NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.1 | 2026-27

Think and Reflect

Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.

Solution:

Taking four consecutive squares, i.e., n², (n + 1)², (n + 2)² and (n + 3)², we observe a pattern.

Here,
n² + (n + 3)² = (n + 1)² + (n + 2)² + 4

For example, when n = 1:

⇒ 1 + 16 = 4 + 9 + 4
⇒ 17 = 17

Think and Reflect

1. What can you say about a and b if (a + b)2 < a2 + b2?

Solution:

Given that (a + b)² < a² + b²

We have, (a + b)² = a² + 2ab + b²

So, a² + 2ab + b² < a² + b²

Subtracting a² + b² from both sides, we get 2ab < 0, which is possible only when a and b have opposite signs.

Therefore, a and b must have opposite signs for (a + b)² < a² + b².

2. What can you say about a and b if (a + b)2 > a2+ b2?

Solution:

Here, (a + b)² > a² + b²

We know, (a + b)² = a² + 2ab + b²

∴ a² + 2ab + b² > a² + b²

⇒ a² + 2ab + b² – a² – b² > a² + b² – a² – b²
[Subtracting (a² + b²) from both sides]

⇒ 2ab > 0

This is possible only when both a and b have the same sign.

Hence, a and b must have the same sign for (a + b)² > a² + b².

3. When will (a + b)2 be equal to a2+ b2 ?

Solution:

Given that (a + b)² = a² + b²

We have, (a + b)² = a² + b² + 2ab

So, a² + 2ab + b² = a² + b²

Subtracting a² + b² from both sides, we get 2ab = 0, which is possible only when either a = 0 or b = 0.

So, either a = 0 or b = 0 for (a + b)² = a² + b².

Did you observe that (a + b)2 and a2+ b2 are both positive? What term will decide which is larger? Use the expansion of (a + b)2 to decide.

Solution:

Yes, both (a + b)² and a² + b² are positive, as they are squares of numbers.

The term 2ab determines which of the two expressions is larger. If a and b have the same sign, then 2ab > 0, so (a + b)² > a² + b². If they have opposite signs, then 2ab < 0, so (a + b)² < a² + b².

Exercise Set 4.1

Question 1.

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

(i) (7x + 4y)2

Solution:

(7x + 4y)2 = (7x)2 + 2 × 7x × 4y + (4y)2

[∵ (a + b)2 = a2 + 2ab + b2] = 49x2 + 56xy + 16y2

ii) (⁷⁄₅x + ³⁄₂y)²

Solution:

Using the identity:

(a + b)² = a² + 2ab + b²

(⁷⁄₅x + ³⁄₂y)²
= (⁷⁄₅x)² + 2(⁷⁄₅x)(³⁄₂y) + (³⁄₂y)²

= ⁴⁹⁄₂₅x² + ²¹⁄₅xy + ⁹⁄₄y²

Answer: ⁴⁹⁄₂₅x² + ²¹⁄₅xy + ⁹⁄₄y²

(iii) (2.5p + 1.5q)2

Solution:

(2.5p + 1.5q)2

= (2.5p)2 + 2 × (2.5p) × (1.5q) + (1.5q)2

= 6.25p2 + 1.5pq + 2.25 q2

(iv) (³⁄₄s + 8t)²

Solution:

Given, (³⁄₄s + 8t)²

= (³⁄₄s)² + 2(³⁄₄s)(8t) + (8t)²

= ⁹⁄₁₆s² + 12st + 64t²

Answer: ⁹⁄₁₆s² + 12st + 64t²

(v) x+12y²

Solution:

x+12y²= 

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

Here,

a = x, b = ½y

Therefore,

(x + ½y)² = x² + 2(x)(½y) + (½y)²

= x² + xy + ¼y²

Hence,

(x + ½y)² = x² + xy + ¼y²

(vi) 1x +1y²

Solution:

Given:

(1/x + 1/y)²

Using the identity:

(a + b)² = a² + 2ab + b²

Here,

a = 1/x, b = 1/y

Therefore,

(1/x + 1/y)² = (1/x)² + 2(1/x)(1/y) + (1/y)²

= 1/x² + 2/xy + 1/y²

Hence,

(1/x + 1/y)² = 1/x² + 2/xy + 1/y²

Question 2.
Using the same identity, find the values of the following:
(i) (64)2

Solution:

(64)2
Using the identity: (a + b)2 = a2 + 2ab + b2
642 = (60 + 4)2
= 602 + 2 × 60 × 4 + 42
= 3600 + 480 + 16
= 4096

(ii) (105)2

Solution:

(105)2
Using the identity: (a + b)2 = a2 + 2ab + b2
1052 = (100 + 5)2
= 1002 + 2 × 100 × 5 + 52
= 10000+ 1000 + 25
= 11025

(iii) (205)2

Solution:

(205)2

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

2052 = (200 + 5)2

= 2002 + 2 × 200 × 5 + 52

= 40000 + 2000 + 25

= 42025

Why Use Vedantu’s Class 9 Maths Chapter 4 Exercise 4.1 Solutions?

Class 9 Maths Chapter 4 Exercise 4.1 Solutions make it easier to understand the questions and follow the correct steps while solving them. Students can use these solutions to:

• Understand each step: Follow clear, step-by-step methods for solving Exercise 4.1 questions.

• Check your answers: Compare your working with the correct NCERT-based solutions and identify mistakes.

• Revise important concepts: Quickly review the concepts and identities used in EX 4.1, Class 9.

• Improve problem-solving: Learn how to apply mathematical identities and methods correctly.

• Prepare for exams: Practise exercise questions and build confidence with well-explained NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.1.

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