Problem

The roots of the equation $ax^2 + bx + c = 0$ will be reciprocal if:

$\textbf{(A)}\ a = b \qquad\textbf{(B)}\ a = bc \qquad\textbf{(C)}\ c = a \qquad\textbf{(D)}\ c = b \qquad\textbf{(E)}\ c = ab$

Solution

Dividing the equation by $a\quad(a\neq0)$ gives: $x^2+\frac{b}{a}x+\frac{c}{a}$ . Let $r$ and $s$ be the roots of the equation $r=\frac{1}{s}$ $rs=1$

From Vieta's formula, $rs=\frac{c}{a}\Rightarrow\frac{c}{a}=1\Rightarrow\boxed{\text{(C) }c=a}$

1956 AHSME (Problems • Answer Key • Resources)
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