What are the values of b which satisfy b- square root of 1-1996b=1?

b - sqrt(1 - 1996b) = 1,

Isolate the radical:
b - 1 = sqrt(1 - 1996b),
Square both sides:
b^2 - 2b + 1 = 1 - 1996b,
b^2 - 2b + 1996b = 1 - 1,
b^2 + 1994b = 0,
Factor out b:
b(b + 1994) = 0,
b = 0,
b + 1994 = 0,
b = -1994,

Solution set: b = 0, and b = -1994.

To find the values of b that satisfy the given equation b - √(1 - 1996b) = 1, we can follow these steps:

Step 1: Move the term containing the square root to the other side of the equation:
b - 1 = √(1 - 1996b)

Step 2: Square both sides of the equation to eliminate the square root:
(b - 1)^2 = (√(1 - 1996b))^2
(b - 1)^2 = 1 - 1996b

Step 3: Expand the squared term:
b^2 - 2b + 1 = 1 - 1996b

Step 4: Simplify and rearrange the equation:
b^2 + 1994b = 0

Step 5: Factor out b from the left side of the equation:
b(b + 1994) = 0

Step 6: Apply the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero:
b = 0 or b + 1994 = 0

Step 7: Solve for b in each equation:
For b = 0, there is one solution: b = 0.
For b + 1994 = 0, subtract 1994 from both sides: b = -1994.

Therefore, the values of b that satisfy the given equation are b = 0 and b = -1994.