\(\sqrt{1-2b} = 1 + b \\ \\ 1 - 2b = (1 + b)^2 \\ \\ 1 - 2b = 1 + 2b + b^2 \\ \\ -2b = 2b + b^2 \\ \\ 2b + 2b + b^2 = 0 \\ \\ 4b + b^2 = 0 \\ \\ b(4 + b) = 0 \\ \\ b = 0, -4 \\ \\ b = -4 \ is \ false \\ \\ b = 0\)
The final result is, C) 0 only.
We gathered the solutions of b = 0, -4.
b = -4 was not true so we had to remove it from the solution set, and we are left with b = 0 now. You can check this answer by substituting 0 in for b in the original equation. You should get 1 = 1, which is the correct answer.
