How to find the value of this expression Suppose I have a general quadratic equation $ ax^2 + bx + c = 0 $ 
And I have $ \alpha , \beta $ as roots, then what is the simplest way to find $ (a\alpha + b)^{-2} + (a\beta + b)^{-2} $ ?
I know the formula for finding the sum and product of roots, but that doesn't helped me in finding the value.
Any good hint is welcomed.
 A: Some assumptions:
$c, \alpha, \beta \neq 0$ (the answer will be a bit modified without these assumptions)
Since $\alpha$ is a root therefore
\begin{align*}
a\alpha^2+b\alpha+c &=0\\
a\alpha+b & = \frac{-c}{\alpha}\\
\frac{1}{(a\alpha+b)^2} & = \frac{\alpha^2}{c^2}
\end{align*}
Thus
\begin{align*}
\frac{1}{(a\alpha+b)^2} + \frac{1}{(a\beta+b)^2}& = \frac{\alpha^2+\beta^2}{c^2}\\
&= \frac{(\alpha+\beta)^2-2\alpha\beta}{c^2}
\end{align*}
A: Alternative hint:  derive the equation whose roots are $y_1=(a\alpha + b)^{-2}$ and $y_2 = (a\beta + b)^{-2}\,$.
Let $y=(ax+b)^{-2}\,$, then $\;\displaystyle ax+b=\pm \frac{1}{\sqrt{y}} \;\;\iff\;\; x = \frac{-b \pm \frac{1}{\sqrt{y}}}{a} = \frac{-b\sqrt{y}\pm1}{a\sqrt{y}}\,$.
Substituting back into the original equation:
$$\require{cancel}
\begin{align}
a\,\frac{b^2 y \mp 2b\sqrt{y}+1}{a^2 y} + b\,\frac{-b \sqrt{y} \pm 1}{a \sqrt{y}}+c = 0 \;\;&\iff\;\; \cancel{b^2 y} \mp 2b\sqrt{y}+1 -\cancel{b^2 y} \pm b\sqrt{y}+acy = 0 \\
 &\iff\;\; acy \mp b\sqrt{y}+1=0 \\
 &\iff\;\; (acy+1)^2 = b^2y \\
 &\iff\;\; a^2c^2y^2 +(2ac-b^2)y+1=0
\end{align}
$$
Then $\displaystyle(a\alpha + b)^{-2} + (a\beta + b)^{-2} =y_1+y_2=\frac{b^2-2ac}{a^2c^2}$ by Vieta's relations.
A: The roots of the equation $ax^2+bx+c=0$ are:
$$\alpha=\frac{-b-\sqrt{b^2-4ac}}{2a}; \beta=\frac{-b+\sqrt{b^2-4ac}}{2a}.$$
Now substitute these into the expression:
$$\frac{1}{(a\alpha+b)^2} + \frac{1}{(a\beta+b)^2}=$$
$$\frac{1}{(\frac{-b-\sqrt{b^2-4ac}}{2}+b)^2} + \frac{1}{(\frac{-b+\sqrt{b^2-4ac}}{2}+b)^2}=$$
$$\frac{4}{(b-\sqrt{b^2-4ac})^2}+\frac{4}{(b+\sqrt{b^2-4ac})^2}=$$
$$\frac{4(2b^2+2(b^2-4ac))}{(4ac)^2}=\frac{b^2-2ac}{a^2c^2}.$$
