3x²-2(3k+1)x+3k²-1=0
方程有实根,判别式△≥0
[-2(3k+1)]²-4×3×(3k²-1)≥0
24k+16≥0
k≥-2/3
设两根分别为x1,x2,由韦达定理得
x1+x2=2(3k+1)/3
x1x2=(3k²-1)/3
两根均为负,则x1+x2<0,x1x2>0
2(3k+1)/3<03k+1<0k<-1/3
(3k²-1)/3>03k²-1>0k>√3/3或k<-√3/3
综上,得-2/3≤k<-√3/3,k的取值范围为[-2/3,-√3/3].