## College Algebra (11th Edition)

Published by Pearson

# Chapter 1 - Review Exercises - Page 163: 38

#### Answer

$x=\left\{ \dfrac{2-2\sqrt{2}}{3},\dfrac{2+2\sqrt{2}}{3} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $(2-3x)^2=8 ,$ take the square root of both sides (Square Root Principle). Then simplify the resulting radical and use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Taking the square root of both sides (Square Root Principle), the given equation is equivalent to \begin{array}{l}\require{cancel} \sqrt{(2-3x)^2}=\pm\sqrt{8} \\\\ 2-3x=\pm\sqrt{4\cdot2} \\\\ 2-3x=\pm\sqrt{(2)^2\cdot2} \\\\ 2-3x=\pm2\sqrt{2} \\\\ -3x=-2\pm2\sqrt{2} \\\\ x=\dfrac{-2\pm2\sqrt{2}}{-3} \\\\ x=\dfrac{2\pm2\sqrt{2}}{3} .\end{array} Hence, the solutions are $x=\left\{ \dfrac{2-2\sqrt{2}}{3},\dfrac{2+2\sqrt{2}}{3} \right\} .$

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