Given the function f (x) = x2 + (2a-1) x-alnx, G (x) = - 4 / x-alnx, (a ∈ R) The function f (x) = x2 + (2a-1) x-alnx, G (x) = - 4 / x-alnx (a ∈ R) is known (1) When a ＜ 0, find the minimum of F (x); (2) If the image of the function y = f (x) and y = g (x) has two different intersections m, N on X ∈ [1,3], the value range of a is obtained

Given the function f (x) = x2 + (2a-1) x-alnx, G (x) = - 4 / x-alnx, (a ∈ R) The function f (x) = x2 + (2a-1) x-alnx, G (x) = - 4 / x-alnx (a ∈ R) is known (1) When a ＜ 0, find the minimum of F (x); (2) If the image of the function y = f (x) and y = g (x) has two different intersections m, N on X ∈ [1,3], the value range of a is obtained

X2 + (2a-1) x-alnx) = - 4 / x-alnxx ^ 2 + (2a-1) x = - 4 / XX ^ 3 + (2a-1) x ^ 2 + 4 = 0 has two non real roots in X ∈ [1,3]. Let y = x ^ 3 + (2a-1) x ^ 2 + 4, in X ∈ [1,3], it has two different intersections with the X axis. So it must take the extreme value in X ∈ [1,3], y '= 3x ^ 2 + (4a-2) x = 0x = 0 or x = (2-4a)