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
Solution :
(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:
(i) 4x = (52)2–(48)2
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)2 =(3x)2 + 2 × (3x) × (5y)+(5y)2
⇒ (3x+5y)2 = 9×2+30xy+25y2
⇒ 9x2+25y2 = (3x+5y)2 –10xy
⇒ 9x2+25y2 = (11)2–(30×2)
⇒9x2+25y2 = 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
Answer: xy = 8
17. If x2+y2=29and xy = 2, Find the value of
(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
