Question

已知x1，x2是方程x·x+3x+1=0的两个实数根·则x1·x1·x1+8x2+20=？

急~~~要具体解题过程

Answer 1

由题意 x1^2+3x1+1=0
x1^2=-1-3x1
原式=x1*x1^2+8x2+20
=x1(-1-3x1)+8x2+20
=-3x1^2-x1+8x2+20
=-3(-1-3x1)-x1+8x2+20
=8x1+8x2+23
=8(x1+x2)+23
=8(-3)+23
=-1

Answer 2

x²+3x+1=0
x=(-3±√(3²-4))／2=（-3±√5）／2
(x1)³+8x2+20=(（-3+√5）／2)³+8(（-3-√5）／2)+20=-9+4√5-12-4√5+20=-1

(x1)³+8x2+20=(（-3-√5）／2)³+8(（-3+√5）／2)+20=-9-4√5-12+4√5+20=-1

Answer 3

答案=-1
由已知得x*x+3x=-1 x*x=-3x-1
x1·x1·x1=(-3x1-1)*x1=-x1-3x1*x1 x1*x1=-3x1-1
带入可得x1*x1*x1=-x1-3(-3x1-1)=8x1+3
原式可得8x1+8x2+23=8(x1+x2)+23
把X1和X2分别代入原方程x1*x1+3x1+1=0 (1)
x2*x2+3x2+1=0 (2)
用（1）-（2）=x1*x1-x2*x2+3x1-3x2=0
能推出x1+x2=-3
所以8（x1+x2）+23=8*（-3）+23=-1