## Elementary Algebra

Published by Cengage Learning

# Chapter 9 - Roots and Radicals - 9.5 - Solving Radical Equations - Problem Set 9.5 - Page 424: 32

#### Answer

$x=9$

#### Work Step by Step

Squaring both sides of the equation and then using the properties of equality, we obtain: \begin{array}{l}\require{cancel}\left( 2\sqrt{x} \right)^2=\left( x-3 \right)^2 \\\\ 4x=(x)^2+2(x)(-3)+(-3)^2 \\\\ 4x=x^2-6x+9 \\\\ 0=x^2+(-6x-4x)+9 \\\\ x^2-10x+9=0 \\\\ (x-9)(x-1)=0 \\\\ x=\{1,9\} .\end{array} Upon checking, only $x=9$ satisfies the original equation.

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