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• RD Sharma Class 9 Solutions

# Class 9 RD Sharma Solutions – Chapter 4 Algebraic Identities- Exercise 4.1 | Set 1

### Question 1.Evaluate each of the following using identities:

(i) (2x – 1/x)2
(ii) (2x + y) (2x – y)
(iii) (a2b – b2a)2
(iv) (a – 0.1) (a + 0.1)
(v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2)

Solution:

i) (2x – 1/x)2

We know that, (a -b)2 = a2 – 2ab + b2

So, (2x – 1/x)2 = (2x)2 – (2 × 2x × 1/x) + (1/x)2

= 4x2 + 1/x2 – 4

ii) (2x + y) (2x – y)

We know that, (a + b)(a – b) = a2 – b2

So, (2x + y) (2x – y) = (2x)2 – (y)2

= 4x2 – y2

iii) (a2b – b2a)2

We know that, (a -b)2 = a2 – 2ab + b2

So, (a2b – b2a)2 = (a2b)2 – (2 × a2b × b2a) + (b2a)2

= a4b2 + b4a2 – 2a3b3

iv) (a – 0.1) (a + 0.1)

We know that, (a + b)(a – b) = a2 – b2

So, (a – 0.1) (a + 0.1) = (a)2 – (0.1)2

= a2 – 0.01

v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2)

We know that, (a + b)(a – b) = a2 – b2

So, (1.5.x2 – 0.3y2) (1.5.x2 + 0.3y2) = (1.5.x2)2 – (0.3y2)2

= 2.225x4 – 0.09y4

### Question 2. Evaluate each of the following using identities:

(i)(399)2
(ii)(0.98)2
(iii)991 × 1009
(iv) 117 × 83

Solution:

i) (399)2

We can write (399)2 as (400 – 1)2

Also, (a -b)2 = a2 – 2ab + b2

= (400)2 + (1)2 – 2 × 400 ×1

= 160000 + 1 – 800 = 159201

Hence, (399)2 = 159201

ii) (0.98)2

We can write (0.98)2 as (1 – 0.02)2

Also, (a -b)2 = a2 – 2ab + b2

= (1)2 + (0.02)2 – 2 × 0.02 ×1

= 1 + 0.0004 – 0.04 = 0.9604

Hence, (0.98)2 = 0.9604

iii) 991 × 1009

We can write 991 × 1009 as (1000 – 9)(1000 + 9)

Also, (a + b)(a – b) = a2 – b2

= (1000)2 – (9)2

= 1000000 – 81 = 999919

Hence, 991 × 1009 = 999919

iv) 117 × 83

We can write 117 × 83 as (100 + 17)(100 – 17)

Also, (a + b)(a – b) = a2 – b2

= (100)2 – (17)2

= 10000 – 289 = 9711

Hence, 117 × 83 = 9711

### Question 3. Simplify each of the following:

(i) 175 × 175 +2 × 175 × 25 + 25 × 25
(ii) 322 × 322 – 2 × 322 × 22 + 22 × 22
(iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24
(iv) (7.83 × 7.83 – 1.17 × 1.17)/6.66

Solution:

i) 175 × 175 +2 × 175 × 25 + 25 × 25

It can be written as (175)2 + 2(175)(25) + (25)2

And we also know that, (a + b)2 = a2 + b2 + 2ab

So, we can conclude (175)2 + 2(175)(25) + (25)2 as (175 + 25)2

= (200)2 = 40000

Hence, 175 × 175 +2 × 175 × 25 + 25 × 25 = 40000

ii) 322 × 322 – 2 × 322 × 22 + 22 × 22

It can be written as (322)2 – 2(322)(22) + (22)2

And we also know that, (a – b)2 = a2 + b2 – 2ab

So, we can conclude (322)2 – 2(322)(22) + (22)2 as (322 – 22)2

= (300)2 = 90000

Hence, 322 × 322 – 2 × 322 × 22 + 22 × 22 = 90000

iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24

It can be written as (0.76)2 + 2(0.76)(0.24) + (0.24)2

And we also know that, (a + b)2 = a2 + b2 + 2ab

So, we can conclude (0.76)2 + 2(0.76)(0.24) + (0.24)2 as (0.76 + 0.24)2

= (1.0)2 = 1

Hence, 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24 = 1

iv) (7.83 × 7.83 – 1.17 × 1.17)/6.66

It can be written as (7.832 – 1.172)/6.66

And we also know that, (a + b)(a – b) = a2 – b2

So, we can conclude (7.832 – 1.172)/6.66 as [(7.83 + 1.17)(7.83 – 1.17)]/6.66

= (9 × 6.66)/6.66 = 9

Hence, (7.83 × 7.83 – 1.17 × 1.17)/6.66 = 9

### Question 4.If x + 1/x = 11, find the value of x2 +1/x2

Solution:

Given, x + 1/x = 11

So, (x + 1/x)2 = (x)2 + (1/x)2 + 2 × (x) × (1/x)

We also know that, (a + b)2 = a2 + b2 + 2ab

So,

(11)2 = x2 + 1/x2 + 2

121 – 2 = x2 + 1/x2

x2 + 1/x2 = 119

Hence, value of x2 + 1/x2 is 119

### Question 5. If x – 1/x = -1, find the value of x2 +1/x2

Solution:

Given, x – 1/x = -1

So, (x – 1/x)2 = (x)2 + (1/x)2 – 2 × (x) × (1/x)

We also know that, (a – b)2 = a2 + b2 – 2ab

So,

(-1)2 = x2 + 1/x2 – 2

1 + 2 = x2 + 1/x2

x2 + 1/x2 = 3

Hence, value of x2 – 1/x2 is 3

### Question 6. If x + 1/x = √5, find the value of x2 +1/x2 and x4 +1/x4

Solution:

Given, x + 1/x = √5

So, (x + 1/x)2 = (x)2 + (1/x)2 + 2 × (x) × (1/x)

We also know that, (a + b)2 = a2 + b2 + 2ab

So,

(√5)2 = x2 + 1/x2 + 2

5 – 2 = x2 + 1/x2

x2 + 1/x2 = 3

Now, taking the square of x2 + 1/x2

(x2 + 1/x2)2 = x4 + 1/x4 + 2 × (x)2 × (1/x2)

(3)2 = x4 + 1/x4 + 2

x4 + 1/x4 = 7

Hence, value of x2 + 1/x2 is 3 and that of x4 +1/x4 is 7

### Question 7. If x2 +1/x2 = 66, find the value of x – 1/x

Solution:

Given, x2 +1/x2 = 66

Let us take the square of x – 1/x

So, (x – 1/x)2 = (x)2 + (1/x)2 – 2 × (x) × (1/x)

Since, (a – b)2 = a2 + b2 – 2ab

So,

(x – 1/x)2 = 66 – 2

(x – 1/x)2 = 64

x – 1/x = ±8

Hence, the value of x – 1/x is 8

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