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Solve the equation : $\;27x^{2}-10x+1=0\;$

$(a)\;\large\frac{2}{27} \pm \large\frac{4}{27} i\qquad(b)\;\large\frac{5}{27} \pm \large\frac{\sqrt{2}}{27} i\qquad(c)\;\large\frac{5}{3} \pm \large\frac{4}{3} i\qquad(d)\;\large\frac{\sqrt{2}}{27} \pm \large\frac{5}{27} i$

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Answer : $\;\large\frac{5}{27} \pm \large\frac{\sqrt{2}}{27} i$
Explanation :
The given quadratic equation is $\;27x^{2}-10x+1=0$
On comparing the given equation with $\;ax^2 + bx +c\;,$ we obtain
$a=27\;,b=-10\;and \;c=1$
Therefore , the discriminant of the given equation is
$D = b^2 -4ac=(-10)^{2} - 4 \times (27) \times (1) =100-108 = -8$
Therefore , the required solutions are
$\large\frac{-b \pm D}{2a} = -\large\frac{-(-10) \pm \sqrt{-8}}{2 \times 27}$
$= \large\frac{10 \pm 2 \sqrt{2} i}{54} \qquad [\sqrt{-1} =i]$
$= \large\frac{5 \pm \sqrt{2} i}{27}$
$=\large\frac{5}{27} \pm \large\frac{\sqrt{2}}{27} i \;.$
answered Apr 11, 2014 by yamini.v
 
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