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If one root of the equation $a x^3+b x+c=0$ is twice another root, then
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Verified Answer
The correct answer is:
$36 b^3+343 \mathrm{ac}^2=0$
Let roots of equation be $2 \alpha, \alpha, \beta$
So, $3 \alpha+\beta=0 \Rightarrow \beta=-3 \alpha$
and $2 \alpha^2 \beta=\frac{-c}{a} \Rightarrow-6 \alpha^3=\frac{-c}{a} \Rightarrow \alpha^3=\frac{c}{6 a}$
Now, $a\left(\frac{c}{6 a}\right)+b\left(\frac{c}{6 a}\right)^{1 / 3}+c=0$
$\begin{aligned}
& \Rightarrow b\left(\frac{c}{6 a}\right)^{1 / 3}=\frac{-c}{6}-c \\
& \Rightarrow \frac{b^3 c}{6 a}=\frac{-343 c^3}{36 \times 6} \Rightarrow 36 b^3+343 c^2 a=0
\end{aligned}$
So, $3 \alpha+\beta=0 \Rightarrow \beta=-3 \alpha$
and $2 \alpha^2 \beta=\frac{-c}{a} \Rightarrow-6 \alpha^3=\frac{-c}{a} \Rightarrow \alpha^3=\frac{c}{6 a}$
Now, $a\left(\frac{c}{6 a}\right)+b\left(\frac{c}{6 a}\right)^{1 / 3}+c=0$
$\begin{aligned}
& \Rightarrow b\left(\frac{c}{6 a}\right)^{1 / 3}=\frac{-c}{6}-c \\
& \Rightarrow \frac{b^3 c}{6 a}=\frac{-343 c^3}{36 \times 6} \Rightarrow 36 b^3+343 c^2 a=0
\end{aligned}$
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