Question

Find the roots.
$x^2-4\sqrt{2x}+6=0$

Answer 1

$x^2-4\sqrt{2}x+6=0$

$x^2-3\sqrt{2}x-\sqrt{2}x+6=0$

$x(x-3\sqrt{2})-\sqrt{2}(x-3\sqrt{2})=0$

$(x-\sqrt{2})(x-3\sqrt{2})=0$

$x=\sqrt{2},3\sqrt{2}$.