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1.

Of which of the binomials given below is the m2n2+ 14mnpq + 49p2q​​​​​​​2 the expansion?(A) (m + n) (p + q) (B) (mn – pq) (C) (7mn + pq)(D) (mn + 7pq)

Answer»

(D) (mn + 7pq)

Here, square root of the first term = mn 

Square root of the last term = 7pq 

∴ Required binomial = (mn + 7pq)2
Discussion
2.

Which of the options given below is the square of the binomial(A) 64 - 1/x2(B) 64 + 1/x2(C) 64 - 16/x + 1/x2(D) 64 + 16/x + 1/x2

Answer»

(C) 64 - 16/x + 1/x2

= (8 - 1/x)2 = 82 - 2 x 8 x 1/2 + (1/x)2 …[(a – b)2 = a2 – 2ab + b2 ]

= 64 - 16x + 1/x2

Discussion
3.

Expand:i. (5a + 6b)2ii. (a/2 + b/3)2iii. (2p - 3q)2iv. (x - 2/x)2

Answer»

i. (5a + 6b)2

Here, A = 5a and B = 6b 

(5a + 6b)2 = (5a)2 + 2 × 5a × 6b + (6b)2 …. [(A + B)2 = A2 + 2AB + B2 ] 

∴ (5a + 6b)2 = 25a2 + 60ab + 36b2

ii. (a/2 + b/2)2

Here A = a/2 and B = b/3

(a/2 + b/3)2 = (a/2)2 + 2 x a/2 x b/3 + (b/3)2 ...[(A + B) = A2 + 2AB + B2]

∴ (a/2 + b/3)2 = a2/4 + ab/3 + b2/9

iii. (2p – 3q)2

Here, a = 2p and b = 3q 

(2p – 3q)2 = (2p)2 – 2 × (2p) × (3q) + (3q)2 …. [(a – b)2 = a2 – 2ab + b2 ] 

∴ (2p – 3q)2 = 4p2 – 12pq + 9q2

iv. (x - 2/x)2

Here a = x and b = 2/x

(x - 2/x)2 = x2 - 2 x (x) x 2/x + (2/x)2 ...[(a - b)2 = a2 - 2ab + b2]

(x - 2/x)2 = x2 - 4 + 4/x2
Discussion
4.

Use an expansion formula to find the values of: i. (997)2ii. (102)2iii. (97)2iv. (1005)2

Answer»

i. (997)2 = (1000 – 3)2

Here, a = 1000 and b = 3 

(1000 – 3)2 = (1000)2 – 2 x 1000 x 3 + 32 …. [(a – b)2 = a2 – 2ab + b2 ] 

= 1000000 – 6000 + 9 

= 994009 

∴ (997)2 = 994009

ii. (102)2 = (100 + 2)2 

Here, a = 100 and b = 2 

(100 + 2)2 = (100)2 + 2 x 100 x 2 + 22 …. [(a + b)2 = a2 + 2ab + b2 ] 

= 10000 + 400 + 4 

= 10404 

∴ (102)2 = 10404

iii. (97)2 = (100 – 3)2 

Here, a = 100 and b = 3 

(100 – 3)2 = (100)2 – 2 x 100 x 3 + 32 …. [(a – b)2 = a2 – 2ab + b2 ] 

= 10000 – 600 + 9 

= 9409 

∴ (97)2 = 9409

iv. (1005)2 = (1000 + 5)2 

Here, a = 1000 and b = 5 

(1000 + 5)2 = (1000)2 + 2 x 1000 x 5 + 52 …. [(a + b)2 = a2 + 2ab + b2 ]

= 1000000+ 10000 + 25 

= 1010025 

∴ (1005)2 = 1010025
Discussion
5.

Use the given values to verify the formulae for squares of binomials. i. a = -7, b = 8 ii. a = 11,b = 3 iii. a = 2.5,b = 1.2

Answer»

i. (a + b)2 = (-7 + 8)2 

= 12 

= 1 

a2 + 2ab + b2 = (-7)2 + 2 x (-7) x 8 + 82 

= 49 – 112 + 64 

= 1 

∴(a + b)2 = a2  + 2ab + b2 

(a – b)2 = (-7 – 8)2 

= (-15)2 

= 225 

a2 – 2ab + b2 = (-7)2  – 2 x (-7) x 8 + (8)2 

= 49 + 112 + 64 

= 225 

∴(a – b)2 = a2  – 2ab + b2 

ii. (a + b)2 = (11 + 3)2 

= 142 

= 196 

a2 + 2ab + b2 = 112 + 2 x 11 x 3 + 32 

= 121 + 66 + 9 

= 196 

∴(a + b)2 = a2 + 2ab + b2 

(a – b)2 = (11 – 3)2 = 82 

= 64 

a2 – 2ab + b2 = 112 – 2 x 11 x 3 + 32 

= 121 – 66 + 9 

= 64 

∴(a – b)2 = a2  – 2ab + b2 

iii. (a + b)2 = (2.5 + 1.2)2  

= 3.72 

= 13.69 

a2 + 2ab + b2 = (2.5)2 + 2 x 2.5 x 1.2 + (1.2)2 

= 6.25 + 6 + 1.44 

= 13.69 

∴(a + b)2  = a2 + 2ab + b2 

(a – b)2 = (2.5 – 1.2)2 

= 1.32 

= 1.69 

a2 – 2ab + b2 = (2.5)2 – 2 x 2.5 x 1.2 + (1.2)2 

= 6.25 – 6 + 1.44 

= 1.69 

∴(a – b)2 = a2 – 2ab + b2 
Discussion
6.

Use the formula to multiply the following: i. (x + y)(x – y) ii. (3x – 5)(3x + 5) iii. (a + 6)(a – 6iv. (x/5 + 6) (x/5 - 6)

Answer»

i. Here, a = x, b = y 

(x + y)(x – y) = x2 – y2  …. [(a + b)(a – b) = a2  – b2 ]

ii. Here, a = 3x, b = 5 

(3x – 5) (3x + 5) = (3x)2  – 52  …. [(a + b)(a – b) = a2  – b2 ]

= 9x2  – 25

iii. Here, A = a, B = 6 

(a + 6) (a – 6) = a2  – 62  …. [(A + B)(A – B) = A2  – B2 ] 

= a2  – 36

iv. Here, a = x/2 , b = 6 

(x/5 + 6) (x/5 - 6) = (x/5)2 - (6)2…. [(a + b)(a – b) = a2  – b2 ]

= x2 /25 - 36
Discussion
7.

Factorize the following expressions: i. p2 – q2ii. 4x2 – 25y2 iii. y2 – 4 iv. P2 - 1/25v. 9x2 - 1/16y2

Answer»

i. p2 – q2 

Here, a = p, b = q 

∴ p2 – q2 = (p + q)(p – q) ….[(a2 – b2 ) = (a + b)(a – b)] 

ii. 4x2 – 25y2 

= (2x)2 – (5y)2 

Here, a = 2x, b = 5y 

∴ (2x)2 – (5y)2 = (2x + 5y)(2x – 5y) ….[(a2 – b2 ) = (a + b)(a – b)] 

iii. y2 – 4 

= y2 – 22 

Here, a = y, b = 2 

∴ y2 – 22 = (y + 2)(y – 2) ….[(a2 – b2 ) = (a + b)(a – b)]

iv P2 - 1/25

Here a = 1/25, b = 1/5

P2  - (1/5)2  = (P + 1/5) (p - 1/5)  ... [(a2 - b2 ) = (a + b)(a - b)]

v. 9x2 - 1/16y2

Here a = 3x, b= 1/4y

∴ (3x)2 - (1/4y)2 = (3x + 1/4y)(3x - 1/4y) ….[(a2 – b2) = (a + b)(a – b)]
Discussion
8.

Factorize the following expressions and write them in the product form. i. 201a3b2ii. 91xyt2 iii. 24a2b2iv. tr2s3

Answer»

i. 201a3b2 

= 3 × 67 × a3 × b2  

= 3 × 67 × a × a × a × b × b 

ii. 91xyt​​​​​​​2 

= 7 × 13 × x × y × t​​​​​​​2 

= 7 × 13 × x × y × t × t 

iii. 24a​​​​​​​2 b​​​​​​​2 

= 2 × 2 × 2 × 3 × a​​​​​​​2 × b​​​​​​​2 

= 2 × 2 × 2 × 3 × a × a × b × b

iv. tr​​​​​​​2 s3

= t × r​​​​​​​2 × s3

= t × r × r × s × s × s
Discussion
9.

Use the formula to find the values: i. 502 × 498 ii. 97 × 103iii. 54 × 46 iv. 98 × 102

Answer»

i. 502 × 498 = (500 + 2) (500 – 2) 

Here, a = 500, b = 2 

∴ (500 + 2) (500 – 2) = 5002 – 22 …. [(a + b)(a – b) = a2 – b2 ] 

= 250000 – 4 

= 249996 

∴ 502 × 498 = 249996 

ii. 97 × 103 = (100 – 3) (100 + 3) 

Here, a = 100, b = 3 

∴ (100 – 3) (100 + 3) = 1002 – 32 …. [(a + b)(a – b) = a2 – b2 ] 

= 10000 – 9 

= 9991

∴ 97 × 103 = 9991

iii. 54 × 46 = (50 + 4) (50 – 4) 

Here, a = 50, b = 4 

∴ (50 + 4) (50 – 4) = 502 – 42 …. [(a + b)(a – b) = a2 – b2 ] 

= 2500 – 16 

= 2484 

∴ 54 × 46 = 2484 

iv. 98 × 102 = (100 – 2) (100 + 2) 

Here, a = 100, b = 2 

∴ (100 – 2) (100 + 2) = 1002 – 22 …. [(a + b)(a – b) = a2 – b2 ] 

 10000 – 4 

= 9996 

∴ 98 × 102 = 9996
Discussion