It is known that the equation x ^ 2-4x + 2a-6 = 0 about X has a negative solution, so we can find the value range of the real number a

It is known that the equation x ^ 2-4x + 2a-6 = 0 about X has a negative solution, so we can find the value range of the real number a

First, the equation must have a solution, that is, the discriminant = (- 4) &# 178; - 4 (2a-6) ≥ 0
The solution is a ≤ 5
Let x ^ 2-4x + 2a-6 = 0 be x 1, x 2
According to Weida's theorem X1 + x2 = 4 > 0
So X1 and X2 are not all negative
Then x1