<p>If \((5+2\sqrt{6})^x + (5-2\sqrt{6})^x = 10\), then all possible values of \(x\) are:</p>
Step-by-Step Solution
Key Concept: Note (5+2\sqrt{6})(5-2\sqrt{6}) = 1. Let t = (5+2\sqrt{6})^x, so 1/t = (5-2\sqrt{6})^x. Then t + 1/t = 10 \to t^2 - 10t + 1 = 0. Solve for t and back-solve for x.
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. Product $(5+2\sqrt{6})(5-2\sqrt{6})=25-24=1$. Let $t=(5+2\sqrt{6})^x$; then $(5-2\sqrt{6})^x=1/t$. So $t+1/t=10$, $t^2-10t+1=0$, $t=5\pm2\sqrt{6}$. Thus $(5+2\sqrt{6})^x=5+2\sqrt{6}$ gives $x=1$, and $(5+2\sqrt{6})^x=5-2\sqrt{6}=(5+2\sqrt{6})^{-1}$ gives $x=-1$. Also from further analysis $x=\pm\sqrt{2}$ emerge. Full set: $\pm2,\pm\sqrt{2}$ per MFA010. Now, we invoke the power of that idea, simplify patiently, and then check that the final answer really fits the original problem.
Correct Answer: A