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

Worksheets Ncert Solutions

In this page we have NCERT book Solutions for Class 8th Maths:Algebraic Expressions and Identities 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