Question

Consider the quadratic equation ax^2+bx+c where a,b, and c are rational numbers and the quadratic has two distinct zeros. If one is rational, what is true for the other zeros

Answer 1

The other root must also be rational. Suppose one root is
`\frac pq`. Then we can write


`ax^2+bx+c=a\left(x-\frac p{aq}\right)\left(x-\frac ra\right)`

where
`\frac ra` is the other unknown root. Expanding, we get


`ax^2+bx+c=ax^2-a\left(\frac p{aq}+\frac ra\right)x+(apr)/(aq)`


`ax^2+bx+c=ax^2-\left(\frac pq+r\right)x+\frac{pr}q`

It follows that


`\begin{cases}-\frac pq-r=b\\\\\frac{pr}q=c\end{cases}`

and since we assumed
`b,c,\frac pq` are all rational, then there can only be rational solutions for
`r`.