HMMT 二月 2015 · 冲刺赛 · 第 25 题

英文解析

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
    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 distinctreal 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 = 0 − 2( − 4) = 8 .2
    Guts
    √ √
    √