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).