AMC10 2003 A

AMC10 2003 A · Q18

AMC10 2003 A · Q18. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

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

方程 $\frac{2003}{2004x + 1} + \frac{1}{x} = 0$ 的根的倒数之和是多少？

(A) $-\frac{2004}{2003}$ $-\frac{2004}{2003}$

(B) $-1$ $-1$

(D) 1 1

(E) $\frac{2004}{2003}$ $\frac{2004}{2003}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Let $a=2003/2004$. The given equation is equivalent to $ax^2+x+1=0$. If the roots of this equation are denoted $r$ and $s$, then $rs=\frac{1}{a}$ and $r+s=-\frac{1}{a}$, so $\frac{1}{r}+\frac{1}{s}=\frac{r+s}{rs}=-1$.

（B）令 $a=2003/2004$。所给方程等价于 $ax^2+x+1=0$。若此方程的根记为 $r$ 和 $s$，则 $rs=\frac{1}{a}$ 且 $r+s=-\frac{1}{a}$，所以 $\frac{1}{r}+\frac{1}{s}=\frac{r+s}{rs}=-1$。

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