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Difference between revisions of "1959 AHSME Problems/Problem 38"
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If <math>4x+\sqrt{2x}=1</math>, then <math>x</math>
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== Problem ==
  
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If <math>4x+\sqrt{2x}=1</math>, then <math>x</math>:
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<math>\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} </math>
  
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(A) is an integer
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== Solution ==
  
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(B) is fractional
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Subtract 4x from both sides so you get:
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<math>\sqrt{2x}=1-4x</math>
  
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(C) is irrational
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Then just square and simplify to get:
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<math>x=\frac{1}{8}</math>
  
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(D) is imaginary
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This is answer choice <math>\boxed{B}</math>.
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(E) may have two different values
 

Latest revision as of 14:04, 16 July 2024

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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