NCERT Solutions of Algebraic Expressions and Identities Class 8
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Algebraic Expressions

Exercise 5 (NCERT)

1. Use a suitable identity to get each of the following products.

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

Answer: Using (a + b)2 = a2 + 2ab + b2 we get the following equation:

= x2 + 6x + 9

(ii) (2y + 5) (2y + 5)

Answer: 4y2 + 20y + 25

(iii) (2a – 7) (2a – 7)

Answer: Using (a - b)2 = a2 - 2ab + b2 we get the following equation:

= 4a2 - 28a + 49

algebra 1

(v) (1.1m – 0.4) (1.1m + 0.4)

Answer: Using (a - b)(a + b) = a2 - b2

= 1.21m2 - 0.16

(vi) (a2+ b2) (– a2+ b2)

Answer: = (a2 - a2)

= (b2 + a2 ) (b2 - a2)

= a4 - b4

(vii) (6x – 7) (6x + 7)

Answer: 36x2 - 49

(viii) (– a + c) (– a + c)

= c2 - a2

algebra 2

(x) (7a – 9b) (7a – 9b)

= 49a2 - 126ab + 81b2

2. Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products.

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

Answer: 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 + (-5-1)4x + 5

= 16x2 - 20x + 5

(iv) (4x + 5) (4x – 1)

= 16x2 + (5-1)4x - 5

= 16x2 +16x - 5

(v) (2x + 5y) (2x + 3y)

= 4x2 + (5y + 3y)4x + 15y2

= 4x2 + 32xy + 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

= b2 - 14b + 49

(ii) (xy + 3z)2

= x2y2 + 6xyz + 9z2

(iii) (6x2– 5y)2

= 36x4 - 60x2y + 25y2

algebra 3

(v) (0.4p – 0.5q)2

= 0.16p2 - 0.4pq + 0.25q2

(vi) (2xy + 5y)

= 4x2y2 + 20xy2 + 25y2

4. Simplify.

(i) (a2– b2)2

= a4 - b4

(ii) (2x + 5)2 – (2x – 5)2

= 4x2 + 20x +25 - (4x2 - 20x + 25)

= 4x2 + 20x + 25 - 4x2 + 20x - 25

= 40

(iii) (7m – 8n)2+ (7m + 8n)2

= 49m2 - 112mn + 64n2 + 49m2 + 112mn + 49n2

= 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

LHS = 9x2 + 42x + 49 - 84x

= 9x2 - 42x + 49

RHS = 9x2 - 42x + 49

LHS = RHS

(ii) (9p – 5q)2+ 180pq = (9p + 5q)2

LHS = 91p2 - 90pq + 25q2 + 180pq

= 91p2 + 90pq + 25q2

RHS = 91p2 + 90pq + 25q2

algebra 4

(iv) (4pq + 3q)2– (4pq – 3q)2 = 48pq2

= 16p2q2 + 24pq2 + 9q2 - 16p2q2 + 24pq2 - 9q2

= 48pq2

v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

= a2 - b2 + b2 - c2 + c2 - a2

= 0

6. Using identities, evaluate.

(i) 71²

Answer: 712 = (70+1)2

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

= 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 + 0.4

= 27.4

(vi) 297 x 303

= (300 - 3 )(300 + 3)

Using (a - b)(a + b) = a2 - b2

= 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

= (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

= (100 + 3)(100 + 4)

= 1002 + (3 + 4)100 + 12

= 10000 + 1200 + 12

= 11212

(ii) 5.1 x 5.2

= (5 + 0.1)(5 + 0.2)

= 52 + (0.1+0.2)5 + 0.02

= 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 + 0.63

= 81 + 13.5 + 0.63

= 95.13


NCERT Solutions of Algebraic Expressions and Identities Class 8
<< Solution of NCERT Exercise 9.4Solution of NCERT Exercise 10.1 >>

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