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} = a^{2} + 2ab + b^{2}
Here
a =5a
b = 6b
(5a + 6b) ^{2} = (5a)^{ 2} + 2 x (5a) x (6b) + (6b) ^{2}
(5a + 6b) ^{2} = 25 a^{2} + 60ab + 36 b^{2}
Expansion of (5a + 6b) ^{2 }is 25 a^{2} + 60ab + 36 b^{2}
(ii) (a /2 + b/3) ^{2}
ANS:
We know formula,
(a + b) ^{2} = a^{2} + 2ab + b^{2}
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} = a^{2}/4 + ab/3+ b^{2}/9
Expansion of (a /2 + b/3) ^{2 }is a^{2}/4 + ab/3+ b^{2}/9
(iii) (2p – 3q)^{ 2}
ANS:
We know formula,
(a – b) ^{2} = a^{2} – 2ab + b^{2}
Here
a =2p
b = 3q
(2p – 3q)^{ 2} = (2p)^{ 2} – 2 x (2p) x (3q) + (3q)^{ 2}
(2p – 3q)^{ 2} = 4p^{2} – 12pq +9q^{ 2}
Expansion of (2p – 3q)^{ 2} is 4p^{2} – 12pq +9q^{ 2}
(iv) (x – 2 / x) ^{2}
ANS:
We know formula,
(a – b) ^{2} = a^{2} – 2ab + b^{2}
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} = x^{2} – 2x +4/x^{2}
Expansion of (x – 2/x)^{ 2} is x^{2} – 2x +4/x^{2}
(v) (ax + by)^{2}
ANS:
We know formula,
(a + b) ^{2} = a^{2} + 2ab + b^{2}
Here
a =ax
b = by
(ax + by) ^{2} = (ax)^{ 2} + 2 x (ax) x (by) + (by) ^{2}
(ax + by) ^{2} = a^{2} x^{2}+ 2axby + b^{2}y^{2}
Expansion of (ax + by) ^{2} is a^{2} x^{2}+ 2axby + b^{2}y^{2}
(vi) (7m – 4)^{2}
ANS:
We know formula,
(a – b) ^{2} = a^{2} – 2ab + b^{2}
Here
a =7m
b = 4
(7m – 4)^{ 2} = (7m)^{ 2} – 2 x (7m) x (4) + (4)^{ 2}
(7m – 4)^{ 2} = 49m^{2} – 56m + 16
Expansion of (7m – 4)^{ 2} is 49m^{2} – 56m + 16
(vii) (x + 1/ 2)^{ 2}
ANS:
We know formula,
(a + b) ^{2} = a^{2} + 2ab + b^{2}
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} = x^{2}+ 2x + 1/4
Expansion of (x + 1/2) ^{2} is x^{2}+ 2x + ¼.
(viii) (a – 1/a)^{ 2}
ANS:
We know formula,
(a – b) ^{2} = a^{2} – 2ab + b^{2}
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}= a^{2} – 2 + 1/a^{2}
Expansion of (a – 1/a)^{ 2} is a^{2} – 2 + 1/a^{2}
2.) Which of the options given below is the square of the binomial (8 – 1/ x)?
(i) 64 – 1/ x^{2} (ii) 64 + 1/ x^{2} (iii) 64- 16/ x + 1 / x^{2}(iv) 64 + 16 /x + 1/ x^{2}
ANS:
Given that binomial (8 – 1/ x).
We have to find square of (8 – 1/ x).
We know formula,
(a – b) ^{2} = a^{2} – 2ab + b^{2}
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/x^{2}
Expansion of (8 – 1/x)^{ 2} is 64 – 16/x + 1/x^{2}.
(iii) 64- 16/ x + 1 / x^{2}
3.) Of which of the binomials given below is m^{2} n^{2 }+ 14mnpq + 49p^{2} q^{2} the expansion?
(i) (m + n) (p + q) (ii) (mn – pq) (iii) (7mn + pq) (iv) (mn + 7pq)
ANS:
Here given that expansion of a binomial m^{2} n^{2 }+ 14mnpq + 49p^{2} q^{2}
We have to find for which binomial this expansion exist.
m^{2} n^{2 }+ 14mnpq + 49p^{2} q^{2}
In given expansion all sign are positive for that we know formula,
(a + b) ^{2} = a^{2} + 2ab + b^{2}
Here,
a^{2 }= m^{2} n^{2}
2ab = 14mnpq
b^{2 }= 49p^{2} q^{2}
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} = a^{2} – 2ab + b^{2}
997 we can write it as (1000 – 3)
a=1000
b =3
(1000 – 3) ^{2} = 1000^{2} – 2 x 1000 x 3 + 3^{2}
(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} = a^{2} + 2ab + b^{2}
102 we can write it as (100 + 2)
a=100
b = 2
(100 + 2) ^{2} = 100^{2} + 2x 100 x 2 + 2^{2}
(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} = a^{2} – 2ab + b^{2}
97 we can write it as (100 – 3)
a=100
b =3
(1000- 3) ^{2} = 100^{2} – 2 x 100 x 3 + 3^{2}
(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} = a^{2} + 2ab + b^{2}
1005 we can write it as (1000 + 5)
a=1000
b = 5
(1000 + 5) ^{2} = 1000^{2} + 2x 1000 x 5 + 5^{2}
(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) = a^{2} – b^{2}
Here,
a= x
b= y
(x + y) (x – y) = x^{2} – y^{2}
Expansion of (x + y) (x – y) is x^{2} – y^{2}
(ii) (3x – 5) (3x + 5)
ANS:
We know formula,
(a- b) (a + b) = a^{2} – b^{2}
Here,
a= 3x
b= 5
(3x – 5) (3x + 5)= (3x)^{2} – 5^{2}
(3x – 5) (3x + 5) = 9x^{2} – 25
Expansion of (3x – 5) (3x + 5) is 9x^{2} – 25
(iii) (a + 6) (a – 6)
ANS:
We know formula,
(a+ b) (a – b) = a^{2} – b^{2}
Here,
a= a
b= 6
(a + 6) (a – 6)= a^{2} – 6^{2}
(a + 6) (a – 6) = a^{2} – 36
Expansion of (a + 6) (a – 6) is a^{2} – 36
(iv) (x /5 + 6) (x /5 – 6)
ANS:
We know formula,
(a + b) (a – b) = a^{2} – b^{2}
Here,
a= x/5
b= 6
(x/5 + 6) (x/5 – 6) = (x/5)^{2} – 6^{2}
(X/5 + 6) (X/5 – 6) = x^{2}/25 – 36
Expansion of (X/5 + 6) (X/5 – 6) is x^{2}/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) = a^{2} – b^{2}
Here,
a= 500
b= 2
(500 + 2) (500 – 2) = 500^{2} – 2^{2}
(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) = a^{2} – b^{2}
Here,
a= 100
b= 3
(100 – 3) (100 + 3) = 100^{2} – 3^{2}
(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) = a^{2} – b^{2}
Here,
a= 50
b= 4
(50 + 4) (50 – 4) = 50^{2} – 4^{2}
(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) = a^{2} – b^{2}
Here,
a= 100
b= 2
(100 – 2) (100 + 2) = 100^{2} – 2^{2}
(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 a^{3} b^{2}
ANS:
Here we have to factorise 201 a^{3} b^{2} monomial.
In this first we factorise coefficient of monomial.i.e.201
201 = we factorise it as 3 x 67
Now,
We factorise
a^{3 }= a x a x a
Now we factorise,
b^{2} = b x b
We combine all factors,
201 a^{3} b^{2} = 3 x 67 x a x a x a x b x b
(ii) 91 xyt^{2}
ANS:
Here we have to factorise 91 xyt^{2} 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,
t^{2} = t x t
We combine all factors,
91 xyt^{2} = 7 x 13 x xX y x t x t
(iii) 24 a^{2 }b^{2}
ANS:
Here we have to factorise 24 a^{2 }b^{2} monomial.
In this first we factorise coefficient of monomial.i.e.24
24 = 2 x 2 x 2 x 3
Now,
We factorise
a^{2} = a x a
Now we factorise,
b^{2} = b x b
We combine all factors,
24 a^{2 }b^{2} = 2 x 2 x 2 x 3 x a x a x b x b
(iv) tr^{2} s^{3}
ANS:
Here we have to factorise tr^{2} s^{3} monomial.
We factorise
t = t
Now we factorise,
r^{2} = r x r
Now we factorise,
s^{3} = s x s x s
We combine all factors,
tr^{2} s^{3 }= t x r x r x s x s x s
Practice Set 53
¤ Factorise the following expressions.
(i) p^{2} – q^{2}
ANS:
We have to factorise p^{2} – q^{2}
We know formula,
a^{2} – b^{2} = (a+ b) (a – b)
Here,
a^{2} = p^{2}
b^{2}= q^{2}
p^{2} – q^{2} = (p+ q) (p – q)
Factorisation of p^{2} – q^{2} is (p+ q) (p – q)
(ii) 4x^{2} – 25y^{2}
ANS:
We have to factorise 4x^{2} – 25y^{2}
We convert it as (2x)^{2} – (5y)^{2}
We know formula,
a^{2} – b^{2} = (a+ b) (a – b)
Here,
a^{2} = 4x^{2}
b^{2}= 25y^{2}
4x^{2} – 25y^{2}= (2x+ 5y) (2x – 5y)
Factorisation of 4x^{2} – 25y^{2} is (2x+ 5y) (2x – 5y).
(iii) y^{2} – 4
ANS:
We have to factorise y^{2} – 4
We convert it as (y)^{2} – (2)^{2}
We know formula,
a^{2} – b^{2} = (a+ b) (a – b)
Here,
a = y
b= 2
y^{2} – 2^{2}= (y+ 2) (y – 2)
Factorisation of y^{2} – 4 is (y+ 2) (y – 2).
(iv) p^{2}– 1/ 25
ANS:
We have to factorise p^{2 }– 1/ 25
We convert it as (p)^{2} – (1/5)^{2}
We know formula,
a^{2} – b^{2} = (a+ b) (a – b)
Here,
a = p
b= 1/5
p^{2} – (1/5)^{2}= (p+ 1/5) (p – 1/5)
Factorisation of p^{2 }– 1/ 25 is (p+ 1/5) (p – 1/5)
(v) 9x^{2 }– 1 /16 y^{2}
ANS:
We have to factorise 9x^{2 }– 1 /16 y^{2}
We convert it as (3x)^{2} – (1/4 x y)^{2}
We know formula,
a^{2} – b^{2} = (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 9x^{2 }– 1 /16 y^{2} is (3x + 1/4 x y) (3x – 1/4 x y)
(vi) x^{2} – 1/ x^{2}
ANS:
We have to factorise x^{2 }– 1 /x^{2}
We convert it as (x)^{2} – (1/ x)^{2}
We know formula,
a^{2} – b^{2} = (a+ b) (a – b)
Here,
a = x
b= 1/x
(x)^{ 2} – (1/x)^{2} = (x + 1/x) (x – 1/x)
Factorisation of x^{2} – 1/ x^{2} is (x + 1/x) (x – 1/x)
(vii) a^{2} b – ab
ANS:
We have to factorise a^{2} b – ab
We taking ab common from a^{2} b – ab
We get,
a^{2} b – ab = ab x (a – 1)
Factorisation of a^{2} b – ab is ab x (a – 1)
(viii) 4x^{2 }y – 6x^{2}
ANS:
We have to factorise 4x^{2 }y – 6x^{2}
We taking 2x^{2} common from 4x^{2 }y – 6x^{2}
We get,
4x^{2 }y – 6x^{2}=2x^{2}(2y -3)
Factorisation of4x^{2 }y – 6x^{2}is 2x^{2 }(2y -3)
(ix) 1/ 2 y^{2 }– 8z^{2}
ANS:
We have to factorise 1/ 2 y^{2 }– 8z^{2}
We taking ½ common from 1/ 2 y^{2 }– 8z^{2}
We get,
1/ 2 y^{2 }– 8z^{2}= ½ (y^{2} -16 z^{2})
Factorisation of 1/ 2 y^{2 }– 8z^{2}is ½ (y^{2} -16 z^{2}).
(x) 2x^{2} – 8y^{2}
ANS:
We have to factorise 2x^{2} – 8y^{2}
We taking 2 common from 2x^{2} – 8y^{2}
We get,
2x^{2} – 8y^{2}= 2 (x^{2} – 4y^{2})
Factorisation of 2x^{2} – 8y^{2}is 2 (x^{2} – 4 y^{2}).