1956 AHSME Problems/Problem 15

Problem

The root(s) of $\frac {15}{x^2 - 4} - \frac {2}{x - 2} = 1$ is (are):

$\textbf{(A)}\ -5\text{ and }3\qquad \textbf{(B)}\ \pm 2\qquad \textbf{(C)}\ 2\text{ only}\qquad \textbf{(D)}\ -3\text{ and }5\qquad \textbf{(E)}\ 3\text{ only}$

Solution

Use a common denominator: $\frac{15 - 2(x+2)}{x^2-4} = 1$ . After moving the $1$ to the left side, notice that you can factor the numerator: $\frac{(x-5)(x+3)}{x^2-4} = 0$ The roots of this equation are $\boxed{\textbf{(D)} \quad -3, 5}.$

1956 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1956 AHSME Problems/Problem 15

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