Chapter 9 Algebraic Expressions and Identities - Exercise 9.5 | Set 1
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
Summary
Algebraic expressions and identities in Class 8 typically cover expanding brackets, factorizing expressions, and using standard algebraic identities like (a + b)² and (a - b)². These concepts are fundamental to algebra and form the basis for more advanced mathematical operations. Understanding these principles helps students simplify complex expressions, solve equations more efficiently, and develop problem-solving skills applicable to various mathematical and real-world scenarios.
