Question

已知x1、x2为方程x²+3x+1=0的两实根，则x1³+8x2+20=？

Answer 1

由题意得：x1+x2=-3
x1²+3x1+1=0
两边同悔洞时乘以x1得 两边敏前雹同时乘以3得
x1³+3x1²+x1=0 ········式一 3x1²+9x1+3=0···········式二
式二带入式一 得
x1³-8x1-3=0
∴x1³=8x1+3
∴桥帆8x1+8x2+3+20=8（x1+x2）+23=-1

Answer 2

x1,x2为慎旦方程x^2+3x+1=0的两实根
==> x1^2 + 3x1 + 1 = 0
==> x1^2 = -3x1 - 1
==> x1^3 = -3x1^2 - x

x1^3+8x2+20
= (-3x1^2 - x1) + 8x2 + 20
= -3x1^2 - x1 + 8x2 + 20
= -3 * (-3x1 -1) - x1 + 8x2 + 20
= 9x1 + 3 - x1 + 8x2 + 20
= 8x1 + 8x2 + 23
= 8 * (x1 + x2) + 23

因为 x1,x2为方程x^2+3x+1=0的两宽猛扰实根, x1 + x2 = -3, x1 * x2 = 1
8 * (x1 + x2) + 23
= 8 * (-3) + 23
= -24 + 23
= -1
所以知虚x1³+8x2+20=-1

Answer 3

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