Difference between revisions of "2007 Alabama ARML TST Problems/Problem 12"
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Since the [[Degree_of_a_polynomial|degree]] of <math>p(x)</math> is <math>2007</math>, <math>p(x)</math> has exactly <math>2007</math> roots. So, the problem is asking for the sum of the roots of <math>p(x)</math>. | Since the [[Degree_of_a_polynomial|degree]] of <math>p(x)</math> is <math>2007</math>, <math>p(x)</math> has exactly <math>2007</math> roots. So, the problem is asking for the sum of the roots of <math>p(x)</math>. | ||
− | Using the [[Binomial Theorem]], <math>(1-x)^{2007} = -x^{2007} + 2007x^{2006} | + | Using the [[Binomial Theorem]], <math>(1-x)^{2007} = -x^{2007} + 2007x^{2006} - \ldots + 1</math>. |
− | So, <math>p(x) = (1-x)^{2007} - x^{2007} = -2x^{2007} + 2007x^{2006} | + | So, <math>p(x) = (1-x)^{2007} - x^{2007} = -2x^{2007} + 2007x^{2006} - \ldots + 1</math>. |
Therefore, by [[Vieta's Formulas]], the sum of the roots of <math>p(x)</math> is <math>-\dfrac{2007}{-2} = \boxed{\dfrac{2007}{2}}</math>. | Therefore, by [[Vieta's Formulas]], the sum of the roots of <math>p(x)</math> is <math>-\dfrac{2007}{-2} = \boxed{\dfrac{2007}{2}}</math>. |
Latest revision as of 13:54, 29 March 2009
Problem
If and , then evaluate
Express your answer as a fraction in lowest terms.
Solution
For all :
is a root of .
is a root of .
is a root of .
is a root of , since .
Since the degree of is , has exactly roots. So, the problem is asking for the sum of the roots of .
Using the Binomial Theorem, .
So, .
Therefore, by Vieta's Formulas, the sum of the roots of is .
See also
2007 Alabama ARML TST (Problems) | ||
Preceded by: Problem 11 |
Followed by: Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |