Difference between revisions of "2005 AIME I Problems/Problem 8"
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== Problem == | == Problem == | ||
| − | The [[equation]] <math> 2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1 </math> has three [[real]] [[root]]s. Given that their sum is <math> | + | The [[equation]] <math> 2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1 </math> has three [[real]] [[root]]s. Given that their sum is <math>m/n</math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math> |
== Solution == | == Solution == | ||
Revision as of 19:50, 2 January 2022
Problem
The equation 
has three real roots. Given that their sum is 
where 
and 
are relatively prime positive integers, find 
Solution
Let 
. Then our equation reads 
or 
. Thus, if this equation has roots 
and 
, by Vieta's formulas we have 
. Let the corresponding values of 
be 
and 
. Then the previous statement says that 
so taking a logarithm of that gives 
and 
. Thus the answer is 
.
See also
| 2005 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
