Question

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

Answer 1

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

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

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$⟹x=\frac{-1\pm \sqrt{1^2-4\left(1\right)\left(\frac{1}{\sqrt{2}}\right)}}{2\left(1\right)}$

$⟹x=\frac{-1\pm \sqrt{1-2\sqrt{2}}}{2}$

$\therefore x=\frac{-1+\sqrt{1-2\sqrt{2}}}{2}$ ( or ) $x=\frac{-1-\sqrt{1-2\sqrt{2}}}{2}$

Hence solved.