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2003 AMC 10A Problem 18

Below is the professionally curated solution for Problem 18 of the 2003 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10A solutions, or check the answer key.

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Concepts:Vieta’s Formulasquadratic

Difficulty rating: 1440

18.

What is the sum of the reciprocals of the roots of the equation 20032004x+1+1x=0?\dfrac{2003}{2004}x + 1 + \dfrac{1}{x} = 0?

−20042003-\dfrac{2004}{2003}

−1-1

20032004\dfrac{2003}{2004}

11

20042003\dfrac{2004}{2003}

Solution:

Let a=20032004.a = \dfrac{2003}{2004}. Multiplying the equation by xx gives ax2+x+1=0.ax^2 + x + 1 = 0.

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

The sum of reciprocals is 1r+1s=r+srs=−1/a1/a=−1.\dfrac{1}{r} + \dfrac{1}{s} = \dfrac{r + s}{rs} = \dfrac{-1/a}{1/a} = -1.

Thus, the correct answer is B.

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