1967 AMC 12 第 35 题

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35.

方程 64x3144x2+92x15=064x^3-144x^2+92x-15=0 的各根成等差数列。最大根与最小根之差为:

The roots of 64x3144x2+92x15=064x^3-144x^2+92x-15=0 are in arithmetic progression. The difference between the largest and smallest roots is:

22

11

12\dfrac12

38\dfrac38

14\dfrac14

答案:B
知识点:多项式等差数列韦达定理
难度评级:1750
小提示:

将三个根写成 tdt-dttt+dt+d

Write the roots as td,t-d, t,t, and t+dt+d

大提示:

先利用根的和求出 tt,再利用根的积求出 d2d^2

Use their sum to find t,t, then use their product to find d2d^2

解答:

设三个根为 tdt-dttt+dt+d。它们的和为 3t=14464=943t=\frac{144}{64}=\frac{9}{4},所以 t=34t=\frac{3}{4}。它们的积为 t(t2d2)=1564 t(t^2-d^2)=\frac{15}{64}\text{。}代入 t=34t=\frac{3}{4}d2=14d^2=\frac{1}{4},所以最大根与最小根之差为 2d=12\lvert d\rvert=1

因此,正确答案是 B

Let the roots be td,t-d, t,t, and t+d.t+d. Their sum is 3t=14464=94,3t=\frac{144}{64}=\frac{9}{4}, so t=34.t=\frac{3}{4}. Their product is t(t2d2)=1564. t(t^2-d^2)=\frac{15}{64}. Substituting t=34t=\frac{3}{4} gives d2=14,d^2=\frac{1}{4}, so the difference between the extreme roots is 2d=1.2\lvert d\rvert=1.

Therefore, the correct answer is B.

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