📘 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
- Rationalizing denominators with single surds
- Rationalizing denominators with binomial surds
- Simplifying expressions with exponents
- Applying laws of exponents
- Using algebraic identities
- Finding values of expressions using given conditions
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
Rationalize the denominator of $\frac{13}{4+\sqrt{3}}$
$\frac{13}{4+\sqrt{3}} \times \frac{4-\sqrt{3}}{4-\sqrt{3}}$
$\frac{13(4-\sqrt{3})}{(4)^2-(\sqrt{3})^2} = \frac{13(4-\sqrt{3})}{16-3}$
$\frac{13(4-\sqrt{3})}{13} = 4-\sqrt{3}$
Final Answer: $4-\sqrt{3}$
Simplify $\left(\frac{81}{16}\right)^{-\frac{3}{4}}$
$\left(\frac{81}{16}\right)^{-\frac{3}{4}} = \left(\frac{16}{81}\right)^{\frac{3}{4}}$
$\left(\frac{2^4}{3^4}\right)^{\frac{3}{4}} = \frac{(2^4)^{\frac{3}{4}}}{(3^4)^{\frac{3}{4}}}$
$\frac{2^{4 \cdot \frac{3}{4}}}{3^{4 \cdot \frac{3}{4}}} = \frac{2^3}{3^3} = \frac{8}{27}$
Final Answer: $\frac{8}{27}$
Created by Hira Science Academy | Aligned with PECTA 2025 Syllabus