Difference between revisions of "1955 AHSME Problems/Problem 42"
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| − | == Solution == | + | == Solution == |
| − | a+b | + | Here, we are told that the two quantities are equal. |
| − | + | ||
| − | + | Squaring both sides, we get: <math>a+\frac{b}{c}=a^2*\frac{b}{c}</math>. | |
| − | + | ||
| − | (a^2-1) | + | Multiply both sides by <math>c</math>: <math>ac+b=ba^2</math>. |
| + | |||
| + | Looking at all answer choices, we can see that <math>ac=ba^2-b=b(a^2-1)</math>. | ||
| + | |||
| + | This means that <math>c=\frac{b(a^2-1)}{a}</math>, and this is option C. Therefore chose <math>\boxed{C}</math> as your answer. | ||
| + | |||
| + | ~hastapasta | ||
== See Also == | == See Also == | ||
Latest revision as of 17:38, 18 February 2022
Problem
If 
, and 
are positive integers, the radicals 
and 
are equal when and only when:
Solution
Here, we are told that the two quantities are equal.
Squaring both sides, we get: 
.
Multiply both sides by : 
.
Looking at all answer choices, we can see that 
.
This means that 
, and this is option C. Therefore chose 
as your answer.
~hastapasta
See Also
| 1955 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 41 |
Followed by Problem 43 | |
| 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. 
