2003 AMC 10A Problem 18

Problem 18 of 25IntermediateAlgebra

What is the sum of the reciprocals of the roots of the equation

\dfrac{2003}{2004}x + 1 + \dfrac{1}{x} = 0?

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Solution

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{-\frac{1}{a}}{\frac{1}{a}} = -1.

Thus, the correct answer is B.