Question

Solve the following equation:
$x^2-4x+6=\sqrt{2x^2-8x+12}$

Answer 1

$x^2-4x+6=\sqrt{2(x^2-4x+6)}$

Let $x^2-4x+6=t$

$\Rightarrow t=\sqrt{2t}$

$\Rightarrow 2t=t^2$

$\Rightarrow t^2-2t=0$

$\Rightarrow t(t-2)=0$

$\Rightarrow t=0,t=2$

$\Rightarrow x^2-4x+6=0$ and $x^2-4x+6=2$

$x=\frac{4\pm \sqrt{16-24}}{2}$ $x^2-4x+4=0$

$=\frac{4\pm \sqrt{-8}}{2}$ ${(x-2)}^2=0$

$=\frac{4\pm 2i\sqrt{2}}{2}$ $x=2$

$=2\pm i\sqrt{2}$

Hence, solved.