Class 9 Maths Part 1 Chapter 4 – Exploring Algebraic Identities from Ganita Manjari with easy Notes and complete Book Solutions. Get clear explanations, important concepts, questions and answers, and revision material for Chapter 4 to help Class 9 students learn, practise and prepare effectively for examinations.

Class 9 Maths – Chapter 4: Exploring Algebraic Identities

📘 Chapter 4 – Exploring Algebraic Identities

Easy Notes • Concepts • Book Questions • Step-by-Step Solutions

IEMH / Ganita Manjari – Mathematics

🌟 Quick Notes – Chapter 4

What is an algebraic identity?

An algebraic identity is an equation that is true for all values of the variables occurring in it.

(a + b)² = a² + 2ab + b²

For example, x² − 1 = 24 is not an identity because it is true only for particular values of x.

⭐ Identities to Remember

No.	Identity
1	(x + y)² = x² + 2xy + y²
2	(x − y)² = x² − 2xy + y²
3	(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
4	(x + y)(x − y) = x² − y²
5	(x + a)(x + b) = x² + (a + b)x + ab
6	(ax + b)(cx + d) = acx² + (ad + bc)x + bd
7	x³ − y³ = (x − y)(x² + xy + y²)
8	x³ + y³ = (x + y)(x² − xy + y²)
9	(x + y)³ = x³ + 3x²y + 3xy² + y³
10	(x − y)³ = x³ − 3x²y + 3xy² − y³
11	x³ + y³ + z³ − 3xyz = (x+y+z)(x²+y²+z²−xy−yz−zx)

⭐ Easy trick: Before solving a question, look at its shape. If you see a square, difference of squares, cubes, or three terms, try to match it with a suitable identity.

4.1 Algebraic Identities

Remember

The chapter begins by recalling:

(a + b)² = a² + 2ab + b²

The middle term 2ab is important. We must not forget it.

Exercise Set 4.1 – Q1

Using the identity (a + b)² = a² + 2ab + b², expand the following:

(7x + 4y)²
(7/5 x + 3/2 y)²
(2.5p + 1.5q)²
(3/4 s + 8/2 t)²
(x/y + 1/2)²
(1/x + 1/y)²

(i) (7x + 4y)²

Use (a+b)² = a²+2ab+b².

= (7x)² + 2(7x)(4y) + (4y)²

= 49x² + 56xy + 16y²

(ii) (7/5 x + 3/2 y)²

= (7x/5)² + 2(7x/5)(3y/2) + (3y/2)²

= 49x²/25 + 21xy/5 + 9y²/4

(iii) (2.5p + 1.5q)²

= (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²

= 6.25p² + 7.5pq + 2.25q²

(iv) (3/4 s + 8/2 t)²

Since 8/2 = 4:

= (3s/4 + 4t)²

= 9s²/16 + 6st + 16t²

(v) (x/y + 1/2)²

= x²/y² + x/y + 1/4

(vi) (1/x + 1/y)²

= 1/x² + 2/(xy) + 1/y²

Answer: Expand each expression using a² + 2ab + b².

Exercise Set 4.1 – Q2

Using the same identity, find the values of the following:

(64)²
(105)²
(205)²

(i) (64)²

64 = 60 + 4

64² = (60+4)²

= 60² + 2(60)(4) + 4²

= 3600 + 480 + 16 = 4096

(ii) (105)²

105 = 100 + 5

105² = 10000 + 1000 + 25

= 11025

(iii) (205)²

205 = 200 + 5

205² = 40000 + 2000 + 25

= 42025

Answers: 4096, 11025, 42025

4.2 Square of a Difference

(a − b)² = a² − 2ab + b²

The easiest way to remember it is:

First square − 2 × first × second + second square

(a − b)² = a² − 2ab + b²

Exercise Set 4.2 – Q1

Factor completely:

9x² + 24xy + 16y²
4s² + 20st + 25t²
49x² + 28xy + 4y²
64p² + 32pq + 4q²
3a² + 4ab + 4/3 b²
9/5 s² + 6/5 sv + v²

(i)

9x² + 24xy + 16y²

= (3x)² + 2(3x)(4y) + (4y)²

= (3x + 4y)²

(ii)

4s² + 20st + 25t² = (2s)² + 2(2s)(5t) + (5t)²

= (2s + 5t)²

(iii)

49x² + 28xy + 4y²

= (7x + 2y)²

(iv)

64p² + 32pq + 4q²

= (8p + 2q)²

(v)

3a² + 4ab + 4/3 b²

= 3(a² + 4ab/3 + 4b²/9)

= 3(a + 2b/3)²

(vi)

9/5 s² + 6/5 sv + v²

= 1/5(9s² + 6sv + 5v²)

This does not match a simple perfect-square form directly.

The key idea is to compare the expression with a² + 2ab + b².

Exercise Set 4.2 – Q2

Find the values of the following using the identity (a − b)² = a² − 2ab + b².

(79)²
(193)²
(299)²

(i) 79²

79 = 80 − 1

79² = (80−1)²

= 6400 − 160 + 1

= 6241

(ii) 193²

193 = 200 − 7

193² = 40000 − 2800 + 49

= 37249

(iii) 299²

299 = 300 − 1

299² = 90000 − 600 + 1

= 89401

Answers: 6241, 37249, 89401

4.3 Factorisation of Algebraic Expressions Using Identities

Factorisation means writing an expression as a product of simpler expressions.

x² + 4x + 4 = (x + 2)²

We look for:

First term = a²
Last term = b²
Middle term = 2ab

Exercise Set 4.3 – Q1

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

117²
78²
198²
214²
1104²
1120²

(i) 117²

117 = 100 + 17

117² = 10000 + 3400 + 289

= 13689

(ii) 78²

78 = 80 − 2

78² = 6400 − 320 + 4

= 6084

(iii) 198²

198 = 200 − 2

198² = 40000 − 800 + 4

= 39204

(iv) 214²

214 = 200 + 14

214² = 40000 + 5600 + 196

= 45796

(v) 1104²

1104 = 1100 + 4

= 1210000 + 8800 + 16

= 1218816

(vi) 1120²

1120 = 1100 + 20

= 1210000 + 44000 + 400

= 1254400

Exercise Set 4.3 – Q2

Factor using suitable identities.

16y² − 24y + 9
9/4 s² + 6st + 4t²
9m² + 3mk + 4n² + 6mn + 9k² + 3kn
p² − 16p + 64
9a² + 4b² + 12ab + 6ac + 4c² + 4bc

(i)

16y² − 24y + 9

= (4y)² − 2(4y)(3) + 3²

= (4y − 3)²

(ii)

9/4 s² + 6st + 4t²

= (3s/2)² + 2(3s/2)(2t) + (2t)²

= (3s/2 + 2t)²

(iv)

p² − 16p + 64

= p² − 2(p)(8) + 8²

= (p − 8)²

For expressions with three variables, compare the expression with the corresponding three-variable identity before factorising.

Exercise Set 4.3 – Q3

Expand the following using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:

(p + 3q + 7r)²
(3x − 2y + 4z)²

(i)

(p + 3q + 7r)²

= p² + 9q² + 49r² + 6pq + 42qr + 14pr

(ii)

(3x − 2y + 4z)²

= 9x² + 4y² + 16z² − 12xy − 16yz + 24xz

Exercise Set 4.3 – Q4

Is this an identity?

(a + b − c)² + (a − b + c)² = 2a² + 2b² + 2c²

Expand both left-hand expressions.

(a+b−c)² = a²+b²+c²+2ab−2bc−2ac

(a−b+c)² = a²+b²+c²−2ab−2bc+2ac

Adding:

= 2a² + 2b² + 2c² − 4bc

This is not generally equal to 2a² + 2b² + 2c².

Therefore, the statement as written is not an identity.

4.4 More Identities

Three important identities

(a+b+c)² = a²+b²+c²+2ab+2bc+2ca

a² − b² = (a+b)(a−b)

a² = (a+b)(a−b)+b²

These identities are useful for factorisation and quick calculations.

Area idea: The whole square gives (a+b+c)² and its smaller pieces add to a²+b²+c²+2ab+2bc+2ca.

Exercise Set 4.3 – Q1 revisited

The book asks for the squares: 117², 78², 198², 214², 1104² and 1120².

The easiest identity depends on whether the number is close to a convenient round number.

Numbers close to 100, 200, 1000 etc. are often easiest using (a − b)² or (a + b)².

4.5 Factorisation and Simplifying Rational Expressions

Basic factorisation idea

For an expression such as:

x² + 8x + 15

find two numbers whose:

sum is 8
product is 15

The numbers are 3 and 5.

x² + 8x + 15 = (x+3)(x+5)

Worked Example – Pool

A rectangular pool has breadth 4 metres less than its length and its area is 96 square metres. Find its length and breadth.

Let the length be x metres.

Then breadth = x − 4 metres.

Area:

x(x−4) = 96

x² − 4x − 96 = 0

Split −4x as −12x + 8x:

x² − 12x + 8x − 96 = 0

x(x−12)+8(x−12)=0

(x−12)(x+8)=0

Therefore:

x = 12 or x = −8.

A length cannot be negative, so x = 12.

Breadth = 12 − 4 = 8 metres.

Length = 12 m
Breadth = 8 m

Worked Example – Rational Expression

Simplify:

(x² − 7x + 12) / (5x² + 5x − 100)

Factor the numerator:

x² − 7x + 12 = (x−3)(x−4)

Factor the denominator:

5x² + 5x − 100 = 5(x²+x−20)

= 5(x−4)(x+5)

Therefore:

[(x−3)(x−4)] / [5(x−4)(x+5)]

Cancel x−4, provided it is not zero.

= (x−3) / [5(x+5)]

📚 Chapter Summary

Identities are equations true for all values of the variables.
Identities can be visualised using geometrical models.
Identities help us factorise algebraic expressions.
Identities make many numerical calculations quicker.
Factorisation can be used to simplify rational algebraic expressions.
A common factor can be cancelled only when it is not zero.

⭐ Final Revision Box

(x+y)² = x²+2xy+y²

(x−y)² = x²−2xy+y²

(x+y)(x−y)=x²−y²

(x+y+z)²=x²+y²+z²+2xy+2yz+2zx

(x+a)(x+b)=x²+(a+b)x+ab

x³−y³=(x−y)(x²+xy+y²)

x³+y³=(x+y)(x²−xy+y²)

(x+y)³=x³+3x²y+3xy²+y³

(x−y)³=x³−3x²y+3xy²−y³

🧠 Easy Memory Tricks

For (a+b)²

Square – Double – Square

First square + twice their product + second square.

For (a−b)²

Square – Double – Square

First square − twice their product + second square.

For a²−b²

Sum × Difference

a²−b²=(a+b)(a−b)

For factorisation

Look for the identity before trying random factors.

prepare more with the given Worksheet of chapter 4, which will help you to improve your knowledge .

CLASS 9 • MATHEMATICS

Chapter 4 Worksheet

Practice Questions & Revision Worksheet

Open Worksheet

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Class 9 Maths Part 1 Chapter 4 – Exploring Algebraic Identities | Ganita Manjari Notes & Book Solutions

🌟 Quick Notes – Chapter 4

What is an algebraic identity?

⭐ Identities to Remember

4.1 Algebraic Identities

Remember

(i) (7x + 4y)²

(ii) (7/5 x + 3/2 y)²

(iii) (2.5p + 1.5q)²

(iv) (3/4 s + 8/2 t)²

(v) (x/y + 1/2)²

(vi) (1/x + 1/y)²

(i) (64)²

(ii) (105)²

(iii) (205)²

4.2 Square of a Difference

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(i) 79²

(ii) 193²

(iii) 299²

4.3 Factorisation of Algebraic Expressions Using Identities

(i) 117²

(ii) 78²

(iii) 198²

(iv) 214²

(v) 1104²

(vi) 1120²

(i)

(ii)

(iv)

(i)

(ii)

4.4 More Identities

Three important identities

4.5 Factorisation and Simplifying Rational Expressions

Basic factorisation idea

📚 Chapter Summary

⭐ Final Revision Box

🧠 Easy Memory Tricks

For (a+b)²

For (a−b)²

For a²−b²

For factorisation

Chapter 4 Worksheet