Question

已知X1，X2为方程X^2+3X+1=0的两实根，则X1^3+8X2+20等于多少

Answer 1

解由X1，X2为方程X^2+3X+1=0的两实根
知x1^2+3x1+1=0且x1+x2=-3
由x1^2+3x1+1=0
得x1^2=-(3x1+1)
故X1^3+8X2+20
=x1x1^2+8X2+20
=x1(-3x1-1)+8x2+20
=-3x1^2-x1+8x2+20
=-3(-3x1-1)-x1+8x2+20
=9x1-x1+8x2+23
=8(x1+x2)+23
=8*(-3)+23
=-1

Answer 2

X^2+3X+1=0
x1+x2=-3,x1x2=1
x1-x2=√[(x1+x2)^2-4x1x2]=√[(-3)^2-4]=√5
设，X1^3+8X2+20＝A.................(1)
X2^3+8X1+20＝B..........................(2)
(1)+(2)得
A+B=(X1^3+X2^3)+8(x1+x2)+40
A+B=(x1+x2)^3-3x1x2(x1+x2)+8(x1+x2)+40.
A+B=(-3)^3+3*3-8*3+40=-8
A+B=-8.........................(3)
(1)-(2)得
A-B=(X1^3-X2^3)-8(x1-x2).....................(4)
A-B=(x1-x2)^3-3x1x2(x1-x2)-8(x1-x2)
A-B=(√5)^3-3√5-8√5=6√5
A-B=-6√5.....................(4)
(3)+(4)得

2A=-8-6√5
A=-4-3√5
即有，X1^3+8X2+20＝A
X1^3+8X2+20＝-4-3√5