quadratic equation problem - proving a statement 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.

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 $$\left(c + \frac{a}{x^2} + \frac{2b}{x}\right)(ax^2 + 2bx + c) = 0.$$