If $\sqrt{(x-6)^2}=6-x$, then which of the following must be true?

Tricky one! I initially thought it was just x=6, but then realized the absolute value. The equation means |x-6| = 6 - x, which forces x ≤ 6. So the answer is C. But it's easy to miss the case where x < 6 because then the right side is positive and it works. Good question!

Why is x=6 not the only solution? If you square both sides, you get (x-6)^2 = (6-x)^2, which is true for any x. But the original equation requires the square root to equal 6-x, so we need 6-x ≥ 0. That gives x ≤ 6. So x=6 is included, but so are all numbers less than 6. Thus C is correct.

I got this wrong because I thought the square root of a square is just the absolute value, so |x-6| = 6-x. That equation is equivalent to x-6 ≤ 0, so x ≤ 6. But I mistakenly picked B because I thought it had to be strictly less, but x=6 also works. So C is the must be true.

This is a classic GMAT trap. The answer is C. But I can see why it's rated 705-805. It tests whether you understand that sqrt(a^2) = |a| and then you have to solve an absolute value equation with a condition. Definitely not a walk in the park.