Given a = {x ∈ R | X & # 178; - 2x-8 = 0}, B = {x ∈ R | X & # 178; + ax + A & # 178; - 12 = 0}, a ∩ B = B, find the value set of A

Given a = {x ∈ R | X & # 178; - 2x-8 = 0}, B = {x ∈ R | X & # 178; + ax + A & # 178; - 12 = 0}, a ∩ B = B, find the value set of A

The solution of X & # 178; - 2x-8 = 0 is x = - 2 or x = 4
That is, a = {- 2,4}
A∩B=B
So B = null or B = {- 2,4} or {- 2} or {4}
When B is empty
a²-4(a²-12)＜0
The solution is a ＞ 4 or a ＜ - 4
When B is a single element set
A & # 178; - 4 (A & # 178; - 12) = 0, that is, a = ± 4
When a = 4, the equation is X & # 178; + 4x + 4 = 0, and the solution of x = - 2 satisfies B = {- 2}
When a = - 4, the equation is X & # 178; - 4x + 4 = 0, and the solution x = 2 is not satisfied
When B = {- 2,4}
We can get - a = - 2 + 4, that is a = - 2
A & # 178; - 12 = - 2 * 4, that is, a = ± 2
That is, when a = - 2, the condition can be satisfied at the same time
In conclusion, the value range of a is a ≥ 4 or a < - 4 or a = - 2

Set a = {x | X & # 178; - 2x-8 = 0}, B = {x | X & # 178; + ax + A & # 178; - 12 = 0}, if B ∪ a ≠ a, find the range of real number a

Using the method of complement to solve the problem
A = {- 2,4}, if B ∪ a = a, then B is a subset of a, divided into the following three cases
(1) B is an empty set, △ = A & # 178; - 4 * (A & # 178; - 12) = - 3 * A & # 178; + 484 or a

① X & # 178; - 6x-3 = 0, ② (x + 1) & # 178; = (2x-1) & # 178; steps and solutions are needed

The first problem X & # 178; - 6x-3 = 0 X & # 178; - 6x + 9 = 12 (x-3) & # 178; = 12 x-3 = ± 2 √ 3. The second problem (x + 1) & # 178; = (2x-1) & # 178; (x + 1) & # 178; - (2x-1) & # 178; = 0 (x + 1 + 2x-1) (x + 1-2x + 1) = 0 by (...)

(2x-1)²-x²=0 x²-6x+9=(5-2x)²

(2x-1)²-x²=0（2x-1+x）（2x-1-x）=0(3x-1)(x-1)=0x1=1/3 x2=1x²-6x+9=(5-2x)² (x-3)²-(5-2x)²=0(x-3+5-2x)(x-3-5+2x)=0(2-x)(3x-8)=0x1=2 x2=8/3