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Q2(i):
Simplify each of the following expressions:
(i) $(3 + \sqrt{3})(2 + \sqrt{2})$
Solution :
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: $6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}$
More Questions from Class 9 Mathematics Number Systems EXERCISE 1.4
- Q1(i): Classify the following numbers as rational or irrational: (i) $2 - \sqrt{5}$
- Q1(ii): Classify the following numbers as rational or irrational: (ii) $(3 + \sqrt{23}) - \sqrt{23}$
- Q1(iii): Classify the following numbers as rational or irrational: (iii) $\frac{2\sqrt{7}}{7\sqrt{7}}$
- Q1(iv): Classify the following numbers as rational or irrational: (iv) $\frac{1}{\sqrt{2}}$
- Q1(v): Classify the following numbers as rational or irrational: (v) $2\pi$
- Q2(ii): Simplify each of the following expressions: (ii) $(3 + \sqrt{3})(3 - \sqrt{3})$
- Q2(iii): Simplify each of the following expressions: (iii) $(\sqrt{5} + \sqrt{2})^2$
- Q2(iv): Simplify each of the following expressions: (iv) $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$
- Q3: Recall, $\pi$ is defined as the ratio of the circumference (say $c$) of a circle to its diameter (say $d$). That is, $\pi = \frac{c}{d}$. This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?
- Q4: Represent $\sqrt{9.3}$ on the number line.
- Q5(i): Rationalise the denominators of the following: (i) $\frac{1}{\sqrt{7}}$
- Q5(ii): Rationalise the denominators of the following: (ii) $\frac{1}{\sqrt{7} - \sqrt{6}}$
- Q5(iii): Rationalise the denominators of the following: (iii) $\frac{1}{\sqrt{5} + \sqrt{2}}$
- Q5(iv): Rationalise the denominators of the following: (iv) $\frac{1}{\sqrt{7} - 2}$
CBSE Solutions for Class 9 Mathematics Number Systems
Chapters in CBSE - Class 9 Mathematics
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