It is known that the image of y = (a + 3A + 2) x + (a + 1) x + 0.25 and X-axis always have focus (a is a constant) 1. The range of value of a is obtained. 2. Let the graph of function be There are two different focal points a (x1,0) B (x2,0) between the image and the x-axis. When 1 / 1 of X1 + 1 / 2 of x2 = the square-3 of a, the value of a is obtained

It is known that the image of y = (a + 3A + 2) x + (a + 1) x + 0.25 and X-axis always have focus (a is a constant) 1. The range of value of a is obtained. 2. Let the graph of function be There are two different focal points a (x1,0) B (x2,0) between the image and the x-axis. When 1 / 1 of X1 + 1 / 2 of x2 = the square-3 of a, the value of a is obtained

1. When a = - 1, the function is y = 0.25, and there is no intersection with X axis, so it is omitted
When a = - 2, the function is y = - x + 0.25, which has an intersection with the X axis
When a ≠ - 1 and a ≠ - 2, the function is quadratic and has intersection with X axis. The discriminant = (a + 1) &# 178; - (A-1) (A-2) ≥ 0
The solution is a ≤ - 1
Therefore, the value range of a is: a < - 1
2.1/x1 + 1 / x2 = (x1 + x2) / (x1x2) = [- (a + 1) / (a + 3A + 2)] / [0.25 / (a + 3A + 2)] = the square of A-3
That is: A & # 178; + 4A + 1 = 0
The solution is: a = - 2 ± √ 3
Because a ＜ - 1, a = - 2 - √ 3