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## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1

Question 1.

Fill in the blanks.

1. (p – q)^{2} = _______

2. The product of (x + 5) and (x – 5) is _______

3. The factors of x^{2} – 4x + 4 are _______

4. Express 24ab^{2}c^{2} as product of its factors is _______

Answers:

1. p^{2} – 2pq + q^{2}

2. x^{2} – 25

3. (x – 2) and (x – 2)

4. 2 × 2 × 2 × 3 × a × b × b × c × c

Question 2.

Say whether the following statements are True or False.

(i) (7x + 3) (7x – 4) = 49 x^{2} – 7x – 12

(ii) (a – 1)^{2} = a^{2} – 1.

(iii) (x^{2} + y^{2})(y^{2} + x^{2}) = (x^{2} + y^{2})^{2}

(iv) 2p is the factor of 8pq.

Answers:

(i) True

(ii) False

(iii) True

(iv) True

Question 3.

Express the following as the product of its factors.

(i) 24ab^{2}c^{2}

(ii) 36 x^{3}y^{2}z

(iii) 56 mn^{2}p^{2}

Solution:

(i) 24ab^{2}c^{2} = 2 × 2 × 2 × 3 × a × b × b × c × c

(ii) 36 x^{3}y^{2}z = 2 × 2 × 3 × 3 × x × x × x × y × y × z

(iii) 56 mn^{2}p^{2} = 2 × 2 × 2 × 7 × m × n × n × p × p

Question 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 x^{2} + x (a + b) + ab

(x + 3) (x + 7) = x^{2} + x (3 + 7) + (3 × 7) = x^{2} + 10x + 21

(ii) (6a + 9) (6a – 5)

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

In (x + a) (x + b) = x^{2} + x (a + b) + ab, we get

(6a + 9)(6a – 5) = (6a)^{2} + 6a (9 + (-5)) + (9 × (-5))

6^{2} a^{2} + 6a (4) + (-45) = 36a^{2} + 24a – 45

(6a + 9) (6a – 5) = 36a^{2} + 24a – 45

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

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

(x + a) (x + b) = x^{2} + x (a + b) + ab, we get

(4x + 3y) (4x – 5y) = (4x)^{2} + 4x (3y + 5y) + (3y) (5y)

= 4^{2} x^{2} + 4x (8y) + 15y^{2} = 16x^{2} + 32xy + 15y^{2}

(4x + 3y) (4x + 5y) = 16x^{2} + 32xy + 15y^{2}

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

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

(x + a) (x + b) = x^{2} + x (a + b) + ab, we get

(pq + 8) (pq + 7) = (pq)^{2} + pq (8 + 7) + (8) (7)

= p^{2} q^{2} + pq (15) + 56

(8 + pq) (pq + 7) = p^{2} q^{2} + 15pq + 56

Question 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} = a^{2} + 2ab + b^{2}

(2x + 5)^{2} = (2x)^{2} + 2(2x) (5) + 5^{2} = 2^{2} x^{2} + 20x + 25

= 2^{2} x^{2} + 20x + 25

(2x + 5)^{2} = 4x^{2} + 20x + 25

(ii) (b – 7)^{2}

Comparing (b – 7)^{2} with (a – b)^{2} we have a = b and b = 7

(a – b)^{2} = a^{2} – 2ab + b^{2}

(b – 7)^{2} = b^{2} – 2(b) (7) + 7^{2}

(b – 7)^{2} = b^{2} – 14b + 49

(iii) (mn + 3p)^{2}

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

(a + b)^{2} = a^{2} + 2ab + b^{2}

(mn + 3p)^{2} = (mn)^{2} + 2(mn) (3p) + (3p)^{2}

(mn + 3p)^{2} = m^{2} n^{2} + 6mnp + 9p^{2}

(iv) (xyz – 1)^{2}

Comparing (xyz – 1)^{2} with (a – b)^{2} we have = a + xyz and b = 1

a = xyz and b = 1

(a – b)^{2} = a^{2} – 2ab + b^{2}

(xyz – 1)^{2} = (xyz)^{2} – 2 (xyz) (1) + 1^{2}

(xyz -1)^{2} = x^{2} y^{2} z^{2} – 2 xyz + 1

Question 6.

Using the identity (a + b)(a – b) = a^{2} – b^{2}, 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) = a^{2} – b^{2}, we get

(p + 2) (p – 2) = p^{2} – 2^{2}

(ii) (1 + 3b)(3b – 1)

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

Substituting a = 36 and b = 1 in the identity

(a + b) (a – b) = a^{2} – b^{2}, we get

(3b + 1)(3b – 1) = (3b)^{2} – 1^{2} = 3^{2} × b^{2} – 1^{2}

(3b + 1) (3b – 1) = 9b^{2} – 1^{2}

(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) = a^{2} – b^{2}, we get

(4 + mn) (4 – mn) = 4^{2} – (mn)^{2} = 16 – m^{2} n^{2}

(iv) (6x + 7y) (6x – 7y)

Substituting a = 6x and b = 7y in

(a + b) (a – b) = a^{2} – b^{2}, We get

(6x + 7y) (6x – 7y) = (6x)^{2} – (7y)^{2} = 6^{2}x^{2} – 7^{2}y^{2}

(6x + 7y) (6x – 7y) = (6x)^{2} – (7y)^{2} = 6^{2}x^{2} – 7^{2}y^{2}

(6x + 7y) (6x – 7y) = 36x^{2} – 49y^{2}

Question 7.

Evaluate the following, using suitable identity.

(i) 51^{2}

(ii) 103^{2}

(iii) 998^{2}

(iv) 47^{2}

(v) 297 × 303

(vi) 990 × 1010

(vii) 51 × 52

Solution:

51^{2}

= (50 + 1)^{2}

Taking a = 50 and b = 1 we get

(a + b)^{2} = a^{2} + 2ab + b^{2}

(50 + 1)^{2} = 50^{2} + 2 (50) (1) + 1^{2} = 2500 + 100 + 1

51^{2} = 2601

(ii) 103^{2}

103^{2} = (100 + 3)^{2}

Taking a = 100 and b = 3

(a + b)^{2} = a^{2} + 2ab + b^{2} becomes

(100 + 3)^{2} = 100^{2} + 2 (100) (3) + 3^{2} = 10000 + 600 + 9

103^{2} = 10609

(iii) 998^{2}

998^{2} = (1000 – 2)^{2}

Taking a = 1000 and b = 2

(a – b)^{2} = a^{2} + 2ab + b^{2} becomes

(1000 – 2)^{2} = 1000^{2} – 2 (1000) (2) + 2^{2}

= 1000000 – 4000 + 4

998^{2} = 10,04,004

(iv) 47^{2}

47^{2} = (50 – 3)^{2}

Taking a = 50 and b = 3

(a – b)^{2} = a^{2} – 2ab + b^{2} becomes

(50 – 3)^{2} = 50^{2} – 2 (50) (3) + 3^{2}

= 2500 – 300 + 9 = 2200 + 9

47^{2} = 2209

(v) 297 × 303

297 × 303 = (300 – 3) (300 + 3)

Taking a = 300 and b = 3, then

(a + b) (a – b) = a^{2} – b^{2} becomes

(300 + 3) (300 – 3) = 300^{2} – 3^{2}

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) = a^{2} – b^{2} becomes

(1000 – 10) (1000 + 10) = 1000^{2} – 10^{2}

990 × 1010 = 1000000 – 100

990 × 1010 = 999900

(vii) 51 × 52

= (50 + 1) (50 + 1)

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

then (x + a) (x + b) = x^{2} + (a + b) x + ab becomes

(50 + 1) (50 + 2) = 50^{2} + (1 + 2) 50 + (1 × 2)

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

51 × 52 = 2652

Question 8.

Simplify: (a + b)^{2} – 4ab

Solution:

(a + b)^{2} – 4ab = a^{2} + b^{2} + 2ab – 4ab = a^{2} + b^{2} – 2ab = (a – b)^{2}

Question 9.

Show that (m – n)^{2} + (m + n)^{2} = 2(m^{2} + n^{2})

Solution:

Taking the LHS = (m – n)^{2} + (m + n)^{2
}

Question 10.

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

Solution:

We have (a + b)^{2} = a^{2} + 2ab + b^{2}

(a + b)^{2} = a^{2} + b^{2} + 2ab

given a + b = 0 and ab = 18

10^{2} = = a^{2} + b^{2} + 2(18)

100 = = a^{2} + b^{2} + 36

100 – 36 = a^{2} + b^{2}

a^{2} + b^{2} = 64

Question 11.

Factorise the following algebraic expressions by using the identity a^{2} – b^{2} = (a + b)(a – b).

(i) z^{2} – 16

(ii) 9 – 4y^{2}

(iii) 25a^{2} – 49b^{2}

(iv) x^{4} – y^{4}

Solution:

(i) z^{2} – 16

z^{2} – 16 = z^{2} – 4^{2}

We have a^{2} – b^{2} = (a + b) (a – b)

let a = z and b = 4,

z^{2} – 4^{2} = (z + 4) (z – 4)

(ii) 9 – 4y^{2}

9 – 4y^{2} = 3^{2} – 2^{2} y^{2 }= 3^{2} – (2y)^{2}

let a = 3 and b = 2y, then

a^{2} – b^{2} = (a + b) (a – b)

∴ 3^{2} – (2y)^{2} = (3 + 2y) (3 – 2y)

9 – 4y^{2} = (3 + 2y) (3 – 2y)

(iii) 25a^{2} – 49b^{2}

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

let A = 5a and B = 7b

A^{2 }B^{2}

(5a)^{2} – (7b)^{2} = (5a + 7b) (5a – 7b)

(iv) x^{4} – y^{4}

Let x^{4} – y^{4} = (x^{2})^{2} – (y^{2})^{2}

We have a^{2} – b^{2} = (a + b) (a – b)

(x^{2})^{2} – (y^{2})^{2} = (x^{2} + y^{2}) (x^{2} – y^{2})

x^{4} – y^{4} = (x^{2} + y^{2}) (x^{2} – y^{2})

Again we have x^{2} – y^{2} = (x + y) (x – y)

∴ x^{4} – y^{4} = (x^{2} + y^{2}) (x + y) (x – y)

Question 12.

Factorise the following using suitable identity.

(i) x^{2} – 8x + 16

(ii) y^{2} + 20y + 100

(iii) 36m^{2} + 60m + 25

(iv) 64x^{2} – 112xy + 49y^{2}

(v) a^{2} + 6ab + 9b^{2} – c^{2}

Solution:

(i) x^{2} – 8x + 16

x^{2} – 8x + 16 = x^{2} – (2 × 4 × x) + 4^{2}

This expression is in the form of identity

a^{2} – 2ab + b^{2} = (a – b)^{2}

x^{2} – 2 × 4 × x + 4^{2} = (x – 4)^{2}

∴ x^{2} – 8x + 16 = (x – 4) (x – 4)

(ii) y^{2} + 20y + 100

y^{2} + 20y + 100 = y^{2} + (2 × (10)) y + (10 × 10)

= y^{2} + (2 × 10 × y) + 10^{2}

This is of the form of identity

a^{2} + 2 ab + b^{2} = (a + b)^{2}

y^{2} + (2 × 10 × y) + 10^{2} = (y + 10)^{2}

y^{2} + 20y + 100 = (y + 10)^{2}

y^{2} + 20y + 100 = (y + 10) (y + 10)

(iii) 36m^{2} + 60m + 25

36m^{2} + 60m + 25 = 62 m^{2} + 2 × 6m × 5 + 5^{2}

This expression is of the form of identity

a^{2} + 2ab + b^{2} = {a + b)^{2}

(6m)^{2} + (2 × 6m × 5) + 5^{2}

= (6m + 5)^{2}

36m^{2} + 60m + 25 = (6m + 5) (6m + 5)

(iv) 64x^{2} – 112xy + 49y^{2}

64x^{2} – 112xy + 49y^{2} = 82 x^{2} – (2 × 8x × 7y) + 7^{2}y^{2}

This expression is of the form of identity

a^{2} – 2ab + b^{2} = (a- b)^{2}

(8x)^{2} – (2 × 8x × 7y) + (7y)^{2} = (8x – 7y)^{2}

64x^{2} – 112xy + 49y^{2} = (8x – 7y) (8x – 7y)

(v) a^{2} + 6ab + 9b^{2} – c^{2}

a^{2} + 6ab + 9b^{2} – c^{2} = a^{2} + 2 × a × 3b + 3^{2} b^{2} – c^{2}

= a^{2} + (2 × a × 3b) + (3b)^{2} – c^{2}

This expression is of the form of identity

[a^{2} + 2ab + b^{2}] – c^{2} = (a + b)^{2} – c^{2}

a^{2} + (2 × a × 36) + (3b)^{2} – c^{2} = (a + 3b)^{2} – c^{2}

Again this RHS is of the form of identity

a^{2} – b^{2} = (a + b) (a – b)

(a + 3b)^{2} – c^{2} = [(a + 3b) + c] [(a + 3b) – c]

a^{2} + 6ab + 9b^{2} – c^{2} = (a + 3b + c) (a + 3b – c)

Objective Type Questions

Question 1.

If a + b = 5 and a^{2} + b^{2} = 13, then ab = ?

(i) 12

(ii) 6

(iii) 5

(iv) 13

Answer:

(ii) 6

Hint: (a + b)^{2} = 25

13 + 2ab = 25

2ab = 12

ab = 6

Question 2.

(5 + 20)(-20 – 5) = ?

(i) -425

(ii) 375

(iii) -625

(iv) 0

Answer:

(iii) -625

Hint: (50 + 20) (-20 – 5) = -(5 + 20)^{2} = – (25)^{2} = – 625

Question 3.

The factors of x^{2} – 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)

Hint: x^{2} – 6x + 9 = x^{2} – 2(x) (3) + 3^{2}

a^{2} – 2ab + b^{2} – (a- b)^{2} = (x – 3)^{2} = (x – 3) (x – 3)

Question 4.

The common factors of the algebraic expression ax^{2}y, bxy^{2} and cxyz is

(i) x^{2}y

(ii) xy^{2}

(iii) xyz

(iv) x

Ans :

(iv) xy

Hint: ax^{2}y = a × x × x × y

bxy^{2} = b × x × y × y

cxyz = C × x × y × z

Common factor = xy