Difference between revisions of "1959 AHSME Problems/Problem 38"
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== Solution == | == Solution == | ||
| − | Subtract 4x from both sides | + | Subtract <math>4x</math> from both sides of the initial equation and solve for <math>x</math>: |
| − | <math>\sqrt{2x}=1-4x | + | \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>. | |
| − | <math>x=\frac{1}{8}</math> | ||
| − | + | == See also == | |
| + | {{AHSME 50p box|year=1959|num-b=37|num-a=39}} | ||
| + | {{MAA Notice}} | ||
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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
