Given the set a = {χ|χ & # 178; + 3 χ - 4 = 0}, B = {χ & # 178; + (a + 1) χ - A-2 = 0}, if a ∪ B = a, find the value of real number a

Given the set a = {χ|χ & # 178; + 3 χ - 4 = 0}, B = {χ & # 178; + (a + 1) χ - A-2 = 0}, if a ∪ B = a, find the value of real number a

x^2+3x-4=0
(x+4)(x-1)=0
A={-4,1}
x^2+(a+1)x-a-2=0
(x+a+2)(x-1)=0
X = - A-2 or x = 1
A ∪ B = a, then B is a subset of A
Because x ^ 2 + (a + 1) x-a-2 = 0 has a solution x = 1,
When B = {1},
x=-a-2=1
a=-3
When B = {- 4,1},
x=-a-2=-4
a=2