The subject experts of Maths have solved each question meticulously, which will help students to solve any question from NCERT Class 8 Maths Chapter 9 Exercise 9.5. Click now and download the free NCERT solutions to prepare for the final examination, do homework or learn a concept. This exercise explains how to simplify binomials and numbers using algebraic identities. Download free NCERT Solutions for Maths Chapter 9 prepared by BYJU’S subject experts and practise offline.
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5
Access answers of Maths NCERT Class 8 Chapter 9 Algebraic Expressions and Identities Exercise 9.5 Page number 151
Exercise 9.5 Page No: 151
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)
(iv) (3a – 1/2)(3a – 1/2)
(v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a2+ b2) (- a2+ b2)
(vii) (6x – 7) (6x + 7)
(viii) (- a + c) (- a + c)
(ix) (1/2 x + 3/4 y) (1/2 x + 3/4 y)
(x) (7a – 9b) (7a – 9b)
Solution:
(i) (x + 3) (x + 3) = (x + 3)2
= x2 + 6x + 9
Using (a+b) 2 = a2 + b2 + 2ab
(ii) (2y + 5) (2y + 5) = (2y + 5)2
= 4y2 + 20y + 25
Using (a+b) 2 = a2 + b2 + 2ab
iii) (2a – 7) (2a – 7) = (2a – 7)2
= 4a2 – 28a + 49
Using (a-b) 2 = a2 + b2 – 2ab
iv) (3a – 1/2)(3a – 1/2) = (3a – 1/2)2
= 9a2 -3a+(1/4)
Using (a-b) 2 = a2 + b2 – 2ab
v) (1.1m – 0.4) (1.1m + 0.4)
= 1.21m2 + 0.44 – 0.44m – 0.16
= 1.21m2 – 0.16
Using (a – b)(a + b) = a2 – b2
vi) (a2+ b2) (– a2+ b2)
= (b2 + a2 ) (b2 – a2)
= -a4 + b4
Using (a – b)(a + b) = a2 – b2
vii) (6x – 7) (6x + 7)
=36x2 – 49
Using (a – b)(a + b) = a2 – b2
viii) (– a + c) (– a + c) = (– a + c)2
= c2 + a2 – 2ac
Using (a-b) 2 = a2 + b2 – 2ab
= (x2/4) + (9y2/16) + (3xy/4)
Using (a+b) 2 = a2 + b2 + 2ab
x) (7a – 9b) (7a – 9b) = (7a – 9b)2
= 49a2 – 126ab + 81b2
Using (a-b) 2 = a2 + b2 – 2ab
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 + 21
= x2 + 10x + 21
ii) (4x + 5) (4x + 1)
= 16x2 + (5 + 1)4x + 5
= 16x2 + 24x + 5
iii) (4x – 5) (4x – 1)
= 16x2 – 4x – 20x + 5
= 16x2 – 24x + 5
iv) (4x + 5) (4x – 1)
= 16x2 + (5-1)4x – 5
= 16x2 +16x – 5
v) (2x + 5y) (2x + 3y)
= 4x2 + (5y + 3y)2x + 15y2
= 4x2 + 16xy + 15y2
vi) (2a2+ 9) (2a2+ 5)
= 4a4 + (9+5)2a2 + 45
= 4a4 + 28a2 + 45
vii) (xyz – 4) (xyz – 2)
= x2y2z2 + (-4 -2)xyz + 8
= x2y2z2 – 6xyz + 8
3. Find the following squares by using the identities.
(i) (b – 7)2
(ii) (xy + 3z)2
(iii) (6x2 – 5y)2
(iv) [(2m/3) + (3n/2)]2
(v) (0.4p – 0.5q)2
(vi) (2xy + 5y)2
Solution:
Using identities:
(a – b) 2 = a2 + b2 – 2ab
(a + b) 2 = a2 + b2 + 2ab
(i) (b – 7)2 = b2 – 14b + 49
(ii) (xy + 3z)2 = x2y2 + 6xyz + 9z2
(iii) (6x2 – 5y)2 = 36x4 – 60x2y + 25y2
(iv) [(2m/3}) + (3n/2)]2 = (4m2/9) +(9n2/4) + 2mn
(v) (0.4p – 0.5q)2 = 0.16p2 – 0.4pq + 0.25q2
(vi) (2xy + 5y)2 = 4x2y2 + 20xy2 + 25y2
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 = a4 + b4 – 2a2b2
(ii) (2x + 5)2 – (2x – 5)2
= 4x2 + 20x + 25 – (4x2 – 20x + 25)
= 4x2 + 20x + 25 – 4x2 + 20x – 25
= 40x
iii) (7m – 8n)2 + (7m + 8n)2
= 49m2 – 112mn + 64n2 + 49m2 + 112mn + 64n2
= 98m2 + 128n2
iv) (4m + 5n)2 + (5m + 4n)2
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 80mn + 41n2
v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
= 6.25p2 – 7.5pq + 2.25q2 – 2.25p2 + 7.5pq – 6.25q2
= 4p2 – 4q2
vi) (ab + bc)2– 2ab²c
= a2b2 + 2ab2c + b2c2 – 2ab2c
= a2b2 + b2c2
vii) (m2 – n2m)2 + 2m3n2
= m4 – 2m3n2 + m2n4 + 2m3n2
= m4 + m2n4
5. Show that.
(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) (9p – 5q)2+ 180pq = (9p + 5q)2
(iii)(4/3 m – 3/4 n)2+ 2mn = 16/9 m2+ 9/16 n2
(iv) (4pq + 3q)2– (4pq – 3q)2 = 48pq2
(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Solution:
i) LHS = (3x + 7)2 – 84x
= 9x2 + 42x + 49 – 84x
= 9x2 – 42x + 49
= (3x – 7)2
= RHS
LHS = RHS
ii) LHS = (9p – 5q)2+ 180pq
= 81p2 – 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
RHS = (9p + 5q)2
= 81p2 + 90pq + 25q2
LHS = RHS
LHS = RHS
iv) LHS = (4pq + 3q)2– (4pq – 3q)2
= 16p2q2 + 24pq2 + 9q2 – 16p2q2 + 24pq2 – 9q2
= 48pq2
RHS = 48pq2
LHS = RHS
v) LHS = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
= a2 – b2 + b2 – c2 + c2 – a2
= 0
= RHS
6. Using identities, evaluate.
(i) 71²
(ii) 99²
(iii) 1022
(iv) 998²
(v) 5.2²
(vi) 297 x 303
(vii) 78 x 82
(viii) 8.92
(ix) 10.5 x 9.5
Solution:
i) 712 = (70+1)2
= 702 + 140 + 12
= 4900 + 140 +1
= 5041
ii) 99²
= (100 -1)2
= 1002 – 200 + 12
= 10000 – 200 + 1
= 9801
iii) 1022= (100 + 2)2
= 1002 + 400 + 22
= 10000 + 400 + 4
= 10404
iv) 9982= (1000 – 2)2
= 10002 – 4000 + 22
= 1000000 – 4000 + 4
= 996004
v) 5.22= (5 + 0.2)2
= 52 + 2 + 0.22
= 25 + 2 + 0.04
= 27.04
vi) 297 x 303
= (300 – 3 )(300 + 3)
= 3002 – 32
= 90000 – 9
= 89991
vii) 78 x 82
= (80 – 2)(80 + 2)
= 802 – 22
= 6400 – 4
= 6396
viii) 8.92= (9 – 0.1)2
= 92 – 1.8 + 0.12
= 81 – 1.8 + 0.01
= 79.21
ix) 10.5 x 9.5
= (10 + 0.5)(10 – 0.5)
= 102 – 0.52
= 100 – 0.25
= 99.75
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)
= 100 x 2
= 200
ii) (1.02)2– (0.98)2
= (1.02 + 0.98)(1.02 – 0.98)
= 2 x 0.04
= 0.08
iii) 1532 – 1472
= (153 + 147)(153 – 147)
= 300 x 6
= 1800
iv) 12.12 – 7.92
= (12.1 + 7.9)(12.1 – 7.9)
= 20 x 4.2= 84
8. Using (x + a) (x + b) = x2 + (a + b) x + ab, find
(i) 103 x 104
(ii) 5.1 x 5.2
(iii) 103 x 98
(iv) 9.7 x 9.8
Solution:
i) 103 x 104
= (100 + 3)(100 + 4)
= 1002 + (3 + 4)100 + 12
= 10000 + 700 + 12
= 10712
ii) 5.1 x 5.2
= (5 + 0.1)(5 + 0.2)
= 52 + (0.1 + 0.2)5 + 0.1 x 0.2
= 25 + 1.5 + 0.02
= 26.52
iii) 103 x 98
= (100 + 3)(100 – 2)
= 1002 + (3-2)100 – 6
= 10000 + 100 – 6
= 10094
iv) 9.7 x 9.8
= (9 + 0.7 )(9 + 0.8)
= 92 + (0.7 + 0.8)9 + 0.56
= 81 + 13.5 + 0.56
= 95.06
Access other exercise solutions of Class 8 Maths Chapter 9 Algebraic Expressions and Identities
Exercise 9.1 Solutions: 4 Questions (Short answers)
Exercise 9.2 Solutions: 5 Questions (Short answers)
Exercise 9.3 Solutions: 5 Questions (Short answers)
Exercise 9.4 Solutions: 3 Questions (Short answers)
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5 is based on the standard algebraic identities. These identities are obtained by multiplying a binomial by another binomial, problems on multiplication of binomial expressions and numbers. Algebraic identities have given us a less tedious method than the direct method of squaring polynomials.
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