A set a = {x | x2 + (2a-3) x-3a = 0, X ∈ r}, B = {x | x2 + (A-3) x + a2-3a = 0, X ∈ r} satisfies a ≠ B, and a ∩ B ≠ & # 8709;, and a ∪ B is represented by an example

A set a = {x | x2 + (2a-3) x-3a = 0, X ∈ r}, B = {x | x2 + (A-3) x + a2-3a = 0, X ∈ r} satisfies a ≠ B, and a ∩ B ≠ & # 8709;, and a ∪ B is represented by an example

solution
Because the set a = {x | x2 + (2a-3) x-3a = 0, X ∈ r}, B = {x | x2 + (A-3) x + a2-3a = 0, X ∈ r} satisfies a ≠ B, and a ∩ B ≠ & # 8709;
Therefore, it is shown that the equations x 2 + (2a-3) x-3a = 0 and x 2 + (A-3) x + a 2-3a = 0 have at least one root of the same real number root
There are three situations:
(1) Both equations have two different real roots, but only one of them is the same;
(2) Equation a has two different real roots, equation B has two same real roots, and only one is the same;
(3) Equation B has two different real roots, equation a has two same real roots, and only one is the same
Because equation a x2 + (2a-3) x-3a = 0 (- 3a) = 4A & # 178; + 9 > 0, equation a has two different real roots
Because B equation x2 + (A-3) x + a2-3a = 0 △ = (A-3) &# 178; - 4 * (a2-3a) = - 3 (a + 1) (A-3) ≥ 0, the solution is - 1 ≤ a ≤ 3
It can be seen that there are only the above two cases (1) (2) and a ∈ [- 1,3]
The same root can be found by simultaneous equations
By solving the equations x2 + (2a-3) x-3a = 0, X2 + (A-3) x + a2-3a = 0, it is shown that
A = 0 or a = 2, obviously a ∈ [- 1,3]
When a = 0, a = 0 is brought into the solution of equations a and B, and a = {0,3} B = {0,3}
Obviously, a = B does not meet the requirement of a ≠ B, so a cannot be equal to 0
When a = 2, a = 2 is introduced into the solutions of equations a and B, and a = {2, - 3} B = {- 1,2}
Obviously, a ≠ B, a ∩ B = {2} ≠ &;, meet the requirements of the topic
So a ∪ B = {- 3, - 1,2}