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If $\overline{ }_{-2}$ is one of the roots of $a x^{2}+b x+c=0$, where $a$,
$b, c$ are real, then what are the values of $a, b, c$ respectively?
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$b, c$ are real, then what are the values of $a, b, c$ respectively?
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Verified Answer
The correct answer is:
6,-4,1
Given quadratic equation is $a x^{2}+b x+c=0$ whose
one root is $\frac{1}{2-\sqrt{-2}}$
Consider $\frac{1}{2-\sqrt{-2}}=\frac{1}{2-\sqrt{2} i} \times \frac{2+\sqrt{2} i}{2+\sqrt{2} i}$
$=\frac{2+\sqrt{2} i}{4+2}=\frac{2+\sqrt{2} i}{6}$
$\therefore$ Another root will be $\frac{2-\sqrt{2} i}{6}$
$(\because$ complex roots always occurs in pairs)
Thus, sum of roots $=\frac{2+\sqrt{2} i}{6}+\frac{2-2 \sqrt{2} i}{6}=\frac{4}{6}$
and product of roots $=\left(\frac{2+\sqrt{2} i}{6}\right)\left(\frac{2-\sqrt{2} i}{6}\right)$
$=\frac{4+2}{36}=\frac{1}{6}$
$\therefore$ Requiredequation is $x^{2}-($ sum of roots $) x+($ product of roots $)=0$
$x^{2}-\frac{4}{6} x+\frac{1}{6}=0$
$\Rightarrow 6 x^{2}-4 x+1=0$
Thus, the values of $a, b, c$ are $6,-4,1$ respectively
one root is $\frac{1}{2-\sqrt{-2}}$
Consider $\frac{1}{2-\sqrt{-2}}=\frac{1}{2-\sqrt{2} i} \times \frac{2+\sqrt{2} i}{2+\sqrt{2} i}$
$=\frac{2+\sqrt{2} i}{4+2}=\frac{2+\sqrt{2} i}{6}$
$\therefore$ Another root will be $\frac{2-\sqrt{2} i}{6}$
$(\because$ complex roots always occurs in pairs)
Thus, sum of roots $=\frac{2+\sqrt{2} i}{6}+\frac{2-2 \sqrt{2} i}{6}=\frac{4}{6}$
and product of roots $=\left(\frac{2+\sqrt{2} i}{6}\right)\left(\frac{2-\sqrt{2} i}{6}\right)$
$=\frac{4+2}{36}=\frac{1}{6}$
$\therefore$ Requiredequation is $x^{2}-($ sum of roots $) x+($ product of roots $)=0$
$x^{2}-\frac{4}{6} x+\frac{1}{6}=0$
$\Rightarrow 6 x^{2}-4 x+1=0$
Thus, the values of $a, b, c$ are $6,-4,1$ respectively
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