Question

已知X1.X2是方程:X的平方+3X+1=0的两个实数根、则X1的三次方+8*X2+20=?

Answer 1

X1.X2是方程:X的平方+3X+1=0的两个实数根
则：
X1²+3X1+1=0
X1²=-3X1-1
由韦达定理得：
X1+X2=-3

X1的三次方+8*X2+20
=X1*X1²+8X2+20
=X1(-3X1-1)+8X2+20
=-3X1²-X1+8X2+20
=-3(-3X1-1)-X1+8X2+20
=9X1+3-X1+8X2+20
=8(X1+X2)+23
=8×(-3)+23
=-24+23
=-1

Answer 2

x^2+3x+1=0
x1+x2=-3
x1x2=1

x1^3+8x2+20
=|x1^3+x2^3|+20
=|(x1+x2)[(x1+x2)^2-3x1x2]|+20
=|-3*(-3)^2-3|+20
=50

Answer 3

X的平方+3X+1=0的两个实数根
x=(-3±根号5）/2

X1的三次方+8*X2+20
=【=(-3+根号5）/2】的立方+8*=(-3-根号5）/2+20

X1的三次方+8*X2+20
=【=(-3-根号5）/2】的立方+8*=(-3+根号5）/2+20

这里用上（a+b）3 （a-b）3 公式

Answer 4

41