Roots and Radical Expressions

Roots and Radical Expressions
ALGEBRA 2 LESSON 7-1 (For help, go to Lesson 5-4.) Write each number as a square of a number. Write each expression as a square of an expression. 4. x10 5. x4y x6y12 4 49 7-1

ALGEBRA 2 LESSON 7-1 1. 25 = = = = 4. x10 = (x5)2 5. x4y2 = (x2y) x6y12 = (13x3y6)2 4 49 22 72 2 7 Solutions 7-1

ALGEBRA 2 LESSON 7-1 Find all the real roots. a. the cube root of 0.027, –125, and 1 64 Since 0.33 = 0.027, 0.3 is the cube root of Since (–5)3 = –125, –5 is the cube root of –125. Since = , is the cube root of 1 4 64 b. the fourth roots of 625, –0.0016, and 81 625 Since 54 = 625 and (–5)4 = 625, 5 and –5 are fourth roots of 625. There is no real number with a fourth power of – Since = and = , and – are fourth roots of 81 625 3 5 –3 7-1

ALGEBRA 2 LESSON 7-1 Find each real-number root. 3 a. –1000 Rewrite –1000 as the third power of a number. 3 = (–10)3 Definition of nth root when n = 3, there is only one real cube root. = –10 b. –81 There is no real number whose square is –81. 7-1

ALGEBRA 2 LESSON 7-1 Simplify each radical expression. a. 9x10 Absolute value symbols ensure that the root is positive when x is negative. 9x10 = 32(x5)2 = (3x5)2 = 3| x5| 3 b. a3b3 (ab)3 = ab 3 Absolute value symbols must not be used here. If a or b is negative, then the radicand is negative and the root must also be negative. 7-1

ALGEBRA 2 LESSON 7-1 (continued) c. x16y4 4 4 x16y14 = (x4)4(y)4 = x4 |y| Absolute value symbols ensure that root is positive when y is negative. They are not needed for x because x2 is never negative. 7-1

ALGEBRA 2 LESSON 7-1 A cheese manufacturer wants to ship cheese balls that weigh from 10 to 11 ounces in cartons that will have 3 layers of 3 cheese balls by 4 cheese balls. The weight of a cheese ball is related to its diameter by the formula w = , where d is the diameter in inches and w is the weight in ounces. What size cartons should be used? Assume whole-inch dimensions. d3 5 Find the diameter of the cheese balls. < – w Write an inequality. Substitute for w in terms of d. d3 5 d Multiply by 5. < – 3 50 d Take cube roots. d The diameters range from 3.68 in. to 3.80 in. < – 7-1

ALGEBRA 2 LESSON 7-1 (continued) The length of a row of 4 of the largest cheese balls is 4(3.80 in.) = 15.2 in. The length of a row of 3 of the largest cheese balls 3(3.80 in.) = 11.4 in. The manufacturer should order cartons that are 16 in. long by 12 in. wide by 12 in. high to accommodate three dozen of the largest cheese balls. 7-1

ALGEBRA 2 LESSON 7-1 pages 366–367 Exercises 1. 15, –15 , –0.07 3. none , – 5. –4 6. 0.5 7. – 9. 2, –2 10. none , –0.3 , – 13. 6 14. –6 15. no real-number root 17. –4 18. –4 19. –3 20. no real-number root 21. 4|x| |x3| 23. x4|y9| 10 3 10 3 24. 8b24 25. –4a 26. 3y2 27. x2|y3| 28. 2y2 in. ft cm mm 33. 10, –10 34. 1, –1 , –0.5 8 13 8 13 1 2 7-1

ALGEBRA 2 LESSON 7-1 , – –64, , – –64, 38. a ft b ft longer 40. 42. 43. 2|c| 44. 3xy2 3 45. 12y2z2x xy 46. y4 2 3 2 3 47. –y4 48. k3 49. –k3 50. |x + 3| 51. (x + 1)2 52. |x| 53. x2 54. |x3| 55. Answers may vary. Sample: –8x6, – 16x8, –32x10 56. a. for all positive integers b. for all odd positive integers 3 6 3 1 3 1 4 3 3 4 5 7-1

ALGEBRA 2 LESSON 7-1 57. Yes, because 10 is really 101. 58. All; x2 is always nonnegative. 59. Some; true only for x > 0. 60. Some; true only for x = –1, 0, 1. 61. Some; true only for x > 0. 62. |m| 63. m2 64. |m3| 65. m4 66. m 67. m2 68. m3 69. m4 70. B 71. I 72. B 73. H 74. [2] x2y4 equals x3y6 whenever y = 0, regardless of the value of x. This is true because x2y4 and x3y6 will be 0 whenever y = x2y4 also equals x3y6 when x > 0. This is true because x2y4 = |x|y2 and x3y6 = xy2, and |x|y2 = xy2 when x > 0. [1] answer only, with no explanation 3 3 3 7-1

ALGEBRA 2 LESSON 7-1 75. x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5 – 96y + 216y2 – 216y3 + 81y4 x6 – 7290x5 + 30,375x4 – 67,500x3 + 84,375x2 – 56,250x + 15,625 a7 – 448a6b + 672a5b2 – 560a4b a3b4 – 84a2b5 + 14ab6 – b7 79. y = x(2x – 7)(2x + 7) 80. y = (9x + 2)2 81. y = 4x(x + 1)2 82. y = 2x(6x + 1)(x + 1) 83. y = 3(x – 0)2 – 7 84. y = –2 – x – – 85. y = (x + 4)2 – 5 1 4 2 79 8 1 4 7-1

ALGEBRA 2 LESSON 7-1 1. Find all the real square roots of each number. a. 121 b. –49 c. 64 d. – 2. Find all the real cube roots of each number. a. –8000 b. 3. Find each real-number root. a b c. – d. –625 4. Simplify each radical expression. a. –8x3 b y4 c x14 5. The formula for the volume of a cone with a base of radius r and height r is V = r 3. Find the radius to the nearest hundredth of a centimeter if the volume is 40 cm3. 1 25 ± 11 none ± 8 none 1 216 1 6 –20 0.7 3 5 –9 4 none 3 6 | x7 | –2x 4y2 1 3 about 3.37 cm 7-1

Multiplying and Dividing Radical Expressions
ALGEBRA 2 LESSON 7-2 (For help, go to page 362.) Find each missing factor. = 52( ) = ( )3(2) 3. 48 = 42( ) 4. x5 = ( )2(x) 5. 3a3b4 = ( )3(3b) 6. 75a7b8 = ( )2(3a) 7-2

ALGEBRA 2 LESSON 7-2 Solutions = 52(6) = (3)32 3. 48 = 42(3) 4. x5 = (x2)2(x) 5. 3a3b4 = (ab)3(3b) 6. 75a7b8 = (5a3b4)2(3a) 7-2

ALGEBRA 2 LESSON 7-2 Multiply. Simplify if possible. a. 3 • 12 3 • 12 = 3 • 12 = 36 = 6 b. –16 • 4 3 3 3 –16 • 4 = –16 • 4 = 64 = –4 c. –4 • 16 The property for multiplying radicals does not apply. –4 is not a real number. 7-2

ALGEBRA 2 LESSON 7-2 Simplify each expression. Assume all variables are positive. a. 50x5 Factor into perfect squares. 50x5 = 52 • 2 • (x2)2 • x = 52 • (x2)2 • 2 • x n a • b = ab = 5x2 2x definition of nth root 3 b. 54n8 3 54n8 = 33 • 2 • (n2)3 • n2 Factor into perfect cubes. 3 = 33(n2)3 • 2n2 n a • b = ab 3 = 3n2 2n2 definition of nth root 7-2

ALGEBRA 2 LESSON 7-2 Multiply and simplify . Assume all variables are positive. 3 25xy8 • 5x4y3 3 25xy8 • 5x4y3 = 25xy8 • 5x4y3 3 n a • b = ab 3 = 53x3(y3)3 • x2y2 Factor into perfect cubes. 3 = 53x3(y3)3 • x2y2 n a • b = ab 3 = 5xy3 x2y2 definition of nth root 7-2

ALGEBRA 2 LESSON 7-2 Divide and simplify. Assume all variables are positive. a. 3 –81 = b. 3x 3 192x8 = = 3 –81 192x8 3x 3 = 3 = –27 3 = 64x7 3 = (–3)3 3 = 43(x2)3 • x = –3 = 4x2 x 3 7-2

ALGEBRA 2 LESSON 7-2 Rationalize the denominator of each expression. Assume that all variables are positive. Method 1: Rewrite as a square root of a fraction. 3 5 = a. 5 3 = 3 • 5 5 • 5 Then make the denominator a perfect square. = 15 52 15 5 = 7-2

ALGEBRA 2 LESSON 7-2 (continued) 3 • 5 5 • 5 Multiply the numerator and denominator by 5 so the denominator becomes a whole number. = 3 5 Method 2: a. 5 3 15 5 = 7-2

ALGEBRA 2 LESSON 7-2 (continued) b. x5 3x2y c. 3 5 4y x5 3x2y • = 3 5 • 42y2 4y • 42y2 5 4y = Rewrite the fraction so the denominator is a perfect cube. 3 80y2 43y3 = 3x7y 3x2y = x xy 3x2y = y2 4y 3 = x xy 3y = 10y2 2y 3 = 7-2

ALGEBRA 2 LESSON 7-2 The distance d in meters that an object will fall in t seconds is given by d = 4.9t 2. Express t in terms of d and rationalize the denominator. d = 4.9t 2 t 2 = d 4.9 t = d 4.9 d • 10 49 = 10d 7 = 7-2

ALGEBRA 2 LESSON 7-2 pages 371–373 Exercises 1. 16 2. 4 3. –9 4. 4 5. not possible 6. not possible 7. –6 8. 6 9. 2x 5x x2 11. 5x2 2x 12. 2a 4a2 13. 3y3 2y 14. 10a3b3 2b 15. –5x2y 2y2 16. 2y 4x3y2 18. 8y3 5y 19. 7x3y4 6y 20. 40xy 3 21. 30y2 2y 22. –2x2y 30x 23. 10 24. 25. 2x2y2 2 26. 5x3 x2y2 27. 28. 29. 30. 31. 32. 5x2 5 33. 3 4x y 3 3 2x 2 4 10x 4x 3 4x 2 3 45x2 3x 3 250 5 4 3 3 3 15y 5y 7-2

ALGEBRA 2 LESSON 7-2 48. 49. 50. 51. – 52. – – 53. 54. mi/h greater cm2 34. 35. r = 36. a. b. c. Answers may vary. Sample: First simplify the denominator. Since = • 49 = , to rationalize the denominator, multiply the fraction by This yields = . x 10 2y Gm1m2F F 6 + 3 15 2 2 • • 2 • 2 14 39. 3x6y5 2y 40. 20x2y3 y 44. 2x 2 45. 3x2 x 46. 47. 3 x 10y 2y 2 x 21x 3x2 3x 6 – 2 4 5 33x 4x 2xy 2 xy x x 7-2

ALGEBRA 2 LESSON 7-2 57. A product of two square roots can be simplified in this way only if the square roots are real numbers. –2 and –8 are not. a5 ft 59. For some values; it is easy to see that the equation is true if x = 0 or x = 1. But when x < 0, x3 is not a real number, although x2 is. 60. Check students’ work. 61. 2xy 62. 2xy2 3 64. 65. 66. 67. a = –2c, b = –6d 68. No changes need to be made; since they are both odd roots, there is no need for absolute value symbols. 69. C 70. H 71. A 72. G yx x 5 x4y3 y 6 x2y xy 3 7-2

ALGEBRA 2 LESSON 7-2 3 2x 4x2 12x2 8x3 < – > 73. D 74. [2] x is a real number if x 0 and –x is a real number if x 0. So the only value that makes x • –x a real number is x = 0. [1] answer only OR error describing value(s) of variables 75. [4] You should multiply by because • = • = = , which has a denominator without a radical. [3] appropriate methods, but with one minor error [2] major error, but subsequent steps consistent with that error [1] correct final expression, but no work shown 7-2

ALGEBRA 2 LESSON 7-2 76. 11|a45| 77. –9c24d32 78. –4a27 79. 2y5 |x3| 81. x2y5 82. 2|x9|y24 x20 84. y2 – 4y + 16, R –128, not a factor 85. x2 – 3x + 9, a factor 86. 3a2 – a, R 4, not a factor 87. 2x3 + x2 + 2x, R 10, not a factor 88. 25 89. 25 90. 91. 92. 94. 95. 121 4 1 36 9 64 100 7-2

ALGEBRA 2 LESSON 7-2 Assume that all variables are positive. 1. Multiply. Simplify if possible. a • b • 2. Simplify. a. 8x5 b. –243x3y10 3. Multiply and simplify. a x3 • 2x2y3 b x2y4 • 4x2y 4. Divide and simplify. a. b. 5. Rationalize the denominator of each expression. 15 3 3 20 2x2 2x 3 3 –3xy3 9y 6x2y xy 3 3 3 2xy 5xy2 3 270x 10xy2 3 3 y y 128x3 2xy 8x y y 7x 3 21x 3 3 x2 4 3 2x2 2 7-2

Binomial Radical Expressions
ALGEBRA 2 LESSON 7-3 (For help, go to Lesson 5-1 or Skills Handbook page 853.) Multiply. 1. (5x + 4)(3x – 2) 2. (–8x + 5)(3x – 7) 3. (x + 4)(x – 4) 4. (4x + 5)(4x – 5) 5. (x + 5)2 6. (2x – 9)2 7-3

ALGEBRA 2 LESSON 7-3 Solutions 1. (5x + 4)(3x – 2) = (5x)(3x) + (5x)(–2) + (4)(3x) + (4)(–2) = 15x2 – 10x + 12x – 8 = 15x2 + 2x – 8 2. (–8x + 5)(3x – 7) = (–8x)(3x) + (–8x)(–7) + (5)(3x) + (5)(–7) = –24x2 + 56x + 15x – 35 = –24x2 + 71x – 35 3. (x + 4)(x – 4) = x2 – 42 = x2 – 16 4. (4x + 5)(4x – 5) = (4x)2 – 52 = 16x2 – 25 5. (x + 5)2 = x2 + 2(5)x + 52 = x2 + 10x + 25 6. (2x – 9)2 = (2x)2 – 2(2x)(9) + 92 = 4x2 – 36x + 81 7-3

ALGEBRA 2 LESSON 7-3 Add or subtract if possible. a xy xy Distributive Property 7 xy xy = (7 + 3) xy = xy Subtract. 3 b x – The radicals are not like radicals. They cannot be combined. 7-3

ALGEBRA 2 LESSON 7-3 The rectangular window shown below is made up of three equilateral triangles and two right triangles. The equilateral triangles have sides of length 4 feet. What is the perimeter of the window? The height of an equilateral triangle with side length 4 ft is ft. So the window’s height is ft. The length of the window is 8 ft. The perimeter is 2( ) ft or ft. 7-3

ALGEBRA 2 LESSON 7-3 Simplify – – = • 5 – • • 5 Factor each radicand. Simplify each radical. = 3 • – • Multiply. = – Use the Distributive Property. = (6 – ) 5 = 7-3

ALGEBRA 2 LESSON 7-3 Multiply ( )(1 – ). Use FOIL. ( )(1 – ) = 2 • 1 – 2 • • 1 – • = 2 + (4 – 10) 3 – 60 Distributive Property = –58 – 7-3

ALGEBRA 2 LESSON 7-3 Multiply ( )(3 – 7). (a + b)(a – b) = a2 – b2 ( )(3 – 7) = 32 – ( 7)2 = 9 – 7 = 2 7-3

ALGEBRA 2 LESSON 7-3 2 – 3 Rationalize the denominator of 2 – 3 = • 4 – 3 4 – 3 is the conjugate of = • 2 – 3 4 – 3 Multiply. 42 – ( 3)2 8 – – ( 3)2 = Simplify. 11 – 13 = 7-3

ALGEBRA 2 LESSON 7-3 pages 376–378 Exercises 3. cannot combine 4. –2 x 5. cannot combine x2 – – 19. 14 20. 4 21. –40 22. –2 23. – 24. 26. x – 31. 33y 6 32. –2 2 33. – 23 4 4 –14 3 3 3 3 3 3 7-3

ALGEBRA 2 LESSON 7-3 36. –36 – 37. x x + 6 38. 8y – y + 30 39. – 2 41. 42. 43. – 45. The reciprocal is , which is one less than 46. a must be twice a perfect square. m 48. Answers may vary. Sample: Without simplifying first, you must estimate three separate square roots, and then add the estimates. If they are first simplified, then they can be combined as 13. Then only one square root need be estimated. 49. Answers may vary. Samples: ( )( 7 – 2); ( )( – 5) s, or about 4.53 s 51. – –239 3 – 7 2 2 3 x x3 x 4 3 3 60 – 7 – 2 1 2 2 7-3

ALGEBRA 2 LESSON 7-3 53. (a = 0 and b > 0) or (b = 0 and a > 0) 54. In the second step the exponent was incorrectly distributed: (a – b)x ax – bx. 55. a. m and n can be any positive integers. b. m must be even or n must be odd. c. m must be even, and n can be any positive integer. 56. B 57. I 58. D 59. I 60. D 61. [2] (1 – 8)( ) = (1 – 2)(1 + 2) = 1 – 4 = –3, which is rational. [1] the value –3 only OR yes only OR minor error in calculation = / 3 3 7-3

ALGEBRA 2 LESSON 7-3 62. [4] – = • – • = – = = [3] minor error, but appropriate method [2] correct method, but finds and gets [1] answer only, no work shown 2 3 5 – 64. x 15 65. 4 66. 67. 2x 68. 7x2 2 69. 70. 71. 2, –1 ± i 3 72. –10, 5 ± 5i 3 , 74. ± 7 75. 76. ± , ± 3 2 5 – ±2 5 5 3 5 – i 3 1 3 93 3 10 – 25 – 4 • 2 25 – 4 • 2 10 – 17 17 10 – – 15 – 17 –5 – 17 3n n 100x2 5x 3 10 – 17 17 17 1 5 –1 ± i 3 10 7-3

ALGEBRA 2 LESSON 7-3 Simplify. – x x – 54 5. Multiply ( ) (5 – 3). 6. Rationalize the denominator in . 3 3 3 13 x 3 3 3 2 3 – 10 – 7 16 – 93 7-3

Rational Exponents Simplify. 1. 2–4 2. (3x)–2 3. (5x2y)–3 4. 2–2 + 4–1
ALGEBRA 2 LESSON 7-4 (For help, go to page 362.) Simplify. 1. 2–4 2. (3x)–2 3. (5x2y)–3 4. 2–2 + 4–1 5. (2a–2b3)4 6. (4a3b–1)–2 7-4

Rational Exponents Solutions 1. 2–4 = = 2. (3x)–2 = = 3. (5x2y)–3 = =
ALGEBRA 2 LESSON 7-4 1 24 16 (3x)2 9x2 (5x2y)3 125x6y3 22 41 4 2 16b12 a8 b2 42a6 16a6 Solutions 1. 2–4 = = 2. (3x)–2 = = 3. (5x2y)–3 = = 4. 2–2 + 4–1 = = = = 5. (2a–2b3)4 = 24a–8b12 = 6. (4a3b–1)–2 = 4–2a–6b2 = = 7-4

Simplify each expression.
Rational Exponents ALGEBRA 2 LESSON 7-4 Simplify each expression. a. 64 1 3 3 64 = Rewrite as a radical. 1 3 = Rewrite 64 as a cube. = 4 Definition of cube root. b. 7 • 7 1 2 7 • 7 = 7 • Rewrite as radicals. 1 2 = 7 By definition, 7 is the number whose square is 7. 7-4

5 • 25 = 5 • 25 Rewrite as radicals.
Rational Exponents ALGEBRA 2 LESSON 7-4 (continued) c. 5 • 25 1 3 5 • 25 = • Rewrite as radicals. 1 3 3 = 5 • 25 property for multiplying radical expressions = 5 By definition, 5 is the number whose cube is 5. 3 7-4

a. Write x and y –0.4 in radical form.
Rational Exponents ALGEBRA 2 LESSON 7-4 a. Write x and y –0.4 in radical form. 2 7 y –0.4 = y = or 2 5 – 1 y2 y 2 2 7 x = x2 or x 2 b. Write the radical expressions and in exponential form. c3 4 b 5 3 = c 3 4 c3 b 5 3 = b 5 7-4

Rational Exponents ALGEBRA 2 LESSON 7-4 The time t in hours needed to cook a pot roast that weighs p pounds can be approximated by using the equation t = 0.89p0.6. To the nearest hundredth of an hour, how long would it take to cook a pot roast that weighs 13 pounds? t = 0.89p0.6 Write the formula. = 0.89(13)0.6 Substitute for p. 4.15 Use a calculator. 7-4

Rational Exponents Simplify each number. a. (–27) Method 1 = (–3)2 = 9
ALGEBRA 2 LESSON 7-4 Simplify each number. a. (–27) 2 3 Method 1 = (–3)2 = 9 (–27) = ((–3)3) 2 3 Method 2 3 = ( (–3)3)2 = (–3)2 = 9 (–27) = ( –27)2 2 7-4

Rational Exponents b. 25–2.5 Method 1 25–2.5 = 25 = (52) = 52 Method 2
ALGEBRA 2 LESSON 7-4 (continued) b. 25–2.5 Method 1 25–2.5 = 25 5 2 – = (52) = 52 Method 2 25–2.5 = 25 5 2 – 1 25 = = 5–5 = 1 55 3125 = 1 55 3125 ( 25)5 7-4

Write (243a–10) in simplest form.
Rational Exponents ALGEBRA 2 LESSON 7-4 Write (243a–10) in simplest form. 2 5 (243a–10) = (35a–10) 2 5 = • a(–10) 2 5 = 32a–4 32 a4 = = 9 a4 7-4

Rational Exponents 7 5 pages 382–384 Exercises 1. 6 2. 3 3. 7 4. 10
ALGEBRA 2 LESSON 7-4 7 5 pages 382–384 Exercises 1. 6 2. 3 3. 7 4. 10 5. –3 6. 6 7. 8 8. 3 9. 3 x x 6 1 2 3 y9 8 y 9 t 3 4 t 22. a 23. a 24. c 25. 25x2y2 m m m m 30. 4 31. 16 32. 4 33. 64 x2 or ( x)2 y2 or ( y)2 or or x3 or ( x)3 y6 or ( y)6 18. (–10) x 20. (7x) 21. (7x) 7-4

Rational Exponents ALGEBRA 2 LESSON 7-4 34. 35. 8 36. 64 38. 39. 40. 41. 42. – 43. –2y3 1 16 x2 x4 3x 2 3 5 x x3 44. 45. x 46. 47. 48. x3y9 49. 50. –7 51. –3 52. 64 y4 y2 x8 y5 x10 13 54. 2,097,152 55. 1,000,000,000 or 109 56. 57. 58. 59. 16 60. – 61. 10 62. 78%, 61%, 37% 4 8 36 81 7-4

Rational Exponents 65. x 66. y 67. x 68. y 69. x y 70. 71. 72. 73. 1 2
ALGEBRA 2 LESSON 7-4 65. x 66. y 67. x 68. y 69. x y 70. 71. 72. 73. 1 2 4 5 6 x y 4x7 9y9 9y8 4x6 x 13 36 74. 75. 76. 77. The cube root of –64 is –4, which equals –(64) . The square root of –64 is not a real number, but –(64) = – = –8. 78. The exponent applies only to the 5, not to the 25. 7 24 3 (xy) 4 – 5 79. a. Answers may vary. Sample: 4 – 5 , 2(4 – 5 ), b. no 80. a. x • x • x • x = x • x = x2, so x2 = x b. x2 = (x2) = x = x = 81. 49 82. 9 83. x2 84. 1 7-4

Rational Exponents 85. 3 2 86. 9 87. 33.13 mi/h 88. B 89. H
ALGEBRA 2 LESSON 7-4 86. 9 mi/h 88. B 89. H 90. [2] x = [1] minor error 91. B 92. A 93. C 94. A – 99. 100. x(x2 – 2x + 4) 102. (x + 2)2 103. (x – 9)2 104. (4a – 3b)(4a + 3b) 105. (5x – 4y)2 106. (3x + 8)2 –41 10 – 7 1 16 3 7-4

1. Simplify each expression. a. 100 b. (–64)
Rational Exponents ALGEBRA 2 LESSON 7-4 1. Simplify each expression. a b. (–64) 2. Write each expression in radical form. a. x b. y c. k1.8 d. a 3. A container with curved sides is 10 ft tall. When water is in the container to a depth of h ft, the number of cubic feet of water can be approximated by using the formula V = 0.95h2.9. Find the amount of water in the container when the depth of the water is 7.5 ft. Round to the nearest hundredth. 4. Simplify each number. a. (–64) b 5. Write x y in simplest form. 1 2 1 3 10 –4 1 7 4 5 y4 5 or y 4 k9 5 or k 9 1 6 – 1 a 6 x 7 ft3 4 3 256 27 y x 8 3 2 3 1 4 – –4 7-4

Solving Radical Equations
ALGEBRA 2 LESSON 7-5 (For help, go to Lesson 5-4.) Solve by factoring. 1. x2 = –x x2 = 5x + 14 3. 2x2 + x = x2 – 2 = 5x 5. 4x2 = –8x x2 = 5x + 6 7-5

ALGEBRA 2 LESSON 7-5 Solutions 1. x2 = –x + 6; x2 + x – 6 = 0 Factors of –6 with a sum of 1: 3 and –2 (x + 3)(x – 2) = 0 x + 3 = 0 or x – 2 = 0 x = –3 or x = 2 3. 2x2 + x = 3; 2x2 + x – 3 = 0 (2x + 3)(x – 1) = 0 2x + 3 = 0 or x – 1 = 0 x = – or x = 1 5. 4x2 = –8x + 5; 4x2 + 8x – 5 = 0 (2x – 1)(2x + 5) = 0 2x – 1 = 0 or 2x + 5 = 0 x = or x = – 2. x2 = 5x + 14; x2 – 5x – 14 = 0 Factors of –14 with a sum of –5: –7 and 2 (x – 7)(x + 2) = 0 x – 7 = 0 or x + 2 = 0 x = 7 or x = –2 4. 3x2 – 2 = 5x; 3x2 – 5x – 2 = 0 (3x + 1)(x – 2) = 0 3x + 1 = 0 or x – 2 = 0 x = – or x = 2 6. 6x2 = 5x + 6; 6x2 – 5x – 6 = 0 (3x + 2)(2x – 3) = 0 3x + 2 = 0 or 2x – 3 = 0 x = – or x = 3 2 1 5 7-5

ALGEBRA 2 LESSON 7-5 Solve – x + 1 = –5. – x + 1 = –5 2x + 1 = 5 Isolate the radical. ( 2x + 1 )2 = 52 Square both sides. 2x + 1 = 25 2x = 24 x = 12 Check: – x + 1 = –5 – (12) –5 – –5 – –5 –5 = –5 7-5

ALGEBRA 2 LESSON 7-5 Solve 3(x + 1) = 24. 3 5 3(x + 1) = 24 3 5 (x + 1) = 8 Divide each side by 3. 3 5 3 5 ((x + 1) ) = 8 Raise both sides to the power. (x + 1)1 = 8 Multiply the exponents and . 5 3 x + 1 = 32 Simplify. x = 31 Check: 3(x + 1) = 24 3(31 + 1) 3(25) 3(2) 24 = 24 3 5 7-5

ALGEBRA 2 LESSON 7-5 An artist wants to make a plastic sphere for a sculpture. The plastic weighs 0.8 ounce per cubic inch. The maximum weight of the sphere is to be 80 pounds. The formula for the volume V of a sphere is V = • r 3, where r is the radius of the sphere. What is the maximum radius the sphere can have? 4 3 4 3 Define: Let r = radius in inches. Relate: volume of sphere • density of plastic maximum weight Write: • r 3 • < – 7-5

ALGEBRA 2 LESSON 7-5 (continued) 4 r 3 3 • < – r 3 3 • 80 4 • • 0.8 < – r 3 75 < – Use a calculator. r < – The maximum radius is about 2.88 inches. 7-5

ALGEBRA 2 LESSON 7-5 Solve x + 2 – 3 = 2x. Check for extraneous solutions. x + 2 – 3 = 2x x + 2 = 2x + 3 Isolate the radical. ( x + 2)2 = (2x + 3)2 Square both sides. 0 = 4x2 + 11x + 7 Combine like terms. 0 = (x + 1)(4x + 7) Factor. x + 2 = 4x2 + 12x + 9 Simplify. x + 1 = 0 or 4x + 7 = 0 Factor Theorem x = –1 or x = – 7 4 7-5

ALGEBRA 2 LESSON 7-5 (continued) Check: x + 2 – 3 = 2x x + 2 – 3 = 2x –1 + 2 – (–1) – 1 – 3 – – 3 –2 = –2 – 3 7 4 – 7 4 – 1 4 7 2 – 1 2 7 2 – – 5 2 7 2 – = / The only solution is –1. 7-5

ALGEBRA 2 LESSON 7-5 Solve (x + 1) – (9x + 1) = 0. Check for extraneous solutions. 2 3 1 3 (x + 1) – (9x + 1) = 0 2 3 1 (x + 1) = (9x + 1) ((x + 1) )3 = ((9x + 1) )3 2 3 1 (x + 1)2 = 9x + 1 x2 + 2x + 1 = 9x + 1 x2 – 7x = 0 x(x – 7) = 0 x = 0 or x = 7 7-5

ALGEBRA 2 LESSON 7-5 (continued) Check: (x + 1) – (9x + 1) = 0 (x + 1) – (9x + 1) = 0 (0 + 1) – (9(0) + 1) (7 + 1) – (9(7) + 1) 0 (1) – (1) (8) – (64 ) 0 (1) – (12) (8) – (82) 0 1 – 1 = – 8 = 0 2 3 1 3 2 3 1 3 2 3 1 3 2 3 1 3 2 3 1 3 2 3 1 3 2 3 1 3 2 3 1 3 2 3 2 3 2 3 2 3 Both 0 and 7 are solutions. 7-5

ALGEBRA 2 LESSON 7-5 pages 388–390 Exercises 1. 16 2. 1 3. 22 4. 4 5. 23 6. 7. 3 8. 29, –27 9. 18 10. 78 11. 8 12. 0 ft 14. 4 in. 15. 3 16. 1 17. –3, –4 18. 9 19. 1 20. 1 21. 3 22. –2 23. 1 24. 6 25. 2 26. –2 27. 5 28. –3 29. 5 30. –2 31. s = A; m, or about 5.7 m 32. a. s = b. about 8.8 in. c. about 15.2 in. 2 3 2 3A 3 7-5

ALGEBRA 2 LESSON 7-5 41. 9, –7 42. 43. 9 44. 2 45. –1, –6 46. 2 47. 7 48. 25 49. 10 50. –1 51. 33. a b c. d. The graph of each pair consists of two straight lines, one of which is horizontal. They intersect at different points, but these points have the same x-value, about 34. 8 35. 4 36. 5 37. 23 38. 1 81 16 5 4 7-5

ALGEBRA 2 LESSON 7-5 52. d = 53. Answers may vary. Sample: x – 3 = 3x + 5 54. 1 55. 2 56. 0 57. Plan 1: Use a calculator to evaluate 2 + 2 and store the result as x. Evaluate x + 2 and store the result as x. Continue this procedure about seven times until it becomes clear that the values are approaching 2. Plan 2: The given equation is equivalent to x = x. Solve this equation to find that x = 2. v2 64 58. a. A counterexample is a = 3, b = –3. b. A counterexample is a = –5, b = 3. 59. 12 60. 79 61. 62. 63. 27 64. 2 65. 16 5 4 1 25 7-5

ALGEBRA 2 LESSON 7-5 70. 72. 25 73. 2 74. 75. 7 77. 60 79. 24 80. 10 81. 21 82. 1 83. 6 84. 7 85. 3, 4 86. 3, 5 87. –5, –4 88. –2, – 89. – , – 90. –2, – 3 1 3 1 106 2 3 1 3 4 3 3 4 7-5

ALGEBRA 2 LESSON 7-5 Solve each equation. Check for extraneous solutions. x – 1 = 10 2. 4(x – 9) = 8 x – 1 = x – 8 4. (4x + 3) = (16x + 44) 5. A circular table is to be made that will have a top covered with material that costs $3.50 per square foot. The covering is to cost no more than $60. What is the maximum radius for the top of the table? 1 3 2 5 17 13 7 4 – , 5 about 2.34 ft 7-5

Find the domain and range of each function.
Function Operations ALGEBRA 2 LESSON 7-6 (For help, go to Lesson 2-1.) Find the domain and range of each function. 1. {(0, –5), (2, –3), (4, –1)} 2. {(–1, 0), (0, 0), (1, 0)} 3. ƒ(x) = 2x – g(x) = x2 Evaluate each function for the given value of x. 5. Let ƒ(x) = 3x + 4. Find ƒ(2). 6. Let g(x) = 2x2 – 3x + 1. Find g(–3). 7-6

Function Operations Solutions
ALGEBRA 2 LESSON 7-6 Solutions 1. {(0, –5), (2, –3), (4, –1)} The domain is {0, 2, 4}. The range is {–5, –3, –1}. 2. {(–1, 0), (0, 0), (1, 0)} The domain is {–1, 0, 1}. The range is {0}. 3. ƒ(x) = 2x – 12 The domain is the set of all real numbers. The range is the set of all real numbers. 4. g(x) = x2 The domain is the set of all real numbers. The range is the set of all nonnegative real numbers. 5. ƒ(x) = 3x + 4; ƒ(2) = 3(2) + 4 = = 10 6. g(x) = 2x2 – 3x + 1; g(–3) = 2(–3)2 – 3(–3) + 1 = 2(9) = = 28 7-6

The domains of ƒ + g and ƒ – g are the set of real numbers.
Function Operations ALGEBRA 2 LESSON 7-6 Let ƒ(x) = –2x + 6 and g(x) = 5x – 7. Find ƒ + g and ƒ – g and their domains. (ƒ + g)(x) = ƒ(x) + g(x) (ƒ – g)(x) = ƒ(x) – g(x) = (–2x + 6) + (5x – 7) = (–2x + 6) – (5x – 7) = 3x – 1 = –7x + 13 The domains of ƒ + g and ƒ – g are the set of real numbers. 7-6

Let ƒ(x) = x2 + 1 and g(x) = x4 – 1. Find ƒ • g and and their domains.
Function Operations ALGEBRA 2 LESSON 7-6 ƒ g Let ƒ(x) = x2 + 1 and g(x) = x4 – 1. Find ƒ • g and and their domains. ƒ g ƒ(x) g(x) (x) = (ƒ • g)(x) = ƒ(x) • g(x) x2 + 1 x4 – 1 = = (x2 + 1)(x4 – 1) = x2 + 1 (x2 + 1)(x2 – 1) = x6 + x4 – x2 – 1 = 1 x2 – 1 The domains of ƒ and g are the set of real numbers, so the domain of ƒ • g is also the set of real numbers. The domain of does not include 1 and –1 because g(1) and g(–1) = 0. ƒ g 7-6

Let ƒ(x) = x3 and g(x) = x2 + 7. Find (g ° ƒ)(2).
Function Operations ALGEBRA 2 LESSON 7-6 Let ƒ(x) = x3 and g(x) = x Find (g ° ƒ)(2). Method 1: (g ° ƒ)(x) = g(ƒ(x)) = g(x3) = x6 + 7 Method 2: (g ° ƒ)(x) = g(ƒ(x)) (g ° ƒ)(2) = (2)6 + 7 = = 71 g(ƒ(2)) = g(23) = g(8) = = 71 7-6

Let x = the original price.
Function Operations ALGEBRA 2 LESSON 7-6 A store offers a 20% discount on all items. You have a coupon worth $3. a. Use functions to model discounting an item by 20% and to model applying the coupon. Let x = the original price. ƒ(x) = x – 0.2x = 0.8x Cost with 20% discount. g(x) = x – 3 Cost with a coupon for $3. b. Use a composition of your two functions to model how much you would pay for an item if the clerk applies the discount first and then the coupon. (g ° ƒ)(x) = g(ƒ(x)) Applying the discount first. = g(0.8x) = 0.8x – 3 7-6

(ƒ ° g)(x) = ƒ(g(x)) Applying the coupon first.
Function Operations ALGEBRA 2 LESSON 7-6 (continued) c. Use a composition of your two functions to model how much you would pay for an item if the clerk applies the coupon first and then the discount. (ƒ ° g)(x) = ƒ(g(x)) Applying the coupon first. = ƒ(x – 3) = 0.8(x – 3) = 0.8x – 2.4 d. How much more is any item if the clerk applies the discount first? (ƒ ° g)(x) – (g ° ƒ)(x) = (0.8x – 2.4) – (0.8x – 3) = 0.6 Any item will cost $.60 more. 7-6

13. 2x2 + 2x – 4; domain: all real numbers
Function Operations ALGEBRA 2 LESSON 7-6 pages 394–398 Exercises 1. x2 + 3x + 5 2. x2 – 3x – 5 3. –x2 + 3x + 5 4. 3x3 + 5x2 5. 6. 7. x2 + 3x + 5 8. –x2 + 3x + 5 9. x2 – 3x – 5 10. 3x3 + 5x2 11. 12. 13. 2x2 + 2x – 4; domain: all real numbers 14. –2x2 + 2; domain: all real numbers 15. 2x2 – 2; domain: all real numbers 16. 2x3 – x2 – 4x + 3; domain: all real numbers 17. 2x + 3; domain: all real numbers except 1 ; domain: all real numbers except and 1 19. 27x2, domain: all real numbers; 3, domain: all real numbers except 0 20. 2x + 3; 9, –1 21. x2 + 5; 14, 9 3x + 5 x2 x2 3x + 5 3x + 5 x2 x2 3x + 5 1 2x + 3 7-6

Function Operations ALGEBRA 2 LESSON 7-6 22. 8 24. 20 25. 16 26. 8 27. 10 28. 12 29. 68 31. 1 32. 25 33. –3 34. 9 38. –2.75 39. c2 – 6c + 9 40. c2 – 3 41. a2 + 6a + 9 42. a2 – 3 43. a. ƒ(x) = 0.9x b. g(x) = x – 2000 c. $14,200 d. $14,400 44. a. (g ° ƒ)(x) = x b pesos 45. x2 – x + 7 46. 6x + 13 47. x2 – 5x – 3 48. –2x2 + 8x + 1 49. –x2 + 5x + 13 50. 2x2 + 2x + 24 51. –3x2 + 2x + 16, domain: all real numbers 52. 3x2 – 12, domain: 53. 3x3 + 8x2 – 4x – 16, domain: 7-6

55. 3x – 4, domain: all real numbers except –2
Function Operations ALGEBRA 2 LESSON 7-6 54. –9x3 – 24x2 + 12x + 48, domain: all real numbers 55. 3x – 4, domain: all real numbers except –2 56. 15x – 20, domain: all real numbers 57. 7; answers may vary. Sample: First evaluate ƒ(3) since the expression is (g ° ƒ)(3), and that means g(ƒ(3)). Then evaluate g(6). 58. 1 59. –4 60. 0 61. 2 62. a ; the area after 2 seconds is about 1963 in.2. b in.2 63. 3x2, 9x2 64. x – 2, x – 2 65. 12x2 + 2, 6x2 + 4 66. x – 3, x – 6 67. –4x – 7, –4x – 28 , 69. Answers may vary. Sample: a. g(x) = 0.12x b. ƒ(x) = 9.50x c. (g ƒ)(x) = 1.14x; your savings will be $1.14 for each hour you work. x2 + 5 2 x2 + 10x + 25 4 7-6

72. a. g(x) is the bonus earned when x is the amount of sales over
Function Operations ALGEBRA 2 LESSON 7-6 70. a. ƒ(x) and g(x) b. 0, 15, 30; 3, 28, 103 c. 3x2 + 9 d. 3*A1^2 + 9, 9, 84, 309 e. 9x2 + 3 f. 9*A1^2 + 3, 3, 228, 903 71. a. P(x) = 5295x – 1000 b. $157,850 72. a. g(x) is the bonus earned when x is the amount of sales over $5000. h(x) is the excess of x sales over $5000. b. (g ° h)(x) because you first need to find the excess sales over $5000 to calculate the bonus. 73. (ƒ+ g)(x) = ƒ(x) + g(x) Def. of Function Add. = 3x – 2 + (x2 + 1) Substitution = x2 + 3x – Comm. Prop. = x2 + 3x – 1 arithmetic 74. (ƒ – g)(x) = ƒ(x) – g(x) def. of function subtraction = 3x – 2 – (x2 + 1) substitution = 3x – 2 – x2 – 1 Opp. of Sum Prop. = –x2 + 3x – 2 – 1 Comm. Prop. = –x2 + 3x – 3 arithmetic 75. (ƒ ° g)(x) = ƒ(g(x)) def. of comp. functions = ƒ(x2 + 1) substitution = 3(x2 + 1) – 2 substitution = 3x2 + 3 – 2 Dist. Prop. = 3x arithmetic 7-6

87. [2] (ƒ • g)(x) means ƒ(x) times g(x) or in this case (3x – 4)
Function Operations ALGEBRA 2 LESSON 7-6 80. x 81. 82. 83. 2 84. 4 85. B 86. F 87. [2] (ƒ • g)(x) means ƒ(x) times g(x) or in this case (3x – 4) (x + 3). The value of (ƒ • g) (x) is 3x2 + 5x – 12. [1] only 3x2 + 5x – 12 OR minor error in explanation 88. A 76. a. ƒ(x) = x + 10; g(x) = 1.09x b. Each grade is increased 9% before adding the 10-point bonus; c. Add the 10-point bonus and then increase the sum by 9%; d. no 77. x7 – x6 – 16x5 + 10x4 + 85x3 – 25x2 – 150x; domain: all real numbers ; domain: all real numbers except 3, 5, and – 5 ; domain: all real numbers except 0, –2, 5, and – 5 1 x 6 – x 8 x2 + 2x x – 3 x – 3 x2 + 2x 7-6

Function Operations 89. C 90. B 91. D 92. 1 93. –3 94. 4 95. 3 96. 2
ALGEBRA 2 LESSON 7-6 89. C 90. B 91. D 92. 1 93. –3 94. 4 95. 3 96. 2 97. 3 98. x8 + 32x x6 + 3584x5 + 17,920x4 + 57,344x ,688x2 + 131,072x + 65,536 99. x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + y6 x4 – 32x3y + 24x2y2 – 8xy3 + y4 x7 – 1344x6y x5y2 – 15,120x4y3 + 22,680x3y4 – 20,412x2y5 + 10,206xy6 – 2187y7 ,049 – 65,610x + 29,160x2 – 6480x3 + 720x4 – 32x5 x5 – 1280x4y + 640x3y2 – 160x2y3 + 20xy4 – y5 104. x8 + 4x7 + 6x6 + 4x5 + x4 105. x x10y3 + 60x8y x6y x4y12 + 192x2y y18 i 107. –2 + 19i 2 i 5 109. –8 – 36i 7-6

ƒ(x) + g(x) = 2x2 + 3x – 1; all reals
Function Operations ALGEBRA 2 LESSON 7-6 Let ƒ(x) = 2x2 and g(x) = 3x – 1. Perform each function operation. Then find the domain. 1. ƒ(x) + g(x) 2. ƒ(x) – g(x) 3. ƒ(x) • g(x) 4. 5. (ƒ ° g)(x) 6. A store is offering a 15% discount on all items. You have a coupon worth $2 off any item. Let x be the original cost of an item. Use a composition of functions to find a function c(x) that gives the final cost of an item if the discount is applied first and then the coupon. Then use this function to find the final cost of an item originally priced at $10. ƒ(x) + g(x) = 2x2 + 3x – 1; all reals ƒ(x) – g(x) = 2x2 – 3x + 1; all reals ƒ(x) • g(x) = 6x3 – 2x2; all reals ƒ(x) g(x) = ƒ(x) g(x) 2x2 3x – 1 all reals except 1 3 (ƒ ° g)(x) = 18x2 – 12x + 2; all reals c(x) = 0.85x – 2; $6.50 7-6

Inverse Relations and Functions
ALGEBRA 2 LESSON 7-7 (For help, go to Lesson 2-1.) Graph each pair of functions on a single coordinate plane. 1. y = x – 6 2. y = 3. y = 3x – 1 y = x + 6 y = 2x + 7 y = 4. y = 0.5x y = –x y = y = 2x – 2 y = y = 5x – 4 x – 7 2 x + 1 3 –x + 4 –1 x + 4 5 7-7

ALGEBRA 2 LESSON 7-7 Solutions 1. y = x – 6 2. y = , or y = x + 6 y = x – y = 2x + 7 3. y = 3x – 1 4. y = 0.5x + 1 y = , or y = 2x – 2 y = x + x – 7 2 1 2 7 2 x + 1 3 1 3 1 3 7-7

ALGEBRA 2 LESSON 7-7 Solutions (continued) 5. y = –x y = , or y = , or y = x + y = x – 4 y = 5x – 4 x + 4 5 –x + 4 –1 1 5 4 5 7-7

ALGEBRA 2 LESSON 7-7 a. Find the inverse of relation m. Relation m x – y –2 –1 –1 –2 Interchange the x and y columns. Inverse of Relation m x –2 –1 –1 –2 y – 7-7

ALGEBRA 2 LESSON 7-7 (continued) b. Graph m and its inverse on the same graph. Reversing the Ordered Pairs Relation m Inverse of m 7-7

ALGEBRA 2 LESSON 7-7 Find the inverse of y = x2 – 2. y = x2 – 2 x = y2 – 2 Interchange x and y. x + 2 = y2 Solve for y. ± x + 2 = y Find the square root of each side. 7-7

ALGEBRA 2 LESSON 7-7 Graph y = –x2 – 2 and its inverse. The graph of y = –x2 – 2 is a parabola that opens downward with vertex (0, –2). The reflection of the parabola in the line x = y is the graph of the inverse. You can also find points on the graph of the inverse by reversing the coordinates of points on y = –x2 – 2. 7-7

ALGEBRA 2 LESSON 7-7 Consider the function ƒ(x) = 2x + 2 . a. Find the domain and range of ƒ. Since the radicand cannot be negative, the domain is the set of numbers greater than or equal to –1. Since the principal square root is nonnegative, the range is the set of nonnegative numbers. b. Find ƒ –1 ƒ(x) = 2x + 2 y = 2x + 2 Rewrite the equation using y. x = 2y + 2 Interchange x and y. x2 = 2y + 2 Square both sides. y = x2 – 2 2 Solve for y. So, ƒ –1(x) = x2 – 2 2 7-7

ALGEBRA 2 LESSON 7-7 (continued) c. Find the domain and range of ƒ –1. The domain of ƒ –1 equals the range of ƒ, which is the set of nonnegative numbers. Note that the range of ƒ–1 is the same as the domain of ƒ. Since x , –1. Thus the range of ƒ–1 is the set of numbers greater than or equal to –1. x2 – 2 2 > – d. Is ƒ –1 a function? Explain. For each x in the domain of ƒ–1, there is only one value of ƒ –1(x). So ƒ –1 is a function. 7-7

ALGEBRA 2 LESSON 7-7 The function d = 16t 2 models the distance d in feet that an object falls in t seconds. Find the inverse function. Use the inverse to estimate the time it takes an object to fall 50 feet. d = 16t 2 t 2 = d 16 Solve for t. Do not interchange variables. t = d 4 Quantity of time must be positive. t = 1 4 The time the object falls is 1.77 seconds. 7-7

ALGEBRA 2 LESSON 7-7 For the function ƒ(x) = x + 5, find (ƒ–1 ° ƒ)(652) and (ƒ ° ƒ–1)(– 86). 1 2 Since ƒ is a linear function, so is ƒ –1. Therefore ƒ –1 is a function. So (ƒ –1 ° ƒ)(652) = 652 and (ƒ ° ƒ –1)(– 86) = – 7-7

ALGEBRA 2 LESSON 7-7 pages 404–406 Exercises 1. 2. 5. y = x – ; yes 6. y = x + ; yes 7. y = – x + ; yes 8. y = y = ± no 9. y = ± x – 4; no 10. y = ± no 11. y = ± x – 1; no 12. y = ± no 13. y = ± ; no 1 3 1 3 3. 4. 1 2 1 2 1 3 4 3 5 – x 2 x + 5 3 x + 4 2 x – 5 – 1 2 7-7

ALGEBRA 2 LESSON 7-7 14. 15. 16. 17. 18. 19. 20. 21. 22. 7-7

ALGEBRA 2 LESSON 7-7 x – 4 3 3 – x2 3 3 2 23. ƒ–1(x) = , and the domain and range for both ƒ and ƒ–1 are all real numbers; ƒ–1 is a function. 24. ƒ–1(x) = x2 + 5, x 0, domain ƒ: {x|x > 5}, range ƒ: {y|y 0}, domain ƒ–1: {x|x 0}, and range ƒ–1: {y|y 5}; ƒ–1 is a function. 25. ƒ–1(x) = x2 – 7, x 0, domain ƒ: {x|x –7}, range ƒ: {y|y 0}, domain ƒ–1: {x|x 0}, and range ƒ–1: {y|y –7}; ƒ–1 is a function. 26. ƒ–1(x) = , x 0, domain ƒ: {x|x }, range ƒ: {y|y 0}, domain ƒ–1: {x|x 0}, and range ƒ–1: {y|y }; ƒ–1 is a function. 27. ƒ–1(x) = ± , x 2, domain ƒ: all reals, range ƒ: {y|y > 2}, domain ƒ–1: {x|x 2}, and range ƒ–1: all reals; ƒ–1 is not a function. 28. ƒ–1(x) = ± 1 – x , x 1, domain ƒ: all reals, range ƒ: {y|y 1}, domain ƒ–1: {x|x 1}, > – < – > – > – 3 2 < – > – x – 2 2 > – > – > – > – > – > – < – > – > – < – < – > – > – 7-7

ALGEBRA 2 LESSON 7-7 3 29. a. F = (C – 32); yes b. –3.89°F 30. a. r = ; yes b ft 31. 10 32. –10 34. d 35. ƒ–1(x) = ± ; no 36. ƒ–1(x) = ± ; no 37. ƒ–1(x) = ± ; yes 3V 4 2x + 8 x x2 – 6x + 10 2 38. ƒ–1(x) = ± x – 1; no 39. ƒ–1(x) = ; no 40. ƒ–1(x) = –1 ± x + 1; no 41. ƒ–1(x) = x; yes 42. ƒ–1(x) = ± x; no 43. ƒ–1(x) = ± ; no 44. x = ; 25 ft, 6.25 ft 45. The range of the inverse is the domain of ƒ, which is x 1. 46. 2 and 5 5x – 5 V 2 64 1± x > – 7-7

ALGEBRA 2 LESSON 7-7 47. ƒ–1(x) = x2, x 0, domain of ƒ: {x|x 0}, range of ƒ: {y|y 0}, domain of ƒ–1: {x|x 0}, range of ƒ–1: {y|y 0}, and ƒ–1 is a function. 48. ƒ–1(x) = (x – 3)2, x 3, domain of ƒ: {x|x 0}, range of ƒ: {y|y 3}, domain of ƒ–1: {x|x 3}, 49. ƒ–1(x) = 3 – x2, x 0, domain of ƒ: {x|x 3}, range of ƒ: {y|y 0}, range of ƒ–1: {y|y 3}, < – 50. ƒ–1(x) = x2 – 2, x 0, domain of ƒ: {x|x –2}, range of ƒ: {y|y 0}, domain of ƒ–1: {x|x 0}, range of ƒ–1: {y|y –2}, and ƒ–1 is a function. 51. ƒ–1(x) = ± 2x , x 0, domain of ƒ: all reals, range of ƒ: {y|y 0}, range of ƒ–1: all reals, and ƒ–1 is not a function. 52. ƒ–1(x) = ± , x > 0, domain of ƒ:{x|x = 0}, range of ƒ:{y|y > 0}, domain of ƒ–1: {x|x > 0}, range of ƒ–1: {y|y = 0}, > – > – > – > – > – > – > – > – > – > – > – > – > – > – > – > – > – 1 x > – < – > – / > – < – / 7-7

ALGEBRA 2 LESSON 7-7 x – 4 2 2 53. ƒ–1(x) = ± x + 4 x 0, domain of ƒ: all reals, range of ƒ: {y|y 0}, domain of ƒ–1: {x|x 0}, range of ƒ–1: all reals, and ƒ–1 is not a function. 54. ƒ–1(x) = 7 ± x x 0, domain of 55. ƒ–1(x) = ± – 1 x > 0, domain of ƒ: {x|x = –1}, range of ƒ: {y|y > 0}, domain of ƒ–1: {x|x > 0}, range of ƒ–1: {y|y = –1}, < – 56. ƒ–1(x) = – x 4, domain of ƒ: {x|x 0}, range of ƒ: {y|y 4}, domain of ƒ–1: {x 4}, range of ƒ–1: { y|y 0}, and ƒ–1 is a function. 57. ƒ–1(x) = x 0, domain of ƒ: {x|x > 0}, range of ƒ: {y|y > 0}, domain of ƒ–1: {x|x > 0}, range of ƒ–1: {y|y > 0}, 58. ƒ–1(x) = – x > 0, domain of ƒ: {x|x < 0}, range of ƒ: {y|y > 0}, range of ƒ–1: {y|y < 0}, < – < – > – < – < – < – > – > – 3 x 2 > – > – > – 1 x 1 2 1 x 2 / / 7-7

ALGEBRA 2 LESSON 7-7 59. a-b. Answers may vary. Sample: a. b. 60. r is not a function because there are two y-values for one x-value. r –1 is a function because each of its x-values has one y-value. 61. h = s 2; in in. 62. Check students’ work. 63. ƒ–1(x) = 5x; yes 64. ƒ–1(x) = x3 + 5; yes 3 65. ƒ–1(x) = 27x3; yes 66. ƒ–1(x) = x; yes 67. ƒ–1(x) = x4; yes 68. ƒ–1(x) = ± no 69. B 70. G 71. C 72. [2] x = 4y2 + 5, so 4y2 = x – 5, y2 = , and y = ± The inverse has values that are real numbers when x 5. [1] y = ± OR x 5 OR minor error 5x 6 4 x – 5 2 > – 7-7

ALGEBRA 2 LESSON 7-7 73. [4] x = y2 – 2y + 1 or x = (y – 1)2. Then y – 1 = ± x or y = ± x + 1. It is not a function because each positive value of x gives two values of y. [3] minor error in finding inverse [2] attempt to find inverse and a statement that the inverse is not a function [1] attempt to find inverse OR a 74. 2x + 28 75. 2x + 7 76. |–x – 10| 77. x + 14 78. |–4x – 10| 79. 2x |–2x + 4| 80. 2 81. –2 82. not a real number 83. 3 84. –3 85. –3 87. 30 88. ±1, ±2, ±3, ±4, ±6, ±12, ± , ± ; roots are – , ±2. 1 2 3 7-7

ALGEBRA 2 LESSON 7-7 89. ±1, ±2, ±4, ± , ± , ± ; roots are , 2, –1. 90. ±1, ±2, ±3, ±4, ±6, ±12, ± , ± , ± ; roots are 1, – , -3. 91. ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30, ± , ± , ± , ± ; roots are 5, – , 2. 92. ±1, ±2, ±3, ±6; roots are 1, 2, 3. 93. ±1, ±2, ±3, ±4, ±6, ±12; roots are –2, 2, –3. 1 3 2 4 5 15 7-7

ALGEBRA 2 LESSON 7-7 1. Find the inverse of the function y = 4x – 7. Is the inverse a function? 2. Find the inverse of y = x + 9. Is the inverse a function? 3. Graph the relation y = 2x2 – 2 and its inverse. 4. For the function ƒ(x) = 2x2 – 1, find ƒ–1, and the domain and range of ƒ and ƒ–1. Determine whether ƒ–1 is a function. 5. If ƒ(x) = 5x + 8, find (ƒ–1 ° ƒ)(59) and (ƒ ° ƒ–1)(3001). 6. A right triangle has a leg of length x and a hypotenuse of length 2x. Write an equation to find the length s of the other leg, and use it to estimate s when the hypotenuse has a length of 5 cm. 7-7

ALGEBRA 2 LESSON 7-7 1. Find the inverse of the function y = 4x – 7. Is the inverse a function? 2. Find the inverse of y = x + 9. Is the inverse a function? 3. Graph the relation y = 2x2 – 2 and its inverse. y = x + 7 4 ; yes y = x2 – 9; yes 7-7

ALGEBRA 2 LESSON 7-7 4. For the function ƒ(x) = 2x2 – 1, find ƒ–1, and the domain and range of ƒ and ƒ–1. Determine whether ƒ–1 is a function. 5. If ƒ(x) = 5x + 8, find (ƒ–1 ° ƒ)(59) and (ƒ ° ƒ–1)(3001). 6. A right triangle has a leg of length x and a hypotenuse of length 2x. Write an equation to find the length s of the other leg, and use it to estimate s when the hypotenuse has a length of 5 cm. ƒ–1(x) = ± x + 2; domain and range of ƒ: all numbers, range of ƒ: all numbers greater than or equal to –1, domain of ƒ–1: all numbers greater than or equal to –1, range of ƒ–1: all numbers; not a function 1 2 59, 3001 s = x 3; about 4.33 cm 7-7

Graphing Radical Functions
ALGEBRA 2 LESSON 7-8 (For help, go to Lesson 5-3.) Graph each equation. 1. y = (x + 2)2 2. y = (x – 3)2 3. y = –(x + 4)2 4. y = –x2 – 1 5. y = –(x + 1) y = 3x2 + 3 7-8

ALGEBRA 2 LESSON 7-8 Solutions 1. y = (x + 2)2 parent function: y = x2 translate 2 units left 2. y = (x – 3)2 translate 3 units right 3. y = –(x + 4)2 parent function: y = –x2 translate 4 units left 7-8

ALGEBRA 2 LESSON 7-8 Solutions (continued) 4. y = –x2 – 1 parent function: y = –x2 translate 1 unit down 5. y = –(x + 1)2 + 1 translate 1 unit left and 1 unit up 6. y = 3x2 + 3 parent function: y = 3x2 translate 3 units up 7-8

ALGEBRA 2 LESSON 7-8 Graph y = x + 5 and y = x – 7. The graph of y = x + 5 is the graph of y = x shifted up 5 units. The graph y = x – 7 is the graph of y = x shifted down 7 units. The domains of both functions are the set of nonnegative numbers, but their ranges differ. 7-8

ALGEBRA 2 LESSON 7-8 Graph y = x + 7 and y = x – 5. The graph of y = x + 7 is the graph of y = x shifted left 7 units. The graph y = x – 5 is the graph of y = x shifted right 5 units. The ranges of both functions are the set of nonnegative numbers, but their domains differ. 7-8

ALGEBRA 2 LESSON 7-8 Graph y = – x, y = –2 x, and y = – x. 1 3 Each y-value of y = –2 x is times the corresponding y-value of y = – x. The domains of the three functions are the set of nonnegative numbers, and the ranges are the set of negative numbers. 7-8

ALGEBRA 2 LESSON 7-8 Graph y = 3 x + 2 – 1. so translate the graph of y = 3 x left 2 units and down 1 unit. y = 3 x – (–2) + (–1), 7-8

ALGEBRA 2 LESSON 7-8 The function h(x) = x models the height h in meters of one group of male giraffes with a body mass of x kilograms. 3 Graph the model with a graphing calculator. Use the Intersect feature to find that x when y = 3. 3 Graph y = x and y = 3. Adjust the window so the graphs intersect. Use the graph to estimate the body mass of a young male giraffe with a height of 3 meters. The giraffe has a mass of about 216 kg. 7-8

ALGEBRA 2 LESSON 7-8 Graph y = 2 x + 2 – 2. 3 3 The graph of y = 2 x + 2 – 2 is the graph of y = 2 x translated 2 units left and 2 units down. 7-8

ALGEBRA 2 LESSON 7-8 Rewrite y = 9x + 18 to make it easy to graph using a translation. Describe the graph. y = 9x + 18 = 9(x + 2) = 3 x + 2 = 3 x – (–2) The graph of y = 9x + 18 is the graph of y = 3 x translated 2 units left. 7-8

ALGEBRA 2 LESSON 7-8 pages 411–413 Exercises 1. 2. 3. 4. 5. 6. 7. 8. 9. 7-8

ALGEBRA 2 LESSON 7-8 10. 11. 12. 13. 14. 15. 16. 17. 18. 7-8

ALGEBRA 2 LESSON 7-8 19. 20. 21. 22. 23. a. b ft, ft, ft 24. 25. 26. 27. 7-8

ALGEBRA 2 LESSON 7-8 28. 29. 30. y = 3 x – 1; the graph is the graph of y = 3 x translated 1 unit to the right. 31. y = –4 x + 2; the graph is the graph of y = –4 x translated 2 units to the left. 32. y = –14 x + 1; the graph is the graph of y = –14 x translated 1 unit to the left. 33. y = 4 x + 2; the graph is the graph of y = 4 x translated 2 units 34. y = 8 x – 2 – 3; the graph is the graph of y = 8 x translated 2 units to the right and 3 units down. 35. y = 3 x – 2 + 1; the graph is the graph of y = 3 x translated 2 units to the right and 1 unit up. 3 3 3 3 7-8

ALGEBRA 2 LESSON 7-8 36. D: x 0, R: y 7 37. D: x 0, R: y –6 38. D: x 6, R: y 0 39. D: x 0, R: y 2 40. D: x 0, R: y 0 41. D: x , R: y 7 1 2 > – < 7-8

ALGEBRA 2 LESSON 7-8 42. D: all real numbers, R: all real numbers 43. D: x 1, R: y 3 44. 45. D: x – , R: y 0 1 2 46. 47. > – < 7-8

ALGEBRA 2 LESSON 7-8 48. D: x –5, R: y –1 49. D: all real numbers, R: all real numbers 50. D: x , R: y 7 51. a. b s; s 52. a. b. D: x 2, R: y –2 c. (2, –2) d. The domain is based on the x-coordinate of that point, and the range is based on the y-coordinate. > – > – > – 3 4 < – > – < – 7-8

ALGEBRA 2 LESSON 7-8 56. y = x ; the graph is the graph of y = 6 x translated 3 units to the left and 4 up. 57. y = – 2 x – ; the graph is the graph of y = –2 x translated unit to the right. 58. y = x – 1 – 2; the graph is the same as y = x translated 1 unit right and 2 down. 59. y = 10 – x + 3; the graph is the same as y = – x translated 3 units to the left and 10 up. 1 3 2 4 > – < 53. a. y = x – 5 – 2 b. y = x – 1 – 5 54. a. b. Both domains are x 2. The range of y = x – 2 + 1 is y 1. The range of y = – x – is y 1. 55. y = x – 4 – 1; the graph is the same as y = 5 x, translated 4 units to the right and 1 down. 7-8

ALGEBRA 2 LESSON 7-8 3 60. y = x ; the graph is the same as y = x, translated 9 units to the left and 5 up. 61. Answers may vary. Sample: y = x – 2 + 4 62. a. b. 20 in. 63. If a > 0, the graph is stretched vertically by a factor of a. If a < 0, the graph is reflected over the x-axis and stretched vertically by a factor of |a|. 1 64. y = – x + 4; the graph is the graph of y = – 2x translated 4 units to the left; domain: x –4, range: y 0. 65. y = – x – ; the graph is the graph of y = – 8x translated units to the right; domain: x , 66. y = • x – ; the graph is the graph of y = 3x translated units to the right and 6 units up; domain: x , range: y 6. 4 5 > – < 7-8

ALGEBRA 2 LESSON 7-8 72. B 73. H 74. [2] Both graphs have the same shape as the graph of y = x. The graph of ƒ(x) = x – 1 is the graph of y = x moved 1 unit to the right, and the graph of g(x) = x – 1 is the graph of y = x moved 1 unit down. [1] minor error in either direction 67. y = – • x – – 3; the graph is the graph of y = – 12x translated units to the left and 3 units down; domain: x – , range: y –3. 68. a. b. The graph of y = h – x is a reflection of the graph of y = x – h in the line x = h. 69. for all odd positive integers 70. A 71. I 3 2 3 2 3 2 > – < – 7-8

ALGEBRA 2 LESSON 7-8 81. ƒ–1(x) = ; yes 82. x 3 83. 84. 85. 86. 87. 88. 89. –5 ± 90. 6 ± , – 48x3y4 2y 3 9 ± 2 –3 ± 75. [4] For ƒ(x) = x – 1 the domain is x 1 and the range is y 0. For g(x) = x – 1 the domain is x 0 and the range is y –1. [3] minor error in one of the descriptions of domains or ranges [2] correct domains or ranges [1] some attempt to describe the domains 76. ƒ–1(x) = ; yes 77. ƒ–1(x) = ; yes 78. ƒ–1(x) = ± ; no 79. ƒ–1(x) = (x + 4)2 – 3; yes 80. ƒ–1(x) = ; no x + 1 4 x – 1 2.4 –1 ± x 3(x + 3) x –1 ± 61 10 12xy2 5 9xy2 3y > – 7-8

ALGEBRA 2 LESSON 7-8 Graph each function on the same graph. 1. y = x – 6 2. y = x + 8 3. y = x 4. y = x – 2 + 5 5. The formula t = can be used to estimate the number of seconds t it takes a pendulum of length L meters to make one complete swing. Graph the equation on a graphing calculator. Then use the graph to estimate the values of t for pendulums of lengths 1.5 meters and 2.5 meters. 6. Rewrite y = x – 27 to make it easy to graph using a translation. 2 3 3 L 9.8 About 2.46 s, about 3.17 s y = 6 x – 3 7-8

Radical Functions and Radical Exponents
ALGEBRA CHAPTER 7 2 3 x xy 3y 6x – 3x 2 6x 5 1 25 4x4 x2 9y 6 24. x – 1, domain {x|x = 2} 25. –x3 + 5x2 – 8x + 4, domain: all reals 26. 16x2 + 8x – 1, 4x2 – 7 27. 18x2 + 9x – 6, –6x2 – 3x + 20 28. The sixth power of a real number is always nonnegative. 29. a. ƒ(x) = 0.5x b. g(x) = 0.75x c. g(g(x)) = x d. The cashier’s solution is too high by 6.25% of the original price. / Page 418 1. –0.3 2. 3|x|y2 6xy 3. –2x2y4 2x4 4. (x – 2)2 5. 7x2 2 6. 6x3y4 2y 7. 8. – 5 11. –17 – 13. – 1 – 15. 16. x 17. 18. 7 19. 2 20. –43 21. 8 22. x2 – 4x + 4; domain: all reals 23. –2x2 + 7x – 6; domain: all reals 7-A

ALGEBRA CHAPTER 7 30. y = 4 x + 5 – 1; y = 4 x translated 5 units left and 1 unit down 31. y = 3 x + ; y = 3 x translated unit left 32. D: {x|x 0}, R: {y|y 3} 33. D: {x|x – }, R: {y|y 0} 34. D: {x|x 4}, 35. D: {x|x – }, R: {y|y –4} 36. –5831 38. –63 1 3 1 3 3 2 3 2 > – > – > – > – > – > – > – < – 7-A

ALGEBRA CHAPTER 7 39. ƒ–1(x) = ; yes 40. g–1(x) = (x + 1)2 – 3; yes 41. g–1(x) = ; yes 42. ƒ–1(x) = ± 4x; no 43. a. V = b. r = c. r in. 44. Check students’ work. seconds; seconds 3 x + 2 3 x2 – 1 2 4 256 3 3V 4 1 3 7-A

Roots and Radical Expressions

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