AMC10 2014 B

AMC10 2014 B · Q20

AMC10 2014 B · Q20. It mainly tests Quadratic equations, Vieta / quadratic relationships (basic).

For how many integers $x$ is the number $x^4 - 51x^2 + 50$ negative?

有整数$x$多少个，使得$x^4-51x^2+50$为负数？

(A) 8 8

(B) 10 10

(D) 14 14

(E) 16 16

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Note that $x^4 - 51x^2 + 50 = (x^2 - 50)(x^2 - 1)$, so the roots of the polynomial are $\pm 1$ and $\pm \sqrt{50}$. Arranged from least to greatest, these roots are approximately $-7.1$, $-1$, $1$, $7.1$. The polynomial takes negative values on the intervals $(-7.1, -1)$ and $(1, 7.1)$, which include 12 integers: $-7$, $-6$, $-5$, $-4$, $-3$, $-2$, $2$, $3$, $4$, $5$, $6$, $7$.

答案（C）：注意 $x^4 - 51x^2 + 50 = (x^2 - 50)(x^2 - 1)$，所以该多项式的根为 $\pm 1$ 和 $\pm \sqrt{50}$。按从小到大排列，这些根约为 $-7.1$、$-1$、$1$、$7.1$。该多项式在区间 $(-7.1, -1)$ 和 $(1, 7.1)$ 上取负值，这些区间包含 12 个整数：$-7$、$-6$、$-5$、$-4$、$-3$、$-2$、$2$、$3$、$4$、$5$、$6$、$7$。

Topics