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If a root of the equation $a x^2+b x+c=0$ be reciprocal of a root of the equation then $a^{\prime} x^2+b^{\prime} x+c^{\prime}=0$, then
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The correct answer is:
$\left(c c^{\prime}-a a^{\prime}\right)^2=\left(b a^{\prime}-c b^{\prime}\right)\left(a b^{\prime}-b c^{\prime}\right)$
Let $\alpha$ be a root of first equation, then $\frac{1}{\alpha}$ be a root of second equation.
Therefore $a \alpha^2+b \alpha+c=0$ and $a^{\prime} \frac{1}{\alpha^2}+b^{\prime} \frac{1}{\alpha}+c^{\prime}=0$
or $c^{\prime} \alpha^2+b^{\prime} \alpha+a^{\prime}=0$
Hence $\frac{\alpha^2}{b a^{\prime}-b^{\prime} c}=\frac{\alpha}{c c^{\prime}-a a^{\prime}}=\frac{1}{a b^{\prime}-b c^{\prime}}$
$\left(c c^{\prime}-a a^{\prime}\right)^2=\left(b a^{\prime}-c b^{\prime}\right)\left(a b^{\prime}-b c^{\prime}\right)$
Therefore $a \alpha^2+b \alpha+c=0$ and $a^{\prime} \frac{1}{\alpha^2}+b^{\prime} \frac{1}{\alpha}+c^{\prime}=0$
or $c^{\prime} \alpha^2+b^{\prime} \alpha+a^{\prime}=0$
Hence $\frac{\alpha^2}{b a^{\prime}-b^{\prime} c}=\frac{\alpha}{c c^{\prime}-a a^{\prime}}=\frac{1}{a b^{\prime}-b c^{\prime}}$
$\left(c c^{\prime}-a a^{\prime}\right)^2=\left(b a^{\prime}-c b^{\prime}\right)\left(a b^{\prime}-b c^{\prime}\right)$
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