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?
 A: Let $r_1,r_2$ be the two solutions of the quadratic. Then we have 
$$ax^2+bx+c=a(x-r_1)(x-r_2)=ax^2-a(r_1+r_2)x+ar_1r_2$$
so that $c=ar_1r_2$. Since one solution is the reciprocal of the other, we have $r_1r_2=1$, so that $a=c$. Hence, the roots are reciprocals of one another only when $a=c$.
A: 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 $a=c$.
A: Here's one way using the quadratic formula:
The roots of $ax^2+bx+c=0$ can be written as $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Let $x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}$ and let $x_2$ be the reciprocal of this, so $$x_2 = \frac{2a}{-b+\sqrt{b^2-4ac}}.$$
However, since $x_2 = \frac{-b-\sqrt{b^2-4ac}}{2a}$, we can set these equal:
$$\frac{2a}{-b+\sqrt{b^2-4ac}} = \frac{-b-\sqrt{b^2-4ac}}{2a}.$$
Multiplying through, we get
$$(2a)^2 = (-b+\sqrt{b^2-4ac})(-b-\sqrt{b^2-4ac}) \\
4a^2 = b^2-b^2+4ac \\
4a^2 = 4ac \\
a = c.
$$

Here's another way:
If $x$ solves $ax^2+bx+c=0$, then it also solves $x^2+\frac{b}{a}x+\frac{c}{a} = 0$, which we obtain by just dividing through by $a$.
Let the roots of this polynomial be $x_1$ and $\frac{1}{x_1}$. Writing the polynomial in terms of factors, we get
$$(x-x_1)(x-\frac{1}{x_1}) = x^2+\frac{b}{a}x+\frac{c}{a}.$$
Expanding on the left hand side, this becomes
$$x^2-\left(x_1+\frac{1}{x_1}\right)x+1 - x^2+\frac{b}{a}+\frac{c}{a}.$$
Therefore, we must have $\frac{c}{a} = 1$, implying $c = a$. We do not need to care about $b$.
