Difference between revisions of "2001 AIME I Problems/Problem 3"
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== Solution 2 == | == Solution 2 == | ||
| − | We find that the given equation has a <math>2000th</math> degree polynomial. Note that there are no multiple roots. Thus, if <math>\frac{1}{2} - x</math> is a root, <math>x</math> is also a root. Thus, we pair up <math>1000</math> pairs of roots that sum to <math>\frac{1}{2}</math> to get a sum of <math>500</math>. | + | We find that the given equation has a <math>2000th</math> degree polynomial. Note that there are no multiple roots. Thus, if <math>\frac{1}{2} - x</math> is a root, <math>x</math> is also a root. Thus, we pair up <math>1000</math> pairs of roots that sum to <math>\frac{1}{2}</math> to get a sum of <math>/boxed{500}</math>. |
== See also == | == See also == | ||
Revision as of 18:45, 6 October 2016
Contents
Problem
Find the sum of the roots, real and non-real, of the equation 
, given that there are no multiple roots.
Solution 1
From Vieta's formulas, in a polynomial of the form 
, then the sum of the roots is 
.
From the Binomial Theorem, the first term of 
is 
, but 
, so the term with the largest degree is 
. So we need the coefficient of that term, as well as the coefficient of 
.
Applying Vieta's formulas, we find that the sum of the roots is 
.
Solution 2
We find that the given equation has a 
degree polynomial. Note that there are no multiple roots. Thus, if 
is a root, 
is also a root. Thus, we pair up 
pairs of roots that sum to 
to get a sum of 
.
See also
| 2001 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
