Roots, Radicals, and Root Functions Chapter 9 Roots, Radicals, and Root Functions

Solving Equations with Radicals 9.6 Solving Equations with Radicals

9.6 Solving Equations with Radicals Objectives Solve radical equations using the power rule. Solve radical equations that require additional steps. Solve radical equations with indexes greater than 2. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals Power Rule for Solving Equations with Radicals Power Rule for Solving Equations with Radicals If both sides of an equation are raised to the same power, all solutions of the original equation are also solutions of the new equation. Read the power rule carefully; it does not say that all solutions of the new equation are solutions of the original equation. They may or may not be. Solutions that do not satisfy the original equation are called extraneous solutions; they must be discarded. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals Caution on Checking Solutions CAUTION When the power rule is used to solve an equation, every solution of the new equation must be checked in the original equation. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 1 Using the Power Rule Solve 5x – 6 = 7. Use the power rule and square both sides to get 5x – 6 = 7 2 5x – 6 = 49 5x = 55 Add 6. x = 11 Divide by 5. To check, substitute the potential solution in the original equation. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 1 Using the Power Rule Solve 5x – 6 = 7. 5x – 6 = 7 5 · 11 – 6 = 7 ? Let x = 11. 7 = 7 True Since 11 satisfies the original equation, the solution set is { 11 }. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals Solving an Equation with Radicals Solving an Equation with Radicals Step 1 Isolate the radical. Make sure that one radical term is alone on one side of the equation. Step 2 Apply the power rule. Raise both sides of the equation to a power that is the same as the index of the radical. Step 3 Solve. Solve the resulting equation; if it still contains a radical, repeat Steps 1 and 2. Step 4 Check all potential solutions in the original equation. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals Caution on Incorrect Solution Sets CAUTION Remember Step 4 or you may get an incorrect solution set. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 2 Using the Power Rule Solve 2x – 9 + 1 = 0. Step 1 To isolate the radical on one side, subtract 1 from each side. 2x – 9 = –1 Step 2 Now square both sides. 2x – 9 = (–1) 2 Step 3 Solve. 2x – 9 = 1 2x = 10 x = 5 Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 2 Using the Power Rule Solve 2x – 9 + 1 = 0. Step 4 Check the potential solution, 5, by substituting it in the original equation. 2x – 9 + 1 = 0 2 · 5 – 9 + 1 = 0 ? Let x = 5. 1 + 1 = 0 False This false result shows that 5 is not a solution of the original equation; it is extraneous. The solution set is ∅. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals Note on Identifying No Solution NOTE We could have determined after Step 1 that the equation in Example 2 has no solution because the expression on the left cannot be negative. EXAMPLE 2 2x – 9 + 1 = 0. 2x – 9 = –1 Step 1 To isolate the radical on one side, subtract 1 from each side. Solve The expression on the left cannot be negative. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals Finding the Square of a Binomial Finding the Square of a Binomial Recall that (x + y)2 = x2 + 2xy + y2. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 3 Using the Power Rule; Squaring a Binomial Solve 9 – x = x + 3. Step 1 The radical is alone on the left side of the equation. Step 2 Now square both sides. 9 – x = (x + 3) 2 9 – x = x2 + 6x + 9 Step 3 The new equation is quadratic, so get 0 on one side. Twice the product of 3 and x. 0 = x2 + 7x Subtract 9 and add x. 0 = x(x + 7) Factor. x = 0 or x + 7 = 0 Zero-factor property x = –7 Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 3 Using the Power Rule; Squaring a Binomial Solve 9 – x = x + 3. Step 4 Check each potential solution, 0 and –7, in the original equation. If x = 0, then If x = –7, then 9 – x = x + 3 9 – x = x + 3 9 – 0 = 0 + 3 ? 9 – (–7) = (–7) + 3 ? 9 = 3 ? 16 = –4 ? 3 = 3 True 4 = –4 False The solution set is { 0 }. The other potential solution, –7, is extraneous. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals Caution on the Middle Term CAUTION When a radical equation requires squaring a binomial as in Example 3, remember to include the middle term. (x + 3)2 = x2 + 6x + 9 Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 4 Using the Power Rule; Squaring a Binomial Solve x2 – 8x + 3 = x – 5. Square both sides. 2 x2 – 8x + 3 = (x – 5) x2 – 8x + 3 = x2 – 10x + 25 Twice the product of 5 and x. 2x = 22 Subtract x2 and 3; add 10x. x = 11 Divide by 2. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 4 Using the Power Rule; Squaring a Binomial Solve x2 – 8x + 3 = x – 5. Check: x2 – 8x + 3 = x – 5 112 – 8 · 11 + 3 = 11 – 5 ? Let x = 11. 6 = 6 True The solution set of the original equation is { 11 }. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 5 Using the Power Rule; Squaring Twice Solve x + 7 + –x + 6 = 5. Start by isolating one radical on one side of the equation by subtracting from each side. Then square both sides. –x + 6 x + 7 + –x + 6 = 5 x + 7 = 5 – –x + 6 2 x + 7 = 5 – –x + 6 = 25 – 10 –x + 6 + (–x + 6) x + 7 Twice the product of 5 and . –x + 6 Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 5 Using the Power Rule; Squaring Twice Solve x + 7 + –x + 6 = 5. This equation still contains a radical, so square both sides again. Before doing this, isolate the radical term on the right. = 25 – 10 –x + 6 + (–x + 6) x + 7 = 31 – x – 10 –x + 6 x + 7 = –10 –x + 6 2x – 24 Subtract 31 and add x. = –5 –x + 6 x – 12 Divide by 2. = –5 –x + 6 (x – 12)2 Square both sides again. 2 Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 5 Using the Power Rule; Squaring Twice Solve x + 7 + –x + 6 = 5. This equation still contains a radical, so square both sides again. = –5 –x + 6 (x – 12)2 Square both sides again. 2 = (–5)2 –x + 6 x2 – 24x + 144 (ab)2 = a2 b2 2 = 25 ( –x + 6 ) x2 – 24x + 144 = –25x + 150 x2 – 24x + 144 Distributive property = 0 x2 + x – 6 Standard form = 0 (x + 3)(x – 2) Factor. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 5 Using the Power Rule; Squaring Twice Solve x + 7 + –x + 6 = 5. Now finish solving the equation. = 0 (x + 3)(x – 2) x + 3 = 0 or x – 2 = 0 Zero-factor property x = –3 or x = 2 Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 5 Using the Power Rule; Squaring Twice Solve x + 7 + –x + 6 = 5. Check each potential solution, –3 and 2, in the original equation. If x = –3, then If x = 2, then x + 7 + –x + 6 = 5 x + 7 + –x + 6 = 5 –3 + 7 + –(–3) + 6 = 5 2 + 7 + –(2) + 6 = 5 4 + 9 = 5 9 + 4 = 5 5 = 5 5 = 5 The solution set is { –3, 2 }. Copyright © 2010 Pearson Education, Inc. All rights reserved.

9.6 Solving Equations with Radicals EXAMPLE 6 Using the Power Rule for a Power Greater than 2 3x – 2 = 2x + 6. 3 Solve Raise both sides to the third power. 3x – 2 = 2x + 6 3 3x – 2 = 2x + 6 x = 8 Check this result in the original equation. 3x – 2 = 2x + 6 3 3 · 8 – 2 = 2 · 8 + 6 ? Let x = 8. 3 22 = 22 True 3 The solution set is { 8 }. Copyright © 2010 Pearson Education, Inc. All rights reserved.