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# Solve the equation : $\;x^{2}+x+\large\frac{1}{\sqrt{2}}=0\;$

$(a)\;\large\frac{-1 \pm( \sqrt{3\sqrt{2}-1})i}{2}\qquad(b)\;\large\frac{-1 \pm( \sqrt{2\sqrt{2}-1})i}{3}\qquad(c)\;\large\frac{-1 \pm( \sqrt{2\sqrt{2}-1})i}{2}\qquad(d)\;\large\frac{1 \pm( \sqrt{2\sqrt{2}+1})i}{2}$

Can you answer this question?

## 1 Answer

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Answer : $\;\large\frac{-1 \pm( \sqrt{2\sqrt{2}-1})i}{2}$
Explanation :
The given quadratic equation is $\;x^2+x+\large\frac{1}{\sqrt{2}}=0$
This equation can also be written as $\sqrt{2} x^{2} + \sqrt{2} x + 1 =0$
On comparing the given equation with $\;ax^2 + bx +c\;,$ we obtain
$a=\sqrt{2}\;,b=\sqrt{2}\;and \;c=1$
Therefore , the discriminant of the given equation is
$D = b^2 -4ac=(\sqrt{2})^{2} - 4 \times (\sqrt{2}) \times (1) =2 - 4 \sqrt{2}$
Therefore , the required solutions are
$\large\frac{-b \pm D}{2a} = -\large\frac{-\sqrt{2} \pm \sqrt{2 (1-2 \sqrt{2})} }{2 \times (\sqrt{2})}$
$=( \large\frac{- \sqrt{2} \pm \sqrt{2} (\sqrt{2 \sqrt{2}-1})i }{2 \sqrt{2}})$
$=\large\frac{-1 \pm( \sqrt{2\sqrt{2}-1})i}{2} \qquad [\sqrt{-1} =i]$
answered Apr 11, 2014 by

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