# Solve the following

Question:

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

Solution:

Given equation: $x^{2}+x+\frac{1}{\sqrt{2}}=0$

Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=1, b=1$ and $c=\frac{1}{\sqrt{2}}$.

Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get:

$\alpha=\frac{-1+\sqrt{1-4 \times \frac{1}{\sqrt{2}}}}{2}$ and $\beta=\frac{-1-\sqrt{1-4 \times \frac{1}{\sqrt{2}}}}{2}$

$\Rightarrow \alpha=\frac{-1+\sqrt{1-2 \sqrt{2}}}{2}$ and $\beta=\frac{-1-\sqrt{1-2 \sqrt{2}}}{2}$

$\Rightarrow \alpha=\frac{-1+i \sqrt{2 \sqrt{2}-1}}{2}$ and $\beta=\frac{-1-i \sqrt{2 \sqrt{2}-1}}{2}$

Hence, the roots of the equation are $\frac{-1 \pm i \sqrt{2 \sqrt{2}-1}}{2}$.