Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1
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 x2 – 4x + 4 are _______
4. Express 24ab2c2 as product of its factors is _______
Answers:
1. p2 – 2pq + q2
2. x2 – 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 x2 – 7x – 12
(ii) (a – 1)2 = a2 – 1.
(iii) (x2 + y2)(y2 + x2) = (x2 + y2)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) 24ab2c2
(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
Question 4.
Using the identity (x + a)(x + b) – x2 + 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
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 = 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 (a – b)2 we have a = b and b = 7
(a – b)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 – 1)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 = x2 y2 z2 – 2 xyz + 1
Question 6.
Using the identity (a + b)(a – b) = a2 – 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) = a2 – b2, 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 = 36 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) = a2 – b2, 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) = a2 – b2, We get
(6x + 7y) (6x – 7y) = (6x)2 – (7y)2 = 62x2 – 72y2
(6x + 7y) (6x – 7y) = (6x)2 – (7y)2 = 62x2 – 72y2
(6x + 7y) (6x – 7y) = 36x2 – 49y2
Question 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:
512
= (50 + 1)2
Taking a = 50 and b = 1 we get
(a + b)2 = a2 + 2ab + b2
(50 + 1)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) = a2 – b2 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) = a2 – b2 becomes
(1000 – 10) (1000 + 10) = 10002 – 102
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) = x2 + (a + b) x + ab becomes
(50 + 1) (50 + 2) = 502 + (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 = a2 + b2 + 2ab – 4ab = a2 + b2 – 2ab = (a – b)2
Question 9.
Show that (m – n)2 + (m + n)2 = 2(m2 + n2)
Solution:
Taking the LHS = (m – n)2 + (m + n)2
Question 10.
If a + b = 10 , and ab = 18, find the value of a2 + 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
Question 11.
Factorise the following algebraic expressions by using the identity a2 – b2 = (a + b)(a – b).
(i) z2 – 16
(ii) 9 – 4y2
(iii) 25a2 – 49b2
(iv) x4 – y4
Solution:
(i) z2 – 16
z2 – 16 = z2 – 42
We have a2 – b2 = (a + b) (a – b)
let a = z 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 = (5a)2 – (7b)2
let A = 5a and B = 7b
A2 B2
(5a)2 – (7b)2 = (5a + 7b) (5a – 7b)
(iv) x4 – y4
Let x4 – y4 = (x2)2 – (y2)2
We have a2 – b2 = (a + b) (a – b)
(x2)2 – (y2)2 = (x2 + y2) (x2 – y2)
x4 – y4 = (x2 + y2) (x2 – y2)
Again we have x2 – y2 = (x + y) (x – y)
∴ x4 – y4 = (x2 + y2) (x + y) (x – y)
Question 12.
Factorise the following using suitable identity.
(i) x2 – 8x + 16
(ii) y2 + 20y + 100
(iii) 36m2 + 60m + 25
(iv) 64x2 – 112xy + 49y2
(v) a2 + 6ab + 9b2 – c2
Solution:
(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 + 2 ab + 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
36m2 + 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 + 9b2 – c2
a2 + 6ab + 9b2 – c2 = a2 + 2 × a × 3b + 32 b2 – c2
= a2 + (2 × a × 3b) + (3b)2 – c2
This expression is of the form of identity
[a2 + 2ab + b2] – c2 = (a + b)2 – c2
a2 + (2 × a × 36) + (3b)2 – c2 = (a + 3b)2 – c2
Again this RHS is of the form of identity
a2 – b2 = (a + b) (a – b)
(a + 3b)2 – c2 = [(a + 3b) + c] [(a + 3b) – c]
a2 + 6ab + 9b2 – c2 = (a + 3b + c) (a + 3b – c)
Objective Type Questions
Question 1.
If a + b = 5 and a2 + b2 = 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 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)
Hint: x2 – 6x + 9 = x2 – 2(x) (3) + 32
a2 – 2ab + b2 – (a- b)2 = (x – 3)2 = (x – 3) (x – 3)
Question 4.
The common factors of the algebraic expression ax2y, bxy2 and cxyz is
(i) x2y
(ii) xy2
(iii) xyz
(iv) x
Ans :
(iv) xy
Hint: ax2y = a × x × x × y
bxy2 = b × x × y × y
cxyz = C × x × y × z
Common factor = xy
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