In the following, determine whether the given values are solutions of the given equation or not:
$x^2\ –\ \sqrt{2}x\ –\ 4\ =\ 0,\ x\ =\ -\sqrt{2},\ x\ =\ -2\sqrt{2}$

Given:

The given equation is $x^2\ –\ \sqrt{2}x\ –\ 4\ =\ 0$.

To do:

We have to determine whether $x=-\sqrt{2}, x=-2\sqrt{2}$ are solutions of the given equation.

Solution:

If the given values are the solutions of the given equation then they should satisfy the given equation.

Therefore,

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

LHS$=x^2 − \sqrt{2}x-4$

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

        $=2+2-4$

       $=0$

       $=$RHS

Hence, $x=-\sqrt{2}$ is a solution of the given equation.

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

LHS$=x^2 − \sqrt{2}x - 4$

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

        $=8+4-4$

        $=8$

RHS$=0$

LHS$≠$RHS

Hence, $x=-2\sqrt{2}$ is not a solution of the given equation.