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 $\boxed{\textbf{(D)} \quad -3, 5}.$

See Also

1956 AHSME (ProblemsAnswer KeyResources)
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png