Question

If $\alpha ,\beta$ are roots of the equation $a{x}^2+bx+c=0$ then evaluate $\frac{1}{(a\alpha {)}^2}+\frac{1}{(a\beta {)}^2}$

Answer 1

We have,

$a{x}^2+bx+c=0$

Since,

$\alpha ,\beta$ are the roots of the equation.

So,

$\alpha +\beta =-\frac{b}{a}$

$\alpha \beta =-\frac{c}{a}$

Since,

$\frac{1}{(a\alpha {)}^2}+\frac{1}{(a\beta {)}^2}$

$\Rightarrow \frac{1}{a^2{\alpha }^2}+\frac{1}{a^2{\beta }^2}$

$\Rightarrow \frac{{\alpha }^2+{\beta }^2}{a^2{\alpha }^2{\beta }^2}$

$\Rightarrow \frac{(\alpha +\beta {)}^2-2\alpha \beta }{a^2{\alpha }^2{\beta }^2}$

$\Rightarrow \frac{\frac{b^2}{a^2}-\frac{2c}{a}}{a^2\frac{c^2}{a^2}}$

$\Rightarrow \frac{\frac{b^2-2ac}{a^2}}{c^2}$

$\Rightarrow \frac{b^2-2ac}{a^2{c}^2}$

Hence, this is the answer.