## College Algebra 7th Edition

$x=2$
$\sqrt{x+2}=x$ $(\sqrt{x+2})^{2}=x^{2}$ $x+2=x^{2}$ $0=x^{2}-x-2$ $x^{2}-x-2=0$ $(x+1)(x-2)=0$ $x+1=0$ or $x-2=0$ $x=-1$ or $x=2$ We need to confirm the solutions. First, we confirm $x=-1$: Left side=$\sqrt{-1+2}=\sqrt{1}=1$ Right side=$-1$ The sides disagree, so $x=-1$ is not a solution. Next, we confirm $x=2$: Left side=$\sqrt{2+2}=\sqrt{4}=2$ Right side=$2$ The sides agree, so $x=2$ is a solution.