What is the solution of the equation? \( \sqrt{2 x-1}+2=x \) Enter your answer in the box. \( x=\square \)

To solve the equation \( \sqrt{2x - 1} + 2 = x \), first isolate the square root by subtracting 2 from both sides: \[ \sqrt{2x - 1} = x - 2 \] Next, square both sides to eliminate the square root: \[ 2x - 1 = (x - 2)^2 \] This expands to: \[ 2x - 1 = x^2 - 4x + 4 \] Rearranging gives us a quadratic equation: \[ x^2 - 6x + 5 = 0 \] Factoring the equation, we get: \[ (x - 5)(x - 1) = 0 \] Thus, the solutions are \( x = 5 \) and \( x = 1 \). However, we need to check if these solutions satisfy the original equation. For \( x = 5 \): \[ \sqrt{2(5) - 1} + 2 = \sqrt{10 - 1} + 2 = \sqrt{9} + 2 = 3 + 2 = 5 \quad \text{(Valid)} \] For \( x = 1 \): \[ \sqrt{2(1) - 1} + 2 = \sqrt{2 - 1} + 2 = \sqrt{1} + 2 = 1 + 2 = 3 \quad \text{(Not Valid)} \] Thus, the only valid solution is: \( x = 5 \)