7 5 Solving Square Root and Other Radical

7 -5 Solving Square Root and Other Radical Equations

Objectives Solving Radical Equations

Vocabulary If , then and

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

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

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 4 is V = 3 • r 3, where r is the radius of the sphere. What is the maximum radius the sphere can have? Relate: volume of sphere • density of plastic < – maximum weight Define: Let r = radius in inches. 4 Write: 3 • r 3 • 0. 8 < – 80

Continued (continued) 4 r 3 3 • 0. 8 < – 80 3 • 80 r 3 < – 4 • • 0. 8 75 r 3 < – r < – 2. 88 Use a calculator. The maximum radius is about 2. 88 inches.

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

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

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

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

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