Question

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

`x^2-sqrt2x-4=0`, `x=-sqrt2`, `x=-2sqrt2`

Answer 1

We have been given that,

`x^2-sqrt2x-4=0`, `x=-sqrt2`, `x=-2sqrt2`

Now if `x=-sqrt2` is a solution of the equation then it should satisfy the equation.

So, substituting `x=-sqrt2` in the equation, we get

`x^2-sqrt2x-4`

`=(-sqrt2)^2-sqrt2(-sqrt2)-4`

= 2 + 2 - 4

= 0

Hence `x=-sqrt2` is a solution of the quadratic equation.

Also, if `x=-2sqrt2` is a solution of the equation then it should satisfy the equation.

So, substituting `x=-2sqrt2` in the equation, we get

`x^2-sqrt2x-4`

`(-2sqrt2)^2-sqrt2(-2sqrt2)-4`

= 8 + 4 - 4

= 8

Hence `x=-2sqrt2`is not a solution of the quadratic equation.

Therefore, from the above results we find out that `x=-sqrt2` is a solution but `x=-2sqrt2`is not a solution of the given quadratic equation.