In the following, determine whether the given values are solutions of the given equation or not:
$x^2\ −\ 3\sqrt{3}x\ +\ 6\ =\ 0,\ x\ =\ \sqrt{3}$  and  $x\ =\ −2\sqrt{3}$

Given:

The given equation is $x^2\ −\ 3\sqrt{3}x\ +\ 6\ =\ 0$.

To do:

We have to determine whether $x=\sqrt{3}, x=-2\sqrt{3}$ 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 $x=\sqrt{3}$,

LHS$=x^2 − 3\sqrt{3}x + 6$

        $=(\sqrt{3})^2-3\sqrt{3}(\sqrt{3})+6$

        $=3-9+6$

       $=0$

       $=$RHS

Hence, $x=\sqrt{3}$ is a solution of the given equation.

For $x=-2\sqrt{3}$,

LHS$=x^2 − 3\sqrt{3}x + 6$

        $=(-2\sqrt{3})^2-3\sqrt{3}(-2\sqrt{3})+6$

        $=12+18+6$

        $=36$

RHS$=0$

LHS$≠$RHS

Hence, $x=-2\sqrt{3}$ is not a solution of the given equation.