Exercise 1.2

Class 9 Mathematics Notes | Unit 1 – Real Numbers | Class 9 Mathematics (New Syllabus)

📘 Exercise 1.2

Learn and practice Class 9 Mathematics Exercise 1.2 on Real Numbers with step-by-step solutions. This page includes free downloadable notes and solved examples aligned with the PECTAA 2025 syllabus.

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Exercise 1.2 - Real Numbers

This exercise focuses on rationalizing denominators, simplifying expressions with exponents, and working with algebraic identities.

Exercise 1.2 provides practice problems that help students master the technique of rationalizing denominators, simplify complex expressions using laws of exponents, and apply algebraic identities to solve problems.

The exercise covers rationalizing denominators with single and binomial surds, simplifying expressions with fractional exponents, and applying algebraic identities to find values of expressions.

Key Topics Covered

Important Note:

Remember that rationalizing the denominator means eliminating radicals from the denominator of a fraction. For binomial denominators containing surds, we multiply numerator and denominator by the conjugate of the denominator.

Important Formulas Used in This Exercise

Square Identities

$(x+y)^2 = x^2 + 2xy + y^2$

$(x-y)^2 = x^2 - 2xy + y^2$

$x^2 - y^2 = (x+y)(x-y)$

Cube Identities

$x^3 + y^3 = (x+y)(x^2 - xy + y^2)$

$x^3 - y^3 = (x-y)(x^2 + xy + y^2)$

Exponent Rules

$a^m \cdot a^n = a^{m+n}$

$(a^m)^n = a^{mn}$

$\frac{a^m}{a^n} = a^{m-n}$

$a^0 = 1$

Sample Problems with Solutions

Problem 1: Rationalize the Denominator

Rationalize the denominator of $\frac{13}{4+\sqrt{3}}$

1
Multiply numerator and denominator by the conjugate

$\frac{13}{4+\sqrt{3}} \times \frac{4-\sqrt{3}}{4-\sqrt{3}}$

2
Apply the formula $(a+b)(a-b) = a^2 - b^2$

$\frac{13(4-\sqrt{3})}{(4)^2-(\sqrt{3})^2} = \frac{13(4-\sqrt{3})}{16-3}$

3
Simplify

$\frac{13(4-\sqrt{3})}{13} = 4-\sqrt{3}$

Final Answer: $4-\sqrt{3}$

Problem 2: Simplify Exponential Expression

Simplify $\left(\frac{81}{16}\right)^{-\frac{3}{4}}$

1
Apply negative exponent rule

$\left(\frac{81}{16}\right)^{-\frac{3}{4}} = \left(\frac{16}{81}\right)^{\frac{3}{4}}$

2
Express numbers as powers

$\left(\frac{2^4}{3^4}\right)^{\frac{3}{4}} = \frac{(2^4)^{\frac{3}{4}}}{(3^4)^{\frac{3}{4}}}$

3
Apply power of a power rule

$\frac{2^{4 \cdot \frac{3}{4}}}{3^{4 \cdot \frac{3}{4}}} = \frac{2^3}{3^3} = \frac{8}{27}$

Final Answer: $\frac{8}{27}$

Created by Hira Science Academy | Aligned with PECTA 2025 Syllabus

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