2003 AMC 10A Problem 18

We want the sum of $1/x$ over the roots, so let $y = 1/x$. Then $x = 1/y$ and the equation becomes
$$\frac{2003}{2004} \cdot \frac{1}{y} + 1 + y = 0.$$
Multiplying through by $y$ (nonzero, since $x$ is finite):
$$y^2 + y + \frac{2003}{2004} = 0.$$
The roots of this quadratic are exactly the reciprocals of the original roots, and by Vieta their sum is $-\frac{1}{1} = -1$.

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

A question about the reciprocals of the roots is a question about the roots of the "reversed" polynomial: substituting $y = 1/x$ reverses the coefficient order. Vieta then reads off the desired sum without solving. (The roots here are non-real, since the discriminant $1 - 4 \cdot \frac{2003}{2004}$ is negative, but Vieta's formulas hold regardless.)

Multiply the original equation by $x$: $\frac{2003}{2004}x^2 + x + 1 = 0$ with roots $a, b$. Then $\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab} = \frac{-b/a}{c/a} = -\frac{b}{c} = -\frac{1}{1} = -1$ in terms of the coefficients $a, b, c$ of $ax^2 + bx + c$.

• Clearing denominators and reporting the sum of the roots themselves, -2004/2003, which is choice (A).
• Panicking about non-real roots; the sum of reciprocals is still determined by the coefficients.