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 1

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 $\frac {-b}{c}$ => $\boxed{E}$ .

Solution 2

Let $f(x)=ax^2+bx+c$ . $f(x)$ has roots $r$ and $s$ , and so $x^2f(\frac{1}{x})$ has roots $\frac{1}{r}$ and $\frac{1}{s}$ . Because $x^2f(\frac{1}{x})=cx^2+bx+a$ , by Vieta's formulas, $\frac{1}{r}+\frac{1}{s}=\boxed{\frac{-b}{c}}$ , which is answer choice $\fbox{E}$ .

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1965 AHSME Problems/Problem 7

Contents

Problem

Solution 1

Solution 2

See Also