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.
On
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.
On
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: $$\frac{1}{(a\alpha+b)^2} + \frac{1}{(a\beta+b)^2}=$$ $$\frac{1}{(\frac{-b-\sqrt{b^2-4ac}}{2}+b)^2} + \frac{1}{(\frac{-b+\sqrt{b^2-4ac}}{2}+b)^2}=$$ $$\frac{4}{(b-\sqrt{b^2-4ac})^2}+\frac{4}{(b+\sqrt{b^2-4ac})^2}=$$ $$\frac{4(2b^2+2(b^2-4ac))}{(4ac)^2}=\frac{b^2-2ac}{a^2c^2}.$$
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*}