AMC10 2002 B

AMC10 2002 B · Q10

AMC10 2002 B · Q10. 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 $x^2 + ax + b = 0$ 的解是 $a$ 和 $b$。则对 $(a, b)$ 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

(C) The given conditions imply that \[ x^2+ax+b=(x-a)(x-b)=x^2-(a+b)x+ab, \] so \[ a+b=-a \quad \text{and} \quad ab=b. \] Since $b\ne 0$, the second equation implies that $a=1$. The first equation gives $b=-2$, so $(a,b)=(1,-2)$.

（C）给定条件推出 \[ x^2+ax+b=(x-a)(x-b)=x^2-(a+b)x+ab, \] 因此 \[ a+b=-a \quad \text{且} \quad ab=b. \] 由于 $b\ne 0$，第二个方程推出 $a=1$。第一个方程得到 $b=-2$，所以 $(a,b)=(1,-2)$。

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