AMC12 2002 A

AMC12 2002 A · Q13

AMC12 2002 A · Q13. It mainly tests Quadratic equations, Exponents & radicals.

Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1$. What is $a+b$?

两个不同的正数 $a$ 和 $b$ 各自与其倒数之差均为 $1$。求 $a+b$。

(A) 1 1

(B) 2 2

(D) $\sqrt{6}$ $\sqrt{6}$

(E) 3 3

Answer

Correct choice: (C)

正确答案：（C）

Solution

Each of the numbers $a$ and $b$ is a solution to $\left| x - \frac 1x \right| = 1$. Hence it is either a solution to $x - \frac 1x = 1$, or to $\frac 1x - x = 1$. Then it must be a solution either to $x^2 - x - 1 = 0$, or to $x^2 + x - 1 = 0$. There are in total four such values of $x$, namely $\frac{\pm 1 \pm \sqrt 5}2$. Out of these, two are positive: $\frac{-1+\sqrt 5}2$ and $\frac{1+\sqrt 5}2$. We can easily check that both of them indeed have the required property, and their sum is $\frac{-1+\sqrt 5}2 + \frac{1+\sqrt 5}2 = \boxed{(C) \sqrt 5}$.

数 $a$ 与 $b$ 都满足 $\left| x - \frac 1x \right| = 1$。因此它要么满足 $x - \frac 1x = 1$，要么满足 $\frac 1x - x = 1$。于是它必为 $x^2 - x - 1 = 0$ 或 $x^2 + x - 1 = 0$ 的解。这样的 $x$ 一共有四个，分别为 $\frac{\pm 1 \pm \sqrt 5}2$。其中有两个为正：$\frac{-1+\sqrt 5}2$ 与 $\frac{1+\sqrt 5}2$。容易验证它们都满足条件，且它们的和为 $\frac{-1+\sqrt 5}2 + \frac{1+\sqrt 5}2 = \boxed{(C) \sqrt 5}$。

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