If one root of the equation $a x^3+b x+c=0$ is twice another root, then

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Aicharya · Accepted Answer

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, $a\left(\frac{c}{6 a}ight)+b\left(\frac{c}{6 a}ight)^{1 / 3}+c=0$ $\begin{aligned} & \Rightarrow b\left(\frac{c}{6 a}ight)^{1 / 3}=\frac{-c}{6}-c \ & \Rightarrow \frac{b^3 c}{6 a}=\frac{-343 c^3}{36 	imes 6} \Rightarrow 36 b^3+343 c^2 a=0 \end{aligned}$

If one root of the equation $a x^3+b x+c=0$ is twice another root, then

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