Under what conditions will one solution of $ax^2+bx+c = 0$ be the reciprocal of the other? Under what conditions will one solution of 
$ax^2+bx+c = 0$ be the reciprocal of the other?

By the Quadratic Formula, given $ax^2+bx+c=0$ with $a\ne0$, we have $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$  Let us assume that $$\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{2a}{-b-\sqrt{b^2-4ac}}.$$  Cross-multiplying, we have $$(2a)^2=\left(-b+\sqrt{b^2-4ac}\right)\left(-b-\sqrt{b^2-4ac}\right)$$ $$\implies 4a^2=\left(-b\right)^2-\left(\sqrt{b^2-4ac}\right)^2\implies 4a^2=b^2-\left(b^2-4ac\right)\implies 4a^2=4ac.$$  Since $a\ne 0$, we can divide both sides of this equation by $4a$ to obtain $a=c$.  Therefore, the two roots of a quadratic will be reciprocals of one another precisely when $a=c$.
