1954 AHSME Problems/Problem 2

Problem 2

The equation $\frac{2x^2}{x-1}-\frac{2x+7}{3}+\frac{4-6x}{x-1}+1=0$ can be transformed by eliminating fractions to the equation $x^2-5x+4=0$ . The roots of the latter equation are $4$ and $1$ . Then the roots of the first equation are:

$\textbf{(A)}\ 4 \text{ and }1 \qquad \textbf{(B)}\ \text{only }1 \qquad \textbf{(C)}\ \text{only }4 \qquad \textbf{(D)}\ \text{neither 4 nor 1}\qquad\textbf{(E)}\ \text{4 and some other root}$

Solution

We plug $x=4, x=1$ into the original equation $x=4\implies\frac{8}{-3}+\frac{15}{3}+\frac{-20}{-3}\implies\frac{15+20-8}{4}+1=0$ $x=1\implies\frac{2}{0}-\frac{9}{3}+\frac{-2}{0}+1=0$ , which is undefined because $\frac{k}{0}$ is undefined.

Thus the only solution to the equation is $x=4$ , $\fbox{C}$

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1954 AHSME Problems/Problem 2

Problem 2

Solution