Question

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

Answer 1

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

Comparing the given quadratic equation with $a{x}^2+bx+c=0$, we have

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

$\therefore x=\frac{-1\pm \sqrt{(1{)}^2-4\times 1\times \frac{1}{\sqrt{2}}}}{2\times 1}$

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

Thus $x=\frac{-1+\sqrt{(2\sqrt{2}-1)i}}{2}$

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