Step-by-Step Solution
Key Concept: Let x = \sqrt{6+x}. Then x^2 = x+6, so x^2-x-6 = 0, (x-3)(x+2) = 0. Since x > 0, x = 3.
Notice that the best first move is to reveal the hidden structure in the expression. A clever move here is to rewrite the problem in the form where the standard theorem or identity applies cleanly. Let $x = \sqrt{6+\sqrt{6+\cdots}}$. Then $x = \sqrt{6+x}$, so $x^2 = 6+x$, giving $x^2-x-6=0$, i.e., $(x-3)(x+2)=0$. Since $x>0$, $x=3$. Now, we invoke the power of that idea, simplify patiently, and then check that the final answer really fits the original problem.
Correct Answer: 1