Question

设x1、x2是方程2x方-6x+3=0的两个根，分别求x1方+x2方和x1三次方+x2三次方的值

初三的知识啊~~要详细的步骤和结果~~明天交作业了~~急啊

Answer 1

解：由韦达定理得
x1+x2=3,x1*x2=3/2
则x1²+x2²=(x1+x2)²-2x1*x2=9-3=6
x1³+x2³=(x1+x2)(x1²-x1*x2+x2²)
=3[(x1+x2)²-3x1*x2]
=3*(9-9/2)
=27/2

Answer 2

x1+x2=6/2=3 x1*x2=3/2
x1方+x2方
=(x1+x2)方-2x1x2
=3*3-2*3/2
=6
x1三次方+x2三次方
=(x1+x2)(x1方+x2方-x1x2)
=3*（6-3/2)
=13.5

Answer 3

x1方+x2方+2x1x2-2x1x2=(x1+x2)方-2x1x2
x1三次方+x2三次方=(x1+x2)(x1方+x2方+2x1x2-3x1x2)=(x1+x2)((x1+x2)方-3x1x2)
x1+x2=3
x1x2=3/2