1969 AHSME Problems/Problem 5

Problem

If a number $N,N \ne 0$ , diminished by four times its reciprocal, equals a given real constant $R$ , then, for this given $R$ , the sum of all such possible values of $N$ is

$\text{(A) } \frac{1}{R}\quad \text{(B) } R\quad \text{(C) } 4\quad \text{(D) } \frac{1}{4}\quad \text{(E) } -R$

Solution

Write an equation from the given information. $N - \frac{4}{N} = R$ $N^2 - 4 = RN$ $N^2 - RN - 4 = 0$ By Vieta's Formulas, the sum of all possible values of $N$ for a given $R$ is $R$ , so the answer is $\boxed{\textbf{(B)}}$ .

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1969 AHSME Problems/Problem 5

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