# NCERT Solutions for class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5

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## Exercise 9.5 Page No: 151

1.Â Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3)
(ii) (2y + 5) (2y + 5)
(iii) (2a â€“ 7) (2a â€“ 7)
(iv) (3a â€“ 1/2)(3a â€“ 1/2)
(v) (1.1m â€“ 0.4) (1.1m + 0.4)
(vi) (a2+ b2) (- a2+ b2)
(vii) (6x â€“ 7) (6x + 7)
(viii) (- a + c) (- a + c)
(ix) (1/2 x + 3/4 y)Â (1/2 x + 3/4 y)
(x) (7a â€“ 9b) (7a â€“ 9b)

Solution:

(i) (x + 3) (x + 3) = (x + 3)2
= x2Â + 6x + 9
Using (a+b) 2 = a2 + b2 + 2ab

(ii) (2y + 5) (2y + 5) = (2y + 5)2
= 4y2Â + 20y + 25
Using (a+b) 2 = a2 + b2 + 2ab

iii) (2a â€“ 7) (2a â€“ 7) = (2a â€“ 7)2
= 4a2Â â€“ 28a + 49

Using (a-b) 2 = a2 + b2 â€“ 2ab

iv)Â Â Â Â Â (3a â€“ 1/2)(3a â€“ 1/2) = (3a â€“ 1/2)2

=Â  9a2Â -3a+(1/4)

Using (a-b) 2 = a2 + b2 â€“ 2ab

v) Â  (1.1m â€“ 0.4) (1.1m + 0.4)

= 1.21m2Â + 0.44 â€“ 0.44m â€“ 0.16

= 1.21m2Â â€“ 0.16

Using (a â€“ b)(a + b) = a2 â€“ b2

vi) (a2+ b2) (â€“ a2+ b2)
= (b2Â + a2Â ) (b2Â â€“ a2)
= -a4Â + b4Â

Using (a â€“ b)(a + b) = a2 â€“ b2

vii) (6x â€“ 7) (6x + 7)
=36x2Â â€“ 49
Using (a â€“ b)(a + b) = a2 â€“ b2

viii) (â€“ a + c) (â€“ a + c) = (â€“ a + c)2
= c2Â + a2 â€“ 2ac
Using (a-b) 2 = a2 + b2 â€“ 2ab

= (x2/4) + (9y2/16) + (3xy/4)

Using (a+b) 2 = a2 + b2 + 2ab

x) (7a â€“ 9b) (7a â€“ 9b) = (7a â€“ 9b)2

= 49a2Â â€“ 126ab + 81b2

Using (a-b) 2 = a2 + b2 â€“ 2ab

2.Â Use the identity (x + a) (x + b) = x2Â + (a + b) x + ab to find the following products.
(i) (x + 3) (x + 7)
(ii) (4x + 5) (4x + 1)
(iii) (4x â€“ 5) (4x â€“ 1)
(iv) (4x + 5) (4x â€“ 1)
(v) (2x + 5y) (2x + 3y)
(vi) (2a2 + 9) (2a2 + 5)
(vii) (xyz â€“ 4) (xyz â€“ 2)

Solution:

(i) (x + 3) (x + 7)

= x2Â + (3+7)x + 21
= x2Â + 10x + 21

ii) (4x + 5) (4x + 1)
= 16x2Â + (5 + 1)4x + 5
= 16x2Â + 24x + 5

iii) (4x â€“ 5) (4x â€“ 1)
= 16x2Â â€“ 4x â€“ 20x + 5
= 16x2Â â€“ 24x + 5

iv) (4x + 5) (4x â€“ 1)
= 16x2Â + (5-1)4x â€“ 5
= 16x2Â +16x â€“ 5

v) (2x + 5y) (2x + 3y)
= 4x2Â + (5y + 3y)2x + 15y2
= 4x2Â + 16xy + 15y2

vi) (2a2+ 9) (2a2+ 5)
= 4a4Â + (9+5)2a2Â + 45
= 4a4Â + 28a2Â + 45

vii) (xyz â€“ 4) (xyz â€“ 2)
= x2y2z2Â + (-4 -2)xyz + 8
= x2y2z2Â â€“ 6xyz + 8

3.Â Find the following squares by using the identities.
(i) (b â€“ 7)2
(ii) (xy + 3z)2
(iii) (6x2 â€“ 5y)2
(iv) [(2m/3) + (3n/2)]2
(v) (0.4p â€“ 0.5q)2
(vi) (2xy + 5y)2

Solution:

Using identities:

(a â€“ b) 2 = a2 + b2 â€“ 2ab
(a + b) 2 = a2 + b2 + 2ab

(i) (b â€“ 7)2 = b2Â â€“ 14b + 49

(ii) (xy + 3z)2 = x2y2Â + 6xyz + 9z2

(iii) (6x2 â€“ 5y)2 = 36x4Â â€“ 60x2y + 25y2

(iv) [(2m/3}) + (3n/2)]2 = (4m2/9) +(9n2/4) + 2mn

(v) (0.4p â€“ 0.5q)2 = 0.16p2Â â€“ 0.4pq + 0.25q2

(vi) (2xy + 5y)2 = 4x2y2Â + 20xy2Â + 25y2

4.Â Simplify.
(i) (a2 â€“ b2)2
(ii) (2x + 5)2Â  â€“ (2x â€“ 5)2
(iii) (7m â€“ 8n)2 + (7m + 8n)2
(iv) (4m + 5n)2Â + (5m + 4n)2
(v) (2.5p â€“ 1.5q)2Â â€“ (1.5p â€“ 2.5q)2
(vi) (ab + bc)2â€“ 2abÂ²c
(vii) (m2Â â€“ n2m)2Â + 2m3n2

Solution:

(i) (a2â€“ b2)2 = a4Â + b4 â€“ 2a2b2

(ii) (2x + 5)2Â  â€“ (2x â€“ 5)2
= 4x2Â + 20x + 25 â€“ (4x2Â â€“ 20x + 25)
= 4x2Â + 20x + 25 â€“ 4x2Â + 20x â€“ 25
= 40x

iii) (7m â€“ 8n)2 + (7m + 8n)2
= 49m2Â â€“ 112mn + 64n2Â + 49m2Â + 112mn + 49n2
= 98m2Â + 128n2

iv) (4m + 5n)2Â + (5m + 4n)2
= 16m2Â + 40mn + 25n2Â + 25m2Â + 40mn + 16n2
= 41m2Â + 80mn + 41n2

v) (2.5p â€“ 1.5q)2Â â€“ (1.5p â€“ 2.5q)2
= 6.25p2Â â€“ 7.5pq + 2.25q2Â â€“ 2.25p2Â + 7.5pq â€“ 6.25q2
= 4p2Â â€“ 4q2

vi) (ab + bc)2â€“ 2abÂ²c
= a2b2Â + 2ab2c + b2c2Â â€“ 2ab2c
= a2b2Â + b2c2

vii) (m2Â â€“ n2m)2Â + 2m3n2
= m4Â â€“ 2m3n2Â + m2n4Â + 2m3n2
= m4Â + m2n4

5.Â Show that.
(i) (3x + 7)2Â â€“ 84x = (3x â€“ 7)2
(ii) (9p â€“ 5q)2+ 180pq = (9p + 5q)2
(iii)(4/3 m â€“ 3/4 n)2+ 2mn = 16/9 m2+ 9/16 n2

(iv) (4pq + 3q)2â€“ (4pq â€“ 3q)2Â = 48pq2
(v) (a â€“ b) (a + b) + (b â€“ c) (b + c) + (c â€“ a) (c + a) = 0

Solution:

i) LHS = (3x + 7)2Â â€“ 84x

= 9x2Â + 42x + 49 â€“ 84x
= 9x2Â â€“ 42x + 49

= (3x â€“ 7)2
= RHS
LHS = RHS

ii) Â LHS = (9p â€“ 5q)2+ 180pq

= 81p2Â â€“ 90pq + 25q2Â + 180pq
= 81p2Â + 90pq + 25q2
RHS = (9p + 5q)2

= 81p2Â + 90pq + 25q2
LHS = RHS

LHS = RHS

iv) Â LHS = (4pq + 3q)2â€“ (4pq â€“ 3q)2

= 16p2q2Â + 24pq2Â + 9q2Â â€“ 16p2q2Â + 24pq2Â â€“ 9q2

= 48pq2

RHS = 48pq2

LHS = RHS

v) LHS = (a â€“ b) (a + b) + (b â€“ c) (b + c) + (c â€“ a) (c + a)

= a2Â â€“ b2Â + b2Â â€“ c2Â + c2Â â€“ a2
= 0
= RHS

6. Using identities, evaluate.
(i) 71Â²
(ii)Â 99Â²
(iii) 1022
(iv) 998Â²
(v) 5.2Â²
(vi) 297 x 303
(vii) 78 x 82
(viii) 8.92
(ix) 10.5 x 9.5

Solution:

i)Â 712Â = (70+1)2
= 702Â + 140 + 12
= 4900 + 140 +1

= 5041

ii) 99Â²
= (100 -1)2
= 1002Â â€“ 200 + 12
= 10000 â€“ 200 + 1
= 9801

iii) 1022= (100 + 2)2
= 1002Â + 400 + 22
= 10000 + 400 + 4
= 10404

iv) 9982= (1000 â€“ 2)2
= 10002Â â€“ 4000 + 22
= 1000000 â€“ 4000 + 4
= 996004

v) 5.22= (5 + 0.2)2
= 52Â + 2 + 0.22
= 25 + 2 + 0.4
= 27.4

vi) 297 x 303
= (300 â€“ 3 )(300 + 3)
= 3002Â â€“ 32
= 90000 â€“ 9
= 89991

vii) 78 x 82
= (80 â€“ 2)(80 + 2)
= 802Â â€“ 22
= 6400 â€“ 4
= 6396

viii) 8.92= (9 â€“ 0.1)2
= 92Â â€“ 1.8 + 0.12
= 81 â€“ 1.8 + 0.01
= 79.21

ix) 10.5 x 9.5

= (10 + 0.5)(10 â€“ 0.5)
= 102Â â€“ 0.52
= 100 â€“ 0.25
= 99.75

7.Â Using a2 â€“ b2Â = (a + b) (a â€“ b), find
(i) 512â€“ 492
(ii) (1.02)2â€“ (0.98)2
(iii) 1532â€“ 1472
(iv) 12.12â€“ 7.92

Solution:

i) 512â€“ 492

= (51 + 49)(51 â€“ 49)
= 100 x 2
= 200

ii) (1.02)2â€“ (0.98)2

= (1.02 + 0.98)(1.02 â€“ 0.98)
= 2 x 0.04
= 0.08

iii) 1532 â€“ 1472

= (153 + 147)(153 â€“ 147)
= 300 x 6
= 1800

iv) 12.12 â€“ 7.92

= (12.1 + 7.9)(12.1 â€“ 7.9)
= 20 x 4.2= 84

8.Â 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)
= 1002Â + (3 + 4)100 + 12
= 10000 + 700 + 12
= 10712

ii) 5.1 x 5.2

= (5 + 0.1)(5 + 0.2)
= 52Â + (0.1 + 0.2)5 + 0.1 x 0.2
= 25 + 1.5 + 0.02
= 26.52

iii) 103 x 98

= (100 + 3)(100 â€“ 2)
= 1002Â + (3-2)100 â€“ 6
= 10000 + 100 â€“ 6
= 10094

iv) 9.7 x 9.8
= (9 + 0.7 )(9 + 0.8)
= 92Â + (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

Algebraic Expressions and Identities Exercise 9.5 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.