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If the roots of the equation $x^3-a x^2+b x-c=0$ are in HP, then the mean of the roots is
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Verified Answer
The correct answer is:
$\frac{3 c}{b}$
Let $\alpha, \beta, \gamma$ are roots of equation
$$
\begin{array}{r}
x^3-a x^2+b x-c=0 \\
\alpha+\beta+\gamma=a \\
\alpha \beta+\beta \gamma+\gamma \alpha=b \\
\alpha \beta \gamma=c
\end{array}
$$
It is given that $\alpha, \beta, \gamma$ are in HP.
$\therefore \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in AP.
Mean of roots $=\frac{3}{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}}=\frac{3 \alpha \beta \gamma}{\alpha \beta+\beta \gamma+\gamma \alpha}=\frac{3 c}{b}$
$$
\begin{array}{r}
x^3-a x^2+b x-c=0 \\
\alpha+\beta+\gamma=a \\
\alpha \beta+\beta \gamma+\gamma \alpha=b \\
\alpha \beta \gamma=c
\end{array}
$$
It is given that $\alpha, \beta, \gamma$ are in HP.
$\therefore \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in AP.
Mean of roots $=\frac{3}{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}}=\frac{3 \alpha \beta \gamma}{\alpha \beta+\beta \gamma+\gamma \alpha}=\frac{3 c}{b}$
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