# Exercise 6.6

## 1. Write the following squares of binomials as trinomials:

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

(a–b)2 = a2–2ab+b2

### (i) (x + 2) 2

#### Solution:

(x+2) 2 is in the form of (a + b)2 = a2+2ab+b2

here, a = x , b = 2

⇒  x2 + 2 × x × 2 +  b2

⇒  x2 + 4x + b2

### (ii)  (8a + 3b) 2

#### Solution:

(8a+3b) 2 is in the form of (x + y) 2 = x2 + 2xy + y2

Here, x = 8a, y = 3b

⇒ (8a)2 + 2 × (8a) × (3b) + (3b)2

⇒ 642 + 48ab + 36b2

### (iii) (2m+1)2

#### Solution:

(2m+1)2 is in the form of (a + b) 2 = a2 + 2ab + b2

here, a =2m, b = 1

⇒ (2m)2 + 2 × (2m) × (1)+(1) 2

⇒ 4m2 + 4m +1

## 2. Find the product of the following binomials:

### (i)  (2x + y) (2x + y)

#### Solution:

(2x + y) (2x + y) can be written as (2x+y)2

⇒ (2x+y)2

⇒ (2x)2+2×(2x)×(y)+y2

⇒ 4x2+4xy+y2

### (ii) (a + 2b) (a – 2b)

#### Solution:

(a + 2b) (a – 2b) can be written as a2–(2b)2

⇒ a2– 4b2

### (iii) (a2+bc) (a2–bc)

#### Solution:

(a2 + bc) (a2 – bc)

= (a2)2 – (bc)2

= a4 – b2c2

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

### (i)  (102)2

#### Solution:

102 can be written as (100 + 2)

Using identity, (a + b)2  =a2+2ab+b2, we have

(102)2 = (100+2)2

Here, a = 100, b = 2

⇒ (100+2)2

⇒ (100)2 + 2 × (100) × 2 + 22

⇒ 10000 + 400 + 4

⇒ 10404

### (ii) (99)2

#### Solution:

(99)2 can be written as (100–1)2

Using identity, (a–b)2=a2–2ab+b2

here, a = 100, b = 1

⇒ (100–1)2

⇒ (100)2 – 2 × (100) ×1+12

⇒ 10000 – 200 + 1

⇒ 9801

### (iii) (1001)2

#### Solution:

(1001)2 can be written as (1000+1)2

Using identity, (a + b)2 = a2 + 2ab  +b2

here, a = 1000, b = 1

⇒ (1000+1)

⇒ (1000)2 + 2 × (1000) ×1+12

⇒ 1000000 + 2000+1

⇒ 1002001

### (iv) (999)2

#### Solution:

(999)2 can be written as (1000–1)2

we know that, (a–b) = a2–2ab+b2

here, a = 1000, b = 1

⇒ (1000–1)2

⇒ (1000)2 – 2 × (1000) × 1+12

⇒ 1000000 – 2000 + 1

⇒ 998001

### (v) (703)2

#### Solution:

(703)2 can be written as (700+3)2

we know that, (a+b) 2 = a2+2ab+b2

here, a = 700, b = 3

⇒ (700+3)2

⇒ (700)2 + 2 × (700) × 3+32

⇒ 490000 + 4200 + 9

⇒ 494209

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

### (i) (82)2–(18)2

#### Solution:

(82)2 – (18)2

here, a = 82, b = 18

⇒ (82 + 18) (82 – 18)

⇒ 100×64

⇒ 6400

### (ii) (467)2–(33)2

#### Solution:

(467)2–(33)2

here, a = 467, b = 33

⇒ (467 + 33) (467 – 33)

⇒ 500×434

⇒ 217000

### (iii) (79)2–(69)2

#### (79)2–(69)2

here, a = 79, b = 69

⇒ (79 + 69) (79 – 69)

⇒ 148×10

⇒ 1480

### (iv) 197×203

#### Solution:

197×203 can be written as (200 – 3) (200 + 3)

⇒ (200 – 3) (200 + 3)

⇒ (200)2–(3)2

⇒ 40000 – 9

⇒ 39991

### (v) 113×87

#### Solution:

113×87 can be written as (100 + 13) (100 – 13)

⇒ (100 + 13) (100 – 13)

⇒ (100)2–(13)2

⇒ 10000 – 169

⇒ 9831

### (vi) 95×105

#### Solution :

95×105 can be written as (100 + 5) (100 – 5)

⇒ (100 + 5) (100 – 5)

⇒ (100)2–(5)2

⇒ 10000 – 25

⇒ 9975

### (vii) 1.8×2.2

#### Solution:

1.8 × 2.2 can be written as (2 + 0.2) (2 – 0.2)

⇒ (2 + 0.2) (2 – 0.2)

⇒ (2)2–(0.2)2

⇒ 4 – 0.04

⇒ 3.96

### (viii) 9.8×10.2

#### Solution:

9.8×10.2 can be written as (10 + 0.2) (10 – 0.2)

⇒ (10 + 0.2) (10 – 0.2)

⇒ (10)2–(0.2)2

⇒ 100 – 0.04

⇒ 99.96

5. Simplify the following using identities

### (ii) (178×178)–(22×22)

#### Solution:

we know that, (a + b) (a – b) = a2–b2

⇒ (178×178)–(22×22) = (178)2–(22)2

⇒ (178×178)–(22×22) = (178 + 22) (178 – 22)

⇒ (178×178)–(22×22) = 200×156

⇒ (178×178)–(22×22) = 31200

(iv) (1.73×1.73)–(0.27×0.27)

Solution:

we know that, (a + b) (a – b) = a2–b2

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

⇒ (1.73×1.73)–(0.27×0.27) = (1.73 + 0.27) (1.73 – 0.27)

⇒ (1.73×1.73)–(0.27×0.27) = 2×1.46

⇒ (1.73×1.73)–(0.27×0.27) = 2.92

## 6. Find the value of x, if:

### Solution:

we know that, (a + b) (a – b) = a2–b2

⇒ 4x = (52)2–(48)2

⇒ 4x = (52 + 48) (52 – 48)

⇒ 4x = 100×4

⇒ 4x = 400

⇒ x =

⇒ x = 100

### (ii) 14x=(47)2–(33)2

#### Solution:

we know that, (a + b) (a – b) = a2–b2

⇒ 14x = (47)2–(33)2

⇒ 14x = (47 + 33) (47 – 33)

⇒ 14x = 80×14

⇒ 14x = 1120

⇒ x =

⇒ x = 80

### (iii) 5x=(50)2–(40)2

#### Solution:

we know that, (a + b) (a – b) = a2–b2

⇒ 5x = (50)2–(40)2

⇒ 5x = (50 + 40) (50 – 40)

⇒ 5x = 90×10

⇒ 5x = 900

⇒ x =

⇒ x = 180

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

Solution:

Given that,

x + y = 4 and xy = 2

we know that, (a + b)2 = a2+2ab+b2

⇒ x2 + y2=(x + y)2–2xy

⇒ x2 + y2 = 42–(2 × 2)

⇒ x2 + y2 =16–4

⇒ x2 + y2=12

### 11.  If x – y = 7 and xy = 9,find the value of x2+y2

#### Solution:

Given that,

x – y = 7 and xy = 9

we know that, (a–b)2 = a2–2ab+b2

⇒ x2 – y2=(x–y)2+2xy

⇒ x2 – y2=72+(2×9)

⇒ x2 – y2=49+18

⇒ x2 – y2=67

### 12. If 3x + 5y = 11 and xy = 2, find the value of 9x2+25y2

#### Solution:

Given that,

3x + 5y = 11 and xy = 2

we know that, (a + b)2 = a2+2ab+b2

(3x+5y)=(3x)2 + 2 × (3x) × (5y)+(5y)2

⇒ (3x+5y)2 = 9×2+30xy+25y2

⇒ 9x2+25y2 = (3x+5y)2 –10xy

⇒ 9x2+25y= (11)2–(30×2)

⇒9x2+25y= 121–60

⇒ 9x2+25y2 = 61

## 13. Find the values of the following expressions

we know that, (a + b)2 =a2+2ab+b2

64x2+81y2+144xy = (8x+9y)2

⇒ 64x2+81y2+144xy = (8(11)+9(43))2

⇒ 64x2+81y2+144xy = (88+12)2

⇒ 64x2+81y2+144xy = (100)2

⇒ 64x2+81y2+144xy = 10000

### 16. If 2x + 3y = 14 and 2x – 3y = 2, find the value of xy

#### Solution:

we know that, (a + b)(a–b) = a2–b2

Given, 2x + 3y = 14 and 2x – 3y = 2

squaring 2x + 3y = 14 and 2x – 3y = 2 and then subtracting them, we get

(2x + 3y)^2 = 14^2 and (2x – 3y)^2 = 2^2

Subtract 2nd term form first, we have

(2x + 3y)^2 – (2x – 3y)^2 = 14^2 – 2^2 …….(1)

Solve L.H.S. first

(2x+3y)2 – (2x–3y)2 = [(2x+3y)+(2x–3y)][(2x+3y)–(2x–3y)]

⇒ (2x+3y)2–(2x–3y)2=4x×6y

⇒ (2x+3y)2 – (2x–3y)2=24xy

Now, from equation (1)

(14)2–(2)2 = 24xy

⇒ (14 + 2) (14 – 2) = 24xy

⇒ 16×12=24xy

⇒ 24xy = 192

⇒ xy = 1928

⇒ xy = 8

### (i) x + y

#### Solution:

Given,

x2+y2=29 and xy = 2

squaring the (x + y)

(x+y)2=x2+2×x×y+y2

⇒ (x+y)2=29+(2×2)

⇒ (x+y)2=29+4

⇒ (x+y)2=33

⇒ x+y= ±

### (ii) x – y

#### Solution:

Given,

x2 +y2=29 and xy = 2

squaring the (x – y)

(x–y)2=x2–2×x×y+y2

⇒ (x–y)2=29–(2×2)

⇒ (x–y)2=29–4

⇒ (x–y)2=25

⇒ x–y=±25−−√

⇒ x+y=±5

### (iii) x4+y4

#### Solution:

Given,

x2+y2=29 and xy = 2

(x2+y2)2=x4+2x2y2+y4

⇒ x4+y4=(x2+y2)2–2x2y2

⇒ x4+y4=(x2+y2)2–2(xy)2

⇒ x4+y4=(29)2–2(2)2

⇒ x4+y4=841–8

⇒ x4+y4=833

Updated: November 11, 2019 — 4:42 pm