HMMT 十一月 2014 · 冲刺赛 · 第 28 题
HMMT November 2014 — Guts Round — Problem 28
题目详情
英文原题
- [ 15 ] Let x be a complex number such that x + x is a root of the polynomial p ( t ) = t + t − 2 t − 1.
7 − 7
Find all possible values of x + x .
解析
英文解析
- [ 15 ] Let x be a complex number such that x + x is a root of the polynomial p ( t ) = t + t − 2 t − 1.
7 − 7
Find all possible values of x + x .
− 1
Answer: 2 Since x + x is a root,
( ) ( ) ( )
3 2 − 1 − 1 − 1
0 = x + x + x + x − 2 x + x − 1
3 − 3 − 1 2 − 2 − 1 = x + x + 3 x + 3 x + x + 2 + x − 2 x − 2 x − 1
3 − 3 2 − 2 − 1 = x + x + x + x + x + x + 1
( ) − 3 2 6 = x 1 + x + x + · · · + x .
Since x 6 = 0, the above equality holds only if x is a primitive seventh root of unity, i.e. x = 1 and 7
7 − 7
x 6 = 1. Therefore, the only possible value of x + x is 1 + 1 = 2.