Question

lf $\alpha ,\beta ,\gamma$ are the roots of $x^3+2x^2-3x-1=0$, then ${\alpha }^{-2}+{\beta }^{-2}+{\gamma }^{-2}$ is:

• A. $12$
• B. $13$
• C. $14$
• D. $15$

Answer 1

The correct option is B $13$

Consider the given cubic equation whose roots are $\alpha ,\beta ,\gamma$

Then $(x-\alpha )(x-\beta )(x-\gamma )=0$ implies

$x^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\gamma \alpha )x-\alpha \beta \gamma =0$

Comparing with $a{x}^3+b{x}^2+cx+d=0$ gives us

Sum of roots $=\frac{-b}{a}$

Product of roots $=\frac{-d}{a}$

Product of roots taken two at a time $=\frac{c}{a}$

In the given equation, we have

$\alpha .\beta .\gamma =1$

$\alpha .\beta +\beta .\gamma +\gamma .\alpha =-3$

$\alpha +\beta +\gamma =-2$.

Now $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}$

$=\frac{1}{(\alpha .\beta .\gamma {)}^2}\left[\right(\beta \gamma {)}^2+(\alpha .\gamma {)}^2+(\alpha .\beta {)}^2]$

$=\frac{1}{\alpha \beta \gamma }\left[\right(\alpha .\beta +\beta \gamma +\gamma \alpha {)}^2-2\left(\alpha \beta \gamma \right)(\alpha +\beta +\gamma )]$

Substitution of the values yields $\frac{1}{1}\left[\right(-3{)}^2-2\left(1\right)(-2)]$

$=9-(-4)$

$=13$