (5r 2 -3rs + 6s 2 )(2r-3.s). 9. Division. The Law of Signs for Division is stated in para- graph 2, h. The Law of Exponents. The exponent of any number in a quotient is equal to its exponent in the dividend minus its exponent in the divisor. This law is proved for certain expo- nents in paragraph 115. Rule. — To divide a monomial by a monomial : 1. Find the quotient of their numerical coefficients, using the Law of Signs for Division. (§ 2, h.) 2. Multiply the result by the product of their literal factors, using the Law of Exponents for Division. Rule. — To divide a polynomial by a monomial : 1. Divide each term of the polynomial by the monomial. 2. Unite the results with their signs. Rule . — To divide a polynomial by a polynomial : 1. Arrange the dividend and the divisor in either ascending or descending powers of some common letter. 2. Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. 3. Multiply the whole divisor by the first term of the quotient ; write the product under the dividend and subtract it from the dividend. THE FUNDAMENTAL OPERATIONS 13 4. Consider the remainder a new dividend, and repeat steps, i, 2, and 3. Example. Divide 2 a 4 + 8 a - a? + 15 by 2 a 2 - 3 a + 5. ). (0 (r + 5*)(r-9«). (d) (*-5)( 2 -8). (i) (# + S4)(c?-12d), (e) (w-3s)(w-7s). (A;) 0?/ + 7 2)0*/ - 12 z). (/) (a-5 6)(a + 3 6). (1) (m 3 - 3 r)(m 3 + 11 r). (m) What is the sign of the third term of a product of this type when the signs of the second terms of the binomials are unlike ? 4. To what type do the following belong ? Multiply : (a) + 3)(2a + l). (d) (2aj-3)(2aj-l). (6) (3a? + l)(2a> + l). (•) (3x-4y)(x-2y). (c) (5 x + 2)(* + 3). (/) (5 a; - 3 y)(s - 2 y). 16 ALGEBRA (g) (a + 2d)(3a-&). Q) (llp + 3q)(2p-3q) (h) (c-3cZ)(3c + 5d). (m) (2 xy + 7)(3 ay - $). (i) (3ffl-4w)(m + 2)i). («) (10 rarc-3 «)(2 wm+3*), (j) (7c-2d)(3c + 4d). (o) (5* + 6rs)(6*-7rs). (A;) (2r*-5*)(3r*-i-2*). (/>) (12x 2 -5y)(3 x 2 + 2y). Find mentally the following products : 5. (5m -2n) 2 . 28. (5 a 8 - 6 ?/ 2 )(5 aj 3 + y 2 ). 6. (a + lly)(a+3y). 29. (5a + i)(5a-i). 7. (x + 12y)(x-2y). 30. (4^-3^. 8. (2»-F3)(*+4). 31. (3 y +7)0/ -5). 9. (a 2 -4 2/)(a 2 + 4?/). , 32. (2 - 3 ay) (5 + 2 xy). 10. (2cd-7) 2 . 33. (x*y + 6 z)(x*y - 13 z). 11. (3 a 2 //- 4 2) (3 ^-4 2). 34. (ay+5)(*p~4), 12. (2»i-5)(m + 4). 35. (a 3 + 7)(a 3 - 11). 13. (2p 2 -7)(3i> 2 + 5). 36. (3a + 5)(7*-8). 14. (r 2 -3s)0« 2 + 7s). 37. (l-9r)(l + 8r). 15. (x + i)(x- J). 38. (3 c + c/ 3 )(2 c - d 8 ). 16. (Jm + 5p)(£m — 5p). 39. (5m-3]y)l 17. (a-i)(a-i). 40. (12a-i)(9a-|). 18. (2a + 3)(£a + l). 41. (20 - 16 2) (3 + 2 z). 19. (3a 2 + 4 6)(3a 2 -4 6). 42. (a + 16 ?/ 3 )(a- ?/ 3 ). 20. (y_10)(y + 4). 43. (2/ - 6 a 2 )(?/ + a 2 ). 21. (a — |)(a + f ). 44. (4 r + s*)(4 r — 5 st). 22. (l-6«)(3 + 4«). 45. (6 m 3 -|) 2 . 23. (2^-7^(3^-4^. 46. (1 +23 z)(5 -z). 24. (3 m _|)(3 m _ hi)> 47 . (5 a 2 - 4^(6 a 2 -5 y). 25. (3t-7r)(2* + dr). 48. (a (5 -?/ 6 )(a 6 + 2/ 6 ). 26. (11«*-.1)(12* + 1). 49. (9a+2y)(3a-4y). 27. (z 2 -6)(z 2 + 12). 50. (12c + 5d)(4c-3d). SPECIAL PRODUCTS AND FACTORING 17 FACTORING 11. A monomial is said to be Rational and Integral when it is either an arithmetical number or a single literal number with unity for its exponent, or the product of two or more such numbers ; as, 3, a, or 2 a?bc 2 . 12. A polynomial is said to be rational and integral when each term is rational and integral ; as, 2 a?b — 3c + d 2 . 13. To Factor an algebraic expression is to find two or more expressions which will produce the given expression when they are multiplied together. It is agreed that only rational and integral factors of integral expressions will be considered. 14. A number which has no facfors except itself and unity is called a Prime Number ; as, 3, a, x + y. 15. Type Forms. Skill in factoring depends upon ability to recognize the type forms. The following forms were studied in the first course in algebra : (a) Removing a Monomial Factor. A monomial factor of an expression is a number which will exactly divide each term of the expression. , mx + my + mz = m(x + y + z). Example. 3afy + 9afy + 3a# = 3 xy(x 2 + 3x + l). (b) The Difference of Two Squares equals the product of the sum and the difference of their square roots x 2 -y 2 = (x-y)(x + y). Example. 25 x* — y A = (5 a? + y 2 )(p a 3 - y 2 )- (c) A Perfect Square Trinomial is one which has two terms that are perfect squares and whose remaining term is twice the product of their square roots. To Find the Square Root of a perfect square trinomial, ex- tract the square roots of the two perfect square terms, and connecb them by the sign of the remaining term. 18 ALGEBRA Example 1. 4 a; 2 - 12 x + 25 is not a perfect square ; for V4^" 2 = 2x, V25 = 5, and 2 . 2 x • 5 = 20 x and not 12 a. Example 2. 4 a? — 20 a# + 25 ?/ 2 is a perfect square ; its square root is 2 x — 5. Hence 4 # 2 — 20 a??/ -f- 25 y 2 = (2 # — 5 #) 2 . (d) Trinomials of the Form x 2 +px + q can be factored when two numbers can be found whose product is q and whose sum is p. Example. x 2 -S x- 65 = (x + 5) (x - 13), for 5(-13) = - 65 , and (+ 5) + (- 13) = - 8. (e) Trinomials of the Form ax 2 + bx + c, if factorable, can be factored as in the following example : Example. Factor 15 x 2 -f- 17 x — 4. Solution : For 15 x 2 , try 5 x and 3 as, thus : (5x )(3a; ) . For — 4, try 2 and 2, with unlike signs, arranging the signs so that the cross product with larger absolute value shall be positive ; thus : (5 x — 2 )(3 x + 2). Middle term +4x; incorrect. (See § 10, d.) For — 4, try 4 and 1, arranging the signs as before ; thus : (5 x + 4) (3 x — 1) . Middle term, + 7 x ; incorrect. Try (5x— 1) (3 a + 4). Middle term, + 17 x ; correct. Note. This method of factoring the general trinomial is the most use- ful. While at first it may seem difficult, it is easily mastered. It applies also to each of the preceding forms. (See First Year Algebra, § 103.) (/) The Sum of Two Cubes, x 3 + f = (x + y)(x 2 - xy + y 2 ). Example, m 6 + 27 f = (m 2 ) 3 + (3 y) z = (m 2 + 3 ?/)(ra 4 - 3 mhj + 9 y 2 ). (a) The Difference of Two Cubes, x 3 — # 3 =(x — y){x 2 + xy + y 2 ). Example. 216 a 9 - b* = (6 a 3 )~ - 5 3 = (6a 3 - 6)(36 a 6 + 6 a 3 6 + 6 2 ). SPECIAL PRODUCTS AND FACTORING 19 16. Complete Factoring. First remove any monomial factor present in the expression ; then factor the resulting expres- sions by any of the preceding type forms which apply until prime factors have been obtained. Example 1. 3 of - 3 tf = 3(> 6 - f) = 3(ar 3 + y 3 ) (a? - y z ) = 3{x -f- y) (x 2 — xy + y 2 )(x — y) (x 2 + xy + y 2 ). Example 2. 24 ax 2 + 22 axy - 10 ay 2 = 2 a(12 x 2 + 11 xy - 5 y 2 ) = 2 a(4 x + 5 y)(3 a; - y). Note. Advanced Topics in Special Products and Factoring are dis- cussed in Chapter IX. This chapter, in whole or in part, as far as para- graph 95, may be studied at this time if desired. EXERCISE 6 1. Remove the monomial factor : (a) 2 ax — 6 ay + 7 az. (/) m (a — 2) 4- n(a —2). (6) IS mhi- 15 mn 2 . (g) a(x> - 3) -5(x 2 -3). (c) 4 r 1 - 8 r 2 - 4 r. (h) 5a(x-y)-3 b(x - y). (d) 7n 7 -7 n. (i) a(m 2 - 2) - 3(m 2 - 2). (e) a 6 - 5 a 5 - 2 a 4 + 3 a 3 . (?) z 2 (a - 6) + x\c - d). 2. To what type do the following belong ? Factor : (a) x 2 -36. (/) A^-iV^- (6) 4 m 2 -81. (g) l-64m 2 n 2 . (c) 9 « 2 - 4 2/ 2 . (h) 36 a 2 - 121 y 2 . (d) 25 m 4 - 1. (i) 1-81 a 2 6 2 c 2 . (e) x»-f. (j) 226- f. 3. Supply the missing term so as to make perfect squares of: ( a ) o2+( )+64. ' (c) a 2 -( ) + 9 6 2 . (6) m 2 -( ) + 144. (d) 4x 2 + ( ) + 25. 20 ALGEBRA (e) 9m 2 *i 2 -( )+4. (h) m 4 -14m 2 + ( ). (/) ^ + 8« + ( ). (0 eP-8 3" 28. a — a u a u — u 3a 2 6 ' 2a6 2 ' 26. 1 a?b 3¥c 2c 2 a 6 ' 10 ' 15 * 29. 4 a 2 2 a 2 - 9' 3 a 2 - 9 a 27. 5 4 6 2 mhi 3 mn? 5 mn 30. 5 3n m 2 — 4 m -h 4' m 2 — 4 31. 3 a 2 a 3 + 27' a 2 -a- 12" 32. 1 3 mn 2 m 2 ? i 2 ft) 2 ' m — n' 2(m — n) 2 ' 3(m — 33. 2 4 6 x + 2' a - 2' a 2 - 3' 34. a + 3 6 a — 36 a + 46 a 2 -7«& + 12& 2 ' a 2_ a & -12 6 2 ' a 2 -9 6 2 ' IK 2x + 3 x + 2 a? — 5 a-2 + 3 a. _ io 7 2 a 2 + 7 a - 15 7 2 a 2 - 7 x + 6 26. Addition and Subtraction of Fractions. Rule. — 1. Reduce the fractions, if necessary, to respectively- equivalent fractions having their lowest common denominator. (S 25). 2. For the numerator of the result, combine the numerators of the resulting fractions, in parentheses, preceding each by the sign of its fraction. 28 ALGEBRA 3. For the denominator of the result, write the L. C. D- 4. Simplify the numerator and reduce the fraction to lowest terms. (§22) Example. la 1 a a 2 — x 2 x 2 — a 3 a 2 — x 2 a 3 — x 3 1 a (a — x)(a + x) (a — x) (a 2 + ax + x 2 ) w ! + ax + x 2 ) a(a + x) (a — x) (a + x) (a 2 + ax + a 2 ) (a — x) (a + x) (a 2 + ax + x 2 ) _ (a 2 + ax + x 2 ) - a(a + x) x 2 (a — x)(a + x)(a 2 + ax 4- x 2 ) (a — x)(a + x)(a 2 + ax + x 2 )' EXERCISE 9 Perform the indicated operations : 4 a; + 7 6 a; — 5 1 1 10 15 ' I6x 2 -8x + l 16z 2 -l 3 a - 8 4 a - 9 Q a-1 a 4- 1 , 4 a 2 + 1 /£. — . O. 9 12 3. x—y y—2z.z—Sx xy 2yz 3 zx \ m 2 m — 2 ra 4- 2 5. a+3_a-3 a — 3 a + 3 . 9. 10. 11. 6. « + 3y __ a?-3y a + 1 a _l a 2_l ' 5 a? 4a 2 4-3a-l a-3 a; 2 + x _ 12 * 3^ + 2 3a-2 9 a 2 + 4 9 x 2 - 4' a; 2 2a 3 x 2 — xy -\- y 2 x? + y* 1 4a -1 a_32/ a + Sy a(2a-l) (2 a- l) 3 m — 1 m 4- 1 m — 6 m — 2 ra 4- 2 4 — ra 5 14. * + » a 2 H-4a& + 46 2 46 2 -a 2 15. FRACTIONS 29 1 2a; 2 + 5a; + 3 4z 2 + 8a + 3 n _ a — n 3(i-4m.3«-5m Id. -p 2a-\-2n 3 a -f 3 n 6 a -j- 6 w a a 17. a 2 + 4a-60 a 2 -4a-12 18. -J--^- + . »- 6 n + 4 1 — n w 2 + 3w-4 19. 1 + I + I (a-b)(a - c) (6 - c)(6- a) (c - a)(c - b) 20 . *-3 + ^ + 27 a 2 + 3a + 9 27. Multiplication and Division of Fractions. Rule. — To find the product of two or more fractions : 1. Find the prime factors of the numerators and denominators of the fractions. 2. Divide out (cancel) factors common to a numerator and a denominator. 3. Multiply the remaining factors of the numerators for the numerator of the product, and of the denominators for the denomi- nator of the product. Rule. — To divide one fraction by another : 1. Invert the divisor fraction. 2. Multiply the dividend by the inverted divisor. Note 1. In problems involving both multiplication and division, per- form the operations in order from left to right. (§ 3.) Note 2. Integral or mixed expressions should first be reduced to fractional form. 30 ALGEBRA \v l vx x l J \ x — vj\v X __ f x 2 — 2vx + v 2 \ ( x — v + 2v \ _^_ ( x 2 -v 2 \ \ X 2 ^ 2 J\ X—V J \ XV J _ (x — v) 2 / #-f-?A xv _ 1 x 2 v 2 \x — vj (x — v) (x + v) XV Complex Fractions are a special case of division of fractions. x 2 -y 2 \ 2 J \ 6 J 2 (x- x+y 2 x 2 — y 2 fx+y 6 3 ■y){?+y) x-y EXERCISE 10 Perform the following multiplications : i. *: 42 36 d n 2 -36 In 2 35 4ti 2 n 2 + n-4:2 2 6 am 27 M 2 Qa 5 a 2 -2 a- 35 4 a 3 - 9 a — • 36 w 5 . 5. • - i 5 2 a 3 -3 a 2 7 (a -7) 3 5a ' ' Sb 2 96 3 7 c 4 , 5z + 2 , Q) 10c 3 6a 4 ' 2a 2 + a-10 S ; * 7 4 m 2 4-8ra + 3 6 m 2 -9 m 2 ra 2 - 5 m + 3 4 m 2 - 1 g 16a-4 20a + 5 x 2 -\-2x + l ox — 5 6 # -f 6 16 # 2 — 1 9 cc 3 + 8 y 3 x — 2y x 2 4- 2 xy 4- 4 ?/ 2 a^ — 8 # 3 a; + 2 ?/ a; 2 — 2 xy + 4 ?/ 2 10. 2 n 2 - n - 3 w 2 + 4w + 4 n 2 - w - 2 n 4 -8w 2 + 16 n 2 + ri 2w 2 -3n FRACTIONS 31 Perform the following divisions : 11 4« 2 -25 . /o„ kn 12 a 2 -a6-26 2 . a-26 a*-3xy . ar 2 -10 3;y + 21?/ 2 /jj _ ^3 a? 2 -j- icy H~ 2/ 2 14 8n 3 + 1 . 4n 2 -2n + l , 2?i 2 + 4n * w 2 + 4 n + 4 ' 2 a 2 _ a& _ 3 &2 3 a 2 + a6 _ 2 b 2 15. 9 a 2 - 25 6 2 9 a 2 - 30 ab + 25 6 2 Perform the indicated operations : 6a 2 -«-2 12a 2 -5a-2 4 a 2 - 9 16 17. 4a 2 -16a + 15 8a 2 -18a-5 4a 2 +6a + 2 4 a 2 _ 4 a + 1 2a 2 + a . a 2 -2a 4 a 2 - 1 ' 8 a 3 - 1 a 2 - 4 ' 18 2 x 2 -5xy-3y 2 . / 2a; 2 - 7 xy - 4 y 2 . a 2 - 4 an/ + 4 y " y x 2 — xy — 2y 2 ' \x 2 — Sxy — 4=y 2 x 2 — xy — 6 y 2 19 /q + 2 2 V a 3_\ V « a-3J\a-2 a + SJ 20 . f 2 _^ + 4x-2n /,jl ^ V x* + 2x-8J \x-2 x + V x + 4 J \ x + 4 J „. H . 2 , _J£ 24 . « (X — -L ~r x y 9 a m 32 ALGEBRA 28. 1 — 25. 26. 27 a 2 b 2 _6 a 9a Q 6 o a 1 1 1-a i-f as 1 1 tf 2 1 + X 2 K)K) 29. 1 + tt 2 5a- *•-* 1 _2a L -f_5 3a-2 X 2_tfy_±tf 27. > 1Z^ ZJL. 30. a + 2 + - x-f--^— x %-\-y IV. SIMPLE EQUATIONS 28. An Equation expresses the equality of two numbers. The two parts (numbers) are called respectively the Left Member and the Right Member of the equation. 29. An Identity or Identical Equation is an equation in which the two members may be made to take exactly the same form by performing the indicated operations. As, (a) 10 + 2 = 15 + 2-4-11. (b) 3x(a- b)=3ax-3bx. (a) is a numerical identity ; (b) becomes a numerical identity for any values of the literal numbers. 30. An equation is said to be satisfied by a set of values of the letters involved in it when it becomes a numerical identity if these values are substituted for the letters. 31. A Conditional Equation is an equation involving one or more literal numbers which is not satisfied by all values of the literal numbers. Thus, Sx — 4 = aj + 6 is an equality only when x = 5. The word equation usually refers to a conditional equation. 32. If an equation involves two or more literal numbers, one or more of the literal numbers may be regarded as unknowns and the remaining ones as known numbers. Thus, in ax + by = c, x and y may be unknowns and a, 6, and c may be known numbers. 33. If an equation has only one unknown number, any value of the unknown number which satisfies the equation is called a Root of the equation. 33 34 ' ALGEBRA (a) x 2 -f- 6 = 5 x has the roots 2 and 3. When x = 2, 2 2 + 6 = 5 • 2, for each equals 10. ' When a; = 3, 3 2 + 6 = 5 • 3, f or each equals 15. (b) 2ax — a — x — \ has the root \. When x — i, 2 a • \ — a = | — j- for each equals 0. 34. To Solve an equation is to find its root or roots. Example. 1. Solve the equation 2 x — 3 = x ~*~ • 2. Multiply both members by 2. 4 # — 6 = x + 6. 3. Subtract # from both members. 3 x — 6 = 6. 4. Add 6 to both members. 3 x = 12. 5. Divide both members by 3. # = 4. The problem is to find the number or numbers which satisfy the given equation. To make certain that 4 satisfies the given equation substitute 4 for x in the equation. Does2.4-3 = t±5? Does 8-3 = — ? Yes. 2 2 This test is called checking by substitution. 35. Axioms Used in Solving an Equation. In the solution of the equation in paragraph 34, the following axioms were assumed : (a) The same number may be added to both members of an equation without destroying the equality. (b) The same number may be subtracted from both members of an equation without destroying the equality. (c) Both members of an equation may be multiplied by the same number without destroying the equality. (d) Both members of an equation may be divided by the same number without destroying the equality. As long as the number used as an addend, subtrahend, multi- plier, or divisor in connection with these axioms is an arith- SIMPLE EQUATIONS 35 metical number (or constant other than zero), the equation obtained by applying any one of the axioms has exactly the same roots as the given equation, provided the given equation has a root at all. 36. In this text, symbols A, S, M, and D are used to abbre- viate the explanations of solutions of equations. Thus : A 5 : means " add 5 to both members of the previous equation." S_ 3n : means " subtract — 3 n from both members of the previous equation." M_ x : means "multiply both members of the previous equa- tion by — 1." D 4 : means "divide both members of the previous equation by 4." Example. Solve the equation ^— ? - 2x-3 _ 8_-3z 1 4 2^ 3 5 1. M 30 : 15(a;- 2)- 10(2 3 -3) =6(8 -3s) + 30. 2. .-. 15 x - 30 - 20 x + 30 = 48 - 18 x + 30. 3. .-. -^5 x = 78 - 18 x. 4. Ai 8x : 13ic = 78. 5. D 13 : x = 6. Check: Does «-2 _ 12-3 = 8_^18 2 3 5 Does 2-3=-2 + l? Yes. EXERCISE 11 Determine by substitution which of the numbers 1, 2, —5, and \ are roots of each of the equations. 1. 4r+-5 = 6+-3r. 2. 16s-l=4s + 5. 3. w; 2 -3m?=-2. 5. 4,-t 12 1-t 3-t 2a2-3z+l = 0. Solve the following equations : 6. 10 x- 3 = 4 + 3 x, 8. 7. 21 y- 23 = 51 -16 y. 9. 3(2«-l) = 8(aj-l). 5/_i_4/ — 41 36 ALGEBRA 10. 2a — |a + |a = 10. 11. fz = Jtf-fa; + -V3. 12. 4(2v-7)+ 5 = 5(v-3) + 16. 13. a- 2(4 - 7 a;) = 4 aj- 9(2 -3 a). 14. (l + 3a) 2 = (5-z) 2 + 4(l-a)(3-2a;). 15. 5 (2 r + 7) (r - 2) - 6 (r + 4) 2 = 5 + (2 r + 3) 2 . 37. Discussion of the Axioms. Two equations are said to be Equivalent when the roots of either are the roots of the other. As remarked in paragraph 35, when the number used as an addend, subtrahend, multiplier, or divisor is an arithmetical number, then the given equation and the resulting equation are equivalent. Axioms (a) and (b). If the number added to or subtracted from both members of an equation is an expression involving the unknown number, the given equation and the resulting equation are equivalent, provided the expression has a finite value for the root or roots of the equation. Axiom (c). If the multiplier is an expression involving the unknown number, the new equation may not be equivalent to the given equation. Example. Sx — 2 = x — 1 has the root \. Multiplying both members by x — 2, Sx 2 -8x + 4=x 2 -Sx-\-2, or2x 2 -5x-\-2 = 0. This equation has the root 2 f or 2 . 2 2 - 5 • 2 + 2 = 8 - 10 + 2 = 0. The given equation does not have the root 2, for 3-2 — 2 does not equal 2 — 1. Hence, multiplying the given equation by x — 2 introduces the root 2. When the multiplier is an arithmetical number, or is the lowest common multiple of the denominators of a fractional equation, the resulting equation and the given equation are equivalent. SIMPLE EQUATIONS 37 Axiom (d). If the divisor is an expression involving the un- known number, the new equation and the given equation are not equivalent. Example, x 2 - 4 = x - 2 has the roct 2, for 2 2 - 4 = 2 - 2. Dividing both members by x '• — 2, x + 2 = 1. This equation does not have the root 2, for 2 + 2 does not equal 1. Whenever an equation is divided by an expression involving the unknown number, one or more roots of the given equation are lost. The following example illustrates a common instance of this fact. Example. 3 x 2 — 2 x = 0. B x : 3 x - 2 = 0, or x = }. | is indeed a root of the equation 3 x 2 — 2 x = 0, but it is not the only root, x = is also a root, for 3 • 2 — 2 • = 0. This root is obtained by setting the divisor x equal to zero. In general, if the expression by which both members of an equation is divided is set equal to zero, the roots of the result- ing equatipn are also roots of the given equation. 38. Mechanical Processes of Solving Equations. (a) Transposition. A term may be transposed from one mem- ber of an equation to the other provided its sign is changed. Proof. Let x 4- a = b. S : x = b — a. Axiom (?>), § 35 (b) Cancellation. A term which appears in both members of an equation may be cancelled. Proof. Let x + a — b + a. S a : x = b. Axiom (6), §35 (c) Changing Signs in an Equation. The signs of all of the terms of an equation may be changed. Proof. Let ax — b = c — dx. M_i : — ax -f- b = — c + dx. Axiom (c), § 35 38 ALGEBRA (d) Clearing of Fractions. An equation may be cleared of fractions by multiplying both members by the lowest common denominator of the fractions involved. This process is based upon axiom (c), § 35. 5 8 Example. Solve the equation = 4a-3 7#-3 Solution: 1. M^ x -sh7x-3) : 5(7 x — 3) — 8(4 x — 3) = 0. 2. .-. 86« — 16 — 32* + 24 = 0. 3. .-. 8* = ~9,ory=- 3. Check : Does = ? Does — — = ? _ 12 - 3 - 21 - 3 - 15 - 24 Does (-t)_(_i) B 0f Yes. EXERCISE 12 Solve the following equations : i i_J_ = i_JL 4 1. JL A. X' 2 9 a 9 6af 5< 10* 15 1 12' - 2 _? + A = i_ii - 2a-7 10*-3 3 y y 2 y 6y x 2 — 4 5 #(# + 2) 2iaH-7a + ll = 3 _27_ 8 18 7 a 2_4 a _9 ' ' ^_5 z + 2 Z 2_s z -10' 7 7m 5m _ 12(m 2 -l) m + 3 m — 1 m 2 + 2 m — 3 8. -^ 3U.+ 2 = 0. 2r + l 3r + 2 6r 2 +7r + 2 12* — 5 3s+4 _4s-5 10. •21 3(3«+l) 7 3 w - 5 4w + 2 == 15w-l 7 2 3w + 2 10 5 11 «* + 3 1 = 2a?-l ' 2(s»-8) 6(a>-2) 3(x 2 + 2x + 4) SIMPLE EQUATIONS 39 12. .05 a; -1.82 -.7 a; = .008 a; -.504. 13. 2.88 y - .756 + .62 y - .858 = .81 y. 14 *-3 f + 4 = St + 20 f.+ l I — 2 * 2 -£-2 Solve the following equations for cc: 15 - 4 + r+-= a + 6 + c - «o oc ac 3m 2a; + 5 m 3a; — ■4m 6 a 2 + 7 ma; — 20 m 2 3a; 2x + n .T-f-2 7i + 2a = 2. x + a x-2a x— a » + 3 a 2«a;-19a 2 _ Q ar* + ax — 6 a 2 x(a + 4=b)-b 2 , x — b __ x + a a 2 - -& 2 a-\-b a—b a b a — b x + b x-\- a x + a + b 16. 17. 18. 19. 20. 39. Algebraic Translation. In applications of algebra, number relations expressed in words must be expressed by means of algebraic symbols. This process may be termed " translation." Skill in making such translation depends in part upon care in reading the statement which gives the number relations and in part upon familiarity with a few simple devices. For the elementary instruction preparatory to the solution of the fol- lowing review exercises, see the " First Year Algebra." Example. Express in symbols : the sum of the squares of two given numbers decreased by 4 times their quotient. Solution : 1. Let x and y be the given numbers. 2. Then x 2 and y 2 are the squares of these numbers, and - is their quotient. * x 3. The expression is : x 2 + y' 2 — 4 - • 40 ALGEBRA EXERCISE 13 Express in symbols the following : 1. Five times a certain number. 2. The sum of the cubes of two given numbers. 3. 3 more than five times a given number. 4. The excess of 10 over a given number ; of y over 5. 5. 5 less than three times a given number; 7b diminished by c. 6. The amount by which 15 exceeds twice a given number. 7. The difference between 5 and 13 ; between a and 20. 8. Five per cent of x dollars ; a per cent of D dollars. 9. The simple interest on P dollars for four years at r per cent. 10. The amount to which M dollars accumulates in t years when invested at five per cent simple interest. 11. The number by which 3x exceeds (x— 6). 12. The larger part of 18 if s is the smaller part. 13. The smaller part of 3 x if (x + 5) is the larger part. 14. The larger of two numbers if c is the smaller and if the difference between them is 15. 15. The smaller of two given numbers whose difference is 4 if the larger is y. 16. The smaller part of 35 if the larger part is x. 17. The two integers consecutive to the integer represented by m. 18. The two odd integers consecutive to x: (a) if x is an odd integer ; (6) if x is an even integer. 19. The complement of a degrees ; the supplement. 20. The third angle of a triangle of which the other two are angles of x degrees and (x — 4) degrees respectively. SIMPLE EQUATIONS 41 21. The perimeter of a rectangle whose base exceeds its alti- tude by 3 inches. 22. The ages of A and B 8 years ago if A's age now is twice B's. 23. The difference between one fifteenth and one third of a certain number. 24. The total number of cents in a sum of money consisting of a certain number of dollars, twice as many quarters, and three times as many dimes as quarters. 25. The time required by a train for a trip of A miles : (a) at 30 miles an hour; (b) at r miles at hour. 26. The rate at which an automobile travels if it goes D miles : (a) in 7 hours ; (b) in n hours. 27. If the rate of a river is 3 miles an hour, express the time required by a boat whose rate in still water is x miles an hour : (a) for a trip of 20 miles downstream ; (b) for a trip of 20 miles upstream ; (c) for a trip of 20 miles downstream and back again. 28. The area of a parallelogram whose base is 3 feet less than twice its altitude. 29. The part of a piece of work a man can do : (a) in c days, if he can do all of it in 10 days ; (b) in 3 days, if he can do all of it in x days. 30. The reciprocal of 3 x. 31. The number whose hundreds' digit is a, whose tens' digit is b, and whose units' digit is c. 32. The number whose digits are the same as those of the number in Example 31, but in the reverse order. Express in words the following expressions : 33. \ab. 35. x 2 + y 2 — 2 xy. 37. a 2 -{-b 2 . 34. 2(x + y) + 3(x-y). 36. }m(w-f»). 38. (a 3 + 6 3 ) 3 . 39. *±*. X 2y2 40. fO+32. 42 ALGEBRA 40. A Problem is a statement of one or more relations be- tween one or more unknown numbers and certain known num- bers, from which the unknown numbers are to be determined. In this paragraph certain problems are considered which can be solved by using only one unknown number. Example. The rate of an express train is five thirds of that of a slow train. It travels 75 miles in one hour less time than the slow train. Find the rate of each train. Solution : 1. Let r = the no. of mi. in the rate per hour of the slow train. 2. Then | r = the no. of mi. in the rate per hour of the fast train. 8. Hence for one train for the other the rate is r the distance is 75 75 the time is 75 r 75 + fr 4. .-. ^ = 1. 75 fr r When this equation is solved, r is found to equal 30. Hence the rate of the slow train is 30 mi. an hour, and of the fast train, therefore, 50 mi. an hour. Check : The time for the slow train for 75 mi. is 75 + 30 or 2.5 hr. The time for the fast train for 75 mi. is 75 -f- 50 or 1.5 hr. The time of the latter is one hour less than that of the former. EXERCISE 14 1. The denominator of a certain fraction exceeds its numer- ator by 6. If the numerator be increased by 7 and the denomi- nator be decreased by 5, the fraction becomes i 7 3 -. Find the fraction. 2. Divide 55 into two parts whose quotient shall be 4. 3. Divide 54 into two parts such that twice the smaller shall exceed 29 as much as 143 exceeds four times the greater. SIMPLE EQUATIONS 43 4. The perimeter of a certain rectangle is 330 feet. The altitude is four sevenths of the base. Find the dimensions. 5. The numerator of a certain fraction exceeds the de- nominator by 5. If the numerator be decreased by 9, and the denominator be increased by 6, the sum of the resulting frac- tion and the given fraction is 2. Find the fraction. 6. Divide 197 into two parts such that, when the greater is divided by the smaller, the quotient is 5 and the remainder is 23. 7. The length of a certain lot is three times its width. If the length be decreased by 20 feet and the width be increased by 10 feet, the area will be increased by 200 square feet. Find its present dimensions. 8. If one fifth of the supplement of a certain angle be diminished by two elevenths of the complement of the angle, the result is 19. Find the angle. 9. A passenger train whose rate is 35 miles an hour and a freight train whose rate is 25 miles an hour start at the same time from points which are 100 miles apart, (a) If they travel toward each other, in how many hours will they meet ? (6) If they travel away from each other, in how many hours will they be 150 miles apart ? 10. A freight train runs 6 miles an hour less than a pas- senger train. It runs 80 miles in the same time that the passenger train runs 112 miles. Find the rate of each. 11. A man walks 10 miles and then returns in a carriage whose rate is 3 times as great as his rate of walking. If it took him 4 hours less time returning than going, what was his rate of walking ? 12. A vessel runs at the rate of 12 miles an hour. If it takes as long to run 27 miles upstream as 45 miles downstream, what is the rate of the current ? 44 ALGEBRA 13. A man travels 130 miles in ten hours in an automobile, part of the distance at an average rate of 20 miles an hour and the rest at an average rate of 10 miles an hour. How far does he travel at each rate ? 14. A man receives $ 140 per year as interest on $ 2500. $ 500 is invested at 5 % ; part of the remainder at 5i % ; and the rest at 6 %. How much has he invested at 5^ % ? 15. A man travels 24 miles at the rate of 5 miles an hour. By how many miles an hour must he increase his rate in order to make the trip in one fourth of the time ? 16. Find three consecutive numbers such that the square of the greatest shall exceed the product of the other two by 49. 17. Find the upper base of a trapezoid whose area is 175 square inches, whose altitude is 10 inches, and whose lower base is 20 inches. 18. Two automobilists use gasoline from a tank containing 60 gallons. If one uses gasoline at the rate of live gallons in three days and the other five gallons in seven days, how long will the 60 gallons last ? 19. If A can do a piece of work in 5 days, B in 8 days, and C in 10 days, how many days will it take them to do the work if they work together ? 20. A man has $ 6.90 in dollars, half dollars, and dimes. The number of halves is twice the number of dollars, and the number of dimes is equal to the sum of the number of dollars and the number of half dollars. Find the number of coins of each kind. SUPPLEMENTARY PROBLEMS 21. Divide a into two parts whose quotient shall be m. 22. The numerator of a certain fraction exceeds its denomi- nator by m. If the denominator be increased by n, the fraction becomes f. (a) Find the fraction, (b) Find the value of the fraction when m is 4 and n is 2. SIMPLE EQUATIONS 45 23. The perimeter of a certain rectangle is c feet. The base exceeds the altitude by d feet, (a) Find the dimensions of the rectangle. (6) Find the values of the dimensions when c equals 50, and d equals 5. 24. The length of a certain field is m times its width. If the length be increased by r feet, and the width by s feet, the area will be increased by t square feet, (a) Find the dimen- sions of the field. (b) Find the values of the dimensions when m is 4, and r, s, and t are respectively 2, 3, and 48. 25. Divide c into two parts such that the sum of one mth of the first part and one nth of the second part shall equal d. Find the values of the parts when m is 2 and n is 3. 26. If A can do a piece of work in a days, B in b days, C in c days, and D in d days, how many days will it take them to do the work if all work together ? 27. A sum of money amounting to m dollars consists entirely of quarters and dimes, there being n more dimes than quarters. How many are there of each ? 28. At what time between 8 and 9 o'clock will the hands of a clock be together ? 29. At what time between 2 and 3 o'clock is the minute hand of a watch 15 minute spaces ahead of the hour hand ? 30. In a mixture of sand and cement containing one cubic yard, 16 % is cement. How much sand must be added to the mixture so that the resulting mixture will contain 12 °J of cement ? V. GRAPHICAL REPRESENTATION 41. In the figure below: XX' is called the Horizontal Axis; YY' is called the Vertical Axis ; together they are called the Axes ; the point is called the Origin ; PR, perpendicular to the horizontal axis, is called the Ordinate of the point P; PS, perpendicular to the vertical axis, is called the Abscissa of P; PR and PS together are called the Coordinates of P. Distances on OX are considered positive, on OX' negative, on OF posi- tive, and on OY' negative. The part of the plane within the angle XOY is called the first quadrant; the part within the angle YOX' is called the second quadrant; etc. The abscissa of P, according to the indicated scale, is 3, and the ordinate is 4. The point P is called the Point (3, 4). Y _ 1 1 i" J., r p !, ! „ ! i , L xn 1 O Ik X ! !- fi -4 - 1 + ] ♦4 »4 ♦S + ~t 1 , 1 \ 1 1 &• , 1 1 j* 1 - , 1 ' V 46 GRAPHICAL REPRESENTATION 47 EXERCISE 15 1. What are the coordinates of each of the points in the figure above ? 2. Locate (plot) on a similar diagram the following points : (a) (0,5); (6) (-3,4); (c) (-5,0); ( X 6~ y^ - X -6 -S -4 -3 r9 " + + Q +3>4 + 5 +0 . >^ i^ ^ 2 ^ + ^ 7 ^^i i ^ -^ ;2 jp - E-*i-»r- - H-JS z: 2 EXERCISE 17 Draw the graphs of the following equations : 1. 2x+3y = 12. 3. 3 x + 2y= 0. 2. 3 aj - 5 i/ = 30. 4. 4 a -f 5 y = 24. 5. (a) Draw the graph ofsc+2y = — 1. (b) Multiply both members of the original equation by 3 and draw the graph of the resulting equation. (c) Multiply both members of the original equation by — 5 and again draw the graph of the equation resulting. (d) From the graphs obtained in parts (a), (6), and (c), what do you conclude is the effect upon the graph of an equation of multiplying both members of the equation by the same number ? (e) What effect does it have upon the solutions of the equation ? VI. SIMULTANEOUS LINEAR EQUATIONS 48. Two equations, each containing two (or more) unknowns, are said to be Independent Equations if each has solutions which are not solutions of the other. 49. Two independent equations which have one (or more) common solutions are called Simultaneous Equations. 50. Two independent linear equations which do not have a common solution are called Inconsistent Equations. 51. Graphical Solution of Simultaneous Linear Equations •Having Two Unknowns. Rule. — 1. Draw upon one sheet the graphs of both equations. 2. Find the coordinates of the point common to the two lines. These coordinates give the common solution. The solution may be checked by substitution. ( 5 x — 3 v = 19. (1) Example. Solve the equations J l7«+4jr-** (2) Solution : 1. For equation (1) : if x = 5, y = 2 ; if x = — 1, y = — 8. 2. For equation (2) : if x = - 2, y = 4 ; if x = 4, y = — 6|. 3. The graph follows. • 4. The straight lines intersect in the point (2, — 3). The common solution is x — 2, y = — 3. Check : In (1 ) : does 5 • 2 - 3( - 3) = 19 ? Yes. In (2) : does 7 • 2 + 4(- 3) = 2 ? Yes. Note 1. The solution obtained by this method is usually only an ap- proximate solution owing to the impossibility of determining exactly the coordinates of the point of intersection of the lines. Note 2. If the equations are inconsistent, the lines will be parallel ; if the equations are dependent, the lines will coincide. 51 52 ALGEBRA y ,„ 5 tK) : :~i \ S V t X 7 \^l -/ \I / \ 4 5 / ^o 2 X " r^ cr r X -in -5 \ / + 5 +A) L_ > Sc«: it A-t *,/ V j!Z I it v T J- \ tE \ -An -S 23^ £ it ^ t II Y EXERCISE 18 Study the following pairs of equations graphically ; if simul- taneous, determine their common solutions : 1. 3x + y = 11. 5 x — y = 13. 2 x + 5y = 10. 4« + 10y = 40. 2. 4^ + 3 ?/ = - 1. 5x + 2/ = 7. 6. 3a;-2y = 6. 9s-6y = 18. 3. 7 x + 8y = 26. 7. ; 5 a; — 4 ?/ = 0. 7 as + G?/ = — 29. 5 a? + 3 y = 14. 4 # — 5 ?/ = 26. 8. 9 x + 14 ?/ = - 25. 3 a, _ 4 y = 22. 52. Two simultaneous equations having two variables may be solved by combining them so as to cause one of the variables to disappear. This process is called Elimination. SIMULTANEOUS LINEAR EQUATIONS 53 53. Elimination by Addition or Subtraction. Rule. — 1. Multiply, if necessary, both equations by such numbers as will make the coefficients of one of the variables of equal absolute value. 2. If the coefficients have the same sign, subtract one equation from the other ; if they have opposite signs, add the equations. 3. Solve the equation resulting from step 2 for the other variable. 4. Substitute the value of the variable found in step 3 in any equation containing both variables, and solve for the remaining variable. 5. Check the solution by substituting it in both of the original equations. Example. Solve the equal . (5a> + 3y = -9. ;ions \ n . ._ [ 3 x - 4 y = - 17. (1) (2) Solution: 1. ilf 4 *(l): 20 x+ 12?/ =-30. (3) 2. J6(S)i 9x-12?/=-51. (4) 3. (3) + (4) : 29z=-87. (5) 4. .-. x=-S. 5. Substitute — 3 for x in (1) : -15 + Sy =-9. 6. .-. 3 y = 0, or y = 2. The solution is : x = — 3, y = 2. Check: In (2) : does3(-3)- 42 =-17? Yes. In (1): does 5(- 3) + 3-2=- 9? Yes. 54. Elimination by Substitution. Rule. — 1. Solve one equation for one variable in terms of the other variable. 2. Substitute for this variable in the other equation the value found for it in step 1. 3. Solve the equation resulting in step 2 for the second variable. 4. Substitute the value of the second variable, obtained in step 3, in any equation containing both variables and solve for the first variable. ' 5. Check the solution by substituting it in the original equations. * M 4 (1) : means " Multiply both members of equation (1) by 4." 54 ALGEBRA Example. Solve the equations 11 x- 5 y = 4. (1) 4 x - 3 y = 5. (2) Solution: 1. Solve (1) for y : y — llx — !, (3) 5 2. Substitute in (2) : 4x - 3 [ 1! x ~ 4 > j % 5. (4) 3. Solving (4) f or x : 20 x - 33 x + 12 = 25 ; - 13 x = 13, or se=- 1. 4. Substitute — 1 for x in (1) : — 11 — 5 y = 4. . •. — 5 ?/ = 15, or y = — 3. The solution is : x =— 1, y = — 3. Check : In (1) : does 11(- 1) - 5(- 3) = 4 ? does - 11 + 15 = 4 ? Yes. In (2) : does4(- 1) -3(-3) =5? does - 4 + 9 = 5 ? Yes. EXERCISE 19 Solve the following pairs of equations by addition or sub- traction. (If difficulty is experienced in obtaining a solution, determine graphically whether the equations are inconsistent or dependent.) [ 2 x - 3 y = 19. ( 6 x + 11 y = 31. 1. J 5. * 7# + 4 y = 23. j 6 y - 11 a; = 74. 2. a; - 5 ?/ = - 21. ( 3 x + 2 w = - 31. 6. \ 3 a: -87 =-35. 1 6 a; + 4y = - 62. f 15x + 8y =3. [4 y_8w = -3. 3. •! ^ ,-^ 7. 6z-12?/ = 5. Ill 2*4-5 w== -15. f 13 m - 7 w = 15. [ 3 x - 4 y = - 13. 4. .8. [ 8 m - 4 n = 9. } 6 a,- - 8 ?/ = - 5. Solve the following pairs of equations by substitution: 2x+y = 8. Ja + 26 = 11. Waj-4y = 43. 10 " { 3a -f- 5 6 = 29. SIMULTANEOUS LINEAR EQUATIONS 55 11. 12. 3r4-7s = -12. _6r + 9s = l. 8e-3/=47. 6e-7/=21. 13. 14. 5x -{- 6y = — 5. 10 a + 9 y = - 6. 3 x - 5 y = 38. -5x4-3?/ = - 26. Solve the following pairs of equations in either manner : 15. 16. 17. 18. 19. 6^~ 9 6 ' 2 9 lla-36 = 3a + fe 11 8 8 a -5 6 = 1. 2 e + * + 6 e-2t-3 5 e 3t— 7. I 4r— 3s r-6g 14 9 2r + 3« = -10. 8 a?— 3 y — 5 9 5 2aj-l 3# + 4 d-2n 20. I3 <* 4- n 4. 3 ( d 4- 4 » — 7 = 0. 1 5' 11 21. » y 8_9 = 7 * 2/ Hint: Eliminate y without clear- ing of fractions. 22. 23. \ 9 -+ 14_ 11 2 ' d 21 _ s -7. 8_ X 3^ a 5y 89 30 _5_ 6x _6_ 2/ 59 18 ' 2 3a; 3 4/ 1 12* 5 4 a; 4 3/ 13 72' 24. Solve the following for x and ?/ : [ o x - 6 y = 8 a. 25. i I 4 a; 4- 9 1/ = 7 a. 56 ALGEBRA 26. 27. 28. bx — ay = b 2 . (a — b)x + by = a 2 . ax + by = 1. cx+ dy = 1. ax + by = 2 a. a 2 x — b 2 y = a 2 -\- b 2 . j 2a# — &y 29. 30. a x-\-by 3a + 2 = b. m n n + y m — X m n n+x m—y 55. Equations Containing Three or More Variables. Example. Solve the set of equations : 12 m — 4 n + p = 3. m — n — 2 2~> = - 1. 5m-2n = 0. Solution: 1. M 2 (1) : 2. (2) + (4): 3. Mg (3) : 4. (5) -(6): 5. Substitute 5 for n in 3. 24w»- 8n + 2j> = 6. 25 m — 9 w = 5. 25 m - 10 n = 0. n = 5. 5 m - 10 = 0, or to = 2. (1) (2) (•3) (4) *(5) (6) 6. Substitute 5 for n and 2 for to in (2) : 2_5-2^=-l. .-. — 2p = 2, orp =— 1. Solution : M».= 2, n = 5, jj =s— 1. Check : The solution satisfies each of the three given equations. Rule. — To solve a system of three equations containing three unknowns : 1. From two equations, say the ist and 2d, eliminate one of the unknowns ; mark the resulting equation as equation (4). 2 From another pair of equations, say the ist and 3d, eliminate the same unknown, and mark the resulting equation as equa- tion (5). 3. Equations (4) and (5), containing two unknowns, are readily solved (if the system has a solution;. Then by substitution the third unknown may be determined. SIMULTANEOUS LINEAR EQUATIONS 57 EXERCISE 20 Solve the following sets of equations : 3 x + y — z = 14. x + 3 y — z = 16. _x + y — 3z = — 10. 3p + 4(/ -f 5r = 10. 4 P — 5 c/ — 3 r = 25. 5 p — 3 7 + 3~x~~ = 2. 3 v. -\-x = -5. 4 x-y = 21. 5 y + z = -19. 6 z — u = 39. Solve Examples 6, 7, and 8 for #, y, and 2. 6. 7. 10. ax -{-by = a 3 + & 3 abc by + cz = - 9 + c 3 abc cz + aa? = c 3 + a z abc [1 1 - + - = a. # y i 1 , y z 1 1 -+- = c. Z X b a - + - = c. x y c 6 _ + _ = <*. .2/ » u —x -\-y = 15. x — y + z = -12. y — z + u = 13. .z — u + X = -14 58 ALGEBRA 56. Solution by Formula. Simultaneous linear equations hav- ing two or more unknowns may be solved by means of certain formulae. This method of solution is considered in § 233, page 240, and may be studied at this time if desired. 57. Many problems are solved more conveniently by using two or more unknowns. Example 1. A certain number of two digits exceeds three times the sum of its digits by 4. If the digits be reversed, the sum of the resulting number and the given number exceeds three times the given number by 2. Find the number. Solution : 1. Let t = the tens' digit, and u = the units' digit. 2. .-. 10 £ + u = the original number, and 10 u + t = the number obtained by reversing the digits. 3. . \ 10 1 + u = 3 (t + u) + 4 or 7t-2u = 4. (1) 4. Also (10«+ m)+ (10m +0 =3 (101+ u) + 2, or 19t-8tt = -2. (2) 5. Solving equations (1) and (2) by the usual methods, t = 2 and u = 5. .-. 25 is the number. Check: Does 25 = 3 (2 + 5) + 4 ? Yes. Does 25 + 52 = 3 (25) + 2 ? Yes. EXERCISE 21 1. Divide 79 into two parts such that twice the smaller ex- ceeds the greater by 5. 2. If 3 be added to the numerator of a certain fraction, and 7 be subtracted from the denominator, the value of the frac- tion becomes -^-. If 1 be subtracted from the numerator, and 7 be added to the denominator, the value becomes f. Find the fraction. 3. The units' digit of a certain number of two digits is one third of the tens' digit. If the number is divided by the differ- QUADRATIC EQUATIONS 59 ence of its digits, the quotient is 15 and the remainder is 3. Find the number. 4. Find the three angles of an isosceles triangle if each of the base angles exceeds the vertical angle by 30°. 5. There are two numbers such that when the first is divided by the second the quotient is 3 and the remainder is 1 ; when the second is divided by one fifth of the first, the quotient is 1 and the remainder is 3. Find the numbers. 6. A man has $10,000 invested, part at 5% and part at 6%. The interest for one year on the 5% investment exceeds the interest for one year on the 6 % investment by $ 60. How much does he have invested at each rate ? 7. A's age is six fifths of B's. Fifteen years ago his age was thirteen tenths of B's. Find their present ages. 8. If the numerator of a certain fraction be increased by 4, the value of the fraction becomes 4 ; if the denominator of the fraction is decreased by 3, the value of the fraction be- comes -f. Find the fraction. 9. A certain chord of a circle divides the circumference into two arcs such that three times the minor arc exceeds twice the major arc by 80°. Find the two arcs. 10. The units' digit of a certain number of two figures exceeds the tens' digit by 5. If the number, increased by 6, be divided by the sum of its digits, the quotient is 4. Find the number. 11. A's age is twice the sum of the ages of B and C. Two years ago, A was 4 times as old as B, and, four years ago, A was 6 times as old as C. Find their ages. 12. Find three numbers such that the sum of the first, one half of the second, and one third of the third shall equal 29 ; also such that the sum of the second, one third of the first, and one fourth of the third shall equal 28; and finally such that the sum of the third, one half of the first, and one third of the second shall equal 36. 60 ALGEBRA 13. A certain rectangular field has the same area as another which is 6 rods longer and 2 rods narrower, and also the same area as a third field which is 3 rods shorter and 2 rods wider. Find the dimensions of the field. 14. The sum of the first and third angles of a certain tri- angle is twice the remaining angle; the sum of the first and second angles exceeds the third angle by 20°. Find the three angles of the triangle. 15. The sum of the two digits of a certain number is 14. If 36 be added to the number, the result has the same digits as the original number, but in reverse order. Find the num- ber. 16. Two trains, starting from points 270 miles apart, and traveling toward each other, will meet at 12 o'clock, if one train starts at 7 a.m. and the other at 10 a.m. The rate of the first train exceeds the rate of the second train by 5 miles an hour. Find the rates of the trains. 17. A boy can row 10 miles downstream on a river in two hours, and can return in 3| hours. Find the rate at which he rows in still water and also the rate of the current of the river. 18. A train leaves A for B, 120 miles distant, at 9 a.m., and, one hour later, a train leaves B for A. They meet at noon. If the second train had started at 9 a.m. and the first at 10.30 a.m., they would still have met at noon. Find their rates. 19. The circumference of the hind wheel of a carriage is 55 inches more than that of the fore wheel. The former makes as many revolutions in going 250 feet as the latter in going 140 feet. Find the circumference of each wheel. 20. A man has quarters, dimes, and nickels to the value of $ 1.40, having in all 12 coins. If he had as man y dimes as he has quarters, and as many quarters as he had dimes, the value of the coins would be S1.55. How many coins of each kind has he ? QUADRATIC EQUATIONS 61 SUPPLEMENTARY PROBLEMS 21. The hundreds' digit of a certain number of three figures is f of the tens' digit, and exceeds the units' digit by 2. If the number be divided by the sum of its digits, the quotient is 38. Find the number. 22. r years ago, A was m times as old as B. In s years, A will be 7i times as old as B. (a) What are their present ages ? (b) Find the values of their present ages if r is 10, s is 5, m is 5, and n is 2. 23. A man has $ 14,250 invested in bonds, which give him annually a total income of $700. Part of the money is in- vested in 4 % bonds, bought at $ 90 per share, and the balance in 6 % bonds, bought at $ 105 per share. How much has he invested in each way ? (The income is always computed on the par value of a bond, which in this example is $ 100 per share.) 24. A vessel contains a mixture of wine and water. If 50 gallons of wine be added, there will then be J as much wine as water ; if 50 gallons of water be added, there will be 4 times as much water as wine. Find the number of gallons of water and of wine at first. 25. The chords AB and CD of a circle form, at their inter- section, an angle of 60°. The chords AD and BC, extended, meet at 0, forming an angle of 40°. Find the number of de- grees in the arcs AC and DB. 26. The formula I = a + (n — 1) d occurs in a more advanced topic in algebra. If I is 32 when n is 10, and is 10 when n is 20, find a and d. 27. The numbers d, a, t, and b are assumed to be connected by the formula d = at + b. If d = f when t = J, and d = i| when t — | , find a and b. From the resulting formula for d, determine t when d is f , giving the result to the nearest eighth of an inch. 62 ALGEBRA 28. Assuming that the numbers a, b,d, and W are connected by the formula W=ad-\-b, find a and b if W = 1.5 when d = .75, and if TF= 4.5 when d = 2.5. From the resulting equation, determine IF when d = 2. 29. If a field were made a feet longer and 6 feet wider, its area would be increased by m square feet ; if its length were made c feet less, and its width d feet less, its area would be de- creased by n square feet. Find its dimensions. 30. An automobile made a trip of 145 miles in 8 hours. The average rate within city limits was 15 miles an hour ; the aver- age rate outside of city limits was 20 miles an hour. Find the part of the trip lying within city limits and the part outside. 31. In round numbers, the average rate of the automobile that won a certain long auto race is 21 times the rate of an ordinary passenger train. At these rates, the automobile can go 150 miles in 2 hours less time than the train requires for a trip of 140 miles. Find the rate of the automobile and of the train. 32. A piece of work can be done by A and B working to- gether in 10 days. After working together for 7 days, A leaves, and B finishes the work in 9 days. How long would A alone take to do the work ? 33. A motor boat which can run at the rate of r miles an hour in still water, went downstream a certain distance in n hours ; it took m hours to return. (a) Find the distance and also the rate of the current. (&) Find the values of the two results in part (a) when r is 10, m is 3, and n is 2. 34. A and B can complete a certain piece of work if A works 5 days and B works 4 days at their usual rates. A and C can do the work if they work together for 5 days, and if then C works one day alone. The number of days it would take C to do the work exceeds by 4 days the number required by B. Find how many days it would take each alone to do the work. VII. SQUARE ROOT AND QUADRATIC SURDS 58. A Square Root of a given number is a number whose square equals the given number. 59. Two square roots are obtained for each number. They are of equal absolute value, but have opposite signs ; they are in- dicated by means of the double sign, ±, read " plus or minus." Example 1. Vl6 x i y 2 = ± 4 x*y, since ( ± 4 x 2 y) 2 = 16 x 4 y 2 . Example 2. V4a? - 20xy + 25y 2 = ±(2x- 5y), since \ ±(2x-5y)\ 2 = + (2 x-5 y) 2 = Ax 2 - 20xy + 25y 2 . The positive square root is called the principal square root of a number ; the square root refers always to the positive root. 60. The square roots of a large number may sometimes be found by inspection by factoring the number. Example. V1764 « 4 = V4 . 441 a 4 = ± 2 . 21 a 2 = ± 42 a 2 . EXERCISE 22 Find the square roots of : 1. 25 a 4 . 4 16 a 6 . 6 144^V # 25 rW ' 81 mV 2 - ***»• 5 laW. 7 121^o 3. 49c 4 d 8 . " 49 mV* ' 169 yV* 8. If a monomial is a perfect square, what kind of numbers are the exponents of its prime factors ? What sign does it have ? 9. When is a trinomial a perfect square ? 10. How may the correctness of the square root of a number be checked ? 63 64 ALGEBRA Find the values of : 11. Vtf 2 + 2 xy 1 + y 4 . 13. Vl44 x 6 - 24 x 3 ?/ + f 12. Va 4 -6a 2 6 + 9& 2 . 14. V49 c 4 - 42 c 2 d 2 + 9 d\ 15. V1225. 17. V676a 2 2/ 2 . 19. V17t>4ary. a + b a 2 + 2 a& + & 2 a 2 2a + b 2a + 6 + 2 a& + 6 2 + 2 a6 + 6 2 16. V784m 4 w 2 . 18. Vl089a 4 . 20. V1024 xyz 2 . 61. Square Root found by Long Division. If it is not possible to factor readily the number under the radical sign, the square root, if there is one, may be found by a process like long division. Example 1. Find the square roots of a 2 + 2 ah -j- 6 2 . Solution : 1. \Ta 2 = a. Place a in the root. 2. Square a ; subtract. 3. 2 x a = 2 a. Trial divisor. 2 ab -f- 2 a = b. Add 6 to the trial divisor and to the root. Complete divisor. 4. b x (2 a + b) ; subtract. The square roots are ; +(« + &) and — (a + b). Explanation : 1. Find the square root of the first term, obtaining a, the first term of the root ; place it in the root. 2. Square the first term of the root and subtract it from the given number, obtaining the first remainder, 2 ab + b' 2 . 3. Double the first term of the root, obtaining 2 a, the trial divisor. Divide the first term of the remainder by 2 a, obtaining b, the second term of the root. Add b to the root and to the trial divisor ; the complete divisor is 2 a + b. 4. Multiply the complete divisor by b and subtract. Step 3 is suggested by the process of squaring a binomial. When squaring a binomial, the middle term is obtained by taking twice the product of the first and second terms ; this is equivalent to taking twice the first term and multiplying by the second. Reversing the process, the second term, 6, will be found, if 2 ab is divided by 2 a. After a 2 is sub- tracted from a 2 + 2 ab + b' 2 the remainder 2 ab + b 2 equals 6(2 a + b). This suggests adding b to the trial divisor and multiplying the sum by b. SQUARE ROOT AND QUADRATIC SURDS 6b Example 2. Find the square roots of 20 a,- 3 - 70 a? + 4 a 4 + 49 - 3 a 2 . Solution : 1. Arrange it in descendir Subtract. -7). lg powers of x : 2 x 2 + 5 x -7 2. V4x4 = 2x 2 . 3. (2 x 2 ) 2 = 4 x 4 . 4x 4 +20x 3 - 3x 2 -70x+49 4x 4 4. 2 x (2 &) = 4 x 2 . 20 x 3 + 4 x 2 = 5 x. 5. 5x(4x 2 + 5 a). 4x'< 4x' + 5x J +5x 20 x 3 - 3x 2 -70x+49 20x 3 + 25x 2 6. 2 x (2x 2 + 6«). 7. -28x 2 -4x 2 =-7. 4x 2 +10x -7 — 28 a? -70 a; + 49 8. -7(4x 2 +10x- 4x 2 +10 z-1 -28x 2 -70x+49 The square roots are : + (2 x 2 + 5 x — 7) and — (2 x 2 + 5 x — 7). Rule. — To find the square root of an algebraic expression : 1. Arrange it according to ascending or descending powers of some letter. 2. Write the positive square root of the first term of the given expression as the first term of the root. Square it and subtract the result from the given expression. 3. Double the root already found, for the trial divisor. Divide the first term of the remainder by the first term of the trial divisor. Add the quotient to the root and also to the trial divisor, obtaining the complete divisor. 4. Multiply the complete divisor by the new term in the square root ; subtract the product from the remainder obtained in step 2. 5. Continue in this manner : (a) double the root already found for a new trial divisor ; (b) divide the first term of the remainder by the first term of this product for the new term of the root ; (c) add the new term of the root to the trial divisor, obtaining the complete divisor; (d) multiply the complete divisor by the new term of the root ; (e) subtract. 66 ALGEBRA EXERCISE 23 Find the square roots of the following : 1. 25x 2 -±0xy + 16y 2 . 3. x A + 6x? + 11 x 2 + 6 x + 1. 2. 36 c 4 - 60 c 2 d + 25 d 2 . 4. a 4 + 4 a 3 + 6 a 2 + 4 a + 1. 5. 9 n 4 + 12 ?i 3 - 20 »* - 16 w + 16. 6. a? 2 + ?/ + 4 z 2 — 2 #?/ + 4 #z — 4 yz. 7. 8a 3 -4a-16a 4 + l + 16a 6 + 4a 2 . 8. 12 n - 42 n 3 + 4 - 19 n 2 + 49 »*. 9. a 6 -2a 5 -a 4 + 6a 3 -3a 2 -4a + 4. 10. 4z 2 + 20a; + 29 + — + i. £ iC 2 2 2 aft , 13 6 2 4 6 3 , 4 ft 4 a "3 + 9 9a 9a 2 ' 12. f»«-.| : n*-.#»« + f* + ff 13 0* , «* | 3 a? a; 1 16 4y 20y* 5tf 25^ Find the fourth roots of : 14. a 8 - 16 a 6 6 3 + 96 a 4 5 6 - 256 a 2 b 9 + 256 6 12 . 15. 81 a 8 - 108 a 7 + 162 a 6 - 120 a 5 + 91 a 4 - 40 a 3 + 18 a 2 - 4 a + 1. Find. the first four terms of the square roots of: 16. 1 + aj. 17. 1 — x. 18. 9 — 2ft 62. Square Root of an Arithmetical Number. The square root of 100 is 10; of 10,000 is 100; etc. Hence, the square root of a number between 1 and 100 is between 1 and 10 ; the square root of a number between 100 and 10,000 is between 10 and 100; etc. SQUARE ROOT AND QUADRATIC SURDS 67 That is, the integral part of the square root of a number of one or two figures contains one figure ; of a number of three or four figures, contains two figures ; and so on. Hence, if the given number is divided into groups of two figures each, beginning with the units figure, for each group in the number there will be one figure in the square root. The groups are called Periods. Thus, 2345 becomes 23 45 ; it has two periods and its square root has two figures, a tens' and a units' figure. 34038 becomes 3 40 38 ; it has three periods and its square root has three figures. A number having an odd number of figures will always have only one figure in its left-hand period, as in this case. A decimal number is divided in the same manner, starting from the decimal point in both directions. Thus, 3257.846 becomes 32 57.84 60. The last decimal period is always completed by annexing a zero. The square root of this number has two figures before the decimal point and two after it. 63. The first figure of the square root of a number is found by inspection ; the remaining figures are found in- the same manner as the square root of a polynomial. Example 1. Find the square roots of 4624. Solution. 1. Divide 4624 into periods ; this gives 46 24. There are in the square root a tens' and a units' figure. 2. The tens' figure must be 6 ; 7 is too large for 70 2 = 4900, which is more than 4624. 3. The rest of the square root is found as follows : 3600 is the largest square less than 4600. V3600 = 60 ; place 60 in the root. Square 60 and subtract. Double 60. Trial divisor. 102 -=- 12 = 8+. Place 8 in root and add to trial divisor. Complete divisor. Multiply complete divisor by 8. The square roots are + 68 and — 68. It is customary to abbreviate the solution by omitting the zeros as in the following example. 60 + 8 46 24 36 00 120 10 24 8 128 10 24 23.5 5 52.25 4 40 1 52 3 43 1 29 460 23 25 5 465 23 25 68 ALGEBRA Example 2. Find the square roots of 552.25. Solution. The largest square less than 5 is 4 ; VI = \ Place 2 in the root. 2x2 = 4; annex 0. Trial divisor. 15 -4- 4 = 3+ ; add 3 to the trial divisor. Complete divisor. Multiply by 3. 2 x 23 = 46 ; annex 0. Trial divisor. 230 -=- 46 = 5+. Add 5 to the trial divisor. Complete divisor. Multiply by 5. The square roots are + 23.5 and — 23.5. Rule. — To find the square root of an arithmetical number : 1. Separate the number into periods (§62). 2. Find the greatest square number in the left-hand period ; write its positive square root as the first figure of the root ; subtract the square of the first root figure from the left-hand period, and to the result annex the next period. 1 3. Form the trial divisor by doubling the root already found and annexing zero. 4. Divide the remainder by the trial divisor, omitting the last figure of each. Annex the quotient to the root already found; add it to the trial divisor for the complete divisor. 5. Multiply the complete divisor by the root figure last obtained and subtract the product from the remainder. 6. If other periods remain, proceed as before, repeating steps 3, 4, and 5 until there is no remainder or until the desired number of decimal places has been obtained for the root. Note 1. It sometimes happens that, on multiplying a complete divisor by the figure of the root last obtained, the product is greater than the remainder. In such cases, the figure of the root last obtained is too great, and the next smaller integer must be substituted for it. Note 2. If any figure of the root is 0, annex to the trial divisor and annex to the remainder the next period. SQUARE ROOT AND QUADRATIC SURDS 69 ExAMPLI :3. Fi id the square root 70.32 Solution : 49 44.90 24 49 1400 44 90 3 1403 14060 42 09 2 8124 2 14062 2 8124 The square roots are + 70.32 and — 70.32. The first trial divisor is 140. Since this is greater than 44, the first remainder, annex to the root, obtaining 70. The second trial divisor is 1400; (2x70=140; annex 0, 1400). Bring clown the next period 90, getting for the second remainder 4490. Divide 44 by 14 gives 3+ ; annex 3 to the root and add 3 to 1400, etc. EXERCISE 24 Find the square roots of : 1. 5776. 4. 8427.24. 7. 54.4644. 10. 106 09. 2. 15376. 5. 7974.49. 8. 1488.4164. 11. 529.9204. 3. 67081. 6. 11.6281. 9. 25.6036. 12. 1592.8081. 64. The Approximate Square Roots of a number which is not a perfect square are often desired. Obtain usually the first three figures following the decimal point. Example. Find the approximate square roots of 2. Solution : 1.414 2.00 0000 20 4 2? 1 100 96 2* JO 4 00 2ST 2 81 2820 1 19 00 2 8"' >4 112 96 The square roots are + 1.414+ and 7 04 - 1.414 70 ALGEBRA Note. In order to obtain the desired number of decimal places, annex zeros until there are three periods. EXERCISE 25 Find the approximate square roots of : 1. 3. 3. 6. 5. 10. 7. 13. 9. 15. 11. 19. 2. 5. 4. 7. 6. 11. 8. 14. 10. 17. 12. 21. 65. Table of Square Roots. In the remainder of the course, it will be necessary to use frequently the square roots of some numbers. Retain some of the square roots as they are found, either in a notebook or in some other convenient place. Make a list of the numbers from 1 to 50, and write their square roots beside them, thus : Number Square Root 1 1.000 2 1.414 3 . 1.732 After working Exercise 25, twelve of the numbers of this table may be tabulated. These roots may be used to obtain the square roots of other numbers. Example 1. Find the square roots of 8. Solution : V8 = \'TV2 = 2 x V2 : = 2 x ( ± 1.414+) = ± 2.828+. Example 2. Find the square roots of 12. Solution : Vl2 = V4 x 3 = 2V~3 = 2 x (± 1.732+)= ± 3.464+. EXERCISE 26 1. Find the following square roots to three decimals: (a) Vl8. (6) V20. (c) V24. (d) V27. (e) V28. 2. Complete your table of square roots up to 50. Get as many roots as possible by inspection (§ 60) ; get as many of the remaining roots as possible as in Example 1. Find the others by the long division method (§ 62). SQUARE ROOT AND QUADRATIC SURDS 71 66. The square roots of a fraction which is not a perfect square may be found as follows : X 2 ^4 ' V4 V6 = 2^49+ = >224+> ^2 2 Rule. — To find the square root of a fraction : 1. Change the fraction into an equivalent fraction with a perfect square denominator. 2. The square root of the new fraction equals the square root of its numerator divided by the square root of its denominator. 3. If desired, express the result of step 2 in simplest decimal form, prefixing the double sign, ± . Example. Find the approximate square roots of |. Solution : 1. The smallest square number into which 8 can be changed is 16 ; multiply both terms of the fraction by 2. % JI = J2Z3 = ^ =± ^ =± ^49 1= +< >8 ^2 x 8 * 16 4 4 EXERCISE 27 Find the approximate square roots of : 1. 9 f' 3. ft- 5. 4 7. 2. 9. 5 11. 5 XT' 2. i 4. 1 T 6. 5 2* 8. 3 V 10. h 12. ft- QUADRATIC SURDS 67. The indicated square root of a number which is not a perfect square is called a Quadratic Surd; as, V8, «%/-> Vsc, ^ *+i. 72 ALGEBRA 68. Surds should be simplified as in the following examples : (a) V24=V4T6=2.V5; (6) ^| = ^ = §^?. Thus, a quadratic surd is in its simplest form when the number under the radical sign is an integer which does not contain any perfect square factor. In problems involving surds, it is agreed to consider for each surd only its principal root (§ 59). 69. Addition and Subtraction of Surds. Example 1. Find the sum of V20 and V45. Solution : 1. V20 + V45 = VlTl + ViT~5 = 2V5 + 8VS = 5 VE. This solution assumes that surds may be added like other numbers. The coefficients of V5 are 2 and 3 ; the sum is found by multiplying V5 by the sum of its coefficients (§ 5). The advantage in adding surds in this way is that fewer square roots need be obtained. Thus, the sum of V20 and V45 is 5 V5 or 5 x (2.23(5+ ) or 11.180+. This same result couid be obtained by adding the square roots of 20 and 45. Example 2. Simplify Vf + VJ. , JO , /l /18 , /2 3V2 . V2 3V2 . 2V2 5V2 2 5 V2 Jx (1.414+) _ 7.070+ _ 1 ?67+ 4 4 4 Example 3. Simplify | + VJ. k 1 2 , Jl 2 , ./3 2 , V3 2+V3 2. 2+V3 = 2 + 1.732+ = 3.732+ =12 u+ Note. The results of problems involving surds are often left in the surd form as in step 1 of Examples 2 and 3. There are advantages in finding the approximate decimal value of the result, SQUARE ROOT AND QUADRATIC SURDS 73 EXERCISE 28 Simplify the following: 1. V12 + V2T. 11. i + VA- 2. V20-V5. 12. l+VS- 3. 2V18 + V98. 13. I+VJ. 4. V63-2V28. 14. *-Vf. 5. 3V24-V54. 15. I-VS- 6. 2V2 + VI8- V50. 16. -I + Vf. 7. V8+VJ. 17. -i-vi. 8. Vf-Vf 18. 1 _a/6 9. 1 + V|. 19. 11 "^ V 12 1* 10. f-Vf. 20. 1 1 _|_ a/ 2T 70. The other operations with surds, namely, multiplication, division, involution, and evolution, are considered in a later chapter, which may, if desired, be studied at this time. VIII. QUADRATIC EQUATIONS 71. A Quadratic Equation is an equation of the second degree (§ 45) ; it may have one or more unknowns. A Pure Quadratic Equation is a quadratic equation having only one unknown, which contains only the second power of the unknown, as, ax 2 = b. Example 1. An acre of ground contains 43,560 square feet. How long must the side of a square field be in order that the area of the field shall be one acre ? Solution : 1. Let s = the number of feet in one side. 2. Then s 2 = the number of square feet in the area. 3. Then s 2 = 43,560. Extract the square root of both members of the equation. 4. Then s = ±208.7+. Since this is a field, only the positive root has meaning ; hence the side of the field must be 208.7+ feet. 72. A pure quadratic equation has two roots, because two square roots are obtained in extracting the square roots of the two members of the equation. Rule. — To solve a pure quadratic equation. 1. Clear the equation of fractions, transpose, and combine terms until the equation takes the form x 1 = a number. 2. Extract the square roots of both members of the equation, placing the double sign, ± , before the root in the right member. Note. After extracting the square roots of both members of an equa- tion like x 2 = a 2 , we get ± x = ± a. This gives : + x = + a, + x = — a, — x = + a, and — x = — a. If both members of the last two equations are multiplied by — 1, the equations become + x = - a, and + x = + a. These are the first two of our four equations. Thus, it is clear that, from x 2 = a 2 , we get only two equations, x = + a and x = — a, or x = ± a. 74 QUADRATIC EQUATIONS 75 Example. Solve the equation 1 — = -—-}- — . 3 m 12 m ., 2 to . 3 to . 12 Solution : 1. 1 = 1 3 to 12 to 2. M 12m : 8 to 2 + 36 = to 2 + 144. 3. Simplifying : 7 w 2 = 108. 4. D 7 : m 2 = i^. 5. y/~ ":* w» =±v^=±6Vf = ±fV21. 6. V21= 4.582: to = ± 2 . (4.582) = ± ^|^ = ± 3.927. 7. To! =+ 3.927 ; to 2 =-3.927. " toi " is read " to one." The numeral 1 is called in such cases a sub- script, "wa" is read "to two." These subscripts are used to distin- guish between the two roots of the quadratic. Check : When the roots are complicated, it is better to check by going over the solution a second time. Great care must be taken, however, for it is easy to overlook an error. Note. Get the result in the radical form first ; that is, to = i f V21 ; then it is wise, for many reasons, to get it in decimal form as finally given. EXERCISE 29 Solve the following equations : 1. 5 c 2 -180 = 0. 3. 13 c 2 -135 = 10 c 2 -27. n 0^2 , O- - 2 KQ* A 4* 2 + 3 8* 2 -l 1 2. 2 x 2 + 27 = i x 2 — 53. 4. 10. ^ — -r- 6 . 5(t + 6)-t(t-3)=$t. 7. 9a 2 -5 = 0. 8. lla 2 -6 = 3. 9 A_J1 = _?. ii 4f Sx 2 3 " 2c + l c-1 * The symbol "V ": placed in the left margin will mean, "take the square root of both members of the previous equation." 7 2 14 2) = 10 m. 2x 3 " 4:X Ix 9 21 2x 5 c- -2 3 c — 5 76 ALGEBRA 1 1 a 2 - 17 12. x -f 3 as — 5 a; 2 — 2 x — 15 a? 2 — a; -|- 2 x 2 + x — 3 _. x— 2 x -f-3 14. 3 a a?4-5& _ ^ a - 5 6 3«4-10 6 15. a 2 - 2 ca 2 = 3 b\ Solve for a. Solution : 1. — 2 ex 2 = 3 6 2 - a 2 . 2. M_i: 2cx 2 = a 2 -3 6 2 . 3. z 2 = ^ 2c __ U-3^ / 2c(«*-3fr 2 ) \ 2 ^ X \ 4 ri* ± — V2 a 2 c - 6 6 2 c. 2c PROBLEMS IN PHYSICS All of the following equations occur in the study of physics. Solve them for the numbers which appear with exponent 2. ;. S = ±gt\ 18. F _mM d 2 20. /=^ J R . E = ±mv\ 19. H= C 2 1U. 21- #=ls D 2 73. Geometry Problems. If the following terms from ge- ometry are not familiar to the student, they should be re- viewed : (a) right triangle ; (b) hypotenuse ; (c) isosceles tri- angle ; (d) equilateral triangle ; (e) circle ; (/) theorem ; (g) altitude of a triangle ; (h) base of a triangle. QUADRATIC EQUATIONS 77 EXERCISE 30 Carry out all results in this exercise to one decimal place : 1. State the theorem about the square of the hypotenuse of a right triangle. 2. Find the altitude of a right triangle whose base is 13 feet and whose hypotenuse is 30 feet. Solution : 1. Let x = the number of feet in the altitude. 2. Then x 2 + 13 2 = 30 2 . (why ?) 3. Complete the solution. 3. Find the base of a right triangle whose hypotenuse is 45 feet and whose altitude is 27 feet. 4. If the altitude of a rectangle is h feet and its base is four times its altitude, find the length of the diagonal. 5. Solve the formula h 2 = o 2 + b 2 : (a) for a ; (6) for b. 6. Find the altitude of an isosceles triangle whose equal sides are each 15 inches and whose base is 8 inches. 7. Find the altitude of an isosceles triangle if its equal sides are each 4 b inches and its base is 2 b inches. 8. Find the altitude of an equilateral triangle if its sides are each 8 inches. 9. Find the altitude of an equilateral triangle if its sides are each a inches. 10. (a) What is the formula for the area of a circle ? (b) Find the area of the circle of radius 7 inches. Express the results of the following examples in simplest radical form : 11. Solve the equation A = -n-r 2 for r: (a) letting ir = 3j; (b) without substituting for it its value. 12. The volume of a circular cone is given by the formula V= \ irr 2 h, where r is the number of units in the radius and h is the number in the altitude. Find V when r = 5 feet and ft = 13 feet. 78 ALGEBRA 13. Find the radius of a circular cone whose volume is 528 cubic feet and whose altitude is 14 feet. 14. Solve the formula for the volume of a circular cone for r in terms of V, h, and w. 15. Solve the formula 8 = 4 irr 2 for r. 16. The distance s, in feet, through which an object falls in t seconds is given by the formula a = \ gt 2 , where g = 32. Suppose that a stone is allowed to fall from a tower; how far will it fall in : (a) 3 seconds ? (b) 5 seconds ? 17. How long will it take a ball to fall 300 feet ? 18. Washington's Monument in Washington, D.C., is 555 feet high. How long will it take a ball to fall that distance ? 19. Solve the formula F== 2 ir 2 Rr 2 for r. 20. Solve the formula v .= J irr 2 a for a. COMPLETE QUADRATIC EQUATIONS 74. A Complete Quadratic Equation is a quadratic equation having only one unknown, which contains the first power of the unknown as well as the second power ; as, 2a 2 -3a-5:=0. A complete quadratic equation may be Solved by Factoring. The solution is based upon the fact that if one of the factors of a product is zero, the value of the product is also zero. Thus, 3x0 = 0; (-5) x0 = 0; 2x0x (- 3) = x (-3)=0. Example 1. Solve the equation 4 x 2 — 9 = 0. Solution: 1. Factor: (2se— 3)(2as + 3) =,0. 2. If 2 x - 3 = 0, then (2 x - 3) (2 x + 3) = 0. 2x— 3=0, if 2 ac = 3 or a; = + |. 3. If 2 x + 3 = 0, then (2 x - 3) (2 x + 3) = 0. 2 x + 3 = 0, if 2 x = - 3, or x = - f . . 4. The roots of the equation are + f and — f . QUADRATIC EQUATIONS 79 5. Check: Does 4(|)2_9 = 0? 1 8 Does jl • - - 9 = ? i.e. 9-9 = 0? Yes. i Does 4(- |)2 -9 = 0? 1 9 Does 4- --9 = 0? i.e. 9-9 = 0? Yes. i Rule. — To solve an equation by factoring : 1. Transpose all terms to the left member. 2. Factor the left member completely. 3. Set each factor equal to zero, and solve the resulting equations. 4. The roots obtained in step 3 are the roots of the given equa- tion. Check by substitution in the given equation. Example 2. Solve the equation = — H 3 2 6 Solution : 1. M 6 : * 2 m 2 — 3 m = 35. 2. S35 : 2 m 2 - 3 m - 35 = 0. 3. Factor : (2 m + 7) (w - 5) = 0. 4. 2 m + 7 =0, if ra=-f. m — 5 = 0, if m = + 5. 5. The roots of the equation are + 5 and — |. Check by substitution. EXERCISE 31 Solve the following equations by factoring : 1. x 2 -15x + 51 = 0. 5. 8a 2 -10a:-r-3 = 0. 2. tf + 4a«96. 6 x 2 + 7x = Q. 3. .t 2 =^ + H0. 7< x * + a x-2a 2 = 0. 4. 6 x 2 + 7 x + 2 = 0. (Solve for x.) * For meaning of " M 6 : " see § 36. 80 ALGEBRA 8. 3z 2 - mz-4m 2 = 0. 1E 5-t 12 15. 9. 15x 2 + xk = 2W. 10. 10x 2 + 7mx = 12m 2 . 16. 11. ^!_?_?_*? = o. 17. 10 5 2 12. ?-X«X. 18: 3 — t 6- -£ s 8 -3 s- if*-": V 5 -3 y- 3 1 -4 6 V 2 -3 l 6 6 p — S 3 9# 3z 2 13 A__l = 1 . 19 3 2Q+6) = ro + 6 ^ ' 2x 2 4 a 4* m+5 2 14. 5-lji*. 20. ^ = - 5 8 ^- 6 2 »as * — 2 8 (z-2) 2 75. Graphical Solution of Equations with One Variable. Many facts about equations containing one variable can be discovered by the aid of graphical representation. Example 1. Consider the equation Sx — 12 = 0. The expression 3 x — 12 has a different value for each value of x. Thus, if x = 2, 3 x - 12 = - 6 ; if X = - 3, 3 x - 12 = - 21. The problem is to find the value of x for which the expression 3 x — 12 will equal zero. Graphical Solution : 1. Let y = 3 x — 12. 2. Find values of y for some values of x : if x = 0, y=-\2; tf x = -2, y = - 18 ; if a; = + 5, */ = + 3 ; if x = + 6, y = + 6. 3. Use these pairs of numbers as coordinates of points and draw the graph. QUADRATIC EQUATIONS 81 Y i r / / / / / L ; / ^ * d A/ X -1 p 10 . / u/ _J B r y 4. ^C crosses the x axis at point A. The coordinates of A are : a; = 4, y = o. 5. Hence when x = 4, 3 x — 12 = 0. (y is the expression 3 x — 12.) .'. x = 4 is the desired solution of the equation, for we were looking for a value of x for which 3 x — 12 = 0. Rule. — To solve graphically an equation containing one variable : 1. Simplify the equation as much as possible. 2. Transpose all terms to the left member. 3. Represent by y the expression found in step 2. 4. Find for y the values which correspond to selected values of the variable in the equation. 5. Use the pairs of values obtained in step 4 as coordinates of points ; plot the points ; draw the graph, making the vertical axis the y axis. 6. The graph crosses the horizontal axis at points whose ordi- nates are zero, and whose abscissae are the desired roots of the equation. 82 ALGEBRA Example 2. Solve the equation x 2 — x = 6. Solution : 1 . x 2 — x = 6, or x 2 — x — 6 = 0. 2. Let y = £ 2 — ae — 6. 3. If x=-4, 2/=(_4) 2 -(-4)-6 = 16 + 4-6 =+14. 4. Similarly if x = + 1 + 2 + 4 + 5 - 1 -2 -3 - 4 then ?/ = -6 — 6 -4 + 6 + 14 -4 + 6 + 14 \ y- n 4 3- Lit t J \ r - ' +10 ""' ' j ' ' "_A_ _J it 4 1-7 3 r V -a, . . _L__j r in \ / > t \ /! Xi' \A j O B/ X -7 -6 -5 -4 -3 -2\ - +1 |+? +5 +4 +5 +6 +7 JL J \ / V^ /l >. y *"*** • j _J_y£ L 5. The graph crosses the horizontal axis at the points A and B. Ac- cording to the rule, the abscissae of these points are the two roots of the equation. At A: a: = - 2, y = ; i.e. x 2 - x - 6 = 0. At B: a; = + 3, y = 0; i.e. x 2 - x - 6 = 0. Check: z = - 2 ; does (- 2) 2 - (- 2)- 6 = ? Yes. x = + 3; does (+3) 2 -(+ 3)- 6 = 0? Yes. EXERCISE 32 Solve graphically the equations : 1. #-3 = 0. 3. x 2 -9 = 0. 2. 2x = 9. 4. x 2 -\-3x = 10. 5. a 2 -7a,' + 10 = 0. 6. a 2 + 7 a; + 6 = 0. QUADRATIC EQUATIONS 83 7. Between what two integers does each of the roots of the following equation lie ? 4 x 2 — 4 x — 35 = 0. Obtain the approximate roots of the following equations to the first decimal place. 8. 2 x 2 - x - 11 = 0. 9. x 2 + 3 x - 14 = 0. 76. Solution by Completing the Square. Development 1. Find: (a) (x — 4) 2 ; (b) (# + 5) 2 ; 2. When is a trinomial a perfect square ? (See § 15, c.) 3. Make a perfect square trinomial of x 2 — 10 x. Solution : 1. | of 10 = 5 ; 5 2 = 25 ; add 25. 2. The perfect square is x 2 — 10 x -f 25 or (x — 5) 2 . t 4. Make perfect square trinomials of the following : (a) a 2 -12z; (6) y 2 -Uy, (c) z 2 - 20 2. 5. Solve the equation x 2 — 12 x + 20 = 0. Solution : 1. S 20 : « 2 — 12 £ = — 20. 2. Make the left member a perfect square by adding 36 ; therefore add 36 to both members (§§ 35, 37). A 36 : x 2 - 12 x + 36 = 36 - 20, or (x- 6) 2 = 16. 3. V": z-6=±4. 4. .-. x — 6 = + 4, or x = 6 + 4 = 10, one root, and x — 6 := — 4, or x — Q — 4=2, another root. Check : x = 10 ; does (10) 2 - 12(10) + 20 = ? Yes. x = 2 ; does (2) 2 - 12(2) + 20 = ? Yes. 6. Solve the equation x 2 — 3 # — 5 = 0. Solution : 1. x' 1 — 3 x — 5 = 0. 2. A 6 : x 2 -3x = + 5. 3. £(- 3) = - | ; (- $) 2 = + | ; add f to both members. 4. A9: x 2 -3x + f = 5 + f = -«£. 84 ALGEBRA 5. V~: x -f =± V^ = ±^V29. 2 2 2 7. Radical results, X X = 3 + ^ and x 2 = S ~^ 29 . 2 2 8. Decimal results, x x = S + 5 ' 385 and a; 2 = 8 ~ f' 385 2 2 8.385 _ - 2.885 2 ' 2 = 4.192-^ =-1.192+ Check : To check the solution by substituting the roots in either their decimal or their radical form is a long process, with many opportunities for errors. Persons skillful in algebra check by going over the solution carefully. A quick check, the reason for which will be learned later in algebra, is to find the algebraic sum of the roots ; this result should equal the nega* tive of the algebraic coefficient of x in the equation in which the coefficient of x 2 is 1. Here : + 4.192+ The coefficient of x 2 is 1. The coefficient of x — 1.192+ is — 3. This equals the negative of the algebraic Sum. + 3 sum of the roots. If the coefficient of x 2 is not 1, first imagine the equation divided by that coefficient, and then select the coefficient of x. Rule. — To solve a quadratic equation by completing the square i 1. Simplify the equation ; transpose all terms containing the unknown number to the left member, and all other terms to the right member so that the equation takes the form ax 2 -f- bx = c 2. If the coefficient of x 2 is not i, divide both members of the equation by it, so that the equation takes the form x*+px=q. 3. Find one half of the coefficient of x, square the result ; add the square to both members of the equation obtained in step 2. This makes the left member a perfect square. QUADRATIC EQUATIONS 85 4. Write the left member as the square of a binomial; express the right member in its simplest form. 5. Take the square root of both members, writing the double sign, ± , before the square root in the right member. 6. Set the left square root equal to the + root in the right member of the equation in step 5. Solve for the unknown. This gives one root. 7. Repeat the process, using the - root in step 5. This gives the second root of the equation. 8. Express the roots first in simplest radical form, and then, if desired, in simplest decimal form. EXERCISE 33 Solve by completing the square: 1. x 2 + 4 x- 5 = 0. 10. ra 2 -fl0m = 3. 2. ^2_ 8a ,_33 = o. 11. x 2 + 3#-4 = 0. 3. ^ + 6^-27 = 0. 12. s 2 = 5s + 6\ 4. x 2 -4- 10^ + 21 = 0. 13. y 2 + 3y = 10. 5. a 2 - 12 x- 13 = 0. 14. z 2 + z = 6. 6. y 2 -2y = \l. 15. r 2 -3 = r. 7. « 2 + 6a = 9. 16. z 2 = 2 + 3z. 8. c 2 -4c = l. 17. w 2 -f-5w; + 3 = 0. 9. d 2 -8d-8 = 0. 18. a 2 -7a + 7 = 0. 19. Solve the equation x 2 — %x = 1. Solution : 1. \ of (- f) = - \ ; (- i) 2 = \. 2. A^: x 2 -fx + i = l + i 3. (x - |)« = V». 4. x-\=±^T0. 86 ALGEBRA 5. x-l = l VlO. *-f = -iVTo. .•.* = | + §v1o X = i-iVlo _1+Vl0 1- VlO 3 3 _ 1 + 3.162 _ 1 - 3.162 3 3 = 4 - 162 = 1.387 + . 3 _ - 2.162 _ ? 3 20. * + {«.. |. 33. i_ A_ 2 . 21. 2/ 2 -f2/ = 5. 3x x 2 ' 22. 23. 2 2 _ 6 f _ 1 = 0. a 2 + ia = f. 34. 2x-\-- — 2 4a; 24. h 2 - 3 & _ 5 = 0. 35. 2/ , 3_ 2 25. /2 _ 5 * _ 2 5 fc 6" fc — ~6- 3 2 32/ 1 . 3 _. 5 5 4a 4a 2 26. 3a 2 -2a; = 40. 36. 27. 4 m 2 — 8ra = 45. o* 24 24 , 28. 8r 2 +2r=3. 3 ?- - — 5 = 1- 31. 9c 2 -f-18c = -8. a: — z a; 29. 4* 2 -3£ = 3. 38 5 + 8 -3 30. a; 2 + 7a; = 5. 38 ' 5Z^ + 83p-* 39. d-S d+4 3 32. 9# 2 + 4a;=6. d-2 d 2 77. Solution of Literal Quadratic Equations. Example. Solve the equation ax 2 — 3 bx — c = 0. Solution : 1. a# 2 - 3 bx — c = 0. 2. D : X 2_§_^_c =0 a a 3. A c : ^-Ma^i. « a a tion QUADRATIC EQUATIONS 87 4. The coefficient of x is ( z1 ^) ; one half of it is f ~^-\ The square of f — ) is ( ]. Add this to both members of equa- 5. x2 _3_^ + ^ = ^ + c a 4 a' 2 4 a 2 a \('-K) 3ft\ 2 9 6 2 + 4ac 4 a 2 , 1 - = ± -±- V 9 ft 2 + 4 ac. 2a 2a _ + 3 6 ± V9 ft 2 + 4 ac 2a Q . _ + 3 ft + V9 ft 2 + 4 ac . 8 _ + 3 b - V9 ft 2 + 4 ac 2a 2a Check : x x + x 2 = — — - = H — -. Since this is the negative of the 2a a coefficient of x in step 2, the roots are correct. EXERCISE 34 Solve the following equations for x : 1. x 2 + 2mx = l — m 2 . 8. a 2 — 2a# = 9 — 6 a. 2. x 2 + 6ax-5 = 0. 9. z 2 -10ta = -9* 2 . 3. a ,2 -2ax + 6=0. 10. ax 2 + 4 # + 1 = 0. 4. a 2 + 6#-c=0. 11. to 2 + 2 ex- 3 = 0. 5. # 2 +|>a;+ g = 0. 12. cx 2 + 2 d# + e = 0. 6. 2a 2 + 6a n = 0. 13. ax 2 + bx = 0. 7. 2 a: 2 + 4 ax — c = 0. 14. ax 2 + bx + c = 0. 78. Solution of Quadratic Equations by a Formula. All quad- ratic equations having one unknown may be put in the form ax 2 +bx + c = 0. 88 ALGEBRA This equation has been solved as Example 14 of Exercise 34. The roots are : , IT z -. — - — b ± V6 2 — 4 ac 2a This result is used as & formula for solving any quadratic equation of the form ax 2 + bx -f- c = 0. Example 1. Solve the equation 2 x 2 — 3 x — 5 = 0. Solution : 1. Comparing the equation with ax 2 + bx -f c = 0: a = 2, b =— 3, c = — 5. 2. Substitute these values in the formula : 3. Then - b ± Vb 2 - 4 ac 2a -(-3)±V(-3)2- -4(2)(- ■5) 2(2) + 3 ± V9 + 40 4 3 ± V49 _ 3 ± 7 4 4 3 + 7_10 5 4 -4-2'^- _3-7_ 4 ^i=-l. 4 4. /. xi = Check : Xi as |. Does 2(|) 2 - 3(f)- 6 as Of Does 2 • - 2 ^ - V - 5 = ? Does - 2 2 5 - - V- - 5 = ? Yes. x 2 =-l. Does 2(- 1)2 _3(- 1)- 5 = 0? Does 2 + 3-5 = 0? Yes. Example 2. Solve the equation 2^ — 3#— 3 = 0. Solution: 1. a = 2, b=— 3, c =— 3. 2. Substituting in the formula, x = ~ h ± V6 ' 2 ~ 4 ac : 2 a - 3 ± V9 + 24 _ 3 ± V33 _ 3 ± 5.744+ X ~ 4 ~" 4 ~ 4 ,. x, = 8 -^ = 2.186+ ; x 2 = ~ 2744+ = - .680+. 4 4 Check : 2.186+ The coefficient of x is — f, when the coeffi.- - .686+ cientofx 2 = 1 ; 1.5 =-(-§). 1.500 (For this method of checking,, s^e § 76.) QUADRATIC EQUATIONS 89 EXERCISE 35 Solve the following equations by the formula: 1. a* _ 12 x _j_ 32 = 0. 13. 2 _ 13 _ I . 2. f + 7y-3Q = 0. 3 « 9 x 2 18 3. 4. 2z 2 -3z-20 = 0. 3x 2 -x-A = 0. 14, 2 c 11 1 5 10 2c' 5. 6. Ay 2 -5y-21=0. 20m 2 + ra-2 = 0. 15. x 1_ 5 1 6 3 2x 2* 7. 9 w 2 - 13 w 4- 3 = 0. 1 R 6s + 5_4s + 4 6. 20m 2 + ra-2 = 0. 7. 9 w 2 - 13 w + 3 = 0. 8. 6 m 2 + m = 3. 9. 4 r 2 — 7 r = — 3. 0. 5a 2 + 3# = 9. 1. z 2 7z_4 2 6 ~3* 2. 7 1_ 5 6< 2 2 12 r 4s-3 s-3 17. -i- = 2*-5. 7 — £ 2 3 5 18. iv — 1 w 6 19 . 20^+1 _,_ 1 x + 1 (x + 1) 2 79. Summary of Methods of Solving a Quadratic. Four methods of solving a quadratic equation have been given : the graphical, by factoring, by completing the square, and by the formula. The first is useful mainly as a means of illus- tration ; the third is useful mainly in solving the general quad- ratic ax 2 -f- bx -f- c = 0, and, thus, in deriving the formula; the fourth is used whenever the solution is not readily accomplished by factoring. Historical Note. Greek mathematicians as early as Euclid were able to solve certain quadratics by a geometric method, about which the student may learn when studying plane geometry. Heron of Alexandria, about 110 b.c, proposed a problem which leads to a quadratic. His solution is not given, but his result would indicate that he probably solved the equa- tion by a rule which might be obtained from the quadratic by completing 90 ALGEBRA its square in a certain manner. Diophantus, 275 a.d., gave many problems which lead to quadratic equations. The rules by which he solved his equations appear to have been derived by completing the square. He considered three separate kinds of quadratics. He gave only one root for a quadratic, even when the equation had two roots. The Hindu mathematicians, knowing about negative numbers, con- sidered one general quadratic. Cridharra gave a rule much like our formula. The Hindus knew that a quadratic has two roots, but they usu- ally rejected any negative roots. The Arabians went back to the practice of Diophantus in considering three or more kinds of quadratics. Mohammed Ben Musa, 820 a.d., had five kinds. He admitted two roots when both were positive. Alkarchi gave a purely algebraic solution of a quadratic by completing the square, and refers to this method as being a diophantic method. In Europe, mathematicians followed the practice of the Arabians, and by the time of Widmann, 1489, had twenty-four special forms of equations. These were solved by rules which were learned and used in a mechanical manner. Stifel, 1486-1567, finally brought the study of quadratics back to the point that had been reached by the Hindus one thousand years before. He gave only three normal forms for the quadratic ; he allowed double roots when they were both positive. Stevih, 1548-1620, went still farther. He gave only one normal form for the general quadratic, as do we ; he solved this in both a geometric and an algebraic manner, giving the method of completing the square. He allowed negative roots. EXERCISE 36 Supplementary Miscellaneous Examples Solve the following equations by any of the preceding methods. As a rule, solve by factoring if possible ; otherwise by the formula. 1. (3» + 2)(2a-3) = (4a;-l) 2 -14. 2. y(5 y + 22) + 15 = (2 y + 5) 2 . 3. A + ^ = _ll 5. 4 7 2 It 3 6 a 4 4 7-y y 3 r - 2 r - 3 15 3 iv 4 — 5 w 5 4 — 5 w 3 iv 6 QUADRATIC EQUATIONS 91 7. 2x-l X X X — X 5 10. -. x — 2 _6x x + 4 5 X + 4: 8. x-2 ic + 4 7 ~3* 11. a — 5 .J _a a-6 3 x + 5 x-3 a(a - 1) 2 = a - 1 3* 3_ l-4 2a + 5 3 3 4 2 - * + 3' a; 'x—l_x 2 + x — l 16 - :t- — — — — 7. * X — 1 X X 2 — X x x x 2 -\-2 x — 2 14. x + 2 x + 3 z 2 + 5x + 6 15 . ?J1+1 + r-9 =1 . 7-r 3r-l 16 . 3«^13 = _J t 6 — w iv — 4 17 . 1=2, = 8+ 6 " 2v 4^-3 18. 3w + g = 1 | 2ro + 5 2 m — 5 3 m — 5 19 5 7 = 8?; 2 -13v-64 2d + 3 3v-4 6^ + ^-12 20. — !— + — i* — + _15_ = o. a-2 24(z + 2) 4-a 2 92 ALGEBRA 21. Solve the equation 2p 2 x 2 — 3 px — 1 = 0. Solution : 1. Use the formula x = ~ 6±Vft ~ 4ac , 2a 2. a = 2p 2 ; b = -3p; c= -1. 3 A x = 3jp ± V9p* - 4 (2^)(-l) = 3p ± V9 p2 + 8p'^ 4^2 4p2 . ^^ Sp^Viy ^^ SpirpVTT ^ B ±Vl7 4^2 4^2 4^ „ 3 4- Vl7 , 3 - VT7 Hence #i = — ; and x 2 = 4p 4p Solve for cc : 22. .t 2 + 5 mx -f 6 m 2 = 0. 26. .t 2 4- 2 tx = r 2 — £ 2 . 23. 3a 2 — 4rsc + 5s=0. 27. Jf/a 2 = &a + Z. 24. 4 fV + 21 te = 18. 28. a- 2 4- (n -f l)x = — ft. 25. 12a 2 =23^«-5e 2 . 29. a; 8 + (a- h)x - ab =t). 30. '/V.1- 2 — 2(r + *)as + 4 = 0. 31. .T 2 -2da;-5a;= - 10c?. 32. (x - 4) 3 - (aj + 3) 3 = - 217. 1 1 14 33. x 2 — 3 x x 2 + 4 x 15 x 2 34 2^+1 3|-_2 s 17 ' 3^-2^2^ + 1 4 3^472-1 2) 1^3^ + 1 3J 36. 1-^-1 =1+ 3 a 2 -4 3<> + 2) 2-a 37 __a a _4 t 2it- + a 3 a — 4 a 3 QUADRATIC EQUATIONS 93 38. (d 2 -d-2)x 2 -(5d-l)x = -6. __ x — a.x-fa x 2 — 5 a 2 Oa. 1 = • x -\- a a — x x 2 — a 2 40. (m + n)x 2 + (3 m + n)x + 2m= 0. EXERCISE 37 1. Twice the square of a certain number equals the sum of 15 and the number. Find the number. 2. If three times the square of a certain number be in- creased by 10 times the number, the sum is 8. Find the number. 3. Find two consecutive numbers whose product is 462. 4. The sum of the squares of three consecutive integers is 434. Find the integers. 5. The sum of a certain number and its reciprocal is jf. Find the number. 6. Find the dimensions of a rectangle whose area is 352 square feet, if its length exceeds its width by 6 feet. 7. The denominator of a certain fraction exceeds twice the numerator by 2, and the difference between the fraction and its reciprocal is f-J. Find the fraction. 8. Find the base and altitude of a triangle whose area is 60 square inches, if the base exceeds the altitude by 7 inches. 9. Find the dimensions of a rectangle whose area equals that of a square of side 18 feet, if the difference between the base and altitude of the rectangle is 15 feet. 10. Find the dimensions of a rectangle whose area is 3000 square feet if the sum of its base and altitude is 115 feet. 11. Find the base and altitude of a right triangle if the hypotenuse is 13 feet and if the base exceeds the altitude by 7 feet. 94 ALGEBRA 12. Find the base and altitude of a right triangle if the hypotenuse is 17 feet and if the sum of its base and altitude is 23 feet. 13. A fast train runs 8 miles an hour faster than a slow train; it requires 3 hours less for a trip of 288 miles than does the slow train. Find the rate of each train. 14. An automobile party made a trip of 160 miles. By in- creasing their average rate by 4 miles an hour, they can make the return trip in 2 hours less time. Find their average rate going. 15. A crew can row downstream 18 miles and back again in a total time of 1\ hours. The rate of the current is known to be one mile an hour. What is the rate of the crew in still water ? 16. Some boys were canoeing on a river, in part of which the rate of the current is 4 miles an hour and in part 2 miles an hour. If, when going downstream, they go 3 miles where the current is rapid and 6 miles where the current is slow in a total time of If hours, what is their rate of rowing in still water ? 17. A tank can be filled by one pipe in 4 hours less time than by another. If the pipes are open together 11 hours, the tank will be filled. In how many hours can each pipe alone fill the tank ? 18. I have a lawn which is 60 feet by 80 feet. How wide a strip must I cut around it when mowing the grass to have cut half of it ? Hint : Referring to the figure, it is clear that if io = the number of feet in the width of the border cut, then the dimensions of the uncut part of the lawn are (60 — 2 w) and (80 - 2 w). Hence, (60 - 2 w) (80 - 2 w) = £ • 60 • 80. Complete the solution. w w CM 1 O "~80-2w _^ L^_ w w w W QUADRATIC EQUATIONS 95 19. A farmer is plowing a field whose dimensions are 40 rods and 90 rods. How wide a border must lie plow around the field in order to have completed ^ of his plowing ? 20. The numerator of a certain fraction is 2 less than the denominator. The reciprocal of the fraction exceeds the frac- tion itself by -j-jj- . Find the fraction. 21. In the formula s = at + ±gt 2 , let s = 124, a = 30, and g = 32. Find t 22. From the formula S = ^\ 2 a + (n .— 1) d j , determine n when S = 5, a = 5, and d = — 1. 23. The numerator of a certain fraction is 5 less than the denominator. If 6 be added to both the numerator and the denominator, the resulting fraction is f of the original frac- tion. Find the fraction. 24. A picture 15 inches by 20 inches in size is to be sur- rounded by a frame, whose area shall be J of that of the picture inclosed. What must be the width of the frame ? 25. The rate of one train exceeds that of another by 5 miles an hour. The fast train makes a trip of 150 miles in one hour less time than the slow train. Find the rate of each train. 26. A workman -and his assistant can do a piece of work together in 3f days. It would take the assistant 4 days longer to do the work alone than it would take the master work- man. How long would it take each alone to do the work ? 27. The area of a certain trapezoid is 150 square feet. The upper base exceeds the altitude ,by 2 feet and the lower base exceeds the altitude by 8 feet. Find the two bases and the altitude of the trapezoid. 28. Divide 30 into two parts such that the square of the greater shall equal the product of 30 and the smaller. • 29. Replace the number 30 of Example 28 by the number a and solve the resulting problem. .96 ALGEBRA IMAGINARY ROOTS IN A QUADRATIC EQUATION 80. Example. Solve the equation x 2 — 2x + 5 = Q. Solution : 1. Use the formula method of solving the equation. a = 1, b = — 2, c = 5. 2. a; - b ± \/P - 4 a< • _ + 2 ± V4 - 4 . . i . . u 2a 2 2±V^T6 2 + 2 ± V4 - 2 ■20 The question arises, what does V— 16 mean? Is — 4 the square root of -16? No, for (-4) 2 = + 16. Is 4-4? No, for (+4) 2 = 4-16. Thus, no number with which the student is acquainted will produce — 16, when it is squared. 81. No rational number raised to an even power will pro- duce a negative result ; hence an even root of a negative num- ber is impossible up to this point. To avoid this difficulty, a new kind of number is introduced. • An Imaginary Number is an indicated square root of a nega- tive number; as, V— 16; V— 3; V— a 2 . The numbers previously studied are called Real Numbers. 82. Every imaginary number can be expressed as the product of a real number and V— 1. V— 1 is indicate! i, and is called the Imaginary Unit. Thus, V^l6 = V16(- l)= ± 4V^T = ± 4 i. V- a" 2 = Va 2 (— 1)= ±aV^l=±ai. V^l = \/5(-l) = ± V5 • V^T = ±i Vft. Historical Note. The symbol i for V— 1 was introduced by Euler, one of the greatest mathematicians of the eighteenth century. EXERCISE 38 Express the following in terms of the unit i; 1. V-9. 3. V-49.r. 5 V-25c 2 . 2. V-36. 4. V-100m 2 . 6. V-81a 2 6 2 , QUADRATIC EQUATIONS 97 7 V-144r 4 . 8. V-l69a 4 . 9. V=T- 10. V-ff 11. V-A- 12. V-y^Cl 2 . 13. V^6. 14. V-8. 15. V-24. 16. V-44. 17. V-45c 2 . 18. V-20a 2 ^ 19. V-50a 4 . 63 y\ 21. Simplify V-- 2 ^-. Solution: V~T = V" > ^ J f ^ = ±|- V3" V^T = ±^V3- 22. V-|. 25. V--\ -. 28. V-ff' 23. V^-| 26. V- - 2 f. ' 29. V— $f 24. V-|f 27. V~f|. 30. V-iH- 83. Addition and Subtraction of Imaginary Numbers. EXERCISE 39 1. Add V-4 and V-36. Solution : V^i + V- 36 = 2 i + 6 i = 8 i. Note. While every imaginary number, like V— 4, has two values, one positive and one negative, in problems such as the one in this exercise, only the principal root, the positive one, is used, as in the case of surds (§ 68). Simplify : x 2 4- 2. V-16+V-4. 3. V-9+V-49. 5/ V-100-V-64. 6. V-l +V-25 -V-49. 4. V-81+V-25. ' 7. V-« 2 -V-4a 2 - V-9a 2 . 8. V-36 V- K)()x 2 -V-81a; 2 . 9. V- 16 xY - V - 25 xhf + V - 49 xhf. )8 ALGEBRA 13. V^20 + V^5 - V^45. 14. V-24-2V-6 + 3V-54. 15. V-28 + 5V-7-V-63. 16. Simplify + ? ± ^_?L SoitmoH: 1. 5 ± JI27 = 5 VE27 = 5 3V3. y^ry 2 \ 4 2 2 2 2 = 5 31V3 2 2 _ 5±Siy/S 2 The numbers in Examples 1-15 are called Pure Imaginaries. The sum, or difference, of a pure imaginary and a real number, § 81, as in this exercise, is called a Complex Number. Simplify : »lWt s ■»■ i*N© - £*>/# ... §*V3 »■ !W^- - s±vs 84. A further discussion of imaginary numbers, more complete, including a discussion of the other fundamental operations upon imaginary numbers, is given in Chapter XIV. 85. Meaning of Imaginary Roots of a Quadratic on the Graph. Example. Consider the equation x 2 + x -f 2 = 0. Solution : 1. Solve the equation by the formula : a = 1, 6 = 1, c = 2. -1 ±y /TZrs = -l j-V_7 = -l±i\/7 2 2 2 QUADRATIC EQUATIONS 2 2 2. Solve the equation graphically. (Review rule § 75.) Let" y = x 2 + « + 2 : 99 When x = then y = 0+1 + 2 +4 + 2 + 8 + 3 + 14 + 2 +4 -3 + 8 -4 + 14 3. The graph has the same shape as the graphs obtained when solving other quadratic equations ; but the graph does not cross the horizontal axis at all. Hence, y or x 2 + x + 2 is never zero for any real value of x. This is characteristic of the graph of a quad- ratic which has imaginary roots. \ + o fe* 3l EXERCISE 40 Solve the following equations. Express the roots in simplest radical form. Draw the graphs for the first three equations. 1. x 2 + x + l = 0. 2. ic 2 -2z+-3 = 0. 3. x 2 -3x + 4, = 4. 2 a 2 - 2 a + 1 = 0. 5. 3m 2 -2m + 2 = 0. 6 . 4c 2 -5c-f-2 = 0. 7. 9r 2 + 4 = 8r. 8. ^-££ + 1 = 0. 10. m-3 2 2 m 4- 1 m + 7 - 1 m +- 1 u . 2« + l+| = o. 3 a 2 12. 2# 2 -f-6da; + 5d 2 = 0. 13. Sx 2 — 5wx + 3w 2 = 0. 14. 5x- 2 -8to-5* 2 = 0. IX. SPECIAL PRODUCTS AND FACTORING ADVANCED TOPICS 86. In paragraph 10 is the rule: "The product of the sura and the difference of any two numbers equals the difference of their squares " ; thus, (x -f y)(x — y) = x 2 — y 2 for all numbers x and y. If z = 2aandy =36, (2a+3&)(2a-3&) =4a 2 -9&2. If x = 14 and y = 5, (14 + 5)(14 - 5) = 196 - 25 = 171. If x = (a -f 6) and y =(c + d), then similarly [(a + b) + (c + d)][(a + 6) - (c + <*)] = (a + &) 2 - (c + d) 2 . Likewise, in any of the type forms studied in Chapter II, the numbers may be general number expressions. Example 1. Multiply (a + b -f- c) by (a -f & — c). Solution : 1. (a + 6 + c) (a + & - c) = {(a + 6) + c}{(a + &) - c} 2. = (a + &) 2 - c 2 = a 2 + 2 a& + b 2 - c 2 . Here x = (a + 6) and y = c. Example 2. Multiply (r + s -M — w) by (r + s — t -f- n). Solution : 1. (r -f s + £ — ri) (r -f s — t + ft) 2. = {(r -h *)+(*- w)}{(r + «)-(*- n)} s (r + s) 2 - (I- n) 2 3. = r 2 + 2 rs + s 2 - t 2 + 2 *n - ft 2 . Here x = (r + s) and y = (t — w). Note. In such examples, the rules for introducing parentheses (§ 6) are used. The various terms of the expressions may be rearranged, if necessary, so that one factor becomes the sum and the other the difference of the same two numbers, when the terms are grouped. 100 SPECIAL PRODUCTS AJJ i> ' F&OTGRItftf ' " 101 EXERCISE 41 Find the following products mentally : 1. S( a + b-)+5ll(a + b)-5\. 2. \(m + n)-2p\\(m + n)+2p\. 3. {10-(r + s)\110 + (r + s)l. 4. \3p-(c + d)\\3p+(c + d)l. 5. {(c + 2 d) - 11 a]{(c + 2d) +11 ttf. 6. (a — 6 4-c)(a — b — c). 7. (#-2/4-z)(>-2/-z). 8 . ( a 2_ f _ a _i)( a 2_ a + i) - 9. (a 2 + ab + 6 2 ) (a 2 - ab + b 2 ). 10. (a-f-2&-3c)(a-2& + 3c). 11. (?>x + ±y + 2z)(3x-4y-2z). 12. (a- 2 4- a- 2) (a; 2 -a -2). 13. (a-\-r— c + d)(a + r + c — d). 14. (a— b + m-\- n)(a — b — m — n). 15. (2x + z — y + w)(2x — z — y — w). 16. \(a + b) +2(a-b)\\(a + b)-3(a-b)l Solution : Just as (re + 2 y) (x — 3 y) = x 2 — xy — 6 y 2 , so {(a + b) + 2(a -&)}{(«+ 6) -3(a-6)} = (a + ft) 2 - (a + 6)(a - 6) - 6(a - ft) 2 = (a 2 + 2 a& + & 2 ) - (a 2 - ft 2 ) -6(a 2 -2aH 6 2 ) = a 2 + 2 a& + 6 2 - a 2 + &* 2 - 6 a 2 + 12 a& - 6 6 2 = 14 a& - 6 a 2 - 4 & 2 . Here * = (a + 6) and y = {a - b). 17. ](m4-w)-4j|(m4-M)-5S„ 18. \(x-y) + S\\(x-y)-6}. 19. {3aj-(y+»)H2aj-(y + 2)(. 102 ALGEBRA 20. \x + 3y + 15z\ \x + 3y - 10 z\. 21. \r + 2s-3t\ jr-f 2s + lt\> 22. \3p-±(q + r)\ {±p- 5(q + r)\. 23. \x 2 + 2x + l\ j# 2 + 2 as - 5j. 24. [(a + 6)-5] 2 . 26. [2a-(c + d)]l 25. [6-f(ra-w)] 2 . 27. [a + 5 + c] 2 . 28. From the result of Example 27, make a rule for deter- mining by inspection the square of any polynomial. Find the following by the rule made in Example 28 : 29. [« + 36-c] 2 . 31. 2r + s-t + x~] 2 . 30. [ a -b + c-d] 2 . 32. [3« - b + 2 c - d~] 2 . 87. General Problems in Factoring. Example 1. Just as x 2 — 3a; — 88 = (x — 11) (a + 8), so (a - 2 6)2 - 3(a - 2 6) - 88 = {(a -2b)- 11} {(a - 2 6) + 8} = (a-2b- ll)(a-26 + 8). EXERCISE 42 Factor completely the following expressions : 1. (« + &) 2 -c 2 . 7. (x-y) 2 + 2(x-y)-63. 2. (m-n) 2 -x 2 . 8. (x + y) 2 - 5(x + y) - 36. 3. 3»_(y+*)t 9. ( r + s )2_L.4( r _L. s y_5^ 4. m 2 -(n-p) 2 . 10. (p-g) 2 +8(>-g)r-20r 2 . 5. (lx-2y) 2 -y 2 . 11. (a; 2 -4) 2 -(aj + 2) 2 . 6 . ( a + 5)2 + 23(a + 6) 4- 60. 12. 9(m - n) 2 - 12(m-n) -f 4. 13. (x - y) 2 — (m — w) 2 . 14. (a 2 -2a) 2 -f 2(a*-2a) + l. 15. (1 + ?l 2 ) 2 - 4 W 2 . 16. (a 2 + 3a) 2 + 4(z 2 + 3a) + 4. SPECIAL PRODUCTS AND FACTORING' lW 17. (9 a 2 + 4) 2 - 144 a 2 . (18. (a 2 + 7 a) 2 + 20 (a 2 -f 7 a) -96. 19. (m + n) 2 4- 7(m + n) — 144. 20. (a; 2 4-»-9) 2 -9. 21. (a + y) 3 -* 3 . 26. (a + ?/) 3 -f-(a-y) 3 . 22. (r + s) 3 + 8 P. 27. « 6 - (ar> -h l) 3 . 23. .(m + n) 3 - (m - n) 3 . 28. 27 m 3 —(m— n)\ 24. a 3 +(a + l) 3 . 29. (2 a - bf-(a +2b)\ 25. a 3 -8(a + 6) 3 . 30. (a? + 3 yf - (m - 3 yf. 88. Polynomials Reducible to the Difference of Two Squares. Certain polynomials may be put into the form of the differ- ence of two squares by grouping certain terms. Example 1. Factor 2 mn + ra 2 — 1 -f n 2 . Solution : 1. 2 mn + m 2 — 1 + n 2 = (m 2 + 2ra« + n 2 ) — 1 2. = (m + n) 2 - 1 3. = (m + rc + l)0+ w-1). (§87) Example 2. Factor a 2 - c 2 + & 2 - d 2 - 2 cd - 2 a&. Solution : 1. a 2 - c 2 + 6 2 - cP - 2 cd - 2 a& 2. =(a 2 -2aH 6 2 ) - (c 2 + 2 cd + d 2 ) 3. = (a - b) 2 - (c + d)a 4. = {(a - 6) + (c + *)} {(a - 6) - (c + *)} 5. = (a — 5 + c + d) (a — b — c — d) EXERCISE 43 Factor : 1. a 2 -2ab + b 2 -c 2 . 6. 2mn-n 2 + l-m\ 2. m 2 + 2mn + n 2 -p 2 . 7. 9a 2 -24a&-f-16& 2 -4c 2 . 3. a 2 -x 2 -2xy-y 2 . 8. 16 a 2 - 4 ?/ 2 -f 20 #z - 25 z 2 4. a? 2 — y 2 — z 2 -|- 2 yz. 9. 4 n 2 -}- m 2 — x 2 — 4 mn. 5. &2_ 4 + 2 a& + a 2 . 10. 4a 2 -66-9-& 2 . 104 ALGEBRA 11. 10 xy-9z 2 + y 2 + 25 x 2 . 12. a 2 -2 ab + b 2 - c 2 + 2 cd- d 2 . 13. a 2 - b 2 -f x 2 - y 2 + 2 ax + 2 6y. 14. x 2 -\-m 2 — y 2 — n 2 — 2 mx — 2 ny. 15. 2xy-a 2 + x 2 -2ab-b 2 + y 2 . 16. 4a 2 + 4a& + 6 2 -9c 2 + 12c-4. 17. 16?/ 2 --36--8a?/-z 2 -f « 2 -12z. 18. m 2 -97i 2 + 25a 2 -6 2 -10am4-66n. 19. 4 a 2 - c 2 - 12 ab + 2 cd + 9 b 2 - d 2 . 20. 9 x* - 4 x 2 + z 2 - 6 a 2 z - 20 a# — 25 # 2 . 89. Trinomials Reducible to the Difference ot Two Squares. Type Form : x* + ax 2 y 2 + y*. Example 1. Factor a A + a 2 b 2 + 6 4 . Solution : 1. a* + a 2 b 2 + b 4 may be changed into a perfect square by- adding a 2 b 2 . Adding and subtracting a 2 b 2 : a 4 + rt 2 & 2 + & 4 = ( a 4 + 2 «2 & 2 + ft 4)_ a '2 & 2, 2. .-. a 4 + a 2 6 2 + 6 4 = (a 2 + 6 2 ) 2 - a 2 6 2 3. =(o 2 + 6 2 + «6)(a 2 + 6 2 -a6). (§87) Example 2. Factor 64 a 4 - 64 a 2 m 2 + 25 m 4 . Solution: 1. A perfect square containing 64a 4 and 25 to 4 is 64 a 4 — 80 a 2 m 2 + 25 to 4 . The given trinomial may be changed into this perfect square by subtracting 16 a 2 m 2 ; then 64 a 4 - 64 a 2 m 2 + 25 m 4 = (64 a 4 - 80 a 2 m 2 + 25 m 4 ) 4- 16 ahn 1 . But this is the sum of two squares and not factorable in this form. 2. Another perfect square containing 64 a 4 and 25 m 4 is 64 a 4 + 80 a 2 m 2 + 25 to 4 . Adding and subtracting 144 a 2 TO 2 : 64 a 4 - 64 a 2 rw 2 + 25 to 4 = (64 a 4 + 80 a?m 2 + 25 to 4 ) - 144 a 2 m 2 . 3. .-. 64 a 4 - 64 a 2 TO 2 + 25 to 4 = (8 a 2 + 5 w 2 ) 2 - ( 12 awi) 2 = (8 a 2 + 5 to 2 + 12 «to) (8 a 2 + 5 to 2 - 12 am). SPECIAL PRODUCTS AND FACTORING 105 EXERCISE 44 Factor the following trinomials : 1. x* + x 2 + l. 13. 4 r 4 - 32 r 2 £ 2 + 49 Z 4 . 2. a 4 + a 2 m 2 + m 4 . 14. 9 m 4 n s — m 2 n 4 + 16. 3. c 4 + 6c 2 + 25. 15. 4 p*- 24 2>V + 25r*. 4. y4 + 3y 2 + 36. 16. 9 a 4 + 17 a 2 6 2 + 49 b\ 5. 1+2^ + 9^. 17. 4# 4 + 7ay + 16.?/ 8 . 6. 1 - ?- 2 -f- 16 r 4 . 18. 9 * 4 - 31 W + 25 a,- 4 . 7. tf-lZtfif + ly 4 . 19. 16 mV + 15 m 2 n 2 + 25. 8. 9m 4 -19m 2 + l. 20. 25 p 4 + 34pV + 49*/*. 9. 4:y 4 -32y 2 + l. 21. x 4 + 4=. 10. 25i 4 f 14m 2 w 2 + 25n 4 . 24. 4^ + 1. 90. Certain polynomials can be factored by grouping their terms. Type Form : ab + ac+ bd+cd= (a + d)(b -f c). Example 1. Just as ax -\- bx=(a + b)x, (§ 5) so a(x + y)+ b(x + y) = (a + b)(x + y). (§ 5) Example 2. Factor 6 a? — 15 a 2 — 8 x + 20. Solution: 1. 6^-15z 2 -8x+20 = (6£ 8 -15x 2 )-(8a;-20) (§6) 2. =3z 2 (2z-5)-4(2a;-5)(§ 15, a) S. = (3x 2 -4)(2x-5). 106 ALGEBRA EXERCISE 45 Factor : 1. 2a(x + y)-3(x + y). 11. 2 + 3a- 8a 2 - 12a 3 . 2. 5m(r + s) + 2w(r+s). 12. 3 X s + 6 x 2 + x -f 2. 3. 3i?(2a-?/) — r(2x-y). 13. 10ma -15ftz- 2m + 3?i. 4. 8(£ + w) — m(£ + w). 14. a 3 x + a&ca; — a 2 by — b 2 cy. 5. a(6 + c) - cZ(6 + c). 15. a 2 bc - ac?d + ab 2 d - bed 2 . 6. a& + an + 6m + mw. 16. 30 a 3 — 12 a 2 — 55 a + 22. 7. ax-ay-\-bx-by. 17. 56 - 32 a + 21^-12^. 8. ac — ad — be + &d. 18. 3 aa; — a?/ + 9 bx — 3 &y. 9. a 3 + a 2 + a+l. 19. 4 ar s + x 2 y 2 - 4 y 3 - 16 xy. 10. 4^—5^ — 4^ + 5. 20. rt — rn — sn + st. 21. ar + as + 6r + 6s — cr — cs. 22. ax + ay — az + bx — 6z + 6y. 23. am — 6m — cp + ap — cm — 6p. 24. or* — xz 2 + # 2 y + xy 2 + y s — yz 2 . 25. 2 ax + ca + 3 by — 2 ay — 3 bx — cy. 91. The Sum or the Difference of Two Like Powers. Type Form : a n ± b\ By actual division, as in § 9, c : (a* - b*) + (a - b) = a 3 + a 2 6 + ab 2 + 6 8 . (a 4 - b A ) + (a + b)=a s -a 2 b + a& 2 - 6 3 . (a 5 - 6 5 )-*- (a - 6)= a 4 + a 3 6 + a 2 b 2 + a& 3 + 6 4 . (a 6 + 6 6 )-s-(a + 6)= a 4 - a 3 6 + orb 2 - a& 3 + 6 4 . SPECIAL PRODUCTS AND FACTORING 107 The following rule may be verified in the same manner: * Rule. — I. Letting n represent any positive integer: 1. a n — b n is always exactly divisible by a — b. 2. a n — 6" is exactly divisible by a +- b when n is even. 3. a n +- 6* is never exactly divisible by a — b. 4. a n +- b n is exactly divisible by a + b when n is odd. II. In the quotient : 1. The signs are all plus when a —bis the divisor. 2. The signs are alternately plus and minus when a +- b is the divisor. 3. The exponent of a in the first term is 1 less than its expo- nent in the dividend, and decreases by 1 in each succeeding term until it becomes 1. 4. The exponent of & is 1 in the second term, and increases by 1 in each succeeding term until it becomes 1 less than its exponent in the dividend. Example 1. Divide a 7 — b 7 by a — b. Solution : 1. By I, 1, a 7 — 6 7 is exactly divisible by a — 6. 2. By II, 1, 3, and 4, qT "" &7 = «6 + a 6 6 + a 4 & 2 + a 3 6 8 + a 2 & 4 + a& 5 _f_ &6. a — b Example 2. Factor 32 a? + 243. Solution: 1. 32 x*> + 243 =(2 x) 5 + 3 5 . 2. This expression is of the type a n + 6 n , where a = 2 sc, 6 = 3, and w = 5. By I, 4 and II, 2, 3, and 4, 32a: 5 + 243=(2a; + 3)[(2a;)*-(2a:)3.3+(2x) 2 .3 2 -(2a;).88 + 3 4 ] = (2 x + 3) (16 tf - 24 X s + 36 re 2 - 54 x + 81). Check : Let x = 1. Then 32 a 5 + 243 = 32 + 243 = 275 ; also, (2 x + 3) (16 & - 24 x s + 36 sc 2 - 54 sc+81) = (2+3) (16-24+36-54+81) = 5 • 65 = 275. * These rules are proved in § 257. 108 ALGEBRA The method of factoring binomials of the form a n ± b n given in this paragraph must be used frequently. However, if the prime factors (§ 14) of a binomial of this form are desired, proceed as in : Example 3. x 6 - y* = (as 8 f f) (ar 3 - f) = ( x + y)(tf ~xy + y 2 )(x - y)(x? + xy + y 2 ). Note 1. By §91, xt — y* =(x — y) (x5 + x*y -\ f-?/ 5 ). The second factor is not prime however. Note 2. Whenever the binomial is the difference of two even powers, it may be treated as the difference of two squares. Example 4. x 9 + y 9 = (st?f + (iff = (or 3 -f tf)(x« - a?f + f) = ( x + y)( x * -®y + y 2 K^ - ^f + y 6 )- Note 1. By § 91, x? + ?/ 9 = (z + ?/) (a;8 - a;?*, + x6y 2 + ... + ?/8 ). The second factor is not prime however. EXERCISE 46 Find the following quotients : t , ^Z.. 6. 21^. 11. 1- 1( 5« 4 7. *±£ 12. x + z 3 8. i + i 5 . 13. 1 + a 9. |H£ 14 2 — a 10 mT~< 15 a& -f- c m — 71 3 c + 2 d Factor the following expressions, if possible : 16. 27x? — 8y 3 18. a,- 5 -?/ 10 . 20. r e s« — y*. 17. a 4 -2/ 4 . 19. 32 -m 5 . 21. a 7 -f& r . a 4 - -6 4 a + b a 4 - -b* a- ■b ar 5 - y\ a> — ■y m 5 - -i m - -l a 6 6 6 -c 6 l + 2a 81a? 4 -?/ 4 3a? + # 16 -x A 2-x a* - 243 ar 5 a — 3 a? 81c 4 -16d 4 X s + x 2 - -2\x-2 x s - 2 x 2 3x* Zx*- -6x 6x- 2 6z-12 SPECIAL PRODUCTS AND FACTORING 109 22. 32 + r\ 25. x w -y 10 . 28. 64 - a 6 . 23. l + 32m 6 . 26. x 8 + y\ 29. a 8 -256. 24. z 6 + ?/ 6 . 27. m 5 -243. 30. 32 a 5 + 243 b 5 . SUPPLEMENTARY TOPICS 92. The Remainder Theorem makes it possible to find the remainder in certain division problems by a short process. It is known that dividend = divisor x quotient -f- remai?ider. Suppose that x* -J- x 2 — 2 is divided by x — 2 ; then : x s + X 2__ 2 = (x -2) • Q + B. x 2 + 3 x + 6 Let x = 2 ; then : 8+4-2=0. Q + B. .-. 10 = • Q + B. .: 10 = i?. That is, the remainder is 10. Check : See solution on right. 10 = B. At the left, the correct remainder was obtained by substitut- ing 2 for x in the given expression. This suggests the Remainder Theorem. If a rational and integral polynomial (§ 12) involving x be divided by x— a, the remainder may be found by substituting a for x in the given polynomial. Proof : 1. The polynomial (containing x) = (x — a) • Q 4- B. 2. .-. The polynomial (x replaced by a) = (a — a) Q + B. 3. ,\ The polynomial (x replaced by a)= • Q + B = B. Example 1. Find the remainder when x i — 3 x + 5 is di- vided by x — 3. Solution: 1. Comparing x — a and x — 3, a must be 3. 2. Substituting 3 for x in x* - 3 x + 5, i2 = 3 4 -3.3 + 5= 81 -9 + 5 = 77. Example 2. Find the remainder when the divisor is a; + 3. Solution : 1. Comparing x — a and x + 3, a must be — 3. 2. Substituting — 3 for x in x* — 3 x + 5, 22=z(-3) 4 -3- (_3)+5 = 81 +9 + 5 = 95. 110 ALGEBRA EXERCISE 47 Find the remainders when : 1. x 5 + 2 x 4 — 7 is divided by x — 1 ; by x + 1. 2. 2# 3 +3ar J — 4 # -f 5 is divided by a; — 5 ; by x -f 5. 3. a 4 4-2^-6 is divided by x + 2; by x — 2. 4. m 3 — 2 m 2 — 4 is divided by m — 3 ; by ra 4- 4. 5. # 4 — y 3 4- y 2 — y + 6 is divided by ?/ -f- 3 ; by y — 2. 93. Synthetic Division is a short process for finding the quotient as well as the remainder when a polynomial containing x is divided by a binomial of the form x — a. Consider the two solutions : Solution (a): Solution (6): 5 a 3 -11 a 2 - 6x- 5 +15 +12+18 5s 2 + 4s 4 6 aj + 3|5aj3-lla;2- 6 x - 10 « - 3 5 x 8 - 11 x 2 - 6 x -10 5 "^rf fl, 5 4 4+ 6||+ 8. 4 z 2 - 12 g Quotient : 5a; 2 + 4a; + 6. ;r- __ « q Remainder : 4 8. 6x-18 4 8 The method of performing the solution (6) : (1) — 3 of the original divisor is changed to 4 3. (2) 5 X s -f- x = 5 x 2 . Place 5 in the third line. (3) 4 3.4 5=415. Add the product, 4 15, to —11. Place the sum, 4 4, in the third line. (4) + 3 • 4 4 = 4 12. Add the product, 4 12, to - 6. Place the sum, 4 6, in the third line. (5) 4 3.46=418. Add the product, 4 18, to - 10. Place the sum, 4 8, in the third line. (6) The numbers 5, 4 4, and 4 6 are the coefficients of the quotient. Since 5 sc 8 -4- x = 5 z 2 , the full quotient is 5 x 2 4 4 x + 6. The last number of the third line, + 8, is the remainder. SPECIAL PRODUCTS AND FACTORING 111 A partial explanation follows : (1) In step (1), — 3 is changed to + 3. This permits addition in steps (3), (4), and (5) instead of the customary subtraction. Thus, in solution (a), when — 15 x 2 is subtracted from — 11 x 2 , the result is 4 x 2 ; in solu- tion (6), when + 15 ic 2 is added to — 11a; 2 , the result is again 4 a; 2 . (2) In step (3), +4 below the line represents, first, the coefficient of the first term of the remainder as in solution (a). When 4x 2 is divided by x the quotient is + 4 x, so that 4 may properly be considered also the coefficient of the second term of the quotient. Similarly in the case of + 6 in step (4). Example 2. Divide 7 x A - 29 M - 3 t A by x + 2 1. Solution. Change x + 2t to x — 2t. x-2t 7 x 4 + tx 3 - 29 t 2 x 2 + fix - 3 tt 7 -u t -f28« 2 +2*3-4*4 7 _14* _ t 2 +2« 3 ||-7*4 Quotient : 7 x z — 14 tx 2 — t 2 x + 2 t 3 . Remainder : — 7 tf 4 . Note 1. "When powers of x are missing, supply them with coefficients zero, as in this example. EXERCISE 48 Divide by synthetic division : 1. x?-2x 2 + 2x-5byx-l. 2. 2x?-4:X 2 + 6x-15hy x + 1. 3. f-Sy + 10bjy-2. 4. z 4 + 5!? + 15z 2 -25byz + 2. 5. ^-32 by t-2. 6. 3m 4 -25m 2 -18by ra-3. 7. 4a 3 + 18a 2 + 50 by a + 5. 8. 6c 4 + 15^ + 28c + 5by c+3. 9. 3 x 3 + 4 ma; 2 — 2 m 2 a; — 5 m 3 by a; — m. 10. 4 a 4 - 15 5V- 4 6 4 by x + 2b. Note. In §§ 92 and 93, two short processes for finding the remainder in certain division problems are given. Each is important. The second has the further advantage of determining the quotient as well. 112 ALGEBRA z 3 - z 2 -Sz + 2 = 0. 3. ^-15?y 2 -16 = 0. 11.* ^_5f-2 = 0. 4. 4* 4 -17* 2 + 4 = 0. 12. ^-1=0. 5. 9^ + 14^-8 = 0. 13. ^-8 = 0. 6* ^-7^ + 6 = 0. 14. r 3 -5?- 2 = 5-r. 7. aj 8 +2a?-9«-18 = 0. 15. x b -x A - 16 a? + 16 = 0. 8.* f-lSy- 12 = 0. 16.* x 4 -x 3 -5» 2 -.c-6 = 0. Solve for a; : 17. x 4 -m 2 x 2 -n 2 x 2 +m 2 n 2 =0. 19* 2 a 3 - 3a?x + a 8 = 0. 18. a^ + &» 2 -ac 2 z-6c 2 =0. 20. a 4 -r 4 = 0- Remark. The graphical solution of equations of higher degree is consid- ered in § 2(37. X. QUADRATIC EQUATIONS HAVING TWO VARIABLES GRAPHICAL SOLUTION 96. Graph of a Single Equation. Example 1. Draw the graph of y — a? — 0. Solution: 1. Solve the equation for y : y = x 2 . 2. When x s= 1 o :; 4 5 -1 -2 -3 -4 -5 then y = 1 4 9 1(3 25 + 1 + 4 + « + 16 + 25 3. This curve is a Parabola. 4. The coordinates of any point on the parabola satisfy the equation. The coordinates of A are: x = _ 45 and y - 20+. Substituting in y = x 2 : does 20+ =(- 4.5) 2 ? Does 20+ = 20.25 ? Yes, approximately. The coordi- nates should satisfy the equa- tion of the graph. Since the graph cannot be absolutely ac- curate, and since the coordi- nates of a point on the graph cannot be read exactly from the graph, the coordinates de- termined may not exactly sat- isfy the equation. ¥ =F 5C :c T " t t 1 V t i _r : -JL " 40 zL _ 7 t t 3 t jt -C 5 j v i£ 7 i : t — V y :: 5 5 : 5 : .___ \ _£ ± _ _ : V : : ::::? :: : i - ^ 5- \ r . X L_ ~^-'- cj^- 1 --x -5 -4 -3 -2 - 2 3 4 5 *-- 117 118 ALGEBRA Example 2. Draw the graph of x 2 + y 2 = 25. Solution : 1. Solve the equation for y : y = ± V25 — x 2 . 2. When x = + 1 + 2 + 3 + 4 + 5 y = ± V25 ±5 ±V24 ±4.8 ±V21 ±4.5 ±Vlo ±4 ±V9 ±3 ±V0 3. When x is negative, y has the values given by the corresponding positive values of x. Thus, when x is — 3, y is ± V25-(-3) 2 = ± V25-9 = ± VlG = ± 4. Notice that for each value of x, y has two values ; thus, when x is + 4, y is either + 3 or — 3. Hence, both (4, 3) and (4, — 3) are on the graph. For x greater than 5, y is imaginary ; thus, when a: =6, -V X' _Q_ X .-.5 _|~4 =3 -2 - 2 3 4 5" y = ± V25 — 36 = ± V- 11. This means that there are not any points on the graph for values of x greater than 5. The square roots required in step 2 may be obtained either by the method of § 64, or from the table of square roots constructed in §65. 4. This curve is a Circle. Every equation of the form x 2 + y 2 = r 2 , is a circle with its center at the origin and its radius equal to r. Example 3. Draw the graph of 9 x- + 25 f = 225. 225 - 9 x 2 Solution : 1 . y' 2 = 25 y= ±$V226-0X*. 2. When x = 2, y = ± \ V225 - 36 = ± \ V189 = ± $(3 V21) = ±K3x4.5) = ±K13.5) = ±2.7. QUADRATIC EQUATIONS HAVING TWO VARIABLES 119 When x = + 1 + 2 +3 + 4 + 5 + 6 then y = \/227 5 ±¥ ±3 ±^V216 ±2.9 ±iVl89 ±K13.5) ±2.7 ±^Vl44 ±¥ ±2.4 ±1.8 Vo ±^V^99 imag'y imag'y For negative values of x, y has the values given by the corresponding positive values of x. (See Example 2, step 3.) Notice that for each value of x, there are two values of y. ' |) — ►-«, *' / / / 1 y 1 I s _ i -3 -o - 2 il <: 1 k ^ > \ V / \ v "V V S ■«~ -3 ' 3. This curve is an Ellipse. Every equation of the form az 2 + by 2 = c, where a, b. and c are positive, and a not equal to b, has for its graph an ellipse. Example 4. Draw the graph of 9 a 2 — 4 # 2 = 36. 9z 2 -36 Solution : 1. y* as ; .•.w=±|Vz 2 -4. 2. When x = l, y = ±|Vl — 4=±f V— 3 ; .-. y is imaginary. When x = + 1 + 2 +3 +4 45 + then y = ±|V=4 imag'y ±|V"^3 imag'v ±|Vo ±fV5 ±3 3 ±|V12 ±5.1 ±|V21 ±6.8 ±|V32 ±8.4 120 ALGEBRA For negative values of x, y has the values given by the corresponding positive values of x. (See Example 2.) Y s, s _ s \ ^ A L _- - 6 _ -t t -4 - ') A \ f -j \ \ j / I i 3. This curve is a Hyperbola. Every equation of the form ax 2 — by 2 = c, where a, 6, and c are positive numbers, is a hyperbola. EXERCISE 52 Draw the graphs for the following equations; name the curves obtained. 1. a* + y* = 36. 2. y = Sx 2 . 3. x 2 = 6y. 4. o^ + 4?/ 2 = 36. 5. x 2 -4,y 2 = 36. 6. xy = 4. 7. x 2 + tf = 55. ' 8. 4z 2 + 2/ 2 = 16. 97- Solution of a Pair of Simultaneous Quadratic Equations. Example 1. Solve the pair of equations a»+y*=25. (1) x-y+l = 0. (2) Solution : 1. The graph of equation (1) was drawn in Example 2 of §96; it is the circle of radius 5, with center at the origin. QUADRATIC EQUATIONS HAVIN T G TWO VARIABLES 121 2. The graph of equation (2) is found as in § 46. It is a straight line (§ 4(5). When as = 0, y =, I j when x = 2, y = 3. 3. Since points A and B are on both graphs, their coordinates should satisfy both equations. A — (3, 4) ; B =(— 4, — 3). When the coordinates of ^4 and of B are substituted in the equations, it is found that they satisfy the equations. .\ x = 3, y = 4, and x = — 4, y = — 3 are solutions of the pair of equations. Nun --«5 -4^3 i^2L 2,3456 it IL^ L Note. Since the graph of every linear equation having two variables is a straight line (§ 47), and since, as the student's subsequent courses in mathe- matics will show, the graph of every quadratic equation having two variables must be one of the curves discussed in § 9b', it is clear that, as a r.ule, a quadratic and a linear equation with two variables will have two common solutions, for a straight line will, in general, meet such curves in two points. The straight line might touch the curve at only one point, thus giving only one solution ; or it might not touch the curve at all, thus not giving any real solution. Example 2. Solve the pair of equations : I , « 2 )o\ Solution : 1. The graph of equation (1) is the circle of radius 5 (see Ex. 2, § 96). The graph of equation (2) is the ellipse of the figure. 2. The points of intersection of the graphs are : A: (4,3); B: (-4,3); C: (-4, -3);D: (4, -3). 122 ALGEBRA 3. Substituting these values of x and y in equations (1) and (2), it be- comes clear that the equations have four common solutions. . *% /'^•""*" n -Ss jtr I ^s rf B A 1 ^Sr VX " It SSit" 1 > i -'- X - Q X -R -7 -fi -5 -4 -3 r 2 "I 1 9 3 < 5 6 7 f ..i. / Vl ' J V ~ \A -2 IZZ ^ ^4 : N ^c ... 1 ? ^v *<£ ££-^.-4 „ --*£ ^s. ■-* S ^ - -5 7 Note. Two quadratic equations, having two variables, will have four common solutions, in general. This becomes clear when the graphs of § 96, which result from such equations, are combined in pairs. For example: However, there are other possibilities. Thus, the ellipse might intersect only one branch of the hyperbola in such manner as to give only two real solutions ; or it might not intersect it at all, giving no real solutions. 2. QUADRATIC EQUATIONS HAVING TWO VARIABLES 123 EXERCISE 53 Solve the following pairs of equations graphically : faj»4-y»=100. 5 (x 2 + y 2 = 50. \x-y + 2 = 0.' [xy=-7. ±x 2 + y 2 = 61. t3x 2 + 4y 2 = 76. 2x-y = l. ' l32/ 2 -lla; 2 = 4. r^_ ? / = _ 9 . 7 |^-2/ 2 = 16. ' \2x-y = 3. I ' \y 2 + x = ±. 4 f«+f--«L |4^ + 2/ 2 = 36. * ljcy=-7. ' \x 2 -y 2 = -16. 9. Draw the graph of — - + ^- = 1. On the same sheet, draw the three graphs obtained from the equation x 2 = y + 7c, when k is made successively 6, 2, and — 6. Remark. The four curves studied in this chapter, the circle, the parab- ola, the ellipse, and the hyperbola are called conic sections, for each may be derived by intersecting a circular cone of two nappes by a plane. A special study of these curves is made in a later course in mathematics, analytic geometry. XI. SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS 98. A set of equations having two or more variables are called Simultaneous Equations if each equation is satisfied by the same set, or sets, of values of the variables. 99. A set of equations that are solved as simultaneous equations will be called a system of equations. 100. A pair of simultaneous linear equations (§ 47) having two variables have one common solution (§ 51). The com- mon solution is readily obtained by the addition or subtrac- tion method (§ 53), or by the substitution method (§ 54) *of elimination. 101. Pairs of simultaneous equations occur of which one or both are of degree higher than the first. Thus, in (a) below, equation (1) is of the first degree and (2) is of the second ; in (6), equation (1) is of the second degree and (2) is of the third. , , f3z+4*, = 5. (1) j^-2^6. (1) (a) \2x^-3xy = 7. (2) W \x* + yS=5. (2) Many such combinations, even with two variables, are possi- ble. Only in special cases, however, are the common solutions readily obtained. A few such cases will be considered. 102. Case I. One Linear and One Quadratic Equation. r a .2 + 2/ 2_|_6a-16 = 0. (1) Example. Solve the system : ]i_i_2a; i/ = (2) Solution : 1. From (2), V = 2 x + *• ( 8 ) 2. Substituting in (1), z 2 + (2a;+ l) 2 + 6* - 16 =0. (4) 124 SIMULTANEOUS EQUATIONS 125 (5) (6) 3. Simplifying (4), x 2 + 2x-3 = 0. 4. Solving for x, x = 1, or x = — 3. 5. Substitute in (2) : when x = 1, 1 + 2 — y = ; .-. y = 3. when £=— 3, 1 — 6 — ?/ = ; .-. y = — 5. The solutions are : x = 1, ?/ = 3 ; £ = — 3, y = — 5. The solutions may be checked by substitution. Note. One linear and one quadratic equation having two variables have, in general, two common solutions. The graphical solution of a particular pair of equations of this type is given in Ex. 1, § i>7. 5. 8. 9. m 2 + mn m— n t 2 = — 19. -7. x + y = -3. 54. f*+Jf \xy = 6a \x*-xy + y 2 \x-y = -3. r 0^ + 2/2 = 101. \x + y = -9. + 5xy-y 2 = -7. SIMULTANEOUS EQUATIONS 127 Check : These four solutions are readily checked by substitution in equations (1) and (2). Note 1. In case one equation does not have a constant term, solve it im- mediately for one variable in terms of the other as the equation (5) in step 2. Note 2. A system consisting of two quadratic equations has, in general, four solutions. Note 3. The graphical solution of a particular pair of equations of this type is given in Ex. 2, § 97. EXERCISE 55 Solve the following systems : (3cd + 2d 2 = -7. *— %m — K(5 Example. Solve the system : I „ J \ & + X y + y 2 = 28. (2) Solution: 1. Dividing (1) by (2): X — y = 2. (3) (i) 2. Form the new system : x 2 + xy + y* = 28. (2) x - y = 2. (3) 3. Solve the new system by the methods of Case I : z = 4, y = 2 ; and x =— 2 ; y =— 4. 128 ALGEBRA Check : These two solutions are readily checked by substitution in the equations (1) and (2). Note 1. Whenever possible, divide one equation of the given system by the other, member by member, and form a new system consisting of the quo- tient equation and the divisor equation. Note 2. The full theory underlying this type of example belongs in a more advanced text and is therefore omitted. 108. Number of Solutions. In Case I (§ 102) two solutions and in Case II (§ 105) four solutions are generally obtained. The following rule for determining the number of solutions of any system of equations having two variables is given without proof : Rule. — Two integral equations, having two variables, whose degrees are m and n respectively, have , in general mn common solutions. Thus, a cubic (third degree) equation and a quadratic equation would have six common solutions. If, however, the system could be reduced to a simpler system, as in the example of § 107, then the number of solutions would be determined by the degrees of the equations forming the new system. EXERCISE 56 Solve the following systems of equations : \x 2 -y 2 = 56. „ \ a 3 + b 3 = -335. 1 x + y m 14. I a 2 - ab + b 2 = 67. fa; 4 -?/ = 240. „ fm 3 -w 3 = -117. 2 ' \x 2 + y 2 = 20. " lm-w = -3. lx>-f = 133. jTSe + tf-i *' \x-y = 7. ' I 27 . J ro + w = 5 + 1 row = — 3. 130 ALGEBRA 4. Solving A : m = 2, n = — 3 ; or m = - Solving B : m — 3, n = — 1 ; or m = - Check : The four solutions check when and (2). EXERCISE 57 Solve the following systems : xy = 12. - 3, n = 2. - 1, n = 3. substituted in equations (1) 2. 6. 8. 10. 11. 12. y 2 = 40. A 2 B 2 +2$AB-m=0. 2A + B = 11. 2w 2 -5t 2 = 13. 15t 2 + w 2 = 2L m 2 +p 2 = 1. 4x 2 + y 2 = 6l. 2ar 2 + 3 2 / 2 = 93. 4v 2 — 5va?=19. ^^+0^ = 6. 3^_5^ + 2i 2 =-3. fSr — on \r-t = l. (c? + cd + d 2 ic-d = 19. 97. x* + tf = 756. t x 2 — xy + y' 2 = 63. > 2 -s 2 = 3. [j» = - 2. V + 26 2 = 47 + 2a. ^ a 2_2b 2 = -7. xy = a 2 — 1. 2 a. 7?, 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. = 10 n m-\-n 3 'm_±_n.m m m 2 -f n 2 = 45. V-2/ 3 = 3a 2 + 3a + l. x~y = l. '7v 2 -5vt-3t 2 =36. >v 2 + 3vt + t 2 =-4. 15 x -f y 2 x V x — y x-\-2y 4 * x-3y«-2. i 1+ m)= m - K 1+ sfo)= 48a f m — v = — 31. [ mi? = — 150. fa^ + /=:7a 8 . [x + y = a. {x 2 + y 2 =2a 2 -2ab + b\ { {2x 2 -y 2 =a 2 + 2ab-b 2 . x 2 + y 2 = 5(a 2 + b*). 4^--y 2 =5a(3a-46). |8ic 2 -ll2/ 2 = 8. il2ic 2 4-13 2/ 2 = 248. f 18 23. r — s r + r-2s = l. SIMULTANEOUS EQUATIONS 30 131 14 = 8. x 2 -y 2 = 16. y 2 — 14 = x. x i -\-x 2 y 2 -{-y 4 = 91. ,x 2 + xy + y 2 = l3. Hint : See § 207. fo 4 + a a 6 2 +6 4 = 133. U 2 - 24. 25. 26 27. a& + 6 2 = 7. y x % x + y = l. Hint : Clear of fractions ; divide (1) by (2). 28. m 2 — mn = 27 n. mn —n 2 = 3m. Hint : M 3 (2) ; add ; factor. [y 2 = 3y-x. Hint: Find (l)-(2). y(x — a) = 2 a&. #(# — 6) = 2 ab. fx 2 y — x= — 14. Uy + a 2 = 148. 5m 2 -9?i 2 =-121. 7n 2 -3m 2 = 105. f mn — (m — n) = 1. lm 2 7i 2 + (w-n) 2 ==13. a 2 -ab-12b 2 = S. a 2 + ab -10 6 2 = 20. 2tf*-Sa#= -4. 4 ## — 5 ?/ 2 = 3. |i 2 + 5^-^=-7. \t 2 + 3tw-2w 2 =-4;. \xy + (x-y) = —5. \xy(x — y) = — 84. ,9x 2 -xy — y = 51. -5xy + y 2 + 3x = $l. 31. 32. 33. 34. 35. 36. 37. EXERCISE 58 1. Find two numbers whose sum is 15 and the sum of whose squares is 113. 2. Find two numbers whose difference is 9 and the sum of whose squares is 221. 3. Find two numbers whose difference is 7 and whose sum multiplied by the greater gives 400. 4. The difference of the squares of two numbers is 16 and the product of the numbers is 15. Find the numbers. 132 ALGEBRA 5. The sura of the squares of two numbers is 52; the difference of the numbers is one fifth of their sum. Find the numbers. 6. The difference of the cubes of two numbers is 218 ; the sum of the squares of the numbers increased by the product of the numbers is 109. Find the numbers. 7. If the product of two numbers be multiplied by their sum, the result is — 6 ; and the sum of the cubes of the num- bers is 19. Find the numbers. 8. Find two numbers whose difference is 4 and the sum of whose reciprocals is f . 9. The sum of the terms of a fraction is 13. If the numera- tor be decreased by 2, and the denominator be increased by 2, the product of the resulting fraction and the original fraction is ^-. Find the fraction. 10. Find the number of two digits in which the units' digit exceeds the tens' digit by 2, and such that the product of the number and its tens' digit is 105. (See § 172, First Yea)- Algebra.) 11. The sum of the squares of the two digits of a number is 58. If 36 be subtracted from the number, the digits of the remainder are the digits of the original number in reverse order. Find the number. 12. Find the number of two digits such that, if the digits be reversed, the difference of the resulting number and the original number is 9, and their product is 736. 13. The area of a rectangular field is 216 square rods, and its perimeter is 60 rods. Find the length and width of the field. 14. The hypotenuse (§ 73) of a certain right triangle is 10 feet, and the area of the triangle is 24 square feet. Find the base and altitude of the triangle. SIMULTANEOUS EQUATIONS 133 15. Find the dimensions of a rectangle whose diagonal is 2V10 inches and whose area is 12 square inches. 16. A rectangular field contains 2\ acres. If its length were decreased by 10 rods, and its width by 2 rods, its area would be less by one acre. Find its length and width. (See p. 145, First Year Algebra.) 17. The altitude of a certain rectangle is 2 feet more than the side of a certain square ; the perimeter of the rectangle is 7 times the side of the square, and the area of the rectangle exceeds twice the area of the square by 32 square feet. Find the side of the square and the base of the rectangle. 18. If the length of a rectangular field be increased by 2 rods and its width be diminished by 5 rods, its area becomes 24 square rods ; if its length be diminished by 4 rods and its width be increased by 3 rods, its area becomes 60 square rods. Find its length and width. 19. A man has two square lots of unequal size, together con- taining 74 square rods. If the lots were side by side, it would require 38 rods of fence to surround them in a single inclosure of six sides. Find the length of the side of each. 20. A and B working together can do a piece of work in 6 days. It takes B 5 days more than A to do the work. Find the number of days it will take each to do the work alone. 21. Find the sides of a parallelogram if the perimeter is 24 inches and the sum of the squares of the number of inches in the long and short sides is 80. 22. One of two angles exceeds the other by 5°. If the num- ber of degrees in each is multiplied by the number in its supplement, the product obtained from the larger of the given angles exceeds the other product by the square of the number of degrees in the smaller of the given angles. Find the angles 134 ALGEBRA 23. Two angles are supplementary. The square of the num- ber of degrees in the larger angle exceeds by 4400 the prod- uct of the number of degrees in one angle by the number in the other angle. Find the number of degrees in each angle. 24. The difference in the rates of a passenger train and a freight train is 10 miles per hour. The passenger train re- quires 1 hour more for a trip of 175 miles than the freight train requires for a trip of 100 miles. Find the rate of each. 25. A crew can row upstream 18 miles in 4 hours more time than it takes them to return. If they row at two thirds of their usual rate, their rate upstream would be 1 mile an hour. Find their rate in still water, and the rate of the stream. 26. The area of one square field exceeds that of another square field by 1008 square yards ; the perimeter of the greater exceeds one half of that of the less by 120 yards. Find the side of each field. 27. The tens' digit of a certain number exceeds the units' digit by 1. If the square of the given number be added to the square of the number with the given digits in reverse order, the sum is 585. Find the number. 28. If the digits of a number of two figures be reversed, the quotient of this number by the given number is If, and their product is 1008. Find the number. XII. THE THEORY OF QUADRATIC EQUATIONS 110. The Sum and the Product of the Roots. The general quadratic equation is : aa?+bx + c = 0. (1) Divide both members by a : ^ + ^.a-f£ = 0. (2) a a The roots of (1) are : 6+ V6»-4oe 5-vV-4«e . ( 2a 2a ri + r 2 = — — = (4) 2a a (_ 6)2 _ ( Vfc 2 - 4 ac) 2 _ 6 2 - (b 2 - 4 oc) _ 4 oc c ... ?i • r 2 = — — — — - — - — - • {o) 4o 2 4 a 2 4 a 2 a Rule. — In the general quadratic equation ax 1 -f bx + c= o : 1. The sum of the roots is • From (4). a 2. The product of the roots is -• From (5). a 3. If the coefficient of x 2 is made 1, the coefficient of x is the negative of the sum of the roots, and the constant term is the product of the roots. From (2), (4), (5). Example 1. Find the sum and the product of the roots of the equation 2a? 2 — 9# — 5 = 0. Solution: 1. a = 2; b=— 9; c = — 5. 2. ... ri + „=_» = _^2 = + | i ™,=5 = ZlS. a 2 2 a 2 Note. The first part of this rule justifies the method of checking solutions of quadratic equations recommended in § 7(5. 135 136 ALGEBRA EXERCISE 59 Find by inspection the sum and the product of the roots; check examples 1, 2, 3, and 7 by finding the roots : 1. a? + 7a + 6-=0. 5. 9r-21r 2 + 7 = 0. 2. ra 2 -ra + 12 = 0. 6. 4- ?/- 6 2/ 2 = 0. 3. 3c 2 -c-6 = 0. 7. 2 a 2 + 3^-5^ = 0. 4. 12 2/ 2 -42/ + 3 = 0. 8. 14 a 2 + 8 tx + 21 * 2 = 0. 9. One root of 4^ — # — 5 = is — 1. Find the other root. Solution : 1. r± =— 1. Let r 2 be the second root. 2. n + r 2 = + J ; /. - 1 + r 2 = i, or r 2 = 1^ = £. Check : Does n- r 2 = ? i.e. does — 1 • - = ? Yes. 4 4 4 10. One root of 3 x 2 + 7 x — 6 = is f . Find the other. 11. One root of 7 q 2 + 20 g + 12 = is - 2. Find the other. 12. One root of 15 ra 2 + 28 ra = 32 is f. Find the other. 13. One root of 3 x 2 - 2 kx = 33 k 2 is - 3 &. Find the other. 14. One root of 4j9 2 — 15 xp — 4 x 2 = is — 4j?. Find the other. 15. Find k so that one root of x 2 — 5 x -f A; = may be 7. Solution : 1. r\ + r 2 == 5 ; .*. r 2 + 7 = 5, or r 2 = — 2. 2. * = ri • r 2 ; .-. A; = 7 • - 2 = - 14. Check : If x 2 — 5 x - 14 = 0, then (at - 7) (x + 2) = 0. .-. as = 7, or -2. 16. Find & so that one root oi2x 2 — Sx—k = may be 3. 17. Find k so that one root of 3a 2 — 7 x — 2k = may be -2. 18. Find w so that one root ofa5 2 -|-7a;-f4ft = may be 5. 19. Find p so that the roots ofor 2 -f-3a;-f-j? = shall be equal. THE THEORY OF QUADRATIC EQUATIONS 137 20. Find r so that the roots of 3x 2 — 5x + r = shall be equal. 111. Formation of Equations Having Given Roots. There are two methods of forming a quadratic equation which shall have given roots. Example 1. Form the equation whose roots shall be i and -f. Solution : 1. Let the coefficient of x 2 be 1 ; then by § 228 the equation x 2 — (ri + r 2 ) x + rir 2 = 0. 3. .-. the equation is : * 2 -(-i)*+(-|) = ' orz* + |-| = 0. (§228) Multiplying both members by 8, 8 x 2 + 2 x - 3 = 0. Check : The given roots, if substituted, will satisfy the equation. Example 2. Form the equation whose roots shall be — 9 and 2. Solution : 1. If x as - 9, then x + 9 = ; K x = 2, then x - 2 = 0. 2. .♦. (x + 9)(x-2) = 0, or x 2 + 7x- 18=0. It is clear that this equation has the given roots. Note. This second method may be used also to form an equation having three or more roots. EXERCISE 60 Form the equations whose roots shall be : 1. 2,3. 4. 12,-5. 7. if. 2. -3,-6. 5. 2,f. 8. 3m, -5m. 3. 6, - 9. 6. 1, - 1 9. 4 1, - 1 1. 10. -fc, fc. 13. 2a-b, 2a + b. 11. 2, 3, - 5. 14. 3 + V5, 3 - V5. *2. a + 3m, a-3m. 15. 2 + 3V2, 2-3V2. 138 ALGEBRA DETERMINATION OF THE CHARACTER OF THE ROOTS 112. Classification of Numbers. The numbers considered in this text to this point are : (A) Eeal numbers. 1. Kational Numbers: (a) integers (positive and negative); (b) fractions whose terms are integers. 2. Irrational numbers : (a) quadratic surds (§ 67) ; (6) surd expressions, such as 2 + V3. (B) Imaginary Numbers : (a) pure imaginaries (§ 83) ; (6) complex numbers (§ 83). 113. It is often necessary to determine the character of the roots of a quadratic. Thus the roots of 2 x 2 - 8 x + 3 = are 4=bv ^ > Since 10 is positive, the roots are real numbers. Since 10 is not a perfect square, the roots are irrational. Since VlO is added in one root and subtracted in the other, the roots are unequal. Hence the roots are real, irrational, and unequal. It is possible to determine the character of the roots how- ever without determining the roots themselves. For the general quadratic ax 2 -+- bx -f c = 0, the roots are _ -& + V52_.4ac -&-V6 2 -4ac • Rule 1. — If ft 2 — 4 ac is positive, the roots are real and unequal. They are rational if fc 2 — 4 ac is a perfect square, and irrational if ft 2 — 4 ac is not a perfect square. 2. If ft 2 — 4 ac equals zero, the roots are real and equal. 3. If ft 2 — 4 ac is less than zero, the roots are imaginary. ft 2 — 4 ac is called the Discriminant of the quadratic. THE THEORY OF QUADRATIC EQUATIONS 139 Example 1. Determine the character of the roots of 2a 2 -5a-18=0. Solution : 1. 6 2 - 4 ac = ( - 5) 2 -4(2) (- 18) =25 + 144 = 169 = 13 2 . 2. By Rule 1, the roots are real, rational, and unequal. Example 2. Determine the character of the roots of 3^ + 2a; + l = 0. Solution : 1. b* - 4 ac = 4 - 4 . 3 • 1 = 4 - 12 = - 8. 2. By Rule 3, the roots are imaginary. Example 3. Determine the character of the roots of 4a 2 -12a; + 9 = 0. Solution : 1. 6 2 - 4 ac = 144 - 4 . 4 . 9 = 144 - 144 = 0. 2. By Rule 2, the roots are real and equal. Note. This type is most easily understood if the quadratic is solved hy factoring. This example becomes (2x — 3) (2* — 3)=0. The roots are then. § and §. It is customary to say that the roots are equal. EXERCISE 61 Determine by inspection the character of the roots of : 1. Gar* + 7a;-5 = 0. 2. 4z 2 -20x + 25 = 0. 3. 3z 2 -8z + 5 = 0. 4. x 2 -9x-\-15=0. 5. 5i* + 7r + 3 = 0. 6. 9s 2 -l = 12s. 7. 5 m — 2 = 4 m 2 . 8. 4 y 2 — y = 6 - 9. 5* 2 + 7 = 8*. 10. 20^-41^ + 20 = = 0. 11. 7a? + 3x = 0. 12. 16 m 2 - 9 = 0. XIII. EXPONENTS 114. In the preceding chapters, only positive integers have been used as exponents. The fundamental definition when m is a positive integer, is : a m = a • a • a ••• a (m factors). (§ 3) 115. There are five fundamental laws of exponents. When m and n are positive integers : I Multiplication Law. Just as a 5 x a 7 = aP, so a m X a n = a m+n . Proof : 1. a m = a • a • a ••• a (tn factors). (§ 114) 2. a n — a • a- a ••• a (n factors). (§114) 3. .•. a m • a n = {a • a • a ••• a («• factors)} • {« • a . a ••• a (« factors)} = a • a • a ••• a {(m + n) factors}. 4. .•. a n • a M - a" l+n . (§ H4) II. Division Law. Just as a 9 -^ a* = a 5 , so a m -=- a n = a m-n . (m greater than n.) Proof: 1. qm = ^' ^' ^ ' 4 "' & ' a ' a '" a (^ factors) , (§114) a n fa • ^•ft-jk ••• ft {n factors) 2. = a • a ••• a {{m — n) factors} as a m ~ n . (§ 114) 3. .*. a m ■«- a n = a m_n . III. Power of a Power. Just as (a 5 ) 3 = a 15 , so (a m ) n = a mn . Proof : 1. (a m ) n - a m a m • a m ••• a m (n factors) (§ 114) 2. = : a m+m+m+ ••• +m (n terms). (Law Ij 3. .*. {a m ) n = a mn {since m .+ m + ••• + m (w terms) as raw} IV. Power of a Product. Just as (ab) 5 = a 5 b 5 , so (ab) n = a n b n . 140 EXPONENTS 141 Proof : 1. («&)• = (ab) - (ab) (g&) ••• (ab) (n factors) (§ 114) 2. = {a • a • a ••• a (n factors)} • {b • b • 6 ••• b (w factors)}. (§ H4) 3. .\ («6) n = a n • b n . V. Power of a Quotient. Just as - ] = -, -tJK-fr '"-'•(!)■-{*)(!)©••(;) <■*■*-> """ „ _ q • g • q ••• g (n factors) 6 • 6 • 6 ••• b (n factors) 3. ,(f)" = ^. (tiu) Involution is the name given to the process of finding a power of a number. (Compare with § 3.) EXERCISE 62 Find the results of the indicated operations in the following examples, using the five laws above; the literal exponents denote positive integers. 1. x™-x. 13. x 15 -j- a^ 3 . 25. O 6 ) 4 - 2. m 12 • m 11 . 14. x l2 + x». 26. G/ 5 ) 7 - 3. y 5 - y n . 15. if n -f- y n . 27. (m 4 ) 8 . 4. m 2a ' m a . 16. m' ,c -+- m e . 28. (- a 5 ** 8 ) 5 . 5. a 3n • a 2n . 17. a 4n -+- a n . 29. (j/Vw) 4 . 6. b r+1 • b 2 . 18. b r+4 + b 2 . 30. (m s nyy. 7. c n ~ 4 • c 5 . 19. c n+5 _j_ J 31. «r. 8. d 2r+1 • d r . 20. rf2r+3_j..^ 32. (6-) 2 . 9. z r+1 • z r ~ l . 21. ar^H-ar* 33. (_C n d m )*. 10. t n-2 . t n+S^ 22. *»+•-*.«•-*. 34. (a?y) m . 11. w m+n • w m ~ n . 23. w w+n _j_ ? y»»-n # 35. (rV)'. 12. gn-r+1 . g r t 24. gln-r + l + g r. 36. (x m y n y. 142 ALGEBRA » ©'• "■ (-?)'■ " (5)" 39. ©'• - (9*- » (£)' 116. Only cube and square roots have been considered in the preceding chapters. More general roots occur in mathe- matics. 117. Just as -y/x indicates the cube root of x (§ 3), so Vx indicates the nth root of x. n is called the Index of the root. The nth root of x is the number whose nth power equals x; thatis > (Vxy=x. Thus, y/x& = x s , since (a 8 ) 4 = x n . y/%™ = a; 4 , since (a 4 ) 5 = x 20 . V- x^y 21 =—x 2 y s , since (— se 2 y 3 ) 7 = - x u y 21 * The number under the radical sign is called the Radicand. Rule. — To find the nth root of a perfect nth power, divide the exponent of each factor of the radicand by n. Every number has n nth roots. Unless something is said to the contrary, the principal root is denoted by the symbol -y/~. If n is even, this root is the positive root ; if n is odd and the radicand is negative, this root is negative. Evolution is the name given to the process of finding the root of a number. (Compare with § 3.) Historical Note. A symbol for extracting a root did not appear until the fifteenth century. In Italian mathematics, the first letter of the word Radix, meaning the root, was used to indicate the square root : thus, R. Presently there were used R.2 a , R.3 a , etc. to indicate the square, cube, and other roots. Chuquet, a French mathematician of about 1500, used R2, R3, etc# EXPONENTS 143 In Germany, a point was placed before a number to indicate that its square root was to be taken. Two points were used to indicate the fourth root, and three the third root. Keise, 1492-1559, replaced the point by the symbol, ^/. to indicate the square root, and Rudolph, 1515, used the symbol, y/^/, for the fourth root. Stevin, 1548-1620, used the better symbols: V®» V(D> etc - Girard, 1590-1632, used: fy fy etc. Des- cartes used the vinculum to indicate what numbers were affected by the root. EXERCISE 63 11. ^64 a 6 b«. 21. VoF. 12. V)* 14. (32a 5 6 20 )i 15. (81 a; 8 ?/ 4 ) *. Express with radical signs: 16. 2* 18. 5i 20. 4a;l 22. 2 abK 24. mM. 17. A 19. (4 ajV> *Yi by: (a) «*; (6) «£; (c) yt; (d) xty. 4. Multiply each of the following numbers : m ~»; n *; m^" 17 ; irti«j m~*w ~*. by: (a) m; (6) m%; (c) w~*; (cQ m"V. Multiply : 5. a* + ah* + 6^ by a* — $. 6. 2a- 1 -7-3aby4a- 1 + 5. 7. of* + 2af* + 4aT* + 8 by af*-2. 8. a?* -|- a;^ 4- y* by a;^ — B*y« + y^. Find : 9. (a~* + &*)(flf*-&*). 12. (»* - 6) (»* + 13). 10. (afi - 1/"^) 2 . 13. (r* - sty. 11. (**-*•)» 14. (a* + 7 6" 1 ) (a* - 8 &- 1 ), 148 ALGEBRA Law II Example, m 2 * -s- m~* = m 2 * ~ ( ~ = m . 15. Express Law II in words. 16. Divide each of the following numbers : by: (a) * 3 ; (6) r 4 ; (c) *j (d) rl 17. Divide each of the following numbers : c- 5 d 4 ; ! >-m "M- -& -4 '-4 »4 »# -4 »■#_ *C -l? *V£- »*? »# -xSI- . ,vn- »^- -^ «y£® 132. To Introduce the Coefficient of a Radical under the Radical Sign. Example. 2aty^=-V(2a)' 3 - •v / 3^='v / 8a 3 .3« 2 =-\/24aV Rule. — To introduce a factor under the radical sign : 1. Raise the factor to the power denoted by the index. 2. Multiply the radicand by the result of step 1. 154 ALGEBRA EXERCISE 71 Introduce under the radical sign the coefficients of : 1. 5V2. 4. 5\/4. 7. 4aV8a. 10. a?y 2 J/tfy\ 2. 8V3. 5. 2^5. 8. 7ajV6a?. 11. 3m* =(5)* = ffA -^ = 1 -^l2S. 4. .*. y/Z is the greatest number. Rule. — To reduce radicals to equivalent radicals of the same order : 1. Express the radicals with fractional exponents. 2. Reduce the exponents to a common denominator. 3. Rewrite the resulting expressions with radical signs. 156 ALGEBRA EXERCISE 73 Reduce to equivalent radicals of the same order: 1. V3 and V5. 6. y/xy, -i/yz, and -fyxz. 2. V2 and y/S. 7. y/Za, ^26, and Vo^". 3. -tyM and v^V 8. %% V8, and Vl3. 4. V2 and Vl2. 9. Vl^a and y/T+x. 5. V4 and V& 10. VcT+6 and V^^. Arrange in order of magnitude : 11. -y/2 and -y/3. 14. V3 and Vl5. 12. y/H and VS. 15. V3, ■&$, and /( + 2. (g) 5x 2 -9. (b) x*-5. (e) a^ + 4. (h) 2x*-5. (c) x> + 9. (/) 3^-4. (f) ax*-b. IRRATIONAL EQUATIONS 148. An Irrational Equation is one in which the unknown number appears under a radical sign or with a fractional exponent. RADICALS 167 149 It is agreed that the radical sign or fractional ex- ponent shall denote the principal root (§ 117) ; thus the square root shall always denote the positive root. 150. The following examples illustrate the methods of solu- tion of irrational equations. Example 1. Solve the equation x — 1 — Vx 2 —5 = 0. Solution: 1. Transposing, x — 1 = Vx 2 — 5. 2. Squaring both members, x 2 — 2 x + 1 = x 2 — 5. 3. .*. - 2 x = — 6, or x = 3. Check : Does 3 - 1 = VP^B ? Does 2 = VI ? Yes. (See § 149.) Note. When a single radical occurs in an equation, transpose the terms until the radical is on one side by itself and the remaining terms are on the other side. Then, if the radical is a square root, square both members of the equation; if it is a cube root, cube both members. Example 2. Solve the equation x — 1 -f V# 2 — 5 = 0. Solution : 1. Transposing, y/x 2 — 5 = 1 — x. 2. Squaring both members, x 2 — 5 = x 2 — 2 x + 1. 3. .-. 2 a; = 6, or x = 3. Check : Does 3 - 1 + V3 2 - 5 = ? Does 2 + V4 = ? No. (§ 149.) Therefore 3 is not a root of the equation. Kecall that in solving an equation a number is sought which will satisfy the equation. The equa- tion may, however, impose an impossible relation upon some numbers, as in this case, and then it is impossible to find a solution. What is the explanation of the solution x — 3 ? If the original equa- tion is compared with the equation of Example 1 , it is noticed that the only difference is in the sign of the radical ; also tbat in step 2, after squaring both members in both examples, the resulting equation is the same. In each example, if the equation of step 1 has a root, that number is a root of the equation of step 2 ; but, since the equation of step 2 is the same in each solution, it cannot be asserted in advance whether its root or roots are roots of the equation of Example 1 or of Example 2. When finally the solution x = 3 is obtained, the question arises, is 3 a root of the equation in Example 1 or m Example 2 ? The root x = 3 satisfies the 168 ALGEBRA equation of Example 1 ; it does not satisfy the equation of Example 2 It is customary to say that, in Example 2, the extraneous root 3 is intro duced by the method of solution. This example makes clear the necessity of checking the solutions oi equations. Example 3. Solve the equation V# — 2 + V2 x -f- 5 = 3. Solution : 1. Transposing, vx — 2= 3— V2 x + 2. Squaring, x — 2 = 9 — 6V2 x + 5 + 2 x + 5. 3. .-. 6V'2x + o = x+ 16. 4. Squaring, 36(2 x + 5) = x 2 + 32 x + 256. 5. . \ a; 2 - 40 z + 76 = 0. .-. a = 2,or38. (§110.) Check: Does v 7 ^ - 2 + V2 . 2 + 5 = 3 ? Does Vo + V9 = 3 ? Yes. Does V38-2 + V2 . 38 + 5 = 3 ? Does V36 + V81 = 3 ? No. (See § 149.) Therefore x = 2 is the only solution of this equation. Note 1. Jt wi ll be f ound that the extraneous root 38 will satisfy the equa- tion V2z + 5— Vcc -2 = 3. Note 2. When there are two radicals in an equation, arrange the terms so that one radical appears alone in one member of the equation. EXERCISE 84 Solve and check the following equations : 1. V3a-5-2 = 0. 9 V^6-fV"^-— - Vz-6 2. V6z + 9 + 8 = 5. , 1A V3r + 1+V3r A 3. V9x* + 5-3x = l. 10- ,-— — = =4 " 4. Vy-V2/-12 = 2. /— — , , 5. V^ + 4 + V^=3. ' VaT^a-VaT^a 6. \/8 x 3 - 12 a,* 2 + 1 = 2 a. V2z-3_ V4a7^4 7. Vs + ll + V^To r =5. V3a; + 2 V6a + 1 ,— . 2 13. VlO + #-VlO-a;=2. 8. Vm+Vm + 4 = — • Vm 14. V6-hl0a-3a 2 =2a-3. RADICALS 169 15. Vc + 2 + V3c + 4 = 2. 17. Vs 8 +8a> 2 +16a>-l=a;+3. 16. Vw-l + V3w + 3 = 4. 18. Vy + 3- V2/+8 = - Vy. 19. Var 2 — V2oj + ! = » — 1. 20. V5+a;+V5 — x — 12 21. Vo-aj V* + 2 V* 6 Solve for a? : 9a 2 -6 2 22. VsB — 12 ct& v« 23. Va + a? — Va — # = Va?. 24. s /(x-2b)(x + Sb)=x + 4:b. 25. V3a+2a— V4 a? — 6 a = V2a. 26. Solve the equation t = «A/-: (a) forZ; (6) for gr. 27. Solve the equation V=V^gs: (a) for # ; (b) for s. 28. Va + ic — V^» = 29. V#-a + V2a+3a= V5 a. 30. V(a + 26)a — 2a& = a — 4 7>. 31 - \^T5 + ^ + 4-2 XV. LOGARITHMS 151. Logarithms are exponents. Every positive number may be expressed, exactly or approxi- mately, as a power of 10. The exponent corresponding to a number so expressed is called its Logarithm to the Base 10. Thus, 10 2 = 100 ; therefore 2 is the logarithm of 100 to the base 10, This is written : logi 100 = 2, or more briefly log 100 = 2. Similarly log 10 35 is read " logarithm of 35 to the base 10." 152. Much difficult computation may be simplified by the use of logarithms. To make this fact clear, the approximate values of some powers of 10 will be computed and some ex- amples will be solved. 10°=1; lO^lO; 10 2 =100; 10 3 =1000, 1. 3. 10-*= 10' = V10 = 3.1623. 10 1J5 = 10 1 x 10 5 = 10 x 3.1623 = 31.623. 10 25 = 10 1 xlO L5 =10 x 31.623=316.23. 10-*= (10 5 )* = V33623 = 1.7782. 10 1 -* 5 = 10 1 X 10 25 = 10 x 1.7782 = 17.782. 10** = 10 1 x 10 125 = 10 X 17.782=177.82. 1 7S = (10 1 - 5 )* = V3T623 = 5.6234. 10 175 = 10 x 10- 75 =10 x 5.6234 = 56.234. 10 2 - 75 = 10 x 10 175 =10 X 56.234=562.34. 170 1.0000 = 10 000 1.7782 = 10 025 3.1623 = 10 050 5.6234 = 10° - 7B 10.0000 = 10 100 17.7820 = 10 1 - 25 31.6230 = 10 150 56.2340 = 10 175 100.0000 = 10 200 177.8200 = 10 225 316.2300 = 10 2 ^° 562.3400 = 10 2 - 75 1000.0000 = 10 300 LOGARITHMS 171 Example 1. Find 3.1623 x 17.782. Solution: 1. 3.1623 x 17.782 2. = 10- 50 x 10 1 - 25 = 10 1 - 7 *. 3. .-. 3.1623 x 17.782 = 56.234. This is approximately correct. Example 2. Find 1000 ■*• 56.234. Solution : 1. 1000 -r- 56.234 2. = 10 3 -f- 10 L75 as 103-1-75 = 101^5. 3. .-. 1000 -*- 56.234 = 17.782. The solution is correct. Check : Check: 3.1623 17.782 63246 252984 221361 221361 31623 66.232+ 17.78 56.234)1000.00000 562 34 437 660 393 638 44 0220 39 3638 4 65820 4 49872 Example 3. Find (5.6234) 2 x 316.23 -s- 177.82. Solution : 1. (5.6234) 2 x 316.23 -=- 177.82 2. = (10-75) 2 X 102.50^.102.25 3. = 101.50+2.50-2.25 — 101-75. 4. .-. (5.6234)2 x 316.23 ~ 177.82 = 56.234. This example also may be checked by ordinary computation. 153. From the examples of § 152 it is clear that a more complete list of exponents (logarithms) and ability to use them must be of great advantage, for in each case the solution by exponents is the simpler. The following paragraphs teach the methods of using logarithms. 154. Logarithms of numbers to the base 10 are called Common Logarithms, and form, collectively, the Common System of Logarithms. 172 ALGEBRA 155. If a number is not an exact power of 10, its logarithm can be given only approximately ; a four-place logarithm is one given correct to four decimal places. Thus the logarithm of 13 is 1.1139 ; i.e. 13 = 10M», approximately. The integral part of the logarithm is called the Characteristic and the decimal part, the Mantissa. The characteristic of log 13 is 1 and the mantissa is .1139. Note 1. The plural of mantissa is mantissx. Note 2. A negative number does not have a logarithm. 156. The Characteristic of the Logarithm of a Number Greater than 1. It is known that 3.53 = 10- 5478 , or log 3.53 = .5478. Multiplying both members of 3.53 = 10 5478 by 10, 35.3 = 10 5478 x 10 1 = 10 1 - 5478 , or log 35.3 = 1.5478. Similarly, 353 = 10 1 x 10 15478 = 10 2 - 5478 , or log 353 = 2.5478. The numbers 3.53, 35.3, and 353 have the same significant figures; they differ only in the location of the decimal point. Their logarithms differ only in the characteristics. These two facts indicate a connection between the location of the decimal point and the characteristic. 3.53 has one figure to the left of the decimal point; its logarithm has as characteristic 1 less than 1, or 0. 35.3 has two figures to the left of the decimal point ; its logarithm has as characteristic 1 less than 2, or 1. 353 has three figures to the left of the decimal point ; its logarithm has as characteristic 1 less than 3, or 2. Rule. — The characteristic of the common logarithm of a number greater than 1 is one less than the number of significant figures to the left of the decimal point. Thus, the characteristic of log 357.83 is 2 ; of log 70390.5 is 4. LOGARITHMS 173 EXERCISE 85 What are the characteristics of the logarithms of : 1. 365. 4. 7. 7. 6.35. 10. 300506.7. 2. 2000. 5. 16.1. 8. 60907.03. 11. 300.506. 3. 50698. 6. 123.05. 9. 500.005. 12. 1000000. Tell the number of significant figures preceding the decimal point when the characteristic of the logarithm is : 13. 4. 14. 2. 15. 0. 16. 1. 17. 3. 18. 5. 157. The Characteristic of the Logarithm of a Number less than 1. Dividing both members of 3.53 = 10 5478 (§ 156) by 10, .353 = 10 5478 -- 10 1 = 10- 5478 " 1 . .-. log .353 = .5478 - 1. Dividing both members of .353 = 10 ,5478_1 , by 10, .0353 = 10- 5478 " 1 -- 10 1 = 10- 5478 " 2 . .-. log .0353 = .5478 - 2. Similarly, .00353 = lO- 5478 " 3 . .-. log .00353 = .5478 - 3. Between the decimal point and the first significant figure of : .353 there are no zeros ; the characteristic of log .353 is — 1. .0353 there is one zero ; the characteristic of log .0353 is —2. .00353 there are two zeros ; the characteristic of log .00353 is -3. Rule. — The characteristic of the common logarithm of a number less than 1 is negative ; numerically it is one more than the number of zeros between the decimal point and the first significant figure. Thus, the characteristic of log .0045 is - 3 ; of log .00027, is - 4. EXERCISE 86 What are the characteristics of the logarithms of : 1. .05. 3. .00064. • 5. .00007. 7. .3. 2. .0032. 4. .0586. 6. .08375. 8. .33759. Tell the number of zeros preceding the first significant figure when the characteristic of the logarithm is : 9. -3. 10. -1. 11. —5. 12. —2. 13. -4. 174 ALGEBRA 158. Method of Writing a Negative Characteristic. In § 157 log .353 = .5478 - 1. Actually, therefore, log .353 is - .4522, a negative number. For many reasons, however, the positive mantissa, and the negative characteristic are retained. .5478 — 1 is written : 9.5478 — 10. Numerically the two ex- pressions have equal value. Note that 9 — 10 = — 1. The process in general is to decide upon the characteristic by the rule in § 157 ; then, if it is — 1, write it 9 — 10 ; if — 2, write it 8 -<- 10 ; etc. Thus, log .02 is 3010 - 2, or 8.3010 - 10. Note. The negative characteristic is often written thus: log .02=2.3010; again, log .353 = 1.5478. The minus sign is written over the characteristic to indicate that it alone is negative, the mantissa heiug positive. EXERCISE 87 1-12. Tell how each of the characteristics of the examples of Exercise 86 should be written. 159. Mantissa of a Logarithm. From §§ 156 and 157 : log 3.53 = .5478 ; log .353 = 9.5478 - 10 ; log 35.3 = 1.5478; log .00353 = 7.5478 - 10. The numbers 3.53, 35.3, .353, and .00353 have the same significant figures. Their common logarithms have the same mantissas. This is an example of the Rule. — The common logarithms of all numbers having the same significant figures have the same mantissae. Thus, the logarithms of 2506, 2.506, 250.6, etc., all have the same mantissae. 160. A Table of Logarithms consists of the mantissas of the logarithms of certain numbers. The characteristics of the logarithms may be determined by the rules given in §§ 156 and 157. The table given on pages 176 and 177 gives the mantissae of all integers from 100 to 999 inclusive, calculated LOGARITHMS 175 to four decimal places. The decimal point is omitted. Such a table is called a four-place table. While a five or six place table would be more accurate, this table is sufficiently ac- curate for all ordinary purposes. 161. To find the Logarithm of a Number of Three Significant Figures. Example 1. Find the logarithm of 16.8. Solution : 1. In the column headed "No." find 16. On the hori- zontal line opposite 16, pass over to the column headed by the figure 8. The mantissa .2253 found there, is the required mantissa. 2. The characteristic is 1, by the rule in § 156. 3. .-.log 16.8 is 1.2253. Rule. — To find the logarithm of a number of three figures : 1. Look in the column headed " No." for the first two figures of the given number. The mantissa will be found on the horizontal line opposite these two figures and in the column headed by the third figure of the given number. 2. Prefix the characteristic according to § § 156 and 157. Example 2. Find log .304. Solution : 1. Opposite 30 in the column headed by 4 is the mantissa .4829. The characteristic is - 1 or 9 - 10. (§§ 157 and 158.) 2. /.log .304 = 9.4829 -10. Note. The logarithm of a number of one or two significant figures may be found by using the column headed 0. Thus the mantissa of log 8.3 is the same as the mantissa of log 8.30 ; of log 9, the same as of log 900. EXERCISE 88 Find the logarithms of : 1. 235. 5. 72. 9. 56.2. 13. .00465. 2. 769. 6. 8. 10. 7.83. 14. 8690. 3. 843. 7. 3.2. 11. .924. 15. 24700. 4. 900. 8. 620. 12. .0326. 16. 60.7. 176 ALGEBRA No. 1 2 3 4 5 6 7 8 9 IO 0000 0043 0086 0128 0170 0212 0253 0294 0334 o374 ii 0414 0453 0492 0531 0569 0607 0645 0682 0719 o755 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1 106 13 "39 "73 1206 1239 1271 1303 1335 1367 1399 143° 14 1461 1492 1523 x 553 1584 1614 1644 1673 1703 1732 15 1 761 1790 1818 1847 1875 1903 1931 1959 1987 2014 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 20 3010 3032 3054 3075 3096 3118 3U9 3160 3181 3201 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 22 3424 3444 3464 3483 3502 3522 354i 3560 3579 3598 23 3617 3636 3655 3674 3692 37" 3729 3747 3766 3784 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 26 415° 4166 4183 4200 4216 4232 4249 4265 4281 4298 27 43*4 4330 4346 4362 4378 4393 4409 4425 4440 445 6 28 4472 4487 4502 4518 4533 4548 45 6 4 4579 4594 4609 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 32 5051 5065 5°79 5092 5105 5"9 5132 5M5 5*59 5 r 72 33 5185 5198 521 1 5224 5237 5250 5263 5270 5289 5302 34 5315 5328 5340 5353 5366 5378 5391 5403 54i6 5428 35 544i 5453 5465 5478 549o 5502 55H 5527 5539 555i 36 5 §o 3 5575 5587 5599 5611 5623 5 6 35 5 6 47 5658 5670 37 5682 5 6 94 5705 5717 5729 5740 5752 5763 5775 5786 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 39 59" 5922 5933 5944 5955 5966 5977 5988 5999 6010 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 4i 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 44 6435 6444 6454 6464 6474 6484 6493 6503 65U 6522 45 6532 6542 655i 6561 6571 6580 6590 6599 6609 6618 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 49 6902 691 1 6920 6928 6937 6946 6955 6964 6972 6981 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7^67 5i 7076 7084 7°93 7101 7110 7118 7126 7135 7*43 7152 52 7160 7168 7177 7185 7*93 7202 7210 7218 7226 7235 53 7243 7251 7259 7267 7275 7284 7292 7300 73o8 73i6 54 7324 7332 7340 7348 735 6 7364 7372 7380 7388 7396 No. 1 2 3 4 5 6 7 8 9 LOGARITHMS 177 No. 1 2 3 | 4 5 6 7 8 9 55 7404 7412 7419 7427 7435 7443 745i 7459 7466 7474 56 7482 7490 7497 7505 75*3 7520 7528 7536 7543 755i 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 7709 7716 77 2 3 773i 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 793i 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 8i95 8202 8209 8215 8222 8228 8235 8241 8248 8254 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 83S8 8395 8401 8407 8414 8420 8426 8432 8439 8445 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 71 8513 8519 8 o S P 8531 8537 8543 8549 8555 8561 8567 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 82 9138 9143 9149 9i54 9159 9165 9170 9175 9180 9186 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 85 9294 9299 9304 93°9 9315 9320 9325 9330 9335 9340 86 9345 935° 9355 9360 9365 937° 9375 9380 9385 939o 87 9395 9400 9405 9410 94i5 9420 9425 943° 9435 9440 88 9445 945° 9455 9460 9465 9469 9474 9479 9484 9489 89 9494 9499 95°4 95°9 95^ 9518 9523 9528 9533 9538 90 9542 9547 9552 9557 9562 9566 9571 9576 958i 9586 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 93 9685 9689 9694 9699 97°3 9708 97 J 3 9717 9722 9727 94 9731 9736 9741 9745 975° 9754 9759 9763 9768 9773 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 993o 9934 9939 9943 9948 9952 99 995 6 9961 9965 9969 9974 9978 9983 9987 9991 9996 No. 1 2 3 4 5 6 7 8 9 178 ALGEBRA 162. To find the Logarithm of a Number of More than Three Significant Figures. Example 1. Find log 327.5. Solution: 1. From the table : log 327 =2.51451 _,_ log 328 = 2.5159J * w * 2. Since 327.5 is between 327 and 328, its logarithm must be between their logarithms. An increase of one unit in the number (from 327 to 328) produces an increase of .0014 in the mantissa. It is assumed there- fore that an increase of .5 in the number (from 327 to 327.5) produces an increase of .5 of .0014, or of .0007, in the mantissa. 3. .-. log 327.5 = 2.5145 + .5 x .0014 . =2.5145 + .0007 =2.5152. This result is obtained in practice as follows. The difference between any mantissa and the next higher mantissa as written in the table (neglect- ing the decimal point) is called the tabular difference. The tabular dif- ference for this example is 14(5159-5145). .5 of the tabular difference is 7. Adding this to 5145 gives 5152, the required mantissa of log 327.5. Similarly to find log 327.25, the tabular difference is 14. .25 x 14=3.5. Hence the mantissa of log 327.25 is 5145 + 3.5 or 5148.5. .-. log 327.25 = 2.5149. Note 1. The process of determing a mantissa which is between two mantissae of the table is called Interpolation. Note 2. The assumption made in step 2 is not warranted by the facts. Nevertheless, for ordinary purposes, the results obtained in this manner are sufficiently correct. This is the common method of interpolating. Note 3. When interpolating, it is customary to cut down all decimals so that the mantissa will again be a four-place decimal. Thus 3.5 is called 4. 3.4 would be called 3. Rule. — To find the logarithm of a number of more than three significant figures : 1. Find the mantissa for the first three figures, and the tabular difference for that mantissa. 2. Multiply the tabular difference by the remaining figures of the given number, preceded by a decimal point. 3. Add the result of step 2 to the mantissa obtained in step 1. 4. Prefix the proper characteristic by §§ 156 and 157. LOGARITHMS 179 Example 2. Find log 34.652. Solution : 1. Mantissa of log 346 = 5391. Mantissa of log 347 = 5403. 2. Tabular difference = 12. .52 x 12 = 6.24 = 6. 3. .-. Mantissa for log 34652 = 5391 + 6 = 5397. 4. .-. log 34.652 = 1.5397. Example 3. Find log .021508. Solution : 1. Mantissa of log 215 = 3324. Mantissa of log 216 = 3345. 2. Tabular difference =± 21. .08 x 21 = 1.68 = 2. 3. . •. mantissa of log 21508 = 3324 + 2 = 3326. 4. .-. log .021508 = .3326 - 2, or 8.3326 - 10. EXERCISE 89 Find the tabular difference for the mantissae: 1. 3222. 3. 6590. 5. 8982. 7. 7076. 2. 4166. 4. 7364. 6. 5340. 8. 8692. Find the logarithms of : 11. 325.5. 12. 263.1. 13. 786.3. 14. 492.2. 15. 703.4. 16. 32.16. 17. 1.608. 18. 7.961. 19. .8462. 20. .05375. 21. 327.11. 22. 243.25. 23. 62.721. 24. 803.75. 25. 6.2534. 9. 4728. 10. 7435. 26. 3.1416. 27. 1.0453. 28. .22735. 29. .063457. 30. .004062. 163. To find the Number Corresponding to a Given Logarithm. Example 1. Find the number whose logarithm is 1.6571. Solution : 1. Find the mantissa 6571 in the table. 2. In the column headed "No." on the line with 6571 is 45. These are the first two figures of the number. At the head of the column con- taining 6571 is 4, the third figure of the number. Hence the number sought has the figures 454. 180 ALGEBRA 3. The characteristic being 1, the number must have two figures to the left of the decimal point. (§ 156.) .-. the number is 45.4. Rule. — To find the number corresponding to a given logarithm when the mantissa appears in the table : 1. Find the three figures corresponding to this mantissa, as in the example. 2. Place the decimal point according to the rules in § § 156 and 157. EXERCISE 90 Find the numbers whose logarithms are : 1. 2.6138. 4. 2.9542. 7. 1.7404. 10. 9.8000-10, 2. 1.3365. 5. 3.9289. 8. 4.7024. 11. 8.5378-10. 3. 3.6972. 6. 0.8162. 9. 0.8893. 12. 7.4133-10. Example 2. Find the number whose logarithm is 1.3934. Solution : 1. The mantissa 3934 does not appear in the table. The next less mantissa is 3927, and the next greater is 3945. The correspoHding numbers are 247 and 248. That is : mantissa of log 247 = 3927 1 Diff. ] Tabular mantissa of log x = 3934 J = 7. i difference mantissa of log 248 = 3945 j = 18. 2. Since an increase of 18 in the mantissa produces an increase of 1 in the number, it is assumed that an increase of 7 in the mantissa must produce an increase of ^g or .38 in the number. Hence the number has the figures 247.38. 3. Since the characteristic is 1, the number must be 24.738. Rule. — To find the number corresponding to a given logarithm when the mantissa does not appear in the table : 1. Find in the table the next less mantissa. Find the correspond- ing number of three figures, and the tabular difference. 2. Subtract the next less mantissa from the given mantissa and divide the remainder by the tabular difference. LOGARITHMS 181 3. Annex the quotient to the number of three figures obtained in step 1. 4. Place the decimal point according to the rules in §§ 156 and 157. EXERCISE 91 Find the numbers whose logarithms are : 1. 1.8079. 6. 0.8744. 11. 2.5369. 2. 3.3565. 7. 9.9108-10. 12. 9.7022-10. 3. 2.6639. 8. 8.8077-10. 13. 2.4644. 4. 0.7043. 9. 7.5862-10. 14. 3.1634. 5. 2.5524. 10. 8.2998-10. 15. 2.9310. PROPERTIES OF LOGARITHMS 164. The preceding discussion relates entirely to the Com- mon System of Logarithms. (§ 154.) Certain properties of logarithms to any base will be considered now. Note. The base may be any positive number different from 1. 165. Just as log 10 3.053 = .4847 means that 10- 4847 = 3.053, so \og a ]¥—x means that J¥=a x . Log a N is read " the logarithm of N to the base a." 166. Logarithm of a Product Assume that a x = and a y X== ^l;then (* = j°&.^ Also a* -a"=MN, or a x+v =MN. .\\og a MN=x+y. (§165) Therefore log a MN= log a M + log a N. Rule. — In any system, the logarithm of a product is equal to the sum of the logarithms of its factors. Example 1. Given log 2 = .3010, and log 3 = .4771, find log 72. 182 ALGEBRA Solution : 1. log 72 = log 2 • 2 . 2 . 3 . 3. 2. = log2 + log 2 + log 2 + log 3 + log3. 3. .-. Iog72 = 31og2 + 2 1og3 = 3(.3010) +2 (.4771) = .9030 + .9542 = 1.8572. EXERCISE 92 Given log 2 = .3010, log 3 = .4771, and log 7 = .8451. Find the following logarithms as in Example 1 ; check the solutions by finding the same logarithms in the table : 1. log 21. 4. log 126. 7. log 324. 2. log 42. 5. log 128. 8. log 378. 3. log 36. 6. log 252. 9. log 168. 10. Find by logarithms the value of 35.2 X 2.35 x 6.43. Solution : 1. Let v = 35.2 x 2.35 x 6.43. 2. . \ log v = log 35.2 + log 2.35 + log 6.43. 3. .-. log v- 2.7258. 4. .-. t> = 531.87. (§163.) log 35.2 log 2.35 log 6.43 1.5465 0.3711 0.8082 2.7258 Find by logarithms the values of : 11. 32.5x27.8. 14. 34.55x29.9. 17. 3.142x6039. 12. 2.49x65.7. 15. 678.1x37. 18. 541.2x1.523. 13. .289 x 365. 16. 1.732 x 580. 19. 43.65 x 865.25. 20. Find by logarithms the value of .0631 X 7.208 x .51272. Solution : 1. log a = log .0631 + log 7.208 + log .51272. log. 0631 = 8.8000-10 log 7. 208= 0.8578 log .51272 = 9.7099 3. /.log = 9.3677 -10. 19.3677 - 20 = 9.3677 - 10 ,2332. (§ 163.) Note. If the sum of the logarithms is a negative number, the result should be written so that the negative part of the characteristic is — 10. LOGARITHMS 183 Eind by logarithms the values of: 21. .0235x3.14. 24. 84.75 x .00368. 22. .5638 x .0245. 25. .0273 x .00569 x .684. 23. .7783x6.282. 26. .2908 X .0305 x .0062. 167. Logarithm of a Quotient. Assume that a x = M\ and o* = N ;then 1* = }°*- x = log a 3f, Also, a x -7- a y = M -f- JVJ or a x - y = M-^K .'. \o% a (M+N) = x — y. Therefore, log a (Jf -~ N) = \og a M— log a iV. Rule. — In any system, the logarithm of the quotient of two num- bers is equal to the logarithm of the dividend minus the logarithm of the divisor. Example 1. Given log 2 = .3010 and log 3 =.4771, find Solution : 1. log f = log 3 - log 2 = .4771 - .3010 = .1761. Example 2. Eind log -J. 2 • 2 • 2 Solution : 1. log f = log • 3 • 3 2. =(log2 +log2 + log 2) -(log 3 + log 3). 3. = 3 (.3010)- 2 (.4771) = .9030 - .9542. 4. .-. log f = 9.9488- 10. .9030 = 10.9030 - 10 .9542 = .9542 9.9488 - 10 Note 1. To find the logarithm of a fraction, add the logarithms of the factors of the numerator, and from the result subtract the sum of the loga- rithms of the factors of the denominator. Note 2. To subtract a greater logarithm from a less, or to subtract a negative logarithm from a positive, increase the characteristic of the minuend by 10, writing — 10 after the mantissa to compensate. Thus, in this example, .9542 is greater than .9030 ; therefore, .9030 is written 10.9030 — 10, after which the subtraction is performed. 184 ALGEBRA EXERCISE 93 Given log 2 =.3010, log 3 = .4771, and log 7 = .8451, find: 1. logf 3. logf 5. log |f 7. logf 2. log-V-. 4. log ty. 6. log-V-. 8. logji Find by logarithms the values of: 9. 255-48. 12. 630.5-402. 15. 2865-1.045. 10. 376-83. 13. 300.25-3.14. 16. 7.835-23.75. 11. 299-99. 14. 230.56-1.06. 17. 9.462-85.64. 3.14 x 25 2 .0036 x 2.35 365 .0084 23.5 x 1.05 2 287.5 x .096 3785 ' ' 3.1416 24.75 x .0058 25.6 x .738 x .0535 1.41 ' ' 265x432 16.08 x 256 1.405 x 207 x .00392 17 ' " 508 x. 6354 168. The Logarithm of a Power of a Number. Assume that a x = M; then x = log d M. Also, (a x ) p = M p , or a px = M p . .-. log M p = px. Therefore, log M p =p log a M. Rule. — In any system, the logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent indicating the power. Example 1. Given log 7 = .8451, find log 7 5 . Solution : log 7 5 = 5 log 7 = 5 x .8451 = 4.2255. Example 2. Find by logarithms 1.04 10 . Solution : 1. log 1.04i° == 10 log 1.04 = 10 x .0170 = .1700. 2. The number whose logarithm is .1700 is 1.479. (§ 163) 3. .-. 1.04 13 = 1.479. LOGARITHMS 185 Example 3. Find by logarithms V365. Solution : 1. log ^365 = log 365? = | log 365. 2. /. log ^365 = i x 2.5623 = 0.8541. 3. The number whose logarithm is 0.8541 is 7.146. (§ 163) 4. /. ^365=7.146. When finding a cube root, the logarithm of the radicand is divided by 3 ; when finding a square root, the logarithm of the radicand is divided by 2. This suggests the Rule. — In any system, the logarithm of a root of a number is the logarithm of the radicand divided by the index of the root. Example 4. Find by logarithms a^-0359. Solution : 1. log \Z.0359 = \ log .0359 = \ (8.5551 - 10). 2. /. log -t/AJSW = \ (38.5551 - 40). (See note.) 3. .-. log \/.0359= 9.6387 -10. 4. The number whose logarithm is 9.6387 - 10 is .4352. (§ 163) 5. /. v'^0359 = .4352. Note. To divide a negative logarithm, write it in such form that the negative part of the characteristic may be divided exactly by the divisor, and give — 10 as quotient. Thus 8.5551 — 10 is changed to 38.5551—40 since the divisor is 4. If the divisor were 3, it would be changed to 28.5551 — 30. EXERCISE 94 Given log 2 = .3010, log 3 = .4771, and log 7 = .8451 ; find : 1. log 3 7 . 3. log 7 4 . 5. log (21)i 7. log \/6. 2. log 2 5 . 4. log 27 3 . 6. log a/7. 8. log a^H. Find by logarithms the values of the following: 9. 235 2 . 13. 3.1416 x 18 2 . 17. V^ftp. 10. 2.045 3 . 14. 7.795 4 . 18. V25 x 19.6 x 17.3. 11. -y/KSE. 15. 12 2 . 19. V3 x a/5. 12. a/97863. 16. | x 3.1416 x 5 3 . 20. (UwrY- 186 ALGEBRA 21. The volume of a right circular cylinder is given by the formula V=irB 2 H. Find the volume (by logarithms); (a) if P = 10.5 and H= 26.5. (b) if i2 = 8.2 and # = 33.1. 22. The volume of a sphere is given by the formula V= f irR 3 . Find the volume : (a) ifP = 12; (b) HE = 6.2. 23. The interest on P dollars at r % for t years is given by the formula /= — . Find 7: 100 (a) if P= $ 765, r = 5, and t = 6.5 years. (6) if P = $ 1250, r = 4.5, and £ = 8 years and 3 months. 24. The amount to which P dollars will accumulate at r% compound interest in n years is given by the formula, (a) if P= $ 250, r = 4, and n = 10. (6) if P = $ 75, r = 3.5, and n = 15. 25. A cylindrical cistern has for its diameter 5 feet. Find the number of barrels of water this cistern has in it when the water is 9 feet deep. (One cubic foot of water is about 1\ gallons ; one barrel contains 31^- gallons.) Historical Note. Logarithms were introduced by John Napier (1550- 1617), a Scotch gentleman who studied mathematics and science as a pastime. The Napier logarithms were not the common logarithms. Briggs (1556-1631) , an English mathematician, computed the first table of Common Logarithms. XVI. PROGRESSIONS ARITHMETIC PROGRESSION 169. An Arithmetic Progression (A. P.) is a sequence of numbers, called terms, each of which after the first is derived from the preceding by adding to it a fixed number, called the Common Difference. Thus, 1, 3, 5, 7, ••• is an A. P. Each term is derived from the preced- ing by adding 2. The next two terms are 9 and 11. 2 is the common difference. Again, 9, 6, 3, ••• is an A. P. The common difference is — 3. The next two terms are and — 3. Note. The common difference may be found by subtracting any term from the one following it. EXERCISE 95 Determine which of the following are arithmetic progres- sions; determine the common difference and the next two terms of the arithmetic progressions : 1. 4, 7, 10, 13, .... 6. 5 m, 7.5 m, 10m, .... 2. 1, 3, 7, 9, 15, .... 7. 4p, 1.5 p, -jp, .... 3. 10, 7, 4, 1, .... 8. 1.06, 1.12, 1.18, .... 4. 25,20,15,10,.... 9. a + b,a + 2b, a + 36, .... 5. 21, 3£, 4, 4f, .... 10. 5r+6s,6H-4s,7r+2s,.... Write the first five terms of the A. P. in which : the first term is the common difference is 11 12 13 14 15 a d 15 6 25 -8 7.5 3.5 X -4 187 188 ALGEBRA 170. The nth Term of an Arithmetic Progression. It is possi- ble to determine a particular term of an arithmetic progression without rinding all of the preceding terms. Given the first term a, the difference d, and the number of the term n, of an arithmetic progression, find the nth term I. The progression is a, a-\-d, a-\-2d, a-\-3d, • • •. The coeffi- cient of d in each term is 1 less than the number of the term. Thus, the 10th term would be a + 9 d. Therefore the coeffi- cient of d in the nth term must be (n — 1). .-. l=a+(n-l)d. Example. Find the 10th term of 8, 5, 2, •••. Solution: 1. a = 8 ; d =- 3 ; n = 10 ; I = ? 2. l = a+(n-l)d. /. Z = 8+(10-l)(-3)=8-27=-19. EXERCISE 96 Find: 1. The 12th term of 3, 9, 15, ••• ; also the 20th. 2. The 15th term of 16, 12, 8, ... j also the 25th. 3. The 13th term of - 7, — 12, - 17, ••• ; also the 31st. 4. The 16th term of 2, 2}, 3, ••• ; also the 51st. 5. The 11th term of 1.05, 1.10, 1.15, ••• ; also the 26th. 6. What term of the progression 5, 8, 11, ••• is 86 ? Solution :1. a = 5 ; d = 3 ; I = 86 ; find n. 2. l = a +(n- l)d. /. 86 = 5 +(n- 1)3. 3. Solving for n, n = 28. .-. 86 is the 28th term. 7. What term of the progression 8, 5, 2, •••is —70? 8. What term of the progression \, -f, |, ••• is 20^? 9. What term of the progression —75, —67, —59, ••• is 197? 10. What term of the progression 1, 1.05, 1.10, ... is 2 ? PROGRESSIONS 189 11. If the first term of an A. P. is 15, and the 11th term is 35, what is the common difference ? Hint: 35 = 15 +(11 - l)d. Find the common difference : 12. If the first term is 5 and the 22d term is 173. 13. If the first term is — 20 and the 33d term is —4 14. If the first term is 325 and the 31st term is 25. 15. Find the 10th term of the arithmetic progression whose first term is 7 and whose 16th term is 97. 16. A man is paying for a lot on the installment plan. His payments the first three months are $10.00, $10.05, and $10.10. What will his 20th and 25th payments be ? 171. The terms of an arithmetic progression between any two other terms are called the Arithmetic Means of those two terms. Thus, the three aiithmetic means of 2 and 14 are 5, 8, 11, since 2, 5, 8, 11, 14 form an arithmetic progression. A single arithmetic mean of two numbers is particularly im- portant. It is called The Arithmetic Mean of the numbers. When two numbers are given, any specified number of arithmetic means may be inserted between them. Example. Insert five arithmetic means between 13 and -11. Solution : 1. There results an arithmetic progression of 7 terms, in which a = 13, I =— 11, and n = 7. Find d. 2. l = a+(n—l)d. .*. — 11 = 13 + 6 d, or d =- 4. 3. The progression is : 13, 9, 5, 1, - 3, — 7, — 11. Check : There is an A. P. with five terms between 13 and — 11. 190 ALGEBRA EXERCISE 97 1. Insert three arithmetic means between 3 and 19. 2. Insert four arithmetic means between — 10 and 20. 3. Insert nine arithmetic means between 3 and- 28. 4. Insert five arithmetic means between -J and 5. 5. Insert five arithmetic means between — j and —5. 6. Find the arithmetic mean of 7 and 15. 7. Find the arithmetic mean of V2 and Vl8. 8. Find the arithmetic mean of x -f 7 and x — 7. 9. Find the arithmetic mean of a and b. From the result, make a rule for finding the arithmetic mean of any two numbers. Note. The arithmetic mean of two numhers is commonly called their average. 10. Find the common difference if two arithmetic means are inserted between r and s. 11. Find the common difference if k arithmetic means are inserted between m and p. 172. The Sum of the First n Terms of an Arithmetic Pro- gression. Given the first term a, the nth term I, and the number of terms n; find the sum of the terms S. Solution: 1. S=a+(a + d) + (a+2d)+ ••• +(l-2d) + (l-d)+L (1) 2. Writing the terms in reverse order, S = l+(l-d) + (l-2d)+... + (a + 2d) + (« + <*)+ a. (2) 3. Adding the equations (1) and (2), term for term, 2fl = (a + Z) + (a + ?) + (a + 0+ ••• +(a + + («+0 + (« + 0- (3) 4. There were n terms in the right member of (1) ; from each, there results a sum (a + I) in (3). /. 2 S = n{a + I) , or S = £ (a + /) (4) PROGRESSIONS 191 6. In § 288, . = a +(n — l)d; substituting this value of I in (4), S = ^{a +(a +0 - 1), etc. Over what distance will it pass in the 10th second ? in the tth. second. 19. Through what total distance does it pass in 5 seconds ? in 10 seconds ? in t seconds ? PROGRESSIONS 193 20. Experiment has shown that an object will fall during successive seconds the following distances : 1st second, 16.08 ft. ; 3d second, 80.40 ft. ; 2d second, 48.32 ft. ; 4th second, 112.56 ft. Find the distance through which the object will fall during the 7th second ; the tth second. 21. Find the total distance through which the object falls in 5 seconds ; in t seconds. 22. Substitute g for 32.16 in the final result of Example 21 and simplify the result. 173. In an arithmetic progression, there are five elements, a, d, I, n, S. Two independent formulae connect these ele- ments, the formula for the sum and the formula for the term I. Hence if any three of the elements are known, the other two may be found. Note. Remember that the formula for the sum is given in two ways. Example 1. Given a — — f, w = 20, S =— f ; find d and I. Solution: 1. S = -(a+l). .'.--= 10 f --+ A ; whence Z_-. 2 V ' 3 V 3 ) 2 2. l = a + (n-l)d. .-. f =- § + (19) • d\ whence d = |. Example 2. Given a = 7, d = 4, S = 403 ; find n and I. Solution : 1. S = -{2 a +(»- l)d}. .'. 403 = -{14 + (n - 1) • 4}. 2 2 I. .-. 806 = n{4 n + 10} ; 4 n 2 + 10 n - 806 = ; 2 n 2 + 5 n - 400 = 0. . „_-5±V25 + 3224 5± v3249 _5.fc.57 62 nrt , 1Q • . /i — — — ' — — — — , or -j- i-o* 4 4 4 4' Since n is the number of terms, n must be 13. 3. I = a + (n - \)d. .-. I = 7 + 12 • 4 = 55. Note. A negative or a fractional value of n is inapplicable, and must be rejected together with all other values depending upon it. 194 ALGEBRA Example 3. The sixth term of an arithmetic progression is 10 and the 16th term is 40. Find the 10th term. Solution : 1. By the formula I = a 4- {n — l)d : a + 5 d - 10. a + 15^=40. 2. Solving the system of equations in step 1, d — 3 and a = — 6. 3. The 10th term : Z = - 5 + 9 • 3 =- 5 + 27 = 22. EXERCISE 99 1. Given d = 5, 1=71, n = 15 ; find a and S. 2. Given a = - 9, n = 23, I = 57; find d and & 3. Given a = J, Z = *£, 5 =»*$*} find d and ». 4. Given a = \, Z = — yV> d = — -jj ; find .w and #. 5. Given d = J, n = 17, S = 17 ; find a and Z. 6. Given a = f, n»15, #=*±$*j find d and I 7. Given a = -f, Z = -- 2 /-, # = -91; find d and n. 8. Given a = - 1 /, d = - § , 5 -* If* • find n and & 9. Given a, I, and n ; derive a formula for d. 10. Given a, d, and Z; derive a formula for n. 11. Given a, n, and £; derive a-formula for Z. 12. Given d, n, and # ; derive a formula for a. 13. Given d, I, and n ; derive formulae for a and #. 14. The 8th term of an arithmetic progression is 10, and the 14th term is - 14. Find the 23d term. 15. The 7th term of an arithmetic progression is — J, the 16th term is J, and the last term is - 1 /-. Find the number of terms. 16. The sum of the 2d and 6th terms of an arithmetic pro- gression is — j-, and the sum of the 5th and 9th terms is — 10. Find the first term. 17. Find four numbers in arithmetic progression such that the sum of the first two shall be 12, and the sum of the last two - 20. PROGRESSIONS 195 18. Find five numbers in arithmetic progression such that the sum of the second, third, and fifth shall be 10, and the product of the first and fourth — 36. 19. Find three numbers in arithmetic progression such, that the sum of their squares is 347, and one half the third number exceeds the sum of the first and second by 4i 20. Find three integers in arithmetic progression such that their sum shall be 12, and their product — 260. GEOMETRIC PROGRESSION 174. A Geometric Progression (G-. P.) is a sequence of num- bers, called terms, each of which, after the first, is derived by multiplying the preceding term by a fixed number called the Ratio. Thus, 2, 6, 18, 54, ••• is a geometric progression. Each term is ob- tained by multiplying the preceding term by 3. The ratio is 3. Again, 15, — 5, -f f, — §, ••• is a G. P. The ratio is — \. The next two terms are + ^ 7 and — / T . Note. The ratio may be found by dividing any term by the one preced- ing it. EXERCISE 100 Determine which of the following are geometric progres- sions ; determine the ratio and also the next two terms of the geometric progressions : 1. 4, 8, 16, 32, •-. 6. 3x, 6x 2 , 12 a?, — 2. 200, 50, 25, 10, .... 7. 2, -4, - 8, 16, -32, .... 3.81,27,9,.... 8 ' (1+^(1 + ^(1+^-. 9 1 1 1 4. -2, +6, -18, +54, .... y - m > — 2 > m s> * K 5m 5m tn 10 2 2 5. 5m,— ,—,.... 10. _,-,_,.... 196 ALGEBRA Write the first five terms of the G. P. in which : 11 12 13 14 15 The first term is : The ratio is : -5 -2 100 1 3 2 a r 175. The nth Term of a Geometric Progression. It is possible to determine a particular term of a geometric progression without finding all of the preceding terms. Given the first term a, the ratio r, and the number of terms n, of a geometric progression, determine the nth term I The progression is a, ar, ar 2 , ar 3 , •••. The exponent of r in each term is 1 less than the number of the term. Hence the 10th term would be ar 9 . Therefore the exponent of r in the nth term must be (n — 1). .-. l=ar n ~\ Example. What is the 7th term of 9, 3, 1, ••• ? Solution : 1. a = 9;r = £;n = 7;Z=? W 36 3* 81 EXERCISE 101 1. Find the 6th term of 1, 3, 9, .... 2. Find the 7th term of 6, 4, f, .... 3. Find the 5th term of - 2, 10, - 50, .... 4. Find the 9th term of 3, f, |, ••• 5. Find the 10th term of -•§, +5, - 10, .... 6. Find the 8th term of %-, ~, f , .... 32 lb 8 7. Indicate the 11th term of 1, (1 + r), (1 + r) 2 , PROGRESSION'S 197 8. Indicate the 15th term of 1 } i, h h "* a ^ so tne ^ fcn term. Ttl TYl 7YL 9. Indicate the 13th term of ra, — , — , — , •••; also the o 9 21 (n + l)th term. 10. What term of the progression 3, 6, 12, 24, ... is 384 ? 11. What term of the progression 5, 10, 20, ••• is 160? 12. What term of the progression 18, 6, 2, ••• is ^? 13. If the first term of a geometric progression is 5, and the 6th term is ^j wna ^ is the ratio ? Find the ratio of the geometric progression if: 14. The first term is -^ and the 5th term is -f. 15. The first term is f, and the 7th term 24. 176. The terms of a geometric progression between any two other terms are called the Geometric Means of those two terms. Thus, the three geometric means of 2 and 162 are 6, 18, and 54, since 2, 6, 18, 54, 162, form a geometric progression. A single geometric mean of two numbers is particularly important. It is called The Geometric Mean of the numbers. When two numbers are given, any specified number of geo- metric means may be inserted, between them. Example. Insert three geometric means between 9 and J^-. Solution : 1. There results a geometric progression of 5 terms, in which a = 9, I = - x & ^, and n = 5. Find r. 2 .l = ar „- K ... | = 9 . r 4,or^=15. ,., = ^? = ± |. 3. The progression is : 9, 6, 4, f , ^-, or 9, — 6, 4, — |, *$-. Check : There is a G. P. with three terms between 9 and - 1 /. 198 ALGEBRA EXERCISE 102 1. Insert 4 geometric means between 3 and 729. 2. Insert 5 geometric means between 2 and 128. 3. Insert 2 geometric means between J and 3. 4. Find the geometric mean of 8 and 32. 5." Find the geometric mean of 3 t and • A.Z t 6. Find the geometric mean between 2 x and 8 x 5 . 7. Find the geometric mean between — and — • x m 8. Find the geometric mean between a and b. 9. Insert 3 geometric means between 3 and 12. 10. Insert 2 geometric means between a and b. 177. The Sum of the First n Terms of a Geometric Progression. Given the first term a, the ratio r, and the number of terms n, of a geometric progression, find the sum of the terms S. Solution: 1. S = a + ar + ar 2 + ••• + ar n ~ 2 + ar n_1 . (1) 2. Multiplying both members of (1) by r, rS — ar + ar 2 + ar 3 + ••• + ar n_1 -f 4. Seven terms of the progression -Jg, — |, i, .... 5. Five terms of the progression —2, 10, —50, •••. 6. Fifteen terms of the progression 3 m, 3 m 3 , 3 m 5 , •••. 7. Ten terms of the progression 1, m 2 , m 4 , m 6 , .... 8. Find the sum of 15 terms of 1, (1 4- r), (1 + rf, .... 9. Find the sum of the first 10 powers of 2. 10. Find the sum of the first 10 powers of 3. 11. Each year a man saves half as much again as he saved the preceding year. If he saved $128 the first year, to what sum will his savings amount at the end of seven years ? 12. Find the sum of the terms from the 11th to the 15th inclusive in the progression y 1 ^, \, \, «... 13. A father agrees to give his son 5 ^ on his fifth birthday, 10^ on his sixth, and each year up to the 21st inclusive to double the gift of the preceding year. How much will he have given him altogether after his 21st birthday? 178. Infinite Geometric Progression. By an infinite geo- metric progression is meant one the number of whose terms increases indefinitely. If the ratio is greater than one, the terms become larger and larger. For example, the progression 3, 6, 12, 24, •••. If S n represents the sum of the first n terms of a progression, then, when r is greater than 1, S n increases in- definitely as n increases indefinitely. Thus, in the progression 3, 6, 12, •••, as n increases indefinitely, 200 ALGEBRA S n increases indefinitely. Hence the sum of an infinite num- ber of terms of the progression must be an indefinitely large number. When the numerical value of the ratio is less than 1, the progression has special interest. Example 1. Consider the progression 5, -f, f, .... Solution : 1. The ratio r is |. 2. When n is : l = at— 1 is: n 1-r is : 4 5Q) 3 = A 10 KW = iwhi 1-i 1-i 3. Clearly, as n increases, I decreases ; also the terra rl of S n decreases. If 11 increases indefinitely, I will become approximately zero, the term rl will become approximately zero, and S n will become approximately 5 _5 35 1-1 t 2* Consider now any geometric progression in which r is less than 1 in absolute value (§ 21). The sum of the first n terms 1S : o a — ar n 1 — r Now as n increases indefinitely, r n decreases indefinitely, becoming approximately zero. Hence the term a • r n becomes approximately zero, a, and 1 — r remain the same. .•. S n becomes approximately ~ or • Hence, the sum of an infinite number of terms of a geo- metric progression in which r is numerically less than 1, is given by the formula S = 1 — r PROGRESSIONS 201 Example. Find the sum to infinity of the progression 8 16 ... 4, -A Solution: 1. a = 4;r==— |. 2. Since r is numerically less than 1 , S = — - — 1 - r 1 + 1 3 EXERCISE 104 Find the sums to infinity of : 6. x, 1. 6,2,1,.... fi m x ? 2' 4' 9 1 1 1 ... " * a. * ■■'-•. 3. 16, 4, 1, .... ' ' 10' 100' '"• 4 5 I « ... 8 - 1 >~h+^,-- *• u > lTF> lT0"> * q_5_10 20 W * 3? ~9 > 2T > * 5. 1, .1, .01, .001, .... 10. t-^+A...... 11. Find the value of the repeating decimal .8181 .... Solution: 1. .8181 ... = fa + fa^ + etc. .... 2. This is a G. P. in which a = fa ; r = j^- The value of the decimal if an infinite number of decimal places is considered is given by the formula S=-^- (§296). l — r . 8 = ^ = 81 x 100 = 81 = 9 1 ~xk 10 ° " " n Find the values of the following repeating decimals : 12. .3333 .... 14. .5333 .... 16. .212121 .... 13. .7777-... 15. .6444-... 17. .151515-... XVII. THE BINOMIAL THEOREM 179. The Binomial Theorem is a formula for determining by inspection the expansion of any power of a binomial. By actual multiplication: (a + xf = a 2 + 2 ax + x 2 . (1) (a + xy = a 3 + 3a 2 x + 3ax 2 + x 3 . (2) (a + x) A = a 4 + 4 A + 6 a V + 4 aa 3 + x\ (3) Rule. — To expand any power of a binomial, like (a + jr) n : 1. The exponent of a in the first term is n and decreases by 1 in each succeeding term until it becomes 1. The last term does not contain a. 2. The first term does not contain *. The exponent of x in the second term is 1 and increases by 1 in each succeeding term until it is n in the last term. 3. The coefficient of the first term is 1 ; of the second is n. 4. If the coefficient of any term be multiplied by the exponent of a in that term, and the product be divided by the number of the term, the quotient is the coefficient of the next term. Note 1. The number of terms is n + 1. Note 2. The coefficients of terms "equidistant from the ends" are the same ; for example, the second and the next to the last. Example 1. Expand (a + x) 5 . Solution : 1. The exponents of a are 5, 4, 3, 2, 1. The exponents of £, starting with 1 in the second term, are 1, 2, 3, 4, and 5. Writing the terms without the coefficients gives : o 5 + a*x + a s $ + a 2 x 3 + a«* + x b . 2. The coefficient of the first term is 1, and of the second term is 5 (Rule 3). Multiplying 5, the coefficient of the second term, by 4, the 202 THE BINOMIAL THEOREM 203 exponent of a in the second term, and dividing by 2, the number of the term, gives 10, the coefficient of the third term ; and so on. Filling in the coefficients in this manner gives : (a + x) 5 = a 5 + 5 a 4 x + 10 a s x 2 + 10 a?x 8 + 5 as 4 + ^ij « j fl + o . a — o _ c + a . c — a b b d d a + b c + d 4. Simplifying step 3: b c-d Example. Since — = — , then, ™-±2 should equal 15 + 3 . Does it ? 2 3' '10-2 H 15-3 196. In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. If 2^*1, etc., prove a ± c ± e ± etc ' ,1 b d f ' * 6 + d+/+etc. b Proof. 1. Let v = the common value of the equal ratios - , -, -, etc. b d f 2. Then since - = v, a — bv. b - = v, c = dv. d j = v,e=fv. 3. Then (a + c + e) = bv + dv +fv = v(b + d+f), *"**■» b + d+f b + d+f b d f Example. Since 1 = - = — , 1+3 + 5 should equal -. Does it ? 2 6 10 2+6 + 10 2 Historical Note. All of these properties of a proportion were known to Euclid, 300 b.c 212 ALGEBRA 197. There are several other properties of a proportion which follow directly from properties of an equation or of a fraction. (a) If -=- then — = — • Raise both members to the third b cC 6 3 d 3 power. (K\ If - = - then va = ^° Extract the cube root of both b~ d" f/l jfa members. nc Multiply numerator and denomi- (c) If - = -, then — - = — • nator of the first ratio by m, and b d mb nd ma mc (d) If - = -, then b d nb nd equation by - lb of the second by n. Multiply both members of the m 198. In the preceding paragraphs, some of the simple prop- erties of a proportion have been given. There are many others which may be derived by means of these simple properties. -™ t* a c 2 a 4- Sb 2a — Sb Example. If - = -, prove -^ — -=— -. b d F 2c + 3d 2c -3d Proof. 1. Since 2 = -, then ?* =?-S. (§ 197, d) b d Sb 3d vs J 2. Then |«±M = |£±|ij. (By § m) 2a — 6b 2 c — 6d 3. Then 2a + 8 b = 2 a ~ 3 h > (By § 191) 2c+3d 2c-3d EXERCISE 110 1. Write by inversion : M 3 15 /7A 2 ra , v a a? (a) i= 2 -o- (6) rr (c) i=v 2. Write these same three proportions by alternation. 3. Write these same three proportions by composition. 4. Write these same three proportions by division. 5. Write the proportion (c) in Example 1 : (a) by inversion and the result by composition ; RATIO, PROPORTION, AND VARIATION 213 (6) by alternation and the result by division ; (c) by composition and the result by alternation ; (d) by division and the result by inversion. 6. If — = - , prove that — :L — = - • n y x+y y 7. If - = -, prove that —*—-= — • s b r a T /? a c ,-1 , a — b b 8. If - = - , prove that = - • b d c — d d rt T o x w , i . x 2 + u 2 w 2 -\-t 2 9. If - = — , prove that — ^- — = — 4—* u t' F u 2 t 2 tn T£ a c „„„ TTni .^ n4 .2a — 3b 2c — 3d 10. it - = — , prove that = -• b d } p b d EXERCISE 111 Proportion in Geometry 1. In a triangle in which BE is parallel to BC, m:r = n:s. To test this truth : (a) measure wi, n, r, and 8 ; (b) find the value of the ratio m : r and of n : s ; (c) compare these two ratios. This truth may be tested in any triangle. It may be expressed thus : the upper segment on one side is to the lower segment on that side as the upper segment on the other is to the lower segment on the other. 2. Write the proportion — = - by alternation. Express the resulting proportion in words as in Example 1. 3. Write the proportion of Example 1 by composition and express it in words. 4. Write the proportion of Example 1 by inversion and express it in words. 5. If AD = 7, DB = 4, and AE = 8, find EC. 214 ALGEBRA 6. If AB = 12, AD = 5, and AC= 14, find AE. Hint. Let AE = x, and CE = 14 - a. 7. If ^4Z> as ZXB, how does AE compare with EC? 8. If AD = 20, 1)5 = 8, and AC= 30, find JLE and EC. 9. If two perpendicular lines BO and Di? are drawn from one side of an angle to the other, then BC:AC=DE: AE. Test this statement by measuring the lines in the figure and finding the value of the ratios. 10. Draw any other perpendicular, as XY. Find the ratio of XY to ^l^and compare the ratio with those found in Example 9. What do you conclude about all ratios obtained by dividing the length of the perpendicular by the distance from A to the foot of the per- pendicular (like AY)? 11. Using the fact stated in Ex- ^i 5 -_- ample 9, tell how to find the height • „ > of the tree in the figure, if the height of the rod and the lengths on the shadows of the tree and the rod are as indicated. 12. Suppose that EF and AC are perpendicular to OC in the adjoining figure. Suppose that EF= 10 feet, OF =12 feet, 0(7=150 feet, and BO as 20 feet. Determine AB. 13. Suppose that CD and AB are perpendicular to AE in the adjoining figure; that AX = 5 feet, YB = 8 feet, AE= 750 feet, CE = 25 feet, and CD = 30 feet. Eind XY. RATIO, PROPORTION, AND VARIATION 215 VARIATION 199. Some quantities change or vary and are called Variable Quantities. Thus, the distance between a moving train and its destination varies, — that is, it decreases j the age of an individual varies from moment to moment, — that is, it increases. 200. A quantity which is fixed in any given problem is called a Constant. Thus, if a workman receives a fixed sum per day, the total wages due him changes from day to day if he works and remains unpaid. His daily wage is a constant ; his total wages is a variable. 201. A change or Variation in one quantity usually produces a variation in one or more other quantities. Such variables are called Related Variables. For each value of one variable there is a corresponding value of the other variable, or variables. Thus, if the side of a square is increased, the perimeter and the area of the square are also increased. 202. Variation is the study of some of the laws connecting related variables. Instead of the quantities themselves, their measures in terms of certain units of measure are used. Thus, distance is expressed as a number of miles, rods, or other units of length ; weight is expressed as a number of units of weight ; area is expressed as a number of units of area. 203. One quantity varies directly as another when the ratio of any value of the one to the corresponding value of the other is constant. Thus, the ratio of the perimeter of a square to the side of the square is always 4, because the perimeter is 4 times the length of the side ; there- fore the perimeter varies directly as the side of the square. 204. The symbol, oc, is read " varies as " ; thus, a oc b is read " a varies as b." 216 ALGEBRA x If xazy, then - = m, where m is a constant, expresses the relation between any two corresponding values of x and y. (See § 203.) x Since - = m, then x = my. Either equation may be used to express direct variation. 205. One quantity is said to vary inversely as another when the product of any value of the one and the corresponding value of the other is constant. Thus, the time and rate of a train going a distance d are connected by the equation rt = d. If the distance remains fixed, then the time varies inversely as the rate ; for example, if the rate is doubled, the time is halved. If x varies inversely as y, then xy = m, where mis a con- stant, expresses the relation between them. If xy = m, then also x = — . Either equation may be used to express inverse variation. 206. One quantity is said to vary jointly as two others when it varies directly as their product. If x varies jointly as y and z, then — = m, where m is a constant, expresses the relation between the variables. Thus, the wages of a workman varies jointly as the amount he receives per day and the number of days he works ; for, letting W equal his total wages, w his daily pay, and n the number of days he works, then W—nw. Here m = 1. Again, the formula for the area of a triangle is A = I ab. This shows that the area of a triangle varies jointly as the base and altitude. (Here m = J.) RATIO, PROPORTION, AND VARIATION 217 207. One quantity may vary directly as a second and inversely as a third. Let x vary directly as y and inversely as z ; then mil # = — z expresses the relation between the variables. Notice that this combines the equation for direct variation of y and inverse variation of z. 208. Variation of more complicated related variables needs to be expressed sometimes. Example 1. x oc y 2 may be written x = my 2 . Example 2. a?ccy 2 may be written x s = my 2 . Example 3. The volume of a circular cylinder varies jointly as the altitude and as the square of the radius. This may be expressed : v oc ar 2 , or v = har 2 Example 4. a varies directly as q, and inversely as d\ This may be expressed : a = -f • EXERCISE 112 Express the following relations both by means of the symbol oc and by an equation : 1. The area of a rectangle varies jointly as the base and altitude. 2. The area of a circle varies as the square of the diameter. 3. The volume of a rectangular prism varies jointly as the length, width, and height. 4. The distance a body falls from a position of rest varies as the square of the number of seconds in which it falls. 5. The interest varies jointly as the principal, the rate, and the time. 218 ALGEBRA Express the following relations by means of equations: 6. The rate of a train varies inversely as the time, if the distance is constant. 7. The rate of gain varies inversely as the capital invested, if the total gain is constant. 8. The weight of an object above the surface of the earth varies inversely as the square of the distance from the center of the earth. 9. The per capita cost of instruction for pupils in a school room varies directly as the salary of the teacher and inversely as the number of the pupils. 10. The volume of a circular cone varies jointly as the alti- tude and the square of the radius. 11. If z varies jointly as x and y, and equals § when y = -| and x = f , find z when y = £ and x — \. Solution. 1. According to the conditions z = mxy. 2. .-. - = m • - • * , or m = -, since z — \ when x = f and y = f. 6^5 3 3. .'.0 = | xy, substituting § for m. 4. .-. z = | • | • f = ^, when x = f and y = \. Note. In such problems, first find the constant, as in step 2. 12. If y oc x and is equal to 40 when B s \ \ I* O i A -a + a -ai« b' GRAPHICAL REPRESENTATION OF NUMBERS 221 In general, pure imaginaries are represented by points on the line YY'. ai is represented by the point a units above 0, and — ai by the point a units below 0. YY' is called the axis of pure imaginaries. 212. Graphical Repre- sentation of Complex Num- bers. To represent a + bi : Let A represent the number a, and B the number bi. Draw AC equal and parallel to OB. Then it is agreed to con- sider point C the repre- sentation of a-\-bi. Thus D represents — 4 — 3 i. ■5 -4 -3 -2 -I 0« -2i — 3i t -4i I Y I I +2 +3 +4 +5 7. 2.5 + ?'. 8. 4-2.5*. EXERCISE 113 Represent on a diagram the numbers : 1. 2 + *. 3. -4 + 2t. 5. -3 + 2*. 2. 2-3 i, 4. -3-2*. 6. 3-5L 9. Represent a + &a and — (a -j- fri). 213. Graphical Represen- tation of Addition of Complex Numbers. (a) To represent graphi- cally the sum of a -+- 6* and c + cfa*. Let M represent a -{-bi and AT, c + di. Construct OJlf and ON, and then construct the par- allelogram OMPN, thus locating point P. Y di — r bi t ' 1 c a X y' 222 ALGEBRA Point P represents (a 4- bi) 4 (c 4 di). Point P may be determined readily without constructing the complete parallelogram, by drawing from M the line MP equal and parallel to ON, thus adding (c 4- di) to (a 4 bi). Note. The correctness of the construction is readily proved by plane geometry. The proof is omitted from the text. EXERCISE 114 Represent graphically the sum of : 3 + i and 2 + 5 i. 2 - 3 i and 1 4 4 L 2 + 4 i and 5 — i. - 6 + 2 i and - 4 - 7 5. 3 and 4 Y bi f' o 1 j-a a Y' 6. — 5 t and 6 -f- 2 & 7. — 3 + 4 i and + 5. 8. 4 5 — 3 i and — 4 & 214. Graphical Represen- tation of Subtraction of Complex Numbers. In the adjoining diagram, if M represents the number a + bi, then M' represents — (a 4 bf), for M' repre- sents the number — a — bi. To subtract (a 4 bi) from (c 4 di), one may add — (a 4 bi) to (c 4- di). To represent this difference graphically : 1. Locate M representing a + bi and M ' representing — a — 6i. 2. Locate JV representing c 4 d£. 3. Draw from N a line equal and parallel to OJ/ 7 , thus locating R. This con- struction represents adding -(a+W) to (c+di). (§213.) 4. i£ represents c4-^0 -(a4&0- . GRAPHICAL REPRESENTATION OF NUMBERS 223 EXERCISE 115 1-8. Represent graphically the result of subtracting the second number from the first number in Examples 1-8 of Exercise 114. 215. It is possible to represent graphically other operations with complex numbers, but such topics are beyond the scope of this text. XX. EQUATIONS IN THE QUADRATIC FORM 216. An equation is in the quadratic form : 1. if it has three terms ; 2. if two of the terms contain the unknown number ; 3. if the exponent of the unknown number in one of these two terms is twice its exponent in the other. Note. The unknown number may be an algebraic expression. Example 1. Solve the equation 16 x* — 22 af * — 3 = 0. _3 _3 Solution : 1. Let y = x ¥ and therefore y 2 = x 2 . 2. Hence the equation becomes 16 y 2 — 22 y — 3 = 0. 3. .-. (8y + l)(2y-3)=0. 4. .VJf=— 1, orff = §; _3 _3 that is x ? = — £, or x ? = § . 5. From x~* = ->, x~* = V(^fj. If the principal cube root (§ 117) of — 4 is taken, x = = — = 16, There are also two imaginary roots, obtained by taking the fourth powers of the two imaginary cube roots of ( — |) . (§ 95.) 6. From x~* = f , af * = \/(|). x again has one real value and two imaginary values. Altogether there are 6 roots for the equation ; the principal roots are 16 and (v^) 4 . 224 EQUATIONS IN THE QUADRATIC FORM 225 EXERCISE 116 Solve the equations : 1. a 4 - 29 a 2 = - 100. 6. 2s- 8 -3os- 4 + 48 = 0. 2. 27s 6 + 46a 8 -16 = 0. 7. 67i-2 = llV/L 3. 16 a 8 -33 a 4 - 243 = 0. 8. a;* + 33a;$ = -32. 4. 161 a 5 + 5 + 32 a 10 = 0. 9. ^-10^ + 9 = 0. 5. 3 x-' 2 + 14 x- 1 = 5. 10. 2a + 3Vz = 27. 217. An equation may sometimes be solved with reference to an expression, by regarding the expression as the unknown number. Example 1. Solve the equation x 2 - 6 x + 5 V x 2 - 6 x + 20 = 46. Solution : 1. Let y = Vx 2 ^6~aT+ 20, and therefore y 2 = x 2 — 6 x + 20. 2. The equation becomes y 2 + 5 ?/ = 66, or */ 2 + 5 y - 66 = 0. 3. .-. (?/ + ll)(j/-6) = 0, or */ = -ll and y = 6. 4. When ?/ = 6, Vx 2 - 6 x + 20 = 6. .-. x 2 - 6 a; + 20 = 36, x 2 - 6 x - 16 = 0. .-. (x-8)(x + 2) = 0, or x = 8 and x=- 2. 5. When y =- 11, Vx 2 - 6x + 2~0 =- 11. . •. x 2 - 6 x + 20 = 121, or x 2 - 6 x - 101 = 0. ^ = 6±V36+404 = 6iV410 = 6±2ViT0 :=3 ^ 2 2 2 Note . If ?/, or Vx 2 — 6 x + 20, is restricted to the principal root, these last two values are not admissible. 226 ALGEBRA EXERCISE 117 Solve the following equations : 1. (2x 2 -3xy-$(2x i -3x)=9. (Hint : Let y - 2 x 2 - 3 as.) 2. 5 x + 12 + 5 V5 x + 12 = - 4. (Lety =V5z + 12.) ' 2a; + a^-3 4 * d 2 + 2 2 d - 5 35 5. 2d-5 d 2 + 6. ar J + 7Va^-4a; + ll=4a;-23. 7. V»i 2 -3m-3 = m 2 -3m-23. ?-5t + l t 2 -2t + 2 8 8. ?-2t + 2 t 2 -5t + l 9. 2r 2 + 4r + Vr 2 + 2r-3 = 9. 10. c 2 + l-fVc 2 -8c + 37 = 8(c + 12). 11. 25(a; + l)- 1 -15(a; + l)^ = -2. - #?->P 2/ 2 + 3 2 XXI. THE BINOMIAL THEOREM (Continued) 218. The Binomial Theorem was formulated in general form (for positive integral exponents), in § 179, after special cases of the general theorem were exhibited. The theorem was not proved; it was arrived at by the process of pure induction. 219. Proof of the Binomial Theorem for Positive Integral Powers. Assume, as in paragraph 179, that (a+x) n =a n +na n ~ l x ± n ( n - l \ »- 2 x* + n ( n - l )( n ~ 2 K n -*3<*+ ... (1) 1 • 2 1 • a • o Multiply both members of (1) by a+x. Then ( a + X )n+l^ a n+I + m n a . + ^(w-l) a n-l x 2 + ^-1Xn-2) aW _ 2a;8+ _ I - 1 • It • o + a n x + ja n ~W + wfo-l) a n-2& + .... 1 1 • '_ .-. {a + x) n + l = a n + l + {n+\)a n x+n\ 1 ?^+\\i n - 1 x 2 1.2 L 3 J = a n+l + ( n+ l) a n x + n , «±1 ^n-la-2 + n(ft-l) > rc±I a n- 2;K 3 + _ 2 1-2 o = a «+i+ („+!)„"*+ ( w + 1 )- w a"-^+ ( w + 1 )- n -( w - i y a >^+ .-. 1*2 1 • 2 • o It will be observed that the expansion on the right is in accordance with the rules of § 179. This proves that if the . rules of § 179 are assumed for any particular positive integer, n, they hold, true, also, for the next greater integer, n -f- 1. 227 228 ALGEBRA But the rules are known to be satisfactory in the case of (a + xy-, hence they hold for (a+#) 5 . Since they hold for (a + x) b , then they hold also for (a + x) 6 ; and so on. Therefore the binomial theorem is true for any positive integer. Note. The above method of proof is known as mathematical induction. 220. Fractional and Negative Exponents. In the expan- sion of § 219, if n is a positive integer, there is ultimately a term (the (n + 2)nd), for and after which the coefficient n(n — l)(n-2) • •• • ™ & - — is zero. 1.2-3... When n is a negative integer or a fraction, there is no term for which the coefficient is zero. Hence the terms continue indefinitely. The resulting expansion has an infinite number of terms. In this case also, the expansion on the right in § 219 has a sum, and this sum is (a + x) n for any rational (§ 112) value of 71, provided the absolute value of a is greater than the absolute value of x. This theorem is proved in a more ad- vanced course in mathematics. Assuming the theorem, the following examples may be solved : Example 1. Expand (a + x)~* to four terms. Solution : 1. Substitute § for n in the formula. vy 3 1-2 1.2-3 = I + 2 a ~W f(-i) a -f X 2 + KHK-f) . a -j X 3 + ... o 2i 1 • 52 • o = J -ff a~%x-\ a~^ 2 +/ T «~ 3 ^ 8 + •••• Example 2. Find the 7th term of (a — 3 &\)~t. Solution: 1. The 7th term may be found by applying the rule in § 180. THE BINOMIAL THEOREM 229 2. Substitute (— 3 x 2 ) for at, and (- |) for n. The exponent of (— 3x z ) is 7 - 1 or 6. The exponent of a is — ^ — or — Xf . The denominator of the coefficient is 1 • 2 • 3 • 4 • 5 • 6. The numerator of the coefficient is (— j).{— J — 1)... until there are six factors. Hence the seventh term is : (-^(-#)(-l)(-^)(-¥)(-V) (a -V )( _ 3 afty 1.2- 3-4- 5- 6 v JK J 3 8 9 EXERCISE 118 Find the first four terms of : 1. (a + x)K 2. (1+z)- 8 . 3. (l~x)~\ 4. V( b. Find the 9. 6th term of (a + a;)i 10. 5th term of (a — &)"&. 11. 7th term of (1 -f x)~ 7 . 12. 8th term" of (1 — xft. 6. (a* + 2 6)*. 7. (a 3 - 4 a;*)*. 8. _!_. 13. 9th term of (a-a)' 3 . 14. 11th term of V(m + w) 5 . 15. 7th term of (cr 2 -2 &*)- 16. 8th term of — . 221. Extraction of Roots. The Binomial Theorem may some- times be used to find the approximate root of a number which is not a perfect power of the same degree as the index of the root. Example. Find V'25 approximately to five decimal places. Solution : 1. The nearest perfect cube to 25 is 27. 230 ALGEBRA 2. .-. v / 25 = ^ / 27-2=[(33) + (-2)]^ = (3B)^+i(33)-|(_2)-|(33)-f(_2)2 +A (33)-t(_2)3... =3 3 • 3 2 9 • 3 5 81 • 3 8 = 3 - .07407 - .00183 - .00008 ••• = 2.92402. Rule. — Separate the given number into two parts, the first of which is the nearest perfect power of the same degree as the required root, and expand the result by the Binomial Theorem. EXERCISE 119 Find the approximate values of the following to five decimal places : 1. VTT. 2. V51. 3. c , f = n(»-l)(n-2)...(»-r + l). (3) T ! 230. Multiply both terms of the fraction (3) by the product of the natural numbers from 1 to n — r inclusive ; then p _ n(n — 1) • • • (n — r 4- 1) • (n — r) - • 2 • 1 _ n ! n r ~ r! xl .2..."(n"-r) ~r\(n-r)\' which is another form of the result. 231. The number of combinations of n different things taken r at a time equals the number of combinations taken n — r at a time. For, for every selection of r things out of n, we leave a selec- tion of n — r things. The theorem may also be proved by substituting n — r for r, in the result of § 230. Example 1. How many different combinations can be formed with 16 letters, taking 12 at a time ? Solution : By § 231, the number of combinations of 16 different things, taken 12 at a time, equals the number of combinations of 16 different things, taken 4 at a time. Putting n = 16, r = 4, in (3), § 229, p _ 16.15.14.13 ,q O0 16 ° 4 " 1.2.3-4 - 1820, Example 2. How many different words, each consisting of 4 consonants and 2 vowels, can be formed from 8 consonants and 4 vowels ? The number of combinations of the 8 consonants, taken 4 at a time, is 8 7 - 6 - 5 ,or70. 1-2-3-4' PERMUTATIONS AND COMBINATIONS 237 The number of combinations of the 4 vowels, taken 2 at a time, is i^,or6. 1-2' Any one of the 70 sets of consonants may be associated with any one of the 6 sets of vowels ; hence, there are in all 70 x 6, or 420 sets, each containing 4 consonants and 2 vowels. But each set of 6 letters may have 6 !, or 720 different permutations (§ 226). Therefore, the whole number of different words is 420 x 720, or 302400. EXERCISE 122 1. How many combinations can be formed from 15 things, taken 5 at a time ? 2. How many combinations can be formed from 17 things, taken 11 at a time ? 3. How many different committees, of 8 persons each, can be selected from 14 persons ? 4. There are 5 points in a plane, no 3 being in 'the same straight line. How many straight lines are determined by them? 5. How many different words, each having 5 consonants and 1 vowel, can be formed from 13 consonants and 4 vowels ? EXERCISE 123 Miscellaneous Examples 1. There are 11 points in a plane, no 3 in the same straight line. How many different quadrilaterals can be formed, hav- ing 4 of the points for vertices ? 2. From a pack of 52 cards, how many different hands of 6 cards each can be dealt ? 3. How many different numbers of 7 figures each can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, if the first, fourth, and last digits must be odd numbers ? 238 ALGEBRA 4. Out of 10 soldiers and 15 sailors, how many different parties can be formed, each consisting of 3 soldiers and 3 sailors ? 5. Out of 3 capitals, 6 consonants, and 4 vowels, how many different words of 6 letters each can be formed, each beginning with a capital, and having 3 consonants and 2 vowels ? 6. How many points of intersection are determined by 6 straight lines if no 3 of the lines pass through the same point, and if no 2 are parallel ? 7. How many different words of 8 letters each can be formed from 8 letters, if 4 of the letters cannot be separated ? 8. In how many ways can a committee of 2 teachers and 3 students be selected from 5 teachers and 10 students ? 9. In how many different ways may 10 students be seated in 15 seats ? (Leave result in factored form.) 10. How many games will* be played in a baseball league of 8 teams if each team plays 10 games with each of the other teams ? 11. How many signals can be made with 1 red, 1 white, and 1 blue flag, using them either 1 at a time, 2 at a time, or all together, if the order in which the flags are shown con- stitutes a part of the signal ? 12. In how many different ways can a captain of a baseball team arrange his batting list of 9 men if he wishes certain 3 men to bat in the order 1, 4, and 7 ? 13. How many different multiplication facts are involved in the multiplication table from lxl up to 9x9, if (for example) 5x4 and 4x5 are considered one fact ? XXIII. DETERMINANTS 232. The symbol 3 4 2 7 is called a determinant. Its value is defined to be 3 • 7 — 2 • 4, which, equals 21 — 8, or 13. a c In general b d and is defined thus : is called a Determinant of the Second Order a c b d = ad — be. The numbers a, b, c, and d are called the elements of the de- terminant. Clearly, any difference such as rs — mn may be arranged as r n a determinant : thus rs — mn m s Example 1. Example 2. = 2-3- 4(- 5) = 6 + 20 = 26. L5 = 2- 13-3. 5= 2 5 |. 3 13 EXERCISE 124 Find the values of : 2. 6 5 • 3. 4 -2 • 5. -5 3 • 7. 2m — p 4 2 6 9 2 6 2n r 5 3 2 -7 • 4. 3 -4 -2 7 • 6. 3a 4 2c 1 8. Sa Id 2 c 5e ' Express as determinants : 9. mn — xy. 11. 33 — 14. 10. 2ab cd. 12. 6c — 5 d. 239 13. cd +pq. 14. 3 mn + 2 rs. 240 ALGEBRA 233. Determinants make it possible to solve simultaneous linear equations by inspection. Solving the following pair of equations, ax + by = c clx+ ey=f ce — bf ae — bd H and y = af— cd ae — bd x = c b a c f e and y = d f a b a b d e d e Notice that the two solutions may be expressed as the quo- tients of determinants whose terms are the coefficients of the equations. Rule. — To solve two simultaneous linear equations having two unknowns by determinants : 1. Arrange the equations in the form : J ' ~ ' \dx -\-ey = f. its denominator is the deter- a b d e 2. The value of x is a fraction minant formed by the coefficients of x and / its numera- tor is the determinant obtained by replacing the coefficients of x in the denominator determinant by the corresponding absolute terms, c b f e 3. The value of / is a fraction with the same denominator as x ; its numerator is the determinant obtained by replacing the coeffi- cients of / in the denominator determinant by the absolute terms, a c d f Example. Solve the pair of equations : J ^ „ 16. DETERMINANTS 241 Solution : x = -16 -5 7 1 2 -5 3 7 _ -i6-7-5(-5) 2-7-3(-5) ■112 + 25 _ -87 14 + 15 29 -3. 2 -16 1 3 jjj = 10- 3(- 16) ^ 10+48 ^58^ 2 Z5] 2.7-3(-6) 14 + 15 29 ~ ' Check: In (1) : Does 2(- 3)- 5(2) =- 16 ? Does - 6 - 10 =-16? Yes. In (2): Does 3(- 3) +7(2) = 6? Does -9 + 14 = 5? Yes. EXERCISE 125 Solve the following equations by determinants : 1. 2. 4. f6a + 5?/ = 28. J4#+ y = 14. f 7 x — 9 ?/ = 15. |-5oj + 8y = -17. f5^-67/ = -9. (3a;- 5 ?/ = — 4. 8 m - 15 v = 18. 12 m+ 6v = -ll. 5. 6. 7. [5p + 2r = -4. Ui>- 11 r = -45 j4r- 3* = - 18. 5s=- 7. f3«- 4^/ = - 11. ^ + 5 5 2/ + 1 -0. | ma? - ???/ as mn. \ m'sc +- n'y = m V. 234. Determinants are especially useful in solving simul- taneous linear equations with more than two unknowns. <*1 « 2 «3 b x b 2 63 C\ c 2 c 3 is called a determinant of the third order. Its value is defined to be: &1& 2 C 3 + «2^3 C 1 + «3C 2 &1 — C!& 2 «3 ~ &1^2 C 3 — UfilPl- 242 ALGEBRA The adjoining diagram aids in recalling this value. Take the product aiboCs along the di- agonal and add to it the two products formed by starting with a 2 and a 3 respectively and fol- lowing the arrows which point in the direction of this diagonal ; then subtract the product ci&2«3 along the other diagonal, and also sub- tract the two other products formed by starting with b\ and a,i respectively and following the arrows which point in the direction of this second diagonal. Example. = 1. 7.6 + 5-3. 2 + 2(-3). 4 -2.7.2-4.5.6-I.3. (-3) = 42+30-24-28-120+9 = -91. Find the values of : EXERCISE 126 1 2 3 2 4 6 2 2 3 2 12 2. 3 - -2 3 '3. -2 -4 -11 3 3 1 1 5 4 5 -6 2 4. Solve the equations : 3 x + y — z — 14. x + 3 y — z = 16. x + y — 3 z = — 10. Solution : A rule similar to that of § 345 applies for linear equations with more than two unknowns. Hence : 14 1 - 1 10 3 -1 10 1 -3 3 1 -1 13-1 1 1 -3 _ 126 + 10 - 16 - 30 + 48 + 14 _ - 100 _ g _ 27 -1-1+3+3+3 -20 DETERMINANTS 243 3 14 -1 1 16 -1 1 -10 -3 - 120 3 1 - 1 -20 1 3 -1 1 1 -3 =6. 3 1 14 1 8 16 1 1 -10 3 1 - 1 1 3 -1 1 1 -3 140 -20 = 7. Check : The solution checks when substituted in the three equations. Note. The equations must be arranged first in the form ax+by+cz = d. Thus the equation 2 x — .3 z = 7 would be written 2x -\- y — S z = 1. Solve the following equations by determinants : 5. x + y — z = 24. 4 x -4- 3 y — x = 61. 6x— 5y— z ™ 11. 5a!— y + 4 z = — 5. 6. , 3x + 5y+ 6z = -20. x + 3y-$z = -27. Aa-5b-6c= 22. 7 - 1 a— b + c = — 6. 9 a + c= 12. 8. 9. 4 x -3*/ = 1. 4,y -3z = -15. 4z -3a? = 10. 9a; + 5z = -7. 3a? + -5.y = :1. 9y + 3z = 2. f 2# + 5?/ + 3z = -7. 10- 2?/-4z = 2-3a\ [5a; + 92/ = 5-f-7z. DETERMINANTS OF ANY ORDER 235. If the numbers 1, 2, 3, 4, 5, ••• n are arranged in any other order, each instance when a greater number precedes a less is called an Inversion. Thus, for the numbers 1, 2, 3, 4, 5, the arrangement 51432 has 7 inver- sions : 5 before 1, before 2, before 3, and before 4 ; 4 before 3, and before 2 ; and 3 before 2. (Ill Q>\2 0t I3 ••• 0£ ln a 2 i d'22 #23 '" a 2n 236. The symbol #nl <*«2 «n3 , having n rows of 244 ALGEBRA elements, each row consisting of n elements is a Determinant of the nth Order. Note 1. The first number of the subscript of an element denotes the row in which the element lies, and the second denotes the column. Thus, a 35 , read " a-three-five," is in the third row and fifth column. 237. Definition of the Value of a Determinant. 1. Form all possible products of the elements of the deter- minant, such that each product shall have as factors one and only one element from each row, and one and only one from each column. 2. Arrange the elements in each product so that the first subscripts are in the order 1, 2, 3, ••• n. 3. Make the product positive or negative according as the number of inversions in the second subscripts is even or odd. Thus, a n «12 «13 «21 «22 «23 «31 «32 «33 = <2ll#22#33 — «11«23«32 — «12«21«33 + «12«23«31 + ^13^21^32 — «13#22#31^ Note 1. This value agrees with that found as in § 234. Note 2. The elements lying in the diagonal joining the upper left hand element with the lower right hand element form the principal diag- onal. The product of these elements is always positive. 238. Consider any term of a determinant of the fourth order, as a u a 2l a zz a A2 . The number of inversions in the second subscripts is 4, an even number. Consider any other arrange- ment of these factors ; as, a zz a^a u a. 2v The total number of in- versions among the first and the second subscripts is 8, again an even number. In general, if the number of inversions for any arrangement of the elements in a term of an expanded determinant is even, the number of inversions remains even for any other arrangement of the elements in that term; similarly, if the number of inversions is odd, it remains odd. DETERMINANTS 245 (a) As a consequence of this fact, the expansion of a deter- minant may have the elements of each product arranged so that the second subscripts are in the order, 1, 2, 3, •••, n, giving each product the plus or minus sign according as the number of inversions in the first subscripts is even or odd. Thus, «11«22«33 — «11«23«32 — «12«521«33 + #12#23#31 + «13«21«32 — ai3#22«31 may be written «11«22«33 — «11«32«23 — «21«12«33 + «31«12«23 + «21«32«13 ~ «31«22«13- Examination will show that the signs are correct. (b) A second immediate consequence is that the elements of each product may be arranged in any manner, provided the sign of each term is determined by the total number of inver- sions among both the first subscripts and the second subscripts of the term. PROPERTIES OF A DETERMINANT 239. A determinant is not altered in value if its rows are changed to columns, and its columns to rows. Thus, it will be proved that an ai2 «13 a2i a 2 2 a 2 3 = a 3 i a32 «33 an a2i a3i ai2 «22 a 3 2 ai3 a 2 3 a33 Proof. The second subscripts of the first determinant are the same as the first subscripts of the second determinant. The number of rows and columns in each is the same. If the first determinant is expanded by the rule in § 237 and the second by the remark (a) in § 238, the results are the same. Thus, the determinants are respectively : aii«22a33 — ana23a32 + ai2«23a3i — ai2a2i«33 + ai3a2ia32 — ai3«22a3i 246 ALGEBRA and «11«22«33 — «11«32«23 _ «21«12«33 + «21«32«13 + #31#12#23 — «31«22«13- These are equal except for the arrangement of the elements in the terms. 240. A determinant is changed in sign if any two consecu- tive rows, or any two consecutive columns, are interchanged. Thus, it will be proved that an «12 «13 an «12 ai3 «21 #22 «23 = - a 3 i «32 ^33 «31 «32 «33 «21 a22 a 2 3 Proof. Consider any term of the first determinant; as, a 12 a 2 ia33. The sign of this term in the first determinant is minus, as there is one inversion. This same term occurs in the expansion of the second de- terminant, as the term has one and only one element from each row and each column, and the rows and columns of the second determinant are the same, except for order, as those of the first determinant. In fact, this term is the product of the ele- ments printed in black type. Oi2 a 2i a 33) considered a term of the second determinant ex- panded according to the remark (p) of § 238, has the same inversions among its second subscripts ( 213 ) as when it is considered a term of the first determinant. Its first sub- scripts ( 123 ) show one inversion, namely 2 before 8, as the proper order of the rows in the second determinant is 132. Hence there is one more inversion among the subscripts in this case, making two, and hence the sign is plus. In a similar manner, it may be proved that every term of each of the determinants is also a term of the other determi- nant if the sign of the term is changed. This proves the the- orem stated. 241. A determinant is changed in sign if any two rows, or any two columns, are interchanged. DETERMINANTS 247 Thus, it will be proved that an «12 an «21 a22 «23 = — «31 «32 «33 «31 «32 «33 «21 «22 «23 an «12 «13 Proof. To change the first determinant into the second, interchange the first row and second row of the first determi- #2i #22 nant, getting the determinant «, , and then inter- change the second row of this new determinant with the third row, getting 21 #22 #2 31 #32 #3 11 Now interchange the first and #11 #12 #13 second rows of the last determinant, and the desired determi- nant #31 «32 #21 #22 #33 #1 a, is obtained. All together three interchanges of consecutive rows have been made. Each causes a change in the sign of the determinant. The resulting determinant is the negative of the given deter- minant. A similar proof may be given for a determinant of any order. 242. Cyclical Interchange of Rows or Columns. It will be proved that #22 = (-1) (n-1) #21 #22 "• «2» #31 #32 '•• #3„ «»1 «n2 •• a nn #11 #12 ... # ln Proof. The first row has been made to occupy the position of the nth row. This may be accomplished by interchanging 248 ALGEBRA the first row with the second, then with the third, and so on up to the nth inclusive. This makes all together (n — 1) inter- changes of adjacent rows, and causes (n — 1) changes in sign of the determinant. Hence the fact stated above is true. 243. If two rows, or two columns, of a determinant are identical, the value of the determinant is zero. Proof. Let D be the value of the original determinant hav- ing two rows identical. If these two rows are interchanged, the value of the result- ing determinant is — D (§ 241). But the two determinants are actually identical, since the rows that were interchanged are identical. Hence D = - D, or 2 D = 0. Therefore D = 0. 244. If each element of one row, or of one column, is a bi- nomial, the determinant can be expressed as the sum of two determinants. Thus, it will be proved that 1>1 + Ci «12 «13 62 + C2 «22 «23 63 + C3 «32 «33 h «12 «13 1>2 ^22 «23 + h. «32 033 <*4 «12 «13 di «22 «23 c s «23 «33 Proof. Each term of the first determinant is the product of a binomial from column one and a monomial from each of the other columns. Consider (b l + c l )a 22 a 33 . This equals b l a 22 a zz -\-c l a 22 a 3S . The result is obviously the sum of the two corresponding terms of the other two determinants. In a similar manner, it may be proved that each term of the first determinant is the sum of the two corresponding terms of the other two determinants. Hence, the first determinant is the sum of the other two. Note. If any column, or any row, consists of the sum of n terms, then the determinant may be expressed as the sum of n determinants. DETERMINANTS 249 245. If all of the elements in one row, or in one column, are multiplied by the same number, the determinant is multi- plied by that number. Thus, it will be proved that a n r Op #13 a 2 \r #22 «23 = r azir #32 #33 a n #12 #13 #21 #22 «23 «31 «32 a33 Proof. Since each term of the expanded determinant on the left contains one and only one factor from the first column, each term must have one and only one factor r. Hence r is a common factor of the terms of the determinant. This proves the theorem. 246. If all of the elements of one column, or of one row, of a determinant be multiplied by the same number, and either added to or subtracted from the corresponding elements of another column, or row, the value of the determinant is not changed. Thus, it will be proved that #ii+&#u #12 ai3 On #12 #13 «21 + ##23 #22 #23 = #21 #22 #23 #31 + &a33 #32 #33 #81 #32 #33 Proof. Oa+kOu #12 #13 #11 #12 «M k,a n a 12 a 13 a 2l + ka 23 #22 #23 = #21 #22 "'J:! + 23 #2^ 2* #31 4" ^#33 #32 #33 #31 #32 ":a ^#33 #32 #33 #11 #12 «13 #13 #12 #13 = #21 #22 hz + k #23 #22 #23 #31 # 32 '-.a #33 #32 #33 #12 #13 #21 #22 + k • 0. (§ 244) (§ 245) (§ 243) This clearly proves the theorem. 250 ALGEBRA Thus, in 247. Minors. If the elements of the row and of the column of a determinant, in which any particular element lies, are omitted from the determinant, the resulting determinant is called the Complementary Minor of that particular element. an ', where D r is the new determinant obtained in step 1. (§ 240) 3. By (j — 1) interchanges of adjacent columns, the jth column can be made the first. 4. Then D = (— l)*'- 14 ' -1 !)", where D" is the new determinant obtained in step 3. (§ 240) That is an ai2 • • Ctln «21 a22 • • «2n «nl «n2 • . a nn = (- iy+s-* <*H an aii • . * . •• a in a\j an «i2 •• . * . • a u a»j 0*1 (*22 • * • «2„ * * * . . * . .. * a ni a n \ a n i • . * a nn Note. The *'s indicate the places formerly occupied by the ith row and jth column. a iX «12 ... * . "• a in a 21 (^22 ... * •• «2n # * ... * . . * «nl «n2 ... * . ' *m 5. .-. the coefficient of ci t .= (-l) <+ >'- 2 = (- lY*4r Note. (-l)*+> = (-l)»'+>- 2 . 250. A determinant of the fourth order may be written : a,, a,, a. U "-12 "13 "14 GE21 ^22 ^23 ^24 a 31 of 32 c-f a DETERMINANTS 255 7. 1 1 1 1 6 11 25 7 8 7 1 3 1 6 6 6 13. a a b b b b b a a b a b 8. 3 1 5 2 4 10 14 6 8 9 1 4 B 15 21 9 14. 1 1 1 1 a 2 6 2 1 a 2 c 2 1 b 2 c 2 9. 10. 6 15 11 10 5 16 12 9 7 14 10 11 8 18 9 12 2 3 5 9 3 7 8 11 6 10 4 2 8 4 5 10 15. 16. 7 10 13 3 14 19 27 6 24 33 41 10 31 47 64 15 5 - ■3 -2 4 1 -6 -1 4 3 6 -4 11. a 6 c 6 -a c — a c 6 17. Ill y n X X y n m X 71 y m lit n y X I X X X X 12. X X y X X y X X X X X y 18. 1 1 1 b c d Ir c 2 d 2 V' c 3 d* XXIV. SUPPLEMENTARY TOPICS CUBE ROOT 252. Cube Root of a Polynomial. By the binomial formula (§ 179), (a+b) 3 =a 3 +3 a 2 b + 3 ab 2 +b 3 . Any polynomial which may be put in this form is a perfect cube. Its cube root may be found by inspection. Example. Find v8 r 3 + 36 r 2 +• 54 r + 27. Solution: 1. 8r 3 +36r 2 + , 54r+27 = (2r) 3 +3(2r) 2 • 3 + 3(2 r) • 3 2 +3 3 . 2. .-. \/8 r 8 + 36 r 2 + 54 r + 27 = 2 r + 3. Notice that " a " is 2 r and " b " is 3. Note. If b is negative, the form is a 3 — 3 d 2 b + 3 a& 2 — 6 3 . EXERCISE 128 Find by inspection the cube roots of : 1. 8ar 3 + 12x 2 + 6x+-l. 4. 8t 6 -60t 4 -125 + 150?. 2. l-12a + 48a*-64a3. & y 125000 = 50. Place 50 in the root. Subtract. 3. Trial divisor : 3 a 2 =3(50) 2 = 7500 b = 324 + 75 = 4+. Place 4 in the root. 4. Complete divisor : 3 ab = 3 • 50 . 4 = 600 6 2 = 42 = _16 5. Multiply by 4. 3 a 2 + 3 a b + 6 2 = 8116 157 464 125 000 32 464 32 464 Rule. — To find the cube root of an arithmetical number : 1. Separate the number into periods (§ 62) of three figures each. 2. Find the greatest cube number in the left hand period ; write its cube root as the first figure of the root ; subtract the cube of the first root figure from the left hand period, and to the result annex the next period. 3. Form the trial divisor by taking three times the square of the part of the root already found and annexing two zeros. 4. Divide the remainder (step 2) by the trial divisor and annex the integral part of the quotient to the root already found. 5. Form the complete divisor by adding to the trial divisor three times the product of the new root figure by the part of the root 260 ALGEBRA already found, with one zero annexed, and also the square of the new root figure. 6. Multiply the complete divisor by the new root figure and subtract the product from the remainder. 7. Continue in this manner until the cube root or the desired number of decimal places for the root has been obtained. Note 1. Note 1, p. 68, applies with equal force to the above rule. Note 2. If any root figure is zero, annex two zeros to the trial divisor and annex the next period to the remainder. Example 2. Find the cube root of 8144.865728. The solution may be arranged as follows : 20.12 8 144.865 728 8 120000 600 1 120601 121203 120 144 865 120 601 90 ;o 4 24 264 728 121323 54 24 264 728 Since 1200 is not contained in 144, the second root figure is zero ; we then annex two zeros to the trial divisor 1200, and annex to the remainder the next period. EXERCISE 130 Find the cube roots of the following numbers : 1. 19683. t 4. 2515456. 7. 187149.248. 2. 148877. 5. 857.375. 8. 444.194947. 3. 59.319. 6. 46.268279. 9. 788889.024. SUPPLEMENTARY TOPICS 261 DETACHED COEFFICIENTS 256. Detached Coefficients. Solutions of examples in " long " multiplication and division may be abbreviated as in the fol- lowing examples. Example 1. Multiply 3a? + 2x-4:by 3 x-2. Solution: (a) Solution : (6) 3x 3 + 0-x 2 + 2a; — 4 3x 3 + 0-x 2 + 2x- 4 Sx -2 Sx -2 9 z* + • x s + 6 x 2 - 12 x 9 +0 +6-12 -6x 3 -0-x' 2 - 4x + 8 _6 -0-4 +8 9 x 4 - 6 x 3 + 6 x 2 - 16 z + 8 9 -6 +6 -16 +8 .\ Kesult = 9 z 4 - 6 a 3 + 6 s 2 - 16 a; + 8. Note that in solution (6) only the coefficients are written in the partial and total products ; that the multiplier and multiplicand are arranged in the same order of powers of x\ that is supplied for the missing powers. Solution (6) is by " detached coefficients." Example 2. Divide 12 a 3 - 25 a - 3 by 2 a - 3. Solution : (a) Solution : (6) 6 a? + 9 a +1 -3 2a- 6+9+1 2 a - 3 1 12 a 3 + • a. 2 - 25 a - -3|12a 3 + 0-25a-3 12 a 3 - 18 a 2 12 - 18 18 a 2 - 25 a 18-25 18 a 2 - 27 a 18-27 2a- -3 2 -3 2a- -3 2 -3 .*. Result = 6 a 2 + 9 a + 1. Solution (b) is by " detached coefficients." EXERCISE 131 Solve by detached coefficients : 1-5. Examples 21-25 on page 12. 6-10. Examples 16-20 on page 13. Note. The same device may be used to abbreviate addition and subtrac- tion exercises. 262 ALGEBRA PROOFS OF THE RULES FOR THE DIVISIBILITY OF a» ± 6» 257. In § 91, the rules for the divisibility of a n ± b n were determined by inspection. These rules may be proved by means of the factor theorem. Proof of I, 1. If b be substituted for a in a n — b n , the result is b n — b n , or 0. Then, by § 94, a n — b n has a — b as a factor. Proof of I, 2. If — b be substituted for a in a* — b n , the result is (— b) n — b n . When n is even, (— b) n — b n = b n — b n = 0. Then, by § 94, a n — & n has a — ( — b) or a + & as a factor, w;^ew Proof of I, 3. If b be substituted for a in a n ■+- b n , the result is 6 n + 6 n , or 2 6". This result is not zero unless tr is zero. Then, by § 94, a n + b n never has a — b as a factor. Proof of I, 4. If — b be substituted for a in o n -f 6 n , the result is (- b) n + & n . When n is odd, (— 6) n ^6 M = — b n + 6 n =0. Then, by § 9.4, a n + 6 n has a — ( — b) or a + 6 as a factor, wAew n is odd. 258. The Highest Common Factor of Polynomials which can- not be Readily Factored. The rule in arithmetic for finding the H. C. F. of two numbers is : 1. Divide the greater number by the less. 2. If there is a remainder, divide the divisor by it. Continue thus to make the remainder the divisor and the preceding divisor the dividend, until there is no remainder. 3. The last divisor is the H. C. F. required. SUPPLEMENTARY TOPICS 263 Example. Find the H. C. F. of 169 and 546. 169)546(3 507 39)169(4 156 the H. C. F. of 169 and 546 is 13. 13)39(3 39 A similar process serves for polynomials. Let A and B be two polynomials, the degree (§ 18) of A being equal to or greater than that of B. Suppose that B is contained in A p times, with a remainder (7; that C is contained in B q times, with a remainder D ; and that D is contained in G exactly r times. B) A(p pB C) B(q qC D) C(r rD Then D is a common factor of A and B. Proof. Since dividend = divisor x quotient + remainder : A=pB+C. (1) B=qC+D. (2) C = rD. Substitute the value of C in (2) ; then, B = qrD +D = D(qr + 1). (3) Substitute the values of B and C in (1) ; then, A = pD(qr + 1) + rD = D(pqr +p + r). (4) From (3) and (4), D is a common factor of A and B. Further, every common factor of A and B is a factor of D. Proof. Let F be any common factor of A and B ; and let A = mF and B = nF. Then : from (1) C = A - pB = mF - pnF, (5) from (2) D = B-qO. (6) 264 ALGEBRA Substituting in (6) the values of B and C, D = nF - q(mF- pnF) = F(n — qm + qpn). (7) Hence F is a factor of D. Then, since every common factor of A and B is a factor of D, and since D itself is a common factor of A and B, it follows that 2) is the highest common factor of A and B. In applying the process to polynomials the following notes should be observed. Note 1. Each division should be continued until the remainder is of a lower degree than that of the divisor. Note 2. If the terms of one expression have a common factor which is not a common factor of the terms of the other expression, the factor may be removed, for it evidently cannot form part of the common factor of the two expressions. In like manner, any remainder may be divided by a factor which is not a factor of the preceding divisor. Note 3. If the given expressions have a common factor which may be seen by inspection, remove it and find the H. C. F. of the resulting expres- sions. The result multiplied by the common factor that has been removed is the H. C. F. of the given expressions. Note 4. If the first term of the dividend, or of any remainder, is not divisible by the first term of the divisor, it may be made so by multiplying the dividend by any number which is not a factor of the divisor. Example 1. Find the H. C. F. of 6a?-25x 2 + Ux and 6 ax 2 + 11 ax- 10 a. Solution : 1. Kemove X from the first expression and a from the second. (See Note 2.) Then continue as below. 6 s 2 - 25 x + 14 1 6 s 2 + 1 1 x - 10 1_1 6 s 2 - 25s + 14 Divide by 12. (Note 2.) 12 |36s-24 3x- 2[ 6s 2 -25s + 14 12s-7 6 s 2 - 4x -21s+ 14 .-. 3s-2istheH.C.F. - 21 x + 14 SUPPLEMENTARY TOPICS 265 Example 2. Find the H. C. F. of 2 m 3 - 3 m 2 - 8 m - 3, and 3 m 4 — 7 m 3 — 5 ra 2 — m — 6. Solution : Since 3 m 4 does not contain 2 m 3 , multiply the second ex- pression by 2. (See Note 4.) 8 m 4 — 7 m 8 — 5 m 2 - m — 6 2 2m 8 -3m 2 -8m-3| 6m 4 -14m 3 — 10m 2 -2m- 12 [3m 6m 4 - 9m 3 -24m 2 -9m - 5m 3 + 14m 2 + 7m -12 --2 10 m 8 - 28 m 2 - 14 m + 24 |£ 10m 8 -15m 2 -40m -15 - 13 | - 13m 2 + 26m + 39 m 2 - 2 m - 3 w 2_2m-31 2m 3 -3m 2 -8m-3 [2ro-l 2 m 3 - 4 m 2 - 6 m m 2 -2m-8 m? — 2 m — 3 .-. m 2 - 2m - 3 is the H.C. F. Notice that —5 m 8 of the first remainder does not contain 2 m 8 , and that the remainder is therefore multiplied by — 2. Notice also that the divisor — 13 is removed from the second remainder, thus making the first term of the new divisor positive. EXERCISE 132 Find the H.C.F. of: 1. ar>+5a>-24 and ar* + 4 a 2 - 26 a> -f 15. 2. 3x 2 -±x-4:im&3x i -7x B + 6x l -9x + 2. 3. 2 m 4 + 5 m 3 — 2 m 2 -f 3 ra and 6 ra 3 w — 7 m 2 n 4- 5 mn — 2 ^. 4. x*y—6xy — 27y and x^y — 2x?y — 8xy + 21 #. 5 . 4 £y - 15 xy 2 + 9 ?/ 3 an d 8 x A - 18 arfy + 25 afy 2 - 1 2 a^/ 8 . 6. 3 n 8 + 8 w 2 - 9 n + 2 and 6rc 4 + 23 w 3 + 2 w 2 - 13 n + 2. 266 ALGEBRA 7. 6a 6 +5a 5 -6a 4 -3a 3 4-2a 2 and9a 4 + 18a 3 +5a 2 -8a-4. 8. 3& 4 -13& 3 + 3& 2 + 4&and9& 3 + 12& 2 -86-5. 9. 12 a 3 -5 a 2 x~ll ax*+6x 3 and 15 a 3 + 11 a 2 x- 8 aa 2 - 4 x 3 . 10. 2ar } -3ar + 2x-8 and 3^-7^ + 4^-4. 259. The L. C. M. of Two Polynomials which cannot be readily- Factored. Let A and B be two polynomials ; let F be their H. C. F. and M their L. C. M. Let A = aF and B = bF. Since F is the highest common factor of aF and bF, a and 6 cannot have any common factors. Hence, the L.C.M. of aF and bF is abF. That is, M=abF=a(bF) = aB; or j*f = abF= b (aF) = bA. Rule. — To find the L. C. M. of two polynomials : Divide one of the polynomials by their H. C.F. and multiply the quotient by the other polynomial. EXERCISE 133 Find the L.C.M. of: 1. 3a 2 -13a + 4and3a 2 + 14a-5. 2. 6a 2 + 25«Z> + 246 2 and 12a 2 + 16a& -36 2 . 3. 12 m 2 — 21 m — 45 and 4 m 3 — 11 m 2 — 6 m -f- 9. 4. 2a 3 -5a 2 -18a-9and3a 3 -14a 2 -a + 6. 5. 6a 3 -7a 2 + 5x-2and4» 4 -5a 2 + 4a-3. INDETERMINATE FORMS — becomes - for .. x-3 x-3 260. The fraction - becomes -• for # = 3; - comes -• Neither has any meaning, for division by zero is not allowed (§ 1, a). Results like these, however, must be interpreted at times. The following paragraphs show how to give the interpretation. 261. A constant is a number which always has the same value in a particular mathematical discussion. SUPPLEMENTARY TOPICS 267 A variable is a number which assumes different values in a particular mathematical discussion. * Thus n may assume the values .1, .01, .001, •••, etc. A limit of a variable is a constant the difference between which and the variable may be made to become and remain less than any assigned positive number, however small. Thus, the variable n above is evidently approaching the value ; or, the limit of n is zero. The symbol = is read " approaches the limit." Thus, n == means " n approaches the limit zero." 262. If a number becomes and remains greater than any positive number which may be assigned, it is said to become infinitely large or to approach infinity as limit. The symbol oo is called " infinity." Thus, if n represents any positive integer (assuming therefore the values 1, 2, 3, •••, etc.), it approaches infinity as limit ; i.e. limit of n = oo, or n = oo. Note, co is not a symbol for some definite value. It is a symbol for the limit of a number which " becomes and remains larger than any assigned positive number." Evidently as n = oo, also n 2 = oo. ^^ t n 2 = cc is read " the limit of n 2 as n, approaches oo is infinity." 263. Interpretation of -. To determine the meaning of -, replace - by - and consider limit - as x = 0. K) x x If x becomes .1, .01, .001, ..., etc., 1 becomes 10, 100, 1000, •••, etc. x Evidently, then, - increases indefinitely. That is, hmit I = ». Then, x * =M) x to the otherwise meaningless form - , give the value oo. 270 ALGEBRA EXERCISE 134 Find the values of the following as x = 0*. 1. t 2- £• 3 J_- 4. 7;»« . 5. * a; a 2 /1\ a; (a + 5) #(# + 1) Tind the values of the following as # = oo : 6. x 2 . 7. 2*. 8. ?• 9. 2 + -. 10. i« a; a; 2 X Find the values of the following mit/ - 2 v^ 2 11. limit/' ^~ 4 V 14 1imit/ ^-^-6 Y 12. limit (fit&\. 15. limit/1 1., , g V • »-°° \ 2/ / '^ V n n(n — 5)J 13. limit fld^Y 16 li^t A 2 4-2tt-7 \ 17. The equations ?/ = 2 sc -f 3 and y = 2 x -+- 5 have no com- mon solution according to § 50. Consider y = 2 a; -\- 3 and ?/ = aaj-}-5. Solve them as simultaneous equations, and hnd the values of x and y as a = 2. 18. Solve 2^ + 3?y = 6 and £x + by = 7 as simultaneous equations, and find the values of x and y as 6 = 6. § 267. Graphical Solution of Equations. In § 95, the state- ment was made that an equation of the nth degree, having one unknown, has n roots, but that these roots are not readily found for equations of degree above the second. Such equations may be solved graphically as in § 75. SUPPLEMENTARY TOPICS 271 Example. Solve the equation x? — 4 # 2 — 2 x + 8 = 0. Solution : 1. Let y = x 3 — 4 x 2 — 2 x 4- 8. When x = 1 2 34 5-1 -2 -8 then y = 83 -4 -70 23 5 -12 -49 X7~ *^ s ^ \ y X v -IE V 7- 5 ^ -/ \ -, r /-- I. L 3 a-C ^ _ C7 £ A/~ ^JB -fcf- r zl -? 7 - ^ ? "fr "!x — — t \ t V 4 1 V £ / 75 S^ / it t % ^ s ^ta5* X dfc f. -iy. J / T 7 t is.r 2. The curve crosses the horizontal axis at points A, B, and C. Hence its roots are, approximately, — 1.42, + 1.42, and + 4. Check : This equation may be solved as in § 95. Using the factor theorem : X s - 4x 2 - 2x + 8 = (x - 4) (x 2 - 2) = 0. .-. x = 4; also x 2 = 2, or x = ± V2 = ± 1.414. Clearly the results ± 1.42 obtained graphically are close to the roots ± 1.414. EXERCISE 135 Solve the following equations graphically : 1. a; 3 -3a; 2 -#4-3 = 0. 4. a; 4 - 10 x* -f 9 = 0. 2. or 5 -4 a 2 -7 a + 10 = 0. 5. x? + x 2 - 10 £- 10 = 0. 3. ot' + jb 2 — 6a = 0. 6. a? — ^-8 a + 8 = 0. 272 ALGEBRA 268. Horner's Synthetic Division. Synthetic division in the case when the divisor is a binomial is considered in § 93. A similar process of division may be employed in other cases of division of a polynomial by a polynomial. Consider : 3 x 2 — 2 x + 4 2x 2 +x 6x* 6x* - X 3 - + 3x 3 - -3x 2 -9x 2 + 10 x- 12 - 4 x 3 + 6 x 2 - 4 x 3 - 2 x 2 + 10 x + 6x 8x 2 8x 2 + + 4x- 4x- 12 12 6x 4 - x 3 - S x 2 + 10 x - 12 - 3 x 3 + 9 x 2 - 6x + 2x 2 - 4x + 12 When performing the subtractions, the signs of the terms subtracted are changed and the results are added to the minu- ends. If + x and — 3 of the divisor are changed in advance to — x and 4- 3, the various partial products may be added to the respective remainders. Following this suggestion and pro- ceeding in a manner entirely similar to that in § 93, the solu- tion may be arranged as follows : 2x 2 — x + 3 3x 2 -2x + 4 || 0+0 Steps of the process. 1. Change the signs of all terms of the divisor excepting the first term. 2. Divide 6X 4 by 2 x 2 , getting 3x 2 , the first term of the quotient. Multiply — x + 3 by 3 x 2 , getting — 3 x 3 + 9 x 2 , which are written in their proper columns. 3. Add the second column, and divide the result by 2 x 2 , getting — 2 x, the second term of the quotient. Place this second term of the quotient at the foot of the second column below the line. 4. Multiply — x + 3 by — 2 x, getting 2 x 2 — 6 x, which are written in their proper columns. 5. Add the third column and divide the result by 2 x 2 , getting 4, the third term of the quotient. Place this third term at the foot of the third column below the line. 12 x* - 11 x 2 y + Oxy* -9y3 + 8 - 16 - 2 + 4 SUPPLEMENTARY TOPICS 273 6. Multiply — x + 3 by 4, getting — 4 £ + 12, which are written in their proper columns. Add the remaining columns and thus find the re- mainder. In this example, the remainder is zero. The quotient is 3 x 2 — 2 x -f 4. This is observed to agree with that found previously. Example 2. Divide 12 x* - 11 x 2 y - 9 f by 3 x 2 - 2 xy+4 y 2 . Solution : 3x* + 2xy 4 x - 1 y || - 18 - 5 Quotient : Ax— y. Remainder : - \&xy' z — 5j/ 3 . EXERCISE 136 Divide the following by synthetic division : 1. 12z 3 -7.T 2 -23a-3by 4.t 2 -5z-3. 2. 4 a 4 - 9 a 2 + 30 a - 25 by 3 a 2 + 2 a - 5. 3. 2 a 4 - a 3 + 8 a - 25 by 2 a 2 - 3 a + 5. 4. 4 m 2 + 1 4- 16 m 4 by 2 m + 4 m 2 + 1. 5. 6 a 5 - 13 x A - 20 or 5 + 55 a 2 - 14 x - 19 by 2 a 2 - 7 a + 6. 6. 8a 5 -4a 4 2/-8xy-18a2/ 4 + 2l2/ 5 by ±x? -2x 2 y + 6xy 2 -If. , 7. 37a 2 + 50 + a 5 -70aby 2a 2 + 5 + a 3 -6a. MISCELLANEOUS EXAMPLES A. The Four Fundamental Operations. 1. Add 3(a - 6) 2 - 9, 4(a - b) 2 - 5 (a - b), and -7(a-6) 2 + 8(a -6). 2. Simplify 6 ran + 5 — ([ — 7 m* — 3] — \ — 5 mn — 11 j ). 3. Simplify 7x-(5x-[- 12a; + 6 as - 11]). 4. Subtract (2 a - 3 % 2 from (5 a - 4 % 2 . 5. Subtract (jp -+- (?)# from mx. 274 ALGEBRA 6. Multiply a 3p b q , b 4 c m , and c n a 2p . 7. Multiply 7 x m y 4n — 8 x h y p by — 3 x*y n . 8. Multiply x 2p +*y — x 7 y q by a; 2 "- 1 + y q ~ x . 9. Multiply 4 a w+5 6 2 - 3 a 4 b n by a w+2 6 - 2 aft"" 1 . 10. Multiply a m + b n ~ c p by a w — b n + c p . 11. Simplify (a + 2 fc) 2 - 2 (a + 2 6)(2 a + 6) + (2o + b)\ 12. Simplify (a; + y + z) 3 - 3(# -J- z)(z + a-)(a; + y)- 13. Divide a m+1 6 n+3 by — ab 2 . 14. Divide x p+q y n+2 — a? + y by a 5 ?/ 3 - 15. Divide a 7p b q c ir — a hp b Zq c 2r — a 3p 6 5 «c 3r by — a Zp b q c 2r . 16. Divide a; 3w+2 + 8 a 3 ™- 1 by x 2m+l -2x 2m + ± x 2m ~K 17. Divide x im + a? 2m ?/ 4n -f y Sn by # 2m - sc m y 2B + 2/ 4n - 18. Divide x 3 + (a — 6 — c)x 2 + (— ab + bc — ac)x + abc bj K x 2 -f (a — b)x — ab. 19. Divide a(a — b)x 2 + (— ab + 6 2 + 6c)a: - c(6 + c) by (a - &)# -f c 20. Divide x? - (3 a + 2 & - 4 0)^+ (6 a& - 8 6c + 12 ca)a - 24 abc by x — 2 b. Set B. Fractions. Simplify the following : s« + 2g'-8a;-16 4 (2 x 2 + 5 a?-2) 2 -25 " ^^a^ + Saj-lG* .' (3 a; 2 -4 a;- 3) 2 - 16* "2^ + 37 ¥i~27. 5. ^"J^tl^i" 1 ' • 2 ' 4^-92/ 2+ 9a: 2 -42/ 2 " + " + Q a; 2 — 5x — 84 # a? -f- 7 fl y — z z — x o. 27 a; 3 -8 3 a; -2 tf-ty-z)* (x-z) 2 - : SUPPLEMENTARY TOPICS 275 3 3 5n 2 5n 2 ' 2» + l 2»--l 8n 3 + l 8?i 3 - a 1 3 _ a 2 + 2 a a + 3 a-3~a 2 -9~~ a 2 + 9 ' _ 3 a 3 a 6 a 2 12 a 4 y - — 7~r H r + o . ro + "3 a + 6 a — ft a 2 + 6 2 a 4 + 6 4 10 3 1 a- 2 2 a 3 + 4 2(a-l) 2(a + l) a 2 + l a 4 -l 11.— 1 t^+ 1 2 a 2 + 3 a - 2 3 a 2 + o a - 2 1 + a - 6 a 2 _2 1 1 2__ ?/ 2 # 2 2a + ?/ » + 2y __ a^-2 a; 2 -4a;4-8 . ^c 2 + 9 a? + 14 „ s 2 -4 3? + ^ lo. — i 7z — ;; tcz Tcz — ~ I — i ~ tc~ X x 4 + 3ar } -27a;-81 V aj2 + 6i « + 9 ^ 2 ■' 6 a 2 - a - 2 ^ 8 a 2 - 18 a - 5 v , 4 a 2 - ■1-4. t t~t ~ /n t~t ~ ~ ~ X 4a + 4\ -9 ; 15. 4 a 2 - 16 a + 15 12a 2 -5a-2 4 a 2 + 8 a + 3 fx±l\ 2 _ f x-l \ 2 Set C. Linear Equations. (One Unknown.) 1. Solve the equation + — — - = 2. Divide each numerator by its corresponding denominator ; then 1 + -^— + 1 - ^^ = 2 , or _?_ - »±± = . 2x-3 z 2 + 4 2x-3 & + 4 Complete the solution. 274 ALGEBRA 6. Multiply a 3p 6«, b 4 c m , and c n « 2p . 7. Multiply 7 x m y** — 8 ^p by — 3 afy B . 8. Multiply a^+fy — a?y by a^" 1 + if~ x . 9. Multiply 4 a w+5 6 2 — 3 a 4 6* by cr +2 6 - 2 aM" 1 . 10. Multiply a m + b n — c p by a w — b n + c p . 11. Simplify (a + 2 6) 2 - 2 (a + 2 6)(2 a + 6) + (2a + 6) 2 . 12. Simplify (a; + # + zf - S(y + z)(z + x)(x + 2/). 13. Divide a m+1 b n+ * by — db 2 . 14. Divide x*+ q if +2 — x*+hf by a*?/. 15. Divide d Ip b*c 4r — a^b^c 27 — a*?b 5 *x-2) 2 -25 ' x*-2x* + Hx~16' ' (Xx 2 ~-4rX-'S) 2 -16' ~7» — f~ o — "o 71 — a? — 2a^-f-2a; — 1 2 a; 3 y 3 a? 2 y . 5- -j — ~— : Z ' 4a?- x*-9 J 6a»-g-2 vy 8a 2 -18a-5 w 4a 2 -9 4a 2 -16a + 15 12a 2 -5a-2 4a 2 + 8a + 3 f x+i \* As-i y >Se£ C Linear Equations. (One Unknown.) 1. Solve the equation - — ^— + ~ , =2. 2»-3 a^H-4 Divide each numerator by its corresponding denominator ; then l + -^_ + l-^±i=2,or-l_-^±l = 0. 2x-S z 2 + 4 2x-3 a? + 4 Complete the solution. 276 ALGEBRA Solve the following equations : 2. 3. 5. 6. 11. 12. x + t x+2 _ x + 5 x + 6 x+2 x+3 x + 6 a- + 7 8 3 10 5 x+3 x-7 x + 9 x-2 x + ?> x + 4 x+2 x+3 , x + 2 Q 3 + 2 = X + 9 X + 4: = * + 4 . a + 3 a; + 18 2« + 3 2x- 3 36 _ „ a jo -f u oart ^* •*' ~r - 1 «? -i 2aj-3 2cc + 3 4ar>-9 2a; + 5 3ar> + 24a; + 19 a; + 7 x 2 + 8 a; + 7 ar* a^ -2a? -2a> + 5 a£ + 3»-7_ -3 ar> + 3a; + l~ 5 1 10 g> ~ ^-ri, . «, t^ L = 2 . 9. 2a>-l 6z+5 3a;-4 4x + l q + 6 , a — 26_(2a— 6)a? + 3a6 10. r_L^ + x x + a xr — a bx a 2 + b 2 a 2 a a 2 b 2 x(a — b) b a b a -b a x 13 a(x-a) + b(x-b) = a + b x — b x — a 14. (a + b)(x-a + b)-(a-b)x + a 2 - b 2 =2a(x + a-b). SUPPLEMENTARY TOPICS 15. (x + p + q) (x — p + q)+ q 2 =(x — p) (x + q). 117 3 277 16. 17. x — 2 a 6 a? -+- a 3 a; — 8 a 2x — Sa 4 1 _4 1 « — 4 n x + n a; + 4w « + 3w ar* — 2 aa; + ct 2 ar * -f ax — 2 a 2 _ o a~> _ 2 a x - 3 a 2 ic 2 + az + 2 a 2 ff-f-a , a; + 5 , x — a — & _q 19. ( - -| - - - — o. x— a x — o x-{-a-\-b 20. ar 3 + (a; - a) 3 + (x - bf = 3x(x-a)(x- 6). #e£ Z>. Linear Equations. (Two or More Unknowns.) Solve the following pairs of equations : x + y 4. X - 1 y + 4 = ^ 2. 3. a? + H y-6 = d 7 5 a; — 2 10 11 a; — y 10 3x + 8 6a;-l 9 a^+5_ o ?/ 2 ^~-~ 3 ^ 2a;— 3y 4a;+6y __ 1 4 3 ' ~ 2' 5a + 2y 7y-3a; = 39 2 5 10* 5. 2/_4 2 2/ + ^ 3 5 (a; - 4 y) 4(a? -ff) =16 6 9 8a?-3 y-5 ^y 4 3 6 a; — y- < |(2a;-l)( 2 /-4)-(a;-o)(2^4-5)=121. '1 4a*-3y=-29. 278 ALGKBRA 8 fK/>-<7)-Hl>-3<7) = 9-3. 1 f(f»-S?)+f(P + ?)=18. 5 a; — 4 y _ _ 3 5x + 42/ _ 13* 5 * — s + .y-j. a?-4,y +6 1 8aj-2y-18 4* 10. Solve the following for x and 7/ : f a&(a — &)« + ab(a + b)y = a 2 + 2 a& — 6 2 . 12. 13. 14. 15. 16. ' ax + by = 2. ra(a + y) + &(« - 2/ ) = 2. 7tt 2 (# + y) — n 2 (x — y)=m — n. (a + b)x+(a - b)y = 2(a 2 + b 2 ). b a x — a — b y - a + b (a + b)x + (a-b)y = 2a 2 -2b\ y x 4 ab a — b a + b a 2 — b 2 bx + ay = 2. + b)x - ab(a - b)y = a 2 + & 2 . {a6(a y — a + b _y — a x + a+b x + b | ay — bx = a 2 + b 2 . a-x b ' {(a + b)x + (a-b)y=2a 2 -2b 2 . y-b a x-\-(a- b 2 )y = 2a 2 + 2b 2 . 18 ( (a + &>* + («-%= 2 a. l(a 2 -& 2 )a;+(a 2 SUPPLEMENTARY TOPICS 279 Set E. TJieory of Exponents. Square the following by the rule of § 10, b : l. 3 a* + 4 6""*. 2. 5m-¥-8m 2 r 4 . 3. Square a?b~* - 2 at — a~ l b%. 4. Expand (4:X*y~% + 7 z~ 2 )(4:X*y~* -7 z~ 2 ) by the rule of § 10, a. Find the value of : 5. 25n ^- 49wl , by the rule of § 15, 6. 5 a- 3 - 7 m* 6. 8g2 + 27 ^ , by the rule of § 15, /. 2 x* + 3 jf* a"-*-, 8 _ al-H. (See §91: 1,2.) #* — x 5 a 3 -t- b 6 9. (3^-4^) 3 . 10. (cr 2 6 3 + 2a 3 6- 2 ) 3 . Find the square roots of the following : 11. 16a-«mK 12. 49. 20. 27 »* + 54 a^ + 36 a?V"* + 8 2T 2 - 21. a?* - 6 a;* + 2.1 x~* - 44 af * + 63 aH - 54 af* + 27 af i 280 ALGEBRA Simplify the following, expressing all the results with posi- tive exponents : 22. [^(^2/- 2 )-^(^V)] T1 . 23. -_^ X -X^_ V6 4 Vc Va 8 24. [a"" 1 x (a -1 n+1 ] X [(a~ n ) _1 X (a"- 2 )- 1 ]. 25. (a? « Xas^)«. si. fL±A_ _«!+&!• a* + &1 a — b m+n\ 2m / n 2m\ m—n 26. r=— 1 ?-) . i i _i ._i / a m+n\2m/ a 2m\ n+1 n-1 n-1 32. Z-2IS- + a* — $ a 2 — 6 2- 27. (a n_1 -j-a n+1 ) 2n . ** &» 2„ _ 1 X 2n + 1 1 1 mn 1 33. p-q p 28. (a;" , -s-a;" , ) ,,l+H -*-a! n - *R + 1 x 2 »-l 34 <»*+&* x °~* + 6 ~* 11 29. [^(•M)]» • ^ a -l_^ X a i_ 6 i + ai -i- y \ x + y a*+2 6* 7 a*6* + 6 6* a-i — yi a? - ?/ a*-2 6* a*+aM— 6b* Set F. Quadratic Equations. Solve the following equations: 1. (a; + !)(*;+ 3) =12+(a> + 7) V2. 2. V5 x 2 - 3 a; - 41 = 3 a; - 7. 3. a/8 r 3 - 35 r 2 + 55 r - 57 = 2 r - 3. 4 4. 3VaT-^i - — = — = 4. Vx-1 5 28 (3 m + 10) 25 _ Q 8 m 3 -27 m(2ro-3) 6. 2V3^ + 4 + 3V37+7= 8 V3* + 4 7. SUPPLEMENTARY TOPICS 281 2 s 2 - 4s-3 = s 2 -4s + 2 2s 2 -»2s + 3~s 2 -3s + 2' (See Example 1, Set C.) g m+1 m + 2 ^ + 3 = 3> m — 1 m — 2 m — 3 Solve the following equations for x. 9. or 2 — m 2 nx + mn 2 cc = m 3 w 3 . 10. tf-kax— 10 x = — 40 a. 2a 11. Va + #-V2 Va +