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Simplifying Radicals

Simplifying Radicals. ALGEBRA 1 LESSON 11-1. (For help, go to Lessons 8-3 and 10-3.). Complete each equation. 1. a 3 = a 2 • a 2. b 7 = b 6 • b 3. c 6 = c 3 • c 4. d 8 = d 4 • d Find the value of each expression. 5. 4 6. 169 7. 25 8. 49. 11-1.

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Simplifying Radicals

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  1. Simplifying Radicals ALGEBRA 1 LESSON 11-1 (For help, go to Lessons 8-3 and 10-3.) Complete each equation. 1.a3 = a2 • a2.b7 = b6 • b 3.c6 = c3 • c4. d8 = d4 • d Find the value of each expression. 5. 4 6. 169 7. 25 8. 49 11-1

  2. 1.a3 = a(2 + 1) = a2 • a12.b7 = b(6 + 1) = b6 • b1 3.c6 = c(3 + 3) = c3 • c34.d8 = d(4 + 4) = d4 • d4 5. 4 = 2 6. 169 = 13 7. 25 = 5 8. 49 = 7 Simplifying Radicals ALGEBRA 1 LESSON 11-1 Solutions 11-1

  3. Simplify 243. 243 = 81 • 3 81 is a perfect square and a factor of 243. = 81 • 3 Use the Multiplication Property of Square Roots. = 9 3 Simplify 81. Simplifying Radicals ALGEBRA 1 LESSON 11-1 11-1

  4. Simplify 28x7. 28x7 = 4x6 • 7x4x6 is a perfect square and a factor of 28x7. = 4x6 • 7xUse the Multiplication Property of Square Roots. = 2x3 7x Simplify 4x6. Simplifying Radicals ALGEBRA 1 LESSON 11-1 11-1

  5. a. 12 • 32 12 • 32 = 12 • 32 Use the Multiplication Property of Square Roots. = 384 Simplify under the radical. = 64 • 6 64 is a perfect square and a factor of 384. = 64 • 6 Use the Multiplication Property of Square Roots. = 8 6 Simplify 64. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. 11-1

  6. 7 5x • 3 8x = 21 40x2Multiply the whole numbers and use the Multiplication Property of Square Roots. = 21 4x2 • 10 4x2 is a perfect square and a factor of 40x2. = 21 4x2 • 10 Use the Multiplication Property of Square Roots. = 21 • 2x 10 Simplify 4x2. = 42x 10 Simplify. Simplifying Radicals ALGEBRA 1 LESSON 11-1 (continued) b. 7 5x • 3 8x 11-1

  7. Suppose you are looking out a fourth floor window 54 ft above the ground. Use the formula d = 1.5h to estimate the distance you can see to the horizon. d = 1.5h = 1.5 • 54Substitute 54 for h. = 81 Multiply. = 9 Simplify 81. Simplifying Radicals ALGEBRA 1 LESSON 11-1 The distance you can see is 9 miles. 11-1

  8. 49 x4 13 64 49 x4 13 64 a. 13 64 49 x4 = Use the Division Property of Square Roots. = Use the Division Property of Square Roots. 13 8 = Simplify 64. b. 7 x2 = Simplify 49 and x4. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. 11-1

  9. 120 10 a. 120 10 = 12 Divide. = 4 • 3 4 is a perfect square and a factor of 12. = 4 • 3 Use the Multiplication Property of Square Roots. = 2 3 Simplify 4. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. 11-1

  10. b. = Use the Division Property of Square Roots. 75x5 48x 75x5 48x 25x4 16 25 • x4 16 = Divide the numerator and denominator by 3x. = Use the Multiplication Property of Square Roots. 5x2 4 25x4 16 = Simplify 25, x4, and 16. Simplifying Radicals ALGEBRA 1 LESSON 11-1 (continued) 11-1

  11. a. 3 7 7 7 3 7 3 7 7 7 = • Multiply by to make the denominator a perfect square. 3 7 49 = Use the Multiplication Property of Square Roots. 3 7 7 = Simplify 49. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. 11-1

  12. 11 12x3 11 12x3 11 12x3 b. 3x 3x 3x 3x = • Multiply by to make the denominator a perfect square. 33x 36x4 = Use the Multiplication Property of Square Roots. 33x 6x2 = Simplify 36x4. Simplifying Radicals ALGEBRA 1 LESSON 11-1 (continued) Simplify the radical expression. 11-1

  13. 21 7 3 3 2 2 30 11 5 3a 7 4c 5 3a 7 Simplifying Radicals ALGEBRA 1 LESSON 11-1 26. 12 mi 27. 17 mi 28. 29. 30. 31. 32. 33. 34. pages 581–583  Exercises 1. 10 2 2. 7 2 3. 5 3 4. –4 5 5. –6 30 6. 40 5 7. 2n 7 8. 6b2 3 9. 6x 3 10. 2n n 11. 2a2 5a 12. –4b2 3 13. 20 14. 18 15. 11 2 16. 84 3 17. 7 3 18. –30 3 19. 6n 2 20. 14t 2 21. 3x2 17 22. 80t3 23. 3a3 2 24. –6a2 2 25. 3 mi 5 2 11-1

  14. 2n 2n 9 55y 2y 21x 7x 2 10n 5n 2b b 3 2 4 9 2 4 3 2 2 a2 3 2 Simplifying Radicals ALGEBRA 1 LESSON 11-1 45. 5 46. 47. 48. 49. 2 3 50. 51. 52. not simplest form; radical in the denominator of a fraction 53. not simplest form; radical in the denominator of a fraction 35. 36. 3 37. 38. 2 3 39. –2 5 40. 2x 7 41. 42. 43. 44. 54. Simplest form; radicand has no perfect-square factors other than 1. 55. Simplest form; radicand has no perfect-square factors other than 1. 56.a. 18 • 10 = 180 = 36 • 5 = 6 5 b. Answers may vary. Sample: a = 36, b = 5; a = 9, b = 20 57. 30 58. 2 13 59. 3 2 s 3 3 y 11-1

  15. 3m 4m –2 a a2 2 ± 10 3 xy y2 8 6a 3a Simplifying Radicals ALGEBRA 1 LESSON 11-1 77. 30a4 78. seconds 79. C 80. F 81. B 82. I 83.[2]A = 96 ft2 s = 96 = 16 • 6 = 4 6 ft [1] correct answer, without work shown 84. quadratic; y = 0.2x2 85. exponential; y = 4(2.5)x 86. linear; y = –4.2x + 7 69. –3 ± 3 2 70. 1 ± 5 71. 72. a. 50 = 25 • 2 = 25 • 2 = 5 2 b. The radicand has no perfect-square factors other than 1. 73. Answers may vary. Sample: 12, 27, 48. 74.a. 2 6 in. b. 4.90 in. 75. 12x 76. 10b2 60. 61. 2 15 62. 63. – 3 64. 4 5 65. 2ab 5b 66. ab2cabc 67. 68. 11-1

  16. Simplifying Radicals ALGEBRA 1 LESSON 11-1 89. 90. 3n2 + 5n + 5 91. 3v2 – v – 9 92. 5t3 + 8t2 – 14t – 11 93. –3b2 – 23b – 21 87. 88. 11-1

  17. 12 36 3 3 8 2 2 a5 2 a a3 3x 15x3 5 5x Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. 1. 16 • 8 2. 4 144 3. 4.5. 48 11-1

  18. The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 (For help, go to Lesson 10-4.) Simplify each expression. 1. 52 + 622. 92 – 423. (3t)2 + (4t)2 Solve each equation. 4.c2 = 36 5. 24 + b2 = 49 6.a2 + 16 = 65 7. 12 + b2 = 32 8. 80 = c29. 100 = a2 + 52 11-2

  19. Solutions 1. 52 + 62 = (5 • 5) + (6 • 6) = 25 + 36 = 61 2. 92 – 42 = (9 • 9) – (4 • 4) = 81 – 16 = 65 3. (3t)2 + (4t)2 = (32 • t2) + (42 • t2) = 9t2 + 16t2 = 25t2 4.c2 = 36 c = ± 36 = ±6 5. 24 + b2 = 49 6.a2 + 16 = 65 b2 = 49 – 24 a2 = 65 – 16 b2 = 25 a2 = 49 b = ± 25 = ±5 a = ± 49 = ±7 7. 12 + b2 = 32 8. 80 = c2 b2 = 32 – 12 c = ± 80 = ± 16 • 5 = ± 16 • 5 = ±4 5 b2 = 20 b = ± 20 = ± 4 • 5 = ± 4 • 5 = ± 2 5 9. 100 = a2 + 52 100 – 52 = a2 48 = a2 a = ± 48 = ± 16 • 3 = ± 16 • 3 = ± 4 3 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 11-2

  20. 289 = c2Find the principal square root of each side. The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 What is the length of the hypotenuse of this triangle? a2 + b2 = c2Use the Pythagorean Theorem. 82 + 152 = c2Substitute 8 for a and 15 for b. 64 + 225 = c2Simplify. 17 = cSimplify. The length of the hypotenuse is 17 m. 11-2

  21. The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 A toy fire truck is near a toy building on a table such that the base of the ladder is 13 cm from the building. The ladder is extended 28 cm to the building. How high above the table is the top of the ladder? Define: Let b = height (in cm) of the ladder from a point 9 cm above the table. Relate: The triangle formed is a right triangle. Use the Pythagorean Theorem. 11-2

  22. b2 = 615 Find the principal square root of each side. b 24.8 Use a calculator and round to the nearest tenth. The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 (continued) Write: a2 + b2 = c2 132 + b2 = 282Substitute. 169 + b2 = 784 Simplify. b2 = 615 Subtract 169 from each side. The height to the top of the ladder is 9 cm higher than 24.8 cm, so it is about 33.8 cm from the table. 11-2

  23. 52 + 52 72Determine whether a2 + b2 = c2, where c is the longest side. 25 + 25 49 Simplify. 50 = 49 / 102 + 242 262Determine whether a2 + b2 = c2, where c is the longest side. 100 + 576 676 Simplify. 676 = 676 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 Determine whether the given lengths are sides of a right triangle. a. 5 in., 5 in., and 7 in. This triangle is not a right triangle. b. 10 cm, 24 cm, and 26 cm This triangle is a right triangle. 11-2

  24. 502 + 1202 1302Determine whether a2 + b2 = c2 where c is the greatest force. 2500 + 14,400 16,900 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 If two forces pull at right angles to each other, the resultant force is represented as the diagonal of a rectangle, as shown in the diagram. The diagonal forms a right triangle with two of the perpendicular sides of the rectangle. For a 50–lb force and a 120–lb force, the resultant force is 130 lb. Are the forces pulling at right angles to each other? 16,900 = 16,900 The forces of 50 lb and 120 lb are pulling at right angles to each other. 11-2

  25. 4 15 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 pages 587–590  Exercises 1. 10 2. 25 3. 17 4. 26 5. 2.5 6. 1 7. 4 8. 5 9. 12 10. 7.1 11. 7.5 24. no 25. yes 26. 1.5 27. or 0.3 28. 3 29. 6 30. 2.6 31. 7.0 32.a. 6 5 ft b. 80.5 ft2 33. yes 34. no 12. 0.6 13. 1.2 m 14. about 15.5 ft 15. about 5.8 km 16. yes 17. no 18. no 19. yes 20. no 21. yes 22. yes 23. no 11-2

  26. The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 35. yes 36. yes 37. 4.2 cm 38. 1000 lb 39. 559.9 40. 9.0 41. 9.7 42.a. These lengths could be 2 legs or one leg and the hypotenuse. b. about 12.8 in. or 6 in. 43.a. 62 + 82 = 36 + 64 = 100 = 102 b. 5; 12; 7; 41 c. Answers may vary. Sample: 10, 24, 26 44.a. 6.9 ft b. 89.2 ft2 c. 981 watts 45. 12.8 ft 46.a. Answers may vary. Sample: 5, 20, 5 b. 5 units2 47.a. 13.4 ft b. 17.0 ft c. 10.6 ft d. No; the hypotenuse d must be longer than each leg. 48. An integer has 2 as a factor; the integer is even; if an integer is even, then it has 2 as a factor; true. 11-2

  27. The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 54. 10 55. 4 3 56. 5 57.n2 + (n + 1)2 = (n + 2)2; 3, 4, 5 58.a. b. 74 59.a.a2 + 2ab + b2 b.c2 c. d. (a + b)2 = c2 + 4 ab ; a2 + 2ab + b2 = 2ab + c2; a2 + b2 = c2 e. This equation is the same as the Pythagorean Theorem. 49. A figure is a square; the figure is a rectangle; if a figure is a rectangle then the figure is a square; false. 50. You are in Brazil; you are south of the equator; if you are south of the equator you are in Brazil; false. 51. An angle is a right angle; its measure is 90°; if the measure of an angle is 90°, then it is a right angle; true. 52. 52 units2 53. 6 in. ab 2 1 2 11-2

  28. – – – – > > 2 6v v4 2 2x2 6 3 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 69. 2b2 10b 70. 71. 72. 3 and 4 73. 8 and 9 74. –8 and –7 75. 11 and 12 76. rational 77. irrational 78. irrational 79. rational 60. D 61. H 62. B 63. C 64. A 65.[2] It is a right triangle. Substitute 17, the longest side, for c and substitute the other lengths for a and b in the Pythagorean Theorem 82 + 152 172 64 + 225 289 289 = 289 [1] incorrect equation OR incorrect explanation 66. 4 3 67. 68. 5 2 11-2

  29. The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 80. 8x2 – 4x 81. 12a2 + 15a 82. 18t3 – 6t2 83. –10p4 + 26p3 84. 15b3 + 5b2 – 45b 85. –7v4 + 42v2 – 7v 11-2

  30. The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 1. Find the missing length 2. Find the missing length to the nearest tenth. to the nearest tenth. 3. A triangle has sides of lengths 12 in., 14 in., and 16 in. Is the triangle a right triangle? 4. A triangular flag is attached to a post. The bottom of the flag is 48 in. above the ground. How far from the ground is the top of the flag? 16.6 5.7 no 57 in. 11-2

  31. The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 (For help, go to Lessons 11-2 and 2-7.) Find the length of the hypotenuse with the given leg lengths. If necessary, round to the nearest tenth. 1.a = 3, b = 4 2.a = 2, b = 5 3.a = 3, b = 8 4.a = 7, b = 5 For each set of values, find the mean. 5.x1 = 6, x2 = 14 6.y1 = –4, y2 = 8 7.x1 = –5, x2 = –7 8. y1 = –10, y2 = –3 11-3

  32. 1.a2 + b2 = c22.a2 + b2 = c2 32 + 42 = c2 22 + 52 = c2 9 + 16 = c2 4 + 25 = c2 25 = c2 29 = c2 c = 25 = 5 c = 29 5.4 The length of the hypotenuse is 5. The length of the hypotenuse is about 5.4. 3.a2 + b2 = c24.a2 + b2 = c2 32 + 82 = c2 72 + 52 = c2 9 + 64 = c2 49 + 25 = c2 73 = c2 74 = c2 c = 73 8.5 c = 74 8.6 The length of the hypotenuse is The length of the hypotenuse is about 8.5. about 8.6. –12 2 6 + 14 2 20 2 –5 + (–7) 2 5. mean = = = 10 6. mean = = = 2 7. mean = = = –6 8. mean = = = –6.5 –10 + (–3) 2 –13 2 –4 + 8 2 4 2 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 Solutions 11-3

  33. d = ( x2 – x1)2 + (y2 – y1)2Use the distance formula. d = (9 – 6)2 + [–4 – (–9)]2Substitute (9, –4) for (x2, y2) and (6, –9) for (x1, y1). d = 32 + 52 Simplify within parentheses. d = 34 Simplify to find the exact distance. d 5.8 Use a calculator. Round to the nearest tenth. The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 Find the distance between F(6, –9) and G(9, –4). The distance between F and G is about 5.8 units. 11-3

  34. EF = [4 – (–1)]2 + (3 + 5)2 = 52 + (–2)2 = 25 + 4 = 29 FG = (3 – 4)2 + (–2 – 3)2 = (–1)2 + (–5)2 = 1 + 25 = 26 EH = [–2 – (–1)]2 + (–2 – 5)2 = (–1)2 + (–7)2 = 1 + 49 = 50 The perimeter = 29 + 26 + 5 + 50 22.6 units. The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 Find the exact lengths of each side of quadrilateral EFGH. Then find the perimeter to the nearest tenth. GH = |–2 – 3| = 5 11-3

  35. Find the midpoint of CD. , = , Substitute (–3, 7) for (x1, y1) and (5, 2) for (x2, y2). 2 2 9 2 = , Simplify each numerator. 1 2 9 2 = 1, 4 Write as a mixed number. (–3) + 5 2 x1+ x2 2 7 + 2 2 y1+ y2 2 1 2 The midpoint of CD is M 1, 4 . The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 11-3

  36. 5 + (–3) 2 , = , Substitute (–3, 5) for (x1, y1) and (4, –3) for (x2, y2). 1 2 2 2 1 2 = , = , 1 (–3) + 4 2 x1+ x2 2 y1+ y2 2 1 2 The center of the circle is at , 1 . The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 A circle is drawn on a coordinate plane. The endpoints of the diameter are (–3, 5) and (4, –3). What are the coordinates of the center of the circle? 11-3

  37. The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 pages 594–597  Exercises 1. 15 2. 14 3. 10 4. 11.3 5. 2.8 6. 8.1 7. 16 8. 21.6 9. (1, 6) 10. (–1, 12) 11. (0, 0) 22.MN 5.1; NP 3.6; MP = 5 23.JK = 4; KL 2.2; LJ 3.6 24.TU 7.6; UV 16.5; VT 13.3 25. a.OR = 29, ST = 29 b. ; c. yes 26.a. (20, 80), (–40, 30) b. 78.1 ft c. (–10, 55) or 55L10 12. (–2, 3) 13. 5, –5 14. – 2 , –1 15. (–4, 4) 16. 8 , –9 17. 10.6 18. 9.4 19.AB 4.1; BC 3.2; AC = 5 20.DE 6.1; EF 3.6; DF 5.7 21.RS 3.2; ST 5.7; RT 5.1 1 2 1 2 5 2 5 2 1 2 11-3

  38. The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 28. Check students’ work. 29.a. b. (0, – ) c. one half mile south of the substation 30. about 9.5 km apart 31.a. 38.1 mi b. 20 mi, 21.2 mi c. 15 min, 16 min 32. Yes; all sides are congruent. 27. Answers may vary. Sample: Suppose you have (1, 1) and (4, –3). To find the distance, square the difference between the x-coordinates. Square the difference between the y-coordinates. Find the sum and take the square root, so 9 + 16 = 5. To find the midpoint, add x-coordinates together and divide by 2. Repeat for y-coordinates. So, 1 2 1 – 3 2 5 2 1 + 4 2 , , –1 = 11-3

  39. The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 47. 11.7 48. 11.2 49. 10.2 50. 26 51. 3.5 52. 8 53. 4.1 54. –14, 14 55. –10, 10 56. no solution 57. –5, 5 58. no solution 33.a.R(–27, –5) b.PR = 13 3.6 RQ = 13 3.6 34.a. about 4.3 mi b. about 17.4 mi 35.a.M(–0.5, 3); N(5.5, 3) b. They are equal. 36. (x, y) 37. They are opposites. 38. a. 5 units b. Answers may vary. Sample: (5, 0), (0, 5), (–5, 0), (0, –5), (3, –4), (–3, 4), (–3, –4) c. circle 4 3 39.a.y = – x + 1 b. c. 3 40. Yes; the distance from each point to the center is 5. 41. 19.1 42. 5 43. 22.9 44. –5 45. 3 46. –8 9 5 7 5 , – 11-3

  40. The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 59. – , 60. k2 + 11k + 24 61.v2 + 2v – 35 62. 2p2 – 17p – 9 63. 8w4 + 19w2 + 11 64. 7t3 + 5t2 + 5t – 2 65. 6c3 – 27c2 + 33c + 24 2 3 2 3 11-3

  41. 1. Find the distance between M(2, –1) and N(–4, 3) to the nearest tenth. 2. Find the distance between P(–2.5, 3.5) and R(–7.5, 8.5) to the nearest tenth. 3. Find the midpoint of AB, A(3, 6) and B(0, 2). 4. Find the midpoint of CD, C(6, –4) and D(12, –2). 5. Find the perimeter of triangle RST to the nearest tenth of a unit. 1 2 (1 , 4) The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 7.2 7.1 (9, – 3) 9.5 units 11-3

  42. 15 2x 3 11 5 8 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 (For help, go to Lesson 11-1.) Simplify each radical expression. 1. 52 2. 200 3. 4 54 4. 125x2 Rationalize each denominator. 5. 6. 7. 11-4

  43. Solutions 1. 52 = 4 • 13 = 4 • 13 = 2 13 2. 200 = 100 • 2 = 100 • 2 = 10 2 3. 4 54 = 4 9 • 6 = 4 • 9 • 6 = 4 • 3 • 6 = 12 6 4. 125x2 = 25 • 5 • x2 = 25 • 5 • x2 = 5x 5 3 11 3 11 5. = • = = 6. = • = = = = = = 7. = • = = 3 • 11 33 11 • 11 11 15 • 2x 2x • 2x 11 11 15 2x 15 2x 2x 2x 5 • 8 8 • 8 4 • 10 8 40 8 10 4 4 • 10 8 2 10 8 8 8 5 8 5 8 30x 2x Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 11-4

  44. Simplify 4 3 + 3. 4 3 + 3 = 4 3 + 1 3 Both terms contain 3. = (4 + 1) 3 Use the Distributive Property to combine like radicals. = 5 3 Simplify. Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 11-4

  45. Simplify 8 5 – 45. 8 5 – 45 = 8 5 + 9 • 5 9 is a perfect square and a factor of 45. = 8 5 – 9 • 5 Use the Multiplication Property of Square Roots. = 8 5 – 3 5 Simplify 9. = (8 – 3) 5 Use the Distributive Property to combine like terms. = 5 5 Simplify. Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 11-4

  46. Simplify 5( 8 + 9). 5( 8 + 9) = 40 + 9 5 Use the Distributive Property. = 4 • 10 + 9 5 Use the Multiplication Property of Square Roots. = 2 10 + 9 5 Simplify. Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 11-4

  47. Simplify ( 6 – 3 21)( 6 + 21). ( 6 – 3 21)( 6 + 21) = 36 + 126 – 3 126 – 3 441 Use FOIL. = 6 – 2 126 – 3(21) Combine like radicals and simplify 36 and 441. = 6 – 2 9 • 14 – 63 9 is a perfect square factor of 126. = 6 – 2 9 • 14 – 63 Use the Multiplication Property of Square Roots. = 6 – 6 14 – 63 Simplify 9. = –57 – 6 14 Simplify. Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 11-4

  48. 8 7 – 3 Simplify . 8 7 – 3 7 + 3 7 + 3 = • Multiply the numerator and denominator by the conjugate of the denominator. 8( 7 + 3) 7 – 3 8( 7 + 3) 4 = Multiply in the denominator. = Simplify the denominator. = 2( 7 + 3) Divide 8 and 4 by the common factor 4. = 2 7 + 2 3 Simplify the expression. Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 11-4

  49. Define: 51 = length of painting x = width of painting Relate: (1 + 5) : 2 = length : width (1 + 5) 2 Write: = x (1 + 5) = 102 Cross multiply. = Solve for x. 51 x 102 (1 + 5) x(1 + 5) (1 + 5) Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 A painting has a length : width ratio approximately equal to the golden ratio (1 + 5 ) : 2. The length of the painting is 51 in. Find the exact width of the painting in simplest radical form. Then find the approximate width to the nearest inch. 11-4

  50. 102 (1 + 5) (1 – 5) (1 – 5) x = • Multiply the numerator and the denominator by the conjugate of the denominator. 102(1 – 5) 1 – 5 x = Multiply in the denominator. 102(1 – 5) –4 x = Simplify the denominator. – 51(1 – 5) 2 – 51(1 – 5) 2 x = Divide 102 and –4 by the common factor –2. x = 31.51973343Use a calculator. x 32 The exact width of the painting is inches. The approximate width of the painting is 32 inches. Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 (continued) 11-4

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