若X1,X2是方程X^2+2X-2006=0的两根,求下列各式的值:(1)X1^2+X2^2 (2)(X1-5)(X2-5) (3)X1^3+X2^3 (4)|X1-X2|越详细越好,时间紧迫

问题描述:

若X1,X2是方程X^2+2X-2006=0的两根,求下列各式的值:
(1)X1^2+X2^2
(2)(X1-5)(X2-5)
(3)X1^3+X2^3
(4)|X1-X2|
越详细越好,时间紧迫

由韦达定理
x1+x2=-2,x1*x2=-2006
x1^2+x2^2=(x1+x2)^2-2x1*x2=4+2*2006=4016
(x1-5)(x2-5)
=x1*x2-5x1-5x2+25
=x1*x2-5(x1+x2)+25
=-2006+5*2+25
=-1971
x1^3+x2^3
=(x1+x2)(x1^2-x1x2+x2^2)
=-2*[4016-(-2006)]
=-12044
|x1-x2|
=√(x1-x2)^2
=√[(x1+x2)^2-4x1x2]
=√(4+4*2006)
=√8028
=6√223