160 CHAPTER 1 Equations and Inequalities EXAMPLE 6 Solving an Equation Containing a Radical (Cube Root) Solve 23 4x2 - 4x + 1 - 23 x = 0. SOLUTION 2 3 4x2 - 4x + 1 - 23 x = 0 Step 1 23 4x2 - 4x + 1 = 23 x Isolate a radical. Step 2 A 23 4x2 - 4x + 1 B 3 = A 23 x B 3 Cube each side. 4x2 - 4x + 1 = x Apply the exponents. Step 3 4x2 - 5x + 1 = 0 Write in standard form. 14x - 121x - 12 = 0 Factor. 4x - 1 = 0 or x - 1 = 0 Zero-factor property x = 1 4 or x = 1 Proposed solutions Step 4 Both are valid solutions, so the solution set is E1 4 , 1F. S Now Try Exercise 69. CHECK 23 4x2 - 4x + 1 - 23 x = 0 Original equation 3 3 4A1 4B 2 - 4A1 4B + 1 - 3 3 1 4 ≟0 Let x = 1 4 . 3 3 1 4 - 3 3 1 4 ≟0 0 = 0 ✓ True 23 41122 - 4112 + 1 - 23 1≟0 Let x = 1. 2 3 1 - 23 1≟0 0 = 0 ✓ True Equations with Rational Exponents An equation with a rational exponent contains a variable, or variable expression, raised to an exponent that is a rational number. For example, the radical equation A 2 5 x B 3 = 27 can be written with a rational exponent as x3/5 = 27 and solved by raising each side to the reciprocal of the exponent, with care taken regarding signs as seen in Example 7(b). EXAMPLE 7 Solving Equations with Rational Exponents Solve each equation. (a) x3/5 = 27 (b) 1x - 422/3 = 16 SOLUTION (a) x3/5 = 27 1x3/525/3 = 275/3 Raise each side to the power 5 3 , the reciprocal of the exponent of x. x = 243 275/3 = A 23 27 B 5 = 35 = 243
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