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+8X2+20=?

Answer 1

X1,X2为方程x²+3x+1=0的两根
那么 x1²+3x1+1=0
x1²=-3x1-1
x1（-3x1-1）+8x2+20
=-3x1²-x1+8x2+20
=-3（-3x1-1）-x1+8x2+20
=8（x1+x2）+23
=8×（-3） +23
=-1

Answer 2

X1.X2是方程的两个实数根
x1²+3x1+1=0
x1³+3x1²+x1=0 两边乘以x1
x1³+3x1²+9x1+3-8x1-3=0
x1³+3(x1²+3x1+1)=8x1+3
x1³=8x1+3
原式=8x1+8x2+23
=8(x1+x2)+23
=8*(-3)+23
=-1

Answer 3

X1=（-3+更号5）/2时 原式=4×更号5+27
X1=（-3-更号5）/2时 原式=-4×更号5+27

Answer 4

X1^3+8X2+20
=x1^3+3x1^2+x1-3x1^2-x1+8x2+20
=x1(x1^2+3x1+1)-3x1^2-x1+8x2+20
=-3x1^2-x1+8x2+20
=-3(x1^2+3x1+1)+9x1+3-x1+8x2+20
=8(x1+x2)+23
=-3*8+23
=-1

Answer 5

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