⊿=16k²-8(k+1)(3k-2)≥0
且 x1·x2=(3k-2)/2(k+1)>0
即k²+k-2≤0 (1)
(3k-2)(k+1)>0 (2)
解(1)得 -2≤k≤1
解(2)得k>2/3或 k