2001 AIME I Problems/Problem 3
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 
.
Solution 3
Note that if 
is a root, then 
is a root and they sum up to 
We make the substitution 
so ![\[(\frac{1}{4}+y)^{2001}+(\frac{1}{4}-y)^{2001}=0.\]](http://latex.artofproblemsolving.com/8/6/f/86fa1781f8716422b3c8b9ec4d0a199e2089c8f6.png)
Expanding gives ![\[2\cdot\frac{1}{4}\cdot\binom{2001}{1}y^{2000}-0y^{1999}+\cdots\]](http://latex.artofproblemsolving.com/e/f/e/efef370e2b35868055004c3c3deab8e120e39e74.png)
so by Vieta, the sum of the roots of 
is 0. Since has a degree of 2000, then has 2000 roots so the sum of the roots is ![\[2000(\sum_{n=1}^{2000} y+\frac{1}{4})=2000(0+\frac{1}{4})=\boxed{500}.\]](http://latex.artofproblemsolving.com/e/1/b/e1b2d6060af31856da4477487536ba14644996d6.png)
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 | ||
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