1955 AHSME Problems/Problem 49

The graphs of $y=\frac{x^2-4}{x-2}$ and $y=2x$ intersect in:

$\textbf{(A)}\ \text{1 point whose abscissa is 2}\qquad\textbf{(B)}\ \text{1 point whose abscissa is 0}\\ \textbf{(C)}\ \text{no points}\qquad\textbf{(D)}\ \text{two distinct points}\qquad\textbf{(E)}\ \text{two identical points}$


We can simplify $y=\frac{x^2-4}{x-2}$ in order to help solve this problem: $y=\frac{x^2-4}{x-2}=\frac{(x+2)(x-2)}{x-2}=x+2$.

In order to find solutions, we have to have $x+2=2x$. $x=2$, but this turns out to be an extraneous solution, as, when put into the two equations, one of them turns out an undefined value. The answer is $\boxed{(C)}$.

See Also

Go back to the rest of the 1955 AHSME Problems

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

Invalid username
Login to AoPS