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

• Last Updated : 02 Feb, 2021

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

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

Solution:

i) (2x â€“ 1/x)2

We know that, (a -b)2 = a2 â€“ 2ab + b2

So, (2x â€“ 1/x)2 = (2x)2 â€“ (2 Ã— 2x Ã— 1/x) + (1/x)2

= 4x2 + 1/x2 â€“ 4

ii) (2x + y) (2x â€“ y)

We know that, (a + b)(a â€“ b) = a2 â€“ b2

So, (2x + y) (2x â€“ y) = (2x)2 â€“ (y)2

= 4x2 â€“ y2

iii) (a2b â€“ b2a)2

We know that, (a -b)2 = a2 â€“ 2ab + b2

So, (a2b â€“ b2a)2 = (a2b)2 â€“ (2 Ã— a2b Ã— b2a) + (b2a)2

= a4b2 + b4a2 â€“ 2a3b3

iv) (a â€“ 0.1) (a + 0.1)

We know that, (a + b)(a â€“ b) = a2 â€“ b2

So, (a â€“ 0.1) (a + 0.1) = (a)2 â€“ (0.1)2

= a2 â€“ 0.01

v) (1.5.x2 â€“ 0.3y2) (1.5x2 + 0.3y2)

We know that, (a + b)(a â€“ b) = a2 â€“ b2

So, (1.5.x2 â€“ 0.3y2) (1.5.x2 + 0.3y2) = (1.5.x2)2 â€“ (0.3y2)2

= 2.225x4 â€“ 0.09y4

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

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

Solution:

i) (399)2

We can write (399)2 as (400 â€“ 1)2

Also, (a -b)2 = a2 â€“ 2ab + b2

= (400)2 + (1)2 â€“ 2 Ã— 400 Ã—1

= 160000 + 1 â€“ 800 = 159201

Hence, (399)2 = 159201

ii) (0.98)2

We can write (0.98)2 as (1 â€“ 0.02)2

Also, (a -b)2 = a2 â€“ 2ab + b2

= (1)2 + (0.02)2 â€“ 2 Ã— 0.02 Ã—1

= 1 + 0.0004 â€“ 0.04 = 0.9604

Hence, (0.98)2 = 0.9604

iii) 991 Ã— 1009

We can write 991 Ã— 1009 as (1000 â€“ 9)(1000 + 9)

Also, (a + b)(a â€“ b) = a2 â€“ b2

= (1000)2 â€“ (9)2

= 1000000 â€“ 81 = 999919

Hence, 991 Ã— 1009 = 999919

iv) 117 Ã— 83

We can write 117 Ã— 83 as (100 + 17)(100 â€“ 17)

Also, (a + b)(a â€“ b) = a2 â€“ b2

= (100)2 â€“ (17)2

= 10000 â€“ 289 = 9711

Hence, 117 Ã— 83 = 9711

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

(i) 175 Ã— 175 +2 Ã— 175 Ã— 25 + 25 Ã— 25
(ii) 322 Ã— 322 â€“ 2 Ã— 322 Ã— 22 + 22 Ã— 22
(iii) 0.76 Ã— 0.76 + 2 Ã— 0.76 Ã— 0.24 + 0.24 Ã— 0.24
(iv) (7.83 Ã— 7.83 â€“ 1.17 Ã— 1.17)/6.66

Solution:

i) 175 Ã— 175 +2 Ã— 175 Ã— 25 + 25 Ã— 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 Ã— 175 +2 Ã— 175 Ã— 25 + 25 Ã— 25 = 40000

ii) 322 Ã— 322 â€“ 2 Ã— 322 Ã— 22 + 22 Ã— 22

It can be written as (322)2 â€“ 2(322)(22) + (22)2

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

So, we can conclude (322)2 â€“ 2(322)(22) + (22)2 as (322 â€“ 22)2

= (300)2 = 90000

Hence, 322 Ã— 322 â€“ 2 Ã— 322 Ã— 22 + 22 Ã— 22 = 90000

iii) 0.76 Ã— 0.76 + 2 Ã— 0.76 Ã— 0.24 + 0.24 Ã— 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 Ã— 0.76 + 2 Ã— 0.76 Ã— 0.24 + 0.24 Ã— 0.24 = 1

iv) (7.83 Ã— 7.83 â€“ 1.17 Ã— 1.17)/6.66

It can be written as (7.832 â€“ 1.172)/6.66

And we also know that, (a + b)(a â€“ b) = a2 â€“ b2

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

= (9 Ã— 6.66)/6.66 = 9

Hence, (7.83 Ã— 7.83 â€“ 1.17 Ã— 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 Ã— (x) Ã— (1/x)

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

So,

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

121 â€“ 2 = x2 + 1/x2

x2 + 1/x2 = 119

Hence, value of x2 + 1/x2 is 119

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

Solution:

Given, x â€“ 1/x = -1

So, (x â€“ 1/x)2 = (x)2 + (1/x)2 â€“ 2 Ã— (x) Ã— (1/x)

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

So,

(-1)2 = x2 + 1/x2 â€“ 2

1 + 2 = x2 + 1/x2

x2 + 1/x2 = 3

Hence, value of x2 â€“ 1/x2 is 3

### Question 6. If x + 1/x = âˆš5, find the value of x2 +1/x2 and x4 +1/x4

Solution:

Given, x + 1/x = âˆš5

So, (x + 1/x)2 = (x)2 + (1/x)2 + 2 Ã— (x) Ã— (1/x)

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

So,

(âˆš5)2 = x2 + 1/x2 + 2

5 â€“ 2 = x2 + 1/x2

x2 + 1/x2 = 3

Now, taking the square of x2 + 1/x2

(x2 + 1/x2)2 = x4 + 1/x4 + 2 Ã— (x)2 Ã— (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 â€“ 1/x

Solution:

Given, x2 +1/x2 = 66

Let us take the square of x â€“ 1/x

So, (x â€“ 1/x)2 = (x)2 + (1/x)2 â€“ 2 Ã— (x) Ã— (1/x)

Since, (a â€“ b)2 = a2 + b2 â€“ 2ab

So,

(x â€“ 1/x)2 = 66 â€“ 2

(x â€“ 1/x)2 = 64

x â€“ 1/x = Â±8

Hence, the value of x â€“ 1/x is 8

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