1957 AHSME Problems/Problem 27

Problem

The sum of the reciprocals of the roots of the equation $x^2 + px + q = 0$ is:

$\textbf{(A)}\ -\frac{p}{q} \qquad \textbf{(B)}\ \frac{q}{p}\qquad \textbf{(C)}\ \frac{p}{q}\qquad \textbf{(D)}\ -\frac{q}{p}\qquad\textbf{(E)}\ pq$

Solution 1

Let $f(x)=x^2+px+q$ . Then, $x^2f(\tfrac1 x)$ equals $qx^2+px+1$ and has roots which are the reciprocals of those of $f(x)$ . Thus, by Vieta's Formulas, the sum of the roots of $x^2f(\tfrac1 x)$ (and thus the sum of the reciprocated roots of $f(x)$ ) is $\boxed{\textbf{(A) }\frac{-p}{q}}$ .

Solution 2

One approach is to plug in some roots.

We have $x^{2}-5x+6=0$

The roots are $x=2$ and $x=3$ .

The sum of the reciprocals of the roots is $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ .

In this case, $p$ and $q$ are $-5$ and $6$ .

Thus, the answer is $\boxed{\textbf{(A) }\frac{-p}q}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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1957 AHSME Problems/Problem 27

Contents

Problem

Solution 1

Solution 2

See Also