Question

Solve the following equation:
$x^2-4x+32=16\sqrt{x}$

Answer 1

$x^2-4x+32=16\sqrt{x}$

$x(x-4)=16\sqrt{x}-32$

$x(\sqrt{x}-2)(\sqrt{x}+2)=16(\sqrt{x}-2)$

Let $\sqrt{x}-2=t$

$(t+2{)}^2(t\left)\right(t+4)=16t$

$t=0$ is a solution

$(t^2+4t+4)(t+4)=16$

$t^3+8t^2+20t+16=16$

$t(t^2+8t+20)=0$

$\therefore t=0$ is only solution

$\sqrt{x}-2=0$

$\sqrt{x}=2$

$x=4$