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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

thiagohubes
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2 Answers2

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Hint: use factor form of quadratic equation

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

and Vieta formulas.

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

nonuser
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  • Hello ChristianF,

    Thank you for your answer, but I'm still doing something wrong, if you could take a look:

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

    $x^2-xx_1^2-xx_2^2+x_1x_2=0$ \

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

    $x^2-x(\frac{b^2}{a^2}-\frac{2c}{a})+\frac{c^2}{a^2}=0$ \

    $\frac{a^2x^2-b^2x-2acx+c^2}{a^2} = 0$ \

    Which is still not the correct answer, I'm still stuck, maybe some silly mistake that you could point it out.

    Thanks

     – thiagohubes Jun 10 '18 at 16:29
  • Last line, you forgot one − after getting rid of a bracket – nonuser Jun 10 '18 at 19:16
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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.

lulu
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