Question

If $\alpha, \beta, \gamma$ are the roots of the cubic $2010 x^3+4 x^2+1=0$, then the value of $\left(\alpha^{-2}+\beta^{-2}+\gamma^{-2}\right)$ is equal to

• A. 8
• B. -8
• C. 4
• D. -4

Answer 1

Answer: (B)

Given $\alpha+\beta+\gamma=\frac{-4}{2010}, \sum \alpha \beta=0$ and $\alpha \beta \gamma=\frac{-1}{2010}$
Now $\alpha^{-2}+\beta^{-2}+\gamma^{-2}=\frac{\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2}{\alpha^2 \beta^2 \gamma^2}=\frac{(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2 \alpha \beta \gamma(\alpha+\beta+\gamma)}{\alpha^2 \beta^2 \gamma^2}$
$=\frac{(0)^2-2\left(\frac{-1}{2010}\right)\left(\frac{-4}{2010}\right)}{\left(\frac{-1}{2010}\right)^2}=-8$