Class 8 RD Sharma Solutions- Chapter 6 Algebraic Expressions And Identities - Exercise 6.6 | Set 1

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