Let's break down and solve the problems presented in the image:

1. Given $A = 4 - 3\sqrt{5}$ and $B = -1 + 2\sqrt{5}$, calculate:

• $A + B$
• $A - B$
• $AB$

(i) $A + B$:

$A + B = (4 - 3\sqrt{5}) + (-1 + 2\sqrt{5}) = 4 - 1 + (-3\sqrt{5} + 2\sqrt{5})$ $A + B = 3 - \sqrt{5}$

(ii) $A - B$:

$A - B = (4 - 3\sqrt{5}) - (-1 + 2\sqrt{5}) = 4 + 1 + (-3\sqrt{5} - 2\sqrt{5})$ $A - B = 5 - 5\sqrt{5}$

(iii) $AB$:

$AB = (4 - 3\sqrt{5})(-1 + 2\sqrt{5})$ Applying distributive property: $AB = 4(-1) + 4(2\sqrt{5}) - 3\sqrt{5}(-1) - 3\sqrt{5}(2\sqrt{5})$ $AB = -4 + 8\sqrt{5} + 3\sqrt{5} - 6 \times 5$ $AB = -4 + 11\sqrt{5} - 30$ $AB = -34 + 11\sqrt{5}$

2. Factorize $9 + 4\sqrt{5}$ and deduce a simplification for $\sqrt{9 + 4\sqrt{5}}$.

We attempt to express $9 + 4\sqrt{5}$ as a square of some binomial, say $(a + b\sqrt{5})^2$: $(a + b\sqrt{5})^2 = a^2 + 2ab\sqrt{5} + 5b^2$ Matching terms with $9 + 4\sqrt{5}$: $a^2 + 5b^2 = 9 \quad \text{and} \quad 2ab = 4$ From $2ab = 4$, we get $ab = 2$, so $b = \frac{2}{a}$.

Substitute this into $a^2 + 5b^2 = 9$: $a^2 + 5\left(\frac{2}{a}\right)^2 = 9$ $a^2 + \frac{20}{a^2} = 9$ Multiply through by $a^2$: $a^4 + 20 = 9a^2$ $a^4 - 9a^2 + 20 = 0$ Let $x = a^2$, so: $x^2 - 9x + 20 = 0$ Solve using the quadratic formula: $x = \frac{9 \pm \sqrt{9^2 - 4(1)(20)}}{2} = \frac{9 \pm \sqrt{81 - 80}}{2} = \frac{9 \pm 1}{2}$ Thus, $x = 5$ or $x = 4$. Since $x = a^2$, this gives $a^2 = 4$, so $a = 2$.

Substitute back $b = \frac{2}{a} = 1$. Therefore, we have: $9 + 4\sqrt{5} = (2 + \sqrt{5})^2$ Thus: $\sqrt{9 + 4\sqrt{5}} = 2 + \sqrt{5}$

3. Show that $\frac{2}{\sqrt{3} - 1} + \frac{-2}{\sqrt{3} + 1}$ is a natural number.

First, rationalize both denominators.

(i) Rationalizing $\frac{2}{\sqrt{3} - 1}$:

Multiply numerator and denominator by $\sqrt{3} + 1$: $\frac{2}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{2(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = \frac{2(\sqrt{3} + 1)}{3 - 1} = \frac{2(\sqrt{3} + 1)}{2} = \sqrt{3} + 1$

(ii) Rationalizing $\frac{-2}{\sqrt{3} + 1}$:

Multiply numerator and denominator by $\sqrt{3} - 1$: $\frac{-2}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{-2(\sqrt{3} - 1)}{(\sqrt{3})^2 - 1^2} = \frac{-2(\sqrt{3} - 1)}{3 - 1} = \frac{-2(\sqrt{3} - 1)}{2} = -(\sqrt{3} - 1)$

(iii) Adding the results:

$\left( \sqrt{3} + 1 \right) + \left( -\sqrt{3} + 1 \right) = 1 + 1 = 2$ Thus, the result is a natural number: $2$.

Would you like more details on any of these steps or further clarifications?

Here are 5 related questions:

1. How do you simplify other complex algebraic expressions involving square roots?
2. What are the general steps to rationalize any denominator?
3. Can you show how to factor binomials using different methods?
4. How do you apply the quadratic formula to solve more complex algebraic equations?
5. What is the significance of simplifying surds in practical applications?

Tip: When rationalizing denominators, always multiply by the conjugate of the denominator to simplify the square root.