Question

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

Answer 1

Given equation: 

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

Comparing the given equation with the general form of the quadratic equation 

\[a x^2 + bx + c = 0\] , we get

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

Substituting these values in

\[\alpha = \frac{- b + \sqrt{b^2 - 4ac}}{2a}\] and \[\beta = \frac{- b - \sqrt{b^2 - 4ac}}{2a}\], we get:

\[\alpha = \frac{- 1 + \sqrt{1 - 4 \times \frac{1}{\sqrt{2}}}}{2}\] 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 \[\frac{- 1 \pm i\sqrt{2\sqrt{2} - 1}}{2}\] .