**Problem:** Given $A = \sqrt{5} + 2$ and $B = \sqrt{5} - 2$, find $A^2$, $B^2$, $A \times B$, and verify that $\frac{A}{B} + \frac{B}{A}$ is a number. 1. Calculate $A^2$: $$A^2 = (\sqrt{5} + 2)^2 = (\sqrt{5})^2 + 2 \times \sqrt{5} \times 2 + 2^2 = 5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5}$$ 2. Calculate $B^2$: $$B^2 = (\sqrt{5} - 2)^2 = (\sqrt{5})^2 - 2 \times \sqrt{5} \times 2 + 2^2 = 5 - 4\sqrt{5} + 4 = 9 - 4\sqrt{5}$$ 3. Calculate $A \times B$: $$A \times B = (\sqrt{5} + 2)(\sqrt{5} - 2) = (\sqrt{5})^2 - (2)^2 = 5 - 4 = 1$$ 4. Verify that $\frac{A}{B} + \frac{B}{A}$ is a number: We rewrite using a common denominator: $$\frac{A}{B} + \frac{B}{A} = \frac{A^2 + B^2}{AB}$$ From steps 1 and 2: $$A^2 + B^2 = (9 + 4\sqrt{5}) + (9 - 4\sqrt{5}) = 18$$ From step 3: $$AB = 1$$ Therefore: $$\frac{A}{B} + \frac{B}{A} = \frac{18}{1} = 18$$ Since this is a rational number, the expression $\frac{A}{B} + \frac{B}{A}$ is indeed a number. **Final answers:** - $A^2 = 9 + 4\sqrt{5}$ - $B^2 = 9 - 4\sqrt{5}$ - $A \times B = 1$ - $\frac{A}{B} + \frac{B}{A} = 18$ which is a number.