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Class 8 RD Sharma Solutions- Chapter 6 Algebraic Expressions And Identities – Exercise 6.6 | Set 1
  • Last Updated : 07 Apr, 2021
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Question 1. Write the following squares of binomials as trinomials:

(i) (x + 2)2

Solution:

x2 + 2 (x) (2) + 22

x2 + 4x + 4

(ii) (8a + 3b)2

Solution:

(8a)2 + 2 (8a) (3b) + (3b)



64a2 + 48ab + 9b2

(iii) (2m + 1)2

Solution:

(2m)2 + 2 (2m) (1) + 12

4m2 + 4m + 1

(iv) (9a + 1/6)2

Solution:

(9a)2 + 2 (9a) (1/6) + (1/6)2

81a2 + 3a + 1/36

(v) (x + x2/2)2

Solution:



(x)2 + 2 (x) (x2/2) + (x2/2)2

x2 + x3 + 1/4x4

(vi) (x/4 – y/3)2

Solution:

(x/4)2 – 2 (x/4) (y/3) + (y/3)2

1/16x2 – xy/6 + 1/9y2

(vii) (3x – 1/3x)2

Solution:

(3x)2 – 2 (3x) (1/3x) + (1/3x)2

9x2 – 2 + 1/9x2

(viii) (x/y – y/x)2

Solution:

(x/y)2 – 2 (x/y) (y/x) + (y/x)2

x2/y2 – 2 + y2/x2

(ix) (3a/2 – 5b/4)2

Solution:

(3a/2)2 – 2 (3a/2) (5b/4) + (5b/4)2

9/4a2 – 15/4ab + 25/16b2

(x) (a2b – bc2)2

Solution:

(a2b)2 – 2 (a2b) (bc2) + (bc2)2

a4b4– 2a2b2c2 + b2c4

(xi) (2a/3b + 2b/3a)2

Solution:

(2a/3b)2 + 2 (2a/3b) (2b/3a) + (2b/3a)2

4a2/9b2 + 8/9 + 4b2/9a2

(xii) (x2 – ay)2

Solution:

(x2)2 – 2 (x2) (ay) + (ay)2

x4 – 2x2ay + a2y2

Question 2. Find the product of the following binomials:

i) (2x + y) (2x + y)

Solution:

2x (2x + y) + y (2x + y)

4x2 + 2xy + 2xy + y2

4x2 + 4xy + y2

(ii) (a + 2b) (a – 2b)

Solution:

a (a – 2b) + 2b (a – 2b)

a2 – 2ab + 2ab – 4b2

a2 – 4b2

(iii) (a2 + bc) (a2 – bc)

Solution:

a2 (a2 – bc) + bc (a2 – bc)

a4 – a2bc + bca2 – b2c2

a4 – b2c2

(iv) (4x/5 – 3y/4) (4x/5 + 3y/4)

Solution:

4x/5 (4x/5 + 3y/4) – 3y/4 (4x/5 + 3y/4)

16/25x2 + 12/20yx – 12/20xy – 9y2/16

16/25x2 – 9/16y2

(v) (2x + 3/y) (2x – 3/y)

Solution:

2x (2x – 3/y) + 3/y (2x – 3/y)

4x2 – 6x/y + 6x/y – 9/y2

4x2 – 9/y2

(vi) (2a3 + b3) (2a3 – b3)

Solution:

2a3 (2a3 – b3) + b3 (2a3 – b3)

4a6 – 2a3b3 + 2a3b3 – b6

4a6 – b6

(vii) (x4 + 2/x2) (x4 – 2/x2)

Solution:

= x4 (x4 – 2/x2) + 2/x2 (x4 – 2/x2)

= x8 – 2x2 + 2x2 – 4/x4

= (x8 – 4/x4)

(viii) (x3 + 1/x3) (x3 – 1/x3)

Solution:

= x3 (x3 – 1/x3) + 1/x3 (x3 – 1/x3)

= x6 – 1 + 1 – 1/x6

= x6 – 1/x6

Question 3. Using the formula for squaring a binomial, evaluate the following:

(i) (102)2

Solution:

We can rewrite 102 as 100 + 2

(102)2 = (100 + 2)2

By simplification ,

(100 + 2)2 = (100)2 + 2 (100) (2) + 22

= 10000 + 400 + 4 = 10404

(ii) (99)2

Solution:

We can rewrite 99 as 100 – 1

 (99)2 = (100 – 1)2

On simplification,

(100 – 1)2 = (100)2 – 2 (100) (1) + 12

= 10000 – 200 + 1 = = 9801

(iii) (1001)2

Solution:

We can rewrite 1001 as 1000 + 1

(1001)2 = (1000 + 1)2

On simplification ,

(1000 + 1)2 = (1000)2 + 2 (1000) (1) + 12

= 1000000 + 2000 + 1 = 1002001

(iv) (999)2

Solution:

We can rewrite 999 as 1000 – 1

(999)2 = (1000 – 1)2

By simplification,

(1000 – 1)2 = (1000)2 – 2 (1000) (1) + 12

= 1000000 – 2000 + 1 = 998001

(v) (703)2

Solution:

We can rewrite 700 as 700 + 3

(703)2 = (700 + 3)2

By simplification,

(700 + 3)2 = (700)2 + 2 (700) (3) + 32

= 490000 + 4200 + 9 = 494209

Question 4. Simplify the following using the formula: (a – b) (a + b) = a2 – b2 :

(i) (82)2 – (18)2

Solution:

Here we will use the formula 

(82)2 – (18)2 = (82 – 18) (82 + 18)

= 64 × 100

= 6400

(ii) (467)2 – (33)22

Solution:

We will using the formula (a – b) (a + b) = a2 – b2

(467)2 – (33)2 = (467 – 33) (467 + 33)

= (434) (500)

= 217000

(iii) (79)2 – (69)2

Solution:

We will using the formula (a – b) (a + b) = a2 – b2

(79)2 – (69)2 = (79 + 69) (79 – 69)

= (148) (10)

= 1480

(iv) 197 × 203

Solution:

We can rewrite 203 as 200 + 3 and 197 as 200 – 3

We will using the formula (a – b) (a + b) = a2 – b2

197 × 203 = (200 – 3) (200 + 3)

= (200)2 – (3)2

= 40000 – 9

= 39991

(v) 113 × 87

Solution:

We can rewrite 113 as 100 + 13 and 87 as 100 – 13

We can using the formula (a – b) (a + b) = a2 – b2

113 × 87 = (100 – 13) (100 + 13)

= (100)2 – (13)2

= 10000 – 169

= 9831

(vi) 95 × 105

Solution:

We can rewrite 95 as 100 – 5 and 105 as 100 + 5

We will  using the formula (a – b) (a + b) = a2 – b2

95 × 105 = (100 – 5) (100 + 5)

= (100)2 – (5)2

= 10000 – 25

= 9975

(vii) 1.8 × 2.2

Solution:

We can rewrite 1.8 as 2 – 0.2 and 2.2 as 2 + 0.2

We will using the formula (a – b) (a + b) = a2 – b2

1.8 × 2.2 = (2 – 0.2) ( 2 + 0.2)

= (2)2 – (0.2)2

= 4 – 0.04

= 3.96

(viii) 9.8 × 10.2

Solution:

We can rewrite 9.8 as 10 – 0.2 and 10.2 as 10 + 0.2

We will using the formula (a – b) (a + b) = a2 – b2

9.8 × 10.2 = (10 – 0.2) (10 + 0.2)

= (10)2 – (0.2)2

= 100 – 0.04

= 99.96

Question 5. Simplify the following using the identities:

(i) ((58)2 – (42)2)/16

Solution:

We will using the formula (a – b) (a + b) = a2 – b2

((58)2 – (42)2)/16 = ((58-42) (58+42)/16)

= ((16) (100)/16)

= 100

(ii) 178 × 178 – 22 × 22

Solution:

We will using the formula (a – b) (a + b) = a2 – b2

178 × 178 – 22 × 22 = (178)2 – (22)2

= (178-22) (178+22)

= 200 × 156

= 31200

(iii) (198 × 198 – 102 × 102)/96

Solution:

We using the formula (a – b) (a + b) = a2 – b2

(198 × 198 – 102 × 102)/96 = ((198)2 – (102)2)/96

= ((198-102) (198+102))/96

= (96 × 300)/96

= 300

(iv) 1.73 × 1.73 – 0.27 × 0.27

Solution:

We will using the formula (a – b) (a + b) = a2 – b2

1.73 × 1.73 – 0.27 × 0.27 = (1.73)2 – (0.27)2

= (1.73-0.27) (1.73+0.27)

= 1.46 × 2

= 2.92

(v) (8.63 × 8.63 – 1.37 × 1.37)/0.726

Solution:

We will using the formula (a – b) (a + b) = a2 – b2

(8.63 × 8.63 – 1.37 × 1.37)/0.726 = ((8.63)2 – (1.37)2)/0.726

= ((8.63-1.37) (8.63+1.37))/0.726

= (7.26 × 10)/0.726

= 72.6/0.726

= 100

Question 6. Find the value of x, if:

(i) 4x = (52)2 – (48)2

Solution:

We will using the formula (a – b) (a + b) = a2 – b2

4x = (52)2 – (48)2

4x = (52 – 48) (52 + 48)

4x = 4 × 100

4x = 400

x = 100

(ii) 14x = (47)2 – (33)2

Solution:

We will using the formula (a – b) (a + b) = a2 – b2

14x = (47)2 – (33)2

14x = (47 – 33) (47 + 33)

14x = 14 × 80

x = 80

(iii) 5x = (50)2 – (40)2

Solution:

We using the formula (a – b) (a + b) = a2 – b2

5x = (50)2 – (40)2

5x = (50 – 40) (50 + 40)

5x = 10 × 90

5x = 900

x = 180

Question 7. If x + 1/x =20, find the value of x2 + 1/ x2.

Solution:

Given equation in the x + 1/x = 20

when squaring both sides, we get

(x + 1/x)2 = (20)2

x2 + 2 × x × 1/x + (1/x)2 = 400

x2 + 2 + 1/x2 = 400

x2 + 1/x2 = 398

Question 8. If x – 1/x = 3, find the values of x2 + 1/ x2 and x4 + 1/ x4.

Solution:

Given in the question x – 1/x = 3

 when squaring both sides,

(x – 1/x)2 = (3)2

x2 – 2 × x × 1/x + (1/x)2 = 9

x2 – 2 + 1/x2 = 9

x2 + 1/x2 = 9+2

x2 + 1/x2 = 11

Now again when we square on both sides ,

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

x4 + 2 × x2 × 1/x2 + (1/x2)2 = 121

x4 + 2 + 1/x4 = 121

x4 + 1/x4 = 121-2

x4 + 1/x4 = 119

x2 + 1/x2 = 11

x4 + 1/x4 = 119

Question 9. If x2 + 1/x2 = 18, find the values of x + 1/ x and x – 1/ x.

Solution:

Given in the question x2 + 1/x2 = 18

When adding 2 on both sides, 

x2 + 1/x2 + 2 = 18 + 2

x2 + 1/x2 + 2 × x × 1/x = 20

(x + 1/x)2 = 20

x + 1/x = √20

When subtracting 2 from both sides, 

x2+ 1/x2 – 2 × x × 1/x = 18 – 2

(x – 1/x)2 = 16

x – 1/x = √16

x – 1/x = 4

Question 10. If x + y = 4 and xy = 2, find the value of x2 + y2

Solution:

We know that x + y = 4 and xy = 2

Upon squaring on both sides of the given expression, we get

(x + y)2 = 42

x2 + y2 + 2xy = 16

x2 + y2 + 2 (2) = 16   (since x y=2)

x2 + y2 + 4  = 16

x2 + y2 = 16 – 4

x2 + y2 =12

Chapter 6 Algebraic Expressions And Identities – Exercise 6.6 | Set 2

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