Question

Solve:
$x^2-\sqrt{2}x-4=0,x=-\sqrt{2},x=-2\sqrt{2}$

Answer 1

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

$\Rightarrow x^2-(2\sqrt{2}-\sqrt{2})x-4=0$

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

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

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

$\Rightarrow x-2\sqrt{2}=0,x+\sqrt{2}=0$

$x=2\sqrt{2},x=-\sqrt{2}$

Hence, this is the answer.