Question

已知x1,x2是方程x²+3x+1的两个根,求x1的三次方+8x2+20=0的值？

Answer 1

X1、X2是方程X^2+3X+1=0的两实数根
韦达定理得:
X1+X2=-3
X1X2=1
X1^2+3X1+1=0
x1^2=-(3x1+1)
x1^3+8x2+20
=-x1*(3x1+1)+8x2+20
=-3x1^2-x1+8x2+20
=-3*(x1^2)-x1+8x2+20
=3*(3x1+1)-x1+8x2+20
=8(x1+x2)+23
=-24+23
=-1,8,
x1是方程的根，x1满足方程：x1²+3x1+1=0
x1²=-3x1-1 x1²+3x=-1
再由韦达定理，得
x1+x2=-3
x2=3-x1
x1³+8x2+20
=x1³+8(3-x1)+20
=x1³-8x1+44
=x1(x1²-8)+44
=x1(-3x1-1-8)+44
=-3(x1²+3x1)+44
=(-3)(-1)+44
=3+44
=47,2,x=x1代入
x1²=-3x1-1
x1³=x1*x1²
=x1(-3x1-1)
=-3x1²-x1
=-3(-3x1-1)-x1
=8x1+3
韦达定理
x1+x2=-3
所以原式=8x1+3+8x2+20
=8(x1+x2)+23
=8×(-3)+23
=-1,2,x=x1代入
x1²=-3x1-1
x1³=x1*x1²
=x1(-3x1-1)
=-3x1²-x1
=-3(-3x1-1)-x1
=8x1+3
韦达定理
x1+x2=-3
把x1²=8x1+3代入原式
原式=8x1+3+8x2+20
=8(x1+x2)+23
=8×(-3)+23
=-1,0,