AMC12 2002 B

AMC12 2002 B · Q6

AMC12 2002 B · Q6. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

Suppose that $a$ and $b$ are nonzero real numbers, and that the equation $x^2 + ax + b = 0$ has solutions $a$ and $b$. Then the pair $(a,b)$ is

假设 $a$ 和 $b$ 是非零实数，且方程 $x^2 + ax + b = 0$ 的解为 $a$ 和 $b$。则有序对 $(a,b)$ 是

(A) (-2, 1) (-2, 1)

(B) (-1, 2) (-1, 2)

(D) (2, -1) (2, -1)

(E) (4, 4) (4, 4)

Answer

Correct choice: (C)

正确答案：（C）

Solution

Since $(x-a)(x-b) = x^2 - (a+b)x + ab = x^2 + ax + b = 0$, it follows by comparing coefficients that $-a - b = a$ and that $ab = b$. Since $b$ is nonzero, $a = 1$, and $-1 - b = 1 \Longrightarrow b = -2$. Thus $(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}$.

由于 $(x-a)(x-b) = x^2 - (a+b)x + ab = x^2 + ax + b = 0$，比较系数可得 $-a - b = a$ 且 $ab = b$。因为 $b$ 非零，所以 $a = 1$，并且 $-1 - b = 1 \Longrightarrow b = -2$。因此 $(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}$。

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