Question

If $\alpha ,\beta$ and $\gamma$ are the roots of $x^3+2x^2-3x-1=0$, then ${\alpha }^{-2}+{\beta }^{-2}+{\gamma }^{-2}$ is equal to

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

Answer 1

The correct option is B

$13$

Explanation for the correct option:

Find the value of ${\alpha }^{-2}+{\beta }^{-2}+{\gamma }^{-2}$:

Given, $\alpha ,\beta$ and $\gamma$ are the roots of $x^3+2x^2-3x-1=0$

$\Rightarrow$$\sum \alpha =–2,\sum \alpha \beta =–3,\alpha \beta \gamma =1$

As we know,

$\begin{array}{rcl}\sum \frac{1}{\alpha }& =& \sum \frac{{\left(\alpha \beta \right)}^2}{{\left(\alpha \beta \gamma \right)}^2}\\ & =& \frac{\sum {\left(\alpha \beta \right)}^2–2\left(\alpha \beta \gamma \right)(\sum \alpha )}{{\left(\alpha \beta \gamma \right)}^2}\\ & =& \frac{{(-3)}^2–2(-2)}{1}\\ & =& 9+4\\ & =& \mathbf{13}\end{array}$

$\therefore$The value of ${\alpha }^{-2}+{\beta }^{-2}+{\gamma }^{-2}$ is $\mathbf{13}$

Hence, Option ‘B’ is Correct.