Difference between revisions of "1959 AHSME Problems/Problem 38"
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− | + | == Problem == | |
− | + | ||
− | (A) is an integer | + | If <math>4x+\sqrt{2x}=1</math>, then <math>x</math>: |
− | (B) is fractional | + | <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> |
− | (C) is irrational | + | |
− | (D) is imaginary | + | == Solution == |
− | (E) may have two different values | + | |
+ | Subtract <math>4x</math> from both sides of the initial equation and solve for <math>x</math>: | ||
+ | \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 <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>. | ||
+ | |||
+ | == See also == | ||
+ | {{AHSME 50p box|year=1959|num-b=37|num-a=39}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 15:48, 21 July 2024
Problem
If , then :
Solution
Subtract from both sides of the initial equation and solve for : \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 , but plugging in into the original equation reveals that it is extraneous. Thus, , which is only consistent with answer choice .
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 37 |
Followed by Problem 39 | |
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