(1)由题意可知:△=[-(2k-3)]2-4(k2+1)>0,
即-12k+5>0
∴k<
5
12.
(2)∵
x1+x2=2k?3<0
x1x2=k2+1>0,
∴x1<0,x2<0.
(3)依题意,不妨设A(x1,0),B(x2,0).
∴OA+OB=|x1|+|x2|=-(x1+x2)=-(2k-3),
OA?OB=|-x1||x2|=x1x2=k2+1,
∵OA+OB=2OA?OB-3,
∴-(2k-3)=2(k2+1)-3,
解得k1=1,k2=-2.
∵k<
5
12,
∴k=-2.