What is the sum of the reciprocals of the roots of the equation (2003)/(2004)x + 1 + (1)/(x) = 0?

2003 AMC 10A Problem 18 — Solution

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18.

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

$-\dfrac{2004}{2003}$

$-1$

$\dfrac{2003}{2004}$

$1$

$\dfrac{2004}{2003}$

Let $a = \dfrac{2003}{2004}.$ Multiplying the equation by $x$ gives $ax^2 + x + 1 = 0.$

If the roots are $r$ and $s,$ then by Vieta's formulas $r + s = -\dfrac{1}{a}$ and $rs = \dfrac{1}{a}.$

The sum of reciprocals is $\dfrac{1}{r} + \dfrac{1}{s} = \dfrac{r + s}{rs} = \dfrac{-1/a}{1/a} = -1.$

Thus, the correct answer is B.