解:(1)依题意,得:△≥0即[-2(k-1)]2-4k2≥0,解得 (2)解法一:依题意,得:x1+ x2 = 2(k-1),x1x2 = k2· 以下分两种情况讨论: ①当x1+x2≥0时,则有x1+ x2 = x1x2-1,即2 (k- 1) =k2 - 1 解得:k1= k2 =1 ∴ k1=k2 =1不合题意,舍去。 ②x1+x2<0时,则有x1+ x2 =-(x1x2-1),即 2 (k - 1 ) = - (k2 - 1 ) 解得:kl = 1,k2 =-3 ,∴k=-3. 综合①、②可知 k= -3. 解法二:依题意可知 x1+x2 =2(k-1). 由(1)可知∴ 2(k-1)<0,即x1+x2<0∴ - 2 ( k - 1 ) = k2 - 1 解得:k1= 1,k2 =-3 ,∴k=-3. |