An algebra problem by Luke Tan

We prove by contradiction. Let the rational solution be $\frac{p}{q}$ where p and q have no common factors. Then the equation becomes a( $\frac{p}{q}$ )^2+b( $\frac{p}{q}$ )+c=0 Since 0 is an even number and c is an odd number, a( $\frac{p}{q}$ )^2+b( $\frac{p}{q}$ ) must be odd too, therefore one of a( $\frac{p}{q}$ )^2 and b( $\frac{p}{q}$ ) is divisible by 2 however, since a and b are both odd, either ( $\frac{p}{q}$ )^2 or $\frac{p}{q}$ is divisible by 2. However, ( $\frac{p}{q}$ )^2 only has factors that $\frac{p}{q}$ has, thus they can only both have factor 2 or both not have factor 2, causing a( $\frac{p}{q}$ )^2+b( $\frac{p}{q}$ ) to be even, a contradiction