## MP Board Class 8th Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.5

Question 1.
Use a suitable identity to get each of the following products.
(i) (x + 3)(x + 3)
(ii) (2y + 5)(2y + 5)
(iii) (2a – 7)(2a – 7)

(v) (1.1m – 0.4)(1.1m + 0.4)
(vi) (a2 + b2)(-a2 + b2)
(vii) (6x – 7)(6x + 7)
(viii) (-a + c)(-a + c)

(x) (7a – 9b)(7a – 9b)
Solution:
(i) (x + 3) (x + 3) = (x + 3)2
= (x)2 + 2(x)(3) + (3)2
[Using identity : (a + b)2 = a2 + 2ab + b2] = x2 + 6x + 9

(ii) (2y + 5) (2y + 5) = (2y + 5)2
= (2y)2 + 2(2y)(5) + (5)2
[Using identity : (a + b)2 = a2 + 2ab + b2] = 4a2 + 20y + 25

(iii) (2a – 7)(2a – 7) = (2a – 7)2
= (2a)2 – 2(2a)(7) + (7)2
[Using identity : (a – b)2 = a2 – 2ab + b2] = 4a2 – 28a + 49

(v) (1.1m – 0.4) (1.1m + 0.4) = (1.1m)2 – (0.4)2
[Using identity : a2 – b2 = (a + b) (a – b)]
= 1.21m2 – 0.16

(vi) (a2 + b2) (- a2 + b2) = (b2 + a2) (b2 – a2)
= (b2)2 – (a2)2
[Using identity : (a2 – b2) = (a + b) (a – b)] = b4 – a4

(vii) (6x – 7) (6x + 7) = (6x)2 – (7)2 = 36x2 – 49
[Using identity : (a2 – b2) = (a + b) (a – b)]

(viii) (- a + c) (- a + c) = (c – a) (c – a) = (c – a)2 = (c)2 – 2(c)(a) + a2
[Using identity : (a – b)2 = a2 – 2ab + b2]
= c2 – 2ac + a2

(x) (7a – 9b)(7a – 9b) = (7a – 9b)2 = (7a)2 – 2(7a)(9b) + (9b)2
[Using identity : (a – b)2 = a2– 2ab + b2]
= 49a2 – 126ab + 81b2

Question 2.
Use the identity
(x + a) (x + b) = x2 + (a + b)x + ab to find the following products.
(i) (x + 3)(x + 7)
(ii) (4x + 5)(4x + 1)
(iii) (4x – 5)(4x – 1)
(iv) (4x + 5)(4x – 1)
(v) (2x + 5y)(2x + 3y)
(vi) (2a2 + 9)(2a2 + 5)
(vii) (xyz – 4)(xyz – 2)
Solution:
(i)(x + 3)(x + 7) = x2 + (3 + 7)x + 3 × 7
= x2 + 10x + 21

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

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

(iv) (4x + 5) (4x – 1) (4x + 5) (4x + (-1)]
= (4x + (5 – 1)(4x) + (5) × (-1)
= 16x2 + 16x – 5

(v) (2x + 5y)(2x + 3y)
= (2x)2 + (5 + 3)y (2x) + (5y) × (3y)
= 4x2 + 16xy + 15y2

(vi) (2a2 + 9) (2a2 + 5)
= (2a2)2 + (9 + 5) (2a2) + 9 × 5
= 4a4 + 28a2 +45

(vii) (xyz – 4) (xyz – 2)
= [xyz + (-4)] [xyz + (-2)]
= (xyz)2 + (- 4 – 2) (xyz) + (- 4)(-2)
= x2y2z2 – 6xyz + 8

Question 3.
Find the following squares by using the identities.
(i) (b – 7)2
(ii) (xy + 3z)2
(iii) (6x2 – 5y)2

(v) (0.4p – 0.5q)2
(vi) (2xy + 5y)2
Solution:
(i) (b – 7)2 = (b)2 – 2(b)(7) + (7)2
[Using identity : (a – b)2 = a2 – 2ab + b2]
= b2 – 14b + 49

(ii) (xy + 3z)2 = (xy)2 + (3z)2 + 2(xy) (3z)
[Using identity : (a + b)2 = a2 + b2 + 2ab]
= x2y2 + 9z2 + 6xyz

(iii) (6x2 – 5y)2 = (6x2)2 + (5y)2 – 2(6x2) (5y)
[Using identity : (a – b)2 = a2 + b2 – 2ab]
= 36x4 + 25y2 – 60x2y

(v) (0.4p – 0.5p)2
= (0.4p)2 + (0.5q)2 – 2(0.4p) (0.5q)
[Using identity : (a – b)2 = a2 + b2 – 2ab]
= 0.16p2 + 0.25q2 – 0.4pq

(vi) (2xy + 5y)2
= (2xy)2 + (5y)2 + 2(2xy) (5y)
[Using identity : (a + b)2 = a2 + b2 + 2ab] = 4x2y2 + 25 y2 + 20xy2

Question 4.
Simplify.
(i) (a2 – b2)2
(ii) (2x + 5)2 – (2x – 5)2
(iii) (7m – 8n)2 + (7m + 8n)2
(iv) (4m + 5n)2 + (5m + 4n)2
(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
(vi) (ab + bc)2 – 2ab2c
(vii) (m2 – n2m)2 + 2m3n2
Solution:
(i) (a2 – b2)2 = (a2)2 – 2a2b2 + (b2)2
[Using identity : (a – b)2 = a2 – 2ab + b2]
= a4 – 2a2b2 + b4

(ii) (2x + 5)2 – (2x – 5)2
=[2x + 5 – (2x – 5)][2x + 5 + 2x – 5]
[Using identity : a2 – b2 – (a – b) (a + h)]
= (10)(4x) = 40x

(iii) (7m – 8n)2 + (7m + 8n)2
= (7m)2 – 2(7m) (8n) + (8n)2 + (7m)2 + 2(7m)(8n) + (8n)2
[Using identity : (a – b)2 = a2 – 2ab + b2]
and (a + b)2 = a2 + 2ab + b2]
= 49m2 – 112mn + 64n2 + 49m2 + 112mn + 64n2
= 98m2 + 128n2

(iv) (4m + 5n)2 + (5m + 4n)2
(4m)2 + 2(4m) (5n) + (5n)2 + (5m)2 + 2(5m)(4n) + (4n)2
(Using identity : (a + b)2 = a2 + 2ab + b2)
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 80mn + 41n2

(v) (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]
[Using identity: (a – b)2 = a2 – 2ab + b2]
= 6.25p2 – 7.5pq + 2.25q2 – [225p2 – 7.5pq + 6.25q2]
= 6.25p2 – 7.5pq + 2.25q2 – 2.25p2 + 7.5pq – 6.25q2
= 4p2 – 4q2

(vi) (ab + bc)2 – 2ab2c
= (ab)2 + 2(ab)(bc) + (bc)2 – 2ab2c
[Using identity : (a + b)2 = a2 + 2ab + b2]
= a2b2 + b2c2 – 2ab2c + 2ab2c = a2b2 + b2c2

(vii) (m2 – n2n)2 + 2m3n2
=(m2)2 – 2(m2)(n2m) + (n2m)2 + 2m3n2
[Using identity : (a – b)2 = a2 – 2ab + b2]
= m4 – 2m3n2 + 2m3n2 = m4 + n4m2

Question 5.
Show that
(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) (9p – 5q)2 + 180pq = (9p + 5q)2

(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
(v) (a – b)(a + b) + (b – c)(b + c) + (c – a)(c + a) = O
Solution:
(i) To prove: (3x + 7)2 – 84x = (3x – 7)2
LH.S. = (3x + 7)2 – 84x
= (3x)2 + 2(3x)(7) + (7)2 – 84x
= 9x2 + 42x +49 – 84x = 9x2 – 42x + 49
R.H.S. = (3x – 7)2 = (3x)2 – 2(3x)(7) + (7)2 = 9x2 – 42x + 49
∴ L.H.S. = R.H.S.

(ii) To prove : (9p – 5q)2 + 180pq = (9p + 5q)2
L.H.S. = (9p – 5q)2 + 180pq
= (9p)2 – 2(9p)(5q) + (5q)2 + 180pq
= 81p2 – 90pq + 25q2 + 180pq
= 81 p2 + 90 pq + 25 q2
R.H.S. = (9p + 5q)2
= (9p)2 + 2(9p)(5q) + (5q)2
= 81p2 + 90pq + 25q2
∴ L.H.S. = R.H.S.

(iii) To prove :

(iv) To prove : (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
L.H.S. = (4pq + 3q)2 – (4pq – 3q)2
= (4 pq)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
= 48pq2 = R.H.S.

(v) To prove : (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
L.H.S. = (a – b) (a + b) + (b – c) (b + c) + (c – a)(c + a)
= a2 – b2 + b2 – c2 + c2 – a2
[Using identity : (x – y)(x + y) = x2 – y2]
= 0 = R.H.S.

Question 6.
Using identities, evaluate,
(i) 712
(ii) 992
(iii) 1022
(iv) 9982
(v) 5.22
(vi) 297 × 303
(vii) 78 × 82
(viii) 8.92
(ix) 1.05 × 9.5
Solution:
(i) 712 = (70 + 1)2
= (70)2 + 2(70)(1) + 12
[Using identity : (a + b)2 = a2 + 2ab + b2]
= 4900 + 140 + 1 = 5041

(ii) 992 = (100 – 1)2
= (100)2 – 2(100) (1) + 12
[Using identity : (a – b)2 = a2 – 2ab + b2]
= 10000 – 200 + 1 = 9801

(iii) (102)2 = (100 + 2)2
= (100)2 + 2(100)(2) + 22
[Using identity : (a + b)2 = a2 + 2ab + b2]
= 10000 + 400 + 4 = 10404

(iv) (998)2 = (1000 – 2)2
= (1000)2 – 2(1000)(2) + (2)2
[Using identity : (a – b)2 = a2 – 2ab + b2]
= 1000000 – 4000 + 4 = 996004

(v) (5.2)2 = (5 + 0.2)2
= (5)2 + 2(5)(0.2) + (0.2)2
[Using identity : (a + b)2 = a2 + 2ab + b2]
= 15 + 1 + 0.04 = 27.04

(vi) 297 2 × 303 = (300 – 3) (300 + 3)
= (300)2 – 32
[Using identity : (a – b) (a + b) = (a2 – b2)
= 90000 – 9 = 89991

(vii) 78 × 82 = (80 – 2) (80 + 2)
= (80)2 – (2)2
[Using identity : (a -b) (a + b) = a2 – b2]
= 6400 – 4 = 6396

(viii) (8.9)2 = (9 – 0.1)2
= (9)2 – 2(9)(0.1) + (0.1)2
[Using identity : (a – b)2 = a2 – 2ab + b2] = 81 – 1.8 + 0.01 = 79.21

(ix) 1.05 × 9.5

Question 7.
Using a2 – b2 = (a + b)(a – b), find
(i) 512 – 492
(ii) (1.02)2 – (0.98)2
(iii) 1532 – 1472
(iv) 12.12 – 7.92
Solution:
(i) 512 – 492 = (51 – 49) (51 + 49)
= (2)(100) = 200
(ii) (1.02)2 – (0.98)2 = (1.02 – 0.98) (1.02 + 0.98)
= (0.04)(2) = 0.08
(iii) 1532 – 1472 = (153 + 147) (153 – 147)
= (300) (6) = 1800
(iv) (12.1)2 – (7.9)2 = (12.1 + 7.9) (12.1 – 7.9)
= (20.0) (4.2) = 84

Question 8.
Using (x + a)(x + b) = x2 + (a + b)x + ab, find
(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8
Solution:
(i) 103 × 104 = (100 + 3)(100 + 4)
= (100)2 + (3 + 4) 100 + 3 × 4
= 10000 + 700 + 12 = 10712

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

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

(iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2)
= [10 + (- 0.3)[10 + (- 0.2)]
= (10)2 + (- 0 3 – 0 2) × 10 + (- 0.3) (- 0.2)
= 100 – 5 + 0.06 = 95.06