# Ncert Solutions for Algebraic Expressions and Identities Class 8 Maths Chapter 9 CBSE Part 5

In this page we have Ncert Solutions for Algebraic Expressions and Identities Class 8 Maths Chapter 9 for EXERCISE 5. Hope you like them and do not forget to like , social share and comment at the end of the page.
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) (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) [ (x/2)+(3y/4)] [ (x/2)+(3y/4)]
(x) (7a – 9b) (7a – 9b)
We will be using below identities in these question
(a + b)2 = a+ 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a – b)(a + b) = a2 – b2
1) This question can be solved using the first identify
(x + 3) (x + 3)
=x2 + 6x + 9
2) This question can be solved using the first identify
(2y + 5) (2y + 5)
=4y2 + 20y + 25
3) This question can be solved using the second identify
(2a – 7) (2a – 7)
= 4a2 – 28a + 49
4) This question can be solved using the second identify
(3a-1/2)(3a-1/2)=  9a2 -3a+(1/4)
5)  This question can be solved using the third identity
(1.1m – 0.4) (1.1m + 0.4)
= 1.21m2 – 0.16
5) This question can be solved using the third identity
(a2+ b2) (– a2+ b2)
= (b2 + a2 ) (b2 – a2)
= a4 - b4
6) This question can be solved using the third identity
(6x – 7) (6x + 7)
=36x2 - 49
7) This question can be solved using the third identity
= c2 - a2
8) This question can be solved using the first identity
[ (x/2)+(3y/4)] [ (x/2)+(3y/4)]
=(x2/4) + (9y2/16) +(3xy/4)
9) This question can be solved using second identity
= 49a2 – 126ab + 81b2
Question 2
Use the identity (x + a) (x + b) = x+ (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)
1) x2 + (3+7)x + 21
= x2 + 10x + 21
2) (4x + 5) (4x + 1)
= 16x2 + (5 + 1)4x + 5
= 16x2 + 24x + 5
3) (4x – 5) (4x – 1)
= 16x2 + (-5-1)4x + 5
= 16x2 - 20x + 5
4) (4x + 5) (4x – 1)
= 16x2 + (5-1)4x - 5
= 16x2 +16x - 5
5) (2x + 5y) (2x + 3y)
= 4x2 + (5y + 3y)4x + 15y2
= 4x2 + 32xy + 15y2
6) (2a2+ 9) (2a2+ 5)
= 4a4 + (9+5)2a2 + 45
= 4a4 + 28a2 + 45
7) (xyz – 4) (xyz – 2)
= x2y2z2 + (-4 -2)xyz - 8
= 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
iv) [(2m/3}) + (3n/2)]2
(v) (0.4p – 0.5q)2
(vi) (2xy + 5y)
1. b2 - 14b + 49
2. x2y2 + 6xyz + 9z2
3. 36x4 - 60x2y + 25y2
4. (4m2/9) +(9n2/4) +2mn
5. 0.16p2 - 0.4pq + 0.25q2
6. 4x2y2 + 20xy2 + 25y2
Question 4
Simplify.
(i) (a2– b2)2
(ii) (2x + 5)– (2x – 5)2
(iii) (7m – 8n)2+ (7m + 8n)2
(iv) (4m + 5n)+ (5m + 4n)2
(v) (2.5p – 1.5q)– (1.5p – 2.5q)2
(vi) (ab + bc)2– 2ab²c
(vii) (m– n2m)+ 2m3n2
1) a4 - b4
2) (2x + 5)– (2x – 5)2
= 4x2 + 20x +25 - (4x2 - 20x + 25)
= 4x2 + 20x + 25 - 4x2 + 20x - 25
= 40
3) (7m – 8n)2+ (7m + 8n)2
= 49m2 - 112mn + 64n2 + 49m2 + 112mn + 49n2
= 98m2 + 128n2
4) (4m + 5n)+ (5m + 4n)2
= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2
= 41m2 + 80mn + 41n2
5) (2.5p – 1.5q)– (1.5p – 2.5q)2
= 6.25p2 - 7.5pq + 2.25q2 - 2.25p2 + 7.5pq - 6.25q2
= 4p2 - 4q2
6) (ab + bc)2– 2ab²c
= a2b2 + 2ab2c + b2c2 - 2ab2c
= a2b2 + b2c2
7) (m– n2m)+ 2m3n2
= m4 - 2m3n2 + m2n4 + 2m3n2
= m4 + m2n4

Question 5
Show that.
(i) (3x + 7)– 84x = (3x – 7)2
(ii) (9p – 5q)2+ 180pq = (9p + 5q)2
iv) (4pq + 3q)2– (4pq – 3q)= 48pq2
v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
1) LHS = 9x2 + 42x + 49 - 84x
= 9x2 - 42x + 49
RHS = 9x2 - 42x + 49
LHS = RHS
2)  LHS = 91p2 - 90pq + 25q2 + 180pq
= 91p2 + 90pq + 25q2
RHS = 91p2 + 90pq + 25q2
4)  LHS= 16p2q2 + 24pq2 + 9q2 - 16p2q2 + 24pq2 - 9q2
= 48pq2
5) LHS= a2 - b2 + b2 - c2 + c2 - a2
= 0
Question 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
We will be using below identities in these question
(a + b)2 = a+ 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a – b)(a + b) = a2 – b2
1) 712 = (70+1)2
= 702 + 140 + 12
= 4900 + 140 +1= 5041
2) 99²
= (100 -1)2
= 1002 - 200 + 12
= 10000 - 200 + 1
= 9801
3) 1022= (100 + 2)2
= 1002 + 400 + 22
= 10000 + 400 + 4
= 10404
4) 9982= (1000 - 2)2
= 10002 - 4000 + 22
= 1000000 - 4000 + 4
= 996004
5) 5.22= (5 + 0.2)2
= 52 + 2 + 0.22
= 25 + 2 + 0.4
= 27.4
6) = (300 - 3 )(300 + 3)
= 3002 - 32
= 90000 - 9
= 89991
7) = (80 - 2)(80 + 2)
= 802 - 22
= 6400 - 4
= 6396
8) 8.92= (9 - 0.1)2
= 92 - 1.8 + 0.12
= 81 - 1.8 + 0.01
= 79.21
9) = (10 + 0.5)(10 - 0.5)
= 102 - 0.52
= 100 - 0.25
= 99.75
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
1) = (51 + 49)(51 - 49)
= 100 x 2
= 200
2) = (1.02 + 0.98)(1.02 - 0.98)
= 2 x 0.04
= 0.08
3) = (153 + 147)(153 - 147)
= 300 x 6
= 1800
4) = (12.1 + 7.9)(12.1 - 7.9)
= 20 x 4.2
= 84
Question 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
1) = (100 + 3)(100 + 4)
= 1002 + (3 + 4)100 + 12
= 10000 + 1200 + 12
= 11212
2) = (5 + 0.1)(5 + 0.2)
= 52 + (0.1+0.2)5 + 0.02
= 25 + 1.5 + 0.02
= 26.52
3) = (100 + 3)(100 - 2)
= 1002 + (3-2)100 - 6
= 10000 + 100 - 6
= 10094
4) = (9 + 0.7 )(9 + 0.8)
= 92 + (0.7 + 0.8)9 + 0.63
= 81 + 13.5 + 0.63
= 95.13