If x is a solution of the equation, √(2x+1) - √(2x) = 1, (x ≥ 1/2), then √(4x² + 1) is equal to:

√(2x+1) - √(2x) = 1. (1)
→ 2x + 1 + 2x - 2√(4x² + 1) = 1
→ 4x - 2√(4x² + 1) = 0
→ 4x = 2√(4x² + 1)
→ 16x² = 4(4x² + 1)
→ 16x² = 16x² + 4
→ This equation has no solution for x. Let's solve the original equation differently.
√(2x+1) - √(2x) = 1
√(2x+1) = 1 + √(2x)
Squaring both sides:
2x + 1 = 1 + 2√(2x) + 2x
1 = 2√(2x)
1/2 = √(2x)
1/4 = 2x
x = 1/8
Now let's check if this satisfies the original equation:
√(2(1/8) + 1) - √(2(1/8)) = √(5/4) - √(1/4) = √5/2 - 1/2 ≈ 1.118 - 0.5 = 0.618 ≠ 1
There must be a mistake in the problem statement or the given solution. Let's assume the equation was meant to be:
√(2x + 1) + √(2x) = 1
√(2x+1) = 1 - √(2x)
2x + 1 = 1 - 2√(2x) + 2x
2√(2x) = 0
√(2x) = 0
x = 0
However, x ≥ 1/2, so this is not a valid solution. Let's try another approach.
Let's square the original equation:
(√(2x+1) - √(2x))² = 1²
2x + 1 - 2√(2x(2x+1)) + 2x = 1
4x - 2√(4x² + 2x) = 0
2x = √(4x² + 2x)
4x² = 4x² + 2x
2x = 0
x = 0
Again, x=0 is not in the domain x ≥ 1/2. There is likely an error in the problem statement or the provided solution.