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I was given that $ax^2+2bx+c=0$ Using $y=x+\frac{1}{x}$ I need to prove that $acy^2+2b(c+a)y+(a-c)^2+4b^2=0$

Tried to make the pattern $x+\frac{1}{x}$ and to substitute $y$, but couldn't prove it.

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    $\begingroup$ You can use that $ax^2=-c-2bx$ in order to get an expression for $x$ in terms of $y$ (you can assume $a\neq 0$, since otherwise the problem is super easy). $\endgroup$
    – Patricio
    Commented Nov 25, 2015 at 17:49

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Given that $y = x + \frac1x,$ if you make this substitution for $y$ in $acy^2+2b(c+a)y+(a-c)^2+4b^2,$ you get

$$ac\left(x + \frac1x\right)^2 + 2b(c+a)\left(x + \frac1x\right) + (a-c)^2 + 4b^2.$$

Use the binomial theorem, multiplication of polynomials, the distributive law, or any other methods you know in order to convert this to an expression with no parentheses. Then look for collections of terms within that expression that have factors of $ax^2 + 2bx + c.$

Here's a really big hint: since $ax^2 + 2bx + c = 0,$ it follows that $$\left(c + \frac{a}{x^2} + \frac{2b}{x}\right)(ax^2 + 2bx + c) = 0.$$

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