## College Algebra 7th Edition

$x=64$
$x-4\sqrt{x}=32$ We make the substitution: $u=\sqrt{x}$. The equation then becomes: $u^{2}-4u=32$ $u^{2}-4u-32=0$ $(u-8)(u+4)=0$ $u-8=0$ or $u+4=0$ $u=8$ or $u=-4$ We convert the $u$ solutions to $x$ solutions: $u=\sqrt{x}$ $x=u^2$ If $u=8$, then: $x=8^2=64$ If $u=-4$, then that would mean $u=-4=\sqrt{x}$ which has no solution. So the only solution to the original equation is $x=64$