<s>[INST] 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?
 [/INST] 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 $b$.
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