Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.6 - Radical Equations - Exercise Set - Page 560: 21

Answer

$\{16\}$.

Work Step by Step

The given expression is $\Rightarrow \sqrt{x-7}=7-\sqrt{x}$ Square both sides. $\Rightarrow (\sqrt{x-7})^2=(7-\sqrt{x})^2$ Use the special formula $(A-B)^2=A^2-2AB+B^2$ We have $A=7$ and $B=\sqrt x$ $\Rightarrow x-7=(7)^2-2(7)(\sqrt x)+(\sqrt{x})^2$ Simplify. $\Rightarrow x-7=49-14\sqrt x+x$ Add $14\sqrt x-x+7$ to both sides. $\Rightarrow x-7+14\sqrt x-x+7=49-14\sqrt x+x+14\sqrt x-x+7$ Add like terms. $\Rightarrow 14\sqrt x=56$ Divide both sides by $14$. $\Rightarrow \frac{14\sqrt x}{14}=\frac{56}{14}$ Simplify. $\Rightarrow \sqrt x=4$ Square both sides. $\Rightarrow (\sqrt x)^2=(4)^2$ Simplify. $\Rightarrow x=16$ The solution set is $\{16\}$.
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