1. Solving Radical Equations ALGEBRA 2 LESSON 7-5 7-5 Algebra 2 Lesson 7-5 (Page 385)

2. –10 + 2x + 1 = –5 Solve –10 + 2x + 1 = –5. Solving Radical Equations ALGEBRA 2 LESSON 7-5 2x + 1 = 5 Isolate the radical. ( 2x + 1 ) 2 = 5 2 Square both sides. 2x + 1 = 25 2x = 24 x = 12 Check: –10 + 2x + 1 = –5 –10 + 2(12) + 1 –5 –10 + 25 –5 –10 + 5 –5 –5 = –5 7-5

3. Solving Radical Equations ALGEBRA 2 LESSON 7-5 Solve 3(x + 1) = 24. 3535 x + 1 = 32Simplify. x = 31 3(x + 1) = 24 3535 (x + 1) = 8Divide each side by 3. 3535 Check: 3(x + 1) = 24 3(31 + 1) 24 3(2 5 ) 24 3(2) 3 24 24 = 24 3535 3535 3535 3535 5353 ((x + 1) ) = 8 Raise both sides to the power. 5353 5353 (x + 1) 1 = 8 Multiply the exponents and. 5353 5353 3535 7-5

4. Solving Radical Equations 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? 4343 7-5 4343 Define: Let r = radius in inches. Relate: volume of sphere density of plastic maximum weight Write: r 3 0.8 80 < – < –

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

6. Solving Radical Equations ALGEBRA 2 LESSON 7-5 Solve x + 2 – 3 = 2x. Check for extraneous solutions. x + 2 – 3 = 2x x + 2 = 2x + 3Isolate the radical. ( x + 2) 2 = (2x + 3) 2 Square both sides.0 = 4x 2 + 11x + 7Combine like terms. 0 = (x + 1)(4x + 7)Factor. x + 2 = 4x 2 + 12x + 9Simplify. x + 1 = 0 or 4x + 7 = 0Factor Theorem x = –1 orx = – 7474 7-5

7. Check: x + 2 – 3 = 2x x + 2 – 3 = 2x –1 + 2 – 3 2(–1) + 2 – 3 2 1 – 3 –2 – 3 –2 = –2 – 3 7272 – 7474 – 1414 1212 7272 – 7272 – = – 5252 / 7474 – Solving Radical Equations ALGEBRA 2 LESSON 7-5 (continued) 7-5 The only solution is –1.

8. Solving Radical Equations ALGEBRA 2 LESSON 7-5 Solve (x + 1) – (9x + 1) = 0. Check for extraneous solutions. 2323 1313 x = 0 or x = 7 (x + 1) 2 = 9x + 1 x 2 + 2x + 1 = 9x + 1 x 2 – 7x = 0 x(x – 7) = 0 (x + 1) – (9x + 1) = 0 2323 1313 (x + 1) = (9x + 1) 2323 1313 ((x + 1) ) 3 = ((9x + 1) ) 3 2323 1313 7-5

9. Solving Radical Equations ALGEBRA 2 LESSON 7-5 (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) – (1 2 ) 0 (8) – (8 2 ) 0 1 – 1 = 0 8 – 8 = 0 2323 1313 2323 1313 2323 1313 2323 1313 2323 2323 2323 1313 2323 1313 2323 1313 2323 1313 2323 2323 Both 0 and 7 are solutions. 7-5