Difference between revisions of "2003 AMC 10A Problems/Problem 18"
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− | Dividing both sides by <math>x</math>, | + | Dividing both sides by <math>x^2</math>, |
<math>\frac{2003}{2004}+\frac 1x+\frac{1}{x^2}=0</math>, | <math>\frac{2003}{2004}+\frac 1x+\frac{1}{x^2}=0</math>, |
Revision as of 14:17, 28 August 2020
Contents
[hide]Problem
What is the sum of the reciprocals of the roots of the equation ?
Solution 1
Multiplying both sides by :
Let the roots be and .
The problem is asking for
By Vieta's formulas:
So the answer is .
Solution 2
Dividing both sides by ,
,
we see by Vieta's formulas that the sum of the roots is .
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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