1974 AMC 12 第 10 题

Question

使方程

2x(kx-4)-x^2+6=0

没有实根的最小整数

k

是多少？

What is the smallest integral value of

k

such that

2x(kx-4)-x^2+6=0

has no real roots?

-1

2

3

4

5

先将方程改写为标准二次方程形式

First rewrite the equation in standard quadratic form

没有实根意味着判别式为负

No real roots means the discriminant is negative

方程可写成

(2k-1)x^2-8x+6=0

。其判别式为

64-24(2k-1)=88-48k\text{。}

当且仅当

k\gt\frac{11}{6}

时，判别式为负，所以满足条件的最小整数是

2

。

因此，正确答案是 B。

The equation is

(2k-1)x^2-8x+6=0.

Its discriminant is

64-24(2k-1)=88-48k.

This is negative exactly when

k\gt\frac{11}{6},

so the least integer possible is

2.

Therefore, the correct answer is B.

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