So 0; 0, so the following two inequalities are obtained:

1)2x^2+2kx+k>； 0, discriminant < 0, that is, 4k 2-8k < 0, and we get 0.

2)2x^2+2kx+k<； 3x 2+6x+4, x 2+2 (3-k) x+4-k > 0 is a constant, that is, the discriminant < 0, get: 4 (3-k) 2-4 (4-k) < 0,

Solution: (5-√ 5)/2

Comprehensive 1), 2) the value range of k is:

(5-√5)/2 & lt； k & lt2