HMMT 二月 2006 · CALC 赛 · 第 9 题

英文解析

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
   Answer: −1
   Solution: The carefully worded problem statement suggests that repeated roots 2
   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
   a − a + 1 = 02
   We have used polynomial long division to deduce that any double root must be a root 3
   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 − . In fact the repeated roots were complex.1 2