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Solve the equation : $\;\sqrt{3}x^{2}-\sqrt{2}x+3\sqrt{3}=0\;$

$(a)\;\large\frac{1 \pm \sqrt{34} i}{3\sqrt{2}}\qquad(b)\;\large\frac{-1 \pm \sqrt{3} i}{-2}\qquad(c)\;\large\frac{\sqrt{2} \pm \sqrt{34} i}{2 \sqrt{3}}\qquad(d)\;\large\frac{-1 \pm \sqrt{7} i}{2 \sqrt{2}}$

1 Answer

Answer : $\;\large\frac{\sqrt{2} \pm \sqrt{34} i}{2 \sqrt{3}}$
Explanation :
The given quadratic equation is $\;\sqrt{3}x^2-\sqrt{2}x+3\sqrt{3}=0$
On comparing the given equation with $\;ax^2 + bx +c\;,$ we obtain
$a=\sqrt{3}\;,b=-\sqrt{2}\;and \;c=3\sqrt{3}$
Therefore , the discriminant of the given equation is
$D = b^2 -4ac=(\sqrt{2})^{2} - 4 \times (\sqrt{3}) \times (3\sqrt{3}) =-34$
Therefore , the required solutions are
$\large\frac{-b \pm D}{2a} = -\large\frac{-(-\sqrt{2}) \pm \sqrt{-34}}{2 \times (\sqrt{3})}$
$=\large\frac{\sqrt{2} \pm \sqrt{34} i}{2 \sqrt{3}} \qquad [\sqrt{-1} = i]$
answered Apr 11, 2014 by yamini.v