Thursday, May 28, 2020

NCERT solution class 8 chapter 9 Algebraic Expression and Identities exercise 9.5

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Exercise 9.5


Q 1 - Ex 9.5 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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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) (iv)
(v) (1.1m − 0.4) (1.1 m + 0.4) (vi) (a2 + b2) (− a2 + b2)
(vii) (6x − 7) (6x + 7) (viii) (− a + c) (− a + c)
(ix) (x) (7a − 9b) (7a − 9b)

Sol :
The products will be as follows.
(i) (x + 3) (x + 3) = (x + 3)2
= (x)2 + 2(x) (3) + (3)2 [(a + b)2 = a2 + 2ab + b2]
= x2 + 6x + 9
(ii) (2y + 5) (2y + 5) = (2y + 5)2
= (2y)2 + 2(2y) (5) + (5)2 [(a + b)2 = a2 + 2ab + b2]
= 4y2 + 20y + 25
(iii) (2a ­− 7) (2a − 7) = (2a − 7)2
= (2a)2 − 2(2a) (7) + (7)2 [(a − b)2 = a2 − 2ab + b2]
= 4a2 − 28a + 49
(iv)
[(a − b)2 = a2 − 2ab + b2]

(v) (1.1m − 0.4) (1.1 m + 0.4)
= (1.1m)2 − (0.4)2 [(a + b) (a − b) = a2 − b2]
= 1.21m2 − 0.16
(vi) (a2 + b2) (− a2 + b2) = (b2 + a2) (b2 − a2)
= (b2)2 − (a2)2 [(a + b) (a − b) = a2 − b2]
= b4 − a4
(vii) (6x − 7) (6x + 7) = (6x)2 − (7)2 [(a + b) (a − b) = a2 − b2]
= 36x2 − 49
(viii) (− a + c) (− a + c) = (− a + c)2
= (− a)2 + 2(− a) (c) + (c)2 [(a + b)2 = a2 + 2ab + b2]
= a2 − 2ac + c2
(ix)
[(a + b)2 = a2 + 2ab + b2]

(x) (7a − 9b) (7a − 9b) = (7a − 9b)2
= (7a)2 − 2(7a)(9b) + (9b)2 [(a − b)2 = a2 − 2ab + b2]
= 49a2 − 126ab + 81b2


Q 2 - Ex 9.5 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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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)

Sol :
The products will be as follows.
(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)
= 16x2 − 24x + 5
(iv)
= 16x2 + 16x − 5
(v) (2x +5y) (2x + 3y) = (2x)2 + (5y + 3y) (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)
=
= x2y2z2 − 6xyz + 8


Q 3 - Ex 9.5 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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Question 3
Find the following squares by suing the identities.
(i) (b − 7)2 (ii) (xy + 3z)2 (iii) (6x2 − 5y)2
(iv) (v) (0.4p − 0.5q)2 (vi) (2xy + 5y)2

Sol :
(i) (b − 7)2 = (b)2 − 2(b) (7) + (7)2 [(a − b)2 = a2 − 2ab + b2]
= b2 − 14b + 49
(ii) (xy + 3z)2 = (xy)2 + 2(xy) (3z) + (3z)2 [(a + b)2 = a2 + 2ab + b2]
= x2y2 + 6xyz + 9z2
(iii) (6x2 − 5y)2 = (6x2)2 − 2(6x2) (5y) + (5y)2 [(a − b)2 = a2 − 2ab + b2]
= 36x4 − 60x2y + 25y2
(iv) [(a + b)2 = a2 + 2ab + b2]

(v) (0.4p − 0.5q)2 = (0.4p)2 − 2 (0.4p) (0.5q) + (0.5q)2
[(a − b)2 = a2 − 2ab + b2]
= 0.16p2 − 0.4pq + 0.25q2
(vi) (2xy + 5y)2 = (2xy)2 + 2(2xy) (5y) + (5y)2
[(a + b)2 = a2 + 2ab + b2]
= 4x2y2 + 20xy2 + 25y2


Q 4 - Ex 9.5 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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Q 4 (Part 2) - Ex 9.5 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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 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

Sol :
(i) (a2 − b2)2 = (a2)2 − 2(a2) (b2) + (b2)2 [(a − b)2 = a2 − 2ab + b2 ]
= a4 − 2a2b2 + b4
(ii) (2x +5)2 − (2x − 5)2 = (2x)2 + 2(2x) (5) + (5)2 − [(2x)2 − 2(2x) (5) + (5)2]
[(a − b)2 = a2 − 2ab + b2]
[(a + b)2 = a2 + 2ab + b2]
= 4x2 + 20x + 25 − [4x2 − 20x + 25]
= 4x2 + 20x + 25 − 4x2 + 20x − 25 = 40x
(iii) (7m − 8n)2 + (7m + 8n)2
= (7m)2 − 2(7m) (8n) + (8n)2 + (7m)2 + 2(7m) (8n) + (8n)2
[(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
[ (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]
[(a − b)2 = a2 − 2ab + b2 ]
= 6.25p2 − 7.5pq + 2.25q2 − [2.25p2 − 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 [(a + b)2 = a2 + 2ab + b2 ]
= a2b2 + 2ab2c + b2c2 − 2ab2c
= a2b2 + b2c2
(vii) (m2 − n2m)2 + 2m3n2
= (m2)2 − 2(m2) (n2m) + (n2m)2 + 2m3n2 [(a − b)2 = a2 − 2ab + b2 ]
= m4 − 2m3n2 + n4m2 + 2m3n2
= m4 + n4m2


Q 5 - Ex 9.5 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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Question 5
Show that
(i) (3x + 7)2 − 84x = (3x − 7)2 (ii) (9p − 5q)2 + 180pq = (9p + 5q)2
(iii)
(iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2
(v) (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0

Sol :
(i) L.H.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) L.H.S = (9p − 5q)2 + 180pq
= (9p)2 − 2(9p)(5q) + (5q)2 − 180pq
= 81p2 − 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
R.H.S = (9p + 5q)2
= (9p)2 + 2(9p)(5q) + (5q)2
= 81p2 + 90pq + 25q2
L.H.S = R.H.S
(iii) L.H.S =

(iv) L.H.S = (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
= 48pq2 = R.H.S
(v) L.H.S = (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a)
= (a2 − b2) + (b2 − c2) + (c2 − a2) = 0 = R.H.S.


Q 6 - Ex 9.5 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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Q 6 - Ex 9.5 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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 Question 6
(i) 712 (ii) 992 (iii) 1022 (iv) 9982
(v) (5.2)2 (vi) 297 × 303 (vii) 78 × 82
(viii) 8.92 (ix) 1.05 × 9.5

Sol :
(i) 712 = (70 + 1)2
= (70)2 + 2(70) (1) + (1)2 [(a + b)2 = a2 + 2ab + b2 ]
= 4900 + 140 + 1 = 5041
(ii) 992 = (100 − 1)2
= (100)2 − 2(100) (1) + (1)2 [(a − b)2 = a2 − 2ab + b2 ]
= 10000 − 200 + 1 = 9801
(iii) 1022 = (100 + 2)2
= (100)2 + 2(100)(2) + (2)2 [(a + b)2 = a2 + 2ab + b2 ]
= 10000 + 400 + 4 = 10404
(iv) 9982 = (1000 − 2)2
= (1000)2 − 2(1000)(2) + (2)2 [(a − b)2 = a2 − 2ab + b2 ]
= 1000000 − 4000 + 4 = 996004
(v) (5.2)2 = (5.0 + 0.2)2
= (5.0)2 + 2(5.0) (0.2) + (0.2)2 [(a + b)2 = a2 + 2ab + b2 ]
= 25 + 2 + 0.04 = 27.04
(vi) 297 × 303 = (300 − 3) × (300 + 3)
= (300)2 − (3)2 [(a + b) (a − b) = a2 − b2]
= 90000 − 9 = 89991
(vii) 78 × 82 = (80 − 2) (80 + 2)
= (80)2 − (2)2 [(a + b) (a − b) = a2 − b2]
= 6400 − 4 = 6396
(viii) 8.92 = (9.0 − 0.1)2
= (9.0)2 − 2(9.0) (0.1) + (0.1)2 [(a − b)2 = a2 − 2ab + b2 ]
= 81 − 1.8 + 0.01 = 79.21
(ix) 1.05 × 9.5 = 1.05 × 0.95 × 10
= (1 + 0.05) (1− 0.05) ×10
= [(1)2 − (0.05)2] × 10
= [1 − 0.0025] × 10 [(a + b) (a − b) = a2 − b2]
= 0.9975 × 10 = 9.975



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

So :
(i) 512 − 492 = (51 + 49) (51 − 49)
= (100) (2) = 200
(ii) (1.02)2 − (0.98)2 = (1.02 + 0.98) (1.02 ­− 0.98)
= (2) (0.04) = 0.08
(iii) 1532 − 1472 = (153 + 147) (153 − 147)
= (300) (6) = 1800
(iv) 12.12 − 7.92 = (12.1 + 7.9) (12.1 − 7.9)
= (20.0) (4.2) = 84


Q 8 - Ex 9.5 - Algebraic Expressions and Identities - NCERT Maths Class 8th - Chapter 9

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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

Sol :
(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)2 + [3 + (− 2)] (100) + (3) (− 2)
= 10000 + 100 − 6
= 10094
(iv) 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 = 100.06 − 5 = 95.06

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