Important Questions Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities

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Ganita Manjari · Grade 9 · Chapter 4

Exploring Algebraic Identities

Important Questions Bank — Short & Long Answer

15 Short Answer Questions
10 Long Answer Questions
Identities & Expansion
Factorisation
Real-Life Applications

A

Part A — Short Answer Questions

2–3 marks each · 15 questions · Theory & Numerical mix

Algebraic Identities — Concept & Expansion

Q1.
What is the difference between an equation and an algebraic identity? Give one example of each.
TheoryEasy

Answer

An equation is a statement that is true only for specific values of the variable. An algebraic identity is an equation that is true for all values of the variables.

Example of equation: x² – 1 = 24 — true only for x = 5 or x = –5.

Example of identity: (x + y)² = x² + 2xy + y² — true for all values of x and y.

Q2.
State and write the expanded form of the identity (a + b)² = ?. Verify it for a = –2 and b = –3.
NumericalEasy

Answer

(a + b)² = a² + 2ab + b²

Verification for a = –2, b = –3:

LHS: (–2 + (–3))² = (–5)² = 25

RHS: (-2)² + 2(–2)(–3) + (–3)² = 4 + 12 + 9 = 25

LHS = RHS = 25 ✓ Identity verified.
Q3.
Expand using a suitable identity: (5x + 2y)²
NumericalEasy

Answer

Using (a + b)² = a² + 2ab + b², with a = 5x and b = 2y:

(5x + 2y)² = (5x)² + 2(5x)(2y) + (2y)²

= 25x² + 20xy + 4y²
Q4.
Calculate 43² using the identity (a + b)² = a² + 2ab + b² without direct multiplication.
NumericalEasy

Answer

Write 43 = 40 + 3, so a = 40 and b = 3.

43² = (40 + 3)² = 40² + 2 × 40 × 3 + 3²

= 1600 + 240 + 9

43² = 1849
Q5.
Factorise: x² + 4x + 4
NumericalEasy

Answer

Observe: x² = (x)², 4 = 2² and 4x = 2(x)(2)

Comparing with a² + 2ab + b² where a = x and b = 2:

x² + 4x + 4 = (x + 2)²

(a – b)² Identity & Numerical Calculations

Q6.
Write the identity for (a – b)². Use it to calculate 29².
NumericalEasy

Answer

(a – b)² = a² – 2ab + b²

Write 29 = 30 – 1, so a = 30, b = 1:

29² = (30 – 1)² = 30² – 2 × 30 × 1 + 1² = 900 – 60 + 1

29² = 841
Q7.
Factorise completely: 9x² + 24xy + 16y²
NumericalMedium

Answer

Observe: 9x² = (3x)², 16y² = (4y)² and 24xy = 2(3x)(4y)

Comparing with a² + 2ab + b² where a = 3x and b = 4y:

9x² + 24xy + 16y² = (3x + 4y)²
Q8.
State the identity for (a + b + c)² and use it to find 119² by writing 119 = 100 + 10 + 9.
NumericalMedium

Answer

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

With a = 100, b = 10, c = 9:

119² = 100² + 10² + 9² + 2(100)(10) + 2(10)(9) + 2(100)(9)

= 10000 + 100 + 81 + 2000 + 180 + 1800

119² = 14161

Difference of Squares & Factorisation

Q9.
State the identity a² – b² = (a + b)(a – b). Use it to calculate 55² (Hint: use the form a² = (a+b)(a–b) + b²).
NumericalMedium

Answer

Using a² = (a + b)(a – b) + b² with a = 55 and b = 5:

55² = (55 + 5)(55 – 5) + 5² = 60 × 50 + 25 = 3000 + 25

55² = 3025
Note: This method was proposed by Śhrīdharāchārya in 750 CE as a quick way to square numbers.
Q10.
Factorise: x² + 11x + 30
NumericalMedium

Answer

Compare with x² + (a + b)x + ab. We need a + b = 11 and ab = 30.

Factors of 30: try 5 × 6 = 30 and 5 + 6 = 11 ✓

Split: x² + 11x + 30 = x² + 5x + 6x + 30 = x(x + 5) + 6(x + 5)

x² + 11x + 30 = (x + 5)(x + 6)
Q11.
Factorise: x² – 5x + 6
NumericalMedium

Answer

We need a + b = –5 and ab = 6.

Try a = –2 and b = –3: (–2) + (–3) = –5 ✓ and (–2)(–3) = 6 ✓

x² – 5x + 6 = (x – 2)(x – 3)

Cubic Identities

Q12.
Write the identity for (a + b)³. Use it to expand (x + 2)³.
NumericalMedium

Answer

(a + b)³ = a³ + 3a²b + 3ab² + b³

With a = x and b = 2:

(x + 2)³ = x³ + 3(x²)(2) + 3(x)(4) + 8

(x + 2)³ = x³ + 6x² + 12x + 8
Q13.
Write the identity for x³ – y³ as a product. Verify it by expanding (x – y)(x² + xy + y²).
TheoryMedium

Answer

x³ – y³ = (x – y)(x² + xy + y²)

Verification:

(x – y)(x² + xy + y²)

= x³ + x²y + xy² – x²y – xy² – y³

= x³ – y³ ✓

Identity verified: x³ – y³ = (x – y)(x² + xy + y²)

Pattern & Consecutive Squares

Q14.
Three consecutive square numbers are taken. Why does adding the smallest and largest and subtracting twice the middle always give 2? Explain using algebra.
TheoryMedium

Answer

Let three consecutive numbers be (n – 1), n and (n + 1). Their squares are (n–1)², n² and (n+1)².

Sum of smallest and largest squares:

(n – 1)² + (n + 1)² = n² – 2n + 1 + n² + 2n + 1 = 2n² + 2

Subtract twice the middle: (2n² + 2) – 2n² = 2

∴ The result is always 2, regardless of the value of n. This is an algebraic proof of the pattern.
Q15.
Simplify the rational expression: (x² – 7x + 12) ÷ (5x² + 5x – 100), assuming the denominator ≠ 0.
NumericalHard

Answer

Factorise numerator: x² – 7x + 12 → need a + b = –7 and ab = 12 → a = –3, b = –4

x² – 7x + 12 = (x – 3)(x – 4)

Factorise denominator: 5x² + 5x – 100 = 5(x² + x – 20)

Need a + b = 1 and ab = –20 → a = 5, b = –4

= 5(x + 5)(x – 4)

Cancel common factor (x – 4):

(x² – 7x + 12) / (5x² + 5x – 100) = (x – 3) / 5(x + 5)

B

Part B — Long Answer Questions

5 marks each · 10 questions · Full step-by-step solutions

L1.
Prove algebraically that for any three consecutive square numbers, the sum of the smallest and largest minus twice the middle is always 2. Then verify with the set 25, 36, 49.
ProofMedium

Solution — Step by step

  1. Represent three consecutive numbers: Let them be (n – 1), n, (n + 1) for any integer n.
  2. Their squares: (n – 1)², n², (n + 1)²
  3. Expand using identity (a – b)²: (n – 1)² = n² – 2n + 1 and (n + 1)² = n² + 2n + 1
  4. Sum of smallest and largest: (n – 1)² + (n + 1)² = n² – 2n + 1 + n² + 2n + 1 = 2n² + 2
  5. Subtract twice the middle: (2n² + 2) – 2 × n² = 2n² + 2 – 2n² = 2
  6. Conclusion: The result is always 2 for any value of n.
∴ The result is always 2. ✓

Verification with 25, 36, 49:

(25 + 49) – (2 × 36) = 74 – 72 = 2 ✓

L2.
Use suitable identities to find the values of: (i) 79² (ii) 193² (iii) 17 × 21 (iv) 104 × 96. Show all steps.
NumericalMedium

Solution — Step by step

  1. 79² = (80 – 1)² using (a – b)² = a² – 2ab + b²:
    = 80² – 2 × 80 × 1 + 1² = 6400 – 160 + 1 = 6241
  2. 193² = (200 – 7)²:
    = 200² – 2 × 200 × 7 + 7² = 40000 – 2800 + 49 = 37249
  3. 17 × 21 = (19 – 2)(19 + 2) using (a – b)(a + b) = a² – b²:
    = 19² – 2² = 361 – 4 = 357
  4. 104 × 96 = (100 + 4)(100 – 4):
    = 100² – 4² = 10000 – 16 = 9984
(i) 6241    (ii) 37249    (iii) 357    (iv) 9984
L3.
Expand (3x – 2y + 4z)² using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
NumericalMedium

Solution — Step by step

  1. Identify: a = 3x, b = –2y, c = 4z
  2. Calculate square terms: a² = 9x², b² = 4y², c² = 16z²
  3. Calculate cross terms: 2ab = 2(3x)(–2y) = –12xy
  4. 2bc = 2(–2y)(4z) = –16yz
  5. 2ca = 2(4z)(3x) = 24xz
  6. Add all terms: 9x² + 4y² + 16z² – 12xy – 16yz + 24xz
(3x – 2y + 4z)² = 9x² + 4y² + 16z² – 12xy – 16yz + 24xz
L4.
Factorise 50p² + 60pq + 18q² completely using the identity (a + b)² = a² + 2ab + b².
NumericalMedium

Solution — Step by step

  1. Look for common factor: 2 is a common factor of all three terms.
    50p² + 60pq + 18q² = 2(25p² + 30pq + 9q²)
  2. Focus on 25p² + 30pq + 9q².
  3. Identify: 25p² = (5p)², 9q² = (3q)², and 30pq = 2(5p)(3q)
  4. Compare with a² + 2ab + b² where a = 5p and b = 3q.
  5. 25p² + 30pq + 9q² = (5p + 3q)²
  6. Substitute back: 50p² + 60pq + 18q² = 2(5p + 3q)²
50p² + 60pq + 18q² = 2(5p + 3q)²
L5.
Derive the identity (a + b)³ = a³ + 3a²b + 3ab² + b³ using the distributive property. Then use it to expand (2n + 3m)³.
ProofHard

Solution — Step by step

Part 1: Derivation

  1. Write: (a + b)³ = (a + b)(a + b)²
  2. We know (a + b)² = a² + 2ab + b²
  3. Multiply: (a + b)(a² + 2ab + b²)
  4. = a(a² + 2ab + b²) + b(a² + 2ab + b²)
  5. = a³ + 2a²b + ab² + a²b + 2ab² + b³
  6. Collect like terms: a³ + 3a²b + 3ab² + b³
(a + b)³ = a³ + 3a²b + 3ab² + b³ ✓

Part 2: Expand (2n + 3m)³ with a = 2n, b = 3m:

  1. a³ = (2n)³ = 8n³
  2. 3a²b = 3(2n)²(3m) = 3 × 4n² × 3m = 36n²m
  3. 3ab² = 3(2n)(3m)² = 3 × 2n × 9m² = 54nm²
  4. b³ = (3m)³ = 27m³
(2n + 3m)³ = 8n³ + 36n²m + 54nm² + 27m³
L6.
Identify the side of a cube whose volume is p³ + 6p²q + 12pq² + 8q³ cubic units.
Word ProblemHard

Solution — Step by step

  1. The volume of a cube with side s is s³, and we want to express the given expression as a perfect cube.
  2. Compare p³ + 6p²q + 12pq² + 8q³ with (a + b)³ = a³ + 3a²b + 3ab² + b³.
  3. Identify a³ = p³ → a = p
  4. Identify b³ = 8q³ = (2q)³ → b = 2q
  5. Verify: 3a²b = 3p²(2q) = 6p²q ✓ and 3ab² = 3p(2q)² = 12pq² ✓
  6. So p³ + 6p²q + 12pq² + 8q³ = (p + 2q)³
∴ The side of the cube is (p + 2q) units.
L7.
A rectangular pool has its breadth 4 metres less than its length and its area is 96 square metres. Find the length and breadth using factorisation.
Word ProblemHard

Solution — Step by step

  1. Set up the equation: Let length = x metres, so breadth = (x – 4) metres.
  2. Area equation: x(x – 4) = 96 → x² – 4x = 96 → x² – 4x – 96 = 0
  3. Split the middle term: Need two numbers that multiply to –96 and add to –4. Try –12 and 8: (–12) × 8 = –96 ✓ and (–12) + 8 = –4 ✓
  4. x² – 12x + 8x – 96 = 0
  5. x(x – 12) + 8(x – 12) = 0
  6. (x – 12)(x + 8) = 0 → x = 12 or x = –8
  7. Since length cannot be negative, x = 12 metres.
  8. Breadth = 12 – 4 = 8 metres.
∴ Length = 12 metres and Breadth = 8 metres.
Check: 12 × 8 = 96 sq. m ✓
L8.
Saira has arranged 1 square of side x units, 8 rectangular strips of sides x units and 1 unit, and 15 unit squares to form a larger rectangle. Find the possible length and breadth of this rectangle in terms of x.
Word ProblemMedium

Solution — Step by step

  1. Calculate total area: Area of square of side x = x²
  2. Area of each x × 1 rectangle = x. Total for 8 such rectangles = 8x
  3. Area of each unit square = 1. Total for 15 unit squares = 15
  4. Total area: x² + 8x + 15 sq. units
  5. Factorise: Need a + b = 8 and ab = 15. So a = 3 and b = 5.
  6. x² + 8x + 15 = (x + 3)(x + 5)
∴ Length = (x + 5) units and Breadth = (x + 3) units.
L9.
The sum of three numbers is 10, their product is 25, and the sum of their squares is 38. Use the identity (x + y + z)(x² + y² + z² – xy – xz – yz) = x³ + y³ + z³ – 3xyz to find the sum of their cubes.
Word ProblemHard

Solution — Step by step

  1. Given: x + y + z = 10, xyz = 25, x² + y² + z² = 38
  2. Find xy + xz + yz: Use identity (x + y + z)² = x² + y² + z² + 2(xy + xz + yz)
  3. 100 = 38 + 2(xy + xz + yz) → xy + xz + yz = 31
  4. Substitute into the sum-of-cubes identity: (10)(38 – 31) = x³ + y³ + z³ – 3(25)
  5. 10 × 7 = x³ + y³ + z³ – 75
  6. 70 = x³ + y³ + z³ – 75
  7. x³ + y³ + z³ = 70 + 75 = 145
∴ Sum of their cubes = 145
Key Identity Used: x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – xz – yz)
L10.
A village playground is a square of side 40 metres. A path of width s metres is made around it. Find an expression for the area of the path in terms of s, using a suitable identity.
Word ProblemHard

Solution — Step by step

  1. Area of the inner square (playground): side = 40 m → area = 40² = 1600 sq. m
  2. Outer square (playground + path): The path of width s metres is added on all sides, so outer side = (40 + 2s) metres.
  3. Area of outer square: (40 + 2s)²
  4. Expand using (a + b)²: (40 + 2s)² = 40² + 2(40)(2s) + (2s)² = 1600 + 160s + 4s²
  5. Area of path = Outer area – Inner area:
    = (1600 + 160s + 4s²) – 1600 = 160s + 4s²
  6. Factorise: = 4s(40 + s)
∴ Area of path = 4s(40 + s) sq. metres or equivalently 160s + 4s² sq. m.
Check: For s = 1 m: Area = 4 × 1 × 41 = 164 sq. m — a thin path around a 40m square makes sense.

Topic Coverage Summary

Identities & Definitions
Q1, Q2 · 2 questions
(a+b)² Expansion & Factorisation
Q3, Q4, Q5, Q7, L4 · 5 questions
(a–b)² Identity & Squares
Q6, L2(i,ii) · 3 questions
(a+b+c)² Identity
Q8, L3 · 2 questions
Difference of Squares a²–b²
Q9, L2(iii,iv) · 3 questions
Factorisation x²+bx+c
Q10, Q11, Q15, L8 · 4 questions
Cubic Identities (a±b)³
Q12, Q13, L5, L6 · 4 questions
Pattern Proof & Applications
Q14, L1, L7, L9, L10 · 5 questions

★

Exam Tips & Strategy

Key reminders for answering questions on this chapter

Choosing the Right Identity

  • Perfect square trinomial → (a±b)²
  • Two squared terms with subtraction → a²–b²
  • Three variables squared → (a+b+c)²
  • Cube of a binomial → (a±b)³
  • Sum/difference of cubes → use x³±y³ forms

Factorising Quadratics x²+bx+c

  • Find two numbers that multiply to c and add to b
  • If b is negative and c positive → both numbers negative
  • If c is negative → one positive, one negative number
  • Always check: expand back to verify factors

Numerical Shortcuts

  • n² near 100: write as (100 ± k)²
  • Products a × b where a + b = 2×mid: use a²–b²
  • Identify “round” nearby numbers to simplify
  • Śhrīdharāchārya: a² = (a+b)(a–b) + b²

Rational Expression Simplification

  • Always factorise numerator and denominator fully
  • Take common factors out first
  • Cancel only factors, never individual terms
  • State the condition that cancelled factor ≠ 0

 

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