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Solve the equation : $\;21x^{2}-28x+10=0\;$

$(a)\;\large\frac{2}{27} \pm \large\frac{4}{27} i\qquad(b)\;\large\frac{5}{3} \pm \large\frac{\sqrt{2}}{21} i\qquad(c)\;\large\frac{2}{3} \pm \large\frac{\sqrt{14}}{21} i\qquad(d)\;\large\frac{\sqrt{2}}{27} \pm \large\frac{5}{27} i$

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Answer : $\;\large\frac{2}{3} \pm \large\frac{\sqrt{14}}{21} i$
Explanation :
The given quadratic equation is $\;21x^{2}-28x+10=0$
On comparing the given equation with $\;ax^2 + bx +c\;,$ we obtain
$a=21\;,b=-28\;and \;c=10$
Therefore , the discriminant of the given equation is
$D = b^2 -4ac=(-28)^{2} - 4 \times (21) \times (10) =784-840 = -56$
Therefore , the required solutions are
$\large\frac{-b \pm D}{2a} = \large\frac{-(-28) \pm \sqrt{-56}}{2 \times 21}$
$ = \large\frac{28 \pm \sqrt{56}i}{42}$
$ = \large\frac{28 \pm 2\sqrt{14}i}{42}$
$=\large\frac{28}{42} \pm \large\frac{2\sqrt{2}}{42} i$
$=\large\frac{2}{3} \pm \large\frac{\sqrt{14}}{21} i$
answered Apr 11, 2014 by yamini.v
 
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