An algebra problem by A Former Brilliant Member

For the roots to be rational, the discriminant $b^2-4ac$ would have to be a square number. But all odd squares are congruent to $1$ modulo $8$ , and $4ac \not\equiv 0 \pmod8$ , so this is not possible.

Assuming the contrary, the discriminant $b^2-4ac$ must be a perfect square. Also, the discriminant is odd, as $b$ is odd. Hence, writing $b=2k+1$ for some integral $k$ , we must have for some integral $n$ : $(2k+1)^2-4ac=(2n+1)^2.$ Simplifying, we have $ac= k(k+1)-n(n+1).$ Note that the LHS of the above equation is odd (since both $a$ and $c$ are odd), but both the terms on the RHS are even (since the product of two consecutive integers is even). Thus, we arrive at an impossibility. QED

Let´s do this by a contradiction, so let the rational root be of the form $p/q$ were p and q are coprime whole numbers. Now if we plug $p/q$ into our original equation we will get:
$a(p/q)^2+b(p/q)+c=0$ ; now if we multiply by $q^2$ we will get:
$ap^2+bpq+cq^2=0$
$p(ap+bq)=-cq^2$
Now we must check 3 cases:
1) p and q are both odd - right hand side of the equation will be even and the left will be odd - contradiction.
2) p is odd and q is even - right hand side of the equation will be odd and the left will be even - contradiction.
3) p is even and q is odd - right hand side of the equation will be even and the left will be odd - contradiction.
We don´t have to check the case when p and q are both even, because in this case they will not be coprime.