Given that there is only one element in the set a = {x | x ∈ R | (A2-1) x2 + (a + 1) x + 1 = 0}, the value of a is obtained

Given that there is only one element in the set a = {x | x ∈ R | (A2-1) x2 + (a + 1) x + 1 = 0}, the value of a is obtained

∵ there is only one element in the set a = {x | x ∈ R | (A2-1) x2 + (a + 1) x + 1 = 0} and there is only one real number root in the equation (A2-1) x2 + (a + 1) x + 1 = 0; ∵ ① when A2-1 = 0, a + 1 ≠ 0, a = 1; ② when A2-1 ≠ 0, (a + 1) 2-4 × (A2-1) = 0, a = - 1 (rounding off) or a = 53; ∵ a = 1 or 53

If a is a root of the equation x & # 178; + 4x-1 = 0, then the value of 2A & # 178; + 8a-5 is

X & # 178; + 4x = 1, then 2A & # 178; + 8A = 2 (X & # 178; + 4x) so 2A & # 178; + 8A = 2, so 2A & # 178; + 8a-5 = - 3

Let a = (Y / y = xsquare - 4x + 6, X belongs to R, y belongs to n), B = (Y / y = - xsquare - 2x + 18, X belongs to R, y belongs to n), the intersection of a is represented by enumeration

x^2-4x+6=(x-2)^2+2>=2
-x^2-2x+18=-(x+1)^2+19=

Set a = {y | y = x ^ 2-4x + 3, X ∈ r}, B = {y | y = - x ^ 2-2x + 2, X ∈ r} then a ∩ B=__
Set a = {y | y = x ^ 2; - 4x + 3, X ∈ r}, B = {y | y = - x ^ 2; - 2x + 2, X ∈ r} then a ∩ B=
The answer is more specific. Also, master, I would like to ask if there is any connection between X in these two questions?
Is it irrelevant? Can two XS take different values at the same time?

A
y=(x-2)²-1>=-1
So a is a set of real numbers greater than or equal to - 1
B
y=-(x+1)²+3