Prove, by contradiction, that a quadratic equation, $ax^2 + bx + $c where $a\neq 0$, has at most two distinct roots. My approach was to show that the discriminant is less than zero, i.e, there are indeed no real roots, which contradicts that there are at most two distinct roots. I'm not exactly sure how to execute this theory however. Should I do so by introducing constants which would make the discriminant less than zero? Proofs is not my strong point :( Help greatly appreciated.

First of all solve the equation for its two roots, hence

$x_1 = \frac{-b -\sqrt{b^2-4ac}}{2a}; x_2 = \frac{-b +\sqrt{b^2-4ac}}{2a}$. (Assume those roots exist in real numbers, otherwise there are less then 2 anyway).

Now you can write your equation as $ax^2 +bx+c = a(x-x_1)(x-x_2)$.

Assume there is a $x_3 \
eq x_2,x_1$ with $ax_3^2+bx_3+c=0$.

A real product is zero if and only if (at least) one of its factors is zero. Therefore

$a(x-x_1)(x-x_2) = 0 \Leftrightarrow (x=x_1 \vee x=x_2)$

It follows that $ax_3^2+bx_3+c \
eq 0$ what contradicts to the assumption.