1. **State the problem:** Given $a=5+2\sqrt{6}$ and $b=\frac{1}{a}$, find $a^2 + b^2$. 2. **Recall the formula:** We want to find $a^2 + b^2$ where $b=\frac{1}{a}$. 3. **Express $b^2$:** Since $b=\frac{1}{a}$, then $b^2=\frac{1}{a^2}$. 4. **Rewrite the expression:** $$a^2 + b^2 = a^2 + \frac{1}{a^2}$$ 5. **Use the identity:** $$a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2$$ 6. **Calculate $a + \frac{1}{a}$:** First, find $\frac{1}{a}$: $$\frac{1}{5 + 2\sqrt{6}}$$ Rationalize the denominator: $$\frac{1}{5 + 2\sqrt{6}} \times \frac{5 - 2\sqrt{6}}{5 - 2\sqrt{6}} = \frac{5 - 2\sqrt{6}}{(5)^2 - (2\sqrt{6})^2} = \frac{5 - 2\sqrt{6}}{25 - 4 \times 6} = \frac{5 - 2\sqrt{6}}{25 - 24} = 5 - 2\sqrt{6}$$ 7. **Sum $a + \frac{1}{a}$:** $$a + \frac{1}{a} = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 5 + 5 + 2\sqrt{6} - 2\sqrt{6} = 10$$ 8. **Calculate $a^2 + b^2$ using the identity:** $$a^2 + b^2 = (10)^2 - 2 = 100 - 2 = 98$$ **Final answer:** $$\boxed{98}$$