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Solve $\; \sqrt{5} x^{2} + x + \sqrt{5}=0 $

$(a)\;\large\frac{-1\pm\sqrt{19}i}{2\sqrt{5}}\qquad(b)\;\large\frac{1+ \sqrt{3}i}{2}\qquad(c)\;\large\frac{-1\pm 19i}{2\sqrt{5}}\qquad(d)\;\large\frac{-1\pm \sqrt{3}i}{2}$

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Answer : $\;\large\frac{-1\pm\sqrt{19}i}{2\sqrt{5}}$
Explanation :
$\; \sqrt{5} x^{2} + x + \sqrt{5}=0 $
Here , the discriminant of the equation is
$b^{2} - 4ac = 1^{2} - 4 \times \sqrt{5} \times \sqrt{5}$
$= 1 -20 = -19$
Therefore , the solutions are given by
$= \large\frac{-1 \pm \sqrt{-19}}{2 \sqrt{5}}$
$= \large\frac{-1 \pm \sqrt{19}i}{2 \sqrt{5}}$
answered Apr 16, 2014 by yamini.v
 
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