Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions
RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions | Mathematics (Maths) Class 8 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
( ) ( ) ( [ ( ]
Page 2
( ) ( ) ( [ ( ]
4x +5y 4x +5y = 4x +5y)
2
= (4x)
2
+(5y)
2
+2 ×4x ×5y using a +b)
2
= a
2
+b
2
+2ab = 16x
2
+25y
2
+40xy
iii
We have:
7a +9b 7a +9b = 7a +9b)
2
= (7a)
2
+(9b)
2
+2 ×7a ×9b using a +b)
2
= a
2
+b
2
+2ab = 49a
2
+81b
2
+126ab
iv
We have:
2
3
x +
4
5
y
2
3
x +
4
5
y =
2
3
x +
4
5
y
2
=
2
3
x
2
+
4
5
y
2
+2 ×
2
3
x ×
4
5
y using a +b)
2
= a
2
+b
2
+2ab =
4
9
x
2
+
16
25
y
2
+
16
15
xy
v
We have:
x
2
+7 x
2
+7 = x
2
+7)
2
= x
2
2
+7
2
+2 ×x
2
×7 using a +b)
2
= a
2
+b
2
+2ab = x
4
+49 +14x
2
vi
We have:
5
6
a
2
+2
5
6
a
2
+2 =
5
6
a
2
+2
2
=
5
6
a
2
2
+(2)
2
+2 ×
5
6
a
2
×2 using a +b)
2
= a
2
+b
2
+2ab =
25
36
a
4
+4 +
10
3
a
2
Q u e s t i o n : 6 3
Find each of the following products:
i
(x - 4)(x - 4)
ii
(2x - 3y)(2x - 3y)
iii
3
4
x -
5
6
y
3
4
x -
5
6
y
iv
x -
3
x
x -
3
x
v
1
3
x
2
-9
1
3
x
2
-9
vi
1
2
y
2
-
1
3
y
1
2
y
2
-
1
3
y
S o l u t i o n :
i
We have:
x -4 x -4 = x -4)
2
= x
2
-2 ×x ×4 +4
2
using a -b)
2
= a
2
-2ab +b
2
= x
2
-8x +16
ii
We have:
2x -3y 2x -3y = 2x -3y)
2
= (2x)
2
-2 ×2x ×3y +(3y)
2
using a -b)
2
= a
2
-2ab +b
2
= 4x
2
-12xy +9y
2
iii
We have:
3
4
x -
5
6
y
3
4
x -
5
6
y =
3
4
x -
5
6
y
2
=
3
4
x
2
-2 ×
3
4
x ×
5
6
y +
5
6
y
2
using a -b)
2
= a
2
-2ab +b
2
=
9
16
x
2
-
15
12
xy +
25
36
y
2
iv
We have:
x -
3
x
x -
3
x
= x -
3
x
2
= (x)
2
-2 ×x ×
3
x
+
3
x
2
using a -b)
2
= a
2
-2ab +b
2
= x
2
-6 +
9
x
2
v
We have:
1
3
x
2
-9
1
3
x
2
-9 =
1
3
x
2
-9
2
=
1
3
x
2
2
-2 ×
1
3
x
2
×9 +(9)
2
using a -b)
2
= a
2
-2ab +b
2
=
1
9
x
4
-6x
2
+81
vi
We have:
1
2
y
2
-
1
3
y
1
2
y
2
-
1
3
y =
1
2
y
2
-
1
3
y
2
=
1
2
y
2
2
-2 ×
1
2
y
2
×
1
3
y +
1
3
y
2
using a -b)
2
= a
2
-2ab +b
2
=
1
4
y
4
-
1
3
y
3
+
1
9
y
2
( ) ( ) ( [ ( ]
( ) ( ) ( [ ( ]
( ) ( ) ( ) ( ) ( ) [ ( ]
( )( ) (
( )
[ ( ]
( ) ( ) ( ) ( ) [ ( ]
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( [ ( ]
( ) ( ) ( [ ( ]
( ) ( ) ( ) ( ) ( ) [ ( ]
( ) ( ) ( ) ( ) [ ( ]
( ) ( ) ( ) ( ) [ ( ]
( ) ( ) ( ) ( ) ( ) [ ( ]
Page 3
( ) ( ) ( [ ( ]
4x +5y 4x +5y = 4x +5y)
2
= (4x)
2
+(5y)
2
+2 ×4x ×5y using a +b)
2
= a
2
+b
2
+2ab = 16x
2
+25y
2
+40xy
iii
We have:
7a +9b 7a +9b = 7a +9b)
2
= (7a)
2
+(9b)
2
+2 ×7a ×9b using a +b)
2
= a
2
+b
2
+2ab = 49a
2
+81b
2
+126ab
iv
We have:
2
3
x +
4
5
y
2
3
x +
4
5
y =
2
3
x +
4
5
y
2
=
2
3
x
2
+
4
5
y
2
+2 ×
2
3
x ×
4
5
y using a +b)
2
= a
2
+b
2
+2ab =
4
9
x
2
+
16
25
y
2
+
16
15
xy
v
We have:
x
2
+7 x
2
+7 = x
2
+7)
2
= x
2
2
+7
2
+2 ×x
2
×7 using a +b)
2
= a
2
+b
2
+2ab = x
4
+49 +14x
2
vi
We have:
5
6
a
2
+2
5
6
a
2
+2 =
5
6
a
2
+2
2
=
5
6
a
2
2
+(2)
2
+2 ×
5
6
a
2
×2 using a +b)
2
= a
2
+b
2
+2ab =
25
36
a
4
+4 +
10
3
a
2
Q u e s t i o n : 6 3
Find each of the following products:
i
(x - 4)(x - 4)
ii
(2x - 3y)(2x - 3y)
iii
3
4
x -
5
6
y
3
4
x -
5
6
y
iv
x -
3
x
x -
3
x
v
1
3
x
2
-9
1
3
x
2
-9
vi
1
2
y
2
-
1
3
y
1
2
y
2
-
1
3
y
S o l u t i o n :
i
We have:
x -4 x -4 = x -4)
2
= x
2
-2 ×x ×4 +4
2
using a -b)
2
= a
2
-2ab +b
2
= x
2
-8x +16
ii
We have:
2x -3y 2x -3y = 2x -3y)
2
= (2x)
2
-2 ×2x ×3y +(3y)
2
using a -b)
2
= a
2
-2ab +b
2
= 4x
2
-12xy +9y
2
iii
We have:
3
4
x -
5
6
y
3
4
x -
5
6
y =
3
4
x -
5
6
y
2
=
3
4
x
2
-2 ×
3
4
x ×
5
6
y +
5
6
y
2
using a -b)
2
= a
2
-2ab +b
2
=
9
16
x
2
-
15
12
xy +
25
36
y
2
iv
We have:
x -
3
x
x -
3
x
= x -
3
x
2
= (x)
2
-2 ×x ×
3
x
+
3
x
2
using a -b)
2
= a
2
-2ab +b
2
= x
2
-6 +
9
x
2
v
We have:
1
3
x
2
-9
1
3
x
2
-9 =
1
3
x
2
-9
2
=
1
3
x
2
2
-2 ×
1
3
x
2
×9 +(9)
2
using a -b)
2
= a
2
-2ab +b
2
=
1
9
x
4
-6x
2
+81
vi
We have:
1
2
y
2
-
1
3
y
1
2
y
2
-
1
3
y =
1
2
y
2
-
1
3
y
2
=
1
2
y
2
2
-2 ×
1
2
y
2
×
1
3
y +
1
3
y
2
using a -b)
2
= a
2
-2ab +b
2
=
1
4
y
4
-
1
3
y
3
+
1
9
y
2
( ) ( ) ( [ ( ]
( ) ( ) ( [ ( ]
( ) ( ) ( ) ( ) ( ) [ ( ]
( )( ) (
( )
[ ( ]
( ) ( ) ( ) ( ) [ ( ]
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( [ ( ]
( ) ( ) ( [ ( ]
( ) ( ) ( ) ( ) ( ) [ ( ]
( ) ( ) ( ) ( ) [ ( ]
( ) ( ) ( ) ( ) [ ( ]
( ) ( ) ( ) ( ) ( ) [ ( ]
Q u e s t i o n : 6 4
Expand:
i
(8a + 3b)
2
ii
(7x + 2y)
2
iii
(5x + 11)
2
iv
a
2
+
2
a
2
v
3x
4
+
2y
9
2
vi
(9x - 10)
2
vii
(x
2
y - yz
2
)
2
viii
x
y
-
y
x
2
ix
3m -
4
5
n
2
S o l u t i o n :
We shall use the identities (a+b)
2
=a
2
+b
2
+2ab and (a-b)
2
=a
2
+b
2
-2ab.
i
We have:
8a +3b)
2
= (8a)
2
+2 ×8a ×3b +(3b)
2
= 64a
2
+48ab +9b
2
ii
We have:
7x +2y)
2
= (7x)
2
+2 ×7x ×2y +(2y)
2
= 49x
2
+28xy +4y
2
iii
We have :
5x +11)
2
= (5x)
2
+2 ×5x ×11 +(11)
2
= 25x
2
+110x +121
iv
We have:
a
2
+
2
a
2
=
a
2
2
+2 ×
a
2
×
2
a
+
2
a
2
=
a
4
2
+2 +
4
a
2
v
We have:
3x
4
+
2y
9
2
=
3x
4
2
+2 ×
3x
4
×
2y
9
+
2y
9
2
=
9x
16
2
+
1
3
xy +
4y
2
81
vi
We have:
9x -10)
2
(9x)
2
-2 ×9x ×10 +(10)
2
= 81x
2
-180x +100
vii
We have:
x
2
y -yz
2
)
2
x
2
y
2
-2 ×x
2
y ×yz
2
+ yz
2
2
= x
4
y
2
-2x
2
y
2
z
2
+y
2
z
4
viii
We have:
x
y
-
y
x
2
=
x
y
2
-2 ×
x
y
×
y
x
+
y
x
2
=
x
2
y
2
-2 +
y
2
x
2
ix
We have:
3m -
4
5
n
2
= (3m)
2
-2 ×3m ×
4
5
n +
4
5
n
2
= 9m
2
-
24mn
5
+
16
25
n
2
Q u e s t i o n : 6 5
Find each of the following products:
i
(x + 3)(x - 3)
( )
( )
( )
( )
(
(
(
( ) ( ) ( )
( ) ( ) ( )
(
(
( ) ( )
( ) ( ) ( )
( ) ( )
Page 4
( ) ( ) ( [ ( ]
4x +5y 4x +5y = 4x +5y)
2
= (4x)
2
+(5y)
2
+2 ×4x ×5y using a +b)
2
= a
2
+b
2
+2ab = 16x
2
+25y
2
+40xy
iii
We have:
7a +9b 7a +9b = 7a +9b)
2
= (7a)
2
+(9b)
2
+2 ×7a ×9b using a +b)
2
= a
2
+b
2
+2ab = 49a
2
+81b
2
+126ab
iv
We have:
2
3
x +
4
5
y
2
3
x +
4
5
y =
2
3
x +
4
5
y
2
=
2
3
x
2
+
4
5
y
2
+2 ×
2
3
x ×
4
5
y using a +b)
2
= a
2
+b
2
+2ab =
4
9
x
2
+
16
25
y
2
+
16
15
xy
v
We have:
x
2
+7 x
2
+7 = x
2
+7)
2
= x
2
2
+7
2
+2 ×x
2
×7 using a +b)
2
= a
2
+b
2
+2ab = x
4
+49 +14x
2
vi
We have:
5
6
a
2
+2
5
6
a
2
+2 =
5
6
a
2
+2
2
=
5
6
a
2
2
+(2)
2
+2 ×
5
6
a
2
×2 using a +b)
2
= a
2
+b
2
+2ab =
25
36
a
4
+4 +
10
3
a
2
Q u e s t i o n : 6 3
Find each of the following products:
i
(x - 4)(x - 4)
ii
(2x - 3y)(2x - 3y)
iii
3
4
x -
5
6
y
3
4
x -
5
6
y
iv
x -
3
x
x -
3
x
v
1
3
x
2
-9
1
3
x
2
-9
vi
1
2
y
2
-
1
3
y
1
2
y
2
-
1
3
y
S o l u t i o n :
i
We have:
x -4 x -4 = x -4)
2
= x
2
-2 ×x ×4 +4
2
using a -b)
2
= a
2
-2ab +b
2
= x
2
-8x +16
ii
We have:
2x -3y 2x -3y = 2x -3y)
2
= (2x)
2
-2 ×2x ×3y +(3y)
2
using a -b)
2
= a
2
-2ab +b
2
= 4x
2
-12xy +9y
2
iii
We have:
3
4
x -
5
6
y
3
4
x -
5
6
y =
3
4
x -
5
6
y
2
=
3
4
x
2
-2 ×
3
4
x ×
5
6
y +
5
6
y
2
using a -b)
2
= a
2
-2ab +b
2
=
9
16
x
2
-
15
12
xy +
25
36
y
2
iv
We have:
x -
3
x
x -
3
x
= x -
3
x
2
= (x)
2
-2 ×x ×
3
x
+
3
x
2
using a -b)
2
= a
2
-2ab +b
2
= x
2
-6 +
9
x
2
v
We have:
1
3
x
2
-9
1
3
x
2
-9 =
1
3
x
2
-9
2
=
1
3
x
2
2
-2 ×
1
3
x
2
×9 +(9)
2
using a -b)
2
= a
2
-2ab +b
2
=
1
9
x
4
-6x
2
+81
vi
We have:
1
2
y
2
-
1
3
y
1
2
y
2
-
1
3
y =
1
2
y
2
-
1
3
y
2
=
1
2
y
2
2
-2 ×
1
2
y
2
×
1
3
y +
1
3
y
2
using a -b)
2
= a
2
-2ab +b
2
=
1
4
y
4
-
1
3
y
3
+
1
9
y
2
( ) ( ) ( [ ( ]
( ) ( ) ( [ ( ]
( ) ( ) ( ) ( ) ( ) [ ( ]
( )( ) (
( )
[ ( ]
( ) ( ) ( ) ( ) [ ( ]
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( [ ( ]
( ) ( ) ( [ ( ]
( ) ( ) ( ) ( ) ( ) [ ( ]
( ) ( ) ( ) ( ) [ ( ]
( ) ( ) ( ) ( ) [ ( ]
( ) ( ) ( ) ( ) ( ) [ ( ]
Q u e s t i o n : 6 4
Expand:
i
(8a + 3b)
2
ii
(7x + 2y)
2
iii
(5x + 11)
2
iv
a
2
+
2
a
2
v
3x
4
+
2y
9
2
vi
(9x - 10)
2
vii
(x
2
y - yz
2
)
2
viii
x
y
-
y
x
2
ix
3m -
4
5
n
2
S o l u t i o n :
We shall use the identities (a+b)
2
=a
2
+b
2
+2ab and (a-b)
2
=a
2
+b
2
-2ab.
i
We have:
8a +3b)
2
= (8a)
2
+2 ×8a ×3b +(3b)
2
= 64a
2
+48ab +9b
2
ii
We have:
7x +2y)
2
= (7x)
2
+2 ×7x ×2y +(2y)
2
= 49x
2
+28xy +4y
2
iii
We have :
5x +11)
2
= (5x)
2
+2 ×5x ×11 +(11)
2
= 25x
2
+110x +121
iv
We have:
a
2
+
2
a
2
=
a
2
2
+2 ×
a
2
×
2
a
+
2
a
2
=
a
4
2
+2 +
4
a
2
v
We have:
3x
4
+
2y
9
2
=
3x
4
2
+2 ×
3x
4
×
2y
9
+
2y
9
2
=
9x
16
2
+
1
3
xy +
4y
2
81
vi
We have:
9x -10)
2
(9x)
2
-2 ×9x ×10 +(10)
2
= 81x
2
-180x +100
vii
We have:
x
2
y -yz
2
)
2
x
2
y
2
-2 ×x
2
y ×yz
2
+ yz
2
2
= x
4
y
2
-2x
2
y
2
z
2
+y
2
z
4
viii
We have:
x
y
-
y
x
2
=
x
y
2
-2 ×
x
y
×
y
x
+
y
x
2
=
x
2
y
2
-2 +
y
2
x
2
ix
We have:
3m -
4
5
n
2
= (3m)
2
-2 ×3m ×
4
5
n +
4
5
n
2
= 9m
2
-
24mn
5
+
16
25
n
2
Q u e s t i o n : 6 5
Find each of the following products:
i
(x + 3)(x - 3)
( )
( )
( )
( )
(
(
(
( ) ( ) ( )
( ) ( ) ( )
(
(
( ) ( )
( ) ( ) ( )
( ) ( )
ii
(2x + 5)(2x - 5)
iii
(8 + x)(8 - x)
iv
(7x + 11y)(7x - 11y)
v
5x
2
+
3
4
y
2
5x
2
-
3
4
y
2
vi
4x
5
-
5y
3
4x
5
+
5y
3
vii
x +
1
x
x -
1
x
viii
1
x
+
1
y
1
x
-
1
y
S o l u t i o n :
i
We have:
x +3 x -3 = x
2
-9 using a +b a -b = a
2
-b
2
ii
We have:
2x +5 2x -5 = 4x
2
-25 using a +b a -b = a
2
-b
2
iii
We have:
8 +x 8 -x = 64 -x
2
using a +b a -b = a
2
-b
2
iv
We have:
7x +11y 7x -11y = 49x
2
-121y
2
using a +b a -b = a
2
-b
2
v
We have:
5x
2
+
3
4
y
2
5x
2
-
3
4
y
2
= 25x
4
-
9
16
y
4
using a +b a -b = a
2
-b
2
vi
We have:
4x
5
-
5y
3
4x
5
+
5y
3
=
16x
2
25
-
25y
2
9
using a +b a -b = a
2
-b
2
vii
We have:
x +
1
x
x -
1
x
= x
2
-
1
x
2
using a +b a -b = a
2
-b
2
viii
We have:
1
x
+
1
y
1
x
-
1
y
=
1
x
2
-
1
y
2
using a +b a -b = a
2
-b
2
ix
We have:
2a +
3
b
2a -
3
b
= 4a
2
-
9
b
2
using a +b a -b = a
2
-b
2
Q u e s t i o n : 6 6
Using the formula for squaring a binomial, evaluate the following:
i
54
2
ii
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( ) [ ( ) ( ) ]
( ) ( ) [ ( ) ( ) ]
( ) ( ) [ ( ) ( ) ]
( ) ( ) [ ( ) ( ) ]
( ) ( ) [ ( ) ( ) ]
( ) ( ) [ ( )( ) )]
( ) ( ) [ ( )( ) ]
( ) ( ) [ ( )( ) ]
( ) ( ) [ ( )( ) ]
Page 5
( ) ( ) ( [ ( ]
4x +5y 4x +5y = 4x +5y)
2
= (4x)
2
+(5y)
2
+2 ×4x ×5y using a +b)
2
= a
2
+b
2
+2ab = 16x
2
+25y
2
+40xy
iii
We have:
7a +9b 7a +9b = 7a +9b)
2
= (7a)
2
+(9b)
2
+2 ×7a ×9b using a +b)
2
= a
2
+b
2
+2ab = 49a
2
+81b
2
+126ab
iv
We have:
2
3
x +
4
5
y
2
3
x +
4
5
y =
2
3
x +
4
5
y
2
=
2
3
x
2
+
4
5
y
2
+2 ×
2
3
x ×
4
5
y using a +b)
2
= a
2
+b
2
+2ab =
4
9
x
2
+
16
25
y
2
+
16
15
xy
v
We have:
x
2
+7 x
2
+7 = x
2
+7)
2
= x
2
2
+7
2
+2 ×x
2
×7 using a +b)
2
= a
2
+b
2
+2ab = x
4
+49 +14x
2
vi
We have:
5
6
a
2
+2
5
6
a
2
+2 =
5
6
a
2
+2
2
=
5
6
a
2
2
+(2)
2
+2 ×
5
6
a
2
×2 using a +b)
2
= a
2
+b
2
+2ab =
25
36
a
4
+4 +
10
3
a
2
Q u e s t i o n : 6 3
Find each of the following products:
i
(x - 4)(x - 4)
ii
(2x - 3y)(2x - 3y)
iii
3
4
x -
5
6
y
3
4
x -
5
6
y
iv
x -
3
x
x -
3
x
v
1
3
x
2
-9
1
3
x
2
-9
vi
1
2
y
2
-
1
3
y
1
2
y
2
-
1
3
y
S o l u t i o n :
i
We have:
x -4 x -4 = x -4)
2
= x
2
-2 ×x ×4 +4
2
using a -b)
2
= a
2
-2ab +b
2
= x
2
-8x +16
ii
We have:
2x -3y 2x -3y = 2x -3y)
2
= (2x)
2
-2 ×2x ×3y +(3y)
2
using a -b)
2
= a
2
-2ab +b
2
= 4x
2
-12xy +9y
2
iii
We have:
3
4
x -
5
6
y
3
4
x -
5
6
y =
3
4
x -
5
6
y
2
=
3
4
x
2
-2 ×
3
4
x ×
5
6
y +
5
6
y
2
using a -b)
2
= a
2
-2ab +b
2
=
9
16
x
2
-
15
12
xy +
25
36
y
2
iv
We have:
x -
3
x
x -
3
x
= x -
3
x
2
= (x)
2
-2 ×x ×
3
x
+
3
x
2
using a -b)
2
= a
2
-2ab +b
2
= x
2
-6 +
9
x
2
v
We have:
1
3
x
2
-9
1
3
x
2
-9 =
1
3
x
2
-9
2
=
1
3
x
2
2
-2 ×
1
3
x
2
×9 +(9)
2
using a -b)
2
= a
2
-2ab +b
2
=
1
9
x
4
-6x
2
+81
vi
We have:
1
2
y
2
-
1
3
y
1
2
y
2
-
1
3
y =
1
2
y
2
-
1
3
y
2
=
1
2
y
2
2
-2 ×
1
2
y
2
×
1
3
y +
1
3
y
2
using a -b)
2
= a
2
-2ab +b
2
=
1
4
y
4
-
1
3
y
3
+
1
9
y
2
( ) ( ) ( [ ( ]
( ) ( ) ( [ ( ]
( ) ( ) ( ) ( ) ( ) [ ( ]
( )( ) (
( )
[ ( ]
( ) ( ) ( ) ( ) [ ( ]
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( [ ( ]
( ) ( ) ( [ ( ]
( ) ( ) ( ) ( ) ( ) [ ( ]
( ) ( ) ( ) ( ) [ ( ]
( ) ( ) ( ) ( ) [ ( ]
( ) ( ) ( ) ( ) ( ) [ ( ]
Q u e s t i o n : 6 4
Expand:
i
(8a + 3b)
2
ii
(7x + 2y)
2
iii
(5x + 11)
2
iv
a
2
+
2
a
2
v
3x
4
+
2y
9
2
vi
(9x - 10)
2
vii
(x
2
y - yz
2
)
2
viii
x
y
-
y
x
2
ix
3m -
4
5
n
2
S o l u t i o n :
We shall use the identities (a+b)
2
=a
2
+b
2
+2ab and (a-b)
2
=a
2
+b
2
-2ab.
i
We have:
8a +3b)
2
= (8a)
2
+2 ×8a ×3b +(3b)
2
= 64a
2
+48ab +9b
2
ii
We have:
7x +2y)
2
= (7x)
2
+2 ×7x ×2y +(2y)
2
= 49x
2
+28xy +4y
2
iii
We have :
5x +11)
2
= (5x)
2
+2 ×5x ×11 +(11)
2
= 25x
2
+110x +121
iv
We have:
a
2
+
2
a
2
=
a
2
2
+2 ×
a
2
×
2
a
+
2
a
2
=
a
4
2
+2 +
4
a
2
v
We have:
3x
4
+
2y
9
2
=
3x
4
2
+2 ×
3x
4
×
2y
9
+
2y
9
2
=
9x
16
2
+
1
3
xy +
4y
2
81
vi
We have:
9x -10)
2
(9x)
2
-2 ×9x ×10 +(10)
2
= 81x
2
-180x +100
vii
We have:
x
2
y -yz
2
)
2
x
2
y
2
-2 ×x
2
y ×yz
2
+ yz
2
2
= x
4
y
2
-2x
2
y
2
z
2
+y
2
z
4
viii
We have:
x
y
-
y
x
2
=
x
y
2
-2 ×
x
y
×
y
x
+
y
x
2
=
x
2
y
2
-2 +
y
2
x
2
ix
We have:
3m -
4
5
n
2
= (3m)
2
-2 ×3m ×
4
5
n +
4
5
n
2
= 9m
2
-
24mn
5
+
16
25
n
2
Q u e s t i o n : 6 5
Find each of the following products:
i
(x + 3)(x - 3)
( )
( )
( )
( )
(
(
(
( ) ( ) ( )
( ) ( ) ( )
(
(
( ) ( )
( ) ( ) ( )
( ) ( )
ii
(2x + 5)(2x - 5)
iii
(8 + x)(8 - x)
iv
(7x + 11y)(7x - 11y)
v
5x
2
+
3
4
y
2
5x
2
-
3
4
y
2
vi
4x
5
-
5y
3
4x
5
+
5y
3
vii
x +
1
x
x -
1
x
viii
1
x
+
1
y
1
x
-
1
y
S o l u t i o n :
i
We have:
x +3 x -3 = x
2
-9 using a +b a -b = a
2
-b
2
ii
We have:
2x +5 2x -5 = 4x
2
-25 using a +b a -b = a
2
-b
2
iii
We have:
8 +x 8 -x = 64 -x
2
using a +b a -b = a
2
-b
2
iv
We have:
7x +11y 7x -11y = 49x
2
-121y
2
using a +b a -b = a
2
-b
2
v
We have:
5x
2
+
3
4
y
2
5x
2
-
3
4
y
2
= 25x
4
-
9
16
y
4
using a +b a -b = a
2
-b
2
vi
We have:
4x
5
-
5y
3
4x
5
+
5y
3
=
16x
2
25
-
25y
2
9
using a +b a -b = a
2
-b
2
vii
We have:
x +
1
x
x -
1
x
= x
2
-
1
x
2
using a +b a -b = a
2
-b
2
viii
We have:
1
x
+
1
y
1
x
-
1
y
=
1
x
2
-
1
y
2
using a +b a -b = a
2
-b
2
ix
We have:
2a +
3
b
2a -
3
b
= 4a
2
-
9
b
2
using a +b a -b = a
2
-b
2
Q u e s t i o n : 6 6
Using the formula for squaring a binomial, evaluate the following:
i
54
2
ii
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( ) [ ( ) ( ) ]
( ) ( ) [ ( ) ( ) ]
( ) ( ) [ ( ) ( ) ]
( ) ( ) [ ( ) ( ) ]
( ) ( ) [ ( ) ( ) ]
( ) ( ) [ ( )( ) )]
( ) ( ) [ ( )( ) ]
( ) ( ) [ ( )( ) ]
( ) ( ) [ ( )( ) ]
82
2
iii
103
2
iv
704
2
S o l u t i o n :
We shall use the identity (a+b)
2
=a
2
+b
2
+2ab.
i
(54)
2
= 50 +4)
2
= (50)
2
+2 ×50 ×4 +(4)
2
= 2500 +400 +16 = 2916
ii
(82)
2
= 80 +2)
2
= (80)
2
+2 ×80 ×2 +(2)
2
= 6400 +320 +4 = 6724
iii
(103)
2
= 100 +3)
2
= (100)
2
+2 ×100 ×3 +(3)
2
= 10000 +600 +9 = 10609
iv
(704)
2
= 700 +4)
2
= (700)
2
+2 ×700 ×4 +(4)
2
= 490000 +5600 +16 = 495616
Q u e s t i o n : 6 7
Using the formula for squaring a binomial, evaluate the following:
i
69
2
ii
78
2
iii
197
2
iv
999
2
S o l u t i o n :
We shall use the identity (a-b)
2
= a
2
+b
2
-2ab.
i
(69)
2
= 70 -1)
2
= (70)
2
-2 ×70 ×1 +1 = 4900 -140 +1 = 4761
ii
(78)
2
= 80 -2)
2
= (80)
2
-2 ×80 ×2 +4 = 6400 -320 +4 = 6084
iii
(197)
2
= 200 -3)
2
= (200)
2
-2 ×200 ×3 +9 = 40000 -1200 +9 = 38809
iv
(999)
2
= 1000 -1)
2
= (1000)
2
-2 ×1000 ×1 +1 = 1000000 -2000 +1 = 998001
Q u e s t i o n : 6 8
Find the value of:
i
82
2
- 18
2
ii
128
2
- 72
2
iii
197 × 203
iv
198×198-102×102
96
v
14.7 ×15.3
(
(
(
(
(
(
(
(
Read More
|
79 videos|408 docs|31 tests
|
Related Exams
About this Document
|
4.63/5 Rating |
|
Nov 24, 2024 Last updated |
Document Description: RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions for Class 8 2024 is part of Mathematics (Maths) Class 8 preparation.
The notes and questions for RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions have been prepared according to the Class 8 exam syllabus. Information about RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions covers topics
like and RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions.
Introduction of RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions in English is available as part of our Mathematics (Maths) Class 8
for Class 8 & RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions in Hindi for Mathematics (Maths) Class 8 course.
Download more important topics related with notes, lectures and mock test series for Class 8
Exam by signing up for free. Class 8: RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions | Mathematics (Maths) Class 8
Description
Full syllabus notes, lecture & questions for RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions | Mathematics (Maths) Class 8 - Class 8 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 8 | Best notes, free PDF download
Information about RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions
In this doc you can find the meaning of RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions defined & explained in the simplest way possible. Besides explaining types of
RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions theory, EduRev gives you an ample number of questions to practice RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions tests, examples and also practice Class 8
tests
|
Explore Courses for Class 8 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions | Mathematics (Maths) Class 8
,Sample Paper
,past year papers
,study material
,Objective type Questions
,MCQs
,Previous Year Questions with Solutions
,practice quizzes
,RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions | Mathematics (Maths) Class 8
,mock tests for examination
,ppt
,Extra Questions
,Summary
,Important questions
,Viva Questions
,video lectures
,shortcuts and tricks
,Free
,RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions | Mathematics (Maths) Class 8
,Exam
,Semester Notes
;
Additional Information about RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions for Class 8 Preparation
RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions Free PDF Download
The RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions is an invaluable resource that delves deep into the core of the Class 8 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions now and kickstart your journey towards success in the Class 8 exam.
Importance of RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions
The importance of RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions cannot be overstated, especially for Class 8 aspirants.
This document holds the key to success in the Class 8 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions Notes
RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions Notes on EduRev are your ultimate resource for success.
RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions Class 8 Questions
The "RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions on the App
Students of Class 8 can study RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RS Aggarwal Solutions: Exercise 6D - Operations on Algebraic Expressions is prepared as per the latest Class 8 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup