Question

已知x1,x2是方程x方+3x+1=0的两实数根，则x1立方+8x+20=

Answer 1

x1,x2是方程x方+3x+1=0的两实数根，
即x1^2+3x1+1=0
x1^2+3x1=-1
x1^2=-3x1-1

根据韦达定理得x1+x2=-3

x1^3+8x2+20
=x1*x1^2+8x2+20
=x1*(-3x1-1)+8x2+20
=-3x1^2-x1+8x2+20
=-3x1^2-9x1+8x1+8x2+20
=-3(x1^2+3x1)+8(x1+x2)+20
=-3*(-1)+8*(-3)+20
=3-24+20
=-1

Answer 2

解：因为 x1^2+3x1=-1 ，又由韦达定理有：x1+x2=-3
x1^3+8x2+20 = x1^3+3x1^2-3x1^2+8x2+20 = x1(x1^2+3x1)-3x1^2+8x2+20
= -x1-3x1^2+8x2+20 =-3x1^2-x1-8x1+8x1+8x2+20
=-3(x1^2+3x1)+8x1+8x2+20 =3+8(x1+x2)+20 = 3+8×(-3)+20
=-1

Answer 3

用求根公式解出x1,x2 再代入 有两个结果