CG Board Class 8 Mathematics Chapter 9 Solutions: Identities (सर्वसमिकाएँ) – 2026-27

(i) (2a + 3)(2a + 3)**

Using Identity 1: (a + b)² = a² + 2ab + b² where first term = 2a, second term = 3

(2a + 3)² = (2a)² + 2(2a)(3) + (3)² = 4a² + 12a + 9

(ii) (2/5m + 3/4)(2/5m + 3/4)

Using Identity 1: (a + b)² = a² + 2ab + b² where a = 2/5m, b = 3/4

(2/5m + 3/4)² = (2/5m)² + 2(2/5m)(3/4) + (3/4)² = 4/25m² + 3/5m + 9/16

(i) (x – 5)(x – 5)**

Using Identity 2: (a – b)² = a² – 2ab + b² where a = x, b = 5

(x – 5)² = x² – 2(x)(5) + (5)² = x² – 10x + 25

(ii) (3/2x – 4/5y)(3/2x – 4/5y)

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

(3/2x – 4/5y)² = (3/2x)² – 2(3/2x)(4/5y) + (4/5y)² = 9/4x² – 12/5xy + 16/25y²

(iii) (2a – 1/2)(2a – 1/2)

Using Identity 2

(2a – 1/2)² = (2a)² – 2(2a)(1/2) + (1/2)² = 4a² – 2a + 1/4

(iv) (x² – y²)(x² – y²)

Using Identity 2 with a = x² and b = y²

(x² – y²)² = (x²)² – 2(x²)(y²) + (y²)² = x⁴ – 2x²y² + y⁴

(i) (4x + 5)(4x – 5)**

Using Identity 3: (a + b)(a – b) = a² – b² where a = 4x, b = 5

(4x)² – (5)² = 16x² – 25

(ii) (x/2 + y/3)(x/2 – y/3)

Using Identity 3

(x/2)² – (y/3)² = (x²)/(4) – (y²)/(9)

(iii) (-a² + b²)(a² + b²)

Rearranging as (b² – a²)(b² + a²) and using Identity 3

(b²)² – (a²)² = b⁴ – a⁴

(iv) (x³ + y³)(x³ – y³)

Using Identity 3 with a = x³, b = y³

(x³)² – (y³)² = x⁶ – y⁶

(i) (2a + 5)²**

Using Identity 1

(2a)² + 2(2a)(5) + (5)² = 4a² + 20a + 25

(ii) (2/3m² + 5/6n²)²

Using Identity 1

(2/3m²)² + 2(2/3m²)(5/6n²) + (5/6n²)² = 4/9m⁴ + 20/18m²n² + 25/36n⁴ = 4/9m⁴ + 10/9m²n² + 25/36n⁴

(iii) (8x³ – 5y³)²

Using Identity 2

(8x³)² – 2(8x³)(5y³) + (5y³)² = 64x⁶ – 80x³y³ + 25y⁶

(i) (41)²**

(ii) (69)²

(iii) (97)²

(iv) (84)²

(i) 105 × 95**

(ii) 92 × 88

(iii) 503 × 497

(i) (28)² – (22)² = 6x**

Using a² – b² = (a + b)(a – b)

(28 + 22)(28 – 22) = 6x ⇒ (50)(6) = 6x ⇒ 6x = 300 ⇒ x = 50

(ii) (31)² – (17)² = 14x

(31 + 17)(31 – 17) = 14x ⇒ (48)(14) = 14x ⇒ x = 48

(iii) (1.9)² – (0.3)² = 1.6x

(1.9 + 0.3)(1.9 – 0.3) = 1.6x ⇒ (2.2)(1.6) = 1.6x ⇒ x = 2.2

(iv) (43)² – (28)² = 15x

(43 + 28)(43 – 28) = 15x ⇒ (71)(15) = 15x ⇒ x = 71

Expansion using Identity 2:**

(3x – 5y)² = (3x)² – 2(3x)(5y) + (5y)² = 9x² – 30xy + 25y²

• Verification for x = 4, y = 2:
• LHS: [3(4) – 5(2)]² = (12 – 10)² = (2)² = 4
• RHS: 9(4)² – 30(4)(2) + 25(2)² = 9(16) – 240 + 25(4) = 144 – 240 + 100 = 4

Since LHS = RHS = 4, verified!

Expansion:**

(x/3 + y/5)² = (x/3)² + 2(x/3)(y/5) + (y/5)² = (x²)/(9) + 2xy/15 + (y²)/(25)

• Verification for x = 9, y = 12:
• LHS: (9/3 + 12/5)² = (3 + 12/5)² = (27/5)² = 729/25
• RHS: 81/9 + (2(9)(12))/(15) + 144/25 = 9 + 72/5 + 144/25 = (225 + 360 + 144)/(25) = 729/25

Since LHS = RHS, verified!

(i) 4x² + 20xy + 25y²**

Form: a² + 2ab + b² = (a + b)² where a = 2x, b = 5y

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

(ii) 25a² + 70ab + 49b²

(iii) 9x² + 6x + 1

(iv) 1 + 18a + 81a²

(v) p² + p + 1/4

(vi) 36a² + 132ab + 121b²

(i) a² – 10ab + 25b²**

Form: a² – 2ab + b² = (a – b)²

(a)² – 2(a)(5b) + (5b)² = (a – 5b)(a – 5b)

(ii) 16x² – 104x + 169

(iii) 121x² – 88xy + 16y²

(iv) x² – 30x + 225

(v) 36a² – 12a + 1

(i) 25a² – 49b²**

Using Identity 3: a² – b² = (a + b)(a – b)

(5a)² – (7b)² = (5a + 7b)(5a – 7b)

(ii) 9x² – 121y²

(iii) 64a² – 1

(iv) 1 – 16b²

(v) 16/25m² – 4/9n²

(i) (x + 4y)² – 49**

Using A² – B² = (A + B)(A – B) where A = (x + 4y) and B = 7

[(x + 4y) + 7][(x + 4y) – 7] = (x + 4y + 7)(x + 4y – 7)

(ii) 100 – (2a + 3b)²

(10)² – (2a + 3b)² = [10 + (2a + 3b)][10 – (2a + 3b)] = (10 + 2a + 3b)(10 – 2a – 3b)

(iii) 4x² + 20xy + 25y² – 36

Group first three terms: (2x + 5y)² – 6²

[(2x + 5y) + 6][(2x + 5y) – 6] = (2x + 5y + 6)(2x + 5y – 6)

(iv) 9x² – 4x – 5y expression form:

(v) x²y² – 16

(i) x² – 25y² = (x + 5y)( )**

Blank = x – 5y.

(ii) x² + 6x + 9 = (x + )(x + )

Blanks = 3, 3 (or 3, x as in key notation).

(iii) 4x² – 28x + 49 = (2x – )(2x – )

Blanks = 7, 7.

(iv) (a + b)² – 4 = (a + b + 2)( )

Blank = a + b – 2.

Side of original square field = x m.

• Width of path = 5 m.
• Side of remaining inner field = (x – 5) m.
• Remaining Area = (x – 5)² = x² – 2(x)(5) + 5² = (x² – 10x + 25) sq. m.

Let the initial number of idols = x.

• Initial total flowers needed = x × x = x².
• New number of idols = x + 2.
• New total flowers needed = (x + 2)².
• According to the question

(x + 2)² – x² = 24

x² + 4x + 4 – x² = 24 ⇒ 4x + 4 = 24 ⇒ 4x = 20 ⇒ x = 5

Therefore, the previous number of idols was 5.

Area = Length × Width

• Width = Area)/(Length) = x² – 25)/(x + 5)
• Factorizing numerator using a² – b² = (a + b)(a – b)

Width = (x + 5)(x – 5))/(x + 5) = (x – 5) m

Let width = x feet, then length = (x + 1) feet.

• According to the question

(x + 1)² – x² = 5

x² + 2x + 1 – x² = 5 ⇒ 2x + 1 = 5 ⇒ 2x = 4 ⇒ x = 2

• Width = 2 feet, Length = 3 feet.