HMMT 二月 2006 · CALC 赛 · 第 9 题

HMMT February 2006 — CALC Round — Problem 9

Compute the sum of all real numbers x such that 6 5 4 3 2 2 x − 3 x + 3 x + x − 3 x + 3 x − 1 = 0 .

解析

Compute the sum of all real numbers x such that 6 5 4 3 2 2 x − 3 x + 3 x + x − 3 x + 3 x − 1 = 0 1 Answer: − 2 Solution: The carefully worded problem statement suggests that repeated roots might be involved (not to be double counted), as well as complex roots (not to be 6 5 4 3 2 counted). Let P ( x ) = 2 x − 3 x + 3 x + x − 3 x + 3 x − 1. Now, a is a double root of ′ the polynomial P ( x ) if and only if P ( a ) = P ( a ) = 0. Hence, we consider the system 6 5 3 3 2 P ( a ) = 2 a − 3 a + 3 a + a − 3 a + 3 a − 1 = 0 ′ 5 4 3 2 P ( a ) = 12 a − 15 a + 12 a + 3 a − 6 a + 3 = 0 4 3 2 = ⇒ 3 a + 8 a − 15 a + 18 a − 7 = 0 3 2 37 a − 57 a + 57 a − 20 = 0 2 a − a + 1 = 0 3 We have used polynomial long division to deduce that any double root must be a root 2 2 2 2 of a − a + 1! With this information, we can see that P ( x ) = ( x − x + 1) (2 x + x − 1). The real roots are easily computed via the quadratic formula, leading to an answer of 1 − . In fact the repeated roots were complex. 2