Question

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

Answer 1

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

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

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

$\alpha =\frac{-{\displaystyle \frac{1}{\sqrt{2}}}+\sqrt{-{\displaystyle \frac{7}{2}}}}{2}$ and $\beta =\frac{-{\displaystyle \frac{1}{\sqrt{2}}}-\sqrt{-{\displaystyle \frac{7}{2}}}}{2}$

$\alpha =\frac{-{\displaystyle \frac{1}{\sqrt{2}}}+i\sqrt{{\displaystyle \frac{7}{2}}}}{2}$ and $\beta =\frac{-{\displaystyle \frac{1}{\sqrt{2}}}-i\sqrt{{\displaystyle \frac{7}{2}}}}{2}$

$\alpha =\frac{-1+i\sqrt{7}}{2\sqrt{2}}$ and $\beta =\frac{-1-i\sqrt{7}}{2\sqrt{2}}$


Hence, the roots of the equation are $\frac{-1\pm i\sqrt{7}}{2\sqrt{2}}$.