Difference between revisions of "2003 AMC 10A Problems/Problem 18"
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== Solution 2 == | == Solution 2 == | ||
Dividing both sides by <math>x</math>, | 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>. | + | |
| + | <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 09:33, 29 July 2014
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 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
