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...


$(ax+b)^2= 0$

$x_1 = -\frac{b}{a}$ (root of multiplicity two)



$x_2 = -\frac{a}{b}$ (root of multiplicity two)


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.


If $r$ is a root of $ax^2+bx+c$, then $ar^2+br+c=0$ and\begin{align}ar^2+br+c=0&\iff \frac{ar^2+br+c}{r^2}=0\\&\iff a++b\times\frac1r+c\times\frac1{r^2}=0\\&\iff\frac1r\text{ is a root of }cx^2+bx+c=0.\end{align}

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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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