If \( a=2 \), what is the solution set of the equation above? (A) \( \{3,6\} \) B) \( \{2\} \) C) \( \{3\} \quad \sqrt{-x-a}=x-4 \) (D) \( \{6\} \)

To solve the equation \( \sqrt{-x - a} = x - 4 \) with \( a = 2 \), we first substitute \( a \) into the equation, yielding \( \sqrt{-x - 2} = x - 4 \). Next, we square both sides to eliminate the square root, resulting in: \[ -x - 2 = (x - 4)^2 \] Expanding the right side gives: \[ -x - 2 = x^2 - 8x + 16 \] Rearranging this leads to: \[ x^2 - 7x + 18 = 0 \] To find the roots, we calculate the discriminant: \[ D = b^2 - 4ac = (-7)^2 - 4(1)(18) = 49 - 72 = -23 \] Since the discriminant is negative, the equation has no real solutions. Therefore, the solution set for this equation is empty. Now, let’s look at interesting aspects related to square roots and equations, shall we? Engaging with equations involving square roots is like playing with fire – a little tricky! These equations often impose restrictions on the possible values of \( x \). In this case, we need \( -x - 2 \geq 0 \), suggesting \( x \leq -2 \). Such insights can significantly narrow down the potential solution space before diving deeper into solving! Don't forget to always check your solutions! When squaring both sides of an equation, it's common to introduce extraneous solutions. Always back-substitute any potential answers into the original equation to ensure they hold true. This step is often where many slip up in problem-solving, leading to incorrect conclusions.