PUMaC 2011 · 代数（B 组） · 第 2 题

PUMaC 2011 — Algebra (Division B) — Problem 2

解析

First Solution : We write x − 2 x + 5 = ( x − a )( x − b ) = x − ( a + b ) x + ab , so ab = 5, a + b = 2 (or we could apply Vieta’s formulas). From these elementary symmetric polynomials, we can find all of the power sums of the roots: 2 2 2 a + b = ( a + b ) − 2 ab = 4 − 10 = − 6 4 4 2 2 2 2 2 a + b = ( a + b ) − 2 a b = 36 − 50 = − 14 8 8 4 4 2 4 4 a + b = ( a + b ) − 2 a b = 196 − 2 · 625 = − 1054 Thus, the answer is 1054 . 2 Second Solution : By the quadratic equation, the roots of x − 2 x + 5 = 0 are given by ( ) √ √ 2 ± 4 − 4 · 5 1 2 i √ √ x = = 1 ± 2 i = 5 · ± . Note that these two roots are complex conjugates 2 5 5 of each other. By De Moivre’s formula, { } √ 8 8 8 4 | a + b | = 2 < 5 cis θ = 2 · 5 cos 8 θ, √ 2 where cos θ = 1 / 5. By three applications of the double-angle formula, cos 8 θ = 2(2(2 cos θ − 2 2 98 − 527 8 8