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Difference between revisions of "1959 AHSME Problems/Problem 38"

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== Solution ==
 
== Solution ==
  
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Subtract 4x from both sides so you get:
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Subtract <math>4x</math> from both sides of the initial equation and solve for <math>x</math>:
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<math>\sqrt{2x}=1-4x</math>
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\begin{align*}
 +
\sqrt{2x} &= 1-4x \\
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2x &= 1-8x+16x^2 \\
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16x^2-10x+1 &= 0 \\
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(8x-1)(2x-1) &= 0
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\end{align*}
  
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Then just square and simplify to get:
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It may look like we have two different solutions for <math>x</math>, but plugging in <math>x=\frac{1}{2}</math> into the original equation reveals that it is [[Extraneous solutions|extraneous]]. Thus, <math>x=\frac{1}{8}</math>, which is only consistent with answer choice <math>\fbox{\textbf{(B)}}</math>.
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<math>x=\frac{1}{8}</math>
 
  
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This is answer choice <math>\boxed{B}</math>.
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== See also ==
 +
{{AHSME 50p box|year=1959|num-b=37|num-a=39}}
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 16:48, 21 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 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 $\fbox{\textbf{(B)}}$.

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 37 		Followed by
Problem 39
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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