$a$,$b$,$c$,$p$,$q$,$r$ are real numbers. If ($ax^2$ + $bx$ + $c$)$y$ + ($px^2$ + $qx$ + $r$) = $0$, and $x$ is a rational function of $y$, then prove that $( $ar$ - $pc$)^2$ = ($aq$ - $pb$)($br$ - $qc$).
This is a question from my book. It basically says that
$y$ = $-$$\frac{(px^2+ax+r)}{(ax^2+by+c)}$
My book starts off by making the given equation a quadratic equation in $x$,
$($p$+$ay$)$$x^2$ + ($q$+$by$)$x$ + ($r$+$cy$) = $0$
and further says that since $y$ has rational roots, so the discriminant of the quadratic equation must be a perfect square, and that’s how the book solves this question.
What I don’t understand is, how can we say for sure that $y$ has rational roots? Can’t it have irrational roots?
For example, how about $y$ = $-$$\frac{(x-\sqrt 3 )(x-2)}{x^2+x+1}$