2014 AMC 10B 第 20 题

Question

有多少个整数

x

使

x^4-51x^2+50

为负？

For how many integers

x

is the number

x^4-51x^2+50

negative?

8

10

12

14

16

把表达式看作关于

x^2

的乘积并因式分解。

Factor the expression as a product in

x^2

找出两个因子异号的范围。

Find where the two factors have opposite signs

首先注意到

x^4-51x^2+50

= (x^2-50)(x^2-1)\text{。}

乘积为负，当且仅当两个因子异号。由于

x^2-50<x^2-1

，必须有

x^2-50<0<x^2-1

。所以

1<x^2<50

，即

2\le |x|\le7

。正整数解有

6

个，负整数解也有

6

个，共有

12

个。

所以正确答案是 C。

First, note that

x^4-51x^2+50

= (x^2-50)(x^2-1).

The product is negative exactly when its two factors have opposite signs. Since

x^2-50<x^2-1

, this requires

x^2-50<0<x^2-1

. Thus

1<x^2<50

, or

2\le |x|\le7

. There are

6

positive and

6

negative integer solutions, for a total of

12

.

Thus, the correct answer is C .

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