Find a quadratic equation with root $x_1^2$ and $x_2^2$

Question

I'm struggling to solve this problem:

"If the equation $ax^2 + bx +c$ $(a \neq 0)$ admits real and not null roots of $x_1$ and $x_2$, obtain the equation which evaluates to roots $(x_1)^2$ and $(x_2)^2$"

I know the answer is $a^2x^2 -(b^2 - 2ac)x + c^2$ but I am not able to get to this answer, I was trying to derive it from the sum and product equations $S=-b/a$ and $P=c/a$ but the answer I'm getting to is a monstruous different thing.

Could someone explain to me how to get to this answer?

Thanks

Answer 1

Hint: use factor form of quadratic equation

$$(x-x_1^2)(x-x_2^2)=0$$

and Vieta formulas.

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Notice that $$x_1^2+x_2^2 = (x_1+x_2)^2-2x_1x_2 = {b^2\over a^2}-{2c\over a}$$

Answer 2

Let $z_i=x_i^2$.

Square your equation to deduce that $$0=a^2z_i^2+(b^2+2ac)z_i+c^2+2abx_i^3+2bcx_i$$

Remark $2bx_i\times$ (your original equation) yields $$0=2abx_i^3+2b^2z_i+2bcx_i$$

Subtract the latter result from the former to get $$0=a^2z_i^2+(b^2+2ac)z_i+c^2-2b^2z_i=a^2z_i^2+(2ac-b^2)z_i+c^2$$ And we are done.