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

$(a)\;\large\frac{2}{3} \pm \large\frac{4}{3} i\qquad(b)\;1\pm \large\frac{2}{3} i\qquad(c)\;\large\frac{5}{3} \pm \large\frac{4}{3} i\qquad(d)\;1 \pm \large\frac{\sqrt{2}}{2} i$

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Answer : $\;1 \pm \large\frac{\sqrt{2}}{2} i$
Explanation :
The given quadratic equation is $\;x^{2}-2x+\large\frac{3}{2}=0$
This equation can also be written as $\; 2x^{2} -4 x + 3 =0$
On comparing the given equation with $\;ax^2 + bx +c\;,$ we obtain
$a=2\;,b=-4\;and \;c=3$
Therefore , the discriminant of the given equation is
$D = b^2 -4ac=(-4)^{2} - 4 \times (2) \times (3) =16-24 = -8$
Therefore , the required solutions are
$\large\frac{-b \pm D}{2a} = -\large\frac{-(-4) \pm \sqrt{-8}}{2 \times 2}$
$= \large\frac{4\pm 2 \sqrt{2} i}{4} \qquad [\sqrt{-1} =i]$
$= \large\frac{2 \pm \sqrt{2}i}{2}$
$=1 \pm \large\frac{\sqrt{2}}{2} i$
answered Apr 11, 2014 by yamini.v
 
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