NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities

Infinity Learn provides NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities, covering Exercise Sets 4.1, 4.2, 4.3, 4.4, and the End-of-Chapter Exercises for the 2026-27 exams. Prepared by expert teachers, these solutions are designed to help students understand every concept clearly and confidently. NCERT Solutions for Class 9 Maths chapter 4 textbook is an important and concept-rich chapter in the new 2026-27 curriculum. It builds on students’ earlier knowledge of linear polynomials and linear equations and introduces algebraic identities special mathematical equalities that remain true for all values of the variables involved. Unlike ordinary equations, which are true only for specific values, algebraic identities represent universal truths in algebra. With step-by-step explanations and clear problem-solving methods, Infinity Learn helps students explore these identities through reasoning, patterns, and structured practice.

NCERT Solutions for Class 9 Maths Chapter 4 Exploring Algebraic Identities

Our NCERT Solutions for Class 9 Maths Chapter 4, Exploring Algebraic Identities, are designed to provide clear, simple, and syllabus-based explanations for students. Prepared by subject experts, these solutions help students understand algebraic identities, factorisation, and related concepts through step-by-step methods, visual learning techniques, and easy examples.

This chapter introduces students to algebraic identities and their use in simplifying and solving algebraic expressions. Students learn how to visualise identities, factorise expressions using identities and algebra tiles, and identify new identities through patterns. The chapter also explains factorisation methods without algebra tiles and covers techniques for simplifying rational expressions. With regular practice and conceptual understanding, students can build strong algebraic reasoning and improve their problem-solving skills.

Topics Covered:

• Introduction
• Visualising Identities
• Factorisation of Algebraic Expressions Using Identities
• More Identities
• Factorisation Using Algebra Tiles
• Factorisation Without Using Algebra Tiles
• Finding New Identities
• Simplifying Rational Expressions

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Ganita Manjari Class 9 Chapter 4 Solutions Exploring Algebraic Identities

 Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.1 Solutions

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

(i) (7x + 4y)²

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

Here, a = 7x and b = 4y

= (7x)² + 2(7x)(4y) + (4y)²

= 49x² + 56xy + 16y²

(ii) [(7/5)x + (3/2)y]²

Solution: Here, a = (7/5)x and b = (3/2)y

= [(7/5)x]² + 2 × (7/5)x × (3/2)y + [(3/2)y]²

= (49/25)x² + 2 × (21/10)xy + (9/4)y²

= (49/25)x² + (42/10)xy + (9/4)y²

Simplify:

= (49/25)x² + (21/5)xy + (9/4)y²

(iii) (2.5p + 1.5q)²

Solution: Here, a = 2.5p and b = 1.5q

= (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²

= 6.25p² + 7.5pq + 2.25q²

(iv) [(3/4)s + 8t]²

Solution:

Here, a = (3/4)s and b = 8t

= [(3/4)s]² + 2 × (3/4)s × 8t + (8t)²

= (9/16)s² + 2 × (24/4)st + 64t²

= (9/16)s² + 12st + 64t²

(v) [x + 1/(2y)]²

Solution:

Here, a = x and b = 1/(2y)

= x² + 2 × x × [1/(2y)] + [1/(2y)]²

= x² + (2x)/(2y) + 1/(4y²)

= x² + x/y + 1/(4y²)

(vi) ( 1/x + 1/y )²

Solution:

Here, a = 1/x and b = 1/y

= (1/x)² + 2(1/x)(1/y) + (1/y)²

= 1/x² + 2/(xy) + 1/y²

2. Using the identity (a + b)² = a² + 2ab + b², find the values of the following:

(i) (64)²

Solution:

Write 64 = 60 + 4

Using (a + b)² = a² + 2ab + b²:

(60 + 4)² = 60² + 2 × 60 × 4 + 4²

= 3600 + 480 + 16

= 4096

(ii) (105)²

Solution:

Writing 105 = 100 + 5

Now (105)² = (100 + 5)²

= 100² + 2 × 100 × 5 + 5² [Using (a + b)² = a² + 2ab + b²]

= 10000 + 1000 + 25

= 11025

(iii) (205)²

Solution:

Write 205 = 200 + 5

So, (205)² = (200 + 5)²

= 200² + 2 × 200 × 5 + 5² [Using (a + b)² = a² + 2ab + b²]

= 40000 + 2000 + 25

= 42025

Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.2 Solutions

1. Factor completely:

(i) 9x² + 24xy + 16y²

Solution:

We know that:

9x² = (3x)²

16y² = (4y)²

24xy = 2 × 3x × 4y

So, 9x² + 24xy + 16y²

= (3x)² + 2 × 3x × 4y + (4y)²

= (3x + 4y)² [Using a² + 2ab + b² = (a + b)²]

(ii) 4s² + 20st + 25t²

Solution:

We can write:

4s² = (2s)²

25t² = (5t)²

20st = 2 × 2s × 5t

So, 4s² + 20st + 25t²

= (2s)² + 2 × 2s × 5t + (5t)²

= (2s + 5t)² [Using a² + 2ab + b² = (a + b)²]

(iii) 49x² + 28xy + 4y²

Solution:

49x² = (7x)²

4y² = (2y)²

28xy = 2 × 7x × 2y

So, 49x² + 28xy + 4y²

= (7x)² + 2 × 7x × 2y + (2y)²

= (7x + 2y)² [Using a² + 2ab + b² = (a + b)²]

(iv) 64p² + (32/3)pq + (4/9)q²

Solution:

64p² = (8p)²

(4/9)q² = (2/3 q)²

Middle term:

2 × 8p × (2/3 q) = 32/3 pq

So, 64p² + (32/3)pq + (4/9)q²

= (8p)² + 2 × 8p × (2/3 q) + (2/3 q)²

= (8p + 2/3 q)² [Using a² + 2ab + b² = (a + b)²]

(v) 3a² + 4ab + (4/3)b²

Solution:

Take common factor 3:

= 3[a² + (4/3)ab + (4/9)b²]

Now, a² = (a)²

(4/9)b² = (2/3 b)²

(4/3)ab = 2 × a × (2/3 b)

So, 3a² + 4ab + (4/3)b²

= 3[a² + (4/3)ab + (4/9)b²]

= 3[a² + 2 × a × (2/3 b) + (2/3 b)²]

= 3(a + 2/3 b)² [Using a² + 2ab + b² = (a + b)²]

(vi) (9/5)s² + 6sv + 5v²

Solution:

Take common factor 1/5:

= (1/5)[9s² + 30sv + 25v²]

Now, 9s² = (3s)²

25v² = (5v)²

30sv = 2 × 3s × 5v

So, (9/5)s² + 6sv + 5v²

= (1/5)[9s² + 30sv + 25v²]

= (1/5)[(3s)² + 2 × 3s × 5v + (5v)²]

= (1/5)(3s + 5v)² [Using a² + 2ab + b² = (a + b)²]

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

(i) (79)²

Solution:

79 = 80 − 1

= (80 − 1)²

= 80² − 2×80×1 + 1² [Using (a − b)² = a² − 2ab + b²]

= 6400 − 160 + 1

= 6241

(ii) (193)²

Solution:

193 = 200 − 7

= (200 − 7)²

= 200² − 2×200×7 + 7² [Using (a − b)² = a² − 2ab + b²]

= 40000 − 2800 + 49

= 37249

(iii) (299)²

Solution:

299 = 300 − 1

= (300 − 1)²

= 300² − 2×300×1 + 1² [Using (a − b)² = a² − 2ab + b²]

= 90000 − 600 + 1

= 89401

Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.3 Solutions

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

(i) 117²

Solution:

117 = 110 + 7

So, 117² = (110 + 7)²

= 110² + + 2(110)(7) + 7² [Using (a + b)² = a² + 2ab + b²]

= 12100 + 1540 + 49

= 13689

(ii) 78²

Solution:

78 = 80 – 2

So, 78² = (80 – 2)²

= 80² – 2(80)(2) + 2² [Using (a − b)² = a² − 2ab + b²]

= 6400 – 320 + 4

= 6084

(iii) 198²

Solution:

198 = 200 – 2

So, 198² = (200 – 2)²

= 200² – 2(200)(2) + 2² [Using (a − b)² = a² − 2ab + b²]

= 40000 – 800 + 4

= 39204

(iv) 214²

Solution:

214 = 200 + 14

So, 214² = (200 + 14)²

= 200² + 2(200)(14) + 14² [Using (a − b)² = a² − 2ab + b²]

= 40000 + 5600 + 196

= 45796

(v) 1104²

Solution:

1104 = 1100 + 4

So, 1104² = (1100 + 4)²

= 1100² + 2(1100)(4) + 4² [Using (a + b)² = a² + 2ab + b²]

= 1210000 + 8800 + 16

= 1218816

(vi) 1120²

Solution:

1120 = 1100 + 20

So, 1120² = (1100 + 20)²

= 1100² + 2(1100)(20) + 20² [Using (a + b)² = a² + 2ab + b²]

= 1210000 + 44000 + 400

= 1254400

2. Factor using suitable identities:

(i) 16y² – 24y + 9

Solution:

16y² = (4y)²

9 = 3²

-24y = -2(4y)(3)

Therefore, 16y² – 24y + 9

= (4y)² -2(4y)(3) + 3²

= (4y – 3)² [Using a² – 2ab + b² = (a – b)²]

(ii) (9/4)s² + 6st + 4t²

Solution:

(9/4)s² = [(3s)/2]²

4t² = (2t)²

6st = 2 × (3s/2) × (2t)

Therefore, (9/4)s² + 6st + 4t²

= [(3s)/2]² + 2 × (3s/2) × (2t) + (2t)²

= [(3s)/2 + 2t]² [Using a² + 2ab + b² = (a + b)²]

(iii) m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²

Solution:

Grouping the terms as:

m²/9 + mk/3 + k²/4 + 2mn + 3nk + 9n²

We have:

m²/9 = (m/3)²

k²/4 = (k/2)²

9n² = (3n)²

Now checking the cross terms:

2 × (m/3) × (k/2) = mk/3

2 × (m/3) × (3n) = 2mn

2 × (k/2) × (3n) = 3nk

So, m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²

= m²/9 + k²/4 + 9n² + + mk/3 + 3nk + 2mn [ Rearranging the terms]

= (m/3)² + (k/2)² + (3n)² + 2 × (m/3) × (k/2) + 2 × (m/3) × (3n) + 2 × (k/2) × (3n)

= (m/3 + k/2 + 3n)² [Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]

(iv) p²/16 – 2 + 16/p²

Solution:

Writing -2 as:

-2 = -2 × (p/4) × (4/p)

Now, p²/16 = (p/4)²

16/p² = (4/p)²

So, p²/16 – 2 + 16/p²

= (p/4)² – 2(p/4)(4/p) + (4/p)²

= (p/4 – 4/p)² [Using a² – 2ab + b² = (a – b)²]

(v) 9a² + 4b² + c² – 12ab + 6ac – 4bc

Solution:

We have:

9a² = (3a)²

4b² = (- 2b)² [Here 4b² = (- 2b)² as in two negative terms – 12ab and –

4bc, b is common]

c² = c²

Now, 9a² + 4b² + c² – 12ab + 6ac – 4bc

= (3a)² + (- 2b)² + c² + 2(3a)(- 2b) + 2(3a)(c) + 2(- 2b)(c)

= (3a – 2b + c)² [Using x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)²]

3. Expand the following using the identity (a + b + c)² = a² +

b² + c² + 2ab + 2bc + 2ca:

(i) (p + 3q + 7r)²

Solution:

Here, a = p, b = 3q, c = 7r

Using the identity:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

So, (p + 3q + 7r)²

= p² + (3q)² + (7r)² + 2(p)(3q) + 2(3q)(7r) + 2(p)(7r)

= p² + 9q² + 49r² + 6pq + 42qr + 14pr

(ii) (3x − 2y + 4z)²

Solution:

Write it as:

[3x + (−2y) + 4z]²

Here,

a = 3x, b = −2y, c = 4z

Using the identity:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

So, (3x − 2y + 4z)²

= (3x)² + (−2y)² + (4z)² + 2(3x)(−2y) + 2(−2y)(4z) + 2(3x)(4z)

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

4. Is this an identity? (a + b − c)² + (a − b + c)² + (a − b − c)² = 2a² + 2b² + 2c²

Solution:

To check whether this is an identity, expand the left-hand side.

First Part: (a + b − c)²

= a² + b² + c² + 2ab − 2ac − 2bc

Second Part: (a − b + c)²

= a² + b² + c² − 2ab + 2ac − 2bc

Third Part: (a − b − c)²

= a² + b² + c² − 2ab − 2ac + 2bc

Now adding all three:

LHS = (a² + b² + c² + 2ab − 2ac − 2bc)

+ (a² + b² + c² − 2ab + 2ac − 2bc)

+ (a² + b² + c² − 2ab − 2ac + 2bc)

= 3a² + 3b² + 3c² − 2ab − 2ac − 2bc

This is NOT equal to 2a² + 2b² + 2c²

Hence, the given statement is not true for all values of a, b and c.

So, it is not an identity.

Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.4 Solutions

1. Fill in the blanks to complete the following identities:

(i) s² − 11s + 24 = ( ____ )( ____ )

Solution:

We need two numbers whose:

Sum = 11 and Product = 24

These are 3 and 8.

So, s² − 11s + 24

= s² − 8s – 3s + 24

= s(s − 8) – 3(s − 8)

= (s − 3)(s − 8)

So, s² − 11s + 24 = = (s − 3)(s − 8)

(ii) ( ____ )(x + 1) = (3x² − 4x − 7)

Solution:

RHS: 3x² − 4x − 7

= 3x² − 7x + 3x − 7

= x(3x − 7) + 1(3x – 7)

= (3x − 7)(x + 1)

So, (3x – 7)(x + 1) = (3x² − 4x − 7)

(iii) 10x² − 11x − 6 = (2x − ____)( ____ + 2)

Solution:

LHS = 10x² − 11x − 6

= 10x² − 15x + 4x − 6

= 5x(2x − 3) + 2(2x − 3)

= (5x + 2)(2x − 3)

(2x − 3)(5x + 2)

So, 10x² − 11x − 6 = (2x − 3)(5x + 2)

(iv) 6x² + 7x + 2 = ( ____ )( ____ )

Solution:

= 6x² + 3x + 4x + 2

= 3x(2x + 1) + 2(2x + 1)

= (3x + 2)(2x + 1)

So, 6x² + 7x + 2 = = (3x + 2)(2x + 1)

2. Select and use the identity that will help you to find the following products without multiplying directly:

(i) (41)²

Solution:

41 = 40 + 1

= (40 + 1)²

= 40² + 2 × 40 × 1 + 1² [Using (a + b)² = a² + 2ab + b²]

= 1600 + 80 + 1

= 1681

(ii) (27)²

Solution:

27 = 30 − 3

So, (27)² = (30 − 3)²

= (30)² – 2 × 30 × 3 + (3)² [Using (a – b)² = a² – 2ab + b²]

= 900 − 180 + 9

= 729

(iii) (23 × 17)

Solution:

Use identity:

23 × 17 = (20 + 3)(20 − 3)

= 20² − 3² [Using (a + b)(a − b) = a² − b²]

= 400 − 9

= 391

(iv) (135)²

Solution:

135 = 100 + 35

So, (135)²

= (100 + 35)²

= (100)² + 2 × 100 × 35 + (35)² [Using (a + b)² = a² + 2ab + b²]

= 10000 + 7000 + 1225

= 18225

(v) (97)²

Solution:

97 = 100 − 3

So, (97)²

= (100 − 3)²

= (100)² – 2 × 100 × 3 + (3)² [Using (a – b)² = a² – 2ab + b²]

= 10000 − 600 + 9

= 9409

(vi) (18 × 29)

Solution:

18 × 29 =

= (20 − 2)(20 + 9)

= 20² + (− 2 + 9) × 20 − 18 [Using (x + a)(x + b) = x² + (a + b)x + a × b]

= 400 + 140 − 18

= 522

(vii) (34 × 43)

Solution:

= (38 − 4)(38 + 5)

= 38² + (− 4 + 5) × 38 − 20 [Using (x + a)(x + b) = x² + (a + b)x + a × b]

= 38² + 38 − 20

= 1444 + 38 − 20

= 1462

(viii) (205)²

Solution:

205 = 200 + 5

= (200 + 5)²

= (200)² + 2 × 200 × 5 + (5)² [Using (a + b)² = a² + 2ab + b²]

= 40000 + 2000 + 25

= 42025

3. Factor the following:

(i) 9a² + b² + 4c² − 6ab + 12ac − 4bc

Solution:

9a² + b² + 4c² − 6ab + 12ac − 4bc

= (3a)² + (b)² + (2c)² + 2(3a)(−b) + 2(−b)(2c) + 2(3a)(2c)

= (3a − b + 2c)² [Using x² + y² + z² + 2xy + 2xz + 2yz = (x + y + z)²]

(ii) 16s² + 25t² − 40st

Solution:

Now, 16s² + 25t² − 40st

= 16s² − 40st + 25t² [Rearranging the terms]

= (4s)² −2(4s)(5t) + (5t)²

= (4s − 5t)² [Using a² − 2ab + b² = (a − b)²]

(iii) r² − r − 42

Solution:

We need two numbers whose product = −42 and Sum = −1.

These numbers are −7 and 6.

So, r² − r − 42

= r² − 7r + 6r − 42

= r(r − 7) + 6(r − 7)

= (r − 7)(r + 6)

(iv) 49g² + 14gh + h²

Solution:

49g² + 14gh + h²

= (7g)² + 2(7g)(h) + (h)²

= (7g + h)² [Using a² + 2ab + b² = (a + b)²]

(v) 64u² + 121v² + 4w² − 176uv − 32uw + 44vw

Solution:

64u² + 121v² + 4w² − 176uv − 32uw + 44vw

= 64u² + 121v² + 4w² − 176uv + 44vw − 32uw [Rearranging the terms]

= (8u)² + (11v)² + (2w)² + 2(8u)(−11v) + 2(−11v)(−2w) + 2(8u)(−2w)

= (8u − 11v − 2w)² [Using x² + y² + z² + 2xy + 2xz + 2yz = = (x + y +z)²

Class 9 Maths Ganita Manjari Chapter 4 Exercise Set 4.5 Solutions

1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:

(i) (3p² − 3pq − 18q²)/(p² + 3pq − 10q²)

Solution:

First factorising the numerator:

3p² − 3pq − 18q²

= 3(p² − pq − 6q²)

= 3[p² − 3pq + 2pq − 6q²]

= 3[p(p − 3q) + 2q(p – 3q)

= 3(p − 3q)(p + 2q)

Now factorising the denominator:

p² + 3pq − 10q²

= p² + 5pq – 2pq − 10q²

= p(p + 5q) – 2q(p + 5q)

= (p + 5q)(p − 2q)

So, (3p² − 3pq − 18q²)/(p² + 3pq − 10q²)

= 3(p − 3q)(p + 2q)/[(p + 5q)(p − 2q)]

No common factor cancels.

(ii) (n³ − 3n²m + 3nm² − m³)/(5m² − 10mn + 5n²)

Solution:

Factorising the numerator using identity:

n³ − 3n²m + 3nm² − m³

= (n − m)³ [Using a³ − 3a²b + 3ab² − b³ = (a − b)³]

Factorising the denominator:

5m² − 10mn + 5n²

= 5(m² − 2mn + n²)

= 5(m − n)² [Using a² – 2ab + b² = (a − b)²]

Now, (n − m)³/[5(m − n)²]

= −(m − n)³/[5(m − n)²] [Since, (n − m) = −(m − n), so, (n − m)³ = −(m −

n)³]

= −(m − n)/5

(iii) (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx)

Solution:

Factorising the numerator:

w³ − v³ + x³ + 3wvx

= w³ + (- v)³ + x³ – 3w(- v)x

= (w − v + x)(w² + v² + x² + wv + vx − wx)

[Using a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)]

Now denominator:

w² + v² + x² − 2wv − 2vx + 2wx

= w² + (- v)² + x² + 2w(-v) + 2(−v)x + 2xw)

= (w − v + x)² [Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]

Therefore, (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx)

= [(w − v + x)(w² + v² + x² + wv + vx − wx)]/(w − v + x)²

= (w² + x² + v² − wx + vx − wv) / (w + x − v) [Canceling the common factor]

(iv) (4y² − 20yz + 25z²)/(25z² − 4y²)

Solution:

Factor numerator:

4y² − 20yz + 25z²

= (2y)² − 2(2y)(5z) + (5z)²

= (2y − 5z)² [Using a² – 2ab + b² = (a − b)²]

Factorising the denominator:

25z² − 4y²

= (5z − 2y)(5z + 2y) [Using a² – b² = (a − b)(a + b)]

So, (4y² − 20yz + 25z²)/(25z² − 4y²)

= (5z − 2y)²/[(5z − 2y)(5z + 2y)] [Since (2y − 5z)² = (−1)²(5z − 2y)² =

(5z − 2y)²]

= (5z − 2y)/(5z + 2y)

(v) [(x² + x − 6)(x² − 7x + 12)]/[(x² − 6x + 8)(x² − 9)]

Solution:

Factorising each polynomial:

x² + x − 6 = x² + 3x – 2x − 6 = x(x + 3) – 2(x + 3) = (x + 3)(x − 2)

x² − 7x + 12 = x² – 3x – 4x + 12 = x(x – 3) – 4(x – 3) = (x − 3)(x − 4)

x² − 6x + 8 = x² – 4x – 2x + 8 = x(x – 4) – 2(x – 4) = (x − 2)(x − 4)

x² − 9 = x² − 3 = (x − 3)(x + 3)

Substituting each polynomial, we get:

[(x + 3)(x − 2)][(x − 3)(x − 4)]/[(x − 2)(x − 4)][(x − 3)(x + 3)]

= 1 [Since all factors cancel]

(vi) (p⁴ − 16) / (p² − 4p + 4)

Solution:

Factorising numerator:

p⁴ − 16

= (p² − 4)(p² + 4) [Using a² − b² = (a + b)(a − b)]

= (p − 2)(p + 2)(p² + 4) [Again using a² − b² = (a + b)(a − b)]

Factorising denominator:

p² − 4p + 4

= p² − 2(p)(2) + 2²

= (p − 2)² [Using a² – 2ab + b² = (a – b)²]

So, [(p − 2)(p + 2)(p² + 4)]/(p − 2)²

= (p + 2)(p² + 4) / (p − 2) [Cancelling common factor (p − 2)].

End-of-Chapter Exercises

1. Use suitable identities to find the following products:

(i) (−3x + 4)²

Solution:

(−3x + 4)²

= (−3x)² + 2(−3x)(4) + 4² [Using the identity (a + b)² = a² + 2ab + b²]

= 9x² − 24x + 16

(ii) (2s + 7)(2s − 7)

Solution:

(2s + 7)(2s − 7)

= (2s)² − 7² [Using the identity (a + b)(a − b) = a² − b²]

= 4s² − 49

(iii) (p² + 1/2)(p² − 1/2)

Solution:

(p² + 1/2)(p² − 1/2)

= (p²)² − (1/2)² [Using the identity (a + b)(a − b) = a² − b²]

= p⁴ − 1/4

(iv) (2n + 7)(2n − 7)

Solution:

(2n + 7)(2n − 7)

= (2n)² − 7² [Using the identity (a + b)(a − b) = a² − b²]

= 4n² − 49

(v) (s − 2t)(s² + 2st + 4t²)

Solution:

(s − 2t)(s² + 2st + 4t²)

= s³ − (2t)³ [Using the identity (a − b)(a² + ab + b²) = a³ − b³]

= s³ − 8t³

(vi) [1/(2r) − 4r]²

Solution:

[1/(2r) − 4r]²

= (1/(2r))² − 2(1/(2r))(4r) + (4r)² [Using the identity (a − b)² = a² − 2ab +

b²]

= 1/(4r²) − 4 + 16r²

(vii) (−3m + 4k − l)²

Solution:

So, (−3m + 4k − l)²

= (−3m)² + (4k)² + (−l)² + 2(−3m)(4k) + 2(4k)(−l) + 2(−3m)(−l)

[Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

= 9m² + 16k² + l² − 24mk − 8kl + 6ml

(viii) (x − 1/3 y)³

Solution:

(x − 1/3 y)³

= x³ − 3x²(y/3) + 3x(y/3)² − (y/3)³

[Using the identity (a − b)³ = a³ − 3a²b + 3ab² − b³]

= x³ − x²y + 3x(y²/9) − y³/27

= x³ − x²y + (1/3)xy² − y³/27

(ix) (7/2 k − 2/3 m)³

Solution:

[Using the identity (a − b)³ = a³ − 3a²b + 3ab² − b³]

Here, a = 7k/2, b = 2m/3

So, a³ = (7k/2)³ = 343k³/8

3a²b = 3 × (49k²/4) × (2m/3) = 49k²m/2

3ab² = 3 × (7k/2) × (4m²/9) = 14km²/3

b³ = (2m/3)³ = 8m³/27

Therefore, (7/2 k − 2/3 m)³

= 343k³/8 − 49k²m/2 + 14km²/3 − 8m³/27.

2. Find the values using suitable identities:

(i) 17 × 21

Solution:

Using identity: (a − b)(a + b) = a² − b²

We have 17 × 21 = (19 − 2)(19 + 2)

= 19² − 2²

= 361 − 4

= 357

(ii) 104 × 96

Solution:

Using identity: (a + b)(a − b) = a² − b²

We have 104 × 96 = (100 + 4)(100 − 4)

= 100² − 4²

= 10000 − 16

= 9984

(iii) 24 × 16

Solution:

Using identity: (a + b)(a − b) = a² − b²

We have 24 × 16 = (20 + 4)(20 − 4)

= 20² − 4²

= 400 − 16

= 384

(iv) 147³

Solution:

We know that 147 = 150 − 3

So, 147³ = (150 − 3)³

= 150³ − 3 × 150² × 3 + 3 × 150 × 3² − 3³ [Using identity (a + b)³ = a³ +

3a²b + 3ab² + b³]

= 3375000 − 202500 + 4050 − 27

= 3176523

(v) 199³

Solution:

199 = 200 − 1

So, 199³ = (200 − 1)³

= 200³ − 3 × 200² × 1 + 3 × 200 × 1² − 1 [Using identity: (a − b)³ = a³ −

3a²b + 3ab² − b³]

= 8000000 − 120000 + 600 − 1

= 7880599

(vi) 127³

Solution:

Here, 127 = 130 − 3

So, 127³ = (130 − 3)³

= 130³ − 3 × 130² × 3 + 3 × 130 × 3² − 3³ [Using identity: (a − b)³ = a³ −

3a²b + 3ab² − b³]

= 2197000 − 152100 + 3510 − 27

= 2048383

(vii) (−107)³

Solution:

(−107)³ = −(107³)

Now, 107 = 100 + 7

107³ = (100 + 7)³

= 100³ + 3 × 100² × 7 + 3 × 100 × 7² + 7³ [Using identity: (a + b)³ = a³ +

3a²b + 3ab² + b³]

= 1000000 + 210000 + 14700 + 343

= 1225043

So, (−107)³ = −1225043

(viii) (−299)³

Solution:

(−299)³ = −(299³)

Here, 299 = 300 − 1

So, 299³ = (300 − 1)³

= 300³ − 3 × 300² × 1 + 3 × 300 × 1² − 1 [Using identity: (a − b)³ = a³ −

3a²b + 3ab² − b³]

= 27000000 − 270000 + 900 − 1

= 26730900 − 1

= 26730899

So, (−299)³ = −26730899

3. Factor the following algebraic expressions:

(i) 4y² + 1 + 1/(16y²)

Solution:

Here, we have 4y² = (2y)²

1/(16y²) = (1/4y)²

and 2 × 2y × 1/(4y) = 1

So, this is of the form: a² + 2ab + b² = (a + b)²

Therefore, 4y² + 1 + 1/(16y²)

= (2y + 1/(4y))²

(ii) 9m² − 1/(25n²)

Solution:

9m² − 1/(25n²)

= (3m)² − (1/5n)²

= (3m + 1/5n)(3m − 1/5n) [Using a² − b² = (a + b)(a − b)]

(iii) 27b³ − 1/(64b³)

Solution:

27b³ − 1/(64b³)

= [3b − 1/(4b)][(3b)² + (3b)(1/4b) + (1/4b)²] [Using a³ − b³ = (a − b)(a² +

ab + b²)]

= (3b − 1/4b)[9b² + 3/4 + 1/16b²]

(iv) x² + 5x/6 + 1/6

Solution:

We need two numbers whose sum is 5/6 and product is 1/6.

These numbers are 1/2 and 1/3.

So, x² + 5x/6 + 1/6

= x² + x/2 + x/3 + 1/6 [Splitting the middle term]

= x(x + 1/2) + 1/3(x + 1/2)

= (x + 1/3)(x + 1/2)

(v) 27u³ − 1/125 − 27u²/5 + 9u/25

Solution:

Given expression:

27u³ − 1/125 − 27u²/5 + 9u/25

= 27u³ − 27u²/5 + 9u/25 − 1/125 [Rearranging the terms]

= (3u)³ − 3(3u)²(1/5) + 3(3u)(1/5)² − (1/5)³

= (3u − 1/5)³ [Using a³ − 3a²b + 3ab² − b³ = (a − b)³]

(vi) 64y³ + 1/125 z³

Solution:

64y³ + 1/125 z³

= (4y)³ + (z/5)³

= (4y + z/5)[(4y)² − (4y)(z/5) + (z/5)²] [Using a³ + b³ = (a + b)(a² − ab +

b²)]

= (4y + z/5)(16y² − 4yz/5 + z²/25)

(vii) p³ + 27q³ + r³ − 9pqr

Solution:

p³ + 27q³ + r³ − 9pqr

= p³ + (3q)³ + r³ − 3(p)(3q)(r)

= (p + 3q + r)[p² + (3q)² + r² − p(3q) − (3q)r − pr)]

[Using a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)]

= (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − pr)

(viii) 9m² − 12m + 4

Solution:

9m² − 12m + 4

= (3m)² − 2(3m)(2) + (2)²

= (3m – 2)² [Using a² – 2ab + b² = (a – b)²]

(ix) 9x³ − 8/3 y³ + z³/3 + 6xyz

Solution:

9x³ − 8/3 y³ + z³/3 + 6xyz

= (1/3)[9x³ − 8y³ + z³ + 18xyz]

= (1/3)[(3x)³ + (- 2y)³ + z³ – 3(3x)(-2y)z]

= (1/3)(3x – 2y + z)[(3x)² + (-2y)² + z² − (3x)(-2y) − (-2y)(z) − (z)(3x)]

[Using a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)]

= (1/3)(3x – 2y + z)[9x² + 4y² + z² + 6xy + 2yz − 3zx]

(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy

Solution:

4x² + 9y² + 36z² + 12xz + 36yz + 24xy

= 4x² + 9y² + 36z² + 24xy + 36yz + 12xz

= (2x)² + (3y)² + (6z)² + 2(2x)(3y) + 2(3y)(6z) + 2(2x)(6z)

= (2x + 3y + 6z)² [Using a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]

(xi) 27u³ − 1/216 − 9u²/2 + u/4

Solution:

27u³ − 1/216 − 9u²/2 + u/4

= 27u³ − 9u²/2 + u/4 − 1/216 [Rearranging the terms]

= (3u)³ − 3(3u)²(1/6) + 3(3u)(1/6)² − (1/6)³

= (3u − 1/6)³ [Using a³ − 3a²b + 3ab² − b³ = (a − b)³].

4. Simplify the following:

Note: Assume that the denominators are not equal to 0.

(i) (4x² + 4x + 1)/(4x² − 1)

Solution:

Factorising numerator:

4x² + 4x + 1

= (2x)² + 2(2x)(1) + 1²

= (2x + 1)² [Using a² + 2ab + b² = (a + b)²]

Factorising denominator:

4x² − 1 = (2x + 1)(2x − 1)

Now the expression: (4x² + 4x + 1)/(4x² − 1)

= (2x + 1)²/[(2x + 1)(2x − 1)]

= (2x + 1)/(2x − 1)

(ii) 9(3a³ − 24b³)/(9a² − 36b²)

Solution:

First simplify:

9(3a³ − 24b³) = 27(a³ − 8b³)

and 9a² − 36b² = 9(a² − 4b²)

So, 9(3a³ − 24b³)/(9a² − 36b²)

= 3(a³ − 8b³)/(a² − 4b²)

= 3(a − 2b)(a² + 2ab + 4b²)/(a² − 4b²) [Since a³ − 8b³ = a³ − (2b)³ = (a −

2b)(a² + 2ab + 4b²)]

= 3(a − 2b)(a² + 2ab + 4b²)/(a − 2b)(a + 2b) [Since a² − 4b² = (a − 2b)(a

+ 2b)]

= 3(a² + 2ab + 4b²)/(a + 2b) [Cancelling the common factor (a − 2b)]

(iii) (s³ + 125t³)/(s² − 2st − 35t²)

Solution:

Factorising numerator:

s³ + 125t³ = s³ + (5t)³

= (s + 5t)(s² − 5st + 25t²) [Using a³ + b³ = (a + b)(a² – ab + b²)]

Factorising denominator:

s² − 2st − 35t²

s² − 7st + 5st − 35t²

= s(s – 7t) + 5t(s − 7t)

= (s + 5t)(s − 7t)

Now the given expression:

(s³ + 125t³)/(s² − 2st − 35t²)

= (s + 5t)(s² − 5st + 25t²)/(s + 5t)(s − 7t)

= (s² − 5st + 25t²)/(s − 7t) [Cancelling common factor (s + 5t)].

5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.

(i) 25a² − 30ab + 9b²

Solution:

This is a perfect square:

25a² − 30ab + 9b²

= (5a)² − 2(5a)(3b) + (3b)²

= (5a − 3b)² [Using a² – 2ab + b² = (a – b)²]

So possible length and breadth are (5a − 3b) and (5a − 3b).

(ii) 36s² − 49t²

Solution:

36s² − 49t²

= (6s)² − (7t)²

= (6s + 7t)(6s − 7t) [Using a² – b² = (a + b)(a – b)]

So possible length and breadth are (6s + 7t) and (6s − 7t).

6. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.

(i) 6a² − 24b²

Solution:

Take common factor:

6a² − 24b²

= 6(a² − 4b²)

= 6[a² − (2b)²]

= 6(a + 2b)(a − 2b) [Using a² – b² = (a + b)(a – b)]

So possible dimensions are 6, (a + 2b) and (a − 2b).

(ii) 3ps² − 15ps + 12p

Solution:

3ps² − 15ps + 12p

= 3p(s² − 5s + 4) [Taking common factor]

= 3p(s² − 4s – 1s + 4)

= 3p[s(s − 4) – 1(s − 4)]

= 3p(s − 1)(s − 4)

So, the possible dimensions are 3p, (s − 1) and (s − 4).

7. The village playground is shaped as a square of side 40

metres. A path of width s metres is created around the

playground for people to walk. Find an expression for the

area of the path in terms of s.

Solution:

Side of playground = 40 m

Since a path of width s metres is made all around the outside, the side of

the outer square becomes:

40 + 2s

Area of outer square = (40 + 2s)²

Area of playground = 40² = 1600

Therefore, area of the path

= (40 + 2s)² − 1600

= (1600 + 160s + 4s²) − 1600

= 4s² + 160s

Hence, the required expression: Area of path = 4s² + 160s square metres.

8. If a number plus its reciprocal equals 10/3, find the number.

Solution:

Let the number be x.

Then, according to question: x + 1/x = 10/3

Multiply both sides by 3x:

3x² + 3 = 10x

⇒ 3x² − 10x + 3 = 0

⇒ 3x² − 9x − x + 3 = 0

⇒ 3x(x − 3) − 1(x − 3) = 0

⇒ (3x − 1)(x − 3) = 0

So, 3x − 1 = 0 or x − 3 = 0

Hence, x = 1/3 or x = 3

Therefore, the number is 3 or 1/3.

9. A rectangular pool has area 2x² + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.

Solution:

Area of pool = 2x² + 7x + 3

Width = 2x + 1

Length = Area/Width

= (2x² + 7x + 3)/(2x + 1)

= (2x² + 6x + x + 3)/(2x + 1)

= [2x(x + 3) + 1(x + 3)]/(2x + 1)

= (2x + 1)(x + 3)/(2x + 1)

= x + 3

Therefore, the length of the pool is x + 3 hastas.

10. If both x − 2 and x − 1/2 are factors of px² + 5x + r, show that p = r.

Solution:

Since x − 2 and x − 1/2 are factors, the quadratic polynomial can be

written as:

px² + 5x + r = k(x − 2)(x − 1/2)

⇒ (x − 2)(x − 1/2) = k[x² − (5/2)x + 1]

⇒ x² − (1/2)x − 2x + 1 = k[x² − (5/2)x + 1]

⇒ x² − (5/2)x + 1 = k[x² − (5/2)x + 1]

Comparing the coefficients on both sides, we get:

Coefficient of x² gives: k = p

Constant term gives: k = r

Therefore, p = r

Hence proved.

11. If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ − 3abc = −25.

Solution:

Given:

a + b + c = 5

ab + bc + ca = 10

First finding a² + b² + c² using:

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

So, 5² = a² + b² + c² + 2(10)

⇒ 25 = a² + b² + c² + 20

⇒ a² + b² + c² = 5

Now, a² + b² + c² − ab − bc − ca = 5 − 10 = −5

Using the identity: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc

− ca)

a³ + b³ + c³ − 3abc

= (5)(−5)

= −25

Hence proved a³ + b³ + c³ − 3abc = −25.

12. By factoring the expression, check that n³ − n is always divisible by 6 for all natural numbers n. Give reasons.

Solution:

We have: n³ − n

= n(n² − 1)

= n(n − 1)(n + 1)

So, n³ − n = n(n − 1)(n + 1)

Thus, n³ − n is the product of three consecutive natural numbers: (n − 1),

n, (n + 1)

Among any three consecutive natural numbers:

i. one is always divisible by 3

ii. at least one is always even, so divisible by 2

Therefore, their product is always divisible by: 2 × 3 = 6

Hence, n³ − n is always divisible by 6 for all natural numbers n.

13. Find the value of:

(i) x³ + y³ − 12xy + 64, when x + y = −4

Solution:

Since x + y = −4, we have:

(x + y)³ = (−4)³ = −64

Using identity: (x + y)³ = x³ + y³ + 3xy(x + y)

So,

−64 = x³ + y³ + 3xy(−4)

⇒ −64 = x³ + y³ − 12xy

Therefore, x³ + y³ − 12xy = −64

Now, x³ + y³ − 12xy + 64 = −64 + 64 = 0

Hence, the value is 0.

(ii) x³ − 8y³ − 36xy − 216, when x − 2y + 6 = 0

Solution:

Given: x − 2y + 6 = 0

So, x − 2y = −6

Now use the identity: (a − b)³ = a³ − 3a²b + 3ab² − b³

We have: x³ − 8y³ − 36xy − 216

= x³ − 8y³ + 3(x)(−2y)(−6) + (−6)³

So it matches: a³ + b³ + c³ − 3abc

with a = x, b = −2y, c = −6

Using identity: a³ + b³ + c³ − 3abc

= (a + b + c)(a² + b² + c² − ab − bc − ca)

Thus, x³ − 8y³ − 36xy − 216

= x³ − 8y³ − 216 − 36xy

= x³ + (-2y)³ + (-6) − 3(x)(-2y)(-6)

= (x − 2y − 6)[x² + 4y² + 36² + 2xy − 12y + 18x]

But from the question:

x − 2y + 6 = 0, which gives x − 2y = −6, so x − 2y − 6 = −12, not zero.

So this expression does not become 0 under the given condition as

written.

Using x = 2y − 6 from the condition, Substitute into the expression:

x³ − 8y³ − 36xy − 216

= (2y − 6)³ − 8y³ − 36(2y − 6)y − 216

= (8y³ − 72y² + 216y − 216) − 8y³ − 72y² + 216y − 216

= −144y² + 432y − 432

= −144(y² − 3y + 3)

Therefore, under the given condition, the value is −144(y² − 3y + 3).

NCERT Solutions for Class 9 Maths Chapter-wise

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• Important Questions for CBSE Class 9 Math
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