Question

Determine whether the given values of variable is a solution of the quadratic equation or not. $y^2+\sqrt{2}y-4=0;y=2\sqrt{2}$

Answer 1

The given quadratic equation is $y^2+\sqrt{2}y-4=0$.

Substitute $y=2\sqrt{2}$ in the equation $y^2+\sqrt{2}y-4=0$ as follows:

$y^2+\sqrt{2}y-4=0\Rightarrow {\left(2\sqrt{2}\right)}^2+(\sqrt{2}\times 2\sqrt{2})-4=0\Rightarrow 8+(2\times \sqrt{2^2})-4=0\Rightarrow 8+(2\times 2)-4=0\Rightarrow 8+4-4=0\Rightarrow 8=0$

which is not true.

Hence, $y=2\sqrt{2}$ is not the solution of the quadratic equation $y^2+\sqrt{2}y-4=0$.