x1,x2是方程x²+3x+1=0的根,由韦达定理,得
x1+x2=-3
又两根均满足方程,有
x1²+3x1+1=0
x1²=-3x1-1
x1²+3x=-1
x1³+8x2+20
=x1(x1²)+8x2+20
=x1(-3x1-1)+8x2+20
=-3x1²-x1+8x2+20
=-3x1²-9x1+8x1+8x2+20
=-3(x1²+3x1)+8(x1+x2)+20
=-3(-1)+8(-3)+20
=3-24+20
=-1
x1,x2是方程x²+3x+1=0的根,由韦达定理,得
x1+x2=-3
又两根均满足方程,有
x1²+3x1+1=0
x1²=-3x1-1
x1²+3x=-1
x1³+8x2+20
=x1(x1²)+8x2+20
=x1(-3x1-1)+8x2+20
=-3x1²-x1+8x2+20
=-3x1²-9x1+8x1+8x2+20
=-3(x1²+3x1)+8(x1+x2)+20
=-3(-1)+8(-3)+20
=3-24+20
=-1