The other root must also be rational. Suppose one root is [tex]\dfrac pq[/tex]. Then we can write
[tex]ax^2+bx+c=a\left(x-\dfrac p{aq}\right)\left(x-\dfrac ra\right)[/tex]
where [tex]\dfrac ra[/tex] is the other unknown root. Expanding, we get
[tex]ax^2+bx+c=ax^2-a\left(\dfrac p{aq}+\dfrac ra\right)x+\dfrac{apr}{aq}[/tex]
[tex]ax^2+bx+c=ax^2-\left(\dfrac pq+r\right)x+\dfrac{pr}q[/tex]
It follows that
[tex]\begin{cases}-\dfrac pq-r=b\\\\\dfrac{pr}q=c\end{cases}[/tex]
and since we assumed [tex]b,c,\dfrac pq[/tex] are all rational, then there can only be rational solutions for [tex]r[/tex].
