Question

If $\alpha \beta$ are the roots of $a{x}^2+bx+c=0$ and $c\ne 0$ then the value of $\frac{1}{{\left(a\alpha +b\right)}^2}+\frac{1}{{\left(a\beta +b\right)}^2}$ in terms of $a,b,c$ is

Answer 1

We have,

$a{x}^2+bx+c=0$ ……. (1)

Since, $\alpha ,\beta$ are root of above equation.

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

$a\alpha +a\beta =-b$

$a\alpha +b=-a\beta$ …….. (2)

$\alpha \beta =\frac{c}{a}$ ……. (3)

Since,

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

From equation $\left(2\right)$,

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

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

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

$\Rightarrow \frac{1}{a^2}\left(\frac{{(\alpha +\beta )}^2-2\alpha \beta }{{\alpha }^2{\beta }^2}\right)$

$\Rightarrow \frac{1}{a^2}\left(\frac{{(-\frac{b}{a})}^2-\frac{2c}{a}}{\frac{c^2}{a^2}}\right)$

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

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

Hence, this is the answer.