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7-5 Solving Square Root and Other Radical Equations

7-5 Solving Square Root and Other Radical Equations. Objectives. Solving Radical Equations. Vocabulary. If , then and . –10 + 2 x + 1 = –5. Isolate the radical. 2 x + 1 = 5. Square both sides. ( 2 x + 1 ) 2 = 5 2.

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7-5 Solving Square Root and Other Radical Equations

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  1. 7-5 Solving Square Root and Other Radical Equations

  2. Objectives Solving Radical Equations

  3. Vocabulary If , then and

  4. –10 + 2x + 1 = –5 Isolate the radical. 2x + 1 = 5 Square both sides. ( 2x + 1 )2 = 52 Check: –10 + 2x + 1 = –5 –10 + 2(12) + 1 –5 –10 + 25 –5 –10 + 5 –5 –5 = –5 Solving Radical Equations Solve –10 + 2x + 1 = –5. 2x + 1 = 25 2x = 24 x = 12

  5. 3 5 3(x + 1) = 24 3 5 (x + 1) = 8 Divide each side by 3. 5 3 3 5 5 3 5 3 ((x + 1) ) = 8 Raise both sides to the power. 3 5 5 3 5 3 (x + 1)1 = 8 Multiply the exponents and . Check: 3(x + 1) = 24   3(31 + 1) 24 3(25) 24 3(2)3 24 24 = 24 3 5 3 5 3 5 Solving Radical Equations with Rational Exponents 3 5 Solve 3(x + 1) = 24. x + 1 = 32 Simplify. x = 31

  6. < Relate:volume of sphere • density of plastic maximum weight – Define: Let r = radius in inches. 4 3 Write:• r3 • 0.8 80 < – Real World Example 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 = • r3, where r is the radius of the sphere. What is the maximum radius the sphere can have? 4 3

  7. < < < < – – – – r3 3 • 80 4 • • 0.8 75 r3 r 2.88 Use a calculator. 4 r3 3 Continued (continued) • 0.8 80 The maximum radius is about 2.88 inches.

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

  9. 7 4 7 4 – – 7 2 7 2 7 2 – – – 5 2 – Continued (continued) Check:x + 2 – 3 = 2xx + 2 – 3 = 2x –1 + 2 – 3 2(–1) + 2 – 3 2 1 – 3 –2 – 3 –2 = –2 – 3 1 4 1 2 = / The only solution is –1.

  10. 2 3 1 3 (x + 1) – (9x + 1) = 0 2 3 1 3 (x + 1) = (9x + 1) 2 3 1 3 ((x + 1) )3 = ((9x + 1) )3 (x + 1)2 = 9x + 1 x2 + 2x + 1 = 9x + 1 x2 – 7x = 0 x(x – 7) = 0 Solving Equations with Two Rational Exponents 2 3 1 3 Solve (x + 1) – (9x + 1) = 0. Check for extraneous solutions. x = 0 or x = 7

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

  12. Homework p 394 # 1, 2, 7, 8, 15, 16

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