1955 AHSME Problems/Problem 49

Revision as of 12:09, 18 November 2020 by Angrybird029 (talk | contribs) (Created page with "The graphs of <math>y=\frac{x^2-4}{x-2}</math> and <math>y=2x</math> intersect in: <math>\textbf{(A)}\ \text{1 point whose abscissa is 2}\qquad\textbf{(B)}\ \text{1 point who...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

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

Solution

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