HMMT 二月 2015 · 冲刺赛 · 第 25 题

HMMT February 2015 — Guts Round — Problem 25

[ 17 ] Let r , . . . , r be the distinct real zeroes of the equation 1 n 8 4 3 2 x − 14 x − 8 x − x + 1 = 0 . 2 2 Evaluate r + · · · + r . 1 n √ √ √

解析

[ 17 ] Let r , . . . , r be the distinct real zeroes of the equation 1 n 8 4 3 2 x − 14 x − 8 x − x + 1 = 0 . 2 2 Evaluate r + · · · + r . n 1 Answer: 8 Observe that 8 4 3 2 8 4 4 3 2 x − 14 x − 8 x − x + 1 = ( x + 2 x + 1) − (16 x + 8 x + x ) 4 2 4 2 = ( x + 4 x + x + 1)( x − 4 x − x + 1) . 15 x 4 2 4 2 2 The polynomial x + 4 x + x + 1 = x + x + ( + 1) has no real roots. On the other hand, 4 2 4 2 let P ( x ) = x − 4 x − x + 1. Observe that P ( −∞ ) = + ∞ > 0, P ( − 1) = − 1 < 0, P (0) = 1 > 0, P (1) = − 3 < 0, P (+ ∞ ) = + ∞ > 0, so by the intermediate value theorem, P ( x ) = 0 has four distinct real roots, which are precisely the real roots of the original degree 8 equation. By Vieta’s formula on P ( x ), ∑ 2 2 2 2 2 r + r + r + r = ( r + r + r + r ) − 2 · ( r r ) 1 2 3 4 i j 1 2 3 4 i<j 2 = 0 − 2( − 4) = 8 . Guts √ √ √