Difference between revisions of "1955 AHSME Problems/Problem 42"
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== Solution == | == Solution == | ||
| − | a+b/c = a^2(b/c) | + | squaring, we get a + b/c = a^2 * (b/c) |
| − | + | simplifying, | |
| − | a=(b/c)(a^2-1) | + | a = (b/c) * (a^2-1) |
| − | a/(a^2-1) = b/c | + | a / (a^2-1) = b/c |
| − | (a^2-1)b/a = c | + | (a^2-1) / a = c/b |
| + | so ((a^2-1)b) / a = c | ||
== See Also == | == See Also == | ||
Revision as of 00:22, 26 August 2021
Problem
If 
, and 
are positive integers, the radicals 
and 
are equal when and only when:
Solution
squaring, we get a + b/c = a^2 * (b/c) simplifying, a = (b/c) * (a^2-1) a / (a^2-1) = b/c (a^2-1) / a = c/b so ((a^2-1)b) / a = c
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. 
