Determining the parity of the coefficients of a quadratic given that there is a rational solution Que.   If the quadratic $ax^2 + bx + c$ has a rational root, and $a$, $b,$ and $c$ are integers, then
A) at least one of $a, b, c$ is even
B) all of $a,b,c$ are even
C) at most one of $a,b,c$ is odd
D) all of $a,b,c$ are odd
So we can say that since the roots are of the form $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, for it to be rational, $b^2 - 4ac$ must be a perfect square. I am not sure now how to determine the parity of the coefficients given this data.
 A: With the quadratic $x^2-6x+5$, you can cross off options $B, C, D$ which leaves only option $A$ by the process of elimination. However, as you stated in the comments, this is not what you wanted.

Let $p/q$ and $r/s$ be the two roots of the quadratic. You can form a quadratic equation which satisfies these conditions as follows:
$$\left(x-\frac{p}{q} \right) \left(x - \frac{r}{s} \right) = 0$$
$$\iff \left(qx-p \right) \left(sx-r  \right) = 0$$
$$\iff (qs)x^2-(qr+ps)x+pr = 0$$
If one root is irrational, then $\Delta = \sqrt{b^2-4ac}$ in the quadratic formula must be irrational. Because of the quadratic formula $x = \frac{-b ± \Delta}{2a}$, either both roots are rational or both roots are irrational. This concept is further explained in this post.
Now that we have confirmed there must be two rational roots, let's check if condition $A$ must always be true. The contrapositive of 'at least one of $a,b,c$ is even is that none of $a,b,c$ are even, which means $qs, qr + ps$ and $pr$ must all be odd. If $qr$ and $ps$ are both even, then at least one of $qs$ and $pr$ is even. However, if $qr$ and $ps$ are both odd, then the resulting sum is also even.
This means that either $qr$ is even and $ps$ is odd, or that $qr$ is odd and $ps$ is even.
Now would be a good time to stop and continue the proof for yourself. If you are interested, the rest of the proof is hidden behind the spoiler effect:

 - Case $1$: $qr$ is even and $ps$ is odd


 Since $qr$ is even, at least one of $q$ and $r$ is even. However, since $\text{even} \times \text{odd} = \text{even}$, either one of $qs$ or $pr$ would be even.


 - Case $2$: $qr$ is odd and $ps$ is even.
 The same logic applies, so either one of $qs$ or $pr$  would be even.


 This means that the quadratic cannot be written in the form $(qs)x^2-(qr+ps)x+pr = 0$. Reversing the order of the steps means that the quadratic does not have two rational roots. By proving the contrapositive of the statement, we have proved that option $A$ is indeed correct.

