Question

If $\alpha ,\beta ,\gamma$ are the roots of the cubic $2010x^3+4x^2+1=0$, 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 ${\alpha }^{-2}+{\beta }^{-2}+{\gamma }^{-2}=\frac{{\alpha }^2{\beta }^2+{\beta }^2{\gamma }^2+{\gamma }^2{\alpha }^2}{{\alpha }^2{\beta }^2{\gamma }^2}=\frac{(\alpha \beta +\beta \gamma +\gamma \alpha {)}^2-2\alpha \beta \gamma (\alpha +\beta +\gamma )}{{\alpha }^2{\beta }^2{\gamma }^2}$
$=\frac{(0{)}^2-2\left(\frac{-1}{2010}\right)\left(\frac{-4}{2010}\right)}{{\left(\frac{-1}{2010}\right)}^2}=-8$