1965 AHSME Problems/Problem 7

Problem

The sum of the reciprocals of the roots of the equation $ax^2 + bx + c = 0$ is:

$\textbf{(A)}\ \frac {1}{a} + \frac {1}{b} \qquad \textbf{(B) }\ - \frac {c}{b} \qquad \textbf{(C) }\ \frac{b}{c}\qquad \textbf{(D) }\ -\frac{a}{b}\qquad \textbf{(E) }\ -\frac{b}{c}$

Solution

Using Vieta's formulas, we can write the sum of the roots of any quadratic equation in the form $ax^2+bx+c = 0$ as $\frac{-b}{a}$ , and the product as $\frac{c}{a}$ .

If $r$ and $s$ are the roots, then the sum of the reciprocals of the roots is $\frac{1}{r} + \frac{1}{s} = \frac{r+s}{rs}$ .

Applying the formulas, we get $\frac{\frac{-b}{a}}{\frac{c}{a}}$ , or $\frac {-b}{c}$ => $\boxed{a}$ .

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1965 AHSME Problems/Problem 7

Problem

Solution