Let a = {x | x2 + 4x = 0}, B = {x | x2 + 2 (a + 1) x + A2-1 = 0}, (1) if a ∩ B = B, find the value of A. (2) if a ∪ B = B, find the value of A

Let a = {x | x2 + 4x = 0}, B = {x | x2 + 2 (a + 1) x + A2-1 = 0}, (1) if a ∩ B = B, find the value of A. (2) if a ∪ B = B, find the value of A

(1) A ∩ B = B, which means that there are all elements a in B, that is, B belongs to a or equals to a
Known by
A={0,-4}
Take 0 into B and get a = 1 or a = - 1
When a = - 1, B = {0}, then there are all elements in a in B, so it is consistent
When a = 1, B = a, then there are all elements a in B, so it is consistent
And then we take - 4 into B and get a = 1 or a = 7
When a = 7, B = {- 4, - 12} is not equal to a, and the elements in B are not all in a, so a = 7 is omitted
When a = 1, it has been proved that
So a = 1 or - 1
(2) If a ∪ B = B, that is to say, there must be all elements B in a, that is, a belongs to B or a = B
From the conclusion in (1)
Here a = 1