Find all real solutions \( \sqrt{x-5}=x-7 \)

To solve the equation \( \sqrt{x-5} = x-7 \), we first note that the square root function is defined only for values where \( x - 5 \geq 0 \). This means that \( x \) must be greater than or equal to 5. Next, we square both sides of the equation to eliminate the square root: \[ x - 5 = (x - 7)^2 \] Expanding the right-hand side gives: \[ x - 5 = x^2 - 14x + 49 \] Rearranging the equation to set it to zero, we have: \[ 0 = x^2 - 15x + 54 \] Now, we can factor or use the quadratic formula. The equation factors as: \[ 0 = (x - 6)(x - 9) \] Setting each factor to zero gives us the potential solutions: \[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \] \[ x - 9 = 0 \quad \Rightarrow \quad x = 9 \] Lastly, we need to check these solutions in the original equation to ensure they are valid. For \( x = 6 \): \[ \sqrt{6 - 5} = \sqrt{1} = 1 \quad \text{and} \quad 6 - 7 = -1 \quad \text{(not valid)} \] For \( x = 9 \): \[ \sqrt{9 - 5} = \sqrt{4} = 2 \quad \text{and} \quad 9 - 7 = 2 \quad \text{(valid)} \] Thus, the only real solution to the equation \( \sqrt{x-5} = x-7 \) is \( x = 9 \).