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 so you get: $\sqrt{2x}=1-4x$

Then just square and simplify to get: $x=\frac{1}{8}$

This is answer choice $\boxed{B}$ .

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1959 AHSME Problems/Problem 38

Problem

Solution