The known set a = {x | x2-4x + 2A + 6 = 0}, B = {x | x}

The known set a = {x | x2-4x + 2A + 6 = 0}, B = {x | x}

solution
The problem can be reduced to the following: 1;
There exists a real number x ＜ 0, such that (X-2) &# 178; + 2A + 2 = 0
∵x＜0
There are: (X-2) &# 178; > 4
There are 4 + 2A + 2 < 0
∴a＜-3

Given the set a = {x | y = x and y = x2 + ax + B}, is there such a real number a, B that - 1 ∈ A and 3 ∈ a hold simultaneously?
If it exists, find out the value of a and B. if it does not exist, explain the reason

X = y, substituting
x=x²+ax+b
That is X & # 178; + (A-1) x + B = 0
-If 1 ∈ A and 3 ∈ a hold at the same time, then - 1 and 3 are the two roots of the equation
1-A = - 1 + 3, a = - 1
B = - 1 * 3, B = - 3
In conclusion, a = - 1, B = - 3

Given the set a = {X / y = x and y = x ^ 2 + ax + B}, is there a real number a, B such that - 1 ∈ A and 3 ∈ a hold simultaneously?
If it exists, calculate the value of a and B; if it does not exist, explain the reason

The set a = {x | y = x and y = x & # 178; + ax + B} represents the abscissa of the solution of the equation system y = x, y = x & # 178; + ax + B, that is, X
Suppose there are such real numbers a and B such that - 1 ∈ A and 3 ∈ a hold simultaneously
Then x = - 1 and x = 3 are the two real roots of the equation x = x & # 178; + ax + B
That is, x = - 1 and x = 3 are the two real roots of X & # 178; + (A-1) x + B = 0
According to Weida's theorem, - 1 + 3 = - (A-1), - 1 * 3 = b
So a = - 1, B = - 3

Given the set a = {x I y = x and y = x ^ 2 + ax + B}, is there such a real number a, B that - 1 ∈ A and 3 ∈ a hold simultaneously? Find a, B?
Solution: a = {x I y = x and y = x ^ 2 + ax + B}, that is, a = {x I x = x ^ 2 + ax + B} = {x I x ^ 2 + (A-1) x + B = 0},
And - 1 ∈ a, 3 ∈ a, that is - 1,3 are the two roots of the quadratic equation x ^ 2 + (A-1) x + B = 0,
{- (A-1) = - 1 + 3, B = - 1 * 3, that is {a = - 1, B = - 3
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And - 1 ∈ a, 3 ∈ a, that is - 1,3 are two roots of the quadratic equation x ^ 2 + (A-1) x + B = 0
Here - 1 ∈ a, 3 ∈ A is x = - 1, x = 3
∴{-（a-1）=-1+3,b=-1*3.
How is this part calculated?
Please explain to me, thanks

-1 ∈ a, 3 ∈ A is X1 = - 1, X2 = 3
According to Weida's theorem, X1 + x2 = - B / A, x1x2 = C / A