Solve for all possible values of x . \[ \sqrt{3-x}=x-3 \]

To solve the equation \( \sqrt{3-x} = x - 3 \), we first square both sides to eliminate the square root: \[ 3 - x = (x - 3)^2 \] Expanding the right side gives: \[ 3 - x = x^2 - 6x + 9 \] Rearranging all terms to one side, we have: \[ 0 = x^2 - 6x + 9 + x - 3 \] This simplifies to: \[ 0 = x^2 - 5x + 6 \] Factoring the quadratic, we find: \[ 0 = (x - 2)(x - 3) \] Setting each factor to zero gives the possible solutions: \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] Next, we need to verify if these solutions satisfy the original equation. For \( x = 2 \): \[ \sqrt{3 - 2} = 2 - 3 \quad \Rightarrow \quad \sqrt{1} = -1 \quad \text{(not valid)} \] For \( x = 3 \): \[ \sqrt{3 - 3} = 3 - 3 \quad \Rightarrow \quad 0 = 0 \quad \text{(valid)} \] Thus, the only solution is: \[ \boxed{3} \]