# Solve the equation

Question:

Solve the equation $x^{2}+x+\frac{1}{\sqrt{2}}=0$

Solution:

The given quadratic equation is $x^{2}+x+\frac{1}{\sqrt{2}}=0$

This equation can also be written as $\sqrt{2} x^{2}+\sqrt{2} x+1=0$

On comparing this equation with $a x^{2}+b x+c=0$, we obtain

$a=\sqrt{2}, b=\sqrt{2}$, and $c=1$

$\therefore$ Discri min ant $(D)=b^{2}-4 a c=(\sqrt{2})^{2}-4 \times(\sqrt{2}) \times 1=2-4 \sqrt{2}$

Therefore, the required solutions are

$\frac{-b \pm \sqrt{D}}{2 a}=\frac{-\sqrt{2} \pm \sqrt{2-4 \sqrt{2}}}{2 \times \sqrt{2}}=\frac{-\sqrt{2} \pm \sqrt{2(1-2 \sqrt{2})}}{2 \sqrt{2}}$

$=\left(\frac{-\sqrt{2} \pm \sqrt{2}(\sqrt{2 \sqrt{2}-1}) \mathrm{i}}{2 \sqrt{2}}\right) \quad[\sqrt{-1}=\mathrm{i}]$

$=\frac{-1 \pm(\sqrt{2 \sqrt{2}-1}) i}{2}$