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Class 8 NCERT Solutions – Chapter 9 Algebraic Expressions and Identities – Exercise 9.5 | Set 2
  • Last Updated : 05 Nov, 2020

Question 5. Show that:

(i) (3x + 7)2 – 84x = (3x – 7)2

Solution:

L.H.S. = (3x + 7)2 – 84x

= 9x2 + 42x + 49 – 84x

= 9x2 – 42x + 49



= (3x – 7)2

= R.H.S.

L.H.S. = R.H.S.

(ii) (9p – 5q)2 + 180pq = (9p + 5q)2

Solution:

LHS = (9p – 5q)2 + 180pq

= 81p2 – 90pq + 25q2 + 180pq

= 81p2 + 90pq + 25q2

RHS = (9p + 5q)2

= 81p2 + 90pq + 25q2

LHS = RHS

(iii) (4/3 m – 3/4 n)2+ 2mn = 16/9 m2+ 9/16 n2

Solution:

LHS = (4/3 m – 3/4 n)2 + 2mn

= 16/9m2 + 9/16n2 – 2nm + 2mn

=16/9 m2 + 9/16 n2

= RHS

LHS = RHS



(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2

Solution:

LHS = (4pq + 3q)2 – (4pq – 3q)2

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

= 48pq2

RHS = 48pq2

LHS = RHS

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

Solution:

LHS = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)

= a2 – b2 + b2 – c2 + c2 – a2

= 0

= RHS

Question 6. Using identities, evaluate.

(i) 71²

Solution:

712 = (70+1)2

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

= 702 + 12 + 140

= 4900 + 140 +1

= 5041

(ii) 99²

Solution:

99² = (100 -1)2

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

= 1002  + 12 – 200

= 10000 – 200 + 1

= 9801

(iii) 1022

Solution:

1022 = (100 + 2)2

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

= 1002 + 400 + 22

= 10000 + 400 + 4

= 10404

(iv) 9982

Solution:

9982 = (1000 – 2)2

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

= 10002 – 4000 + 22

= 1000000 – 4000 + 4

= 996004

(v) 5.2²

Solution:

 5.22 = (5 + 0.2)2

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

= 52 + 2 + 0.22

= 25 + 2 + 0.4

= 27.4

(vi) 297 × 303

Solution:

297 × 303

= (300 – 3 ) (300 + 3)

Using formula (a + b) (a – b) = a2 – b2

= 3002 – 32

= 90000 – 9

= 89991

(vii) 78 × 82

Solution:

78 × 82

= (80 – 2) (80 + 2)

Using formula (a + b) (a – b) = a2 – b2

= 802 – 22

= 6400 – 4

= 6396

(viii) 8.92

Solution:

8.92= (9 – 0.1)2

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

= 92 – 1.8 + 0.12

= 81 – 1.8 + 0.01

= 79.21

(ix) 10.5 × 9.5

Solution:

10.5 × 9.5 = (10 + 0.5) (10 – 0.5)

Using formula (a + b) (a – b) = a2 – b2

= 102 – 0.52

= 100 – 0.25

= 99.75

Question 7. Using a2 – b2 = (a + b) (a – b), find

(i) 512 – 492

Solution:

512 – 492

= (51 + 49) (51 – 49)

= 100 × 2

= 200

(ii) (1.02)2 – (0.98)2

Solution:

(1.02)2 – (0.98)2

= (1.02 + 0.98) (1.02 – 0.98)

= 2 × 0.04

= 0.08

(iii) 1532 – 1472

Solution:

1532 – 1472

= (153 + 147) (153 – 147)

= 300 × 6

= 1800

(iv) 12.12– 7.92

Solution:

12.12 – 7.92

= (12.1 + 7.9) (12.1 – 7.9)

= 20 × 4.2 = 84

Question 8. Using (x + a) (x + b) = x2 + (a + b) x + ab, find

(i) 103 × 104

Solution:

103 × 104

= (100 + 3) (100 + 4)

= 1002 + (3 + 4)100 + 12

= 10000 + 700 + 12

= 10712

(ii) 5.1 × 5.2

Solution:

5.1 × 5.2

= (5 + 0.1) (5 + 0.2)

= 52 + (0.1 + 0.2)5 + 0.1 × 0.2

= 25 + 1.5 + 0.02

= 26.52

(iii) 103 × 98

Solution:

103 × 98

= (100 + 3) (100 – 2)

= 1002 + (3-2)100 – 6

= 10000 + 100 – 6

= 10094

(iv) 9.7 × 9.8

Solution:

9.7 × 9.8

= (9 + 0.7) (9 + 0.8)

= 92 + (0.7 + 0.8)9 + 0.56

= 81 + 13.5 + 0.56

= 95.06

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