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Let's denote the equation as follows:

\[
\left(\sqrt{5 + 2\sqrt{6}}\right)^x + \left(\sqrt{5 - 2\sqrt{6}}\right)^x = 10.
\]

To solve this equation, let's start by simplifying the expressions inside the square roots.

1. Let:
   \[
   a = \sqrt{5 + 2\sqrt{6}}, \quad b = \sqrt{5 - 2\sqrt{6}}.
   \]
   
   So, the equation becomes:
   \[
   a^x + b^x = 10.
   \]

2. Notice that \( a \) and \( b \) are conjugates of each other in a sense because \( (5 + 2\sqrt{6}) \) and \( (5 - 2\sqrt{6}) \) are conjugate pairs. Additionally, 
   \[
   a \cdot b = \sqrt{(5 + 2\sqrt{6})(5 - 2\sqrt{6})} = \sqrt{(5^2 - (2\sqrt{6})^2)} = \sqrt{25 - 24} = \sqrt{1} = 1.
   \]

   Therefore, \( a \cdot b = 1 \), which means:
   \[
   b = \frac{1}{a}.
   \]

3. Substituting \( b = \frac{1}{a} \) into the equation, we get:
   \[
   a^x + \left(\frac{1}{a}\right)^x = 10.
   \]

   Simplifying this further, we have:
   \[
   a^x + \frac{1}{a^x} = 10.
   \]

4. Let \( y = a^x \). Then the equation becomes:
   \[
   y + \frac{1}{y} = 10.
   \]

5. Multiply through by \( y \) to get a quadratic equation:
   \[
   y^2 - 10y + 1 = 0.
   \]

6. Solve this quadratic equation using the quadratic formula:
   \[
   y = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} = \frac{10 \pm 4\sqrt{6}}{2}.
   \]

   Simplifying this, we get:
   \[
   y = 5 \pm 2\sqrt{6}.
   \]

7. Since \( y = a^x \) and \( a = \sqrt{5 + 2\sqrt{6}} > 1 \), we take the positive root:
   \[
   y = 5 + 2\sqrt{6}.
   \]

   So,
   \[
   a^x = 5 + 2\sqrt{6}.
   \]

   Therefore,
   \[
   x = 2.
   \]

Hence, the solution to the equation is:

\[
x = 2.
\]

Would you like more details or have any questions about this solution? 

Here are some related questions for further exploration:

1. What is the process to solve equations involving conjugate pairs?
2. How can the quadratic equation be used in more complex algebraic expressions?
3. What other methods can be used to verify the value of \(x\) in this equation?
4. How do we derive the product of conjugates in terms of their expressions?
5. What role does the quadratic formula play in solving higher-order algebraic equations?

**Tip:** When dealing with expressions involving square roots and conjugates, look for simplifications like products or sum of conjugates to facilitate solving the equation.

(√(5 + 2√6))^x + (√(5 - 2√6))^x = 10

Learn how to solve the equation (√(5 + 2√6))^x + (√(5 - 2√6))^x = 10 step-by-step. This solution covers key algebraic concepts like conjugates and the quadratic formula. Suitable for high school students studying advanced algebra.

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