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

Chapter 4 Exploring Algebraic Identities is among the most substantively complex topics of the new Ganita Manjari textbook 2026-27. The chapter explains the concept of algebraic identities and how they are used to simplify, expand, and factorise algebraic expressions. From scaling the skill of solving general equations to developing the ability to identify patterns, the chapter shows you why some algebraic identities hold true.

The NCERT Class 9 Maths Ganita Manjari Chapter 3 Solutions provide step-by-step answers to the questions in the chapter, covering Exercise Sets 4.1, 4.2, 4.3, 4.4 and End-of-Chapter Exercises. These ch-wise NCERT solution material not only explains identities but also visualise them Geometrically using squares, cubes, and rectangles, helping students understand rather than memorize.

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Topic Coverage in Class 9th Maths Chapter 4

Based on the new CBSE maths syllabus 2026-27, this chapter introduces various identities and demonstrates how they can be utilized in algebraic calculations. These are the main concepts introduced:

• Identity ((a + b)² = a² + 2ab + b²)
• Identity ((a - b )² = a² - 2ab +b²)
• Identity (a² - b² = (a + b)(a - b))
• Identity ((a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca)
• Identity (a³ + b³ + c³ - 3abc)
• Sum and difference of cubes
• Factorisation based on algebraic identities
• Factorisation using algebra tiles
• Factorisation by splitting the middle term
• Finding new algebraic identities
• Recognize algebraic identities by using number patterns
• Difference between an algebraic identity and an equation
• Visualise identities through squares and rectangles
• Simplifying rational algebraic expressions
• Using identities in numerical calculations and in algebraic expressions
• Problems on End of Chapter exercise based on expansion, factorisation and simplification

Exercise-wise NCERT Solutions for Class 9 Maths Chapter 4

The Class 9 Ganita Manjari Chapter 4 solutions explain each step very clearly so that students can understand the use of identities like (a+b)², (a-b)², a² - b², and cube identities in various questions. These solutions can be helpful to you whenever you check your answers, revise your answers before appearing for a test, or practice from the Class 9 Ganita Manjari textbook.

Exercise Set 4.1

1. Expand using (a + b)² = a² + 2ab + b²:

(A)  (7x + 4y)² 

Ans.  (7x)² + 2(7x)(4y) + (4y)² = 49x² + 56xy + 16y²

(B) (7x/5 + 3y/2)²

Ans.  (7x/5)² + 2(7x/5)(3y/2) + (3y/2)²

= 49x²/25 + 21xy/5 + 9y²/4

(C) (2.5p + 1.5q)² 

Ans.  (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²

= 6.25p² + 7.5pq + 2.25q²

(D) (3s/4 + 8t)² 

Ans.  (3s/4)² + 2(3s/4)(8t) + (8t)²

  = 9s²/16 + 12st + 64t²

(E) (x + 1/(2y))² 

Ans.  x² + 2·x·(1/(2y)) + (1/(2y))²

 = x² + x/y + 1/(4y²)

(F) (1/x + 1/y)² 

Ans.  1/x² + 2/(xy) + 1/y²

2. Find the values:

(A)  64² 

Ans. (60 + 4)² = 60² + 2(60)(4) + 4² = 3600 + 480 + 16 = 4096

(B) 105² 

Ans.  (100 + 5)² = 10000 + 1000 + 25 = 11025

(C) 205²

Ans.  (200 + 5)² = 40000 + 2000 + 25 = 42025

Exercise Set 4.2

1. Factor completely:

(A)  9x² + 24xy + 16y² 

Ans.  (3x)² + 2(3x)(4y) + (4y)² = (3x + 4y)²

(B) 4s² + 20st + 25t² 

Ans. (2s)² + 2(2s)(5t) + (5t)² = (2s + 5t)²

(C) 49x² + 28xy + 4y² =

Ans. (7x)² + 2(7x)(2y) + (2y)² = (7x + 2y)²

(D) 64p² + (32/3)pq + (4/9)q²

Ans. (8p)² + 2(8p)(2q/3) + (2q/3)² = (8p + 2q/3)²

(E)  3a² + 4ab + (4/3)b²   

Ans. Take 1/3 common factor:

= (1/3)[9a² + 12ab + 4b²] = (1/3)[(3a)² + 2(3a)(2b) + (2b)²]

= (1/3)(3a + 2b)²

(F) (9/5)s² + 6sv + 5v²

Ans. Take 1/5 common factor:

= (1/5)[9s² + 30sv + 25v²] = (1/5)[(3s)² + 2(3s)(5v) + (5v)²]

= (1/5)(3s + 5v)²

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

(A)  79² 

Ans. (80 − 1)² = 802 – 2*80*1 + 12 = 6400 − 160 + 1 = 6241

(B) 193² 

 Ans.  (200 − 7)² = 2002 – 2*200*7 + 72 = 40000 − 2800 + 49 = 37249

(C) 299² 

Ans.  (300 − 1)² = 3002 – 2*300*1 + 12 = 90000 − 600 + 1 = 89401

Exercise Set 4.3

1. Find the squares using a suitable identity:

(A)  117² = (100 + 17)² 

Ans. Use (a + b)²:

= 10000 + 2(100)(17) + 289 = 10000 + 3400 + 289 = 13689

(B) 78² = (80 − 2)² 

Ans. Use (a − b)²: 

= 6400 − 320 + 4 = 6084

(C) 198² = (200 − 2)²   

Ans. Use (a − b)²:

= 40000 − 800 + 4 = 39204

(D) 214² = (200 + 14)²

Ans. Use (a + b)²:

= 40000 + 5600 + 196 = 45796

(E)  1104² = (1100+ 4)²

Ans. Use (a + b)²:

= 1100² + 2(1100)4 + 4²

= 1210000 + 8800 + 16 = 1218816

(F) 1120² = (1100+ 20)² 

Ans. Use (a + b)²: 

= 11002 + 2(1100)(20) + 202

= 1210000 + 44000 + 400

= 1254400

2. Factor using suitable identities:

(A)  16y² − 24y + 9 

Ans.  16y² − 24y + 9 = (4y)² − 2(4y)(3) + 3² = (4y − 3)²

(B) (9/4)s² + 6st + 4t² 

Ans.  (9/4)s² + 6st + 4t² 

= (3s/2)² + 2(3s/2) (2t) + (2t)²

= (3s/2 + 2t)²

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

Ans. (m/3)² + (k/2)² + (3n)² + 2(m/3)(k/2) + 2(k/2)(3n) + 2(m/3)(3n)

= (m/3 + k/2 + 3n)²

(D) p²/16 − 2 + 16/p² 

Ans.  (p/4)² − 2(p/4)(4/p) + (4/p)² = (p/4 − 4/p)²

(E)  9a² + 4b² + c² − 12ab + 6ac − 4bc

Ans. (3a)² + (−2b)² + c² + 2(3a)(−2b) + 2(−2b)(c) + 2(3a)(c)

= (3a − 2b + c)²

3. Expand using (a + b + c)² = a2 + b2 + c2 + 2ab + 2bc + 2ca:

(A) (p + 3q + 7r)²

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

Using the identity:

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

p² + 9q² + 49r² + 6pq + 42qr + 14pr

(B) (3x − 2y + 4z)²

Ans. Here, a = 3x, b = -2y c = 4z

Using the identity:

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

9x² + 4y² + 16z² − 12xy − 16yz + 24xz

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

Solution:

Expand each:

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

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

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

Sum = 3a² + 3b² + 3c² + (2ab − 2ab − 2ab) + (−2ac + 2ac − 2ac) + (−2bc − 2bc + 2bc)

= 3a² + 3b² + 3c² − 2ab − 2ac − 2bc

This is NOT equal to 2a² + 2b² + 2c². Hence, it is NOT an identity.

(Substitute a = 1, b = 1, c = 0: LHS = 0 + 4 + 0 = 4; RHS = 2 + 2 + 0 = 4. 

Substitute a = b = c = 1: LHS = 1 + 1 + 1 = 3; RHS = 6. Different, so not an identity.)

Exercise Set 4.4

1. Fill in the blanks:

(A)  s² − 11s + 24 = (________) (________) 

Ans. Use a + b = −11, ab = 24 

So, a = −3, b = −8

s² − 11s + 24 = (s − 3)(s − 8)

(B) (___)(x + 1) = 3x² − 4x − 7 

Ans. Divide 3x² − 4x − 7 by (x + 1): 3x² − 4x − 7 = 3x² + 3x − 7x − 7 

= 3x(x + 1) − 7(x + 1)

= (3x − 7)(x + 1) = 3x² − 4x − 7

(C) 10x² − 11x − 6 = (2x – ___) (___ + 2)

Ans. Split −11x using factors of 10 × (−6) = −60:

−15 + 4 = −11 and −15 × 4 = −60 

= 10x² − 15x + 4x − 6 

= 5x(2x − 3) + 2(2x − 3) 

= (2x − 3)(5x + 2)

(D) 6x² + 7x + 2 = (____________) (___________)

Ans.  Split 7x using factors of 12: 3 + 4 = 7, 3 × 4 = 12

= 6x² + 3x + 4x + 2 

= 3x(2x + 1) + 2(2x + 1) 

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

2. Find products using suitable identities:

(A)  41² 

Ans. (40 + 1)² = 1600 + 80 + 1 = 1681

(B) 27² 

Ans.  (30 − 3)² = 900 − 180 + 9 = 729

(C) 23 × 17 

Ans.  23 × 17 = (20 + 3)(20 − 3) = 20² − 3² = 400 − 9 = 391

(D) 135² 

Ans. (100 + 30 + 5)²

= 10000 + 900 + 25 + 6000 + 300 + 1000 = 18225

(E)  97² 

Ans. (100 − 3)² = 10000 − 600 + 9 = 9409

(F) 18 × 29. 

Ans. Since, 18 × 29 is not a perfect square pattern. 

So, 18 × 29 = 18(30 − 1) = 540 − 18 = 522

(G) 34 × 43 

Ans. (Doesn't fit standard (a+b)(a−b)). 

Direct: 34 × 43 = 34 × 40 + 34 × 3 = 1360 + 102 = 1462

(H) 205² 

Ans. (200 + 5)² = 40000 + 2000 + 25 = 42025

3. Factor the following:

(A)  9a² + b² + 4c² − 6ab + 12ac − 4bc

Ans. 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

(B) 16s² + 25t² − 40st 

Ans.  16s² + 25t² − 40st = (4s)² − 2(4s)(5t) + (5t)² = (4s − 5t)²

(C) r² − r − 42.

Ans.  Use a + b = −1, ab = −42 

So, a = −7, b = 6

r² − r − 42 = (r − 7)(r + 6)

(D) 49g² + 14gh + h² 

Ans.  49g² + 14gh + h²  = (7g)² + 2(7g)(h) + h² = (7g + h)²

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

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

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

= (8u − 11v − 2w)²

Exercise Set 4.5

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

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

Ans. 

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.

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

Ans. 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

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

Ans. Factorising the numerator:

w³ - v³ + x³ +3wvx

= w³ + (-v)³ + x³  - 3w(-v)x

= (w - v + x)(w2 + y2 + 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 + y² + 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]

(D) (4y² − 20yz + 25z²) / (25z² − 4y²)

Ans. 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)

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

Ans. x² + x − 6 = (x + 3)(x − 2)

x² − 7x + 12 = (x − 3)(x − 4)

x² − 6x + 8 = (x − 2)(x − 4)

x² − 9 = (x − 3)(x + 3)

= [(x + 3)(x − 2)(x − 3)(x − 4)] / [(x − 2)(x − 4)(x − 3)(x + 3)] = 1

(F) (p⁴ − 16)/(p² − 4p + 4)

Ans. Numerator = (p² − 4)(p² + 4) = (p − 2)(p + 2)(p² + 4)

Denominator = (p − 2)²

Result = [(p + 2)(p² + 4)] / (p − 2)

End of Chapter Exercises

1. Find the following products using identities:

(A)  (−3x + 4)² = (−3x)² + 2(−3x)(4) + 16 = 9x² − 24x + 16

(B) (2s + 7)(2s − 7) = (2s)² − 7² = 4s² − 49

(C) (p² + 1/2)(p² − 1/2) = (p²)² − (1/2)² = p⁴ − 1/4

(D) (2n + 7)(2n − 7) = 4n² − 49

(E) (s − 2t)(s² + 2st + 4t²) = s³ − (2t)³ = s³ − 8t³

[Using a³ − b³ = (a − b)(a² + ab + b²)]

(F) (1/(2r) − 4r)² = (1/(2r))² − 2·(1/(2r))·(4r) + (4r)² = 1/(4r²) − 4 + 16r²

(G) (−3m + 4k − l)² = 9m² + 16k² + l² − 24mk − 8kl + 6ml

Apply a + b + c formula with a = −3m, b = 4k, c = −l:

a² + b² + c² = 9m² + 16k² + l²

2ab = 2(−3m)(4k) = −24mk

2bc = 2(4k)(−l) = −8kl

2ca = 2(−l)(−3m) = 6lm

Sum: 9m² + 16k² + l² − 24mk − 8kl + 6lm

(H) (x − y/3)³ = x³ − 3x²(y/3) + 3x(y/3)² − (y/3)³

= x³ − x²y + xy²/3 − y³/27

(I) (7k/2 − 2m/3)³

Using (a − b)³ = a³ − 3a²b + 3ab² − b³ with a = 7k/2, b = 2m/3:

So, (7k/2 − 2m/3)³ = 343k³/8 − 49k²m/2 + 14km²/3 − 8m³/27

2. Find values using suitable identities:

(A)  17 × 21 = (19 − 2)(19 + 2) = 19² − 4 = 361 − 4 = 357

(B) 104 × 96 = (100 + 4)(100 − 4) = 10000 − 16 = 9984

(C) 24 × 16 = (20 + 4)(20 − 4) = 400 − 16 = 384

(D) 147³ = (150 − 3)³ = 150³ − 3·150²·3 + 3·150·9 − 27

   = 3375000 − 202500 + 4050 − 27 = 3176523

(E) 199³ = (200 − 1)³ = 8000000 − 3(40000) + 3(200) − 1

   = 8000000 − 120000 + 600 − 1 = 7880599

(F) 127³ = (130 − 3)³ = 130³ − 3·130²·3 + 3·130·9 − 27

   = 2197000 − 152100 + 3510 − 27 = 2048383

(G) (−107)³ = −(107)³ = −(100 + 7)³

  107)³ = 1000000 + 3(10000)(7) + 3(100)(49) + 343

  = 1000000 + 210000 + 14700 + 343 = 1225043

So, (−107)³ = −1225043

(H) (−299)³ = −(299)³ = −(300 − 1)³

299³ = 27000000 − 3(90000) + 3(300) − 1 = 27000000 − 270000 + 900 − 1 = 26730899

So, (−299)³ = −26730899

3. Factor:

(A)  4y² + 1 + 1/(16y²) = (2y)² + 2(2y)(1/(4y)) + (1/(4y))² = (2y + 1/(4y))²

(B) 9m² − 1/(25n²) = (3m)² − (1/(5n))² = (3m − 1/(5n))(3m + 1/(5n))

(C) 27b³ − 1/(64b³) = (3b)³ − (1/(4b))³

= (3b − 1/(4b))[(3b)² + (3b)(1/(4b)) + (1/(4b))²]

= (3b − 1/(4b))(9b² + 3/4 + 1/(16b²))

(D) x² + 5x/6 + 1/6. Multiply/divide appropriately; need a + b = 5/6, ab = 1/6:

a = 1/2, b = 1/3.  Check: 1/2 + 1/3 = 5/6 ✓, (1/2)(1/3) = 1/6 ✓

x² + 5x/6 + 1/6 = (x + 1/2)(x + 1/3)

(E) 27u³ − 1/125 − 27u²/5 + 9u/25

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

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

= (3u − 1/5)³

(F) 64y³ + z³/125 = (4y)³ + (z/5)³

= (4y + z/5)[(4y)² − (4y)(z/5) + (z/5)²]

= (4y + z/5)(16y² − 4yz/5 + z²/25)

(G) p³ + 27q³ + r³ − 9pqr = p³ + (3q)³ + r³ − 3·p·(3q)·r

Using a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) with a = p, b = 3q, c = r:

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

(H) 9m² − 12m + 4 = (3m)² − 2(3m)(2) + (2)^2 = (3m − 2)²

(I) 9x³ − (8/3)y³ + z³/3 + 6xyz

Ans. 9x³- 8/3 y³ + z³/3 + 6xyz
= (1/3)[27x³ - 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]   

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

Ans. 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]

(K) 27u³ − 1/216 − 9u²/2 + u/4

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

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

= (3u − 1/6)³

4. Simplify:

(A)  (4x² + 4x + 1)/(4x² − 1)

Numerator = (2x + 1)²

Denominator = (2x − 1)(2x + 1)

Result = (2x + 1)/(2x − 1)

(B) 9(3a³ − 24b³) / (9a² − 36b²)

Numerator = 9*3(a³ − 8b³) = 27(a − 2b)(a² + 2ab + 4b²)

Denominator = 9(a² − 4b²) = 9(a − 2b)(a + 2b)

Result = 27(a² + 2ab + 4b²) / [9(a + 2b)] = 3(a² + 2ab + 4b²)/(a + 2b)

(C) (s³ + 125t³) / (s² − 2st − 35t²)

Numerator = s³ + (5t)³ = (s + 5t)(s² − 5st + 25t^2)

Denominator = s² − 2st − 35t² 

Split: Use a + b = −2t, ab = −35t² 

So, a = −7t, b = 5t. Check: −7t + 5t = −2t, (−7t)(5t) = −35t² 

s² + 5st − 7st − 35t² = s(s + 5t) − 7t(s + 5t) = (s + 5t)(s − 7t)

Result = (s² − 5st + 25t²)/(s − 7t)

5. Possible expressions for length and breadth:

(A)  Area = 25a² − 30ab + 9b² 

= (5a)² − 2(5a)(3b) + (3b)² 

= (5a − 3b)²

Length = breadth = (5a − 3b) units

(B) Area = 36s² − 49t² 

= (6s)² − (7t)² = (6s + 7t)(6s − 7t)

Length = (6s + 7t), Breadth = (6s − 7t)

6. Possible length, breadth, height for cuboid volumes:

(A)  6a² − 24b² = 6(a² − 4b²) = 6(a − 2b)(a + 2b)

Possible dimensions: Length = 6, Breadth = (a − 2b), Height = (a + 2b)

(B) 3ps² − 15ps + 12p = 3p(s² − 5s + 4) = 3p(s − 1)(s − 4)

Possible dimensions: Length = 3p, Breadth = (s − 1), Height = (s − 4)

7. Square playground side 40 m with a path of width s metres around it. Find the area of the path.

Ans. Outer square side = 40 + 2s metres (path on each side).

Area of path = (outer area) − (playground area)

= (40 + 2s)² − 40²

            = (1600 + 160s + 4s²) − 1600

            = 4s² + 160s 

            = 4s(s + 40) sq. m

8. Number + reciprocal = 10/3. Find the number.

Ans.

Let the number be x. Then, x + 1/x = 10/3.

Multiply by 3x:  3x² + 3 = 10x  ⇒  3x² − 10x + 3 = 0

Factor:  Use ab = 9, a + b = −10 → −1 and −9

3x² − 9x − x + 3 = 0  ⇒  3x(x − 3) − 1(x − 3) = 0  ⇒  (3x − 1)(x − 3) = 0

x = 1/3  or  x = 3.

Both are valid (and they are reciprocals of each other). The number is 3 or 1/3.

9. Rectangular pool, area = 2x² + 7x + 3 sq. hastas, width = 2x + 1. Find length.

Ans:

Factor 2x² + 7x + 3. Split 7x using factors of 6: 6 + 1.

2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)

Length = (x + 3) hastas.

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

Ans.

If (x − 2) is a factor:  p(2)² + 5(2) + r = 0  ⇒  4p + 10 + r = 0  ⇒  4p + r = −10   ...(i)

If (x − 1/2) is a factor:  p(1/4) + 5/2 + r = 0  ⇒  p/4 + r = −5/2

Multiply by 4:  p + 4r = −10    ...(ii)

From (i) and (ii),  

4p + r = p + 4r  ⇒  3p = 3r  ⇒  p = r. 

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

Ans.

Identity: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

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

25 = a² + b² + c² + 20  ⇒  a² + b² + c² = 5

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

Therefore, a³ + b³ + c³ − 3abc = 5 × (−5) = −25. 

*12. Show that n³ − n is divisible by 6 for all natural numbers n.

Ans.

Factor: n³ − n = n(n² − 1) = n(n − 1)(n + 1)

This is the product of three consecutive integers.

Among any three consecutive integers, at least one is divisible by 2 (an even number), and exactly one is divisible by 3.

Therefore, the product is divisible by 2 × 3 = 6.

Hence n³ − n is divisible by 6 for every natural number n.

*13. Find the values:

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

Ans. Rewrite: x³ + y³ + 64 − 12xy = x³ + y³ + 4³ − 3·x·y·4

Since, a³ + b³ + c³ − 3abc with a = x, b = y, c = 4.

= (x + y + 4)(x² + y² + 16 − xy − 4y − 4x)

From x + y = −4, we get x + y + 4 = 0.

Therefore, x³ + y³ − 12xy + 64 = 0 × (…) = 0.

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

Rewrite: x³ + (−2y)³ + (−6)³ − 3·x·(−2y)·(−6) = x³ − 8y³ − 216 − 36xy

Since, a³ + b³ + c³ − 3abc with a = x, b = −2y, c = −6.

= (x − 2y − 6)(x² + 4y² + 36 + 2xy − 12y + 6x)

From x = 2y + 6, we get x − 2y − 6 = 0.

Therefore, x³ − 8y³ − 36xy − 216 = 0.

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Algebraic Identities Class 9th Chapter 4 Summary

Here you’ll find quick glimpses of important definitions, key points, and essential formulas that you should remember while solving questions and preparing for exams.

1. Algebraic Identities

Algebraic Identities are equations that are always true for all permissible values you plug in for the variables. An identity is an equation that holds true for all numbers; unlike an equation that might be true for specific values.

2. Visualising Identities

Geometric figures, like squares and rectangles, are used to visually represent identities, aiding students in understanding the concept effectively. This makes it easier to understand why expressions like ((a+b)^2) can be expanded in a particular way.

3. Factorisation Using Identities

Identities can also be used in reverse. For instance, you can write the expression (a^2+2ab+b2) as (a+b)^2. Identifying those patterns can make factorisation quicker and easier.

4. Factorisation Using Algebra Tiles

Algebra tiles offer a visual way to comprehend how algebraic expressions can be set up in factors. This aids in connecting multiplication and factorisation rather than treating factorisation as a formula-based process.

5. Splitting the Middle Term

If the middle term is suitable for an expression to be a quadratic, it can be broken down into two terms, enabling grouping and common factors for completing the factorisation.

6. Simplifying Rational Expressions

Factorisation can make algebraic fractions easier to simplify. The numerator and denominator are first split into factors, and then common factors are cancelled (but not when the denominator is zero).

7. Fundamental Algebraic Identities

• Square of Sum (a+b)²  = a² + 2ab + b² 
• Square of Difference (a - b)²  = a² - 2ab + b²
• Square of a Trinomial (a + b + c)²  = a² + b² + c² + 2ab + 2bc + 2ca
• Difference of Squares (a² - b²) = (a + b)(a - b)
• Sum of Cubes (a³ + b³)= (a + b)(a² - ab + b²)
• Difference of Cubes (a³ - b³)= (a - b)(a² + ab + b²)
• Sum of Three Cubes (a³ + b³ + c³ - 3abc) = (a + b + c)(a² +b² + c² - ab - bc - ca)
• Cube of Sum (a + b)³ = a³ + 3a² b + 3ab² + b³
• Cube of Difference (a - b)³ = a³ - 3a²b + 3ab² - b³

These identities not only facilitate direct expansion, but they also help in factorisation, simplification and mental calculation.

Essential Things to Remember When Solving Chapter 4 Questions

1. Understand the pattern first: Before expanding or factorising, first check whether the expression is a known identity.
2. Look for signs: Note that the middle term is different in ((a+b)²) and ((a-b)²), so it's easy to make errors with the signs.
3. Rational expressions: Note that only common factors can be cancelled; not individual terms.
4. Check the middle term: The two numbers when splitting the middle term should add up to the middle term and multiply to give the leading coefficient in the quadratic.
5. Recognise identities in reverse: Many factorisation problems are easier to solve by recognising standard identities backwards.
6. Demonstrate the steps: Writing the identity before solving the equation helps to make the solution procedure clearer and to avoid calculation errors.

It is advisable that you learn the important examples and concepts given in the Class 9 Maths book and then verify your work with the help of the solutions. Post NCERT questions, utilising a Maths question bank for Class 9 can help in additional practice.

Frequently Asked Questions

1. Do NCERT Class 9 Maths Chapter 4 solutions follow the latest syllabus 2026-27?

Yes, the NCERT solutions for Class 9 Maths Chapter 4 are designed with the latest CBSE-prescribed Ganita Manjari syllabus. It introduces several revised concepts, such as algebraic identities, visualisation of algebraic identities, factorisation, cube identities, splitting the middle term and simplification of rational expressions.

2. What do these Class 9 Ch 4 solutions cover?

Educart’s NCERT solutions provide step-wise solutions for each question across five exercise sets, Exercise 4.1, 4.2, 4.3, 4.4, 4.5, and End-of-Chapter Exercises, in Chapter 4.

3. How do these NCERT solutions help in Class 9 exams?

NCERT Solutions for Class 9 Maths Chapter 4 Exploring Algebraic Identities can be used to verify the questions asked in the chapter and can be used to practice the questions step-by-step.

4. Can algebraic identities help simplify expressions?

Yes, algebraic identities can be used to factor, expand, and simplify rational algebraic expressions and help calculate squares and cubes more easily and promptly.

5. How do you factorise algebraic expressions using identities in Class 9?

Algebraic expressions can be factorised by identifying the common patterns (e.g. a² + 2ab + b² = (a+b)², (a² - b²) = (a + b)(a - b) and other identities).

6. What are the important identities in Class 9 Maths Chapter 4?

The important identities include (a+b)², (a-b)², (a² - b²), (a+b+c)², and the sum and difference of cubes.

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