159 1.6 OtherTypes of Equations and Applications Step 3 x2 + 2x - 15 = 0 Write in standard form. 1x + 521x - 32 = 0 Factor. x + 5 = 0 or x - 3 = 0 Zero-factor property x = -5 or x = 3 Proposed solutions Step 4 CHECK x - 215 - 2x = 0 Original equation -5 - 215 - 21-52≟0 Let x = -5. -5 - 225≟0 -5 - 5≟0 -10 = 0 False 3 - 215 - 2132≟0 Let x = 3. 3 - 29≟0 3 - 3≟0 0 = 0 ✓ True As the check shows, only 3 is a solution, so the solution set is 536. S Now Try Exercise 45. EXAMPLE 5 Solving an Equation Containing Two Radicals Solve 22x + 3 - 2x + 1 = 1. SOLUTION 2 2x + 3 - 2x + 1 = 1 Step 1 22x + 3 = 1 + 2x + 1 Isolate 22x + 3. Step 2 A 22x + 3 B 2 = A 1 + 2x + 1 B 2 Square each side. 2x + 3 = 1 + 22x + 1 + 1x + 12 Be careful: 1a +b22 =a2 +2ab +b2 Step 1 x + 1 = 22x + 1 Isolate the remaining radical. Step 2 1x + 122 = A 22x + 1B 2 Square again. x2 + 2x + 1 = 41x + 12 Apply the exponents. x2 + 2x + 1 = 4x + 4 Distributive property Step 3 x2 - 2x - 3 = 0 Write in standard form. 1x - 321x + 12 = 0 Factor. x - 3 = 0 or x + 1 = 0 Zero-factor property x = 3 or x = -1 Proposed solutions Step 4 Isolate one of the radicals on one side of the equation. 1ab22 = a 2b 2 CHECK 22x + 3 - 2x + 1 = 1 Original equation 22132 + 3 - 23 + 1≟1 Let x = 3. 29 - 24≟1 3 - 2≟1 1 = 1 ✓ True 22 -12 + 3 - 2-1 + 1≟1 Let x = -1. 21 - 20≟1 1 - 0≟1 1 = 1 ✓ True Both 3 and -1 are solutions of the original equation, so 5-1, 36 is the solution set. S Now Try Exercise 57.
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