#### Question

In the following, determine whether the given values are solutions of the given equation or not:

`x^2 - 3sqrt3x+6=0`, `x=sqrt3`, `x=-2sqrt3`

#### Solution

We have been given that,

`x^2 - 3sqrt3x+6=0`, `x=sqrt3`, `x=-2sqrt3`

Now if `x=sqrt3` is a solution of the equation then it should satisfy the equation.

So, substituting `x=sqrt3` in the equation, we get

`x^2 - 3sqrt3x+6`

`=(sqrt3)^2-3sqrt3(sqrt3)+6`

= 3 - 9 + 6

= 0

Hence `x=sqrt3` is a solution of the quadratic equation.

Also, if `x=-2sqrt3` is a solution of the equation then it should satisfy the equation

So, substituting `x=-2sqrt3` in the equation, we get

`x^2 - 3sqrt3x+6`

`=(-2sqrt3)^2-3sqrt3(-2sqrt3)+6`

= 12 - 18 + 6

= 0

Hence `x=-2sqrt3` is a solution of the quadratic equation.

Therefore, from the above results we find out that `x=sqrt3` and `x=-2sqrt3` are the solutions of the given quadratic equation.