Selina Concise Class 9 Maths Chapter 4 Expansions 4A Solutions

Selina Concise Class 9 Maths Chapter 4 Expansions Exercise 4A Solutions

 

EXERCISE – 4A

 

(1) Find the square of

(i) 2a + b

= Solution:

We have to square of (2a + b)

Now, we have to use formula

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

a = 2a and b = b

(2a + b)² = (2a) + 2 × 2a × b + b2

(2a + b)2 = 4a2 + 4ab + b2

 

(ii) 3a + 7b

= Solution :

We have to square of (3a + 7b)

Now, we have to use formula,

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

a = 3a and b = 7b

(3a + 7b) = (3a)2 + 2 × (3a) × (7b) + (7b)2

(3a + 7b)2 = 9a2 + 42ab + 49b2

 

(iii) 3a – 4b

= Solution:-

We have to square of (3a – 4b)

Now, we have to use formula

(a – b)2 = a2 – 2ab + b2

a = 3a and b = 4b

(3a – 4b)2 = (3a)2 – 2 × (3a) × (4b) + (4b)2

(3a – 4b)2 = 9a2 – 24ab + 16b2

 

(iv) 3a/2a – 2b/3a

= Solution:

we have to square of (3a/2b – 2b/3a)

Now, we have to use formula

(a – b)2 = a2 – 2ab + b2

a = 3a/2b and b = 2b/3a

(3a/2b – 2b/3a)2

= (3a/2b)2 – 2 × (3a/2b) x (2b/3a) + (2b/3a)

= 9a2/4b2 – 12ab/6ab + 4b2/9a2

=  9a2/4b2 – 2 + 4b2/9a2

 

(2) Use identities to evaluate

(i) (101)2

= Solution:

We have to evaluate (101)2 first we have to convert in to (a + b)2 so, we have write (101) into

(101)2 = (100 +)2

a = 100 and b = 1

Now, we have to use identity

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

(100 + 1)2 = (100)2 + 2 × (100) × (1)2

= 10,000 + 200 + 1

∴ (100 + 1)2 = 10,201

∴ (101)2 = 10,201

 

(ii) (502)2

= Solution:

We have to evaluate (502)2 first we have to convert in to (a + b)2 so we have write (502) in to

(502)2 = (500 + 2)2

a = 500 and b = 2

Now, we have to use identity

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

(502)2 = (500 + 2)2

= (500)2 + 2 × (500) × (2) + (2)2

= 25,0000 + 20,00 + 4

(502)2 = 25,2004

 

(iii) (97)2

= Solution

We have to evaluate (97)2

First we have to convert into (a + b)2

So, We have write (97) into

(97)2 = (90 + 7)2

a = 90 and b = 7

Now we  have to use identity

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

(97)2 = (90 + 7)2

= (90)2 + 2 × 90 × 7 + (7)2

= 81,00 + 1260 + 49

(97)2 = 9409

 

(iv) (998)2

= solution:

we have to evaluate 998)2 first we have to convert into (a + b)2

So, we have to write (998) into,

(998)2 = (1000 – 2)2

a = 1000 and b = 2

Now, we have to use identity

(a – b)2 = a2 – 2ab + b2

(998)2 = (1000 – 2)2

= (1000)2 – 2× (1000) × (2) + (2)2

= 10,0000 – 4000 + 4

(998)2 = 99,6004

 

(3) Evaluate

(i) (7x/8 + 4y/5)2

= Solution:

we have to evaluate (7x/8 + 4y/5)2

we have to use identity

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

∴ a = 7x/8 and b = 4y/5

∴ (7x/8 + 4y/5)2

= (7x/8)2 + 2 × (7x/8) × (4y/5) + (4y/5)2

(7x/8 + 4y/5)2 = 49x2/64 + 56xy/40 + 16y2/25

(7x/8 + 4y/5)2 = 49x2/64 + 7xy/5 + 16xy2/25

 

(ii) (2x/7 – 7y/4)2

= Solution:

We have to evaluate (2x/7 – 7y/4)

we have to use identity

(a – b)2 = a2 – 2ab + b2

a = 2x/7 and b = 7y/4

(2x/7 – 7y/4)2

= (2x/7)2 – 2× (2x/7) (7y/4) + (7y/4)2

= 4x2/49 – 28x/28 + 49y2/16

(2x/7 – 7y/4)2 = 4x2/49 – xy + 49y2/16

 

(4) Evaluate

(i) (a/2b + 2b/a)2 – (a/2b – 2b/a)2 = 4

= Solution:

We have to Evaluate

(a/2b + 2b/a)2 – (a/2b – 2b/a)2 – 4

Firstly we have to expand the first term of given expansion

So we have to use identity

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

(a/2b + 2b/a)2 = (a/2b)2 + 2 × (a/2b) × (2b/a) + (2b/a)2

= a2/4b2 + 4ab/2ab + 4b2/a2

(a/2b + 2b/a)2 = a2/4b2 + 2 + 4b2/a2

Now, we have to expand second term

(a/2b – 2b/a)2

So, we have to use identity

(a – b)2 = a2 – 2ab + b2

a = a/2b and b = 2b/a

(a/2b – 2b/a)2 = (a/2b)2 – 2 × (a/2b) × (2b/a) + (2b/a)2

= a2/4b2 – 4ab/2ab + 4b2/a2

(a/2b – 2b/a)2 = a2/4b2 – 2 + 4b2/a2

put equation (A) and (B) in given expansion

(a2/4b2 + 2 + 4b2/a2) – (a2/4b2 – 2 + 4b2/a2) – 4

= a2/4b2 + 2 + 4b2/a2 – a2/4b2 + 2 – 4b2/a2 – 4

= 0

 

(ii) (4a + 3b)2 – (4a – 3b)2 + 48ab

= Solution:

We have to Evaluate

(4a + 3b)2 – (4a – 3b)2 + 48ab

Firstly we have to expand the first term of given expansion

So, we have to use identity

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

(4a + 3b)2 = (4a)2 + 2 × (4a) x (3b) + (3b)2

(4a + 3b)2 = 16a2 – 24ab + 9b2

Now, we have to expand second term

(4a – 3b)2 = (4a)2 – 2 × (4a) × (3b) + (3b)2

(4a – 3b)2 = 16a2 – 24ab + 9b2

(a – b)2 = a2 – 2ab + b2

Put equation (A) and (B) in given expansion

(16a2 + 24ab + 9b2) – (16a2 – 24ab + 9b2) + 48ab

= 16a2 + 24ab + 9b2 – 16a2 + 24ab – 9b2 + 48ab

= 48ab + 48ab

= 96ab

 

(5) If a + b = 7 and 10, find a – b

= Solution:

given that, a + b = 7 and ab = 7

To find = (a – b) = ?

Now, we know that,

(a- b)2 = a2 – 2ab + b2

= a2 + b2 – 2ab

= a2 + b2 + 2ab – 4ab {- 2ab = 2ab – 4ab}

(a – b)2 = (a + b)2 – 4ab

Substitute a = b = 7 and ab = 10 in (A)

(a – b)2 = (7)2 – 4 × 10)

= 49 – 40

(a –b)2 = 9

Taking square root on both sides,

a – b = ± 3

 

(6) If a – b = 7 and ab = 18, find a + b

= Solution:

Given that, a – b = 7 and ab = 18

To find = (a + b) = 2

Now, we know that,

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

= a2 + b2 – 2ab + 4ab

(a +b)2 = (a – b)2 + 4ab

Substitute a – b = 7 and ab = 18 in (A)

(a + b)2 = (7)2 + 4 x 18

= 49 + 72

(a + b)2 = 121

Taking square root on both sides

√(a + b)2 = √121

a + b = ± 11

 

(7) If x + y = 7/2 and xy = 5/2 find

(i) x – y  

(ii) x2 – y2

= Solution:

given that, x + y = 7/2 and xy = 5/2

(i) To find = x – y

we have to use identity

(x – y)2 = x2 – 2xy + y2

= x2 + y2 + 2xy – 4xy

(x – y)2 = (x + y)2 – 4xy

Now, we have to substitute

x + y = 7/2 and xy = 5/2

(x – y)2 = (7/2)2 – 4 × 5/2

= 49/4 – 10

= 49 – 40/4

(x – y)2 = 9/4

Taking square root on both side

√(x – y)2 = √9/4

[(x – y) = ± 3/2

(ii) To find = x2 – y2

We know that,

a2 – b2 = (a + b) (a – b)

x2 – y2 = (x + y) (x – y)

= (7/2) (+3/2) from A and from given

x2 – y2 = ±21/4

 

(8) If a – b = 0.9 and ab = 0.36; find

(i) a + b

(ii) a2 – b2

= Solution:

given that, a – b = 0.9 and ab = 0.36

(i) To find = (i) a + b

We know that,

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

= a2 + b2 – 2ab + 4ab

(a + b)2 = (a – b)2 + 4ab

Substitute a – b = 0.9 and ab = 0.36

(a + b)2 = (0.9)2 + 4(0.36)

= 0.81 + 1.44

(a + b)2 = 2.25

Taking square root on both sides

√(a + b)2 = √2.25

(a + b) = ± 1.5

(ii) To find = a2 – b2

We know that,

a2 – b2 = (a + b) (a – b)

= (± 1.5) (.09) { from (A) and from given}

a2 – b2 = ± 1.35

 

(9) If a – b = 4 and a + b = 6, find

(i) a2 + b2

(ii) ab

= Solution:

given that, a – b = 4 and a + b = 6

To find = (ii) ab

Now, we know that,

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

= a2 + b2 – 2ab + 4ab

(a + b)2 = (a – b)2 + 4ab

Substitute, a – b = 4 and (a + b) = 6

(6)2 = (4)2 + 4ab

36 = 16 + 4ab

36 – 16 = 4ab

20 = 4ab

20/4 = ab

5 = ab

∴ ab = 5

 

(ii) To find = a2 + b2

Now, we know that

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

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

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

(6)2 = (a2 + b2) + 2 × 5

36 = (a2 + b2) + 10 {from given and from (A)}

36 – 10 = a2 + b2

26 = a2 + b2

a2 + b2 = 26

 

(10) If a + 1/a = 6 and a ≠ o find:

(i) a – 1/a

(ii) a2 – 1/a2

= solution:

given that, a + 1/a = 6 and a ≠ o

(i) To find = a – 1/a

We have to use identity

(a – b)2 = a2 – 2ab + b2

Now,

(a – 1/a)2 = a2 – 2 × a × (1/a) + (1/a)2     

= a2 – 2 + 1/a2

(a – 1/a)2 = (a2 + 1/a2) – 2

Now, we use another identity,

(a + 1/a)2 = a2 + 2a × 1/a + 1/a2

(a + 1/a)2 = (a2 + 1/a2) + 2

(6)2 = (a2 + 1/a2) + 2 {from given}

36 = (a2 + 1/a2) + 2

36 – 2 = a2 + 1/a2

34 = a2 + 1/a2

a2 + 1/a2 = 34 put in equation (A)

(a – 1/a)2 = (34) – 2

(a – 1/a)2 = 32

Taking square root on both side,

√(a – 1/a)2 = √32

(a – 1/a) = √8 × 4

= 2√8

= 2√4 × 2

(a – 1/a) = 4√2 (B)

 

(ii) To find = a2 – 1/a2

We know that,

(a2 – 1/a2) = (a + 1/a) (a – 1/a)

{ ∴ (a2 – b2) (a – b)}

a2 – 1/a2 = (6) (4√2) {∴ from given and equation (B)

a2 – 1/a2 = 24√2

 

 

(11) If a – 1/a = 8 and a ≠ 0 and

(i) a + 1/a

(ii) a2 – 1/a2

= Solution :

given that, a – 1/a = 8 and a ≠ 0

(i) To find = a + 1/a

We have to use identity,

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

∴ (a + 1/a)2 = a2 + 2 × a × 1/a  + (1/a)2

= a2 + 2 + 1/a2

Now, we have to use another identity,

(a – b)2 = a2 – 2ab + b2

∴ (a – 1/a)2 = a2 – 2 × a × 1/a + (1/a)2

(a – 1/a)2 = (a2 + 1/a2) – 2

(8)2 = (a2 + 1/a2) – 2  (from given)

64 = (a2 + 1/a2) – 2

64 + 2 = (a2 + 1/a2)

66 = a2 = 1/a2

a2 + 1/a2 = 66 put in equation (A)

(a + 1/a)2 = 66 + 2

(a + 1/a)2 = 68

Taking square root on both sides,

√(a + 1/a)2 = √68

a + 1/a = √4 × 17

(a + 1/a)  = ± 2√17 (B)

(ii) To find = a2 – 1/a

We know that,

(a2 – b2) = (a + b) (a – b)

∴ (a2 – 1/a2) = (a + 1/a) (a – 1/a)

= (±2√17) (8) {∴ from given and from equation}

(a2 – 1/a2) = ± 16√17

 

(12) If a² – 3a + 1 = o and a ≠ o; find:

(i) a + 1/a

(ii) a2 = 1/a2

= Solution:

given that, a2 – 3a + 1 = 0 and a ≠ o

(i) To find = a + 1/a

given equation:

a2 – 3a + 1 = 0

divide ‘a’ on both sides

a2/a – 3a/a + 1/a = 0/a

a – 3 + 1/a = 0

(a + 1/a) – 3 = 0

a + 1/a = 3 (A)

(ii) To find = a2 + 1/a2

We know that,

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

(a + 1/a)2 = a2 + 2 × a × 1/a + 1/a2

(a + 1/a)2 = (a2 + 1/a2) + 2

(3)2 = (a2 + 1/a2) + 2 {from A}

9 = (a2 + 1/a2) + 2

9 – 2 = a2 + 1/a2

7 = a2 + 1/a2

a2 + 1/a2 = 7

 

(13) If a2 – 5a – 1 = 0 and a ≠ 0; find

(i) a – 1/a

(ii) a + 1/a

(iii) a2 – 1/a2

= Solution:

given that, a2 – 5a – 1 = 0 and a ≠ 0

(i) To find = a – 1/a

given equation:

a2 – 5a – 1 = 0

divide ‘a’ on both sides;

a2/a – 5a/a – 1/a = 0

a – 5 – 1/a = 0

(a – 1/a) – 5 = 0

(a – 1/a) = 5

 

(ii) To find = a + 1/a

We know that,

(a – b)2 = a2 – 2ab + b2

∴ (a – 1/a)2 = a2 – 2 × a × 1/a + (1/a)2

= a2 – 2 + 1/a2

(a – 1/a)2 = (a2 + 1/a2) – 2

(5)2 = (a2 + 1/a2) – 2

25 = (a2 + 1/a2) – 2

25 + 2 = (a2 + 1/a2)

27 = a2 + 1/a2

∴ a2 + 1/a2 = 27   (A)

Now, we have to use another identity,

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

(a + 1/a)2 = a2 + × a × 1/a + 1/a2

= a2 + 2 + 1/a2

(a + 1/a)2 = (a2 + 1/a2) + 2

= 27 + 2   { from equation (A)}

(a + 1/a)2 = 29

Taking square root on both side,

√ (a + 1/a) = √29

a + 1/a = ± √29 (B)

 

(iii) To find = a2 – 1/a2

We know that,

(a2 – b2) = (a + b) (a – b)

∴ (a2 – 1/a2) = (a + 1/a) (a – 1/a)

(a2 – 1/a2) =   (± √29) (5) {∴ from (B) and from given}

(a2 = 1/a2) = ± 5√29

 

(14) If 3a + 4b = 16 and ab = 4; and the value of 9a2 + 16b2

= Solution:

given that, 3a + 4b = 16 and ab = 4

To find = 9a2 + 16b2

∴ We have to write expansion of (3a + 4b)

(3a + 4b)2 = (3a)2 + 2 × (3a) × (4b) + (4b)2

= 9a2 + 24ab + 16b2

(3a + 4b)2 = (9a2 + 16b2) + 24ab

(16)2 = (9a2 + 16b2) + 24 x (4)

{∴ from given}  

256 = (9a2 + 16b2) + 96

256 – 96 = 9a2 + 16b2

160 = 9a2 + 16b2

9a2 + 16b2 = 16

 

(15) The number a is 2 more than the number b. If the sum of the squares of and b is 34, then find the product of a and b.

= Solution:

given that, The number a is 2 more that the number b

a = b + 2 (A)

and also given that, the sum of the squares of a and b is 34

a2 + b2 = 34

(b + 2)2 = b2 = 34 (from (A))

b2 + 2 × b × 2 + (2)2 + b2 = 34

b2 + 4b + 4 + b2 = 34    {∴ (a + b)2 = a2 + b2 + 2ab}

2b2 + 4b + 4 = 34

2b2 + 4b = 34 – 4

2b2 + 4b = 30

2b2 + 4b – 30 = 0

2b2 + 4b – 30 = 0

divide ‘2’ on both sides,

2b2/2 + 4b/2 – 30/2 = 0

b2 + 2b – 15 = 0

b2 + 5b – 3b – 15 = 0

b (b + 5) – 3 (b + 5) = 0

(b + 5) (b – 3) = 0

b + 5 = 0 or b – 3 = 0

b = – 5 or b = 3

∴ We have to find the product of a and b

If a = -5 then a = – 5 + 2 = -3

If b = 3 then a = 3 + 2 = 5

The product is

ab = (- 3) × (- 5) = ± 15 and

ab = 5 × 3 = 15

 

(16) The difference between two positive numbers is 5 and the sum of their squares is 73 find the product of these numbers.

= Solution:

Let the two positive numbers are x and y

given that, the difference between two positive numbers is 5

x – y = 5  (i)

and also given that the sum of their squares is 73.

x2 + y2 = 73 (2)

equation (i) = x – y = 5

x = 5 + y put in equation (2)

(5 + y)2 + y2 = 73

(5)2 + 2 × 5 × y + y2 + y2 = 73   {∴ (a + b)2 = a2 + b2 + 2ab)

25 + 10y + 2y2 = 73

2y2 + 10y = 73 – 25

2y2 + 10y = 48

2y2 + 10y – 48 = 0

divide ‘2’ on both sides,

2y2/2 + 10y/2 – 48/2 = 0

y2 + 5y – 24 = 0

y2 + 8y – 3y – 24 = 0

y (y + 8) – 3 (y + 8) = 0

(y + 8) (y – 3) = 0

y + 8 = 0 or y – 3 = 0

y = 8 or y = 3

If y = – 8 then

If y = 3 then

x = 5 – 8 = -3

x = 5 + 3 = 8

Now, we have to find product of these two numbers:

xy = 8 × 3 = 24

∴ xy = 24

 

 

Here is your solution of Selina Concise Class 9 Maths Chapter 4 Expansions Exercise 4A

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