Question

Solve

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

Answer 1

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

Comparing the given quadratic equation with

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

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

$=\frac{\frac{-1}{\sqrt{2}}\pm \sqrt{\frac{1}{2}-4}}{2}=\frac{\frac{-1}{\sqrt{2}}\pm \sqrt{\frac{-7}{2}}}{2}$

$=\frac{-1\pm \sqrt{7}i}{2\sqrt{2}}$

Thus $x=\frac{-1+\sqrt{7}i}{2\sqrt{2}}andx=\frac{-1-\sqrt{7}i}{2\sqrt{2}}$