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

$(a)\;\large\frac{-1 \pm \sqrt{7} i}{-3}\qquad(b)\;\large\frac{-1 \pm \sqrt{3} i}{-2}\qquad(c)\;\large\frac{-1 \pm \sqrt{5} i}{2 \sqrt{2}}\qquad(d)\;\large\frac{-1 \pm \sqrt{7} i}{2 \sqrt{2}}$

1 Answer

Answer : $\; \large\frac{-1 \pm \sqrt{7} i}{2 \sqrt{2}}$
Explanation :
The given quadratic equation is $\;\sqrt{2}x^2+x+\sqrt{2}=0$
On comparing the given equation with $\;ax^2 + bx +c\;,$ we obtain
$a=\sqrt{2}\;,b=1\;and \;c=\sqrt{2}$
Therefore , the discriminant of the given equation is
$D = b^2 -4ac=1^{2} - 4 \times (\sqrt{2}) \times (\sqrt{2}) =-7$
Therefore , the required solutions are
$\large\frac{-b \pm D}{2a} = -\large\frac{-1 \pm \sqrt{-7}}{2 \times (\sqrt{2})}$
$=\large\frac{-1 \pm \sqrt{7} i}{2 \sqrt{2}} \qquad [\sqrt{-1} = i]$
answered Apr 11, 2014 by yamini.v
 
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