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Solve the equation : $\;3x^{2}-4x+\large\frac{20}{3}=0\;$

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

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Answer : $\;\large\frac{2}{3} \pm \large\frac{4}{3} i$
Explanation :
The given quadratic equation is $\;3x^{2}-4x+\large\frac{20}{3}=0$
This equation can also be written as $\; 9x^{2} -12 x + 20 =0$
On comparing the given equation with $\;ax^2 + bx +c\;,$ we obtain
$a=9\;,b=-12\;and \;c=20$
Therefore , the discriminant of the given equation is
$D = b^2 -4ac=(-12)^{2} - 4 \times (9) \times (20) =144-720 = -576$
Therefore , the required solutions are
$\large\frac{-b \pm D}{2a} = -\large\frac{-(-12) \pm \sqrt{-576}}{2 \times 9}$
$= \large\frac{12 \pm \sqrt{576}i}{18} \quad [\sqrt{-1} =i]$
$= \large\frac{12 \pm 24 i}{18}$
$= \large\frac{6(2+4i)}{18}$
$=\large\frac{2}{3} \pm \large\frac{4}{3} i$
answered Apr 11, 2014 by yamini.v
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