AMC12 2015 A

AMC12 2015 A · Q18

AMC12 2015 A · Q18. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

The zeros of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $a$?

函数 $f(x)=x^2-ax+2a$ 的零点都是整数。求所有可能的 $a$ 的值之和。

(A) 7 7

(B) 8 8

(D) 17 17

(E) 18 18

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The zeros of $f$ are integers and their sum is $a$, so $a$ is an integer. If $r$ is an integer zero, then $r^2-ar+2a=0$ or $$ a=\frac{r^2}{r-2}=r+2+\frac{4}{r-2}. $$ So $\frac{4}{r-2}=a-r-2$ must be an integer, and the possible values of $r$ are $6,4,3,1,0,$ and $-2$. The possible values of $a$ are $9,8,0,$ and $-1$, all of which yield integer zeros of $f$, and their sum is $16$.

答案（C）：$f$ 的零点都是整数，并且它们的和为 $a$，所以 $a$ 是整数。若 $r$ 是一个整数零点，则 $r^2-ar+2a=0$，即 $$ a=\frac{r^2}{r-2}=r+2+\frac{4}{r-2}. $$ 因此 $\frac{4}{r-2}=a-r-2$ 必须是整数，$r$ 的可能取值为 $6,4,3,1,0,$ 和 $-2$。$a$ 的可能取值为 $9,8,0,$ 和 $-1$，它们都能使 $f$ 的零点为整数，并且这些零点的和为 $16$。

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