---
title: "NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.2"
url: https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-2
dateModified: 2026-10-07T15:43:39+00:00
---

# NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.2

Chapter 4: Exploring Algebraic Identities. Every question from Exercise 4.2, with full working and the final answer.

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## Exercise Set 4.2

### Question 1

*3 marks · Short answer*

Factor completely:
(i) $9x^2 + 24xy + 16y^2$
(ii) $4s^2 + 20st + 25t^2$
(iii) $49x^2 + 28xy + 4y^2$
(iv) $64p^2 + \frac{32}{3}pq + \frac{4}{9}q^2$
*(v) $3a^2 + 4ab + \frac{4}{3}b^2$
*(vi) $\frac{9}{5}s^2 + 6sv + 5v^2$
(Hint: 2 was taken out as a common factor in Example 7. Is it possible to do something similar in Exercises (v) and (vi) above?)

**Part (i)**

1. Rewrite the expression as $(3x)^2 + 2(3x)(4y) + (4y)^2$.
2. Use the identity $(a + b)^2 = a^2 + 2ab + b^2$ where $a = 3x$ and $b = 4y$.
3. Thus, $9x^2 + 24xy + 16y^2 = (3x + 4y)^2$.

Answer (i): $$ (3x + 4y)^2 $$

**Part (ii)**

1. Rewrite the expression as $(2s)^2 + 2(2s)(5t) + (5t)^2$.
2. Use the identity $(a + b)^2 = a^2 + 2ab + b^2$ where $a = 2s$ and $b = 5t$.
3. Thus, $4s^2 + 20st + 25t^2 = (2s + 5t)^2$.

Answer (ii): $$ (2s + 5t)^2 $$

**Part (iii)**

1. Rewrite the expression as $(7x)^2 + 2(7x)(2y) + (2y)^2$.
2. Use the identity $(a + b)^2 = a^2 + 2ab + b^2$ where $a = 7x$ and $b = 2y$.
3. Thus, $49x^2 + 28xy + 4y^2 = (7x + 2y)^2$.

Answer (iii): $$ (7x + 2y)^2 $$

**Part (iv)**

1. Rewrite the expression as $\left(8p\right)^2 + 2(8p)\left(\frac{2}{3}q\right) + \left(\frac{2}{3}q\right)^2$.
2. Use the identity $(a + b)^2 = a^2 + 2ab + b^2$ where $a = 8p$ and $b = \frac{2}{3}q$.
3. Thus, $64p^2 + \frac{32}{3}pq + \frac{4}{9}q^2 = \left(8p + \frac{2}{3}q\right)^2$.

Answer (iv): $$ \left(8p + \frac{2}{3}q\right)^2 $$

**Part (v)**

1. Take $\frac{1}{3}$ as a common factor to get $\frac{1}{3}(9a^2 + 12ab + 4b^2)$.
2. Rewrite the expression inside the bracket as $(3a)^2 + 2(3a)(2b) + (2b)^2$.
3. Apply the identity $(a + b)^2 = a^2 + 2ab + b^2$ to obtain $\frac{1}{3}(3a + 2b)^2$.

Answer (v): $$ \frac{1}{3}(3a + 2b)^2 $$

**Part (vi)**

1. Take $\frac{1}{5}$ as a common factor to get $\frac{1}{5}(9s^2 + 30sv + 25v^2)$.
2. Rewrite the expression inside the bracket as $(3s)^2 + 2(3s)(5v) + (5v)^2$.
3. Apply the identity $(a + b)^2 = a^2 + 2ab + b^2$ to obtain $\frac{1}{5}(3s + 5v)^2$.

Answer (vi): $$ \frac{1}{5}(3s + 5v)^2 $$

**Answer:** Factorised expressions for all six parts.

> Common mistake: Forgetting to take out the common fractional factor in parts (v) and (vi) before applying the identity.

### Question 2

*3 marks · Short answer*

Find the values of the following using the identity $(a - b)^2 = a^2 - 2ab + b^2$.
(i) $(79)^2$
(ii) $(193)^2$
(iii) $(299)^2$

**Part (i)**

1. Given: $(79)^2$.
2. Formula: $(a - b)^2 = a^2 - 2ab + b^2$.
3. Substitution: $(80 - 1)^2 = (80)^2 - 2(80)(1) + (1)^2$.
4. Result: $6400 - 160 + 1 = 6241$.

Answer (i): $$ 6241 $$

**Part (ii)**

1. Given: $(193)^2$.
2. Formula: $(a - b)^2 = a^2 - 2ab + b^2$.
3. Substitution: $(200 - 7)^2 = (200)^2 - 2(200)(7) + (7)^2$.
4. Result: $40000 - 2800 + 49 = 37249$.

Answer (ii): $$ 37249 $$

**Part (iii)**

1. Given: $(299)^2$.
2. Formula: $(a - b)^2 = a^2 - 2ab + b^2$.
3. Substitution: $(300 - 1)^2 = (300)^2 - 2(300)(1) + (1)^2$.
4. Result: $90000 - 600 + 1 = 89401$.

Answer (iii): $$ 89401 $$

**Answer:** Calculated values for all three parts.

> Common mistake: Errors in arithmetic expansion, especially missing the subtraction of the middle term.

## Related pages

- [All Chapter 4 NCERT solutions](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions)
- [Exercise 4.1](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-1)
- [Exercise 4.3](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-3)
- [Exercise 4.4](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-4)
- [Exercise 4.5](https://www.swavid.com/maths/class/9/chapter/exploring-algebraic-identities/ncert-solutions/exercise-4-5)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student