It is known that the function f (x) = x (x ~ 2 + 2aX + 2A + 8) has three zeros x1, X2 and X3

It is known that the function f (x) = x (x ~ 2 + 2aX + 2A + 8) has three zeros x1, X2 and X3

It seems that you are not all the same. First, to find a, you only need to satisfy that there are three zeros. Therefore, you only need to find out the Δ > 0 in the brackets of F (x) expression. The solution is a > 4, a < - 2. Second, to find out the values of zeros, that is, to find out x1, X2, X3 with a when a

It is known that the function f (x) = X2 - (a2-2a-1) x-a-2 is an increasing function on [1, + ∞); (1) find the value range of real number a; (2) compare f (1) with 2F（
2 is to compare the size of F (1) and 2F (1)

(1) Because the function f (x) = X2 - (a2-2a-1) x-a-2 is an increasing function on [1, + ∞), the symmetry axis X of the quadratic function