Difference between revisions of "2003 AMC 10A Problems/Problem 18"
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<math> \mathrm{(A) \ } -\frac{2004}{2003}\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } \frac{2003}{2004}\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } \frac{2004}{2003} </math> | <math> \mathrm{(A) \ } -\frac{2004}{2003}\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } \frac{2003}{2004}\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } \frac{2004}{2003} </math> | ||
− | == Solution == | + | == Solution 1 == |
Multiplying both sides by <math>x</math>: | Multiplying both sides by <math>x</math>: | ||
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So the answer is <math> \frac{a+b}{ab}=\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-1 \Rightarrow\boxed{\mathrm{(B)}\ -1}</math>. | So the answer is <math> \frac{a+b}{ab}=\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-1 \Rightarrow\boxed{\mathrm{(B)}\ -1}</math>. | ||
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+ | == Solution 2 == | ||
+ | Dividing both sides by <math>x</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>. | ||
== See Also == | == See Also == |
Revision as of 09:32, 29 July 2014
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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