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 $\boxed{\textbf{(B)}}$.

See also

1969 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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