# 1964 AHSME Problems/Problem 8

## 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 $\boxed{\textbf{(C)}}$.

## See Also

 1964 AHSC (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions

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