smc

Dividing both sides of the equation by $a(a\ne 0)$ gives $x^2+\frac{b}{a}x+\frac{c}{a}$. Letting $r$ and $s$ be the respective roots to this quadratic, $r=\frac{1}{s}\Rightarrow rs=1$. From Vieta's, $rs=\frac{c}{a}$, so $\frac{c}{a}=1\Rightarrow \boxed{c=a}$