Question

函数的零点都是整数。所有可能的 $a$ 值之和是多少？ $f(x)=x^2-ax+2a$

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

$7$

$8$

$16$

$17$

$18$

答案：C

知识点：韦达定理西蒙最爱的因式分解技巧二次方程

难度评级：1660

解答：

设整数零点为 $r$ 和 $s$ 由韦达定理， $a = r + s$ ，且 $2a = rs$ 。

因此整理并配方得到 $rs = 2(r + s),$ $rs - 2r - 2s = 0$ $rs - 2r - 2s + 4 = 4$ $(r - 2)(s - 2) = 4.$

可行的 $(r - 2, s - 2)$ 为对每一对，都有不同的 $a$ 值为它们的和为 $-1 + 8 + 9 = 16$ 。 $(1, 4), (-1, -4), (4, 1), (-4, -1),$ $(2, 2), (-2, -2).$ $a = r - 2 + s - 2 + 4.$ $-1, 0, 8, 9.$

所以正确答案是 C。

Let the zeros be $r$ and $s.$ Using Vieta's formulas, we have that $a = r + s$ and $2a = rs.$

Then we get that $rs = 2(r + s),$ which rearranges to $rs - 2r - 2s = 0$ $rs - 2r - 2s + 4 = 4$ $(r - 2)(s - 2) = 4.$

The only possible pairs $(r - 2, s - 2)$ that work are $(1, 4), (-1, -4), (4, 1), (-4, -1),$ $(2, 2), (-2, -2).$ For any of these pairs, we have that $a = r - 2 + s - 2 + 4.$ We want all the such unique values of $a.$ We get that they are $-1, 0, 8, 9.$ The sum of these values is $-1 + 8 + 9 = 16.$

Thus, C is the correct answer.

2015 AMC 10A 第 23 题

答案：C

难度评级：1660

解答：

其他年份的第 23 题