<s>[INST] Show that the solutions from one quadratic equation are reciprocal to the solutions of another quadratic equation From Sullivan's Algebra & Trigonometry book: Chapter 1.2; Exercise 116:

Show that the real solutions of the equation $ax^2+bx+c=0$ are the
  reciprocals of the real solutions of the equation $cx^2+bx+a=0$.
  Assume that $b^2-4ac\geq0$.


The most I reached in my attempt to solve the problem was to prove that, as $ax^2+bx+c=0$ can be expressed as $a^2x^2+abx+b^2$, and that $cx^2+bx+a=0$ can be expressed as $b^2x^2+abx+a^2$; thus, the solutions from the one will be the reciprocal of the another...
$a^2x^2+abx+b^2=(ax+b)^2$
$(ax+b)^2= 0$
$x_1 = -\frac{b}{a}$ (root of multiplicity two)
$b^2x^2+abx+a^2=(bx+a)^2$
$(bx+a)^2=0$
$x_2 = -\frac{a}{b}$ (root of multiplicity two)

$x_1x_2=0$

The problem here is that the solution only works for perfect squares; I want to know, and that's what the problem is asking for, to prove it for the $ax^2+bx+c$ form.

Another attempts to solve the problem...
I've tried to solve it using the quadratic formula, I started from the equation $\frac{2a}{-b+\sqrt{b^2-4ac}}=0$ and then try to get to the expression $\frac{-b+\sqrt{b^2-4ac}}{2c}=0$, but I wasn't cappable, don't even know it it's possible to do or correct to try.
I also tried the same from above with the reciprocal of the actual equation: $\frac{1}{ax^2+bx+c}=0$. But for me it was also impossible to do anything with that expression.
 [/INST] To ensure that the equations are quadratic, we need the assumption that $a,c\ne0$.
If $\alpha$ is a root of $ax^2+bx+c=0$, then $a\alpha^2+b\alpha+c=0$.
Note that $\alpha\ne0$, otherwise, $c=0$.
So $\displaystyle c\left(\frac{1}{\alpha}\right)^2+b\left(\frac{1}{\alpha}\right)+a=\frac{a\alpha^2+b\alpha+c}{\alpha^2}=0$.
$\displaystyle \frac{1}{\alpha}$ is a root of $cx^2+bx+a=0$.
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