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$a$,$b$,$c$,$p$,$q$,$r$ 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}$

4d_
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  • This makes no sense to me. You're given that $x$ is a rational function of $y$. What does it mean to say that $y$ has rational roots? – John Hughes Jul 26 '19 at 14:50
  • That's what I couldn't understand too. If $x$ is a rational function of $y$, then $y$ can have rational, or irrational, or imaginary roots as well. I mean, each of these three situations should be possible (I think). And I have typed the question correctly, no additional data, no other information has been given – 4d_ Jul 26 '19 at 14:54
  • Perhaps you need a new book. :) – John Hughes Jul 26 '19 at 14:56
  • Perhaps it was a typo and the book forgot to mention $y$ had rational roots. If $y=-\frac{(x-\sqrt 3 )(x-2)}{x^2+x+1}$ we'd have $(ax^2+bx+c)(-\frac{(x-\sqrt 3 )(x-2)}{x^2+x+1}) + (px^2 + qx + r) = 0$. Not sure what happens then. – fleablood Jul 26 '19 at 15:26
  • That phrase does not make sense, but it's also stated that $x$ is a rational function of $y$ and that's why he discriminant of the quadratic equation must be a perfect square – Vasili Jul 26 '19 at 15:36
  • @fleablood The rational function that I wrote at the end of my post is just a random function, it is not related to the question. I randomly chose that function just to show that rational functions with real coefficients should have irrational roots too – 4d_ Jul 26 '19 at 16:53
  • @Vasya Are you saying that if $y$ is a rational function of $x$, then the discriminant of the quadratic equation would be a perfect square? – 4d_ Jul 26 '19 at 16:54
  • @πtimese: no, the problem states that "...and $x$ is a rational function of $y$" which means $x$ can be expressed as a rational function of $y$ and for that to be true, roots of the quadratic equation for $x$ cannot contain square roots of $y$ – Vasili Jul 26 '19 at 19:59
  • @Vasya "...for that to be true, roots of the quadratic equation for $x$ cannot contain square roots of $y$" Didn't get that. $x$ can have an irrational root, can't it? – 4d_ Jul 28 '19 at 02:25

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