Question

已知x1和x2是方程x2+3x+1=0的两个实数根，则x13+8 x2+20=

Answer 1

x1和x2是方程x²＋3x＋1=0的根，则：x1²＋3x1＋1=0即：x1²=－3x1－1；x1＋x2=－3，x1x2=1
则：
x1³＋8x2＋20
=x1(－3x1－1)＋8x2＋20
=－3x1²－x1＋8x2＋20
=－3(－3x1－1)－x1＋8x2＋20
=8x1＋8x2＋23
=8(x1＋x2)＋23
=－24＋23
=－1

Answer 2

由题意 x1^2+3x1+1=0
x1^2=-1-3x1
原式=x1*x1^2+8x2+20
=x1(-1-3x1)+8x2+20
=-3x1^2-x1+8x2+20
=-3(-1-3x1)-x1+8x2+20
=8x1+8x2+23
=8(x1+x2)+23
=8(-3)+23
=-1

Answer 3

x1和x2是方程x²＋3x＋1=0的根，则：x1²＋3x1＋1=0即：x1²=－3x1－1；x1＋x2=－3，x1x2=1
则：
x1³＋8x2＋20
=x1(－3x1－1)＋8x2＋20
=－3x1²－x1＋8x2＋20
=－3(－3x1－1)－x1＋8x2＋20
=8x1＋8x2＋23
=8(x1＋x2)＋23
=－24＋23
=－1