7-5 Solving Square Root and Other Radical Equations

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Presentation transcript:

7-5 Solving Square Root and Other Radical Equations

Objectives Solving Radical Equations

Vocabulary If , then and

Solving Radical Equations Solve –10 + 2x + 1 = –5. –10 + 2x + 1 = –5 2x + 1 = 5 Isolate the radical. ( 2x + 1 )2 = 52 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

Solving Radical Equations with Rational Exponents 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) 24 3(25) 24 3(2)3 24 24 = 24 3 5

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 = • 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 • 0.8 80 < –

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

Checking for Extraneous Solutions 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

Continued Check: x + 2 – 3 = 2x x + 2 – 3 = 2x –1 + 2 – 3 2(–1) + 2 – 3 2 1 – 3 –2 – 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.

Solving Equations with Two Rational Exponents 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

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 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.

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