Coefficients Flipped!

As $r$ is a root of $ax^2+bx+c$ then $ar^2+br+c=0$

As $r$ is non-zero we can divide both sides of the equation by $r^2$ :

$\frac{ar^2+br+c}{r^2}$ $=$ $\frac{0}{r^2}$

$a+\frac{b}{r}+\frac{c}{r^2}=0$

$c(\frac{1}{r})^2+b(\frac{1}{r})+a=0$

So $\frac{1}{r}$ is a root of $cx^2+bx+a$

Analogously as $s$ is a non-zero root of $ax^2+bx+c$ then $\frac{1}{s}$ is a root of $cx^2+bx+a$

So the roots of $cx^2+bx+a$ are $\frac{1}{r}$ and $\frac{1}{s}$

In the first equation, Vieta's Formula gives the sum of the roots as $-\frac{b}{a} = r+s$ and the product of the roots as $\frac{c}{a} = rs.$

In the second equation, the sum of the roots is (-\frac{b}{c} = \frac{-b/a}{c/a} = \frac{r+s}{rs}) while the product of the roots is $\frac{a}{c} = \frac{1}{c/a} = \frac{1}{rs}.$

It's not too hard to see that the two numbers with sum $\frac{r+s}{rs}$ and product $\frac{1}{rs}$ are $\frac{1}{r}$ and $\frac{1}{s},$ so these are the roots of the second equation!

In the first equation the product of the roots is c/a while in the second it's a/c. The only pair of roots in the solution which have as product $\frac{1}{rs}$ is the first. Of course the sign of the roots must remain the same by Descartes' rule of signs.