Difference between revisions of "2001 AIME I Problems/Problem 3"
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Find the sum of the [[root]]s, real and non-real, of the equation <math>x^{2001}+\left(\frac 12-x\right)^{2001}=0</math>, given that there are no multiple roots. | Find the sum of the [[root]]s, real and non-real, of the equation <math>x^{2001}+\left(\frac 12-x\right)^{2001}=0</math>, given that there are no multiple roots. | ||
− | == Solution == | + | == Solution 1 == |
From [[Vieta's formulas]], in a [[polynomial]] of the form <math>a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0</math>, then the sum of the roots is <math>\frac{-a_{n-1}}{a_n}</math>. | From [[Vieta's formulas]], in a [[polynomial]] of the form <math>a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0</math>, then the sum of the roots is <math>\frac{-a_{n-1}}{a_n}</math>. | ||
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Applying Vieta's formulas, we find that the sum of the roots is <math>-\frac{-2001 \cdot 250}{\frac{2001}{2}}=250 \cdot 2=\boxed{500}</math>. | Applying Vieta's formulas, we find that the sum of the roots is <math>-\frac{-2001 \cdot 250}{\frac{2001}{2}}=250 \cdot 2=\boxed{500}</math>. | ||
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+ | == Solution 2 == | ||
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+ | 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>. | ||
== See also == | == See also == |
Revision as of 05:54, 27 February 2011
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 |