Question

If alpha , beta and gamma are the roots of the polynomial ,x^3-5x^2-2x+24. find the value of alpha^-2+beta^-2+gamma^-2

Answer 1

Answer:  $\bold{\frac{61}{64}}$


We are given a polynomial:

$p(x)=x^3-5x^2-2x+24$

$\alpha$, $\beta$ and $\gamma$ are zeros.

We have:

$\alpha + \beta + \gamma = -\frac{\text{Coefficient of }x^2}{\text{Coefficient of }x^3} = -\frac{(-5)}{1} = 5 \\ \\ \\ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{\text{Coefficient of x}}{\text{Coefficient of }x^3} = \frac{-2}{1} = -2 \\ \\ \\ \alpha\beta\gamma = -\frac{\text{Constant Term}}{\text{Coefficient of }x^3} = -\frac{24}{1} = -24$

To Summarize:

$\boxed{\begin{array}{ccc}\alpha+\beta+\gamma & = & 5 \\ \\ \alpha\beta+\beta\gamma+\gamma\alpha & = & -2 \\ \\ \alpha\beta\gamma & = & -24 \end{array}}$

Now, we can proceed to find value:

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

Thus we see that we need the value of $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$

Let $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=k$


Consider the following:


$\begin{array}{cc}(\alpha\beta+\beta\gamma+\gamma\alpha)^2 = & (\alpha\beta)^2+(\beta\gamma)^2+(\gamma\alpha)^2 + \\ & 2(\alpha\beta)(\beta\gamma)+2(\beta\gamma)(\gamma\alpha)+2(\gamma\alpha)(\alpha\beta)\end{array} \\ \\ \\ \implies (-2)^2 = k+2\alpha\beta^2\gamma+2\beta\gamma^2\alpha+2\gamma\alpha^2\beta \\ \\ \\ \implies 4 = k + 2\alpha\beta\gamma(\beta+\gamma+\alpha) \\ \\ \\ \implies 4 = k + 2(-24)(5) \\ \\ \\ \implies 4 = k - 240 \\ \\ \\ \implies k = 244$

Now, we can continue with our question:

$\alpha^{-2}+\beta^{-2}+\gamma^{-2} \\ \\ \\ = \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} \\ \\ \\ = \frac{\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2}{\alpha^2\beta^2\gamma^2} \\ \\ \\ = \frac{244}{(-24)^2} \\ \\ \\ = \frac{244}{256} \\ \\ \\ = \frac{61}{64} \\ \\ \\ \\ \implies \boxed{\alpha^{-2}+\beta^{-2}+\gamma^{-2}=\frac{61}{64}}$