1. 1. INSTRUCTORS SOLUTIONS MANUAL JUDITH A. PENNA Indiana University Purdue University Indianapolis ALGEBRA AND TRIGONOMETRY THIRD EDITION AND PRECALCULUS THIRD EDITION Judith A. Beecher Indiana University Purdue University Indianapolis Judith A. Penna Indiana University Purdue University Indianapolis Marvin L. Bittinger Indiana University Purdue University Indianapolis bitt_464818_ttl.qxd 2/8/07 11:53 AM Page 1
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3. 3. Contents Chapter R . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . 117 Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . 185 Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 273 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . 329 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . 379 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . 419 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . 471 Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . 569 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . 653
4. 4. 04 35 2 1 1 2 3 4 5 04 4 04 1 0 1 6 02 05 0 3.8 -3 -2 -1 0 1 2 33 [ 0 7 03 Chapter R Basic Concepts of Algebra Exercise Set R.1 1. Whole numbers: 3 8, 0, 9, 25 ( 3 8 = 2, 25 = 5) 2. Integers: 12, 3 8, 0, 9, 25 ( 3 8 = 2, 25 = 5) 3. Irrational numbers: 7, 5.242242224 . . . , 14, 5 5, 3 4 (Although there is a pattern in 5.242242224 . . . , there is no repeating block of digits.) 4. Natural numbers: 3 8, 9, 25 5. Rational numbers: 12, 5.3, 7 3 , 3 8, 0, 1.96, 9, 4 2 3 , 25, 5 7 6. Real numbers: All of them 7. Rational numbers but not integers: 5.3, 7 3 , 1.96, 4 2 3 , 5 7 8. Integers but not whole numbers: 12 9. Integers but not rational numbers: 12, 0 10. Real numbers but not integers: 7, 5.3, 7 3 , 5.242242224 . . ., 14, 5 5, 1.96, 4 2 3 , 3 4, 5 7 11. This is a closed interval, so we use brackets. Interval no- tation is [3, 3]. 12. (4, 4) 13. This is a half-open interval. We use a bracket on the left and a parenthesis on the right. Interval notation is [4, 1). 14. (1, 6] 15. This interval is of unlimited extent in the negative direc- tion, and the endpoint 2 is included. Interval notation is (, 2]. 16. (5, ) 17. This interval is of unlimited extent in the positive direc- tion, and the endpoint 3.8 is not included. Interval nota- tion is (3.8, ). 18. [ 3, ) 19. {x|7 < x}, or {x|x > 7}. This interval is of unlimited extent in the positive direction and the endpoint 7 is not included. Interval notation is (7, ). 20. (, 3) 21. The endpoints 0 and 5 are not included in the interval, so we use parentheses. Interval notation is (0, 5). 22. [1, 2] 23. The endpoint 9 is included in the interval, so we use a bracket before the 9. The endpoint 4 is not included, so we use a parenthesis after the 4. Interval notation is [9, 4). 24. (9, 5] 25. Both endpoints are included in the interval, so we use brackets. Interval notation is [x, x + h]. 26. (x, x + h] 27. The endpoint p is not included in the interval, so we use a parenthesis before the p. The interval is of unlimited ex- tent in the positive direction, so we use the innity symbol . Interval notation is (p, ). 28. (, q] 29. Since 6 is an element of the set of natural numbers, the statement is true. 30. True
5. 5. 2 Chapter R: Basic Concepts of Algebra 31. Since 3.2 is not an element of the set of integers, the state- ment is false. 32. True 33. Since 11 5 is an element of the set of rational numbers, the statement is true. 34. False 35. Since 11 is an element of the set of real numbers, the statement is false. 36. False 37. Since 24 is an element of the set of whole numbers, the statement is false. 38. True 39. Since 1.089 is not an element of the set of irrational num- bers, the statement is true. 40. True 41. Since every whole number is an integer, the statement is true. 42. False 43. Since every rational number is a real number, the state- ment is true. 44. True 45. Since there are real numbers that are not integers, the statement is false. 46. False 47. The sentence 6x = x6 illustrates the commutative prop- erty of multiplication. 48. Associative property of addition 49. The sentence 3 1 = 3 illustrates the multiplicative identity property. 50. Commutative property of addition 51. The sentence 5(ab) = (5a)b illustrates the associative prop- erty of multiplication. 52. Distributive property 53. The sentence 2(a+b) = (a+b)2 illustrates the commutative property of multiplication. 54. Additive inverse property 55. The sentence 6(m + n) = 6(n + m) illustrates the com- mutative property of addition. 56. Additive identity property 57. The sentence 8 1 8 = 1 illustrates the multiplicative inverse property. 58. Distributive property 59. The distance of 7.1 from 0 is 7.1, so | 7.1| = 7.1. 60. 86.2 61. The distance of 347 from 0 is 347, so |347| = 347. 62. 54 63. The distance of 97 from 0 is 97, so | 97| = 97. 64. 12 19 65. The distance of 0 from 0 is 0, so |0| = 0. 66. 15 67. The distance of 5 4 from 0 is 5 4 , so 5 4 = 5 4 . 68. 3 69. | 5 6| = | 11| = 11, or |6 (5)| = |6 + 5| = |11| = 11 70. |0 (2.5)| = |2.5| = 2.5, or | 2.5 0| = | 2.5| = 2.5 71. | 2 (8)| = | 2 + 8| = |6| = 6, or | 8 (2)| = | 8 + 2| = | 6| = 6 72. 15 8 23 12 = 45 24 46 24 = 1 24 = 1 24 , or 23 12 15 8 = 46 24 45 24 = 1 24 = 1 24 73. |12.1 6.7| = |5.4| = 5.4, or |6.7 12.1| = | 5.4| = 5.4 74. | 3 (14)| = | 3 + 14| = |11| = 11, or | 14 (3)| = | 14 + 3| = | 11| = 11 75. 3 4 15 8 = 6 8 15 8 = 21 8 = 21 8 , or 15 8 3 4 = 15 8 + 3 4 = 15 8 + 6 8 = 21 8 = 21 8 76. | 3.4 10.2| = | 13.6| = 13.6, or |10.2 (3.4)| = |10.2 + 3.4| = |13.6| = 13.6 77. | 7 0| = | 7| = 7, or |0 (7)| = |0 + 7| = |7| = 7 78. |3 19| = | 16| = 16, or |19 3| = |16| = 16 79. Provide an example. For instance, 16(82) = 164 = 4, but (16 8) 2 = 2 2 = 1.
6. 6. Exercise Set R.2 3 80. a is a rational number when a is the square of a rational number. That is, a is a rational number if there is a rational number c such that a = c2 . 81. Answers may vary. One such number is 0.124124412444 . . . . 82. Answers may vary. Since 2.01 1.418 and 2 1.414, one such number is 1.415. 83. Answers may vary. Since 1 101 = 0.0099 and 1 100 = 0.01, one such number is 0.00999. 84. Answers may vary. One such number is 5.995. 85. Since 12 + 32 = 10, the hypotenuse of a right triangle with legs of lengths 1 unit and 3 units has a length of 10 units.3 1 c c2 = 12 + 32 c2 = 10 c = 10 Exercise Set R.2 1. 37 = 1 37 am = 1 am , a = 0 2. 1 (5.9)4 = (5.9)4 3. Observe that each exponent is negative. We move each factor to the other side of the fraction bar and change the sign of each exponent. x5 y4 = y4 x5 4. Observe that each exponent is negative. We move each factor to the other side of the fraction bar and change the sign of each exponent. a2 b8 = b8 a2 5. Observe that each exponent is negative. We move each factor to the other side of the fraction bar and change the sign of each exponent. m1 n12 t6 = t6 m1n12 , or t6 mn12 6. Observe that each exponent is negative. We move each factor to the other side of the fraction bar and change the sign of each exponent. x9 y17 z11 = z11 x9y17 7. 18 = 1 (For any nonzero real number, a0 = 1.) 8. 4 3 0 = 1 9. x9 x0 = x9+0 = x9 10. a0 a4 = a0+4 = a4 11. 58 56 = 58+(6) = 52 , or 25 12. 62 67 = 62+(7) = 65 , or 1 65 13. m5 m5 = m5+5 = m0 = 1 14. n9 n9 = n9+(9) = n0 = 1 15. y3 y7 = y3+(7) = y4 , or 1 y4 16. b4 b12 = b4+12 = b8 17. 73 75 7 = 73+(5)+1 = 71 , or 1 7 18. 36 35 34 = 36+(5)+4 = 35 19. 2x3 3x2 = 2 3 x3+2 = 6x5 20. 3y4 4y3 = 3 4 y4+3 = 12y7 21. (3a5 )(5a7 ) = 3 5 a5+(7) = 15a12 , or 15 a12 22. (6b4 )(2b7 ) = 6 2 b4+(7) = 12b11 , or 12 b11 23. (5a2 b)(3a3 b4 ) = 5 3 a2+(3) b1+4 = 15a1 b5 , or 15b5 a 24. (4xy2 )(3x4 y5 ) = 4 3 x1+(4) y2+5 = 12x3 y7 , or 12y7 x3 25. (6x3 y5 )(7x2 y9 ) = 6(7)x3+2 y5+(9) = 42x1 y4 , or 42 xy4 26. (8ab7 )(7a5 b2 ) = 8(7)a1+(5) b7+2 = 56a4 b9 , or 56b9 a4 27. (2x)3 (3x)2 = 23 x3 32 x2 = 8 9 x3+2 = 72x5 28. (4y)2 (3y)3 = 16y2 27y3 = 432y5 29. (2n)3 (5n)2 = (2)3 n3 52 n2 = 8 25 n3+2 = 200n5 30. (2x)5 (3x)2 = 25 x5 32 x2 = 32 9 x5+2 = 288x7 31. b40 b37 = b4037 = b3 32. a39 a32 = a3932 = a7
7. 7. 4 Chapter R: Basic Concepts of Algebra 33. x5 x16 = x516 = x21 , or 1 x21 34. y24 y21 = y24(21) = y24+21 = y3 , or 1 y3 35. x2 y2 x1y = x2(1) y21 = x3 y3 , or x3 y3 36. x3 y3 x1y2 = x3(1) y32 = x4 y5 , or x4 y5 37. 32x4 y3 4x5y8 = 32 4 x4(5) y38 = 8xy5 , or 8x y5 38. 20a5 b2 5a7b3 = 20 5 a57 b2(3) = 4a2 b, or 4b a2 39. (2ab2 )3 = 23 a3 (b2 )3 = 23 a3 b23 = 8a3 b6 40. (4xy3 )2 = 42 x2 (y3 )2 = 16x2 y6 41. (2x3 )5 = (2)5 (x3 )5 = (2)5 x35 = 32x15 42. (3x2 )4 = (3)4 (x2 )4 = 81x8 43. (5c1 d2 )2 = (5)2 c1(2) d2(2) = c2 d4 (5)2 = c2 d4 25 44. (4x5 z2 )3 = (4)3 (x5 )3 (z2 )3 = x15 z6 (4)3 = x15 z6 64 45. (3m4 )3 (2m5 )4 = 33 m12 24 m20 = 27 16m12+(20) = 432m8 , or 432 m8 46. (4n1 )2 (2n3 )3 = 42 n2 23 n9 = 16 8 n2+9 = 128n7 47. 2x3 y7 z1 3 = (2x3 y7 )3 (z1)3 = 23 x9 y21 z3 = 8x9 y21 z3 , or 8y21 z3 x9 48. 3x5 y8 z2 4 = 81x20 y32 z8 , or 81x20 z8 y32 49. 24a10 b8 c7 12a6b3c5 5 = (2a4 b5 c2 )5 = 25 a20 b25 c10 , or b25 32a20c10 50. 125p12 q14 r22 25p8q6r15 4 = (5p4 q20 r37 )4 = 54 p16 q80 r148 , or q80 625p16r148 51. Convert 405,000 to scientic notation. We want the decimal point to be positioned between the 4 and the rst 0, so we move it 5 places to the left. Since 405,000 is greater than 10, the exponent must be positive. 405, 000 = 4.05 105 52. Position the decimal point 6 places to the left, between the 1 and the 6. Since 1,670,000 is greater than 10, the exponent must be positive. 1, 670, 000 = 1.67 106 53. Convert 0.00000039 to scientic notation. We want the decimal point to be positioned between the 3 and the 9, so we move it 7 places to the right. Since 0.00000039 is a number between 0 and 1, the exponent must be negative. 0.00000039 = 3.9 107 54. Position the decimal point 4 places to the right, between the 9 and the 2. Since 0.00092 is a number between 0 and 1, the exponent must be negative. 0.00092 = 9.2 104 55. Convert 234,600,000,000 to scientic notation. We want the decimal point to be positioned between the 2 and the 3, so we move it 11 places to the left. Since 234,600,000,000 is greater than 10, the exponent must be positive. 234, 600, 000, 000 = 2.346 1011 56. Position the decimal point 9 places to the left, between the 8 and the 9. Since 8,904,000,000 is greater than 10, the exponent must be positive. 8, 904, 000, 000 = 8.904 109 57. Convert 0.00104 to scientic notation. We want the deci- mal point to be positioned between the 1 and the last 0, so we move it 3 places to the right. Since 0.00104 is a number between 0 and 1, the exponent must be negative. 0.00104 = 1.04 103 58. Position the decimal point 9 places to the right, between the 5 and the 1. Since 0.00000000514 is a number between 0 and 1, the exponent must be negative. 0.00000000514 = 5.14 109 59. Convert 0.000016 to scientic notation. We want the decimal point to be positioned between the 1 and the 6, so we move it 5 places to the right. Since 0.000016 is a number between 0 and 1, the exponent must be negative. 0.000016 = 1.6 105 60. Position the decimal point 12 places to the left, between the ones. Since 1,137,000,000,000 is greater than 10, the exponent must be positive. 1, 137, 000, 000, 000 = 1.137 1012
8. 8. Exercise Set R.2 5 61. Convert 8.3 105 to decimal notation. The exponent is negative, so the number is between 0 and 1. We move the decimal point 5 places to the left. 8.3 105 = 0.000083 62. The exponent is positive, so the number is greater than 10. We move the decimal point 6 places to the right. 4.1 106 = 4, 100, 000 63. Convert 2.07 107 to decimal notation. The exponent is positive, so the number is greater than 10. We move the decimal point 7 places to the right. 2.07 107 = 20, 700, 000 64. The exponent is negative, so the number is between 0 and 1. We move the decimal point 6 places to the left. 3.15 106 = 0.00000315 65. Convert 3.496 1010 to decimal notation. The exponent is positive, so the number is greater than 10. We move the decimal point 10 places to the right. 3.496 1010 = 34, 960, 000, 000 66. The exponent is positive, so the number is greater than 10. We move the decimal point 11 places to the right. 8.409 1011 = 840, 900, 000, 000 67. Convert 5.41 108 to decimal notation. The exponent is negative, so the number is between 0 and 1. We move the decimal point 8 places to the left. 5.41 108 = 0.0000000541 68. The exponent is negative, so the number is between 0 and 1. We move the decimal point 10 places to the left. 6.27 1010 = 0.000000000627 69. Convert 2.319 108 to decimal notation. The exponent is positive, so the number is greater than 10. We move the decimal point 8 places to the right. 2.319 108 = 231, 900, 000 70. The exponent is negative, so the number is between 0 and 1. We move the decimal point 24 places to the left. 1.67 1024 g = 0.00000000000000000000000167 g 71. (3.1 105 )(4.5 103 ) = (3.1 4.5) (105 103 ) = 13.95 102 This is not scientic notation. = (1.395 10) 102 = 1.395 103 Writing scientic notation 72. (9.1 1017 )(8.2 103 ) = 74.62 1014 = (7.462 10) 1014 = 7.462 1013 73. (2.6 1018 )(8.5 107 ) = (2.6 8.5) (1018 107 ) = 22.1 1011 This is not scientic notation. = (2.21 10) 1011 = 2.21 1010 74. (6.4 1012 )(3.7 105 ) = 23.68 107 = (2.368 10) 107 = 2.368 108 75. 6.4 107 8.0 106 = 6.4 8.0 107 106 = 0.8 1013 This is not scientic notation. = (8 101 ) 1013 = 8 1014 Writing scientic notation 76. 1.1 1040 2.0 1071 = 0.55 1031 = (5.5 101 ) 1031 = 5.5 1030 77. 1.8 103 7.2 109 = 1.8 7.2 103 109 = 0.25 106 This is not scientic notation. = (2.5 101 ) 106 = 2.5 105 78. 1.3 104 5.2 1010 = 0.25 106 = (2.5 101 ) 106 = 2.5 107 79. We multiply the number of AUs from Earth to Pluto by the number of miles in 1 AU. 39 93 million = 39 93, 000, 000 = (3.9 10) (9.3 107 ) Writing scientic notation = (3.9 9.3) (10 107 ) = 36.27 108 = (3.627 10) 108 = 3.627 109 The distance from Earth to Pluto is about 3.627 109 mi. 80. 3.26 5.88 1012 = 19.1688 1012 = (1.91688 10) 1012 = 1.91688 1013 mi
9. 9. 6 Chapter R: Basic Concepts of Algebra 81. We multiply the diameter, in nanometers, by the number of meters in 1 nanometer. 360 0.000000001 = (3.6 102 ) 109 = 3.6 (102 109 ) = 3.6 107 The diameter of the wire is 3.6 107 m. 82. 30 million 365 = 30 106 3.65 102 8.2192 104 On average, about 8.2192104 pieces of luggage were lost each day of the year. 83. The average cost per mile is the total cost divided by the number of miles. $210 106 17.6 = $210 106 1.76 10 $119 105 ($1.19 102 ) 105 $1.19 107 The average cost per mile was about $1.19 107 . 84. 412 9, 600, 000 = 4.12 102 9.6 106 0.43 104 (4.3 101 ) 104 4.3 105 square miles 85. First nd the number of seconds in 1 hour: 1 hour = 1 hr4 60 min3 1 hr4 60 sec 1 min3 = 3600 sec The number of disintegrations produced in 1 hour is the number of disintegrations per second times the number of seconds in 1 hour. 37 billion 3600 = 37, 000, 000, 000 3600 = 3.7 1010 3.6 103 Writing scientic notation = (3.7 3.6) (1010 103 ) = 13.32 1013 Multiplying = (1.332 10) 1013 = 1.332 1014 One gram of radium produces 1.332 1014 disintegrations in 1 hour. 86. 2 93, 000, 000 = 2 9.3 107 58 107 5.8 10) 107 5.8 108 mi 87. 3 2 4 22 + 6(3 1) = 3 2 4 22 + 6 2 Working inside parentheses = 3 2 4 4 + 6 2 Evaluating 22 = 6 16 + 12 Multiplying = 10 + 12 Adding in order = 2 from left to right 88. 3[(2 + 4 22 ) 6(3 1)] = 3[(2 + 4 4) 6 2] = 3[(2 + 16) 6 2] = 3[18 6 2] = 3[18 12] = 3[6] = 18 89. 16 4 4 2 256 = 4 4 2 256 Multiplying and dividing in order from left to right = 16 2 256 = 8 256 = 2048 90. 26 23 210 28 = 23 210 28 = 27 28 = 2 91. 4(8 6)2 4 3 + 2 8 31 + 190 = 4 22 4 3 + 2 8 3 + 1 Calculating in the numerator and in the denominator = 4 4 4 3 + 2 8 4 = 16 12 + 16 4 = 4 + 16 4 = 20 4 = 5 92. [4(8 6)2 + 4](3 2 8) 22(23 + 5) = [4 22 + 4](3 16) 22(8 + 5) = [4 4 + 4](13) 22 13 = [16 + 4](13) 4 13 = 20(13) 52 = 260 52 = 5
10. 10. Exercise Set R.2 7 93. Since interest is compounded semiannually, n = 2. Substi- tute $2125 for P, 6.2% or 0.062 for i, 2 for n, and 5 for t in the compound interest formula. A = P 1 + i n nt = $2125 1 + 0.062 2 25 Substituting = $2125(1 + 0.031)25 Dividing = $2125(1.031)25 Adding = $2125(1.031)10 Multiplying 2 and 5 $2125(1.357021264) Evaluating the exponential expression $2883.670185 Multiplying $2883.67 Rounding to the nearest cent 94. A = $9550 1 + 0.054 2 27 $13, 867.23 95. Since interest is compounded quarterly, n = 4. Substitute $6700 for P, 4.5% or 0.045 for i, 4 for n, and 6 for t in the compound interest formula. A = P 1 + i n nt = $6700 1 + 0.045 4 46 Substituting = $6700(1 + 0.01125)46 Dividing = $6700(1.01125)46 Adding = $6700(1.01125)24 Multiplying 4 and 6 $6700(1.307991226) Evaluating the exponential expression $8763.541217 Multiplying $8763.54 Rounding to the nearest cent 96. A = $4875 1 + 0.058 4 49 $8185.56 97. Yes; nd the results with parentheses and without them. 4 25 (10 5) = 4 25 5 = 100 5 = 20, but 4 25 10 5 = 100 10 5 = 10 5 = 5. 98. No; x2 , or 1 x2 is positive for all x0 and x1 , or 1 x is negative for all x0. Partial conrmation can be ob- tained by inspecting the graphs of y1 = x2 and y2 = x1 for x0. 99. Substitute $250 for P, 0.05 for r and 27 for t and perform the resulting computation. S = P 1 + r 12 12t 1 r 12 = $250 1 + 0.05 12 1227 1 0.05 12 $170, 797.30 100. t =65 25 = 40 S =$100 1 + 0.04 12 1240 1 0.04 12 $118, 196.13 101. Substitute $120,000 for S, 0.06 for r, and 18 for t and solve for P. S = P 1 + r 12 12t 1 r 12 $120, 000 = P 1 + 0.06 12 1218 1 0.06 12 $120, 000 = P (1.005)216 1 0.05 $120, 000 P(387.3532) $309.79 P 102. t =70 30 = 40 $200, 000 =P 1 + 0.045 12 1240 1 0.045 12 P $149.13 103. (xt x3t )2 = (x4t )2 = x4t2 = x8t 104. (xy xy )3 = (x0 )3 = 13 = 1 105. (ta+x txa )4 = (t2x )4 = t2x4 = t8x 106. (mxb nx+b )x (mb nb )x = (mx2 bx nx2 +bx )(mbx nbx ) = mx2 nx2 107. (3xa yb )3 (3xayb)2 2 = 27x3a y3b 9x2ay2b 2 = 3xa yb 2 = 9x2a y2b 108. xr yt 2 x2r y4t 2 3 = x2r y2t x4r y8t 3 = x2r y6t 3 = x6r y18t , or x6r y18t
11. 11. 8 Chapter R: Basic Concepts of Algebra Exercise Set R.3 1. 5y4 + 3y3 + 7y2 y 4 = 5y4 + 3y3 + 7y2 + (y) + (4) Terms: 5y4 , 3y3 , 7y2 , y, 4 The degree of the term of highest degree, 5y4 , is 4. Thus, the degree of the polynomial is 4. 2. 2m3 m2 4m + 11 = 2m3 + (m2 ) + (4m) + 11 Terms: 2m3 , m2 , 4m, 11 The degree of the term of highest degree, 2m3 , is 3. Thus, the degree of the polynomial is 3. 3. 3a4 b 7a3 b3 + 5ab 2 = 3a4 b + (7a3 b3 ) + 5ab + (2) Terms: 3a4 b, 7a3 b3 , 5ab, 2 The degrees of the terms are 5, 6, 2, and, 0, respectively, so the degree of the polynomial is 6. 4. 6p3 q2 p2 q4 3pq2 + 5 = 6p3 q2 + (p2 q4 ) + (3pq2 ) + 5 Terms: 6p3 q2 , p2 q4 , 3pq2 , 5 The degrees of the terms are 5, 6, 3, and 0, respectively, so the degree of the polynomial is 6. 5. (5x2 y 2xy2 + 3xy 5)+ (2x2 y 3xy2 + 4xy + 7) = (5 2)x2 y + (2 3)xy2 + (3 + 4)xy+ (5 + 7) = 3x2 y 5xy2 + 7xy + 2 6. 2x2 y 7xy2 + 8xy + 5 7. (2x + 3y + z 7) + (4x 2y z + 8)+ (3x + y 2z 4) = (2 + 4 3)x + (3 2 + 1)y + (1 1 2)z+ (7 + 8 4) = 3x + 2y 2z 3 8. 7x2 + 12xy 2x y 9 9. (3x2 2x x3 + 2) (5x2 8x x3 + 4) = (3x2 2x x3 + 2) + (5x2 + 8x + x3 4) = (3 5)x2 + (2 + 8)x + (1 + 1)x3 + (2 4) = 2x2 + 6x 2 10. 4x2 + 8xy 5y2 + 3 11. (x4 3x2 + 4x) (3x3 + x2 5x + 3) = (x4 3x2 + 4x) + (3x3 x2 + 5x 3) = x4 3x3 + (3 1)x2 + (4 + 5)x 3 = x4 3x3 4x2 + 9x 3 12. 2x4 5x3 5x2 + 10x 5 13. (a b)(2a3 ab + 3b2 ) = (a b)(2a3 ) + (a b)(ab) + (a b)(3b2 ) Using the distributive property = 2a4 2a3 b a2 b + ab2 + 3ab2 3b3 Using the distributive property three more times = 2a4 2a3 b a2 b + 4ab2 3b3 Collecting like terms 14. (n + 1)(n2 6n 4) = (n + 1)(n2 ) + (n + 1)(6n) + (n + 1)(4) = n3 + n2 6n2 6n 4n 4 = n3 5n2 10n 4 15. (x + 5)(x 3) = x2 3x + 5x 15 Using FOIL = x2 + 2x 15 Collecting like terms 16. (y 4)(y + 1) = y2 + y 4y 4 = y2 3y 4 17. (x + 6)(x + 4) = x2 + 4x + 6x + 24 Using FOIL = x2 + 10x + 24 Collecting like terms 18. (n 5)(n 8) = n2 8n 5n + 40 = n2 13n + 40 19. (2a + 3)(a + 5) = 2a2 + 10a + 3a + 15 Using FOIL = 2a2 + 13a + 15 Collecting like terms 20. (3b + 1)(b 2) = 3b2 6b + b 2 = 3b2 5b 2 21. (2x + 3y)(2x + y) = 4x2 + 2xy + 6xy + 3y2 Using FOIL = 4x2 + 8xy + 3y2 22. (2a 3b)(2a b) = 4a2 2ab 6ab + 3b2 = 4a2 8ab + 3b2 23. (y + 5)2 = y2 + 2 y 5 + 52 [(A + B)2 = A2 + 2AB + B2 ] = y2 + 10y + 25 24. (y + 7)2 = y2 + 2 y 7 + 72 = y2 + 14y + 49 25. (x 4)2 = x2 2 x 4 + 42 [(A B)2 = A2 2AB + B2 ] = x2 8x + 16 26. (a 6)2 = a2 2 a 6 + 62 = a2 12a + 36 27. (5x 3)2 = (5x)2 2 5x 3 + 32 [(A B)2 = A2 2AB + B2 ] = 25x2 30x + 9
12. 12. Exercise Set R.3 9 28. (3x 2)2 = (3x)2 2 3x 2 + 22 = 9x2 12x + 4 29. (2x + 3y)2 = (2x)2 + 2(2x)(3y) + (3y)2 [(A+B)2 = A2 +2AB+B2 ] = 4x2 + 12xy + 9y2 30. (5x + 2y)2 = (5x)2 + 2 5x 2y + (2y)2 = 25x2 + 20xy + 4y2 31. (2x2 3y)2 = (2x2 )2 2(2x2 )(3y) + (3y)2 [(A B)2 = A2 2AB + B2 ] = 4x4 12x2 y + 9y2 32. (4x2 5y)2 = (4x2 )2 2 4x2 5y + (5y)2 = 16x4 40x2 y + 25y2 33. (a + 3)(a 3) = a2 32 [(A + B)(A B) = A2 B2 ] = a2 9 34. (b + 4)(b 4) = b2 42 = b2 16 35. (2x 5)(2x + 5) = (2x)2 52 [(A + B)(A B) = A2 B2 ] = 4x2 25 36. (4y 1)(4y + 1) = (4y)2 12 = 16y2 1 37. (3x 2y)(3x + 2y) = (3x)2 (2y)2 [(AB)(A+B) = A2 B2 ] = 9x2 4y2 38. (3x + 5y)(3x 5y) = (3x)2 (5y)2 = 9x2 25y2 39. (2x + 3y + 4)(2x + 3y 4) = [(2x + 3y) + 4][(2x + 3y) 4] = (2x + 3y)2 42 = 4x2 + 12xy + 9y2 16 40. (5x + 2y + 3)(5x + 2y 3) = (5x + 2y)2 32 = 25x2 + 20xy + 4y2 9 41. (x + 1)(x 1)(x2 + 1) = (x2 1)(x2 + 1) = x4 1 42. (y 2)(y + 2)(y2 + 4) = (y2 4)(y2 + 4) = y4 16 43. No; if the leading coecients of the polynomials are addi- tive inverses, the degree of the sum is less than n. For ex- ample, the sum of the second degree polynomials x2 +x1 and x2 + 4 is x + 3, a rst degree polynomial. 44. Algebraically: Choose specic values for A and B, A = 0, B = 0, and evaluate (A + B)2 and A2 + B2 . For example, (2 + 3)2 = 52 = 25, but 22 + 32 = 4 + 9 = 13. Geometrically: Show that the area of a square with side A + B is not equal to A2 + B2 . See the gure below. A2 AB AB B2 A B B A 45. (an + bn )(an bn ) = (an )2 (bn )2 = a2n b2n 46. (ta + 4)(ta 7) = (ta )2 7ta + 4ta 28 = t2a 3ta 28 47. (an + bn )2 = (an )2 + 2 an bn + (bn )2 = a2n + 2an bn + b2n 48. (x3m t5n )2 = (x3m )2 2 x3m t5n + (t5n )2 = x6m 2x3m t5n + t10n 49. (x 1)(x2 + x + 1)(x3 + 1) = [(x 1)x2 + (x 1)x + (x 1) 1](x3 + 1) = (x3 x2 + x2 x + x 1)(x3 + 1) = (x3 1)(x3 + 1) = (x3 )2 12 = x6 1 50. [(2x 1)2 1]2 = [4x2 4x + 1 1]2 = [4x2 4x]2 = (4x2 )2 2(4x2 )(4x) + (4x)2 = 16x4 32x3 + 16x2 51. (xab )a+b = x(ab)(a+b) = xa2 b2 52. (tm+n )m+n (tmn )mn = tm2 +2mn+n2 tm2 2mn+n2 = t2m2 +2n2 53. (a + b + c)2 = (a + b + c)(a + b + c) = (a + b + c)(a) + (a + b + c)(b) + (a + b + c)(c) = a2 + ab + ac + ab + b2 + bc + ac + bc + c2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
13. 13. 10 Chapter R: Basic Concepts of Algebra Exercise Set R.4 1. 2x 10 = 2 x 2 5 = 2(x 5) 2. 7y + 42 = 7 y + 7 6 = 7(y + 6) 3. 3x4 9x2 = 3x2 x2 3x2 3 = 3x2 (x2 3) 4. 20y2 5y5 = 5y2 4 5y2 y3 = 5y2 (4 y3 ) 5. 4a2 12a + 16 = 4 a2 4 3a + 4 4 = 4(a2 3a + 4) 6. 6n2 + 24n 18 = 6 n2 + 6 4n 6 3 = 6(n2 + 4n 3) 7. a(b 2) + c(b 2) = (b 2)(a + c) 8. a(x2 3) 2(x2 3) = (x2 3)(a 2) 9. x3 + 3x2 + 6x + 18 = x2 (x + 3) + 6(x + 3) = (x + 3)(x2 + 6) 10. 3x3 x2 + 18x 6 = x2 (3x 1) + 6(3x 1) = (3x 1)(x2 + 6) 11. y3 y2 + 3y 3 = y2 (y 1) + 3(y 1) = (y 1)(y2 + 3) 12. y3 y2 + 2y 2 = y2 (y 1) + 2(y 1) = (y 1)(y2 + 2) 13. 24x3 36x2 + 72x 108 = 12(2x3 3x2 + 6x 9) = 12[x2 (2x 3) + 3(2x 3)] = 12(2x 3)(x2 + 3) 14. 5a3 10a2 + 25a 50 = 5(a3 2a2 + 5a 10) = 5[a2 (a 2) + 5(a 2)] = 5(a 2)(a2 + 5) 15. a3 3a2 2a + 6 = a2 (a 3) 2(a 3) = (a 3)(a2 2) 16. t3 + 6t2 2t 12 = t2 (t + 6) 2(t + 6) = (t + 6)(t2 2) 17. x3 x2 5x + 5 = x2 (x 1) 5(x 1) = (x 1)(x2 5) 18. x3 x2 6x + 6 = x2 (x 1) 6(x 1) = (x 1)(x2 6) 19. p2 + 6p + 8 We look for two numbers with a product of 8 and a sum of 6. By trial, we determine that they are 2 and 4. p2 + 6p + 8 = (p + 2)(p + 4) 20. Note that (5)(2) = 10 and 5 + (2) = 7. Then w2 7w + 10 = (w 5)(w 2). 21. x2 8x + 12 We look for two numbers with a product of 12 and a sum of 8. By trial, we determine that they are 2 and 6. x2 8x + 12 = (x 2)(x 6) 22. Note that 1 5 = 5 and 1 + 5 = 6. Then x2 + 6x + 5 = (x + 1)(x + 5). 23. t2 + 8t + 15 We look for two numbers with a product of 15 and a sum of 8. By trial, we determine that they are 3 and 5. t2 + 8t + 15 = (t + 3)(t + 5) 24. Note that 3 9 = 27 and 3 + 9 = 12. Then y2 + 12y + 27 = (y + 3)(y + 9). 25. x2 6xy 27y2 We look for two numbers with a product of 27 and a sum of 6. By trial, we determine that they are 3 and 9. x2 6xy 27y2 = (x + 3y)(x 9y) 26. Note that 3(5) = 15 and 3 + (5) = 2. Then t2 2t 15 = (t + 3)(t 5). 27. 2n2 20n 48 = 2(n2 10n 24) Now factor n2 10n24. We look for two numbers with a product of 24 and a sum of 10. By trial, we determine that they are 2 and 12. Then n2 10n 24 = (n + 2)(n 12). We must include the common factor, 2, to have a factorization of the original trinomial. 2n2 20n 48 = 2(n + 2)(n 12) 28. 2a2 2ab 24b2 = 2(a2 ab 12b2 ) Note that 4 3 = 12 and 4 + 3 = 1. Then 2a2 2ab 24b2 = 2(a 4b)(a + 3b). 29. y4 4y2 21 = (y2 )2 4y2 21 We look for two numbers with a product of 21 and a sum of 4. By trial, we determine that they are 3 and 7. y4 4y2 21 = (y2 + 3)(y2 7) 30. Note that 9(10) = 90 and 9 + (10) = 1. Then m4 m2 90 = (m2 + 9)(m2 10).
14. 14. Exercise Set R.4 11 31. y4 + 9y3 + 14y2 = y2 (y2 + 9y + 14) Now factor y2 + 9y + 14. Look for two numbers with a product of 14 and a sum of 9. The numbers are 2 and 7. Then y2 + 9y + 14 = (y + 2)(y + 7). We must include the common factor, y2 , in order to have a factorization of the original trinomial. y4 + 9y3 + 14y2 = y2 (y + 2)(y + 7) 32. 3z3 21z2 + 18z = 3z(z2 7z + 6) = 3z(z 1)(z 6) 33. 2x3 2x2 y 24xy2 = 2x(x2 xy 12y2 ) Now factor x2 xy 12y2 . Look for two numbers with a product of 12 and a sum of 1. The numbers are 4 and 3. Then x2 xy12y2 = (x4y)(x+3y). We must include the common factor, 2x, in order to have a factorization of the original trinomial. 2x3 2x2 y 24xy2 = 2x(x 4y)(x + 3y) 34. a3 b9a2 b2 +20ab3 = ab(a2 9ab+20b2 ) = ab(a4b)(a5b) 35. 2n2 + 9n 56 We use the FOIL method. 1. There is no common factor other than 1 or 1. 2. The factorization must be of the form (2n+ )(n+ ). 3. Factor the constant term, 56. The possibilities are 156, 1(56), 228, 2(28), 416, 4(16), 7 8, and 7(8). The factors can be written in the opposite order as well: 56(1), 56 1, 28(2), 28 2, 16(4), 16 4, 8(7), and 8 7. 4. Find a pair of factors for which the sum of the out- side and the inside products is the middle term, 9n. By trial, we determine that the factorization is (2n 7)(n + 8). 36. 3y2 + 7y 20 = (3y 5)(y + 4) 37. 12x2 + 11x + 2 We use the grouping method. 1. There is no common factor other than 1 or 1. 2. Multiply the leading coecient and the constant: 12 2 = 24. 3. Try to factor 24 so that the sum of the factors is the coecient of the middle term, 11. The factors we want are 3 and 8. 4. Split the middle term using the numbers found in step (3): 11x = 3x + 8x 5. Factor by grouping. 12x2 + 11x + 2 = 12x2 + 3x + 8x + 2 = 3x(4x + 1) + 2(4x + 1) = (4x + 1)(3x + 2) 38. 6x2 7x 20 = (3x + 4)(2x 5) 39. 4x2 + 15x + 9 We use the FOIL method. 1. There is no common factor other than 1 or 1. 2. The factorization must be of the form (4x+ )(x+ ) or (2x+ )(2x+ ). 3. Factor the constant term, 9. The possibilities are 1 9, 1(9), 3 3, and 3(3). The rst two pairs of factors can be written in the opposite order as well: 9 1, 9(1). 4. Find a pair of factors for which the sum of the out- side and the inside products is the middle term, 15x. By trial, we determine that the factorization is (4x + 3)(x + 3). 40. 2y2 + 7y + 6 = (2y + 3)(y + 2) 41. 2y2 + y 6 We use the grouping method. 1. There is no common factor other than 1 or 1. 2. Multiply the leading coecient and the constant: 2(6) = 12. 3. Try to factor 12 so that the sum of the factors is the coecient of the middle term, 1. The factors we want are 4 and 3. 4. Split the middle term using the numbers found in step (3): y = 4y 3y 5. Factor by grouping. 2y2 + y 6 = 2y2 + 4y 3y 6 = 2y(y + 2) 3(y + 2) = (y + 2)(2y 3) 42. 20p2 23p + 6 = (4p 3)(5p 2) 43. 6a2 29ab + 28b2 We use the FOIL method. 1. There is no common factor other than 1 or 1. 2. The factorization must be of the form (6x+ )(x+ ) or (3x+ )(2x+ ). 3. Factor the coecient of the last term, 28. The pos- sibilities are 1 28, 1(28), 2 14, 2(14), 4 7, and 4(7). The factors can be written in the op- posite order as well: 281, 28(1), 142, 14(2), 7 4, and 7(4). 4. Find a pair of factors for which the sum of the out- side and the inside products is the middle term, 29. Observe that the second term of each bino- mial factor will contain a factor of b. By trial, we determine that the factorization is (3a4b)(2a7b). 44. 10m2 + 7mn 12n2 = (5m 4n)(2m + 3n)
15. 15. 12 Chapter R: Basic Concepts of Algebra 45. 12a2 4a 16 We will use the grouping method. 1. Factor out the common factor, 4. 12a2 4a 16 = 4(3a2 a 4) 2. Now consider 3a2 a 4. Multiply the leading coecient and the constant: 3(4) = 12. 3. Try to factor 12 so that the sum of the factors is the coecient of the middle term, 1. The factors we want are 4 and 3. 4. Split the middle term using the numbers found in step (3): a = 4a + 3a 5. Factor by grouping. 3a2 a 4 = 3a2 4a + 3a 4 = a(3a 4) + (3a 4) = (3a 4)(a + 1) We must include the common factor to get a factor- ization of the original trinomial. 12a2 4a 16 = 4(3a 4)(a + 1) 46. 12a2 14a 20 = 2(6a2 7a 10) = 2(6a + 5)(a 2) 47. m2 4 = m2 22 = (m + 2)(m 2) 48. z2 81 = (z + 9)(z 9) 49. 4z2 81 = (2z)2 92 = (2z + 9)(2z 9) 50. 16x2 9 = (4x 3)(4x + 3) 51. 6x2 6y2 = 6(x2 y2 ) = 6(x + y)(x y) 52. 8a2 8b2 = 8(a2 b2 ) = 8(a + b)(a b) 53. 4xy4 4xz2 = 4x(y4 z2 ) = 4x[(y2 )2 z2 ] = 4x(y2 + z)(y2 z) 54. 5x2 y 5yz4 = 5y(x2 z4 ) = 5y(x + z2 )(x z2 ) 55. 7pq4 7py4 = 7p(q4 y4 ) = 7p[(q2 )2 (y2 )2 ] = 7p(q2 + y2 )(q2 y2 ) = 7p(q2 + y2 )(q + y)(q y) 56. 25ab4 25az4 = 25a(b4 z4 ) = 25a(b2 + z2 )(b2 z2 ) = 25a(b2 + z2 )(b + z)(b z) 57. y2 6y + 9 = y2 2 y 3 + 32 = (y 3)2 58. x2 + 8x + 16 = (x + 4)2 59. 4z2 + 12z + 9 = (2z)2 + 2 2z 3 + 32 = (2z + 3)2 60. 9z2 12z + 4 = (3z 2)2 61. 1 8x + 16x2 = 12 2 1 4x + (4x)2 = (1 4x)2 62. 1 + 10x + 25x2 = (1 + 5x)2 63. a3 + 24a2 + 144a = a(a2 + 24a + 144) = a(a2 + 2 a 12 + 122 ) = a(a + 12)2 64. y3 18y2 + 81y = y(y2 18y + 81) = y(y 9)2 65. 4p2 8pq + 4q2 = 4(p2 2pq + q2 ) = 4(p q)2 66. 5a2 10ab + 5b2 = 5(a2 2ab + b2 ) = 5(a b)2 67. x3 + 8 = x3 + 23 = (x + 2)(x2 2x + 4) 68. y3 64 = (y 4)(y2 + 4y + 16) 69. m3 1 = m3 13 = (m 1)(m2 + m + 1) 70. n3 + 216 = (n + 6)(n2 6n + 36) 71. 2y3 128 = 2(y3 64) = 2(y3 43 ) = 2(y 4)(y2 + 4y + 16) 72. 8t3 8 = 8(t3 1) = 8(t 1)(t2 + t + 1) 73. 3a5 24a2 = 3a2 (a3 8) = 3a2 (a3 23 ) = 3a2 (a 2)(a2 + 2a + 4) 74. 250z4 2z = 2z(125z3 1) = 2z(5z 1)(25z2 + 5z + 1) 75. t6 + 1 = (t2 )3 + 13 = (t2 + 1)(t4 t2 + 1) 76. 27x6 8 = (3x2 2)(9x4 + 6x2 + 4) 77. 18a2 b 15ab2 = 3ab 6a 3ab 5b = 3ab(6a 5b) 78. 4x2 y + 12xy2 = 4xy(x + 3y) 79. x3 4x2 + 5x 20 = x2 (x 4) + 5(x 4) = (x 4)(x2 + 5)
16. 16. Exercise Set R.4 13 80. z3 + 3z2 3z 9 = z2 (z + 3) 3(z + 3) = (z + 3)(z2 3) 81. 8x2 32 = 8(x2 4) = 8(x + 2)(x 2) 82. 6y2 6 = 6(y2 1) = 6(y + 1)(y 1) 83. 4y2 5 There are no common factors. We might try to factor this polynomial as a dierence of squares, but there is no integer which yields 5 when squared. Thus, the polynomial is prime. 84. There are no common factors and there is no integer which yields 7 when squared, so 16x2 7 is prime. 85. m2 9n2 = m2 (3n)2 = (m + 3n)(m 3n) 86. 25t2 16 = (5t + 4)(5t 4) 87. x2 + 9x + 20 We look for two numbers with a product of 20 and a sum of 9. They are 4 and 5. x2 + 9x + 20 = (x + 4)(x + 5) 88. Note that 3(2) = 6 and 3 + (2) = 1. Then y2 + y 6 = (y + 3)(y 2). 89. y2 6y + 5 We look for two numbers with a product of 5 and a sum of 6. They are 5 and 1. y2 6y + 5 = (y 5)(y 1) 90. Note that 7(3) = 21 and 7 + 3 = 4. x2 4x 21 = (x 7)(x + 3) 91. 2a2 + 9a + 4 We use the FOIL method. 1. There is no common factor other than 1 or 1. 2. The factorization must be of the form (2a+ )(a+ ). 3. Factor the constant term, 4. The possibilities are 14, 1(4), and 22. The rst two pairs of factors can be written in the opposite order as well: 4 1, 4(1). 4. Find a pair of factors for which the sum of the out- side and the inside products is the middle term, 9a. By trial, we determine that the factorization is (2a + 1)(a + 4). 92. 3b2 b 2 = (3b + 2)(b 1) 93. 6x2 + 7x 3 We use the grouping method. 1. There is no common factor other than 1 or 1. 2. Multiply the leading coecient and the constant: 6(3) = 18. 3. Try to factor 18 so that the sum of the factors is the coecient of the middle term, 7. The factors we want are 9 and 2. 4. Split the middle term using the numbers found in step (3): 7x = 9x 2x 5. Factor by grouping. 6x2 + 7x 3 = 6x2 + 9x 2x 3 = 3x(2x + 3) (2x + 3) = (2x + 3)(3x 1) 94. 8x2 + 2x 15 = (4x 5)(2x + 3) 95. y2 18y + 81 = y2 2 y 9 + 92 = (y 9)2 96. n2 + 2n + 1 = (n + 1)2 97. 9z2 24z + 16 = (3z)2 2 3z 4 + 42 = (3z 4)2 98. 4z2 + 20z + 25 = (2z + 5)2 99. x2 y2 14xy + 49 = (xy)2 2 xy 7 + 72 = (xy 7)2 100. x2 y2 16xy + 64 = (xy 8)2 101. 4ax2 + 20ax 56a = 4a(x2 + 5x 14) = 4a(x + 7)(x 2) 102. 21x2 y + 2xy 8y = y(21x2 + 2x 8) = y(7x 4)(3x + 2) 103. 3z3 24 = 3(z3 8) = 3(z3 23 ) = 3(z 2)(z2 + 2z + 4) 104. 4t3 + 108 = 4(t3 + 27) = 4(t + 3)(t2 3t + 9) 105. 16a7 b + 54ab7 = 2ab(8a6 + 27b6 ) = 2ab[(2a2 )3 + (3b2 )3 ] = 2ab(2a2 + 3b2 )(4a4 6a2 b2 + 9b4 ) 106. 24a2 x4 375a8 x = 3a2 x(8x3 125a6 ) = 3a2 x(2x 5a2 )(4x2 + 10a2 x + 25a4 )
17. 17. 14 Chapter R: Basic Concepts of Algebra 107. y3 3y2 4y + 12 = y2 (y 3) 4(y 3) = (y 3)(y2 4) = (y 3)(y + 2)(y 2) 108. p3 2p2 9p + 18 = p2 (p 2) 9(p 2) = (p 2)(p2 9) = (p 2)(p + 3)(p 3) 109. x3 x2 + x 1 = x2 (x 1) + (x 1) = (x 1)(x2 + 1) 110. x3 x2 x + 1 = x2 (x 1) (x 1) = (x 1)(x2 1) = (x 1)(x + 1)(x 1), or (x 1)2 (x + 1) 111. 5m4 20 = 5(m4 4) = 5(m2 + 2)(m2 2) 112. 2x2 288 = 2(x2 144) = 2(x + 12)(x 12) 113. 2x3 + 6x2 8x 24 = 2(x3 + 3x2 4x 12) = 2[x2 (x + 3) 4(x + 3)] = 2(x + 3)(x2 4) = 2(x + 3)(x + 2)(x 2) 114. 3x3 + 6x2 27x 54 = 3(x3 + 2x2 9x 18) = 3[x2 (x + 2) 9(x + 2)] = 3(x + 2)(x2 9) = 3(x + 2)(x + 3)(x 3) 115. 4c2 4cd d2 = (2c)2 2 2c d d2 = (2c d)2 116. 9a2 6ab + b2 = (3a b)2 117. m6 + 8m3 20 = (m3 )2 + 8m3 20 We look for two numbers with a product of 20 and a sum of 8. They are 10 and 2. m6 + 8m3 20 = (m3 + 10)(m3 2) 118. x4 37x2 + 36 = (x2 1)(x2 36) = (x + 1)(x 1)(x + 6)(x 6) 119. p 64p4 = p(1 64p3 ) = p[13 (4p)3 ] = p(1 4p)(1 + 4p + 16p2 ) 120. 125a 8a4 = a(125 8a3 ) = a(5 2a)(25 + 10a + 4a2 ) 121. A2 + B2 can be factored when A and B have a common factor. For example, let A = 2x and B = 10. Then A2 + B2 = 4x2 + 100 = 4(x2 + 25). 122. A3 B3 = A3 + (B)3 = (A + (B))(A2 A(B) + (B)2 ) = (A B)(A2 + AB + B2 ) 123. y4 84 + 5y2 = y4 + 5y2 84 = u2 + 5u 84 Substituting u for y2 = (u + 12)(u 7) = (y2 + 12)(y2 7) Substituting y2 for u 124. 11x2 + x4 80 = x4 + 11x2 80 = u2 + 11u 80 Substituting u for x2 = (u + 16)(u 5) = (x2 + 16)(x2 5) Substituting x2 for u 125. y2 8 49 + 2 7 y = y2 + 2 7 y 8 49 = y + 4 7 y 2 7 126. t2 27 100 + 3 5 t = t2 + 3 5 t 27 100 = t + 9 10 t 3 10 127. x2 + 3x + 9 4 = x2 + 2 x 3 2 + 3 2 2 = x + 3 2 2 128. x2 5x + 25 4 = x 5 2 2 129. x2 x + 1 4 = x2 2 x 1 2 + 1 2 2 = x 1 2 2 130. x2 2 3 x + 1 9 = x 1 3 2 131. (x + h)3 x3 = [(x + h) x][(x + h)2 + x(x + h) + x2 ] = (x + h x)(x2 + 2xh + h2 + x2 + xh + x2 ) = h(3x2 + 3xh + h2 ) 132. (x + 0.01)2 x2 = (x + 0.01 + x)(x + 0.01 x) = 0.01(2x + 0.01), or 0.02(x + 0.005) 133. (y 4)2 + 5(y 4) 24 = u2 + 5u 24 Substituting u for y 4 = (u + 8)(u 3) = (y 4 + 8)(y 4 3) Substituting y 4 for u = (y + 4)(y 7)
18. 18. Exercise Set R.5 15 134. 6(2p + q)2 5(2p + q) 25 = 6u2 5u 25 Substituting u for 2p + q = (3u + 5)(2u 5) = [3(2p + q) + 5][2(2p + q) 5] Substituting 2p + q for u = (6p + 3q + 5)(4p + 2q 5) 135. x2n + 5xn 24 = (xn )2 + 5xn 24 = (xn + 8)(xn 3) 136. 4x2n 4xn 3 = (2xn 3)(2xn + 1) 137. x2 + ax + bx + ab = x(x + a) + b(x + a) = (x + a)(x + b) 138. bdy2 + ady + bcy + ac = dy(by + a) + c(by + a) = (by + a)(dy + c) 139. 25y2m (x2n 2xn + 1) = (5ym )2 (xn 1)2 = [5ym + (xn 1)][5ym (xn 1)] = (5ym + xn 1)(5ym xn + 1) 140. x6a t3b = (x2a )3 (tb )3 = (x2a tb )(x4a + x2a tb + t2b ) 141. (y 1)4 (y 1)2 = (y 1)2 [(y 1)2 1] = (y 1)2 [y2 2y + 1 1] = (y 1)2 (y2 2y) = y(y 1)2 (y 2) 142. x6 2x5 + x4 x2 + 2x 1 = x4 (x2 2x + 1) (x2 2x + 1) = (x2 2x + 1)(x4 1) = (x 1)2 (x2 + 1)(x2 1) = (x 1)2 (x2 + 1)(x + 1)(x 1) = (x2 + 1)(x + 1)(x 1)3 Exercise Set R.5 1. Since 3 4 is dened for all real numbers, the domain is {x|x is a real number}. 2. Since 8 x = 0 when x = 8, the domain is {x|x is a real number and x = 8}. 3. 3x 3 x(x 1) The denominator is 0 when the factor x = 0 and also when x 1 = 0, or x = 1. The domain is {x|x is a real number and x = 0 and x = 1}. 4. 15x 10 2x(3x 2) Since 2x = 0 when x = 0 and 3x 2 = 0 when x = 2 3 , the domain is x x is a real number and x = 0 and x = 2 3 . 5. x + 5 x2 + 4x 5 = x + 5 (x + 5)(x 1) We see that x + 5 = 0 when x = 5 and x 1 = 0 when x = 1. Thus, the domain is {x|x is a real number and x = 5 and x = 1}. 6. (x2 4)(x + 1) (x + 2)(x2 1) = (x2 4)(x + 1) (x + 2)(x + 1)(x 1) x + 2 = 0 when x = 2; x + 1 = 0 when x = 1; x 1 = 0 when x = 1. The domain is {x|x is a real number and x = 2 and x = 1 and x = 1}. 7. We rst factor the denominator completely. 7x2 28x + 28 (x2 4)(x2 +3x10) = 7x2 28x + 28 (x+2)(x2)(x+5)(x2) We see that x + 2 = 0 when x = 2, x 2 = 0 when x = 2, and x + 5 = 0 when x = 5. Thus, the domain is {x|x is a real number and x = 2 and x = 2 and x = 5}. 8. 7x2 + 11x 6 x(x2 x 6) = 7x2 + 11x 6 x(x 3)(x + 2) The denominator is 0 when x = 0 or when x 3 = 0 or when x + 2 = 0. Now x 3 = 0 when x = 3 and x + 2 = 0 when x = 2. Thus, the domain is {x|x is a real number and x = 0 and x = 3 and x = 2}. 9. x2 4 x2 4x + 4 = (x + 2)(x2) (x 2)(x2) = x + 2 x 2 10. x2 + 2x 3 x2 9 = (x 1)(x+3) (x+3) (x 3) = x 1 x 3 11. x3 6x2 + 9x x3 3x2 = x(x2 6x + 9) x2(x 3) = x/(x3) (x 3) x/ x(x3) = x 3 x 12. y5 5y4 + 4y3 y3 6y2 + 8y = y3 (y2 5y + 4) y(y2 6y + 8) = y/ y y (y 1)(y4) y/(y 2)(y4) = y2 (y 1) y 2
19. 19. 16 Chapter R: Basic Concepts of Algebra 13. 6y2 + 12y 48 3y2 9y + 6 = 6(y2 + 2y 8) 3(y2 3y + 2) = 2 3/ (y + 4)(y2) 3/(y 1)(y2) = 2(y + 4) y 1 14. 2x2 20x + 50 10x2 30x 100 = 2(x2 10x + 25) 10(x2 3x 10) = 2/(x5) (x 5) 2/ 5 (x5) (x + 2) = x 5 5(x + 2) 15. 4 x x2 + 4x 32 = 1(x4) (x4) (x + 8) = 1 x + 8 , or 1 x + 8 16. 6 x x2 36 = 1(x6) (x + 6)(x6) = 1 x + 6 , or 1 x + 6 17. x2 y2 (x y)2 1 x + y = (x2 y2 ) 1 (x y)2(x + y) = (x+y) (xy) 1 (xy) (x y)(x+y) = 1 x y 18. r s r + s r2 s2 (r s)2 = (r s)(r2 s2 ) (r + s)(r s)2 = (rs) (rs) (r+s) 1 (r+s) (rs) (rs) = 1 19. x2 2x 35 2x3 3x2 4x3 9x 7x 49 = (x7) (x + 5)(x)4 (2x + 3)(2x3)$ x/ x(2x3)$(7)(x7) = (x + 5)(2x + 3) 7x 20. x2 + 2x 35 3x3 2x2 9x3 4x 7x + 49 = (x+7) (x 5)(x)4 (3x + 2)(3x2)$ x/ x(3x2)$(7)(x+7) = (x 5)(3x + 2) 7x 21. a2 a 6 a2 7a + 12 a2 2a 8 a2 3a 10 = (a3) (a + 2)(a4) (a+2) (a4) (a3) (a 5)(a+2) = a + 2 a 5 22. a2 a 12 a2 6a + 8 a2 + a 6 a2 2a 24 = (a4) (a + 3)(a+3)(a2) (a2) (a4) (a 6)(a+4) = (a + 3)2 (a 6)(a + 4) 23. m2 n2 r + s m n r + s = m2 n2 r + s r + s m n = (m + n)(mn) (r+s) (r+s) (m n) = m + n 24. a2 b2 x y a + b x y = a2 b2 x y x y a + b = (a+b) (a b)(xy) (xy) (a+b) 1 = a b 25. 3x + 12 2x 8 (x + 4)2 (x 4)2 = 3x + 12 2x 8 (x 4)2 (x + 4)2 = 3(x+4) (x4) (x 4) 2(x4) (x+4) (x + 4) = 3(x 4) 2(x + 4) 26. a2 a 2 a2 a 6 a2 2a 2a + a2 = a2 a 2 a2 a 6 2a + a2 a2 2a = (a2) (a + 1)(a)4 (2+a) (a 3)(a+2) (a)4 (a2) = a + 1 a 3 27. x2 y2 x3 y3 x2 + xy + y2 x2 + 2xy + y2 = (x + y)(x y)(x2 + xy + y2 ) (x y)(x2 + xy + y2)(x + y)(x + y) = 1 x + y (x + y)(x y)(x2 + xy + y2 ) (x + y)(x y)(x2 + xy + y2) = 1 x + y 1 Removing a factor of 1 = 1 x + y
20. 20. Exercise Set R.5 17 28. c3 + 8 c2 4 c2 2c + 4 c2 4c + 4 = c3 + 8 c2 4 c2 4c + 4 c2 2c + 4 = (c + 2)(c2 2c + 4)(c 2)(c 2) (c + 2)(c 2)(c2 2c + 4) = (c + 2)(c2 2c + 4)(c 2) (c + 2)(c2 2c + 4)(c 2) c 2 1 = c 2 29. (x y)2 z2 (x + y)2 z2 x y + z x + y z = (x y)2 z2 (x + y)2 z2 x + y z x y + z = (x y + z)(x y z)(x + y z) (x + y + z)(x + y z)(x y + z) = (x y + z)(x + y z) (x y + z)(x + y z) x y z x + y + z = 1 x y z x + y + z Removing a factor of 1 = x y z x + y + z 30. (a + b)2 9 (a b)2 9 a b 3 a + b + 3 = (a + b + 3)(a + b 3)(a b 3) (a b + 3)(a b 3)(a + b + 3) = (a + b + 3)(a b 3) (a + b + 3)(a b 3) a + b 3 a b + 3 = a + b 3 a b + 3 31. 5 2x + 1 2x = 5 + 1 2x = 6 2x = 2/ 3 2/ x = 3 x 32. 10 9y 4 9y = 6 9y = 3/ 2 3/ 3y = 2 3y 33. 3 2a + 3 + 2a 2a + 3 = 3 + 2a 2a + 3 = 1 34. a 3b a + b + a + 5b a + b = 2a + 2b a + b = 2(a+b) 1 (a+b) = 2 35. 5 4z 3 8z , LCD is 8z = 5 4z 2 2 3 8z = 10 8z 3 8z = 7 8z 36. 12 x2y + 5 xy2 , LCD is x2 y2 = 12y x2y2 + 5x x2y2 = 12y + 5x x2y2 37. 3 x + 2 + 2 x2 4 = 3 x + 2 + 2 (x + 2)(x 2) , LCD is (x+2)(x2) = 3 x + 2 x 2 x 2 + 2 (x + 2)(x 2) = 3x 6 (x + 2)(x 2) + 2 (x + 2)(x 2) = 3x 4 (x + 2)(x 2) 38. 5 a 3 2 a2 9 = 5 a 3 2 (a + 3)(a 3) , LCD is (a + 3)(a 3) = 5(a + 3) 2 (a + 3)(a 3) = 5a + 15 2 (a + 3)(a 3) = 5a + 13 (a + 3)(a 3) 39. y y2 y 20 2 y + 4 = y (y + 4)(y 5) 2 y + 4 , LCD is (y + 4)(y 5) = y (y + 4)(y 5) 2 y + 4 y 5 y 5 = y (y + 4)(y 5) 2y 10 (y + 4)(y 5) = y (2y 10) (y + 4)(y 5) = y 2y + 10 (y + 4)(y 5) = y + 10 (y + 4)(y 5)
21. 21. 18 Chapter R: Basic Concepts of Algebra 40. 6 y2 + 6y + 9 5 y + 3 = 6 (y + 3)2 5 y + 3 , LCD is (y + 3)2 = 6 5(y + 3) (y + 3)2 = 6 5y 15 (y + 3)2 = 5y 9 (y + 3)2 41. 3 x + y + x 5y x2 y2 = 3 x + y + x 5y (x + y)(x y) , LCD is (x + y)(x y) = 3 x + y x y x y + x 5y (x + y)(x y) = 3x 3y (x + y)(x y) + x 5y (x + y)(x y) = 4x 8y (x + y)(x y) 42. a2 + 1 a2 1 a 1 a + 1 = a2 + 1 (a + 1)(a 1) a 1 a + 1 , LCD is (a + 1)(a 1) = a2 + 1 (a 1)(a 1) (a + 1)(a 1) = a2 + 1 a2 + 2a 1 (a + 1)(a 1) = 2a (a + 1)(a 1) 43. y y 1 + 2 1 y = y y 1 + 1 1 2 1 y = y y 1 + 2 y 1 = y 2 y 1 44. a a b + b b a = a a b + b a b = a b a b = 1 45. x 2x 3y y 3y 2x = x 2x 3y 1 1 y 3y 2x = x 2x 3y y 2x 3y = x + y 2x 3y [x (y) = x + y] 46. 3a 3a 2b 2a 2b 3a = 3a 3a 2b 2a 3a 2b = 5a 3a 2b 47. 9x + 2 3x2 2x 8 + 7 3x2 + x 4 = 9x + 2 (3x + 4)(x 2) + 7 (3x + 4)(x 1) , LCD is (3x + 4)(x 2)(x 1) = 9x + 2 (3x+4)(x2) x 1 x 1 + 7 (3x+4)(x1) x 2 x 2 = 9x2 7x 2 (3x+4)(x2)(x1) + 7x 14 (3x+4)(x1)(x2) = 9x2 16 (3x + 4)(x 2)(x 1) = (3x+4)$(3x 4) (3x+4)$(x 2)(x 1) = 3x 4 (x 2)(x 1) 48. 3y y2 7y + 10 2y y2 8y + 15 = 3y (y 2)(y 5) 2y (y 5)(y 3) , LCD is (y 2)(y 5)(y 3) = 3y(y 3) 2y(y 2) (y 2)(y 5)(y 3) = 3y2 9y 2y2 + 4y (y 2)(y 5)(y 3) = y2 5y (y 2)(y 5)(y 3) = y(y5) (y 2)(y5) (y 3) = y (y 2)(y 3) 49. 5a a b + ab a2 b2 + 4b a + b = 5a a b + ab (a + b)(a b) + 4b a + b , LCD is (a + b)(a b) = 5a a b a + b a + b + ab (a + b)(a b) + 4b a + b a b a b = 5a2 + 5ab (a+b)(ab) + ab (a+b)(ab) + 4ab 4b2 (a+b)(ab) = 5a2 + 10ab 4b2 (a + b)(a b)
22. 22. Exercise Set R.5 19 50. 6a a b + 3b b a + 5 a2 b2 = 6a a b + 3b a b + 5 (a + b)(a b) , LCD is (a + b)(a b) = 6a(a + b) + 3b(a + b) + 5 (a + b)(a b) = 6a2 + 6ab + 3ab + 3b2 + 5 (a + b)(a b) = 6a2 + 9ab + 3b2 + 5 (a + b)(a b) 51. 7 x + 2 x + 8 4 x2 + 3x 2 4 4x + x2 = 7 x + 2 x + 8 (2 + x)(2 x) + 3x 2 (2 x)2 , LCD is (2 + x)(2 x)2 = 7 2 + x (2 x)2 (2 x)2 x + 8 (2 + x)(2 x) 2 x 2 x + 3x 2 (2 x)2 2 + x 2 + x = 2828x+7x2 (166xx2 )+3x2 +4x4 (2 + x)(2 x)2 = 28 28x + 7x2 16 + 6x + x2 + 3x2 + 4x 4 (2 + x)(2 x)2 = 11x2 18x + 8 (2 + x)(2 x)2 , or 11x2 18x + 8 (x + 2)(x 2)2 52. 6 x + 3 x + 4 9 x2 + 2x 3 9 6x + x2 = 6 x + 3 x + 4 (3 + x)(3 x) + 2x 3 (3 x)2 , LCD is (3 + x)(3 x)2 = 6(3 x)2 (x + 4)(3 x) + (2x 3)(3 + x) (3 + x)(3 x)2 = 54 36x + 6x2 + x2 + x 12 + 2x2 + 3x 9 (3 + x)(3 x)2 = 33 32x + 9x2 (3 + x)(3 x)2 , or 9x2 32x + 33 (x + 3)(x 3)2 53. 1 x + 1 + x 2 x + x2 + 2 x2 x 2 = 1 x + 1 + x 2 x + x2 + 2 (x + 1)(x 2) = 1 x + 1 + 1 1 x 2 x + x2 + 2 (x + 1)(x 2) = 1 x + 1 + x x 2 + x2 + 2 (x + 1)(x 2) , LCD is (x + 1)(x 2) = 1 x + 1 x 2 x 2 + x x 2 x + 1 x + 1 + x2 + 2 (x + 1)(x 2) = x 2 (x+1)(x2) + x2 x (x+1)(x2) + x2 + 2 (x+1)(x2) = x 2 x2 x + x2 + 2 (x + 1)(x 2) = 0 (x + 1)(x 2) = 0 54. x 1 x 2 x + 1 x + 2 x 6 4 x2 = x 1 x 2 x + 1 x + 2 x 6 (2 + x)(2 x) = 1 x 2 x x + 1 x + 2 x 6 (2 + x)(2 x) , LCD is (2 + x)(2 x) = (1 x)(2 + x) (x + 1)(2 x) (x 6) (2 + x)(2 x) = 2 x x2 + x2 x 2 x + 6 (2 + x)(2 x) = 6 3x (2 + x)(2 x) = 3(2x) (2 + x)(2x) = 3 2 + x 55. x2 y2 xy x y y = x2 y2 xy y x y = (x + y)(xy) y4 x y4 (xy) = x + y x
23. 23. 20 Chapter R: Basic Concepts of Algebra 56. a b b a2 b2 ab = a b b ab (a + b)(a b) = a b4(ab) b/ (a + b)(ab) = a a + b 57. x y y x 1 y + 1 x = x y y x 1 y + 1 x xy xy , LCM is xy = x y y x (xy) 1 y + 1 x (xy) = x2 y2 x + y = (x+y) (x y) (x+y) 1 = x y 58. a b b a 1 a 1 b = a2 b2 b a Multiplying by ab ab = (a + b)(a b) b a = (a + b)(ab) 1 (ab) = a b 59. c + 8 c2 1 + 2 c = c c2 c2 + 8 c2 1 c c + 2 c = c3 + 8 c2 c + 2 c = c3 + 8 c2 c c + 2 = (c+2) (c2 2c + 4)c/ c/ c(c+2) = c2 2c + 4 c 60. a a b b b a = ab a b ab b a = a(b 1) b a b(a 1) = a2 (b 1) b2(a 1) 61. x2 + xy + y2 x2 y y2 x = x2 + xy + y2 x2 y x x y2 x y y = x2 + xy + y2 x3 y3 xy = (x2 + xy + y2 ) xy x3 y3 = (x2 + xy + y2 )(xy) (x y)(x2 + xy + y2) = x2 + xy + y2 x2 + xy + y2 xy x y = 1 xy x y = xy x y 62. a2 b + b2 a a2 ab + b2 = a3 + b3 ab a2 ab + b2 = (a+b)(a2 ab+b2 ) ab 1 a2 ab + b2 = a + b ab a2 ab + b2 a2 ab + b2 = a + b ab 63. a a1 a + a1 = a 1 a a + 1 a = a a a 1 a a a a + 1 a = a2 1 a a2 + 1 a = a2 1 a a a2 + 1 = a2 1 a2 + 1 64. x1 + y1 x3 + y3 = 1 x + 1 y 1 x3 + 1 y3 = 1 x + 1 y (x3 y3 ) 1 x3 + 1 y3 (x3 y3 ) = x2 y3 + x3 y2 y3 + x3 = x2 y2 (y+x) (y+x) (y2 yx + x2) = x2 y2 y2 yx + x2
24. 24. Exercise Set R.5 21 65. 1 x 3 + 2 x + 3 3 x 1 4 x + 2 = 1 x 3 x + 3 x + 3 + 2 x + 3 x 3 x 3 3 x 1 x + 2 x + 2 4 x + 2 x 1 x 1 = x + 3 + 2(x 3) (x 3)(x + 3) 3(x + 2) 4(x 1) (x 1)(x + 2) = x + 3 + 2x 6 (x 3)(x + 3) 3x + 6 4x + 4 (x 1)(x + 2) = 3x 3 (x 3)(x + 3) x + 10 (x 1)(x + 2) = 3x 3 (x 3)(x + 3) (x 1)(x + 2) x + 10 = (3x 3)(x 1)(x + 2) (x 3)(x + 3)(x + 10) , or 3(x 1)2 (x + 2) (x 3)(x + 3)(x + 10) 66. 5 x + 1 3 x 2 1 x 5 + 2 x + 2 = 5(x 2) 3(x + 1) (x + 1)(x 2) x + 2 + 2(x 5) (x 5)(x + 2) = 5x 10 3x 3 (x + 1)(x 2) x + 2 + 2x 10 (x 5)(x + 2) = 2x 13 (x + 1)(x 2) 3x 8 (x 5)(x + 2) = 2x 13 (x + 1)(x 2) (x 5)(x + 2) 3x 8 = (2x 13)(x 5)(x + 2) (x + 1)(x 2)(3x 8) 67. a 1 a + 1 + a a 1 a a + a 1 + a = a 1 a a a + 1 + a a 1 a 1 a 1 a a 1 + a 1 + a + a 1 + a a a = a2 + (1 a2 ) a(1 a) (1 a2 ) + a2 a(1 + a) = 1 a/(1 a) a/(1 + a) 1 = 1 + a 1 a 68. 1 x x + x 1 + x 1 + x x + x 1 x = 1 x2 + x2 x(1 + x) 1 x2 + x2 x(1 x) = 1 x(1 + x) x(1 x) 1 = x/(1 x) x/(1 + x) = 1 x 1 + x 69. 1 a2 + 2 ab + 1 b2 1 a2 1 b2 = 1 a2 + 2 ab + 1 b2 1 a2 1 b2 a2 b2 a2b2 , LCM is a2 b2 = b2 + 2ab + a2 b2 a2 = (b+a) (b + a) (b+a) (b a) = b + a b a 70. 1 x2 1 y2 1 x2 2 xy + 1 y2 = y2 x2 y2 2xy + x2 Multiplying by x2 y2 x2y2 = (y + x)(yx) (y x)(yx) = y + x y x 71. When the least common denominator is used, the multi- plication in the numerators is often simpler and there is usually less simplication required after the addition or subtraction is performed. 72. When there are three or more dierent binomial denomi- nators, as in Exercise 65, Method 2 is usually preferable. Otherwise, some might prefer Method 1 while others will prefer Method 2. 73. (x + h)2 x2 h = x2 + 2xh + h2 x2 h = 2xh + h2 h = h/(2x + h) h/ 1 = 2x + h
25. 25. 22 Chapter R: Basic Concepts of Algebra 74. 1 x + h 1 x h = x x h x(x + h) h = h x(x + h) 1 h = 1 h/ xh/(x + h) = 1 x(x + h) 75. (x + h)3 x3 h = x3 + 3x2 h + 3xh2 + h3 x3 h = 3x2 h + 3xh2 + h3 h = h/(3x2 + 3xh + h2 ) h/ 1 = 3x2 + 3xh + h2 76. 1 (x + h)2 1 x2 h = x2 x2 2xh h2 x2(x + h)2 h = 2xh h2 x2(x + h)2 1 h = h/(2x h) x2h/(x + h)2 = 2x h x2(x + h)2 77. x + 1 x 1 + 1 x + 1 x 1 1 5 = (x + 1) + (x 1) x 1 (x + 1) (x 1) x 1 5 = 2x x 1 x 1 2 5 = 2/x(x1)3 1 2/(x1)3 5 = x5 78. 1 + 1 1 + 1 1 + 1 1 + 1 x = 1 + 1 1 + 1 1 + 1 x + 1 x = 1 + 1 1 + 1 1 + x x + 1 = 1 + 1 1 + 1 2x + 1 x + 1 = 1 + 1 1 + x + 1 2x + 1 = 1 + 1 3x + 2 2x + 1 = 1 + 2x + 1 3x + 2 = 5x + 3 3x + 2 79. n(n + 1)(n + 2) 2 3 + (n + 1)(n + 2) 2 = n(n + 1)(n + 2) 2 3 + (n + 1)(n + 2) 2 3 3 , LCD is 2 3 = n(n + 1)(n + 2) + 3(n + 1)(n + 2) 2 3 = (n + 1)(n + 2)(n + 3) 2 3 Factoring the num- erator by grouping 80. n(n+1)(n+2)(n+3) 2 3 4 + (n+1)(n+2)(n+3) 2 3 = n(n+1)(n+2)(n+3) + 4(n+1)(n+2)(n+3) 2 3 4 , LCD is 2 3 4 = (n + 1)(n + 2)(n + 3)(n + 4) 2 3 4
26. 26. Exercise Set R.6 23 81. x2 9 x3 + 27 5x2 15x + 45 x2 2x 3 + x2 + x 4 + 2x = (x + 3)(x 3)(5)(x2 3x + 9) (x + 3)(x2 3x + 9)(x 3)(x + 1) + x2 + x 4 + 2x = (x + 3)(x 3)(x2 3x + 9) (x + 3)(x 3)(x2 3x + 9) 5 x + 1 + x2 + x 4 + 2x = 1 5 x + 1 + x2 + x 4 + 2x = 5 x + 1 + x2 + x 2(2 + x) = 5 2(2 + x) + (x2 + x)(x + 1) 2(x + 1)(2 + x) = 20 + 10x + x3 + 2x2 + x 2(x + 1)(2 + x) = x3 + 2x2 + 11x + 20 2(x + 1)(2 + x) 82. x2 + 2x 3 x2 x 12 x2 1 x2 16 2x + 1 x2 + 2x + 1 = x2 + 2x 3 x2 x 12 x2 16 x2 1 2x + 1 x2 + 2x + 1 = (x+3)3 (x1)3 (x + 4)(x4)3 (x4)3 (x+3)3 (x + 1)(x1)3 2x + 1 x2 + 2x + 1 = x + 4 x + 1 2x + 1 (x + 1)(x + 1) = (x + 4)(x + 1) (2x + 1) (x + 1)(x + 1) = x2 + 5x + 4 2x 1 (x + 1)(x + 1) = x2 + 3x + 3 (x + 1)2 Exercise Set R.6 1. (11)2 = | 11| = 11 2. (1)2 = | 1| = 1 3. 16y2 = (4y)2 = |4y| = 4|y| 4. 36t2 = |6t| = 6|t| 5. (b + 1)2 = |b + 1| 6. (2c 3)2 = |2c 3| 7. 3 27x3 = 3 (3x)3 = 3x 8. 3 8y3 = 2y 9. 4 81x8 = 4 (3x2)4 = |3x2 | = 3x2 10. 4 16z12 = |2z3 | = 2|z3 | = 2z2 |z| 11. 5 32 = 5 25 = 2 12. 5 32 = 2 13. 180 = 36 5 = 36 5 = 6 5 14. 48 = 16 3 = 4 3 15. 72 = 36 2 = 36 2 = 6 2 16. 250 = 25 10 = 5 10 17. 3 54 = 3 27 2 = 3 27 3 2 = 3 3 2 18. 3 135 = 3 27 5 = 3 3 5 19. 128c2d4 = 64c2d4 2 = |8cd2 | 2 = 8 2 |c|d2 20. 162c4d6 = 81c4 d6 2 = 9c2 |d3 | 2 = 9 2c2 d2 |d| 21. 4 48x6y4 = 4 16x4y4 3x2 = |2xy| 4 3x2 = 2|x||y| 4 3x2 22. 4 243m5n10 = 4 81m4n8 3mn2 = 3|m|n2 4 3mn2 23. x2 4x + 4 = (x 2)2 = |x 2| 24. x2 + 16x + 64 = (x + 8)2 = |x + 8| 25. 10 30 = 10 30 = 300 = 100 3 = 100 3 = 10 3 26. 28 14 = 28 14 = 14 2 14 = 14 2 27. 12 33 = 12 33 = 3 4 3 11 = 32 4 11 = 3 2 11 = 6 11 28. 15 35 = 15 35 = 3 5 5 7 = 5 21 29. 2x3y 12xy = 24x4y2 = 4x4y2 6 = 2x2 y 6 30. 3y4z 20z = 60y4z2 = 4y4z2 15 = 2y2 z 15 31. 3 3x2y 3 36x = 3 108x3y = 3 27x3 4y = 3x 3 4y 32. 5 8x3y4 5 4x4y = 5 32x7y5 = 2xy 5 x2 33. 3 2(x + 4) 3 4(x + 4)4 = 3 8(x + 4)5 = 3 8(x + 4)3 (x + 4)2 = 2(x + 4) 3 (x + 4)2 34. 3 4(x + 1)2 3 18(x + 1)2 = 3 72(x + 1)4 = 3 8(x + 1)3 9(x + 1) = 2(x + 1) 3 9(x + 1) 35. 6 m12n24 64 = 6 m2n4 2 6 = m2 n4 2 36. 8 m16n24 28 = m2 n3 2
27. 27. 24 Chapter R: Basic Concepts of Algebra 37. 3 40m 3 5m = 3 40m 5m = 3 8 = 2 38. 40xy 8x = 40xy 8x = 5y 39. 3 3x2 3 24x5 = 3 3x2 24x5 = 3 1 8x3 = 1 2x 40. 128a2b4 16ab = 128a2b4 16ab = 8ab3 = 4 2 a b2 b = 2b 2ab 41. 3 64a4 27b3 = 3 64 a3 a 27 b3 = 3 64a3 3 a 3 27b3 = 4a 3 a 3b 42. 9x7 16y8 = 9 x6 x 16 y8 = 3x3 x 4y4 43. 7x3 36y6 = 7 x2 x 36 y6 = x2 7x 36y6 = x 7x 6y3 44. 3 2yz 250z4 = 3 y 125z3 = 3 y 3 125z3 = 3 y 5z 45. 9 50 + 6 2 = 9 25 2 + 6 2 = 9 5 2 + 6 2 = 45 2 + 6 2 = (45 + 6) 2 = 51 2 46. 11 27 4 3 = 11 3 3 4 3 = 33 3 4 3 = 29 3 47. 6 20 4 45 + 80 = 6 4 5 4 9 5 + 16 5 = 6 2 5 4 3 5 + 4 5 = 12 5 12 5 + 4 5 = (12 12 + 4) 5 = 4 5 48. 2 32 + 3 8 4 18 = 2 4 2 + 3 2 2 4 3 2 = 8 2 + 6 2 12 2 = 2 2 49. 8 2x2 6 20x 5 8x2 = 8x 2 6 4 5x 5 4x2 2 = 8x 2 6 2 5x 5 2x 2 = 8x 2 12 5x 10x 2 = 2x 2 12 5x 50. 2 3 8x2 + 5 3 27x2 3 x3 = 4 3 x2 + 15 3 x2 3x x = 19 3 x2 3x x 51. 3 2 3 + 2 = 3 2 2 2 = 3 2 = 1 52. ( 8 + 2 5)( 8 2 5) = 8 4 5 = 12 53. (2 3 + 5)( 3 3 5) = 2 3 3 2 3 3 5 + 5 3 5 3 5 = 2 3 6 15 + 15 3 5 = 6 6 15 + 15 15 = 9 5 15 54. ( 6 4 7)(3 6 + 2 7) = 3 6 + 2 42 12 42 8 7 = 18 + 2 42 12 42 56 = 38 10 42 55. (1 + 3)2 = 12 + 2 1 3 + ( 3)2 = 1 + 2 3 + 3 = 4 + 2 3 56. ( 2 5)2 = 2 10 2 + 25 = 27 10 2 57. ( 5 6)2 = ( 5)2 2 5 6 + ( 6)2 = 5 2 30 + 6 = 11 2 30 58. ( 3 + 2)2 = 3 + 2 6 + 2 = 5 + 2 6 59. We use the Pythagorean theorem to nd b, the airplanes horizontal distance from the airport. We have a = 3700 and c = 14, 200. c2 = a2 + b2 14, 2002 = 37002 + b2 201, 640, 000 = 13, 690, 000 + b2 187, 950, 000 = b2 13, 709.5 b The airplane is about 13, 709.5 ft horizontally from the airport.
28. 28. Exercise Set R.6 25 60. @@@@@@@@@@@@ hhhhhhhhhhhh a b E' c 2 mi = 2 5280 ft = 10, 560 ft 2 mi + 2 ft = 10, 562 ft Then b = 1 2 10, 560 ft = 5280 ft and c = 1 2 10, 562 ft = 5281 ft. a2 = c2 b2 a2 = (5281)2 (5280)2 a2 = 10, 561 a 102.8 ft 61. a) h2 + a 2 2 = a2 Pythagorean theorem h2 + a2 4 = a2 h2 = 3a2 4 h = 3a2 4 h = a 2 3 b) Using the result of part (a) we have A = 1 2 base height A = 1 2 a a 2 3 a 2 + a 2 = a A = a2 4 3 62. c2 = s2 + s2 c2 = 2s2 c = s 2 Length of third side 63. x x x x8 2 d d d d d d d x2 + x2 = (8 2)2 Pythagorean theorem 2x2 = 128 x2 = 64 x = 8 64. d d d d d d d d d d d d S x D x P x B x R x C x Q x A x y y y y (2x)2 = 100 4x2 = 100 x2 = 25 x = 5 A = y2 = x2 + x2 = 52 + 52 = 25 + 25 = 50 ft2 65. 2 3 = 2 3 3 3 = 6 9 = 6 9 = 6 3 66. 3 7 = 3 7 7 7 = 21 49 = 21 7 67. 3 5 3 4 = 3 5 3 4 3 2 3 2 = 3 10 3 8 = 3 10 2 68. 3 7 3 25 = 3 7 3 25 3 5 3 5 = 3 35 3 125 = 3 35 5 69. 3 16 9 = 3 16 9 3 3 = 3 48 27 = 3 48 3 27 = 3 8 6 3 = 2 3 6 3 70. 3 3 5 = 3 3 5 25 25 = 3 75 125 = 3 75 5 71. 6 3 + 5 = 6 3 + 5 3 5 3 5 = 6 3 5 9 5 = 6 3 5 4 = 3 3 5 2 = 9 3 5 2 72. 2 3 1 = 2 3 1 3 + 1 3 + 1 = 2/( 3 + 1) 2/ 1 = 3 + 1
29. 29. 26 Chapter R: Basic Concepts of Algebra 73. 1 2 2 3 6 = 1 2 2 3 6 2 3 + 6 2 3 + 6 = 2 3 + 6 2 6 12 4 3 6 = 2 3 + 6 2 6 2 3 12 6 = 6 6 , or 6 6 74. 5 + 4 2 + 3 7 = 5 + 4 2 + 3 7 2 3 7 2 3 7 = 10 3 35 + 4 2 12 7 2 9 7 = 10 3 35 + 4 2 12 7 61 75. 6 m n = 6 m n m + n m + n = 6( m + n) ( m)2 ( n)2 = 6 m + 6 n m n 76. 3 v + w = 3 v + w v w v w = 3 v 3 w v w 77. 12 5 = 12 5 3 3 = 36 5 3 = 6 5 3 78. 50 3 = 50 3 2 2 = 100 3 2 = 10 3 2 79. 3 7 2 = 3 7 2 49 49 = 3 343 98 = 3 343 3 98 = 7 3 98 80. 3 2 5 = 3 2 5 4 4 = 3 8 20 = 2 3 20 81. 11 3 = 11 3 11 11 = 121 33 = 11 33 82. 5 2 = 5 2 5 5 = 25 10 = 5 10 83. 9 5 3 3 = 9 5 3 3 9 + 5 9 + 5 = 92 ( 5)2 27 + 3 5 9 3 15 = 81 5 27 + 3 5 9 3 15 = 76 27 + 3 5 9 3 15 84. 8 6 5 2 = 8 6 5 2 8 + 6 8 + 6 = 64 6 40 + 5 6 8 2 12 = 58 40 + 5 6 8 2 2 3 85. a + b 3a = a + b 3a a b a b = ( a)2 ( b)2 3a( a b) = a b 3a a 3a b 86. p q 1 + q = p q 1 + q p + q p + q = p q p + q + pq + q 87. x3/4 = 4 x3 88. y2/5 = 5 y2 89. 163/4 = (161/4 )3 = 4 16 3 = 23 = 8 90. 47/2 = ( 4)7 = 27 = 128 91. 1251/3 = 1 1251/3 = 1 3 125 = 1 5 92. 324/5 = 5 32 4 = 24 = 1 16 93. a5/4 b3/4 = a5/4 b3/4 = 4 a5 4 b3 = a 4 a 4 b3 , or a 4 a b3 94. x2/5 y1/5 = 5 x2 y 95. m5/3 n7/3 = 3 m5 3 n7 = 3 m5n7 = mn2 3 m2n 96. p7/6 q11/6 = 6 p7 6 q11 = 6 p7q11 = pq 6 pq5 97. 4 13 5 = 4 135 = 135/4 98. 5 173 = 173/5 99. 3 202 = 202/3 100. 5 12 4 = 124/5 101. 3 11 = 11 1/3 = (111/2 )1/3 = 111/6 102. 3 4 7 = (71/4 )1/3 = 71/12 103. 5 3 5 = 51/2 51/3 = 51/2+1/3 = 55/6 104. 3 2 2 = 21/3 21/2 = 25/6 105. 5 322 = 322/5 = (321/5 )2 = 22 = 4 106. 3 642 = 642/3 = (641/3 )2 = 42 = 16 107. (2a3/2 )(4a1/2 ) = 8a3/2+1/2 = 8a2 108. (3a5/6 )(8a2/3 ) = 24a3/2 = 24a a 109. x6 9b4 1/2 = x6 32b4 1/2 = x3 3b2 , or x3 b2 3
30. 30. Exercise Set R.7 27 110. x2/3 4y2 1/2 = x1/3 41/2y1 = 3 x 2y1 , or y 3 x 2 111. x2/3 y5/6 x1/3y1/2 = x2/3(1/3) y5/61/2 = xy1/3 = x 3 y 112. a1/2 b5/8 a1/4b3/8 = a1/4 b1/4 = 4 ab 113. (m1/2 n5/2 )2/3 = m 1 2 2 3 n 5 2 2 3 = m1/3 n5/3 = 3 m 3 n5 = 3 mn5 = n 3 mn2 114. (x5/3 y1/3 z2/3 )3/5 = xy1/5 z2/5 = x 5 yz2 115. a3/4 (a2/3 + a4/3 ) = a3/4+2/3 + a3/4+4/3 = a17/12 + a25/12 = 12 a17 + 12 a25 = a 12 a5 + a2 12 a 116. m2/3 (m7/4 m5/4 ) = m29/12 m23/12 = 12 m29 12 m23 = m2 12 m5 m 12 m11 117. 3 6 2 = 61/3 21/2 = 62/6 23/6 = (62 23 )1/6 = 6 36 8 = 6 288 118. 2 4 8 = 21/2 (23 )1/4 = 21/2 23/4 = 25/4 = 4 25 = 2 4 2 119. 4 xy 3 x2y = (xy)1/4 (x2 y)1/3 = (xy)3/12 (x2 y)4/12 = (xy)3 (x2 y)4 1/12 = x3 y3 x8 y4 1/12 = 12 x11y7 120. 3 ab2 ab = (ab2 )1/3 (ab)1/2 = (ab2 )2/6 (ab)3/6 = 6 (ab2)2(ab)3 = 6 a5b7 = b 6 a5b 121. 3 a4 a3 = a4 a3 1/3 = (a4 a3/2 )1/3 = (a11/2 )1/3 = a11/6 = 6 a11 = a 6 a5 122. a3 3 a2 = (a3 a2/3 )1/2 = (a11/3 )1/2 = a11/6 = 6 a11 = a 6 a5 123. (a + x)3 3 (a + x)2 4 a + x = (a + x)3/2 (a + x)2/3 (a + x)1/4 = (a + x)26/12 (a + x)3/12 = (a + x)23/12 = 12 (a + x)23 = (a + x) 12 (a + x)11 124. 4 (x + y)2 3 x + y (x + y)3 = (x + y)2/4 (x + y)1/3 (x + y)3/2 = (x + y)2/3 = 1 3 (x + y)2 125. Choose specic values for a and b (a = 0, b = 0) and show that a + b = a+ b. For example let a = 9 and b = 16. Then 9 + 16 = 25 = 5, but 9 + 16 = 3 + 4 = 7. 126. Observe that 2625, so 26 25, or 265. Then 10 26 5010 5 50, or 10 26 500. 127. 1 + x2 + 1 1 + x2 = 1 + x2 1 + x2 1 + x2 + 1 1 + x2 1 + x2 1 + x2 = (1 + x2 ) 1 + x2 1 + x2 + 1 + x2 1 + x2 = (2 + x2 ) 1 + x2 1 + x2 128. 1 x2 x2 2 1 x2 = 1 x2 x2 1 x2 2(1 x2) Rationalizing the denominator = 2(1 x2 ) 1 x2 x2 1 x2 2(1 x2) = (2 3x2 ) 1 x2 2(1 x2) 129. a a a = a a/2 a = aa/2 130. (2a3 b5/4 c1/7 )4 (54a2b2/3c6/5)1/3 = 16a12 b5 c4/7 541/3a2/3b2/9c2/5 = 16 3 54a34/3 b47/9 c34/35 = 24 3 21/3 a34/3 b47/9 c34/35 = 3 213/3 a34/3 b47/9 c34/35 , or 48 21/3 a34/3 b47/9 c34/35 Exercise Set R.7 1. 6x 15 = 45 6x = 60 Adding 15 x = 10 Dividing by 6 The solution is 10. 2. 4x 7 = 81 4x = 88 x = 22
31. 31. 28 Chapter R: Basic Concepts of Algebra 3. 5x 10 = 45 5x = 55 Adding 10 x = 11 Dividing by 5 The solution is 11. 4. 6x 7 = 11 6x = 18 x = 3 5. 9t + 4 = 5 9t = 9 Subtracting 4 t = 1 Dividing by 9 The solution is 1. 6. 5x + 7 = 13 5x = 20 x = 4 7. 8x + 48 = 3x 12 5x + 48 = 12 Subtracting 3x 5x = 60 Subtracting 48 x = 12 Dividing by 5 The solution is 12. 8. 15x + 40 = 8x 9 7x = 49 x = 7 9. 7y 1 = 23 5y 12y 1 = 23 Adding 5y 12y = 24 Adding 1 y = 2 Dividing by 12 The solution is 2. 10. 3x 15 = 15 3x 6x = 30 x = 5 11. 3x 4 = 5 + 12x 9x 4 = 5 Subtracting 12x 9x = 9 Adding 4 x = 1 Dividing by 9 The solution is 1. 12. 9t 4 = 14 + 15t 6t = 18 t = 3 13. 5 4a = a 13 5 5a = 13 Subtracting a 5a = 18 Subtracting 5 a = 18 5 Dividing by 5 The solution is 18 5 . 14. 6 7x = x 14 8x = 20 x = 5 2 15. 3m 7 = 13 + m 2m 7 = 13 Subtracting m 2m = 6 Adding 7 m = 3 Dividing by 2 The solution is 3. 16. 5x 8 = 2x 8 3x = 0 x = 0 17. 11 3x = 5x + 3 11 8x = 3 Subtracting 5x 8x = 8 Subtracting 11 x = 1 The solution is 1. 18. 20 4y = 10 6y 2y = 10 y = 5 19. 2(x + 7) = 5x + 14 2x + 14 = 5x + 14 3x + 14 = 14 Subtracting 5x 3x = 0 Subtracting 14 x = 0 The solution is 0. 20. 3(y + 4) = 8y 3y + 12 = 8y 12 = 5y 12 5 = y 21. 24 = 5(2t + 5) 24 = 10t + 25 1 = 10t Subtracting 25 1 10 = t Dividing by 10 The solution is 1 10 . 22. 9 = 4(3y 2) 9 = 12y 8 17 = 12y 17 12 = y
32. 32. Exercise Set R.7 29 23. 5y (2y 10) = 25 5y 2y + 10 = 25 3y + 10 = 25 Collecting like terms 3y = 15 Subtracting 10 y = 5 Dividing by 3 The solution is 5. 24. 8x (3x 5) = 40 8x 3x + 5 = 40 5x + 5 = 40 5x = 35 x = 7 25. 7(3x + 6) = 11 (x + 2) 21x + 42 = 11 x 2 21x + 42 = 9 x Collecting like terms 22x + 42 = 9 Adding x 22x = 33 Subtracting 42 x = 3 2 Dividing by 22 The solution is 3 2 . 26. 9(2x + 8) = 20 (x + 5) 18x + 72 = 20 x 5 18x + 72 = 15 x 19x = 57 x = 3 27. 4(3y 1) 6 = 5(y + 2) 12y 4 6 = 5y + 10 12y 10 = 5y + 10 Collecting like terms 7y 10 = 10 Subtracting 5y 7y = 20 Adding 10 y = 20 7 Dividing by 7 The solution is 20 7 . 28. 3(2n 5) 7 = 4(n 9) 6n 15 7 = 4n 36 6n 22 = 4n 36 2n = 14 n = 7 29. x2 + 3x 28 = 0 (x + 7)(x 4) = 0 Factoring x + 7 = 0 or x 4 = 0 Principle of zero products x = 7 or x = 4 The solutions are 7 and 4. 30. y2 4y 45 = 0 (y 9)(y + 5) = 0 y 9 = 0 or y + 5 = 0 y = 9 or y = 5 31. x2 8x = 0 x(x 8) = 0 Factoring x = 0 or x 8 = 0 Principle of zero products x = 0 or x = 8 The solutions are 0 and 8. 32. t2 9t = 0 t(t 9) = 0 t = 0 or t 9 = 0 t = 0 or t = 9 33. y2 + 6y + 9 = 0 (y + 3)(y + 3) = 0 y + 3 = 0 or y + 3 = 0 y = 3 or y = 3 The solution is 3. 34. n2 + 4n + 4 = 0 (n + 2)(n + 2) = 0 n + 2 = 0 or n + 2 = 0 n = 2 or n = 2 35. x2 + 100 = 20x x2 20x + 100 = 0 Subtracting 20x (x 10)(x 10) = 0 x 10 = 0 or x 10 = 0 x = 10 or x = 10 The solution is 10. 36. y2 + 25 = 10y y2 10y + 25 = 0 (y 5)(y 5) = 0 y 5 = 0 or y 5 = 0 y = 5 or y = 5 37. x2 4x 32 = 0 (x 8)(x + 4) = 0 x 8 = 0 or x + 4 = 0 x = 8 or x = 4 The solutions are 8 and 4. 38. t2 + 12t + 27 = 0 (t + 9)(t + 3) = 0 t + 9 = 0 or t + 3 = 0 t = 9 or t = 3
33. 33. 30 Chapter R: Basic Concepts of Algebra 39. 3y2 + 8y + 4 = 0 (3y + 2)(y + 2) = 0 3y + 2 = 0 or y + 2 = 0 3y = 2 or y = 2 y = 2 3 or y = 2 The solutions are 2 3 and 2. 40. 9y2 + 15y + 4 = 0 (3y + 4)(3y + 1) = 0 3y + 4 = 0 or 3y + 1 = 0 3y = 4 or 3y = 1 y = 4 3 or y = 1 3 41. 12z2 + z = 6 12z2 + z 6 = 0 (4z + 3)(3z 2) = 0 4z + 3 = 0 or 3z 2 = 0 4z = 3 or 3z = 2 z = 3 4 or z = 2 3 The solutions are 3 4 and 2 3 . 42. 6x2 7x = 10 6x2 7x 10 = 0 (6x + 5)(x 2) = 0 6x + 5 = 0 or x 2 = 0 6x = 5 or x = 2 x = 5 6 or x = 2 43. 12a2 28 = 5a 12a2 5a 28 = 0 (3a + 4)(4a 7) = 0 3a + 4 = 0 or 4a 7 = 0 3a = 4 or 4a = 7 a = 4 3 or a = 7 4 The solutions are 4 3 and 7 4 . 44. 21n2 10 = n 21n2 n 10 = 0 (3n + 2)(7n 5) = 0 3n + 2 = 0 or 7n 5 = 0 3n = 2 or 7n = 5 n = 2 3 or n = 5 7 45. 14 = x(x 5) 14 = x2 5x 0 = x2 5x 14 0 = (x 7)(x + 2) x 7 = 0 or x + 2 = 0 x = 7 or x = 2 The solutions are 7 and 2. 46. 24 = x(x 2) 24 = x2 2x 0 = x2 2x 24 0 = (x 6)(x + 4) x 6 = 0 or x + 4 = 0 x = 6 or x = 4 47. x2 36 = 0 (x + 6)(x 6) = 0 x + 6 = 0 or x 6 = 0 x = 6 or x = 6 The solutions are 6 and 6. 48. y2 81 = 0 (y + 9)(y 9) = 0 y + 9 = 0 or y 9 = 0 y = 9 or y = 9 49. z2 = 144 z2 144 = 0 (z + 12)(z 12) = 0 z + 12 = 0 or z 12 = 0 z = 12 or z = 12 The solutions are 12 and 12. 50. t2 = 25 t2 25 = 0 (t + 5)(t 5) = 0 t + 5 = 0 or t 5 = 0 t = 5 or t = 5 51. 2x2 20 = 0 2x2 = 0 x2 = 10 x = 10 or x = 10 Principle of square roots The solutions are 10 and 10, or 10. 52. 3y2 15 = 0 3y2 = 15 y2 = 5 y = 5 or y = 5
34. 34. Exercise Set R.7 31 53. 6z2 18 = 0 6z2 = 18 z2 = 3 z = 3 or z = 3 The solutions are 3 and 3, or 3. 54. 5x2 75 = 0 5x2 = 75 x2 = 15 x = 15 or x = 15 55. To eliminate a term from one side of an equation, add the opposite of the term. To eliminate a coecient, multiply by its reciprocal. 56. If the solutions of a quadratic equation are 3 and 4, then we have: x = 3 or x = 4 x + 3 = 0 or x 4 = 0. Then the equation can be written (x + 3)(x 4) = 0, or x2 x 12 = 0. 57. 3[5 3(4 t)] 2 = 5[3(5t 4) + 8] 26 3[5 12 + 3t] 2 = 5[15t 12 + 8] 26 3[7 + 3t] 2 = 5[15t 4] 26 21 + 9t 2 = 75t 20 26 9t 23 = 75t 46 66t 23 = 46 66t = 23 t = 23 66 The solution is 23 66 . 58. 6[4(8 y) 5(9 + 3y)] 21 = 7[3(7 + 4y) 4] 6[32 4y 45 15y] 21 = 7[21 + 12y 4] 6[13 19y] 21 = 7[17 + 12y] 78 114y 21 = 119 84y 114y 99 = 119 84y 30y = 20 y = 2 3 59. x {3x [2x (5x (7x 1))]} = x + 7 x {3x [2x (5x 7x + 1)]} = x + 7 x {3x [2x (2x + 1)]} = x + 7 x {3x [2x + 2x 1]} = x + 7 x {3x [4x 1]} = x + 7 x {3x 4x + 1} = x + 7 x {x + 1} = x + 7 x + x 1 = x + 7 2x 1 = x + 7 x 1 = 7 x = 8 The solution is 8. 60. 232[4+3(x1)]+5[x2(x+3)]=7{x2[5(2x+3)]} 23 2[4 + 3x 3] + 5[x 2x 6]=7{x 2[5 2x 3]} 23 2[3x + 1] + 5[x 6]=7{x 2[2x + 2]} 23 6x 2 5x 30=7{x + 4x 4} 11x 9=7{5x 4} 11x 9=35x 28 46x=19 x= 19 46 61. (5x2 + 6x)(12x2 5x 2) = 0 x(5x + 6)(4x + 1)(3x 2) = 0 x = 0 or 5x+6 = 0 or 4x+1 = 0 or 3x2 = 0 x = 0 or 5x = 6 or 4x = 1 or 3x = 2 x = 0 or x = 6 5 or x = 1 4 or x = 2 3 The solutions are 0, 6 5 , 1 4 , and 2 3 . 62. (3x2 + 7x 20)(x2 4x) = 0 (3x 5)(x + 4)(x)(x 4) = 0 3x 5 = 0 or x + 4 = 0 or x = 0 or x 4 = 0 x = 5 3 or x = 4 or x = 0 or x = 4 63. 3x3 + 6x2 27x 54 = 0 3(x3 + 2x2 9x 18) = 0 3[x2 (x + 2) 9(x + 2)] = 0 Factoring by grouping 3(x + 2)(x2 9) = 0 3(x + 2)(x + 3)(x 3) = 0 x + 2 = 0 or x + 3 = 0 or x 3 = 0 x = 2 or x = 3 or x = 3 The solutions are 2, 3, and 3.
35. 35. 32 Chapter R: Basic Concepts of Algebra 64. 2x3 + 6x2 = 8x + 24 2x3 + 6x2 8x 24 = 0 2(x3 + 3x2 4x 12) = 0 2[x2 (x + 3) 4(x + 3)] = 0 2(x + 3)(x2 4) = 0 2(x + 3)(x + 2)(x 2) = 0 x + 3 = 0 or x + 2 = 0 or x 2 = 0 x = 3 or x = 2 or x = 2 Chapter R Review Exercises 1. True 2. For any real number a, a = 0, and any integers m and n, am an = am+n . Thus the given statement is false. 3. True 4. True 5. Integers: 12, 3, 1, 19, 31, 0 6. Natural numbers: 12, 31 7. Rational numbers: 43.89, 12, 3, 1 5 , 1, 4 3 , 7 2 3 , 19, 31, 0 8. Real numbers: All of them 9. Irrational numbers: 7, 3 10 10. Whole numbers: 12, 31, 0 11. [3, 5) 12. | 3.5| = 3.5 13. |16| = 16 14. | 7 3| = | 10| = 10, or |3 (7)| = |3 + 7| = |10| = 10 15. 53 [2(42 32 6)]3 = 53 [2(16 9 6)]3 = 53 [2(1)]3 = 53 [2]3 = 125 8 = 117 16. 34 (6 7)4 23 24 = 34 (1)4 23 24 = 81 1 8 16 = 80 8 = 10 17. The exponent is positive, so the number is greater than 10. We move the decimal point 6 places to the right. 3.261 106 = 3, 261, 000 18. The exponent is negative, so the number is between 0 and 1. We move the decimal point 4 places to the left. 4.1 104 = 0.00041 19. Position the decimal point 2 places to the right, between the 1 and the 4. Since 0.01432 is a number between 0 and 1, the exponent must be negative. 0.01432 = 1.432 102 20. Position the decimal point 4 places to the left, between the 4 and the 3. Since 43,210 is greater than 10, the exponent must be positive. 43, 210 = 4.321 104 21. 2.5 108 3.2 1013 = 2.5 3.2 108 1013 = 0.78125 1021 = (7.8125 101 ) 1021 = 7.8125 1022 22. (8.4 1017 )(6.5 1016 ) = 54.6 1033 = (5.46 10) 1033 = 5.46 1032 23. (7a2 b4 )(2a4 b3 ) = 7(2)a2+(4) b4+3 = 14a2 b7 , or 14b7 a2 24. 54x6 y4 z2 9x3y2z4 = 54 9 x6(3) y42 z2(4) = 6x9 y6 z6 , or 6x9 z6 y6 25. 4 81 = 4 34 = 3 26. 5 32 = 2
36. 36. Chapter R Review Exercises 33 27. b a1 a b1 = b 1 a a 1 b = b a a 1 a a b b 1 b = ab a 1 a ab b 1 b = ab 1 a ab 1 b = ab 1 a b ab 1 = (ab1 )b a(ab1 ) = b a 28. x2 y + y2 x y2 xy + x2 = x3 + y3 xy y2 xy + x2 = x3 + y3 xy 1 y2 xy + x2 = (x + y)(x2 xy + y2 ) xy(y2 xy + x2) = x + y xy x2 xy + y2 x2 xy + y2 = x + y xy 29. ( 3 7)( 3 + 7) = ( 3)2 ( 7)2 = 3 7 = 4 30. (5x2 2)2 = 25x4 10 2x2 + 2 31. 8 5 + 25 5 = 8 5 + 25 5 5 5 = 8 5 + 25 5 5 = 8 5 + 5 5 = 13 5 32. (x + t)(x2 xt + t2 ) = (x + t)(x2 ) + (x + t)(xt) + (x + t)(t2 ) = x3 + x2 t x2 t xt2 + xt2 + t3 = x3 + t3 33. (5a + 4b)(2a 3b) = 10a2 15ab + 8ab 12b2 = 10a2 7ab 12b2 34. (5xy4 7xy2 + 4x2 3)(3xy4 + 2xy2 2y+ 4) = (5xy4 7xy2 + 4x2 3)+(3xy4 2xy2 + 2y 4) = (5 + 3)xy4 + (7 2)xy2 + 4x2 + 2y + (3 4) = 8xy4 9xy2 + 4x2 + 2y 7 35. x3 + 2x2 3x 6 = x2 (x + 2) 3(x + 2) = (x + 2)(x2 3) 36. 12a3 27ab4 = 3a(4a2 9b4 ) = 3a(2a + 3b2 )(2a 3b2 ) 37. 24x + 144 + x2 = x2 + 24x + 144 = (x + 12)2 38. 9x3 + 35x2 4x = x(9x2 + 35x 4) = x(9x 1)(x + 4) 39. 8x3 1 = (2x)3 13 = (2x 1)(4x2 + 2x + 1) 40. 27x6 + 125y6 = (3x2 + 5y2 )(9x4 15x2 y2 + 25y4 ) 41. 6x3 + 48 = 6(x3 + 8) = 6(x + 2)(x2 2x + 4) 42. 4x3 4x2 9x + 9 = 4x2 (x 1) 9(x 1) = (x 1)(4x2 9) = (x 1)(2x + 3)(2x 3) 43. 9x2 30x + 25 = (3x 5)2 44. 18x2 3x + 6 = 3(6x2 x + 2) 45. a2 b2 ab 6 = (ab 3)(ab + 2) 46. 3x2 12 x2 + 4x + 4 x 2 x + 2 = 3x2 12 x2 + 4x + 4 x + 2 x 2 = 3(x+2) (x2) (x+2) (x+2) (x+2) (x2) = 3
37. 37. 34 Chapter R: Basic Concepts of Algebra 47. x x2 + 9x + 20 4 x2 + 7x + 12 = x (x + 5)(x + 4) 4 (x + 4)(x + 3) LCD is (x + 5)(x + 4)(x + 3) = x (x + 5)(x + 4) x + 3 x + 3 x (x + 4)(x + 3) x + 5 x + 5 = x(x + 3) 4(x + 5) (x + 5)(x + 4)(x + 3) = x2 + 3x 4x 20 (x + 5)(x + 4)(x + 3) = x2 x 20 (x + 5)(x + 4)(x + 3) = (x 5)(x+4) (x + 5)(x+4) (x + 3) = x 5 (x + 5)(x + 3) 48. y5 3 y2 = (y5 )1/2 (y2 )1/3 = y5/2 y2/3 = y19/6 = 6 y19 = y3 6 y 49. (a + b)3 3 a + b 6 (a + b)7 = (a + b)3/2 (a + b)1/3 (a + b)7/6 = (a + b)3/2+1/37/6 = (a + b)9/6+2/67/6 = (a + b)2/3 = 3 (a + b)2 50. b7/5 = 5 b7 = b 5 b2 51. 8 m32n16 38 = m32 n16 38 1/8 = m4 n2 3 52. 4 3 5 + 3 = 4 3 5 + 3 5 3 5 3 = 20 4 3 5 3 + 3 25 3 = 23 9 3 22 53. a = 8 and b = 17. Find c. c2 = a2 + b2 c2 = 82 + 172 c2 = 64 + 289 c2 = 353 c 18.8 The guy wire is about 18.8 ft long. 54. 2x 7 = 7 2x = 14 x = 7 55. 5x 7 = 3x 9 2x 7 = 9 2x = 2 x = 1 The solution is 1. 56. 8 3x = 7 + 2x 5x = 15 x = 3 57. 6(2x 1) = 3 (x + 10) 12x 6 = 3 x 10 12x 6 = x 7 13x 6 = 7 13x = 1 x = 1 13 The solution is 1 13 . 58. y2 + 16y + 64 = 0 (y + 8)(y + 8) = 0 y + 8 = 0 or y + 8 = 0 y = 8 or y = 8 59. x2 x = 20 x2 x 20 = 0 (x 5)(x + 4) = 0 x 5 = 0 or x + 4 = 0 x = 5 or x = 4 The solutions are 5 and 4. 60. 2x2 + 11x 6 = 0 (2x 1)(x + 6) = 0 2x 1 = 0 or x + 6 = 0 x = 1 2 or x = 6 61. x(x 2) = 3 x2 2x = 3 x2 2x 3 = 0 (x + 1)(x 3) = 0 x + 1 = 0 or x 3 = 0 x = 1 or x = 3 The solutions are 1 and 3. 62. y2 16 = 0 (y + 4)(y 4) = 0 y + 4 = 0 or y 4 = 0 y = 4 or y = 4
38. 38. Chapter R Test 35 63. n2 7 = 0 n2 = 7 n = 7 or n = 7 The solutions are 7 and 7, or 7. 64. 128 (2)3 (2) 3 = 128 (8) (2) 3 = 16 (2) 3 = 8 3 = 24 Answer B is correct. 65. 9x2 36y2 = 9(x2 4y2 ) = 9[x2 (2y)2 ] = 9(x + 2y)(x 2y) Answer C is correct. 66. Anya is probably not following the rules for order of op- erations. She is subtracting 6 from 15 rst, then dividing the dierence by 3, and nally multiplying the quotient by 4. The correct answer is 7. 67. When the number 4 is raised to a positive integer power, the last digit of the result is 4 or 6. Since the calculator returns 4.398046511 1012 , or 4,398,046,511,000, we can conclude that this result is an approximation. 68. Substitute $98, 000 $16, 000, or $82,000, for P, 0.065 for r, and 12 25, or 300, for n and perform the resulting computation. M = P r 12 1 + r 12 n 1 + r 12 n 1 = $82, 000 0.065 12 1 + 0.065 12 300 1 + 0.065 12 300 1 $553.67 69. Substitute $124, 000 $20, 000, or $104,000 for P, 0.0575 for r, and 12 30, or 360, for n and perform the resulting computation. M = P r 12 1 + r 12 n 1 + r 12 n 1 = $104, 000 0.0575 12 1 + 0.0575 12 360 1 + 0.0575 12 360 1 $606.92 70. P = $135, 000 $18, 000 = $117, 000, r = 0.075, n = 12 20 = 240. M = $117, 000 0.075 12 1 + 0.075 12 240 1 + 0.075 12 240 1 $942.54 71. P = $151, 000 $21, 000 = $130, 000, r = 0.0625, n = 12 25 = 300. M = $130, 000 0.0625 12 1 + 0.0625 12 300 1 + 0.0625 12 300 1 $857.57 72. (xn + 10)(xn 4) = (xn )2 4xn + 10xn 40 = x2n + 6xn 40 73. (ta + ta )2 = (ta )2 + 2 ta ta + (ta )2 = t2a + 2 + t2a 74. (yb zc )(yb + zc ) = (yb )2 (zc )2 = y2b z2c 75. (an bn )3 = (an bn )(an bn )2 = (an bn )(a2n 2an bn + b2n ) = a3n 2a2n bn +an b2n a2n bn +2an b2n b3n = a3n 3a2n bn + 3an b2n b3n 76. y2n + 16yn + 64 = (yn )2 + 16yn + 64 = (yn + 8)2 77. x2t 3xt 28 = (xt )2 3xt 28 = (xt 7)(xt + 4) 78. m6n m3n = m3n (m3n 1) = m3n [(mn )3 13 ] = m3n (mn 1)(m2n + mn + 1) Chapter R Test 1. a) Integers: 8, 0, 36 b) Rational numbers: 8, 11 3 , 0, 5.49, 36, 10 1 6 c) Rational numbers but not integers: 11 3 , 5.49, 10 1 6 d) Integers but not natural numbers: 8, 0 2. 14 5 = 14 5 3. |19.4| = 19.4 4. | 1.2xy| = | 1.2||x||y| = 1.2|x||y| 5. (3, 6]
39. 39. 36 Chapter R: Basic Concepts of Algebra 6. | 7 5| = | 12| = 12, or |5 (7)| = |5 + 7| = |12| = 12 7. 32 23 12 4 3 = 32 8 12 4 3 = 4 12 4 3 = 4 3 3 = 4 9 = 5 8. Position the decimal point 5 places to the right, between the 3 and the 6. Since 0.0000367 is a number between 0 and 1, the exponent must be negative. 0.0000367 = 3.67 105 9. The exponent is positive, so the number is greater than 10. We move the decimal point 6 places to the right. 4.51 106 = 4, 510, 000 10. 2.7 104 3.6 103 = 0.75 107 = (7.5 101 ) 107 = 7.5 106 11. x8 x5 = x8+5 = x3 , or 1 x3 12. (2y2 )3 (3y4 )2 = 23 y6 32 y8 = 8 9 y6+8 = 72y14 13. (3a5 b4 )(5a1 b3 ) = 3 5 a5+(1) b4+3 = 15a4 b1 , or 15a4 b 14. (3x4 2x2 + 6x) (5x3 3x2 + x) = 3x4 2x2 + 6x 5x3 + 3x2 x = 3x4 5x3 + x2 + 5x 15. (x + 3)(2x5) = 2x2 5x + 6x15 = 2x2 + x15 16. (2y 1)2 = (2y)2 2 2y 1 + (1)2 = 4y2 4y + 1 17. x y y x x + y = x y x x y x y y x + y = x2 xy y2 xy x + y = x2 y2 xy x + y = x2 y2 xy 1 x + y = (x+y) (x y) xy(x+y) = x y xy 18. 54 = 9 6 = 9 6 = 3 6 19. 3 40 = 3 8 5 = 3 8 3 5 = 2 3 5 20. 3 75 + 2 27 = 3 25 3 + 2 9 3 = 3 5 3 + 2 3 3 = 15 3 + 6 3 = 21 3 21. 18 10 = 18 10 = 2 3 3 2 5 = 2 3 5 = 6 5 22. (2 + 3)(5 2 3) = 2 5 4 3 + 5 3 2 3 = 10 4 3 + 5 3 6 = 4 + 3 23. y2 3y 18 = (y + 3)(y 6) 24. x3 + 10x2 + 25x = x(x2 + 10x + 25) = x(x + 5)2 25. 2n2 + 5n 12 = (2n 3)(n + 4) 26. 8x2 18 = 2(4x2 9) = 2(2x + 3)(2x 3) 27. m3 8 = (m 2)(m2 + 2m + 4) 28. x2 + x 6 x2 + 8x + 15 x2 25 x2 4x + 4 = (x2 + x 6)(x2 25) (x2 + 8x + 15)(x2 4x + 4) = (x+3) (x2) (x+5) (x 5) (x+3) (x+5) (x2) (x 2) = x 5 x 2 29. x x2 1 3 x2 + 4x 5 = x (x + 1)(x 1) 3 (x 1)(x + 5) LCD is (x + 1)(x 1)(x + 5) = x (x+ 1)(x1) x+ 5 x + 5 3 (x1)(x + 5) x+ 1 x+ 1 = x(x + 5) 3(x + 1) (x + 1)(x 1)(x + 5) = x2 + 5x 3x 3 (x + 1)(x 1)(x + 5) = x2 + 2x 3 (x + 1)(x 1)(x + 5) = (x + 3)(x1) (x + 1)(x1) (x + 5) = x + 3 (x + 1)(x + 5) 30. 5 7 3 = 5 7 3 7 + 3 7 + 3 = 35 + 5 3 49 3 = 35 + 5 3 46
40. 40. Chapter R Test 37 31. t5/7 = 7 t5 32. ( 5 7)3 = (71/5 )3 = 73/5 33. a = 5 and b = 12. Find c. c2 = a2 + b2 c2 = 52 + 122 c2 = 25 + 144 c2 = 169 c = 13 The guy wire is 13 ft long. 34. 7x 4 = 24 7x = 28 x = 4 The solution is 4. 35. 3(y 5) + 6 = 8 (y + 2) 3y 15 + 6 = 8 y 2 3y 9 = y + 6 4y 9 = 6 4y = 15 y = 15 4 The solution is 15 4 . 36. 2x2 + 5x + 3 = 0 (2x + 3)(x + 1) = 0 2x + 3 = 0 or x + 1 = 0 2x = 3 or x = 1 x = 3 2 or x = 1 The solutions are 3 2 and 1. 37. z2 11 = 0 z2 = 11 z = 11 or z = 11 The solutions are 11 and 11, or 11. 38. (x y 1)2 = [(x y) 1]2 = (x y)2 2(x y)(1) + 12 = x2 2xy + y2 2x + 2y + 1
41. 41. 4 2 2 4 4224 x y (1, 4) (3, 5) (0, 2) (4, 0) (2, 2) 4 2 2 4 4224 x y (5, 0) (4, 2) (1, 4) (4, 0) (2, 4) 4 2 2 4 424 x y (5, 1) (2, 3) (0, 1) (5, 1) (2, 1) 4 2 2 4 4224 x y (5, 2) (5, 0) (1, 5) (4, 0) (4, 3) Chapter 1 Graphs, Functions, and Models Exercise Set 1.1 1. To graph (4, 0) we move from the origin 4 units to the right of the y-axis. Since the second coordinate is 0, we do not move up or down from the x-axis. To graph (3, 5) we move from the origin 3 units to the left of the y-axis. Then we move 5 units down from the x-axis. To graph (1, 4) we move from the origin 1 unit to the left of the y-axis. Then we move 4 units up from the x-axis. To graph (0, 2) we do not move to the right or the left of the y-axis since the rst coordinate is 0. From the origin we move 2 units up. To graph (2, 2) we move from the origin 2 units to the right of the y-axis. Then we move 2 units down from the x-axis. 2. 3. To graph (5, 1) we move from the origin 5 units to the left of the y-axis. Then we move 1 unit up from the x-axis. To graph (5, 1) we move from the origin 5 units to the right of the y-axis. Then we move 1 unit up from the x-axis. To graph (2, 3) we move from the origin 2 units to the right of the y-axis. Then we move 3 units up from the x-axis. To graph (2, 1) we move from the origin 2 units to the right of the y-axis. Then we move 1 unit down from the x-axis. To graph (0, 1) we do not move to the right or the left of the y-axis since the rst coordinate is 0. From the origin we move 1 unit up. 4. 5. The rst coordinate represents the year and the corre- sponding second coordinate represents total advertisement spending, in millions of dollars. The ordered pairs are (2000, 310.6), (2001, 311.2), (2002, 348.2), (2003, 361.6), (2004, 435.8), and (2005, 467.7). 6. (1995, 21.6), (2001, 19.0), (2002, 16.9), (2003, 15.8), (2004, 15.6) 7. To determine whether (1, 1) is a solution, substitute 1 for x and 1 for y. y = 2x 3 1 ? 2 1 3 2 3 1 1 TRUE The equation 1 = 1 is true, so (1, 1) is a solution. To determine whether (0, 3) is a solution, substitute 0 for x and 3 for y. y = 2x 3 3 ? 2 0 3 0 3 3 3 FALSE The equation 3 = 3 is false, so (0, 3) is not a solution. 8. For (2, 5): y = 3x 1 5 ? 3 2 1 6 1 5 5 TRUE (2, 5) is a solution.
42. 42. 40 Chapter 1: Graphs, Functions, and Models For (2, 5): y = 3x 1 5 ? 3(2) 1 6 1 5 7 FALSE (2, 5) is not a solution. 9. To determine whether 2 3 , 3 4 is a solution, substitute 2 3 for x and 3 4 for y. 6x 4y = 1 6 2 3 4 3 4 ? 1 4 3 1 1 TRUE The equation 1 = 1 is true, so 2 3 , 3 4 is a solution. To determine whether 1, 3 2 is a solution, substitute 1 for x and 3 2 for y. 6x 4y = 1 6 1 4 3 2 ? 1 6 6 0 1 FALSE The equation 0 = 1 is false, so 1, 3 2 is not a solution. 10. For (1.5, 2.6): x2 + y2 = 9 (1.5)2 + (2.6)2 ? 9 2.25 + 6.76 9.01 9 FALSE (1.5, 2.6) is not a solution. For (3, 0): x2 + y2 = 9 (3)2 + 02 ? 9 9 + 0 9 9 TRUE (3, 0) is a solution. 11. To determine whether 1 2 , 4 5 is a solution, substitute 1 2 for a and 4 5 for b. 2a + 5b = 3 2 1 2 + 5 4 5 ? 3 1 4 5 3 FALSE The equation 5 = 3 is false, so 1 2 , 4 5 is not a solu- tion. To determine whether 0, 3 5 is a solution, substitute 0 for a and 3 5 for b. 2a + 5b = 3 2 0 + 5 3 5 ? 3 0 + 3 3 3 TRUE The equation 3 = 3 is true, so 0, 3 5 is a solution. 12. For 0, 3 2 : 3m + 4n = 6 3 0 + 4 3 2 ? 6 0 + 6 6 6 TRUE 0, 3 2 is a solution. For 2 3 , 1 : 3m + 4n = 6 3 2 3 + 4 1 ? 6 2 + 4 6 6 TRUE 2 3 , 1 is a solution. 13. To determine whether (0.75, 2.75) is a solution, substi- tute 0.75 for x and 2.75 for y. x2 y2 = 3 (0.75)2 (2.75)2 ? 3 0.5625 7.5625 7 3 FALSE The equation 7 = 3 is false, so (0.75, 2.75) is not a solution. To determine whether (2, 1) is a solution, substitute 2 for x and 1 for y. x2 y2 = 3 22 (1)2 ? 3 4 1 3 3 TRUE The equation 3 = 3 is true, so (2, 1) is a solution. 14. For (2, 4): 5x + 2y2 = 70 5 2 + 2(4)2 ? 70 10 + 2 16 10 + 32 42 70 FALSE (2, 4) is not a solution.
43. 43. y x4 2 2 4 4 2 2 4 5x 3y 15 (3, 0) (0, 5) y x4 2 2 4 4 2 2 4 2x 4y 8 (4, 0) (0, 2) y x4 2 2 4 4 2 2 4 (0, 4) 2x y 4 (2, 0) y x4 2 2 4 2 2 4 6 (0, 6) 3x y 6 (2, 0) y x4 2 2 4 4 2 2 4 4y 3x 12 (4, 0) (0, 3) y x4 2 2 4 4 2 2 4 3y 2x 6 (3, 0) (0, 2) Exercise Set 1.1 41 For (4, 5): 5x + 2y2 = 70 5 4 + 2(5)2 ? 70 20 + 2 25 20 + 50 70 70 TRUE (4, 5) is a solution. 15. Graph 5x 3y = 15. To nd the x-intercept we replace y with 0 and solve for x. 5x 3 0 = 15 5x = 15 x = 3 The x-intercept is (3, 0). To nd the y-intercept we replace x with 0 and solve for y. 5 0 3y = 15 3y = 15 y = 5 The y-intercept is (0, 5). We plot the intercepts and draw the line that contains them. We could nd a third point as a check that the intercepts were found correctly. 16. 17. Graph 2x + y = 4. To nd the x-intercept we replace y with 0 and solve for x. 2x + 0 = 4 2x = 4 x = 2 The x-intercept is (2, 0). To nd the y-intercept we replace x with 0 and solve for y. 2 0 + y = 4 y = 4 The y-intercept is (0, 4). We plot the intercepts and draw the line that contains them. We could nd a third point as a check that the intercepts were found correctly. 18. 19. Graph 4y 3x = 12. To nd the x-intercept we replace y with 0 and solve for x. 4 0 3x = 12 3x = 12 x = 4 The x-intercept is (4, 0). To nd the y-intercept we replace x with 0 and solve for y. 4y 3 0 = 12 4y = 12 y = 3 The y-intercept is (0, 3). We plot the intercepts and draw the line that contains them. We could nd a third point as a check that the intercepts were found correctly. 20.
44. 44. 6 2 2 424 x y y 3x 5 2 4 2 4 424 2 x y y 2x 1 4 2 2 4 4224 x y x y 3 6 4 2 4 4224 x y x y 4 y x4 2 2 4 4 2 2 4 3 4y x 3 y x4 2 2 4 4 2 2 4 3y 2x 3 42 Chapter 1: Graphs, Functions, and Models 21. Graph y = 3x + 5. We choose some values for x and nd the corresponding y-values. When x = 3, y = 3x + 5 = 3(3) + 5 = 9 + 5 = 4. When x = 1, y = 3x + 5 = 3(1) + 5 = 3 + 5 = 2. When x = 0, y = 3x + 5 = 3 0 + 5 = 0 + 5 = 5 We list these points in a table, plot them, and draw the graph. x y (x, y) 3 4 (3, 4) 1 2 (1, 2) 0 5 (0, 5) 22. 23. Graph x y = 3. Make a table of values, plot the points in the table, and draw the graph. x y (x, y) 2 5 (2, 5) 0 3 (0, 3) 3 0 (3, 0) 24. 25. Graph y = 3 4 x + 3. By choosing multiples of 4 for x, we can avoid fraction values for y. Make a table of values, plot the points in the table, and draw the graph. x y (x, y) 4 6 (4, 6) 0 3 (0, 3) 4 0 (4, 0) 26. 27. Graph 5x 2y = 8. We could solve for y rst. 5x 2y = 8 2y = 5x + 8 Subtracting 5x on both sides y = 5 2 x 4 Multiplying by 1 2 on both sides By choosing multiples of 2 for x we can avoid fraction values for y. Make a table of values, plot the points in the table, and draw the graph. x y (x, y) 0 4 (0, 4) 2 1 (2, 1) 4 6 (4, 6)
45. 45. y x4 2 2 4 4 2 2 4 5x 2y 8 y x4 2 2 4 4 2 2 4 4 3y 2 x 4 2 2 4 642 x y x 4y 5 y x4 2 2 4 4 2 2 4 6x y 4 4 2 4 226 x y 2x 5y 10 4 2 8 6 4 4224 x y 4x 3y 12 2 8 6 4 4224 x y y x2 8 6 4 2 4224 x y y x2 Exercise Set 1.1 43 28. 29. Graph x 4y = 5. Make a table of values, plot the points in the table, and draw the graph. x y (x, y) 3 2 (3, 2) 1 1 (1, 1) 5 0 (5, 0) 30. 31. Graph 2x + 5y = 10. In this case, it is convenient to nd the intercepts along with a third point on the graph. Make a table of values, plot the points in the table, and draw the graph. x y (x, y) 5 0 (5, 0) 0 2 (0, 2) 5 4 (5, 4) 32. 33. Graph y = x2 . Make a table of values, plot the points in the table, and draw the graph. x y (x, y) 2 4 (2, 4) 1 1 (1, 1) 0 0 (0, 0) 1 1 (1, 1) 2 4 (2, 4) 34.
46. 46. 6 4 2 2 4224 x y y x2 3 2 2 4 44 x y y 4 x2 y x8 4 4 8 8 12 4 4 y x2 2x 3 4 2 4 424 x y y x2 2x 1 44 Chapter 1: Graphs, Functions, and Models 35. Graph y = x2 3. Make a table of values, plot the points in the table, and draw the graph. x y (x, y) 3 6 (3, 6) 1 2 (1, 2) 0 3 (0, 3) 1 2 (1, 2) 3 6 (3, 6) 36. 37. Graph y = x2 + 2x + 3. Make a table of values, plot the points in the table, and draw the graph. x y (x, y) 2 5 (2, 5) 1 0 (1, 0) 0 3 (0, 3) 1 4 (1, 4) 2 3 (2, 3) 3 0 (3, 0) 4 5 (4, 5) 38. 39. Either point can be considered as (x1, y1). d = (4 5)2 + (6 9)2 = (1)2 + (3)2 = 10 3.162 40. d = (3 2)2 + (7 11)2 = 41 6.403 41. Either point can be considered as (x1, y1). d = (6 9)2 + (1 5)2 = (3)2 + (6)2 = 45 6.708 42. d = (4 (1))2 + (7 3)2 = 109 10.440 43. Either point can be considered as (x1, y1). d = (4.2 2.1)2 + [3 (6.4)]2 = (6.3)2 + (9.4)2 = 128.05 11.316 44. d = 3 5 3 5 2 + 4 2 3 2 = 14 3 2 = 14 3 45. Either point can be considered as (x1, y1). d = 1 2 5 2 2 + (4 4)2 = (3)2 + 02 = 9 = 3 46. d = [0.6 (8.1)]2 + [1.5 (1.5)]2 = (8.7)2 = 8.7 47. Either point can be considered as (x1, y1). d = ( 6 3)2 + (0 ( 5))2 = 6 + 2 18 + 3 + 5 = 14 + 2 9 2 = 14 + 2 3 2 = 14 + 6 2 4.742 48. d = ( 20)2 +(1 7)2 = 2+12 7+7 = 10 2 7 2.170 49. Either point can be considered as (x1, y1). d = (0 a)2 + (0 b)2 = a2 + b2 50. d = [r (r)]2 + [s (s)]2 = 4r2 + 4s2 = 2 r2 + s2