①b？ -4ac= 12-8a？ ≦0, that is, a≤-√6/2 or a≦√6/2. There is only one intersection point between the function image and the horizontal axis at most, that is, the function value is constant ≧0 in the function definition domain and the solution set is (a, positive infinity).

② 12-8a？ > 0, i.e. -√ 6/2 < a < √ 6/2. This function has two intersections with the horizontal axis. The rightmost intersection is [2a+√( 12-8a? )]/6。 When [2a+√( 12-8a? )]/6 & gt； A (reflected on the coordinate axis, that is, the intersection point is at the right end of A), that is, -√ 2/2 < a < √ 2/2, and the solution set should be ([2a+√( 12-8a? ) ]/6, positive infinity). When [2a+√( 12-8a? ) ]/6≤a, and the solution set should be (a, positive infinity).