Let set a = {- 1,1} set B = {x ^ 2-2ax + B = 0} if B ≠ empty set B is contained in a, find the value of a and B

Let set a = {- 1,1} set B = {x ^ 2-2ax + B = 0} if B ≠ empty set B is contained in a, find the value of a and B

B is not empty, which means that x ^ 2-2ax + B = 0 has a solution, B is contained in a, and the value of a is the solution of B. (1) B has two elements, substituting - 1,1 into x ^ 2-2ax + B = 0 to get (- 1) ^ 2-2a * (- 1) + B = 0 (1) ^ 2-2a * (1) + B = 0, a = 0, B = - 1 (2) B has only one element, 4 * a ^ 2-4 * b = 0, that is, B = a ^ 2 substituting 1 into 1 to get 1-2a + a ^ 2 = 0, that is, a = 1, B = 1