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# Mathematics – Class 8 – Chapter 9 – Algebraic Expression and Identities – Exercise 9.5 -NCERT Exercise Solution

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

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

Solution:

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

= (x + 3) (x + 3)

= (x + 3)2

= x2 + 6x + 9

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

Solution:

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

= (2y + 5) (2y + 5)

= (2y + 5)2

= 4y2 + 20y + 25

(iii) (2a – 7) (2a – 7)

Solution:

Using (a-b) 2 = a2 + b2 – 2ab

= (2a – 7) (2a – 7) = (2a – 7)2

= 4a2 – 28a + 49

(iv) (3a – 1/2) (3a – 1/2)

Solution:

Using (a-b) 2 = a2 + b2 – 2ab

= (3a – 1/2) (3a – 1/2)

= (3a – 1/2)2

= 9a2 -3a+(1/4)

(v) (1.1m – 0.4) (1.1m + 0.4)

Solution:

Using (a – b) (a + b) = a2 – b2

(1.1m – 0.4) (1.1m + 0.4)

= 1.21m2 – 0.16

(vi) (a2+ b2) (- a2+ b2)

Solution:

Using (a – b) (a + b) = a2 – b2

= (a2+ b2) (– a2+ b2)

= (b2 + a2) (b2 – a2)

= -a4 + b4

(vii) (6x – 7) (6x + 7)

Solution:

Using (a – b) (a + b) = a2 – b2

= (6x – 7) (6x + 7)

=36x2 – 49

(viii) (- a + c) (- a + c)

Solution:

Using (a-b) 2 = a2 + b2 – 2ab

= (– a + c) (– a + c) = (– a + c)2

= c2 + a2 – 2ac

(ix) (1/2x + 3/4y) (1/2x + 3/4y)

Solution:

Using (a-b) 2 = a2 + b2 – 2ab

= (7a – 9b) (7a – 9b) = (7a – 9b)2

= 49a2 – 126ab + 81b2

(x) (7a – 9b) (7a – 9b)

Solution:

Using (a – b)2 = a2 – b2 + 2ab

= (7a – 9b)2

= 49a2 + 81b2 – 126ab

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

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

Solution:

= (x + 3) (x + 7)

= x2 + (3+7)x + 21

= x2 + 10x + 21

(ii) (4x + 5) (4x + 1)

Solution:

= (4x + 5) (4x + 1)

= 16x2 + 4x + 20x + 5

= 16x2 + 24x + 5

(iii) (4x – 5) (4x – 1)

Solution:

= (4x – 5) (4x – 1)

= 16x2 – 4x – 20x + 5

= 16x2 – 24x + 5

(iv) (4x + 5) (4x – 1)

Solution:

= (4x + 5) (4x – 1)

= 16x2 + (5-1)4x – 5

= 16x2 +16x – 5

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

Solution:

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

= 4x2 + (5y + 3y)2x + 15y2

= 4x2 + 16xy + 15y2

(vi) (2a2 + 9) (2a2 + 5)

Solution:

= (2a2+ 9) (2a2+ 5)

= 4a4 + (9+5)2a2 + 45

= 4a4 + 28a2 + 45

(vii) (xyz – 4) (xyz – 2)

Solution:

= (xyz – 4) (xyz – 2)

= x2y2z2 + (-4 -2)xyz + 8

= x2y2z2 – 6xyz + 8

3. Find the following squares by using the identities.

In this question we Use these identities:

(a – b) 2 = a2 + b2 – 2ab

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

(i) (b – 7)2

Solution:

= (b – 7)2

= b2 – 14b + 49

(ii) (xy + 3z)2

Solution:

= (xy + 3z)2

= x2y2 + 6xyz + 9z2

(iii) (6x2 – 5y)2

Solution:

= (6x2 – 5y)2

= 36x4 – 60x2y + 25y2

(iv) [(2m/3) + (3n/2)]2

Solution:

= [(2m/3}) + (3n/2)]2

= (4m2/9) +(9n2/4) + 2mn

(v) (0.4p – 0.5q)2

Solution:

= (0.4p – 0.5q)2

= 0.16p2 – 0.4pq + 0.25q2

(vi) (2xy + 5y)2

Solution:

= (2xy + 5y)2

= 4x2y2 + 20xy2 + 25y2

4. Simplify.

(i) (a2 – b2) 2

Solution:

= a4 + b4 – 2a2b2

(ii) (2x + 5)2 – (2x – 5)2

Solution:

= 4x2 + 20x + 25 – (4x2 – 20x + 25)

= 4x2 + 20x + 25 – 4x2 + 20x – 25

= 40x

(iii) (7m – 8n)2 + (7m + 8n)2

Solution:

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

= 98m2 + 128n2

(iv) (4m + 5n)2 + (5m + 4n)2

Solution:

= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2

= 41m2 + 80mn + 41n2

(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2

Solution:

= 6.25p2 – 7.5pq + 2.25q2 – 2.25p2 + 7.5pq – 6.25q2

= 4p2 – 4q2

(vi) (ab + bc)2– 2ab²c

Solution:

= a2b2 + 2ab2c + b2c2 – 2ab2c

= a2b2 + b2c2

(vii) (m2 – n2m)2 + 2m3n2

Solution:

= m4 – 2m3n2 + m2n4 + 2m3n2

= m4 + m2n4

5. Show that.

(i) (3x + 7)2 – 84x = (3x – 7)2

Solution:

Let us take LHS:

LHS = (3x + 7)2 – 84x

= 9x2 + 42x + 49 – 84x

= 9x2 – 42x + 49

= (3x – 7)2

Hence,

LHS = RHS

(ii) (9p – 5q)2+ 180pq = (9p + 5q)2

Solution:

Let us take LHS:

LHS = (9p – 5q)2+ 180pq

= 81p2 – 90pq + 25q2 + 180pq

= 81p2 + 90pq + 25q2

Now, taking RHS:

RHS = (9p + 5q)2

= 81p2 + 90pq + 25q2

Hence, LHS = RHS

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

Solution:

Let us take LHS:

LHS = (4/3m – 3/4n)2 + 2mn

= 16/9 m2 + 9/16 n2 – 2mn + 2mn

= 16/9 m2 + 9/16 n2

Hence, LHS = RHS

(iv) (4pq + 3q)2– (4pq – 3q)2 = 48pq2

Solution:

Let us take LHS:

LHS = (4pq + 3q)2– (4pq – 3q)2

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

= 48pq2

Hence,

LHS = RHS

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

Solution:

Let us take LHS:

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

= a2 – b2 + b2 – c2 + c2 – a2

= 0

Hence,

LHS = RHS

6. Using identities, evaluate.

(i) 71²

Solution:

According to question:

= (70+1)2

= 702 + 140 + 12

= 4900 + 140 +1

= 5041

(ii) 99²

Solution:

According to question:

= (100 -1)2

= 1002 – 200 + 12

= 10000 – 200 + 1

= 9801

(iii) 1022

Solution:

According to question:

= (100 + 2)2

= 1002 + 400 + 22

= 10000 + 400 + 4 = 10404

(iv) 998²

Solution:

According to question:

= (1000 – 2)2

= 10002 – 4000 + 22

= 1000000 – 4000 + 4

= 996004

(v) 5.2²

Solution:

According to question:

= (5 + 0.2)2

= 52 + 2 + 0.22

= 25 + 2 + 0.04

= 27.04

(vi) 297 x 303

Solution:

According to question:

= (300 – 3) (300 + 3)

= 3002 – 32

= 90000 – 9

= 89991

(vii) 78 x 82

Solution:

According to question:

= (80 – 2) (80 + 2)

= 802 – 22

= 6400 – 4

= 6396

(viii) 8.92

Solution:

According to question:

= (9 – 0.1)2

= 92 – 1.8 + 0.12

= 81 – 1.8 + 0.01

= 79.21

(ix) 10.5 x 9.5

Solution:

According to question:

= (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

Solution:

According to question:

= (51 + 49) (51 – 49)

= 100 x 2 = 200

(ii) (1.02)2– (0.98)2

Solution:

According to question:

= (1.02 + 0.98) (1.02 – 0.98)

= 2 x 0.04

= 0.08

(iii) 1532– 1472

Solution:

According to question:

= (153 + 147) (153 – 147)

= 300 x 6

= 1800

(iv) 12.12– 7.92

Solution:

According to question:

= (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

Solution:

According to question:

= (100 + 3) (100 + 4)

= 1002 + (3 + 4)100 + 12

= 10000 + 700 + 12 = 10712

(ii) 5.1 x 5.2

Solution:

According to question:

= (5 + 0.1) (5 + 0.2)

= 52 + (0.1 + 0.2)5 + 0.1 x 0.2

= 25 + 1.5 + 0.02

= 26.52

(iii) 103 x 98

Solution:

According to question:

= (100 + 3) (100 – 2)

= 1002 + (3-2)100 – 6

= 10000 + 100 – 6

= 10094

(iv) 9.7 x 9.8

Solution:

According to question:

= (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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