1. **State the problem:** Given $a=5+2\sqrt{5}$ and $b=\frac{1}{a}$, find $a^2 + b^2$. 2. **Recall the formula:** We want to find $a^2 + b^2$. Since $b=\frac{1}{a}$, we can write $b^2 = \frac{1}{a^2}$. 3. **Express the sum:** $$a^2 + b^2 = a^2 + \frac{1}{a^2}$$ 4. **Use the identity:** $$a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2$$ 5. **Calculate $a + \frac{1}{a}$:** $$a + \frac{1}{a} = 5 + 2\sqrt{5} + \frac{1}{5 + 2\sqrt{5}}$$ 6. **Rationalize the denominator:** $$\frac{1}{5 + 2\sqrt{5}} \times \frac{5 - 2\sqrt{5}}{5 - 2\sqrt{5}} = \frac{5 - 2\sqrt{5}}{(5)^2 - (2\sqrt{5})^2} = \frac{5 - 2\sqrt{5}}{25 - 4 \times 5} = \frac{5 - 2\sqrt{5}}{25 - 20} = \frac{5 - 2\sqrt{5}}{5}$$ 7. **Simplify:** $$\frac{5 - 2\sqrt{5}}{5} = 1 - \frac{2}{5}\sqrt{5}$$ 8. **Sum $a + \frac{1}{a}$:** $$5 + 2\sqrt{5} + 1 - \frac{2}{5}\sqrt{5} = 6 + \left(2 - \frac{2}{5}\right)\sqrt{5} = 6 + \frac{8}{5}\sqrt{5}$$ 9. **Square the sum:** $$\left(6 + \frac{8}{5}\sqrt{5}\right)^2 = 6^2 + 2 \times 6 \times \frac{8}{5}\sqrt{5} + \left(\frac{8}{5}\sqrt{5}\right)^2 = 36 + \frac{96}{5}\sqrt{5} + \frac{64}{25} \times 5$$ 10. **Simplify the last term:** $$\frac{64}{25} \times 5 = \frac{64}{5}$$ 11. **Sum all terms:** $$36 + \frac{96}{5}\sqrt{5} + \frac{64}{5} = \left(36 + \frac{64}{5}\right) + \frac{96}{5}\sqrt{5} = \frac{180}{5} + \frac{64}{5} + \frac{96}{5}\sqrt{5} = \frac{244}{5} + \frac{96}{5}\sqrt{5}$$ 12. **Recall the identity:** $$a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2 = \frac{244}{5} + \frac{96}{5}\sqrt{5} - 2 = \frac{244}{5} - \frac{10}{5} + \frac{96}{5}\sqrt{5} = \frac{234}{5} + \frac{96}{5}\sqrt{5}$$ **Final answer:** $$a^2 + b^2 = \frac{234}{5} + \frac{96}{5}\sqrt{5}$$