Difference between revisions of "2003 AMC 10A Problems/Problem 18"
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<math>\frac{2003}{2004}+\frac 1x+\frac{1}{x^2}=0</math>, | <math>\frac{2003}{2004}+\frac 1x+\frac{1}{x^2}=0</math>, | ||
− | we see by [[Vieta's formulas]] that the sum of the roots is <math>-1 \Rightarrow\boxed{\mathrm{(B)}\ -1}</math>. | + | we see by [[Vieta's formulas]](with the same reasoning as above) that the sum of the roots is <math>-1 \Rightarrow\boxed{\mathrm{(B)}\ -1}</math>. |
== See Also == | == See Also == |
Revision as of 10:40, 21 June 2020
Contents
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(with the same reasoning as above) 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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