GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

   

Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

Question 1.
Use a suitable identity to get each of the following products?

  1. (x + 3)(x + 3)
  2. (2y + 5)(2y + 5)
  3. (2a – 7)(2a – 7)
  4. (3a – \(\frac{1}{2}\)) (3a – \(\frac{1}{2}\))
  5. (1.1m – 0.4)(1.1m + 0.4)
  6. (a2 + b2)(-a2 + b2)
  7. (6x – 7)(6x + 7)
  8. (-a + c)(-a + c)
  9. (\(\frac{x}{2}\) + \(\frac{3y}{4}\)) (\(\frac{x}{2}\) + \(\frac{3y}{4}\))
  10. (7a – 9b) (7a – 9b)

Solution:
1. Using the identity,
(x + a)(x + b) = x2 + (a + b)x + ab
we have
(x + 3)(x + 3) = x2 + (3 + 3)x + (3 × 3)
= x2 + 6x + 9

2. Using the identity,
(x + a)(x + b) = x2 + (a + b)x + ab,
we have
(2y + 5)(2y + 5)
= (2y)2 + (5 + 5)2y + (5 × 5)
= 4y2 + (10)2y + 5
= 4y2 + 20y + 25

3. Using the identity, (a – b)2 = a2 – 2ab + b2
(2a – 7)(2a – 7) = (2a – 7)2
= (2a)2 – 2(2a)(7) + (7)2
= 4a2 – 28a + 49

4. Using the identity, (a – b)2 = a2 – 2ab + b2,
we have
(3a – \(\frac{1}{2}\)) (3a – \(\frac{1}{2}\)) = (3a – \(\frac{1}{2}\))2
= (3a)2 – 2(3a) (\(\frac{1}{2}\)) + (\(\frac{1}{2}\))2 = 9a2 – 3a + \(\frac{1}{4}\)

5. Using the identity, (a + b)(a -b) = a2 – b2,
we have
(1.1m – 0.4)(1.1 m – 0.4)
= (1.1m)2 – (0.4)2 = 1.21m2 – 0.16

6. ∵ (a2 + b2)(-a2 + b2) = (b2 + a2)(b2 – a2)
Using the identity,
(a + b)(a – b) = a2 – b2, we have
(b2 + a2)(b2 – a2) = (b2)2 – (a2)2 = b4 – a4

7. ∵ (6x – 7)(6x + 7) = (6x + 7)(6x – 7)
∴ Using the identity,
(a + b)(a – b) = a2 – b2,
we have
(6x + 7)(6x – 7) = (6x)2 – (7)2
= 36x2 – 49

8. Using the identity, (a + b)2 = a2 + 2ab + c2, we have
(-a + c)(-a + c) = (-a + c)2
= (-a)2 + 2(-a)(c) + (c)2
= a2 + 2(-ac) + c2
= a2 – 2ac + c2

9. Using the identity, (a + b)2 = a2 + b2 + 2ab, we have
(\(\frac{x}{2}\) + \(\frac{3y}{4}\)) (\(\frac{x}{2}\) + \(\frac{3y}{4}\)) = (\(\frac{x}{2}\) + \(\frac{3y}{4}\))2
= (\(\frac{x}{2}\))2 + 2(\(\frac{x}{2}\))(\(\frac{3y}{4}\)) + (\(\frac{3y}{4}\))2
= \(\frac{x^{2}}{4}\) + \(\frac{3xy}{4}\) + \(\frac{9y^{2}}{16}\)

10. Using the identity (a – b)2 = a2 – 2ab + b2
we have
(7a – 9b)(7a – 9b) = (7a – 9b)2
= (7a)2 – 2(7a)(9b) + (9b)2
= 49a2 – 126ab + 81b2

GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

Question 2.
Use the identity (x + a)(x + b) = x2 + (a + b)x + ab to find the following products.

  1. (x + 3)(x + 7)
  2. (4x + 5)(4x + 1)
  3. (4x – 5)(4x – 1)
  4. (4x + 5)(4x – 1)
  5. (2x + 5y)(2x + 3y)
  6. (2a2 + 9)(2a2 + 5)
  7. (xyz – 4)(xyz – 2)

Solution:
1. (x + 3)(x + 7) = x2 + (3 + 7)x + (3 × 7)
= x2 + 10x + 21

2. (4x + 5)(4x +1)
= (4x)2 + (5 + 1)4x + (5 × 1)
= 16x2 + 6 × 4x + 5
= 16x2 + 24x + 5

3. (4x – 5)(4x – 1)
= (4x)2 + (-5 – 1)4x + [(-5) × (-1)]
= 16x2 + (-6)4x + (5)
= 16x2 – 24x + 5

4. (4x + 5)(4x – 1)
= (4x)2 + (5 – 1)4x + [5 × (-1)]
= 16x2 + (4)4x + (-5)
= 16x2 + 16x – 5

5. (2x + 5y)(2x + 3y)
= (2x)2 + (5y + 3y)2x + (5y × 3y)
= 4x2 + (5 + 3)y × 2x + 15y2
= 4x2 + [8y × 2x] + 15y2
= 4x2 + 16xy + 15y2

6. (2a2 + 9)(2a2 + 5)
= [2a2]2 + (9 + 5)2a2 + 9 × 5
= 4a4 + 14 × 2a2 + 45
= 4a4 + 28a2 + 45

7. (xyz – 4)(xyz – 2)
= (xyz)2 + [(-4) + (-2)]xyz + (-4) × (-2)
= x2y2z2 + [-6]xyz + 8
= x2y2z2 – 6xyz + 8

GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

Question 3.
Find the following squares by using the identities?

  1. (b – 7)2
  2. (xy + 3z)2
  3. (6x2 – 5y)2
  4. (\(\frac{2}{3}\)m + \(\frac{3}{2}\)n)2
  5. (0.4p – 0.5q)2
  6. (2xy + 5y)2

Solution:
1. (b – 7)2
Using (a – b)2 = a2 – 2ab + b2, we have
(b – 7)2 = (b)2 – 2(b)(7) + (7)2
= b2 – 14b + 49

2. (xy + 3z)2
Using (a + b)2 = a2 + 2ab + b2, we have
(xy + 3z)2 = (xy)2 + 2(xy)(3z) + (3z)2
= x2y2 + 6xyz + 9z2

3. (6x2 – 5y)2
Using the identity (a – b)2 = a2 – 2ab + b2,
we have
(6x2 – 5y)2 = (6x2)2 – 2(6x2)(5y) + (5y)2
= 36x4 – 60x2y + 25y2

4. (\(\frac{2}{3}\)m + \(\frac{3}{2}\)n)2
Using (a + b)2 = a2 + 2ab + b2, we have
(\(\frac{2}{3}\)m + \(\frac{3}{2}\)n)2
= (\(\frac{2}{3}\)m)2 + 2(\(\frac{2}{3}\)m)(\(\frac{3}{2}\)n) + (\(\frac{3}{2}\)n)2
= \(\frac{4}{9}\)m2 + 2mn + \(\frac{9}{4}\)n2

5. (0.4p – 0.5q2)
Using (a – b)2 = a2 – 2ab + b2,
we have (0.4p – 0.5q)2
= (0.4p)2 – 2(0.4p)(0.5q) + (0.5q)2
= 0.16p2 – 0.4pq + 0.25q2

6. (2xy + 5y2)
Using (a + b)2 = a2 + 2ab + b2, we have
(2xy + 5y)2 = (2xy)2 + 2(2xy)(5y) + (5y)2
= 4x2y2 + 20xy2 + 25y2

GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

Question 4.
Simplify:

  1. (a2 – b2)2
  2. (2x + 5)2 – (2x – 5)2
  3. (7m – 8n)2 + (7m + 8n)2
  4. (4m + 5n)2 + (5m + 4n)2
  5. (2.5p – 1.5q)2 – (1.5p – 2.5q)2
  6. (ab + bc)2 – 2ab2c
  7. (m2 – n2m)2 + 2m3n2

Solution:
1. (a2 – b2)2 = (a2)2 – 2a2 × b2 + (b2)2
[Using (a – b)2 = a2 – 2ab + b2]
= a4 – 2a2b2 + b4

2. (2x + 5)2 – (2x – 5)2
= [(2x)2 + 2(2x)(5) + (5)2] – [(2x)2 – 2(2x)(5) + (5)2]
= [4x2 + 20x + 25] – [4x2 – 20x + 25]
= 4x2 + 20x + 25 – 4x2 + 20x – 25
= (4 – 4)x2 + (20 + 20)x + (25 – 25)
= (0)x2 + 40x + 0 = 40x

3. (7m – 8n)2 + (7m + 8n)2
= [(7m)2 – 2(7m)(8n) + (8n)2] + [(7m)2 + 2(7m)(8n) + (8n)2]
= [49m2 – 112mn + 64n2] + [49m2 + 112mn + 64n2]
= 49m2 – 112mn + 64n2 + 49m2 + 112mn + 64n2
= (49 + 49)m2 + (112 – 112)mn + (64 + 64)n2
= 98m2 + (0)mn + 128n2 = 98m2 + 128n2

4. (4m + 5n)2 + (5m + 4n)2
= [(4m)2 + 2(4m)(5n) + (5n)2] + [(5m)2 + 2(5m)(4n) + (4n)2]
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= (16 + 25)m2 + (40 + 40)mn + (25 + 16)2
= 41m2 + 80mn + 41n2

5. (2.5p – 1.5q)2 – (1.5p – 2.5q)2
= [(2.5p)2 – 2(2.5p)(1.5q) + (1.5q)2] – [(1.5p)2 – 2(1.5p) (2.5q) + (2.5q)2]
= 6.25p2 – 7.5pq + 2.25q2 – [2.25p2 – 7.5pq + 6.25q2]
= 6.25p2 – 7.5pq + 2.25q2 – 2.25p2 + 7.5pq – 6.25)q2
= 4p2 + 0(pq) – 4q2 = 4p2 = 4p2 – 4q2

6. (ab + bc)2 – 2ab2c
= (ab)2 + 2(ab)(bc) + (bc)2 – 2ab2c
= a2b2 + 2ab2c + b2c2 – 2ab2c
= a2b2 + (2 – 2)ab2c + b2c2
= a2b2 + (0)ab2c + b2c2
= a2b2 + 0 + b2c2 = a2b2 + b2c2

7. (m2 – n2m)2 + 2m3n2
= (m2)2 – 2(m2)(n2m) + (n2m)2 + 2m3n2
= m4 – 2m3n2 + n4m2 + 2m3n2
= m4 + (-2 + 2)m3n2 + n4m2
= m4 + (0)m3n2 + n4m2 = m4 + m2n4

GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

Question 5.
Show that

  1. (3x + 7)2 – 84x = (3x – 7)2
  2. (9p – 5q)2 + 180pq = (9p + 5q)2
  3. (\(\frac{4}{3}\)m – \(\frac{3}{4}\)n)2 + 2mn = \(\frac{16}{9}\)m2 + \(\frac{9}{16}\)n2
  4. (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
  5. (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = 0

Solution:
1. LHS = (3x + 7)2 – 84x
= (3x)2 + 2(3x)(7) + (7)2 – 84x
= 9x2 + 42x + 49 – 84x
= 9x2 + (42 – 84)x + 49
= 9x2 – 42x + 49
RHS = (3x – 7)2
(3x)2 – 2(3x)(7) + (7)2
= 9x2 – 42x + 49 = LHS
Since, LHS = RHS
∴ (3x + 7)2 – 84x = (3x – 7)2

2. LHS = (9p – 5q)2 + 180pq
= (9p)2 – 2(9p)(5q)2 + 180pq
= 81p2 – 90pq + 25q2 + 180pq
= 81p2 + (-90 + 180)pq + 25q2
= 81p2 + 90pq + 25q2

RHS = (9p + 5q)2
= (9p)2 + 2(9p)(5q) + (5q)2
= 81p2 + 90pq + 25q2
Since, LHS = RHS
∴ (9p – 5q)2 + 180pq = (9p + 5q)2

3. LHS = [\(\frac{4}{3}\)m – \(\frac{3}{4}\)n]2 + 2mn
= (\(\frac{4}{3}\)m)2 – 2(\(\frac{4}{3}\)m)(\(\frac{3}{4}\)n) + (\(\frac{3}{4}\)n)2 + 2mn
= \(\frac{16}{9}\)m2 – 2mn + \(\frac{9}{16}\)n2 + 2mn
= \(\frac{16}{9}\)m2 + (-2 + 2)mn + \(\frac{9}{16}\)n2
= \(\frac{16}{9}\)m2 + (0)mn + \(\frac{9}{16}\)n2
= \(\frac{16}{9}\)m2 + \(\frac{9}{16}\)n2 = RHS
Since, RHS = LHS
∴ (\(\frac{4}{3}\)m – \(\frac{3}{4}\)n)2 + 2mn = \(\frac{16}{9}\)m2 + \(\frac{9}{16}\)n2

4. LHS = (4pq + 3q)2 – (4pq – 3q)2
= [(4pq)2 + 2(4pq)(3q) + (3q)2] – [(4pq)2 – 2(4pq)(3q) + (3q)2]
= 16p2q2 + 24pq2 + 9q2 – [16p2q2 – 24pq2 + 9q2]
= 16p2q2 + 24pq2 + 9q2 – 16p2q2 + 24pq2 – 9q2
= (16 – 16)p2q2 + (24 + 24)pq2 + (9 – 9)q2
= (0)p2q2 + 48pq2 + (0)q2 = 48pq2 = RHS
Since, LHS = RHS
∴ (4pq + 3q)2 – (4pq – 3q)2 = 48pq2

5. LHS = (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a)
= (a2 – b2) + (b2 – c2) + (c2 – a2)
= a2 – b2 + b2 – c2 + c2 – a2
= a2 – a2 + b2 – b2 + c2 – c2 = 0 = RHS
Since, LHS = RHS
∴ (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a)= 0

GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

Question 6.
Using identities, evaluate:

  1. 712
  2. 992
  3. 1022
  4. 9982
  5. 5.22
  6. 297 × 303
  7. 78 × 82
  8. 8.92
  9. 1.05 × 9.5

Solution:
1. ∵ 71 = 70+1
∴ (71)2 = (70 + 1)2
= (70)2 + 2(70)(1) + (1)2 [Using (a + b)2]
= a2 + 2ab + b2
= 4900 + 140 + 1 = 5041

2. We have 100 – 1 = 99
∴ (99)2 = (100 – 1)2
= (100)2 – 2(100)(1) + (1)2
[Using (a – b)2 = a2 – 2ab + b2]
= 10000 – 200 + 1
= 9801

3. ∵ 102 = 100 + 2
(102)2 = (100 + 2)2
= (100)2 + 2(100)(2) + (2)2 [Using (a + b)2 = a2 + 2ab + b2]
= 10000 + 400 + 4 = 10404

4. ∵ 998 = 1000 – 2
∴ (998)2 = (1000 – 2)2
= (1000)2 – 2(1000)(2) + (2)2
[Using (a – b)2 = a2 – 2ab + b2]
= 1000000 – 4000 + 4
= 996004

5. ∵ 5.2 = 5 + 0.2
∴ (5.2)2 = (5 + 0.2)2
= (5)2 + 2(5)(0.2) + (0.2)2
[Using (a + b)2 = a2 + 2ab + b2]
= 25 + 2 + 0.04
= 27.04

6. ∵ 297 = 300 – 3 and 303 = 300 + 3
∴ 297 × 303 = (300 – 3)(300 + 3)
= (300)2 – (3)2
[Using (a + b)(a – b) = a2 – b2]
= 90000 – 9
= 89991

7. ∵ 78 = 80 – 2 and 82 = 80 + 2
∴ 78 × 82 = (80 – 2)(80 + 2)
= 802 – (2)2
[Using (a + b)(a – b) = a2 – b2]
= 6400 – 4
= 6396

8. ∵ 8.9 = (9 – 0.1)
∴ (8.9)2 = (9 – 0.1)2
= (9)2 – 2(9)(0.1) + (0.1)2
[Using (a – b)2 = a2 – 2ab + b2]
= 81 – 1.8 + 0.01
= 81.01 – 1.80
= 79.21

9. ∵ 1.05 = 1 + 0.05
∴ (1.05) × 9.5 = (1 + 0. 05) × 9.5
= (1 × 9.5) + (9.5 × 0.05)
= 9.500 + 0.475
= 9.975

GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

Question 7.
Using a2 – b2 = (a + b)(a – b), find:

  1. 512 – 492
  2. (1.02)2 – (0.98)2
  3. 1532 – 1472
  4. 12.12 – 7.92

Solution:
1. 512 – 492 = (51 + 49)(51 – 49)
= (100) × (2)
= 200

2. (1.02)2 – (0.98)2
= (1.02 + 0.98)(1.02 – 0.98)
= (2.0) × (0.04) = 0.08

3. 1532 – 1472
= (153 + 147)(153 – 147)
= (300) × (6) = 1800

4. (12.1)2 – (7.9)2
= (12.1 + 7.9)(12.1 – 7.9)
= 20 × 4.2
= 84

GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

Question 8.
Using (x + a)(x + b) = x2 + (a + b)x + ab, find:

  1. 103 × 104
  2. 5.1 × 5.2
  3. 103 × 98
  4. 9.7 × 9.8

Solution:
1. 103 × 104 = (100 + 3) × (100 + 4)
= (100)2 + (3 + 4) × 100 + (3 × 4)
= 10000 + 700 + 12
= 10712

2. 5.1 × 5.2 = (5 + 0.1)(5 + 0.2)
= (5)2 + (0.1 + 0.2) × 5 + (0.1 × 0.2)
= 25 + (0.3) × 5 + 0.02
= 25 + 1.5 + 0.02
= 26.52

3. 103 × 98 = (100 + 3)(100 – 2)
= (100)2 + (3 – 2)100 + [3 × (- 2)]
= 10000 + 100 – 6
= 10094

4. 9.7 × 9.8 = (10 – 0.3)(10 – 0.2)
= (10)2 + [(-0.3) + (-0.2)] 10 + [(-0.3) × (-0.2)]
= 100 + [- 0.5] × 10 + 0.06
= 100 – 5 + 0.06
= 95.06

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