If $\alpha, \beta$ roots of the equation $ax^2+by+c=0$. Then find the value of

Question

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

I can solve it by simplifying everything, but it will obviously we very long. How should I shorten my calculations?

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

Answer 1

Dividing the equation by $x$ given gives us $$ax + b = -\frac{c}{x} \implies \frac{1}{(ax + b)^2} = \frac{x^2}{c^2}$$ Since $\alpha,\beta$ are the roots we got $$\frac{1}{(a\alpha + b)^2} + \frac{1}{(a\beta + b)^2} = \frac{\alpha^2}{c^2} + \frac{\beta^2}{c^2} = \frac{\alpha^2 + \beta^2}{c^2}$$

You can find $\alpha^2 + \beta^2$ using $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha \beta$, which equals $(b^2 - 2ac)/a^2$

Answer 2

Let $y=ax+b\implies x=?$

$$0=a((y-b)/a)^2+b(y-b)/a+c$$

$$y^2-by+ca=0$$ whose roots are $y_1,y_2$(say)

We need $$1/y_1^2+1/y_2^2=\dfrac{(y_1+y_2)^2-2y_1y_2}{(y_1y_2)^2}=?$$

Answer 3

Another way:

Let $y=\dfrac1{(ax+b)^2}$

$\iff a^2y x^2+2aby x+b^2y-1=0$

Also $ax^2+bx+c=0$

Solve the two simultaneous equations in $x^2,x$

Use $x^2=(x)^2$ to eliminate $x$ to form a quadratic equation in $y$ whose roots are $y_1,y_2$

We need $y_1+y_2$