1
$\begingroup$

Suppose the cofficients of a quadratic are rational. How can we use the discriminant to determine if the roots are also rational?

Firstly the discriminant has to be nonnegative.

I've had a read of Rational root theorem. I kinda get it but i'm not quite sure how to apply it to find a solution for the problem.

$$\sqrt{b^2-4ac} = \frac{p}{q}$$, where $p, q \in \mathbb{Z} $ and are coprime. $$b^2-4ac = \frac{p^2}{q^2}$$ $$ q^2(b^2-4ac) = p^2$$ Am I going in the right direction ? I want reason that some $p/q$ does exist. Please guide me.

$\endgroup$
1
  • 1
    $\begingroup$ I think you have this turned around. It is presumed that $a$, $b$, and $c$ are given. One then computes the single number $d = b^2 - 4ac$. Either this $d$ is or it is not a square rational number -- there is no solving to be done. $\endgroup$ Commented Dec 12, 2021 at 2:01

1 Answer 1

2
$\begingroup$

To find out if the roots are rational, you need to check if the discriminant $b^2-4ac$ (which is a rational number $s/t$) is a square of a rational number. We can assume that the fraction $s/t$ is in the lowest terms, i.e., $s$, $t$ are coprime integers. Then you just need to find out if $s$ and $t$ are perfect squares (of integers).

For example $1/2x^2-x+1/4$ has discriminant $1/2$ but $2$ is not a perfect square, so the roots are not rational. On the other hand, $1/2x^2-3/2x+1$ has discriminant $9/4-2=1/4$, where both $1$ and $4$ are perfect squares, so roots are rational.

$\endgroup$
5
  • $\begingroup$ So we really answer this probelm rhetorically? And end by saying either $s$ and $t$ are perfect squares or not. There is no solving to be done? $\endgroup$
    – user839943
    Commented Dec 12, 2021 at 2:29
  • $\begingroup$ I don't quite understand the first sentence of your comment. Note that $s$ and $t$ are integers, not arbitrary rational numbers. To check that, say, $s$ is a perfect square, one needs, for example, to find its prime decomposition. Does this qualify for "solving"? $\endgroup$
    – markvs
    Commented Dec 12, 2021 at 2:33
  • $\begingroup$ Sorry, I mean the answer you gave is mainly words. Not a long drawn out algebra equations. I assume this is because for example it isn't possible to do a prime decomposition on variable. $\endgroup$
    – user839943
    Commented Dec 12, 2021 at 2:40
  • $\begingroup$ We don't have say much/do much to conclude this solution. $\endgroup$
    – user839943
    Commented Dec 12, 2021 at 2:41
  • 1
    $\begingroup$ To check if $s$ or $t$ is a square, you may need to say a lot. It is quite difficult in general. There are other methods to find rational solutions, and it is not clear which method is faster. $\endgroup$
    – markvs
    Commented Dec 12, 2021 at 3:18

You must log in to answer this question.