NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5

 

Introduction:

In this exercise/article we will learn about multiplication of algebraic expressions. [ using identity = ( a + b )2 = a2 + 2ab + b2 and ( a - b )2 = a2 - 2ab + b2 and ( x + a ) ( x + b ) = x2 + ( a + b ) x + ab and ( a - b ) ( a + b ) = a2 - b2 you learn identity before then do this exercise. In all question put on identity.

 

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities :

Class 8 Maths Exercise 9.5 (Page-151)

 

Q1. Use a suitable identity to get each of the following products.

(i) ( x + 3 ) ( x + 3 )

Solution :

We have ( x + 3 ) ( x + 3 ) = ( x + 3 )2

= ( x + 3 )2

[ using identity = ( a + b )2 = a2 + 2ab + b2 ]

Here a = x and b = 3

So that,

( x )2 + 2 × x × 3 + ( 3 )2

= x2 + 6x + 9

So, answer is x2 + 6x + 9

(ii) ( 2y + 5 ) ( 2y + 5 )

Solution :

We have ( 2y + 5 ) ( 2y + 5 ) = ( 2y + 5 )2

= ( 2y + 5 )2

[ using identity = ( a + b )2 = a2 + 2ab + b2 ]

Here, a = 2y and b = 5

So that,

( 2y )2 + 2 × 2y × 5 + ( 5 )2

= 4y2 + 20y + 25

So, answer is 4y2 + 20y + 25

(iii) ( 2a - 7 ) ( 2a - 7 )

Solution :

We have ( 2a - 7 ) ( 2a - 7 ) = ( 2a - 7 )2

= ( 2a - 7 )2

[ using identity = ( a - b )2 = a2 - 2ab + b2 ]

Here, a = 2a and b = 7

So that,

( 2a )2 - 2 × 2a × 7 + ( 7 )2

= 4a2 - 28a + 49

So, answer is 4a2 - 28a + 49

(iv) \(\displaystyle (\,3a\,-\,\frac{1}{2}\,)\,(\,3a\,-\,\frac{1}{2}\,)\)

Solution :     

We have \(\displaystyle (\,3a\,-\,\frac{1}{2}\,)\,(\,3a\,-\,\frac{1}{2}\,)\)\(\displaystyle ={{(\,3a\,-\,\frac{1}{2}\,)}^{2}}\)

[ using identity = ( a - b )2 = a2 - 2ab + b2 ]

Here, a = 3a and b = \(\displaystyle \frac{1}{2}\)

So that,

( 3a )2 - 2 × 3a × \(\displaystyle \frac{1}{2}\) + ( \(\displaystyle \frac{1}{2}\) )2

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5 Q.1-(iv)

= 9a2 - 3a + \(\displaystyle \frac{1}{4}\)

So, answer is 9a2 - 3a + \(\displaystyle \frac{1}{4}\)

(v) ( 1.1m - 0.4 ) ( 1.1m + 0.4 )

We have ( 1.1m - 0.4 ) ( 1.1m + 0.4 )

[ using identity = ( a - b ) ( a + b ) = a2 - b2 ]

Here, a = 1.1m and b = 0.4

( 1.1m )2 - ( 0.4 )2

= 1.21m2 - 0.16

So, answer is 1.21m2 - 0.16

(vi) ( a2 + b2 ) ( - a2 + b2 )

Solution :

We have ( a2 + b2 ) ( - a2 + b2 )

[ we have  multiply to the first bracket by the second bracket ]

a2 ( - a2 + b2 ) + b2 ( - a2 + b2 )

= - a4 + a2b2 - a2b2 + b4         [ because ( a2b2 - a2b2 ) = 0 ]

= b4 - a4

So, answer is b4 - a4

(vii) ( 6x - 7 ) ( 6x + 7 )

Solution :

We have ( 6x - 7 ) ( 6x + 7 )

[ using identity = ( a - b ) ( a + b ) = a2 - b2 ]

Here, a = 6x and b = 7

So that,

( 6x )2 - ( 7 )2

= 36x2 - 49

So, answer is 36x2 - 49

(viii) ( - a + c ) ( - a + c )

Solution :

We have  ( - a + c ) ( - a + c ) = ( - a + c )2

[ using identity = ( a + b )2 = a2 + 2ab + b2 ]

Here, a = - a and b = c

So that,

( - a )2 + 2 × - a × c + ( c )2

= a2 - 2ac + c2

So, answer is a2 - 2ac + c2

(ix) \(\displaystyle \left( {\frac{x}{2}\,+\,\frac{{3y}}{4}} \right)\) \(\displaystyle \left( {\frac{x}{2}\,+\,\frac{{3y}}{4}} \right)\)

Solution :

We have \(\displaystyle \left( {\frac{x}{2}\,+\,\frac{{3y}}{4}} \right)\) \(\displaystyle \left( {\frac{x}{2}\,+\,\frac{{3y}}{4}} \right)\) = \(\displaystyle {{\left( {\frac{x}{2}+\frac{{3y}}{4}} \right)}^{2}}\)

[ using identity = ( a + b )2 = a2 + 2ab + b2 ]

Here, a = \(\displaystyle {\frac{x}{2}}\) and b = \(\displaystyle {\frac{{3y}}{4}}\)

So that,

= \(\displaystyle {{\left( {\frac{x}{2}} \right)}^{2}}\,+\,2\,\times \,\frac{x}{2}\,\times \,\frac{{3y}}{4}\,+\,{{\left( {\frac{{3y}}{4}} \right)}^{2}}\)

= NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5 Q.1-(ix)

= \(\displaystyle \frac{{{{x}^{2}}}}{4}\,+\,\frac{{3xy}}{4}\,+\,\frac{{9{{y}^{2}}}}{{16}}\)

So, answer is \(\displaystyle \frac{{{{x}^{2}}}}{4}\,+\,\frac{{3xy}}{4}\,+\,\frac{{9{{y}^{2}}}}{{16}}\)

(x) ( 7a - 9b ) ( 7a - 9b )

Solution :

We have ( 7a - 9b ) ( 7a - 9b ) = ( 7a - 9b )2

[ using identity = ( a - b )2 = a2 - 2ab + b2 ]

Here, a = 7a and b = 9b

So that,

( 7a )2 - 2 × 7a × 9b + ( 9b )2

= 49a2 - 126ab + 81b2

So, answer is 49a2 - 126ab + 81b2

Q2. Use the identity ( x + a ) ( x + b ) = x2 + ( a + b ) x + ab to find the following products.

(i) ( x + 3 ) ( x + 7 )

Solution :

We have ( x + 3 ) ( x + 7 )

[ using identity = ( x + a ) ( x + b ) = x2 + ( a + b ) x + ab ]

Here, x = x , a = 3 and b = 7

So that,

( x )2 + ( 3 + 7 ) x + 3 × 7

= x2 + ( 10 ) x + 21

= x2 + 10 × x + 21

= x2 +10x +21

So, answer is x2 +10x +21

(ii) ( 4x + 5 ) ( 4x + 1 )

Solution :

We have ( 4x + 5 ) ( 4x + 1 )

[ using identity = [( x + a ) ( x + b ) = x2 + ( a + b ) x + ab ]

Here, x = 4x , a = 5 and b = 1

So that,

( 4x )2 + ( 5 + 1 ) 4x + 5 × 1

= 16x2 + ( 6 ) 4x + 5

= 16x2 + 6 × 4x + 5

= 16x2 + 24x + 5

So, answer is 16x2 + 24x + 5

(iii) ( 4x - 5 ) ( 4x - 1 )

We have ( 4x - 5 ) ( 4x - 1 )

[ using identity = [( x + a ) ( x + b ) = x2 + ( a + b ) x + ab ]

Here, x = 4x , a = - 5 and b = -1

So that,

[( 4x )2 + {( - 5 ) + ( -1 )} 4x + ( - 5 ) × ( - 1 )]

= 16x2 + (- 6) 4x + ( 5 )

= 16x2 - 6 × 4x + 5

= 16x2 - 24x + 5

So, answer is 16x2 - 24x + 5

(iv) ( 4x + 5 ) ( 4x - 1 )

Solution :

We have ( 4x + 5 ) ( 4x - 1 )

[ using identity = [( x + a ) ( x + b ) = x2 + ( a + b ) x + ab ]

Here, x = 4x , a = 5 and b = - 1

So that,

[( 4x )2 + { 5 + ( -1 ) } 4x + 5 × ( - 1 )]

= 16x2 ( 4 ) 4x - 5

= 16x2 + 4 × 4x - 5

= 16x2 + 16x - 5

So, answer is 16x2 + 16x - 5

(v) ( 2x + 5y ) ( 2x + 3y )

Solution :

We have ( 2x + 5y ) ( 2x + 3y )

[ using identity = [( x + a ) ( x + b ) = x2 + ( a + b ) x + ab ]

Here, x = 2x , a = 5y and b = 3y

So that,

( 2x )2 + ( 5y + 3y ) 2x + 5y × 3y

= 4x2 + ( 8y ) 2x + 15y2

= 4x2 + 8y × 2x + 15y2

= 4x2 + 16xy + 15y2

So, answer is 4x2 + 16xy + 15y2

(vi) ( 2a2 + 9 ) ( 2a2 + 5 )

We have ( 2a2 + 9 ) ( 2a2 + 5 )

[ using identity = [( x + a ) ( x + b ) = x2 + ( a + b ) x + ab ]

Here, x = 2a2 , a = 9 and b = 5

So that,

( 2a )2 + ( 9 + 5 ) 2a2 + 9 × 5

= 4a4 + ( 14 ) 2a2 + 45

= 4a4 + 14 × 2a2 + 45

= 4a4 + 28a2 + 45

So, answer is 4a4 + 28a2 + 45

(vii) ( xyz - 4 ) ( xyz - 2 )

Solution :

We have ( xyz - 4 ) ( xyz - 2 )

[ using identity = [( x + a ) ( x + b ) = x2 + ( a + b ) x + ab ]

Here, x = xyz , a = - 4 and b = - 2

So that,

[( xyz )2 + { ( - 4 ) + ( - 2 ) } xyz + ( - 4 × - 2 )]

= x2y2z2 + ( - 6 ) xyz + 8

= x2y2z2 - 6 × xyz + 8

= x2y2z2 - 6xyz + 8

So, answer is x2y2z2 - 6xyz + 8

Q3. Find the following squares by using the identities.

(i) ( b - 7 )2

Solution :

We have ( b - 7 )2

[ using identity = ( a - b )2 = a2 - 2ab + b2 ]

Here , a = b and b = 7

So that,

b2 - 2 × b × 7 + 72

= b2 - 14b + 49

So, answer is b2 - 14b + 49

(ii) ( xy + 3z )2

Solution :

We have ( xy + 3z )2

[ using identity = ( a + b )2 = a2 + 2ab + b2 ]

Here , a = xy and b = 3z

So that,

( xy )2 + 2 × xy × 3z + ( 3z )2

= xy2 + 6xyz + 9z2

So, answer is xy2 + 6xyz + 9z2

(iii) ( 6x2 - 5y )2

Solution :

We have ( 6x2 - 5y )2

[ using identity = ( a - b )2 = a2 - 2ab + b2 ]

Here , a = 6x2 and b = 5y

So that,

( 6x2 ) - 2 × 6x2 × 5y + ( 5y )2

= 36x2 - 60x2y + 5y2

So, answer is 36x2 - 60x2y + 5y2

(iv) \(\displaystyle {{\left( {\frac{2}{3}m\,+\,\frac{3}{2}n} \right)}^{2}}\)

Solution :

We have \(\displaystyle {{\left( {\frac{2}{3}m\,+\,\frac{3}{2}n} \right)}^{2}}\)

[ using identity = ( a + b )2 = a2 + 2ab + b2 ]

Here , a = \(\displaystyle {\frac{2}{3}m}\) and b = \(\displaystyle {\frac{3}{2}n}\)

So that,

= \(\displaystyle {{\left( {\frac{2}{3}m} \right)}^{2}}\,+\,2\,\times \,\frac{2}{3}m\,\times \,\frac{3}{2}n\,+\,{{\left( {\frac{3}{2}n} \right)}^{2}}\)

= NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5 Q.3-(iv)

= \(\displaystyle \frac{4}{9}{{m}^{2}}\,+\,2mn\,+\,\frac{9}{4}{{n}^{2}}\)

So, answer is \(\displaystyle \frac{4}{9}{{m}^{2}}\,+\,2mn\,+\,\frac{9}{4}{{n}^{2}}\)

(v) ( 0.4p - 0.5q )2

Solution :

We have ( 0.4p - 0.5q )2

[ using identity = ( a + b )2 = a2 + 2ab + b2 ]

Here , a = 0.4p and b = 0.5q

So that,

( 0.4p )2 + 2 × 0.4p × 0.5q + ( 0.5q )2

= 0.16p2 + 0.4pq + 0.25q2

So, answer is 0.16p2 + 0.4pq + 0.25q2

(vi) ( 2xy + 5y )2

Solution :

We have ( 2xy + 5y )2

[ using identity = ( a + b )2 = a2 + 2ab + b2 ]

Here , a = 2xy and b = 5y

So that,

( 2xy )2 + 2 × 2xy × 5y + ( 5y )2

= 4x2y2 + 20xy2 + 25y2

So, answer is 4x2y2 + 20xy2 + 25y2

Class 8 Maths Chapter 9 Exercise 9.5

 

Q4. Simplify.

(i) ( a2 - b2 )2

Solution :

We have ( a2 - b2 )2

[ using identity = ( a - b )2a2 - 2ab + b2 ]

Here , a = a2 and b = b2

So that,

( a2 )2 - 2 × a2 × b2 + ( b2 )2

= a4 - 2a2b2 + b4

So, answer is a4 - 2a2b2 + b4

(ii) ( 2x + 5 )2 - ( 2x - 5 )2

Solution :

We have ( 2x + 5 )2 - ( 2x - 5 )2

[ using identity = ( a + b )2a2 + 2ab + b2 and ( a - b )2a2 - 2ab + b2 ]

Here , a = 2x and b = 5 and a = 2x and b = 5

So that,

[ { ( 2x )2 + 2 × 2x × 5 + ( 5 )2 } - { ( 2x )2 - 2 × 2x × 5 + ( 5 )2 } ]

= ( 2x )2 + 2 × 2x × 5 + ( 5 )2 - ( 2x )2 + 2 × 2x × 5 - ( 5 )2

= 4x2 + 20x + 25 - 4x2 + 20x - 25                       [ because = 4x2 - 4x2 = 0 and 25 - 25 = 0 ]

= 40x

So, answer is 40x

(iii) ( 7m - 8n )2 + ( 7m + 8n )2

Solution :

We have ( 7m - 8n )2 + ( 7m + 8n )2

[ using identity = ( a - b )2a2 - 2ab + b2 and ( a + b )2a2 + 2ab + b2 ]

Here , a = 7m and b = 8n and a = 7m and b = 8n

So that,

[ { ( 7m )2 - 2 × 7m × 8n + ( 8n )2 } +{ ( 7m )2 + 2 × 7m × 8n + ( 8n )2 } ]

= ( 7m )2 - 2 × 7m × 8n + ( 8n )2 + ( 7m )2 + 2 × 7m × 8n + ( 8n )2

= 49m2 - 112mn + 64n2 + 49m2 + 112mn + 64n2       [ because = - 112mn + 112mn = 0 ]

= ( 49m2 + 49m2 ) ( 64n2 + 64n2 )

= 98m2 + 128n2

So, answer is 98m2 + 128n2

(vi) ( 4m + 5n )2 + ( 5m + 4n )2

Solution : 

We have  ( 4m + 5n )2 + ( 5m + 4n )2

[ using identity = ( a + b )2a2 + 2ab + b2 and ( a + b )2a2 + 2ab + b2 ]

Here , a = 4m and b = 5n and a = 5m and b = 4n

So that,

( 4m )2 + 2 × 4m × 5n + ( 5n )2 + ( 5m )2 + 2 × 5m × 4n + ( 4n )2

= 16m2 + 40mn + 25n2 + 25m2 + 40mn + 16n2

= ( 16m2 + + 25m2 ) ( 40mn + 40mn ) ( 25n2 + 16n2 )

= 41m2 + 80mn + 41n2

So, answer is 41m2 + 80mn + 41n2

(v) ( 2.5p - 1.5q )2 - ( 1.5p - 2.5q )2

Solution :

We have ( 2.5p - 1.5q )2 - ( 1.5p - 2.5q )2

[ using identity = ( a - b )2a2 - 2ab + b2 and ( a - b )2a2 - 2ab + b2 ]

Here , a = 2.5p and b = 1.5q and a = 1.5p and b = 2.5q

So that,

[{ ( 2.5p )2 - 2 × 2.5p × 1.5q + ( 1.5q )2 } - { ( 1.5p )2 - 2 × 1.5q × 2.5p + ( 2.5q )2 }]

= 6.25p2 - 7.5pq + 2.25q2 - 2.25p2 + 7.5pq - 6.25q2          [ because = - 7.5pq + 7.5pq = 0 ]

= ( 6.25p2 - 2.25p2 ) - ( 2.25q2 - 6.25q2 )

= 4p2 - 4q2

So, answer is 4p2 - 4q2

(vi) ( ab + bc )2 - 2ab2c

Solution :

We have ( ab + bc )2 - 2ab2c

[ using identity = ( a + b )2a2 + 2ab + b2 ]

Here , a = ab and b = bc

So that,

( ab )2 + 2 × ab × bc + ( bc )2 - 2ab2c

= a2b2 + 2ab2c + b2c2 - 2ab2c                           [ because = 2ab2c - 2ab2c = 0 ]

= a2b2 + b2c2

So, answer is a2b2 + b2c2

(vii) ( m2 - n2m )2 + 2m3n2

Solution :

We have ( m2 - n2m )2 + 2m3n2

[ using identity = ( a - b )2a2 - 2ab + b2 ]

Here , a = m2 and b = n2m

So that,

( m2 )2 - 2 × m2 × n2m + ( n2m )2 + 2m3n2

= m4 - 2m3n2 + n4m2 + 2m3n2                            [ because = - 2m3n2 + 2m3n2 = 0 ]

= m4 + n4m2

So, answer is m4 + n4m2

Q5. Show that.

(i) ( 3x + 7 )2 - 84x = ( 3x - 7 )2

Solution :

We have ( 3x + 7 )2 - 84x = ( 3x - 7 )2

[ using identity = ( a + b )2a2 + 2ab + b2 ]

LHS = ( 3x + 7 )2 - 84x = RHS = ( 3x - 7 )2

= ( 3x )2 + 2 × 3x × 7 + ( 7 ) - 84x

= 9x2 + 42x + 49 - 84x                                      [ because = 42x - 84x = - 42 ]

= 9x2 - 42x + 49

= ( 3x - 7 )2 = ( 3x - 7 )2

= LHS = RHS

Hence proved

(ii) ( 9p - 5q )2 + 180pq = ( 9p + 5q )2

Solution :

We have ( 9p - 5q )2 + 180pq = ( 9p + 5q )2

[ using identity = ( a - b )2a2 - 2ab + b2 ]

LHS = ( 9p - 5q )2 + 180pq = RHS = ( 9p + 5q )2

( 9p )2 - 2 × 9p × 5q + ( 5q )2 + 180pq

81p2 - 90pq + 25q2 + 180pq                             [ because = - 90pq + 180pq = 0 ]

= 81p2 + 90pq + 25q2

= ( 9p + 5q )2 = ( 9p + 5q )2

= LHS = RHS

Hence proved

(iii) \(\displaystyle {{\left( {\frac{4}{3}m\,-\,\frac{3}{4}n} \right)}^{2}}\,+\,2mn\,=\,\frac{{16}}{9}{{m}^{2}}\,+\,\frac{9}{{16}}{{n}^{2}}\)

Solution :

We have \(\displaystyle {{\left( {\frac{4}{3}m\,-\,\frac{3}{4}n} \right)}^{2}}\,+\,2mn\,=\,\frac{{16}}{9}{{m}^{2}}\,+\,\frac{9}{{16}}{{n}^{2}}\)

[ using identity = ( a - b )2a2 - 2ab + b2 ]

LHS = \(\displaystyle {{\left( {\frac{4}{3}m\,-\,\frac{3}{4}n} \right)}^{2}}\,+\,2mn\) = RHS = \(\displaystyle \frac{{16}}{9}{{m}^{2}}\,+\,\frac{9}{{16}}{{n}^{2}}\)

\(\displaystyle {{\left( {\frac{4}{3}m} \right)}^{2}}\,-\,2\,\times \,\frac{4}{3}m\,\times \,\frac{3}{4}n\,+\,{{\left( {\frac{3}{4}n\,} \right)}^{2}}\,+\,2mn\)

= NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5 Q.5-(iii)

= \(\displaystyle \frac{{16}}{9}{{m}^{2}}\,-\,2mn\,+\,\frac{9}{{16}}{{n}^{2}}\,+\,2mn\)                                [ because = \(\displaystyle -\,2mn\) + \(\displaystyle \,2mn\) = 0 ]

= \(\displaystyle \frac{{16}}{9}{{m}^{2}}\,+\,\frac{9}{{16}}{{n}^{2}}\,\) = \(\displaystyle \frac{{16}}{9}{{m}^{2}}\,+\,\frac{9}{{16}}{{n}^{2}}\,\)

= LHS = RHS

Hence proved

(iv) ( 4pq + 3q )2 - ( 4pq - 3q )2 = 48pq2

Solution :

We have ( 4pq + 3q )2 - ( 4pq - 3q )2 = 48pq2

[ using identity = ( a + b )2a2 + 2ab + b2 and ( a - b )2a2 - 2ab + b2 ]

LHS = ( 4pq + 3q )2 - ( 4pq - 3q )2 = RHS = 48pq2

[{ ( 4pq )2 + 2 × 4pq × 3q + ( 3q )2 } - { ( 4pq )2 - 2 × 4pq × 3q + ( 3q )2 }]

= 16p2q2 + 24pq2 + 9q2 - 16p2q2 + 24pq2 - 9q2                  [ because = 16p2q2 - 16p2q2 = o and 9q2 - 9q2 = 0 ]

= 48pq2 = 48pq2

= LHS = RHS

Hence proved

(v) ( a - b ) ( a + b ) + ( b - c ) ( b + c ) + ( c - a ) ( c + a ) = 0

Solution :

We have ( a - b ) ( a + b ) + ( b - c ) ( b + c ) + ( c - a ) ( c + a ) = 0

[ we have to the multiply the first bracket by the second bracket ]

a ( a + b ) - b ( a + b ) + b ( b + c ) - c ( b + c ) + c ( c + a ) - a ( c + a ) = 0

= a2 + ab - ab - b2 + b2 + bc - bc - c2 + c2 + ac - ac - a2 = 0

= ( a2 - a2 ) ( ab - ab ) ( - b2 + b2 ) ( bc - bc ) ( - c2 + c2  ) ( ac - ac ) = 0

= ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) ( 0 ) = 0

= 0 = 0

= LHS = RHS

Hence proved

Q6. Using identities evaluate.

(i) 712

Solution :

We have 712 = ( 70 + 1 )2

[ using identity = ( a + b )2a2 + 2ab + b2 ]

Here, a = 70 and b = 1

So that,

( 70 )2 + 2 × 70 × 1 + ( 1 )2

= 4900 + 140 + 1

= 5041

So, answer is 5041

(ii) 992

Solution :

We have 992 = ( 100 - 1 )2

[ using identity = ( a - b )2a2 - 2ab + b2 ]

Here, a = 100 and b = 1

So that,

( 100 )2 - 2 × 100 × 1 + ( 1 )2

= 10000 - 200 + 1

= 9800 + 1

= 9801

So, answer is 9801

(iii) 1022

Solution :

We have 1022 = ( 100 + 2 )2

[ using identity = ( a + b )2a2 + 2ab + b2 ]

Here, a = 100 and b = 2

So that,

( 100 )2 + 2 × 100 × 2 + ( 2 )2

= 10000 + 400 + 4

= 10404

So, answer is 10404

(iv) 9982

Solution :

We have 9982  = ( 1000 - 2 )2

[ using identity = ( a - b )2a2 - 2ab + b2 ]

Here, a = 1000 and b = 2

So that,

( 1000 )2 - 2 × 1000 × 2 + ( 2 )2

= 1000000 - 4000 + 4

= 996000 + 4

= 996004

So, answer is 996004

(v) 5.22

We have 5.22 = ( 5 + 0.2 )2

[ using identity = ( a - b )2a2 - 2ab + b2 ]

Here, a = 5 and b = 0.2

So that,

( 5 )2 + 2 × 5 × 0.2 + ( 0.2 )2

= 25 + 2 + 0.04

= 27.04

So, answer is 27.04

(vi) 297 × 303

Solution :

We have 297 × 303 = ( 300 - 3 ) ( 300 + 3 )

[ using identity = ( a - b ) ( a  + b ) = a2 - b2 ] 

Here, a = 300 and b = 3

So that,

( 300 )2 - ( 3 )2

= 90000 - 9

= 89991

So, answer is 89991

(vii) 78 × 82

Solution :

We have 78 × 82 = ( 80 - 2 ) ( 80 + 2 )

[ using identity = ( a - b ) ( a  + b ) = a2 - b2 ] 

Here, a = 80 and b = 2

So that,

( 80 )2 - ( 2 )2

= 6400 - 4

= 6396

So, answer is 6396

(viii) 8.92

Solution :

We have 8.92 = ( 9 - 0.1 )2

[ using identity = ( a - b )2a2 - 2ab + b2 ]

Here, a = 9 and b = 0.1

So that,

( 9 )2 - 2 × 9 × 0.1 + ( 0.1 )

= 81 - 1.8 + 0.01

= 79.2 + 0.01

= 79.21

So, answer is 79.21

(ix) 10.5 × 9.5

Solution :

We have 10.5 × 9.5 = ( 10 + 0.5 ) ( 10 - 0.5 )

[ using identity = ( a - b ) ( a  + b ) = a2 - b2 ] 

Here, a = 10 and b = 0.5

So that,

( 10 )2 - ( 0.5 )2

= 100 - 0.25

= 99.75

So, answer is 99. 75

Q7. Using a2 - b2 = ( a + b ) ( a - b ), find

(i) 5l2 - 492

Solution :

We have 512 - 492

[ using identity = a2 - b2 = ( a + b ) ( a - b ) ] 

Here, a = 51 and b = 49

So that,

( 51 + 49  ) ( 51 - 49 )

= ( 100 ) ( 2 )

= 100 × 2

= 200

So, answer is 200

(ii) ( 1.02 )2 - ( 0.98 )2

Solution :

We have ( 1.02 )2 - ( 0.98 )2

[ using identity = a2 - b2 = ( a + b ) ( a - b ) ] 

Here, a = 1.02 and b = 0.98

So that,

( 1.02 + 0.98 ) ( 1.02 - 0.98 )

= ( 2 ) ( 0.04 )

= 2 × 0.04

= 0.08

So, answer is 0.08

(iii) 1532 - 1472

Solution :

We have 1532 - 1472

[ using identity = a2 - b2 = ( a + b ) ( a - b ) ] 

Here, a = 153 and b = 147

So that,

( 153 + 147 ) ( 153 - 147 )

= ( 300 ) ( 6 )

= 300 × 6

= 1800

So, answer is 1800

(iv) 12.12 - 7.92

Solution :

We have 12.12 - 7.92

[ using identity = a2 - b2 = ( a + b ) ( a - b ) ] 

Here, a = 12.1 and b = 7.9

So that,

( 12.1 + 7.9 ) ( 12.1 - 7.9 )

= ( 20 ) ( 4.2 )

= 20 × 4.2

= 84

So, answer is 84

Q8. Using ( x + a ) ( x + b ) = x2 + ( a + b ) x + ab, find

(i) 103 × 104

Solution :

We have 103 × 104

( 100 + 3 ) ( 100 + 4 )

[ using identity = ( x + a ) ( x + b ) = x2 + ( a + b ) x + ab ]

Here, x = 100 , a = 3 and b = 4

So that,

( 100 )2 + ( 3 + 4 ) 100 + 3 × 4

= 10000 + ( 7 ) 100 + 12

= 10000 + 7 × 100 + 12

= 10000 + 700 + 12

= 10712

So, answer is 10712

(ii) 5.1 × 5.2

Solution :

We have 5.1 × 5.2

( 5 + 0.1 ) ( 5 + 0.2 )

[ using identity = ( x + a ) ( x + b ) = x2 + ( a + b ) x + ab ]

Here, x = 5 , a = 0.1 and b = 0.2

So that,

( 5 )2 + ( 0.1 + 0.2 ) 5 + 0.1 × 0.2

= 25 + ( 0.3 ) 5 + 0.02

= 25 + 1.5 + 0.02

= 26.57

So, answer is 26.57

(iii) 103 × 98

Solution :

We have 103 × 98

( 100 + 3 ) ( 100 - 2 )

[ using identity = ( x + a ) ( x + b ) = x2 + ( a + b ) x + ab ]

Here, x = 100 , a = 3 and b = 2

So that,

( 100 )2 + ( 3 - 2 )100 + 3 × ( - 2 )

= 10000 + 1 × 100 - 6

= 10000 + 100 - 6

= 10100 - 6

= 10094

So, answer is 10094

(iv) 9.7 × 9.8

Solution :

We have 9.7 × 9.8

( 10 - 0.3 ) ( 10 - 0.2 )

[ using identity = ( x + a ) ( x + b ) = x2 + ( a + b ) x + ab ]

Here, x = 10 , a = 0.3 and b = 0.2

So that,

( 10 )2 + ( - 0.3 - 0.2 )10 + ( - 0.3 ) × ( - 0.2 )

= 100 - 0.5 × 10 + 0.06

= 100 - 5 + 0.06

= 100.06 - 5

= 95.06

So, answer is 95.06 .

 

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities :

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