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 $x$ : $\begin{aligned} \sqrt{2 x} & = 1 - 4 x \\ 2 x & = 1 - 8 x + 16 x^{2} \\ 16 x^{2} - 10 x + 1 & = 0 \\ (8 x - 1) (2 x - 1) & = 0 \end{aligned}$

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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