1959 AHSME Problems/Problem 38

Problem

If $4x+\sqrt{2x}=1$ , then $x$ : $\textbf{(A)}\ \text{is an integer} \qquad\textbf{(B)}\ \text{is fractional}\qquad\textbf{(C)}\ \text{is irrational}\qquad\textbf{(D)}\ \text{is imaginary}\qquad\textbf{(E)}\ \text{may have two different values}$

Solution

Subtract $4x$ from both sides of the initial equation and solve for $x$ : \begin{align*} \sqrt{2x} &= 1-4x \\ 2x &= 1-8x+16x^2 \\ 16x^2-10x+1 &= 0 \\ (8x-1)(2x-1) &= 0 \end{align*}

It may look like we have two different solutions for $x$ , but plugging in $x=\frac{1}{2}$ into the original equation reveals that it is extraneous. Thus, $x=\frac{1}{8}$ , which is only consistent with answer choice $\fbox{\textbf{(B)}}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 37	Followed by Problem 39
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1959 AHSME Problems/Problem 38

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