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

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