# 2003 AMC 10A Problems/Problem 18

## 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 $\frac{1}{a}+\frac{1}{b}= \frac{a+b}{ab}$ $a+b=(-1)^{1}\frac{1}{\frac{2003}{2004}}=-\frac{2004}{2003}$ $ab=(-1)^{2}\frac{1}{\frac{2003}{2004}}=\frac{2004}{2003}$

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 $-1 \Rightarrow\boxed{\mathrm{(B)}\ -1}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 