Q2(i):

Simplify each of the following expressions: (i) $(3 + \sqrt{3})(2 + \sqrt{2})$

Initial Setup & Mathematical Objective

We are tasked with simplifying the product of two binomial expressions containing irrational numbers (surds). The given expression is:

$(3 + \sqrt{3})(2 + \sqrt{2})$

Step 1: Applying the Distributive Property

To multiply two binomials, we utilize the distributive property of multiplication over addition, commonly referred to as the FOIL method (First, Outer, Inner, Last) in elementary algebra. The general algebraic identity is defined as:

$(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd$

[Per the Axioms of Real Numbers, multiplication distributes over addition].

Mapping our given values to this identity:

• $a = 3$
• $b = \sqrt{3}$
• $c = 2$
• $d = \sqrt{2}$

Step 2: Term-by-Term Multiplication

We will now systematically multiply each term:

• First terms ($ac$): $3 \times 2 = 6$
• Outer terms ($ad$): $3 \times \sqrt{2} = 3\sqrt{2}$
• Inner terms ($bc$): $\sqrt{3} \times 2 = 2\sqrt{3}$
• Last terms ($bd$): $\sqrt{3} \times \sqrt{2} = \sqrt{3 \times 2} = \sqrt{6}$
  [Per the Product Rule for Radicals: $\sqrt{x} \cdot \sqrt{y} = \sqrt{xy}$ for all real numbers $x, y \ge 0$].

Step 3: Aggregating the Expanded Terms

Combining the results from Step 2, we construct the expanded expression:

$6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}$

Step 4: Analyzing for Like Terms

In radical expressions, terms can only be added or subtracted if they are "like surds" (i.e., they share the exact same irrational radicand). Let us evaluate our terms:

• $6$ is a rational integer.
• $3\sqrt{2}$ contains the irrational root $\sqrt{2}$.
• $2\sqrt{3}$ contains the irrational root $\sqrt{3}$.
• $\sqrt{6}$ contains the irrational root $\sqrt{6}$.

Because $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{6}$ are distinct, mutually prime radicands, none of these terms are like terms. Therefore, no further algebraic simplification or combination is mathematically permissible.

Geometric Visualization: Area Model

The multiplication of these binomials can be geometrically represented as calculating the total area of a rectangle with side lengths $(3 + \sqrt{3})$ and $(2 + \sqrt{2})$. The total area is the sum of the four smaller rectangular regions.

Final Solution: $6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}$