已知x1x2是方程x²+3x+1=0的两个根,求x1³+8x2+20的值

已知x1x2是方程x²+3x+1=0的两个根,求x1³+8x2+20的值
已知x1x2是方程x²+3x+1=0的两个根,求x1³+8x2+20的值

已知x1x2是方程x²+3x+1=0的两个根,求x1³+8x2+20的值
x1x2是方程x²+3x+1=0的两个根
所以x1+x2=-3 (韦达定理)
x1³+8x2+20=x1(x1)^2+8x2+20
=x1(-3x1-1)+8x2+20
=-3x1^2-x1+8x2+20
=-3(-3x1-1)-x1+8x2+20
=9x1-x1+8x2+20+3
=8(x1+x2)+20+3
=-24+20+3
=-1

-1

x1+x2=-3
x1x2=1
x1^2+3x1+1=0, x1^3+3x1^2+x1=0, x1^3=-3x1^2-x1=3(3x1+1)-x1=8x1+3
x1^3+8x2+20=8x1+3+8x2+20=8(x1+x2)+23=-24+23=-1