2003 AMC 10A Problem 18

The question asks about $\frac{1}{x}$ for each solution, so it helps to give $\frac{1}{x}$ its own name and write an equation for it directly.

1. Let $y = \frac{1}{x}$. Then $x = \frac{1}{y}$. (Neither can be $0$, because the equation contains $\frac{1}{x}$.)
2. Put these into $\frac{2003}{2004}x + 1 + \frac{1}{x} = 0$:
$$\frac{2003}{2004} \cdot \frac{1}{y} + 1 + y = 0.$$
3. Multiply every term by $y$ to clear the fraction:
$$y^2 + y + \frac{2003}{2004} = 0.$$
4. Each solution $y$ of this equation is the reciprocal of a solution $x$ of the original. So the question is now: what is the sum of the solutions of $y^2 + y + \frac{2003}{2004} = 0$?
5. For an equation $y^2 + by + c = 0$, the two solutions add up to $-b$. (Reason: if the solutions are $p$ and $q$, the left side is $(y - p)(y - q) = y^2 - (p + q)y + pq$, so $b = -(p+q)$.) Here $b = 1$, so the sum is $-1$.

The answer is $\boxed{\textbf{(B)}\ -1}$.

A question about reciprocals of solutions becomes a question about solutions of a new equation once you substitute $y = \frac{1}{x}$; this flips the order of the coefficients. The sum of the solutions can then be read straight off the middle coefficient, with no need to solve. (The solutions here are not real numbers, since $1 - 4 \cdot \frac{2003}{2004}$ is negative, but the sum rule still holds.)

Multiply the original equation by $x$ to get $\frac{2003}{2004}x^2 + x + 1 = 0$, with solutions $p$ and $q$. Then
$$\frac{1}{p} + \frac{1}{q} = \frac{p + q}{pq}.$$
Dividing the equation by $\frac{2003}{2004}$ shows that $p + q = -\frac{2004}{2003}$ and $pq = \frac{2004}{2003}$. So the sum of reciprocals is $-\frac{2004}{2003} \div \frac{2004}{2003} = -1$.