AMC10 2006 B

AMC10 2006 B · Q14

AMC10 2006 B · Q14. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

Let $a$ and $b$ be the roots of the equation $x^2 - mx + 2 = 0$. Suppose that $a + (1/b)$ and $b + (1/a)$ are the roots of the equation $x^2 - px + q = 0$. What is $q$?

设方程 $x^2 - mx + 2 = 0$ 的根为 $a$ 和 $b$。假设 $a + (1/b)$ 和 $b + (1/a)$ 是方程 $x^2 - px + q = 0$ 的根。求 $q$。

(A) $\frac{5}{2}$ $\frac{5}{2}$

(B) $\frac{7}{2}$ $\frac{7}{2}$

(D) $\frac{9}{2}$ $\frac{9}{2}$

(E) 8 8

Answer

Correct choice: (D)

正确答案：（D）

Solution

Since $a$ and $b$ are roots of $x^2 - mx + 2 = 0$, we have $ab = 2$. In a similar manner, the constant term of $x^2 - px + q$ is the product of $a + (1/b)$ and $b + (1/a)$, so $$q = \left(a + \frac{1}{b}\right) \left(b + \frac{1}{a}\right) = ab + 1 + 1 + \frac{1}{ab} = 2 + 2 + \frac{1}{2} = \frac{9}{2}.$$

因为 $a$ 和 $b$ 是 $x^2 - mx + 2 = 0$ 的根，故 $ab = 2$。类似地，$x^2 - px + q$ 的常数项是 $a + (1/b)$ 和 $b + (1/a)$ 的乘积，因此 $$q = \left(a + \frac{1}{b}\right) \left(b + \frac{1}{a}\right) = ab + 1 + 1 + \frac{1}{ab} = 2 + 2 + \frac{1}{2} = \frac{9}{2}。$$

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