Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1959 AHSME Problems/Problem 38: Difference between revisions

Created page with " If 4x + √2x = 1, then x (A) is an integer (B) is fractional (C) is irrational (D) is imaginary (E) may have two different values"
 
Thepowerful456 (talk | contribs)
editor
642 edits
m category
 
(4 intermediate revisions by 3 users not shown)
Line 1: Line 1:
If 4x +
== Problem ==
√2x = 1, then x
 
(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 $4x+\sqrt{2x}=1$, 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1959_AHSME_Problems/Problem_38&oldid=223341"