ICSE Class 9 Maths Chapter 04 Expansions

Unit 3: Algebra

4 Expansions

4.1 Introduction

Expansion is the process in which the contents of brackets are evaluated.

Recall of concepts of expansions learned in earlier classes:

1. Since, \((a + b)^2 = (a + b) (a + b)\)

\(= a(a + b) + b(a + b)\)

\(= a^2 + ab + ab + b^2\)

\(= a^2 + 2ab + b^2\)

\(a^2 + 2ab + b^2\) is the expansion of \((a + b)^2\)

Similarly,

2. \((a - b)^2 = a^2 - 2ab + b^2\)

3. \((a + b)^2 + (a - b)^2 = 2(a^2 + b^2)\) (On adding I and II)

4. \((a + b)^2 - (a - b)^2 = 4ab\) (On subtracting II from I)

If \(a \neq 0\), then:

5. \((a + \frac{1}{a})^2 = a^2 + \frac{1}{a^2} + 2 \Rightarrow a^2 + \frac{1}{a^2} = (a + \frac{1}{a})^2 - 2\)

6. \((a - \frac{1}{a})^2 = a^2 + \frac{1}{a^2} - 2 \Rightarrow a^2 + \frac{1}{a^2} = (a - \frac{1}{a})^2 + 2\)

7. \((a + \frac{1}{a})^2 + (a - \frac{1}{a})^2 = 2(a^2 + \frac{1}{a^2})\)

8. \((a + \frac{1}{a})^2 - (a - \frac{1}{a})^2 = 4\)

Teacher's Note

Algebraic expansions are used in engineering and construction to calculate areas and volumes of complex shapes by breaking them into simpler components.

4.2 Identities

Consider the expansion: \((a + b)^2 = a^2 + 2ab + b^2\)

1. If \(a = 5\) and \(b = 3\)

\((a + b)^2 = (5 + 3)^2 = 8^2 = 64\) and

\(a^2 + 2ab + b^2 = 5^2 + 2 \times 5 \times 3 + 3^2 = 25 + 30 + 9 = 64\)

i.e. \((a + b)^2 = a^2 + 2ab + b^2\)

2. If \(a = -8\) and \(b = 5\)

\((a + b)^2 = (-8 + 5)^2 = (-3)^2 = 9\) and

\(a^2 + 2ab + b^2 = (-8)^2 + 2 \times -8 \times 5 + 5^2 = 64 - 80 + 25 = 9\)

i.e. \((a + b)^2 = a^2 + 2ab + b^2\)

In the same way, if we give any number of values to \(a\) and \(b\); every time \((a + b)^2\) and \(a^2 + 2ab + b^2\) will come same (equal).

An equation, which is true for all values of its variables, is called an identity. Each equation (expansion) given above in article 4.1 is an identity.

Teacher's Note

Identities are fundamental in mathematics because they allow us to verify the correctness of calculations and solve complex problems by recognizing patterns, much like using shortcuts in everyday calculations.

Worked Examples

Example 1

Evaluate: (i) \((a + 2b)^2\) (ii) \((2a - 3b)^2\)

Solution:

(i) \((a + 2b)^2 = (a)^2 + 2 \times a \times 2b + (2b)^2\)

\(= a^2 + 4ab + 4b^2\)

(ii) \((2a - 3b)^2 = (2a)^2 - 2 \times 2a \times 3b + (3b)^2\)

\(= 4a^2 - 12ab + 9b^2\)

Example 2

If \(a + b = 9\) and \(ab = -22\), find: (i) \(a - b\) (ii) \(a^2 - b^2\)

Solution:

(i) \((a + b)^2 - (a - b)^2 = 4ab\)

\(\Rightarrow (a - b)^2 = (a + b)^2 - 4ab\)

\(= (9)^2 - 4 \times -22\)

\(= 81 + 88 = 169\)

\(\therefore a - b = \pm \sqrt{169} = \pm 13\)

OR, \((a + b)^2 = 9^2\)

\(\Rightarrow a^2 + b^2 + 2ab = 81\)

\(\Rightarrow a^2 + b^2 + 2 \times -22 = 81\)

\(\Rightarrow a^2 + b^2 = 125\)

Now, \((a - b)^2 = a^2 + b^2 - 2ab\)

\(= 125 - 2 \times -22\)

\(= 169\)

\(\therefore a - b = \pm 13\)

(ii) \(a^2 - b^2 = (a + b)(a - b) = 9 \times \pm 13 = \pm 117\)

Example 3

If \(x \neq 0\) and \(x + \frac{1}{x} = 2\), find: (i) \(x^2 + \frac{1}{x^2}\) (ii) \(x^4 + \frac{1}{x^4}\)

Solution:

(i) \(x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2\)

\(= (2)^2 - 2\)

\(= 4 - 2 = 2\)

Alternative method:

(i) \(x + \frac{1}{x} = 2\)

\(\Rightarrow (x + \frac{1}{x})^2 = (2)^2\)

\(\Rightarrow x^2 + \frac{1}{x^2} + 2 \times x \times \frac{1}{x} = 4\)

\(\Rightarrow x^2 + \frac{1}{x^2} = 4 - 2 = 2\)

(ii) \(x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2\)

\(= (2)^2 - 2\)

\(= 4 - 2 = 2\)

Alternative method:

(ii) \((x^2 + \frac{1}{x^2})^2 = (2)^2\)

\(\Rightarrow x^4 + \frac{1}{x^4} + 2 = 4\)

\(\Rightarrow x^4 + \frac{1}{x^4} = 2\)

Example 4

Given: \(a^2 + \frac{1}{a^2} = 7\) and \(a \neq 0\), find:

(i) \(a + \frac{1}{a}\) (ii) \(a - \frac{1}{a}\) (iii) \(a^2 - \frac{1}{a^2}\)

Solution:

(i) \(\therefore (a + \frac{1}{a})^2 = a^2 + \frac{1}{a^2} + 2 = 7 + 2 = 9\)

\(\Rightarrow a + \frac{1}{a} = \pm \sqrt{9} = \pm 3\)

(ii) \(\therefore (a - \frac{1}{a})^2 = a^2 + \frac{1}{a^2} - 2 = 7 - 2 = 5\)

\(\Rightarrow a - \frac{1}{a} = \pm \sqrt{5}\)

(iii) \(a^2 - \frac{1}{a^2} = (a + \frac{1}{a})(a - \frac{1}{a}) = (\pm 3) \times (\pm \sqrt{5}) = \pm 3\sqrt{5}\)

Remember:

\((\pm a) \times (\pm b) = (+ a) \times (+ b)\) or \((- a) \times (+ b)\) or \((+ a) \times (- b)\) or \((- a) \times (- b)\)

\(= + ab\) or \(- ab\) or \(- ab\) or \(+ ab\)

\(= \pm ab\)

\(\therefore (\pm a) \times (\pm b) = \pm ab\)

Example 5

If \(a^2 - 5a + 1 = 0\) and \(a \neq 0\), find: (i) \(a + \frac{1}{a}\) (ii) \(a^2 + \frac{1}{a^2}\)

Solution:

(i) \(a^2 - 5a + 1 = 0\)

\(\Rightarrow \frac{a^2}{a} - \frac{5a}{a} + \frac{1}{a} = 0\) (Dividing each term by a)

\(\Rightarrow a - 5 + \frac{1}{a} = 0 \Rightarrow a + \frac{1}{a} = 5\)

(ii) \(a^2 + \frac{1}{a^2} = (a + \frac{1}{a})^2 - 2 = 5^2 - 2 = 25 - 2 = 23\)

Teacher's Note

Working with algebraic identities helps develop problem-solving skills that are essential in fields like physics and finance where relationships between variables must be understood and manipulated.

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