# 1964 AHSME Problems/Problem 8

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## Problem

The smaller root of the equation $\left(x-\frac{3}{4}\right)\left(x-\frac{3}{4}\right)+\left(x-\frac{3}{4}\right)\left(x-\frac{1}{2}\right) =0$ is:

$\textbf{(A)}\ -\frac{3}{4}\qquad \textbf{(B)}\ \frac{1}{2}\qquad \textbf{(C)}\ \frac{5}{8}\qquad \textbf{(D)}\ \frac{3}{4}\qquad \textbf{(E)}\ 1$

## Solution

Note that this equation is of the form $a^2 + ab = 0$, which factors to $a(a + b) = 0$. Plugging in $a = x - \frac{3}{4}$ and $b = x - \frac{1}{2}$ gives:

$(x - \frac{3}{4})(x - \frac{3}{4} + x - \frac{1}{2}) = 0$

$(x - \frac{3}{4})(2x - \frac{5}{4}) = 0$

The roots are $x = \frac{3}{4}$ nad $x = \frac{1}{2} \cdot \frac{5}{4}$. The smaller root is $\frac{5}{8}$, which is option $\boxed{\textbf{(C)}}$.