方程有两根,二次项系数≠0,方程判别式△≥0,两根均为负,两根之和0.
k-2≠0 k≠2
△≥0 [-(3k+6)]²-4(k-2)(6k)≥0
5k²-28k-12≤0
(k-6)(5k+2)≤0
-2/5≤k≤6
设两根分别为x1,x2,由韦达定理得
x1+x2=(3k+6)/(k-2)
x1x2=6k/(k-2)
(3k+6)/(k-2)