1956 AHSME Problems/Problem 7

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

This quadratic is equivalent to $x^2+\frac{b}{a}x+\frac{c}{a}$.

Letting $r$ and $s$ be the respective roots to this quadratic, if $r=\frac{1}{s}$, then $rs=1$.

From Vieta's, $rs=\frac{c}{a}$, but we know that $rs=1$ so $\frac{c}{a}=1$ as well.

Multiply both sides by $a$ to get $\boxed{\text{(C) }c=a}$

See Also

1956 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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