Maharashtra Board Class 7 Math Chapter 14 Algebraic Formulae – Expansion of Squares Solution

Maharashtra Board Class 7 Math Solution Chapter 14 – Algebraic Formulae – Expansion of Squares

Balbharati Maharashtra Board Class 7 Math Solution Chapter 14: Algebraic Formulae – Expansion of Squares. Marathi or English Medium Students of Class 7 get here Algebraic Formulae – Expansion of Squares full Exercise Solution.

Std

Maharashtra Class 7
Subject

Math Solution

Chapter

Algebraic Formulae – Expansion of Squares


Practice Set 50

1.) Expand.

 (i) (5a + 6b) 2

ANS:

We know formula,

(a + b) 2 = a2 + 2ab + b2

Here

a =5a

b = 6b

(5a + 6b) 2 = (5a) 2 + 2 x (5a) x (6b) + (6b) 2

(5a + 6b) 2 = 25 a2 + 60ab + 36 b2

Expansion of (5a + 6b) 2 is 25 a2 + 60ab + 36 b2

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

ANS:

We know formula,

(a + b) 2 = a2 + 2ab + b2

Here

a =a/2

b = b/3

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

(a /2 + b/3) 2 =  a2/4 + ab/3+  b2/9

Expansion of (a /2 + b/3) 2 is a2/4 + ab/3+ b2/9

(iii) (2p – 3q) 2

ANS:

We know formula,

(a – b) 2 = a2 – 2ab + b2

Here

a =2p

b = 3q

(2p – 3q) 2 = (2p) 2 – 2 x (2p) x (3q) + (3q) 2

(2p – 3q) 2 = 4p2 – 12pq +9q 2

Expansion of (2p – 3q) 2 is 4p2 – 12pq +9q 2

(iv) (x – 2 / x) 2

ANS:

We know formula,

(a – b) 2 = a2 – 2ab + b2

Here

a =x

b = 2/x

(x – 2/x) 2 = (x) 2 – 2 x (x) x (x/2) + (2/x) 2

(x – 2/x) 2 = x2 – 2x +4/x2

Expansion of (x – 2/x) 2 is x2 – 2x +4/x2

(v) (ax + by)2

ANS:

We know formula,

(a + b) 2 = a2 + 2ab + b2

Here

a =ax

b = by

(ax + by) 2 = (ax) 2 + 2 x (ax) x (by) + (by) 2

(ax + by) 2 = a2 x2+ 2axby +  b2y2

Expansion of (ax + by) 2 is a2 x2+ 2axby +  b2y2

(vi) (7m – 4)2

ANS:

We know formula,

(a – b) 2 = a2 – 2ab + b2

Here

a =7m

b = 4

(7m – 4) 2 = (7m) 2 – 2 x (7m) x (4) + (4) 2

(7m – 4) 2 = 49m2 – 56m + 16

Expansion of (7m – 4) 2 is 49m2 – 56m + 16

(vii) (x + 1/ 2) 2

ANS:

We know formula,

(a + b) 2 = a2 + 2ab + b2

Here

a =x

b = 1/2

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

(x + 1/2) 2 = x2+ 2x + 1/4

Expansion of (x + 1/2) 2 is x2+ 2x + ¼.

(viii) (a – 1/a) 2

ANS:

We know formula,

(a – b) 2 = a2 – 2ab + b2

Here

a =a

b = 1/a

(a – 1/a) 2 = (a) 2 – 2 x (a) x (1/a) + (1/a) 2

(a – 1/a) 2= a2 – 2 + 1/a2

Expansion of (a – 1/a) 2 is a2 – 2 + 1/a2

2.) Which of the options given below is the square of the binomial (8 – 1/ x)?

(i) 64 – 1/ x2 (ii) 64 + 1/ x2 (iii) 64- 16/ x + 1 / x2(iv) 64 + 16 /x + 1/ x2

ANS:

Given that binomial (8 – 1/ x).

We have to find square of (8 – 1/ x).

We know formula,

(a – b) 2 = a2 – 2ab + b2

Here

a =8

b = 1/x

(8 – 1/x) 2 = (8) 2 – 2 x (8) x (1/x) + (1/x) 2

(8 – 1/x) 2 = 64 – 16/x + 1/x2

Expansion of (8 – 1/x) 2 is 64 – 16/x + 1/x2.

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

3.) Of which of the binomials given below is m2 n2 + 14mnpq + 49p2 q2 the expansion?

(i) (m + n) (p + q) (ii) (mn – pq) (iii) (7mn + pq) (iv) (mn + 7pq)

ANS:

Here given that expansion of a binomial m2 n2 + 14mnpq + 49p2 q2

We have to find for which binomial this expansion exist.

m2 n2 + 14mnpq + 49p2 q2

In given expansion all sign are positive for that we know formula,

(a + b) 2 = a2 + 2ab + b2

Here,

a2 = m2 n2

2ab = 14mnpq

b2 = 49p2 q2

From above we get,

a=mn and b =7pq.

(a + b) = (mn+ 7pq)

(iv) (mn + 7pq)

4.) Use an expansion formula to find the values.

(i) (997)2

ANS:

To find (997)2 we know expansion formula,

(a – b) 2 = a2 – 2ab + b2

997 we can write it as (1000 – 3)

a=1000

b =3

(1000 – 3) 2 = 10002 – 2 x 1000 x 3 + 32

(1000 – 3) 2 =10, 00,000 – 6000 + 9

(1000 – 3) 2 = 9, 94,000 +9

(1000 – 3) 2 = 9, 94,009

(997)2 = 9, 94,009

 (ii) (102)2

ANS:

To find (102)2 we know expansion formula,

(a + b) 2 = a2 + 2ab + b2

102 we can write it as (100 + 2)

a=100

b = 2

(100 + 2) 2 = 1002 + 2x 100 x 2 + 22

(100 + 2) 2 = 10,000 + 400 + 4

(100 + 2) 2 = 10,404

(102)2 =10,404

(iii) (97)2

ANS:

To find (97)2 we know expansion formula,

(a – b) 2 = a2 – 2ab + b2

97 we can write it as (100 – 3)

a=100

b =3

(1000- 3) 2 = 1002 – 2 x 100 x 3 + 32

(100 – 3) 2 =10000 – 600 + 9

(1000 – 3) 2 = 9400 +9

(100 – 3) 2 = 9,409

(97)2 = 9,409

 

(iv) (1005)2

ANS:

To find (1005)2 we know expansion formula,

(a + b) 2 = a2 + 2ab + b2

1005 we can write it as (1000 + 5)

a=1000

b = 5

(1000 + 5) 2 = 10002 + 2x 1000 x 5 + 52

(1000 + 5) 2 = 10, 00,000 + 10,000 + 25

(1000 + 5) 2 = 10, 10,025

(1005)2 = 10, 10,025

 

 

Practice Set 51

1.) Use the formula to multiply the following.

(i) (x + y) (x – y)

ANS:

We know formula,

(a+ b) (a – b) = a2 – b2

Here,

a= x

b= y

(x + y) (x – y) = x2 – y2

Expansion of (x + y) (x – y) is x2 – y2

 

(ii) (3x – 5) (3x + 5)

ANS:

We know formula,

(a- b) (a + b) = a2 – b2

Here,

a= 3x

b= 5

(3x – 5) (3x + 5)= (3x)2 – 52

(3x – 5) (3x + 5) = 9x2 – 25

Expansion of (3x – 5) (3x + 5) is 9x2 – 25

 

(iii) (a + 6) (a – 6)

ANS:

We know formula,

(a+ b) (a – b) = a2 – b2

Here,

a= a

b= 6

(a + 6) (a – 6)= a2 – 62

(a + 6) (a – 6) = a2 – 36

Expansion of (a + 6) (a – 6) is a2 – 36

 

 (iv) (x /5 + 6) (x /5 – 6)

ANS:

We know formula,

(a + b) (a – b) = a2 – b2

Here,

a= x/5

b= 6

(x/5 + 6) (x/5 – 6) = (x/5)2 – 62

(X/5 + 6) (X/5 – 6) = x2/25 – 36

Expansion of (X/5 + 6) (X/5 – 6) is x2/25 – 36.

 

2.) Use the formula to find the values.

(i) 502 × 498

ANS:

We have to do multiplication of 502 × 498

We write this in the form of (a + b) (a – b)

502 we write it as (500 + 2)

498 we write it as (500 – 2)

We know the formula,

(a+ b) (a – b) = a2 – b2

Here,

a= 500

b= 2

(500 + 2) (500 – 2) = 5002 – 22

(500 + 2) (500 – 2) = 2, 50,000 – 4

(500 + 2) (500 – 2) = 2, 49,996

Multiplication of 502 × 498 is 2, 49,996.

 

(ii) 97 × 103

ANS:

We have to do multiplication of 97 × 103

We write this in the form of (a- b) (a + b)

97 we write it as (100 – 3)

103 we write it as (100 + 3)

We know the formula,

(a- b) (a + b) = a2 – b2

Here,

a= 100

b= 3

(100 – 3) (100 + 3) = 1002 – 32

(100 – 3) (100 + 3) = 10000 – 9

(100 – 3) (100 + 3) = 9,991

Multiplication of 97 × 103 is 9,991

 

(iii) 54 × 46

ANS:

We have to do multiplication of 502 × 498

We write this in the form of (a + b) (a – b)

54 we write it as (50 + 4)

46 we write it as (50 – 4)

We know the formula,

(a+ b) (a – b) = a2 – b2

Here,

a= 50

b= 4

(50 + 4) (50 – 4) = 502 – 42

(50 + 4) (50 – 4) = 2500 – 16

(50 + 4) (50 – 4) = 2,484

Multiplication of 54 × 46 is 2,484

 

 (iv) 98 × 102

ANS:

We have to do multiplication of 98 × 102

We write this in the form of (a- b) (a + b)

98 we write it as (100 – 2)

102 we write it as (100 + 2)

We know the formula,

(a- b) (a + b) = a2 – b2

Here,

a= 100

b= 2

(100 – 2) (100 + 2) = 1002 – 22

(100 – 2) (100 + 2) = 10000 – 4

(100 – 2) (100 + 2) = 9,996

Multiplication of 98 × 102 is 9,996.

 

Practice Set 52

 ¤ Factorise the following expressions and write them in the product form.

(i) 201 a3 b2

ANS:

Here we have to factorise 201 a3 b2 monomial.

In this first we factorise coefficient of monomial.i.e.201

201 = we factorise it as 3 x 67

Now,

We factorise

a3 = a x a x a

Now we factorise,

b2 = b x b

We combine all factors,

201 a3 b2 = 3 x 67 x a x a x a x b x b

 

 (ii) 91 xyt2

ANS:

Here we have to factorise 91 xyt2 monomial.

In this first we factorise coefficient of monomial.i.e.91

91 = 7 x 13

Now,

We factorise

xy = x x y

Now we factorise,

t2 = t x t

We combine all factors,

91 xyt2 = 7 x 13 x xX y x t x t

 

  (iii) 24 a2 b2

ANS:

Here we have to factorise 24 a2 b2 monomial.

In this first we factorise coefficient of monomial.i.e.24

24 = 2 x 2 x 2 x 3

Now,

We factorise

a2 = a x a

Now we factorise,

b2 = b x b

We combine all factors,

24 a2 b2 = 2 x 2 x 2 x 3 x a x a x b x b

 

 (iv) tr2 s3

ANS:

Here we have to factorise tr2 s3 monomial.

We factorise

t = t

Now we factorise,

r2 = r x r

Now we factorise,

s3 = s x s x s

We combine all factors,

tr2 s3 = t x r x r x s x s x s

 

Practice Set 53

¤ Factorise the following expressions.

(i) p2 – q2

ANS:

We have to factorise p2 – q2

We know formula,

a2 – b2 = (a+ b) (a – b)

Here,

a2 = p2

b2= q2

p2 – q2 = (p+ q) (p – q)

Factorisation of p2 – q2 is (p+ q) (p – q)

 

 (ii) 4x2 – 25y2

ANS:

We have to factorise 4x2 – 25y2

We convert it as (2x)2 – (5y)2

We know formula,

a2 – b2 = (a+ b) (a – b)

Here,

a2 = 4x2

b2= 25y2

4x2 – 25y2= (2x+ 5y) (2x – 5y)

Factorisation of 4x2 – 25y2 is (2x+ 5y) (2x – 5y).

 

 (iii) y2 – 4

ANS:

We have to factorise y2 – 4

We convert it as (y)2 – (2)2

We know formula,

a2 – b2 = (a+ b) (a – b)

Here,

a = y

b= 2

y2 – 22= (y+ 2) (y – 2)

Factorisation of y2 – 4 is (y+ 2) (y – 2).

 

 

(iv) p2– 1/ 25

ANS:

We have to factorise p2 – 1/ 25

We convert it as (p)2 – (1/5)2

We know formula,

a2 – b2 = (a+ b) (a – b)

Here,

a = p

b= 1/5

p2 – (1/5)2= (p+ 1/5) (p – 1/5)

Factorisation of p2 – 1/ 25 is (p+ 1/5) (p – 1/5)

 

(v) 9x2 – 1 /16 y2

ANS:

We have to factorise 9x2 – 1 /16 y2

We convert it as (3x)2 – (1/4 x y)2

We know formula,

a2 – b2 = (a+ b) (a – b)

Here,

a = 3x

b= ¼ x y

(3x) 2 – (1/4 x y)2 = (3x + 1/4 x y) (3x – 1/4 x y)

 

Factorisation of 9x2 – 1 /16 y2 is (3x + 1/4 x y) (3x – 1/4 x y)

 

(vi) x2 – 1/ x2

ANS:

We have to factorise x2 – 1 /x2

We convert it as (x)2 – (1/ x)2

We know formula,

a2 – b2 = (a+ b) (a – b)

Here,

a = x

b= 1/x

(x) 2 – (1/x)2 = (x + 1/x) (x – 1/x)

Factorisation of x2 – 1/ x2 is (x + 1/x) (x – 1/x)

 

(vii) a2 b – ab

ANS:

We have to factorise a2 b – ab

We taking ab common from a2 b – ab

We get,

a2 b – ab = ab x (a – 1)

Factorisation of a2 b – ab is ab x (a – 1)

(viii) 4x2 y – 6x2

ANS:

We have to factorise 4x2 y – 6x2

We taking 2x2 common from 4x2 y – 6x2

We get,

4x2 y – 6x2=2x2(2y -3)

Factorisation of4x2 y – 6x2is 2x2 (2y -3)

 

(ix) 1/ 2 y2 – 8z2

ANS:

We have to factorise 1/ 2 y2 – 8z2

We taking ½ common from 1/ 2 y2 – 8z2

We get,

1/ 2 y2 – 8z2= ½ (y2 -16 z2)

Factorisation of 1/ 2 y2 – 8z2is ½ (y2 -16 z2).

 

(x) 2x2 – 8y2

ANS:

We have to factorise 2x2 – 8y2

We taking 2 common from 2x2 – 8y2

We get,

2x2 – 8y2= 2 (x2 – 4y2)

Factorisation of 2x2 – 8y2is 2 (x2 – 4 y2).

Updated: August 6, 2021 — 12:32 pm

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