# Chapter 9 - Roots and Radicals - 9.5 - Solving Radical Equations - Problem Set 9.5: 26

$x=\{-7,-6\}$

#### 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( \sqrt{x+7} \right)^2=\left( x+7 \right)^2 \\\\ x+7=(x)^2+2(x)(7)+(7)^2 \\\\ x+7=x^2+14x+49 \\\\ 0=x^2+(14x-x)+(49-7) \\\\ x^2+13x+42=0 \\\\ (x+6)(x+7)=0 \\\\ x=\{-7,-6\} .\end{array} Upon checking, both solutions satisfy the original equation. Hence, $x=\{-7,-6\} .$

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