Question

已知X1,X2是方程X²+3X+1=0的两实根,则X1²+8X2+20=？

Answer 1

x1、 x2是 x^2 + 3x +1=0的两根
有:x1^2+3x1+1=0
x1^2=-3x1-1
且x1,2=1/2(-3+根号5)或者1/2(-3-根号5)

x1^2+8x2+20
=-3x1-1+8x2+20
=-3x1+8x2+19
=-3*1/2(-3+根号5)+8*1/2(-3-根号5)+19
=23/2-11/2*根号5
或者
=-3*1/2(-3-根号5)+8*1/2(-3+根号5)+19
=23/2+11/2*根号5

Answer 2

x1、 x2是一元二次方程 x^2 + 3x +1=0的两根
那么有:x1+x2=-3,x1^2+3x1+1=0
x1^2=-3x1-1

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