Class 9 Maths · Chapter 4
Short answer:
Chapter 4 algebra ka woh "shortcut engine" hai jo har square, cube aur factorisation ko formula-pattern se seconds me solve karta hai — bina lambi multiplication kiye.
Ganita Manjari Part 1 ke 8-chapter naye Class 9 Maths syllabus (session 2026-27) me Chapter 4 "Exploring Algebraic Identities" woh jagah hai jahan pichle chapters ka foundation kaam aata hai. Chapter 2 "Introduction to Linear Polynomials" me jo variable-handling seekhi thi, Chapter 3 "The World of Numbers" me jo number-sense develop hua tha — Chapter 4 unhe milaakar ek naya, powerful toolkit deta hai: algebraic identities.
Ye chapter sirf formulas ratne wala nahi hai. Ganita Manjari ka approach hai identities ko squares, rectangles aur cubes ke through geometrically visualise karna — taaki (a+b)² sirf ek abstract formula na lage, balki ek real area diagram lage. Iske baad ye identities factorisation, rational expressions ki simplification, aur bade numbers (jaise 105×95 ya 998²) ke fast mental calculation me use hoti hain.
Agar tum Ganita Manjari Class 9 Maths NCERT solutions dhundh rahe ho step-by-step working ke saath — is chapter ke saare exercises (4.1 se 4.5 tak) plus End of Chapter Exercise yahan clear step-by-step style me cover kiye gaye hain, taaki ratta lagane ki jagah pattern samajh aaye.
Chapter 4 Summary — 5 Minute Revision
Exploring Algebraic Identities ka core message simple hai: har baar poora multiplication karne ki jagah, ek pehchana hua pattern (identity) use karo aur seconds me answer nikaalo — chahe woh 998² ho ya x³+y³+z³−3xyz ka factorisation. Chapter geometric visualisation se shuru hokar factorisation, rational-expression simplification, aur applied area/volume problems tak jaata hai.
Is chapter ke baad Ganita Manjari Part 1 me chaar aur chapters bache hain: Chapter 5 "I'm Up and Down, and Round and Round" (circles — practice ke liye class 9 maths chapter 5 im up and down and round and round circles important questions zaroor dekhna), Chapter 6 "Measuring Space" (perimeter-area — class 9 maths chapter 6 measuring space perimeter and area important questions), Chapter 7 "Probability" (bilkul naya chapter — class 9 maths chapter 7 probability ncert solutions), aur Chapter 8 "Sequences and Progressions" (bhi naya — class 9 maths chapter 8 sequences and progressions ncert solutions).
Revision ke liye poore syllabus ka overview chahiye ho to class 9 maths new syllabus 2026-27 chapter list pdf aur class 9 maths ganita manjari all chapters pdf download dono ncert.nic.in ke official listing se hi confirm karo — kai third-party sites abhi bhi purane 14-15 chapter wale syllabus ka data dikha rahi hain jo is session ke liye valid nahi hai. Peeche revise karna ho to class 9 maths chapter 3 the world of numbers ncert solutions aur class 9 maths chapter 2 introduction to linear polynomials important questions se foundation pakki kar lo.
In-Text Questions — Solutions
Expand (2x + 3y)² using a suitable identity.
Use (a + b)² = a² + 2ab + b² with a = 2x, b = 3y.
(2x + 3y)² = (2x)² + 2(2x)(3y) + (3y)²
= 4x² + 12xy + 9y²
Expand (5p − 4q)² using a suitable identity.
Use (a − b)² = a² − 2ab + b² with a = 5p, b = 4q.
(5p − 4q)² = (5p)² − 2(5p)(4q) + (4q)²
= 25p² − 40pq + 16q²
Find the value of 98² using an identity (do not multiply directly).
Write 98 = 100 − 2 and use (a − b)² = a² − 2ab + b².
98² = (100 − 2)² = 100² − 2(100)(2) + 2²
= 10000 − 400 + 4 = 9604
Expand (3x + 2y + z)² using the (a + b + c)² identity.
Here a = 3x, b = 2y, c = z. Use (a+b+c)² = a²+b²+c²+2ab+2bc+2ca.
(3x+2y+z)² = 9x² + 4y² + z² + 2(3x)(2y) + 2(2y)(z) + 2(z)(3x)
= 9x² + 4y² + z² + 12xy + 4yz + 6zx
Evaluate 105 × 95 using a suitable identity.
Write as (100 + 5)(100 − 5) and use a² − b² = (a+b)(a−b).
105 × 95 = (100 + 5)(100 − 5) = 100² − 5²
= 10000 − 25 = 9975
Factorise: 49x² − 81y²
Write as a difference of squares.
49x² − 81y² = (7x)² − (9y)²
= (7x + 9y)(7x − 9y)
Find the product (x + 7)(x + 3) using the identity (x+a)(x+b) = x² + (a+b)x + ab.
Here a = 7, b = 3.
(x+7)(x+3) = x² + (7+3)x + (7)(3)
= x² + 10x + 21
Find the value of 998² using an identity.
Write 998 = 1000 − 2.
998² = (1000 − 2)² = 1000² − 2(1000)(2) + 2²
= 1000000 − 4000 + 4 = 996004
Expand (2x + 1)³ using the (a+b)³ identity.
Here a = 2x, b = 1. Use (a+b)³ = a³ + 3a²b + 3ab² + b³.
(2x+1)³ = (2x)³ + 3(2x)²(1) + 3(2x)(1)² + 1³
= 8x³ + 12x² + 6x + 1
Expand (3x − 2y)³ using the (a−b)³ identity.
Here a = 3x, b = 2y. Use (a−b)³ = a³ − 3a²b + 3ab² − b³.
(3x−2y)³ = (3x)³ − 3(3x)²(2y) + 3(3x)(2y)² − (2y)³
= 27x³ − 54x²y + 36xy² − 8y³
Find the value of 102³ using a suitable identity.
Write 102 = 100 + 2 and use (a+b)³.
102³ = (100+2)³ = 100³ + 3(100)²(2) + 3(100)(2)² + 2³
= 1000000 + 60000 + 1200 + 8 = 1061208
Factorise: 27x³ + 64y³
Write as a sum of cubes: (3x)³ + (4y)³, and use a³+b³ = (a+b)(a²−ab+b²).
27x³ + 64y³ = (3x + 4y)((3x)² − (3x)(4y) + (4y)²)
= (3x + 4y)(9x² − 12xy + 16y²)
Factorise: 8x³ − 125
Write as a difference of cubes: (2x)³ − 5³, and use a³−b³ = (a−b)(a²+ab+b²).
8x³ − 125 = (2x − 5)((2x)² + (2x)(5) + 5²)
= (2x − 5)(4x² + 10x + 25)
Factorise x² + 9x + 18 by splitting the middle term.
Find two numbers whose product is 18 and sum is 9 — these are 6 and 3.
x² + 6x + 3x + 18 = x(x + 6) + 3(x + 6)
= (x + 6)(x + 3)
If a + b + c = 0, prove that a³ + b³ + c³ = 3abc, and use this to evaluate 3³ + 4³ + (−7)³.
From the identity a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca), if a+b+c = 0, the right side becomes 0.
⇒ a³ + b³ + c³ − 3abc = 0 ⇒ a³ + b³ + c³ = 3abc
Check: 3 + 4 + (−7) = 0, so the shortcut applies.
3³ + 4³ + (−7)³ = 3(3)(4)(−7) = −252
Simplify (x² − y²) ÷ (x + y) for x = 15, y = 8 using an identity.
Factorise the numerator using a² − b² = (a+b)(a−b):
(x² − y²) ÷ (x + y) = (x+y)(x−y) ÷ (x+y) = x − y
Substituting x = 15, y = 8:
= 15 − 8 = 7
Exercise Questions — Solutions (Q1–Q9)
Evaluate 999³ using a suitable identity.
Write 999 = 1000 − 1 and use (a − b)³ = a³ − 3a²b + 3ab² − b³ with a = 1000, b = 1.
999³ = (1000 − 1)³ = 1000³ − 3(1000)²(1) + 3(1000)(1)² − 1³
= 1000000000 − 3000000 + 3000 − 1
= 997002999
Factorise: 4x² + 12xy + 9y² − 1
Group the first three terms — they form a perfect square trinomial.
4x² + 12xy + 9y² = (2x)² + 2(2x)(3y) + (3y)² = (2x + 3y)²
So the expression becomes a difference of squares:
(2x + 3y)² − 1² = (2x + 3y + 1)(2x + 3y − 1)
If x + 1/x = 5, find the value of (i) x² + 1/x² and (ii) x³ + 1/x³.
(i) Square both sides of x + 1/x = 5:
(x + 1/x)² = x² + 2 + 1/x² = 25
⇒ x² + 1/x² = 25 − 2 = 23
(ii) Use (a+b)³ = a³ + b³ + 3ab(a+b) with a = x, b = 1/x, ab = 1:
5³ = x³ + 1/x³ + 3(1)(5)
125 = x³ + 1/x³ + 15 ⇒ x³ + 1/x³ = 110
If a − b = 4 and ab = 21, find the value of a² + b² and a³ − b³.
From (a − b)² = a² − 2ab + b²:
16 = a² + b² − 2(21) ⇒ a² + b² = 16 + 42 = 58
Now use a³ − b³ = (a − b)(a² + ab + b²):
a³ − b³ = 4 × (58 + 21) = 4 × 79 = 316
Simplify: [(x + y)² − (x − y)²] ÷ 4xy
Expand both squares:
(x + y)² − (x − y)² = (x² + 2xy + y²) − (x² − 2xy + y²) = 4xy
So the expression simplifies to:
4xy ÷ 4xy = 1
A square garden has side (3p + 2q) metres. Write its area using an identity, then find the area when p = 10 m and q = 5 m.
Area = (side)² = (3p + 2q)²
= 9p² + 12pq + 4q²
Substituting p = 10, q = 5:
Area = 9(100) + 12(50) + 4(25) = 900 + 600 + 100 = 1600 m²
A cubical water tank has edge (x + 3) metres. Express its volume using an identity and find the volume when x = 7 m.
Volume = (edge)³ = (x + 3)³
= x³ + 9x² + 27x + 27
At x = 7:
Volume = 343 + 9(49) + 27(7) + 27 = 343 + 441 + 189 + 27 = 1000 m³
(Check: (7+3)³ = 10³ = 1000 m³ ✓)
Factorise x³ + y³ + z³ − 3xyz, and verify the factorisation for x = 1, y = 2, z = 3.
Standard identity:
x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx)
For x = 1, y = 2, z = 3: x+y+z = 6; x²+y²+z² = 1+4+9 = 14; xy+yz+zx = 2+6+3 = 11.
RHS = 6 × (14 − 11) = 6 × 3 = 18
Direct check: 1³+2³+3³ − 3(1)(2)(3) = 1+8+27−18 = 18 ✓
Factorise x² − 2x − 15 by splitting the middle term.
We need two numbers whose product is −15 and sum is −2. These are −5 and 3.
x² − 5x + 3x − 15 = x(x − 5) + 3(x − 5)
= (x − 5)(x + 3)
Important Equations — Ek Nazar Me
| Identity | Formula |
|---|---|
| Square of a sum | (a + b)² = a² + 2ab + b² |
| Square of a difference | (a − b)² = a² − 2ab + b² |
| Difference of squares | a² − b² = (a + b)(a − b) |
| Square of a trinomial | (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca |
| Product of two binomials (common factor form) | (x + a)(x + b) = x² + (a + b)x + ab |
| Cube of a sum | (a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a + b) |
| Cube of a difference | (a − b)³ = a³ − 3a²b + 3ab² − b³ = a³ − b³ − 3ab(a − b) |
| Sum of cubes | a³ + b³ = (a + b)(a² − ab + b²) |
| Difference of cubes | a³ − b³ = (a − b)(a² + ab + b²) |
| Three-variable cube identity | a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) |
| Special case (a+b+c=0) | if a + b + c = 0, then a³ + b³ + c³ = 3abc |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Sign error in (a−b)² : students write a² + 2ab + b² by habit, bhool jaate hain ki middle term negative hota hai — sahi form hai a² − 2ab + b².
- (a+b)² ko a²+b² samajh lena: cross term 2ab poora chhoot jaata hai, jo sabse common aur sabse costly mistake hai.
- Cube expansion me coefficient galat: (a+b)³ ke middle terms me coefficient 3 hota hai, kai students 2 likh dete hain — a³+2a²b+2ab²+b³ jaisi galat expansion.
- Sum aur difference of cubes ke formula swap ho jaana: a³+b³ = (a+b)(a²−ab+b²) hai, lekin students galti se (a+b)(a²+ab+b²) likh dete hain — sign of 'ab' term uda dete hain.
- a³+b³+c³ = 3abc shortcut har jagah laga dena: ye sirf tab valid hai jab a+b+c = 0 ho — bina check kiye is shortcut ko use karna galat answer deta hai.
- Splitting the middle term me galat factor-pair choose karna: khaaskar jab constant term negative ho, sign combination miss ho jaata hai aur factorisation match nahi karta.
Board-Style Important Questions
- If a + b = 10 and a − b = 4, find the value of a² − b² using a suitable identity.
- Expand (x − 2y + 3z)² using the (a+b+c)² identity.
- Factorise 125x³ − y³ completely using an identity.
- If a + b = 8 and ab = 15, find the value of a³ + b³.
- State and prove the identity a³ + b³ + c³ − 3abc = (a+b+c)(a²+b²+c²−ab−bc−ca), and use it to factorise 27x³ + 8y³ + z³ − 18xyz.
Aksar Poochhe Jaane Wale Sawaal
Ganita Manjari class 9 maths NCERT solutions Chapter 4 me kitni exercises hain?
Chapter 4 "Exploring Algebraic Identities" me paanch exercises hain (4.1 se 4.5 tak), plus "Think and Reflect" activities aur ek End of Chapter Exercise set. Exercise 4.1-4.2 square/product identities cover karte hain, 4.3 cube identities, 4.4 factorisation, aur 4.5 teen-variable identity (a³+b³+c³−3abc) aur rational expressions ki simplification.
Class 9 maths Chapter 4 Exploring Algebraic Identities ke extra questions kahan se practice karein?
Upar diye gaye in-text aur End of Chapter examples ke alawa, is chapter ki neev Chapter 2 (Introduction to Linear Polynomials) aur Chapter 3 (The World of Numbers) me hai — inhe revise karke identities ke extra questions solve karna easier ho jaata hai, kyunki dono chapters variable-handling aur number-sense build karte hain.
Class 9 maths ka new syllabus 2026-27 chapter list me kitne chapters hain, aur ye kahan milega?
Session 2026-27 se Class 9 Maths ka poora syllabus badal gaya hai — purani "Mathematics" book ki jagah Ganita Manjari aayi hai. Abhi sirf Part 1 published hai, jisme 8 chapters hain. Poori chapter list PDF ke liye ncert.nic.in ka official textbook listing hi authoritative source hai — kai third-party sites abhi bhi purana 14-15 chapter wala data dikha rahi hain jo is session ke liye valid nahi hai.
Class 9 maths Ganita Manjari Part 2 kab aayega?
4 August 2026 tak Ganita Manjari Part 2 release nahi hui hai aur na hi koi official release date announce hui hai. Part 2 me Linear Equations in Two Variables, Euclid's Geometry, Lines and Angles, Triangles, Quadrilaterals, Surface Area & Volume, aur Statistics jaise topics expected hain — lekin ye sirf secondary-source estimates hain, official confirmation nahi.
Chapter 4 ke baad Ganita Manjari Part 1 me aage kya padhna hai?
Chapter 4 ke baad chaar chapters bache hain: Chapter 5 "I'm Up and Down, and Round and Round" (circles), Chapter 6 "Measuring Space" (perimeter aur area), Chapter 7 "Probability", aur Chapter 8 "Sequences and Progressions" — ye dono aakhri chapters bilkul naye hain jo pehle Class 9 me is form me nahi the.
Algebraic identities real life ya higher classes me kaam kaise aati hain?
Ye identities Class 10 ke Polynomials/Quadratic Equations, Class 11-12 ke Binomial Theorem, aur engineering-level algebra tak use hoti rehti hain. Practical side pe, ye large-number multiplication (jaise 998² ya 105×95) ko seconds me solve karne ka mental-math shortcut bhi deti hain — bina calculator ke.
Class 9 Maths — Saare Chapters
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