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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>
  
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== Solution ==
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== 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

Problem

What is the sum of the reciprocals of the roots of the equation $\frac{2003}{2004}x+1+\frac{1}{x}=0$?

$\mathrm{(A) \ } -\frac{2004}{2003}\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } \frac{2003}{2004}\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } \frac{2004}{2003}$

Solution 1

Multiplying both sides by $x$:

$\frac{2003}{2004}x^{2}+1x+1=0$

Let the roots be $a$ and $b$.

The problem is asking for $\frac{1}{a}+\frac{1}{b}= \frac{a+b}{ab}$

By Vieta's formulas:

$a+b=(-1)^{1}\frac{1}{\frac{2003}{2004}}=-\frac{2004}{2003}$

$ab=(-1)^{2}\frac{1}{\frac{2003}{2004}}=\frac{2004}{2003}$

So the answer is $\frac{a+b}{ab}=\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-1 \Rightarrow\boxed{\mathrm{(B)}\ -1}$.

Solution 2

Dividing both sides by $x$, $\frac{2003}{2004}+\frac 1x+\frac{1}{x^2}=0$, we see by Vieta's formulas that the sum of the roots is $-1 \Rightarrow\boxed{\mathrm{(B)}\ -1}$.

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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