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# Exercise 3.1

7th Maths : Term 3 Unit 3 : Algebra : Introduction to Identities, Geometrical Approach to Multiplication of Monomials, Geometrical proof of Identities, Factorisation using identities : Exercise 3.1 : Text Book Back Exercises Questions with Answers, Solution

Exercise 3.1

1. Fill in the blanks

(i) ( p q)2 = ----------------------------. [Answer: p2–2pq + q2]

(ii) The product of (x + 5) and (x − 5) is ----------------------. [Answer: x2 – 25]

(iii) The factors of x2 4x + 4 are -----------------------------. [Answer: (x – 2) and (x – 2)]

(iv) Express 24ab2c2 as product of its factors is ---------. [Answer: 2 × 2 × 2 × 3 × a × b × b × c × c ]

2. Say whether the following statements are True or False.

(i) (7x + 3)(7x − 4) = 49x2 7x 12 . [Answer: True]

(ii) (a 1)2 =a2 1 . [Answer: False]

(iii) (x 2 + y 2 )( y 2 + x2 ) = ( x2 + y2 )2 . [Answer: True]

(iv) 2p is the factor of 8pq. [Answer: True]

3. Express the following as the product of its factors.

(i) 24 ab2c2.

(ii) 36 x3y2z.

(iii) 56 mn2p2.

Solution:

(i) 24ab2c2 = 2 × 2 × 2 × 3 × a × b × b × c × c

(ii) 36 x3y2z = 2 × 2 × 3 × 3 × x × x × x × y × y × z

(iii) 56 mn2p2 = 2 × 2 × 2 × 7 × m × n × n × p × p

4. Using the identity (x +a)(x +b) = x 2 + x(a + b) +ab , find the following product.

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

(ii) (6a + 9)(6a − 5)

(iii) (4x + 3y)(4x + 5y)

(iv) (8 + pq)(pq + 7)

Solution:

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

Let a = 3;   b = 7, then

(x + 3)(x + 7) is of the form x2 + x (a + b) + ab

(x + 3)(x + 7) = x2 + x(3 + 7) + (3 × 7) = x2+ 10x + 21

(ii) (6a + 9)(6a – 5)

Substituting x = 6a ;   a = 9 and b = –5

In (x + a) (x + b) = x2 + x (a + b) + ab, we get

(6a + 9)(6a – 5) = (6a)2 + 6a (9 + (–5)) + (9 × (–5))

= 62 a2 + 6a (4) + (–45) = 36a2 + 24a – 45

(6a + 9) (6a –5) = 36a2 + 24a – 45

(iii) (4x + 3y)(4x + 5y)

Substituting x = 4x ; a = 3y and b = 5y in

(x + a) (x + b) = x2 + x (a + b) + ab, we get

(4x + 3y)(4x – 5y) = (4x)2 + 4x (3y + 5y) + (3y) (5y)

= 42 x2 + 4x (8y) + 15y2 = 16x2 + 32xy + 15y2

(4x + 3y)(4x + 5y) = 16x2 + 32xy + 15y2

(iv) (8 + pq ) ( pq + 7)

Substituting x = pq ; a = 8 and b = 7 in

(x + a) (x + b) = x2 + x (a + b) + ab, we get

(pq + 8)(pq + 7) = (pq)2+ pq (8 + 7) + (8) (7)

= p2 q2 + pq (15) + 56

(8 + pq)(pq + 7) = p2 +q2 + 15pq + 56

5. Expand the following squares, using suitable identities.

(i) (2x +5)2

(ii) (b -7)2

(iii) (mn +3p)2

(iv) (xyz -1)2

Solution:

(i) (2x + 5)2

Comparing (2x + 5)2 with (a + b)2 we have a = 2x and b = 5

a = 2x and b = 5,

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

(2x + 5)2 = (2x)2 + 2(2x) (5) + 52 = 22 x2 + 20x + 25

= 22 x2 + 20x + 25

(2x + 5)2 = 4x2 + 20x + 25

(ii) (b – 7)2

Comparing (b – 7)2 with (ab)2 we have a = b and b = 7

(ab)2 = a2 – 2ab + b2

(b – 7)2 = b2 – 2(b)(7) + 72

(b – 7)2 = b2 – 14b + 49

(iii) (mn + 3p)2

Comparing (mn + 3p)2 with (a + b)2 we have

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

(mn + 3p)2 = (mn)2 + 2(mn) (3p) + (3p)2

(mn + 3p)2 = m2 n2 + 6mnp + 9p2

(iv) (xyz – 1)2

Comparing (xyz – l)2 with (a – b)2 we have = a + xyz and b = 1

a = xyz and b = 1

(a – b)2 = a2 – 2ab + b2

(xyz – 1)2 = (xyz)2 – 2 (xyz) (1) + 12

(xyz – 1 )2 = x2y2z2 – 2 xyz + 1

6. Using the identity (a + b)(ab) = a 2 b2 , find the following product.

(i) (p + 2)(p − 2)

(ii) (1 + 3b)(3b − 1)

(iii) (4 − mn)(mn + 4)

(iv) (6x + 7y)(6x – 7y)

Solution:

(i) (p + 2)(p – 2)

Substituting a = p ; b = 2 in the identity (a + b) (a– b) = a2b2, we get

(p + 2)(p – 2) = p2 – 22

(ii) (1 + 3b)(3b – 1)

(1 + 3b) (3b –1) can be written as (3b + 1) (3b –1)

Substituting a = 3b and b = 1 in the identity

(a + b) (a – b) = a2 – b2, we get

(3b + 1) (3b – 1) = (3b)2 – 12 = 32 × b2 – 12

(3b + 1) (3b – 1) = 9b2 – 12

(iii) (4 – mn)(mn + 4)

(4 – mn) (mn + 4) can be written as (4 – mn) (4 + mn) = (4 + mn) (4 – mn)

Substituting a = 4 and b = mn is

(a + b) (a – b) = a2b2, we get

(4 + mn) (4 – mn) = 42 – (mn)2 = 16 – m2 n2

(iv) (6x + 7y)(6x – 7y)

Substituting a = 6x and b = 7y in

(a + b) (a – b) = a2b2, we get

(6x + 7y) (6x – 7y). = (6x)2 – (7y)2 = 62x2 – 72y2

(6x + 7y) (6x – 7y) = 36x2 – 49y2

7. Evaluate the following, using suitable identity.

(i) 512

(ii) 1032

(iii) 9982

(iv) 472

(v) 297 × 303

(vi) 990 × 1010

(vii) 51 × 52

Solution:

(i). 512

512 = (50+ l)2

Taking a = 50 and b – 1 we get

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

(50+ l)2 = 502 + 2 (50) (1) + 12 = 2500 + 100 +1

512 = 2601

(ii) 1032

1032 = (100+ 3)2

Taking a = 100 and b = 3

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

(100+ 3)2 = 1002 + 2 (100) (3) + 32 = 10000 + 600 + 9

1032 = 10609

(iii) 9982

9982 = (1000–2)2

Taking a = 1000 and b = 2

(a – b)2 = a2 + 2ab + b2 becomes

(1000 – 2)2 = 10002 – 2 (1000) (2) + 22

= 1000000 – 4000 + 4

9982 = 10,04,004

(iv) 472

472 = (50 – 3)2

Taking a = 50 and b = 3

(a – b)2 = a2 – 2ab + b2 becomes

(50 – 3)2 = 502 – 2 (50) (3) + 32

= 2500 – 300 + 9 = 2200 + 9

472 = 2209

(v) 297 × 303

297 × 303 = (300–3) (300+ 3)

Taking a = 300 and b = 3, then

(a + b) (a – b) = a2b2 becomes

(300+ 3) (300–3) = 3002 – 32

303 × 297 = 90000 – 9

297 × 303 = 89,991

(vi) 990 × 1010

990 × 1010 = (1000– 10) (1000+ 10)

Taking a = 1000 and b = 10 , then

(a – b) (a + b) = a2b2 becomes

(1000 – 10) (1000 + 10) = 10002 – 102

990 × 1010 = 1000000 – 100

990 × 1010 = 999900

(vii) 51 × 52

51 × 52 = (50+ 1) (50 + 2)

Taking x = 50 , a = 1 and b = 2

then (x + a) (x + b) = x2 + (a + b) x + ab becomes

(50 +1) (50 + 2) = 502 + (1 + 2) 50 + (l × 2)

= 2500 + (3) 50 + 2 = 2500 + 150 + 2

51 × 52 = 2652

8. Simplify: (a + b)2 4ab

Solution:  (a + b)2 – 4ab = a2 + b2 + 2ab – 4ab = a2 + b2 – 2ab = (a – b)2

9. Show that (m n)2 +(m + n)2 =2(m2 +n2 )

Solution: Taking the LHS = (m – n)2 + (m + n)2

= m2 – 2mn + n2 + m2 + 2mn + n2 = m2 + n2 + m2 + n2

= 2m2 + 2n2

= 2(m2 + n2) = RHS

(m – n)2 + (m + n)2 = 2(m2 + n2)

10. If a + b =  10 and ab = 18, find the value of a 2 +b2

Solution:

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

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

given a + b = 0 and ab = 18

102 = a2 + b2 + 2(18)

100 = a2 + b2 + 36

100 – 36 = a2 + b2

a2 + b2 = 64

11. Factorise the following algebraic expressions by using the identity a 2 b 2 =(a + b)(a b) .

(i) z2 16

(ii) 9 4y2

(iii) 25a2 49b2

(iv) x 4 − y4

Solution:

(i) z2 – 16

z2 – 16 = z2 – 42

We have a2b2 = (a + b) (a – b)

let a =  and    b = 4,

z2 – 42 = (z + 4)(z – 4)

(ii) 9 – 4y2

9 – 4y2 = 32 – 22 y2 = 32 – (2y)2

let a = 3 and b = 2y, then

a2 b2 = (a + b) (a – b)

32 – (2y)2 = (3 + 2y)(3 – 2y)

9 – 4y2 = (3 + 2y) (3 – 2y)

(iii) 25a2 – 49b2

25a2– 49b2 = 52 a2 – 72 b2 = (5a)2 – (7b)2

let A = 5a and B = 7b

A2 – B2 = (A + B) (A – B) becomes

(5a)2 – (7b)2 = (5a + 7b)(5a –7b)

(iv) x4y4

Let x4y4 = (x2)2 – (y2)2

We have a2b2 = (a + b) (a – b)

(x2)2 – (y2)2 = (x 2+ y2)(x2y2)

x4y4 = (x2 + y2) (x2y2)

Again we have x2y2 = (x + y) (x – y)

x4y4 = (x2 + y2) (x + y) (x – y)

12. Factorise the following using suitable identity.

(i) x2 8 x +16

(ii) y2 + 20y +100

(iii) 36m2 + 60m +25

(iv) 64x2 112xy + 49 y2

(v) a2 +6ab + 9b2 c2

Solution:

(i) (i) x2 – 8x +16

x2 – 8x +16 = x2 – (2 × 4 × x) + 42

This expression is in the form of identity

a2 – 2ab + b2 = (a – b)2

x2 – 2 × 4 × x + 42 = (x – 4)2

x2 – 8x + 16 = (x – 4) (x – 4)

(ii) y2 + 20y +100

y2 + 20y + 100 = y2 + (2 × (10))y + (10 × 10)

= y2 + (2 × 10 × y) + 102

This is of the form of identity

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

y2 + (2 × 10 × y) + 102 = (y + 10)2

y2 + 20y +100 = (y +10)2

y2 + 20y + 100 = (y + 10) (y + 10)

(iii) 36m2 + 60m + 25

36m2 + 60m + 25 = 62 m2 + 2 × 6m × 5 + 52

This expression is of the form of identity

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

(6m)2 + (2 × 6m × 5) + 52 = (6m + 5)2

36 m2 + 60m + 25 = (6m + 5) (6m + 5)

(iv) 64x2 – 112xy + 49y2

64x2 – 112xy + 49y2 = 82 x2 – (2 × 8x × 7y) + 72y2

This expression is of the form of identity

a2 – 2ab + b2 = (a – b)2

(8x)2 – (2 × 8x × 7y) + (7y)2 = (8x – 7y)2

64x2 – 112xy + 49y2 = (8x – 7y) (8x – 7y)

(v) a2 + 6ab + 9b2c2

a2 + 6ab + 9b2c2 = a2 + 2 × a × 3b + 32 b2c2

= a2 + (2 × a × 3b) + (3b)2c2

This expression is of the form of identity

[a2 + 2ab + b2] – c2 = (a + b)2c2

a2 + (2 × a × 3b) + (3b)2c2 = (a + 3b)2c2

Again this RHS is of the form of identity

a2b2 = (a + b) (a – b)

(a + 3b)2c2 = [(a + 36) + c] [(a + 3b) – c]

a2 + 6ab + 9b2c2 = (a +3b + c) (a + 3bc)

Objective Type Questions

13. If a + b = 5 and a2 + b2 =13, then ab = ?

(i) 12

(ii) 6

(iii) 5

(iv) 13

Solution: (a + b)2 = 25

13 + 2ab = 25

2ab =12

ab = 6

14. (5 +20)(−205)=?

(i) −425

(ii) 375

(iii) −625

(iv) 0

Solution: (50 + 20) (–20 – 5) = – (5 + 20)2 = – (25)2 = – 625

15. The factors of x2 6x +9 are

(i) (x − 3)(x − 3)

(ii) (x − 3)(x + 3)

(iii) (x + 3)(x + 3)

(iv) (x − 6)(x + 9)

[Answer: (i) (x – 3)(x –3)]

Solution: x2 – 6x + 9 = x2 – 2(x) (3) + 32

a2 – 2ab + b2 = (a b)2 = (x – 3)2 = (x – 3) (x – 3)

16. The common factors of the algebraic expressions ax2 y , bxy2 and cxyz is

(i) x2 y

(ii) xy2

(iii) xyz

(iv) xy

Solution: ax2y = a × x × x × y

bxy2 = b × x × y × y

cxyz = c × x × y × z

Common factor = xy

Exercise 3.1

1. (i) p2–2pq+q2 (ii) x2–25 (iii) (x–2) and (x–2) (iv) 2 × 2 × 2 × 3 × a × b × b × c × c

2. (i) True (ii) False (iii) True (iv) True.

3. (i) 2 × 2 × 2 × 3 × a × b × b × c × c

(ii) 2 × 2 × 3 × 3 × x × x × x × y × y × z

(iii) 2 × 2 × 2 × 7 × m × n × n × p × p

4. (i) x2 +10x + 21 (ii) 36a2 + 24a – 45 (iii) 16x2 + 32xy+15y2 (iv) p2q2 + 15pq + 56

5. (i) 4x2 + 20x + 25 (ii) b2 – 14b + 49 (iii) m2n2 + 6 mnp + 9p2 (iv) x2y2y2 – 2 xyz + 1

6. (i) p2– 4 (ii) 9b2–1 (iii) 16 – m2n2 (iv) 36x2– 49y2

7. (i) 2, 601 (ii) 10, 609 (iii) 9,96,004 (iv) 2, 209 (v) 89,991 (vi) 9,99,900 (vii) 2,652

8. (a–b)2

10. 64

11. (i) (z + 4) (z – 4) (ii) (3 + 2y) (3 – 2y) (iii) (5a + 7b) (5a – 7b) (iv) (x2 + y2) (x + y) (x – y)

12. (i) (x–4) (x–4) (ii) (y + 10) (y + 10) (iii) (6m + 5) (6m + 5) (iv) (8x – 7y) (8x – 7y) (v) (a + 3b + c) (a + 3b – c)

Objective type questions

13. (ii) 6

14. (iii) – 625

15. (i) (x – 3) (x–3)

16. (iv) xy

Tags : Algebra | Term 3 Chapter 3 | 7th Maths , 7th Maths : Term 3 Unit 3 : Algebra
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7th Maths : Term 3 Unit 3 : Algebra : Exercise 3.1 | Algebra | Term 3 Chapter 3 | 7th Maths