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Roots and Radical Expressions

Write each number as a square of a number. 1. 25 2. 0.09 3. Write each expression as a square of an expression. 4. x 10 5. x 4 y 2 6. 169 x 6 y 12. 4 49. Roots and Radical Expressions. ALGEBRA 2 LESSON 7-1. (For help, go to Lesson 5-4.). 7-1.

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Roots and Radical Expressions

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  1. Write each number as a square of a number. 1. 25 2. 0.09 3. Write each expression as a square of an expression. 4.x105.x4y26. 169x6y12 4 49 Roots and Radical Expressions ALGEBRA 2 LESSON 7-1 (For help, go to Lesson 5-4.) 7-1

  2. 1. 25 = 522. 0.09 = 0.323. = = 2 4.x10 = (x5)25.x4y2 = (x2y)26. 169x6y12 = (13x3y6)2 4 49 22 72 2 7 Roots and Radical Expressions ALGEBRA 2 LESSON 7-1 Solutions 7-1

  3. 1 64 1 64 1 64 a. the cube root of 0.027, –125, and 1 4 1 4 Since 3 = , is the cube root of . 81 625 81 625 81 625 81 625 b. the fourth roots of 625, –0.0016, and –3 5 3 5 3 5 3 5 Since 4 = and 4 = , and – are fourth roots of . Roots and Radical Expressions ALGEBRA 2 LESSON 7-1 Find all the real roots. Since 0.33 = 0.027, 0.3 is the cube root of 0.027. Since (–5)3 = –125, –5 is the cube root of –125. 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 –0.0016. 7-1

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

  5. a. 9x10 9x10 =32(x5)2 = (3x5)2 = 3| x5| Absolute value symbols ensure that the root is positive when x is negative. 3 b.a3b3 3 (ab)3 = ab Roots and Radical Expressions ALGEBRA 2 LESSON 7-1 Simplify each radical expression. 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

  6. c.x16y4 4 x16y14 = (x4)4(y)4 = x4 |y| 4 4 Roots and Radical Expressions ALGEBRA 2 LESSON 7-1 (continued) 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

  7. < < < < < < < < < < – – – – – – – – – – 10 w 11 Write an inequality. d3 5 10 11 Substitute for w in terms of d. 50 d3 55 Multiply by 5. d3 5 50 d3 55 Take cube roots. 3 3 3 3.68 d 3.80 The diameters range from 3.68 in. to 3.80 in. Roots and Radical Expressions 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. Find the diameter of the cheese balls. 7-1

  8. Roots and Radical Expressions 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

  9. Roots and Radical Expressions ALGEBRA 2 LESSON 7-1 pages 366–367  Exercises 1. 15, –15 2. 0.07, –0.07 3. none 4. , – 5. –4 6. 0.5 7. – 8. 0.07 9. 2, –2 10. none 11. 0.3, –0.3 12. , – 13. 6 14. –6 15. no real-number root 16. 0.6 17. –4 18. –4 19. –3 20. no real-number root 21. 4|x| 22. 0.5|x3| 23.x4|y9| 24. 8b24 25. –4a 26. 3y2 27.x2|y3| 28. 2y2 29. 1.34 in. 30. 1.68 ft 31. 0.48 cm 32. 0.08 mm 33. 10, –10 34. 1, –1 35. 0.5, –0.5 10 3 10 3 8 13 8 13 1 2 7-1

  10. Roots and Radical Expressions ALGEBRA 2 LESSON 7-1 36. , – 37. –64, 64, – –64, 64 38.a. 35 ft b. 20 ft longer 39. 0.5 40. 41. 0.2 42. 43. 2|c| 44. 3xy2 3 45. 12y2z2x xy 46.y4 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 2 3 3 6 3 1 3 1 4 3 3 4 5 7-1

  11. Roots and Radical Expressions 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 = 0. 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

  12. Roots and Radical Expressions ALGEBRA 2 LESSON 7-1 75. x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5 76. 16 – 96y + 216y2 – 216y3 + 81y4 77. 729x6 – 7290x5 + 30,375x4 – 67,500x3 + 84,375x2 – 56,250x + 15,625 78. 128a7 – 448a6b + 672a5b2 – 560a4b3 + 280a3b4 – 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 79 8 2 1 4 7-1

  13. 3 4 3 1 25 1 216 Roots and Radical Expressions 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. 0.49 b. 125 c. – 81 d. –625 4. Simplify each radical expression. a. –8x3b. 16y4c. 36x14 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. ± 11 none ± 8 none 1 6 –20 5 –9 0.7 none 6 | x7 | 4y2 –2x 1 3 about 3.37 cm 7-1

  14. Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 (For help, go to page 362.) Find each missing factor. 1. 150 = 52( ) 2. 54 = ( )3(2) 3. 48 = 42( ) 4.x5 = ( )2(x) 5. 3a3b4 = ( )3(3b) 6. 75a7b8 = ( )2(3a) 7-2

  15. Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 1. 150 = 52(6) 2. 54 = (3)32 3. 48 = 42(3) 4.x5 = (x2)2(x) 5. 3a3b4 = (ab)3(3b) 6. 75a7b8 = (5a3b4)2(3a) Solutions 7-2

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

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

  18. 3 3 3 3 3 Factor into perfect cubes. 3 25xy8 • 5x4y3 = 25xy8 • 5x4y3 = 53x3(y3)3 • x2y2 3 3 = 53x3(y3)3 • x2y2 = 5xy3x2y2 definition of nth root 3 a • b = ab a • b = ab n n n n n n Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 Multiply and simplify . Assume all variables are positive. 25xy8 • 5x4y3 7-2

  19. = 192x8 3 –81 3 3 3 3 = = 3x –81 3 3 = 3 3 3 192x8 3x 3 = 64x7 = –27 3 = 43(x2)3 • x = (–3)3 = 4x2x 3 Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 Divide and simplify. Assume all variables are positive. a. b. = –3 7-2

  20. 3 3 Method 1: Rewrite as a square root of a fraction. 3 5 = 5 5 Then make the denominator a perfect square. 3 • 5 5 • 5 = 15 52 = = 5 15 Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 Rationalize the denominator of each expression. Assume that all variables are positive. a. 7-2

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

  22. c. b. x5 x5 x5 3x2y 3x2y 3x2y 3x2y 3x2y • 3x7y x3 3xy x3xy 5 4y 5 4y 3 3 5 • 42y2 4y • 42y2 = = 3 3x2y 3x2y 3y • Rewrite the fraction so the denominator is a perfect cube. = = 10y2 2y 3 2 10y2 4y 80y2 43y3 3 = 3 = = = Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 (continued) 7-2

  23. 10d 7 t2 = t = d 4.9 d 4.9 = d • 10 49 = Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 The distance d in meters that an object will fall in t seconds is given by d = 4.9t2. Express t in terms of d and rationalize the denominator. d = 4.9t2 7-2

  24. 45x2 3x Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 24. 25. 2x2y2 2 26. 5x3 x2y2 27. 28. 29. 30. 31. 32. 5x2 5 33. 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 10. 3 3x2 11. 5x2 2x 12. 2a 4a2 13. 3y3 2y 14. 10a3b3 2b 15. –5x2y 2y2 16. 2y 4x3y2 17. 2 12 18. 8y3 5y 19. 7x3y4 6y 20. 40xy 3 21. 30y2 2y 22. –2x2y 30x 23. 10 4x y 3 3 3 2x 2 4 10x 4x 3 3 4x 2 3 4 250 5 3 3 3 15y 5y 7-2

  25. 34. 35. r = 36. a. b. c. Answers may vary. Sample: First simplify the denominator. Since 98 = 2 • 49 = 7 2, to rationalize the denominator, multiply the fraction by . This yields = . 37. 10 2 38. 4 5 39. 3x6y5 2y 40. 20x2y3y 41. 10 + 7 2 42. 15 + 3 21 43. 5 + 5 3 44. 2x 2 45. 3x2x 46. 47. 3 x 10 2y 3 3 Gm1m2F F 2 25x x 6 + 3 15 3 6 + 3 15 33x 4x 3 3 3 5 + 5 5 2xy2 xy 3x2 3x 3 6 – 2 4 2 2 3 2 • 2 + 3 • 2 7 2 • 2 x 10y 2y2 2 + 6 14 5 14x 21x Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 48. 49. 50. 51. – 52. –4 4 – 6 2 53. 54. 55. 212 mi/h greater 56. 20 22 cm2 7-2

  26. 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. 58. 288a5 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 63. 2 5 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 3 x2y xy 5 yx x 6 x4y3 y 3 Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 7-2

  27. < – 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 > – 3 4x2 4x2 3 2x 3 3 3 3 3 3 3 12x2 8x3 12x2 2x 3 2x 4x2 4x2 4x2 4x2 3 3 2x 3 3 3 3 Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 7-2

  28. 88. 25 89. 25 90. 91. 92. 93. 0.0225 94. 95. 76. 11|a45| 77. –9c24d32 78. –4a27 79. 2y5 80. 0.5|x3| 81.x2y5 82. 2|x9|y24 83. 0.08x20 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 121 4 121 4 1 36 9 64 9 100 Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 7-2

  29. 3 3 –3xy3 9y 2x2 2x 3 3 2xy 5xy2 3 3 3 6x2yxy 3 y y 270x 10xy2 3 3 3 8xy y 128x3 2xy 3 3 x2 4 2x2 2 3 21x 3 7x 3 Multiplying and Dividing Radical Expressions ALGEBRA 2 LESSON 7-2 Assume that all variables are positive. 1. Multiply. Simplify if possible. a. 5 • 45 b. 4 • 2000 2. Simplify. a. 8x5b. –243x3y10 3. Multiply and simplify. a. 18x3 • 2x2y3b. 10x2y4 • 4x2y 4. Divide and simplify. a.b. 5. Rationalize the denominator of each expression. a.b. 15 20 7-2

  30. 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)26. (2x – 9)2 7-3

  31. Binomial Radical Expressions 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

  32. a. 7 xy + 3 xy Distributive Property 7 xy + 3 xy = (7 + 3) xy Subtract. = 10 xy b. 2 x – 2 5 3 3 Binomial Radical Expressions ALGEBRA 2 LESSON 7-3 Add or subtract if possible. The radicals are not like radicals. They cannot be combined. 7-3

  33. The height of an equilateral triangle with side length 4 ft is 2 3 ft. So the window’s height is 2 3 ft. The perimeter is 2(2 3 + 8) ft or 16 + 4 3 ft. Binomial Radical Expressions 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 length of the window is 8 ft. 7-3

  34. Factor each radicand. = 3 • 2 5 – 3 5 + 4 • 4 5 Simplify each radical. 3 20 – 45 + 4 80 = 3 22 • 5 – 32 • 5 + 4 42 • 5 = 6 5 – 3 5 + 16 5 Multiply. = (6 – 3 + 16) 5 Use the Distributive Property. = 19 5 Binomial Radical Expressions ALGEBRA 2 LESSON 7-3 Simplify 3 20 – 45 + 4 80. 7-3

  35. (2 + 4 3)(1 – 5 3) = 2 • 1 – 2 • 5 3 + 4 3 • 1 – 4 3 • 5 3 Use FOIL. = 2 + (4 – 10) 3 – 60 Distributive Property = –58 – 6 3 Binomial Radical Expressions ALGEBRA 2 LESSON 7-3 Multiply (2 + 4 3)(1 – 5 3). 7-3

  36. (3 + 7)(3 – 7) = 32 – ( 7)2 (a + b)(a – b) = a2 – b2 Binomial Radical Expressions ALGEBRA 2 LESSON 7-3 Multiply (3 + 7)(3 – 7). = 9 – 7 = 2 7-3

  37. 4 – 3 4 – 3 2 – 3 4 + 3 2 – 3 4 + 3 2 – 3 4 + 3 2 – 3 4 + 3 4 – 3 4 – 3 = • 4 – 3 is the conjugate of 4 + 3. Multiply. = • 8 – 2 3 – 4 3 + ( 3)2 Simplify. = 42 – ( 3)2 = 11 – 6 3 13 Binomial Radical Expressions ALGEBRA 2 LESSON 7-3 Rationalize the denominator of 7-3

  38. 12 3 + 8 23 11 + 8 2 –14 Binomial Radical Expressions ALGEBRA 2 LESSON 7-3 pages 376–378  Exercises 1. 6 6 2. 4 3 3. cannot combine 4. –2 x 5. cannot combine 6. 5 x2 7. 33 2 8. 13 5 9. 7 2 10. 5 2 11. 9 3 – 6 2 24. 25. 13 + 7 3 26. 27. 13 2 28. 8 3 29. 48 2x 30. 5 3 – 4 2 31. 33y 6 32. –2 2 33. –11 + 21 34. 8 + 10 35. 17 + 31 2 12. 2 2 + 2 3 13. 8 + 4 5 14. 23 + 7 7 15. 63 – 38 2 16. 8 + 2 15 17. 49 + 12 13 18. 38 + 12 10 19. 14 20. 4 21. –40 22. –2 23. –2 + 2 3 4 4 3 3 3 3 3 3 7-3

  39. 89 + 42 3 –239 60 – 20 2 7 3 – 7 2 2 + 3 4 2 x + 5 x3 x –1 + 5 2 1 + 5 2 Binomial Radical Expressions ALGEBRA 2 LESSON 7-3 46.a must be twice a perfect square. 47. 14 7 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)( 7 – 2); (2 2 + 5)(2 2 – 5) 50. s, or about 4.53 s 51. – 52. 4 3 36. –36 – 15 2 37. x + 3 3x + 6 38. 8y – 22 2y + 30 39. 40. 2 3 – 2 41. 42. 43. 44. 2 2 – 12 45. The reciprocal is , which is one less than . 3 4 3 3 1 2 7-3

  40. = / Binomial Radical Expressions 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)xax – 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 + 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

  41. 2 5 + 2 2 2 5 + 2 2 3 5 – 2 2 3 5 – 2 2 5 + 2 2 5 + 2 2 5 – 2 2 5 – 2 2 –5 – 10 2 17 100x2 5x 10 – 4 2 – 15 – 6 2 17 15 + 6 2 17 15 + 6 2 25 – 4 • 2 15 + 6 2 17 25 + 2 2 17 10 – 4 2 25 – 4 • 2 10 – 4 2 17 10 – 4 2 17 Binomial Radical Expressions 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 63. 3 2 64.x 15 65. 4 66. 67. 2x 68. 7x2 2 69. 70. 71. 2, –1 ± i 3 72. –10, 5 ± 5i 3 73. , 74. ± 7 75. 76. ± , ± 3 ±2 5 5 1 3 i 3 93 3 3n n 3 –1 ± i 3 10 1 5 7-3

  42. 3 3 3 3 10 3 2 3 29 + 3 3 16 – 17 7 93 2 10 13 x 3 Binomial Radical Expressions ALGEBRA 2 LESSON 7-3 Simplify. 1. 6 10 – 4 10 2. 8 x + 5 x 3. 2 27 + 48 4. 128 – 54 5. Multiply (7 + 2 3) (5 – 3). 6. Rationalize the denominator in . 3 – 2 7 10 – 7 7-3

  43. Rational Exponents ALGEBRA 2 LESSON 7-4 (For help, go to page 362.) Simplify. 1. 2–42. (3x)–2 3. (5x2y)–34. 2–2 + 4–1 5. (2a–2b3)46. (4a3b–1)–2 7-4

  44. 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 = = Solutions 1 9x2 1 125x6y3 1 (3x)2 1 24 1 16 1 4 1 4 2 4 1 2 1 22 1 41 1 (5x2y)3 16b12 a8 b2 16a6 b2 42a6 Rational Exponents ALGEBRA 2 LESSON 7-4 7-4

  45. 1 3 a. 64 1 3 64 = 64 Rewrite as a radical. 3 3 = 43Rewrite 64 as a cube. 1 2 1 2 b. 7 • 7 1 2 1 2 7 • 7 = 7 • 7 Rewrite as radicals. = 7 By definition, 7 is the number whose square is 7. Rational Exponents ALGEBRA 2 LESSON 7-4 Simplify each expression. = 4 Definition of cube root. 7-4

  46. 1 3 1 3 c. 5 • 25 1 3 1 3 5 • 25 = 5 • 25 Rewrite as radicals. 3 3 3 = 5 • 25 property for multiplying radical expressions 3 = 5 By definition, 5 is the number whose cube is 5. Rational Exponents ALGEBRA 2 LESSON 7-4 (continued) 7-4

  47. 2 5 – 2 7 1 y 2 a. Write x and y –0.4 in radical form. 5 7 7 1 y2 2 7 x = x2 or x2 y –0.4 = y = or 5 b. Write the radical expressions and in exponential form. 3 4 = c c3 c3 4 4 5 3 = b b 5 b 5 3 3 Rational Exponents ALGEBRA 2 LESSON 7-4 7-4

  48. 4.15 Use a calculator. 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 = 0.89p0.6Write the formula. = 0.89(13)0.6Substitute for p. 7-4

  49. 2 3 a. (–27) Method 1 Method 2 2 3 (–27) = ( –27)2 3 2 3 2 3 (–27) = ((–3)3) = ( (–3)3)2 3 = (–3)2 = 9 = (–3)2 = 9 Rational Exponents ALGEBRA 2 LESSON 7-4 Simplify each number. 7-4

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

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