初三数学~~~mx^2-(2m+1)x+m+3=0

(1)Δ=(2m+1)^2-4m*(m+3)
=4m^2+4m+1-4m^2-12m
=-8m+1>0
所以 m<-1/8
(2)把x=-1代入方程，得，m+2m+1+m+3=0 m=-1
方程变为 -x^2+x+2=0 x1+x2=1 x1*x2=-2
(x+1)(x-2)=0
所以另一个根x=2
（x1-根号3）（x2-根号3）=x1*x2-根号3*(x1+x2)+3
=4+2*根号3

1）当m>0 , (2m+1)^2-4m(m+3)>0
<=> -8m+1>0
<=>m<1/8
<=> 0<m<1/8
当m<0 , (2m+1)^2-4m(m+3)<0
<=> -8m+1<0
<=>m>1/8
<=> 无解
2）x=-1解得m=-1
-x^2+x+2=0
解得x2=2
x1-根号3 = -1-根号3
x2-根号3 = 2-根号3