HMMT 十一月 2012 · 冲刺赛 · 第 29 题

HMMT November 2012 — Guts Round — Problem 29

[ 15 ] Find the sum of the real values of x satisfying ( x + 1)(2 x + 1)(3 x + 1)(4 x + 1) = 16 x .

解析

[ 15 ] √ 25+5 17 Answer: − First, note that ( x + 1)(2 x + 1)(3 x + 1)(4 x + 1) = (( x + 1)(4 x + 1))((2 x + 1)(3 x + 8 2 2 2 2 2 2 2 2 4 1)) = (4 x + 5 x + 1)(6 x + 5 x + 1) = (5 x + 5 x + 1 − x )(5 x + 5 x + 1 + x ) = (5 x + 5 x + 1) − x . √ 2 2 4 2 2 Therefore, the equation is equivalent to (5 x + 5 x + 1) = 17 x , or 5 x + 5 x + 1 = ± 17 x . √ √ √ 2 2 2 If 5 x + 5 x + 1 = 17 x , then (5 − 17) x + 5 x + 1 = 0. The discriminant of this is 25 − 4(5 − 17) = √ √ 5 25+5 17 √ 5 + 4 17, so in this case, there are two real roots and they sum to − = − . 8 5 − 17 √ √ √ 2 2 2 If 5 x + 5 x + 1 = − 17 x , then (5 + 17) x + 5 x + 1 = 0. The discriminant of this is 25 − 4(5 + 17) = √ 5 − 4 17. This is less than zero, so there are no real solutions in this case. Therefore, the sum of all √ 25+5 17 real solutions to the equation is − . 8