This document discusses indices and surds. It presents 8 rules of indices related to multiplication, division, exponents, and fractions. It also discusses irrational numbers as numbers that cannot be expressed as a fraction of integers, such as π. Example problems are given to simplify surd expressions using the rules presented.

1. 1
2
JSTS Mathematics
4
0011 0010 1010 1101 0001 0100 1011
Chapter : Indices and Surds
Presented by: Himani Asija
(P.G.T. Math,
Dps Vasant Kunj)
1

2. Indices and Surds
0011 0010 1010 1101 0001 0100 1011
Index, exponential, power
a×a = a 1
2
2
Base
4 2

3. Indices Rule 1
a ×a
2 3
0011 0010 1010 1101 0001 0100 1011
= a×a×a×a×a
=a
1
2
5
4
2+3
=a
m+ n
a ×a = a
m n
3

4. Indices Rule 2
a ÷a
5 3
0011 0010 1010 1101 0001 0100 1011
a×a×a×a×a
2
=
a×a×a
1
4
= a×a
m−n
=a 2 a ÷a = a
m n
5−3
=a 4

5. Indices Rule 3
5 3
(a )
0011 0010 1010 1101 0001 0100 1011
From Rule 1
2
= a ×a ×a
5 5 5
a ×a = a
m n m+ n
1
4
5+ 5+ 5
=a
=a 15
(a ) = a
m n mn
5×3
=a 5

6. Indices Rule 4
a ×b
3 3
0011 0010 1010 1101 0001 0100 1011
= a×a×a×b×b×b
1
2
= (a × b)(a × b)(a × b)
4
= ( a × b) 3
a × b = (ab)
m m m
6

7. Indices Rule 5
a ÷b
3 3
0011 0010 1010 1101 0001 0100 1011
a×a×a
=
2
b×b×b
a a a
1
4
= ( )( )( )
b b b
a 3
=( )
b a ÷ b = ( a ÷ b)
m m m
7

8. Indices Rule 6, 7
3 = 27
3
0011 0010 1010 1101 0001 0100 1011
÷3
3 =9
2 a =10
2
÷3 1
1
−m
3 =3
1 a = m
÷3 a
4
1
3 =?
0
−1 1 ÷3
3 =? 3
−2
3 =?
1 1
= 2
÷3
9 3
−3
3 =?
1 1
= 3 ÷3 8
27 3

9. Indices Rule 8
1 2 1
 
0011 0010 1010 1101 0001 × 2
0100 1011
a 2 =a 2 1
 
  a = a
n n
1
2
2 m
 1 a n
= a
n m
=( a)
n m
4
a2  = a
 
 
1
a = a= a
2 2
9

10. Irrational Number
0011 0010 1010that 0001 0100 1011
•Number 1101 cannot be expressed as a fraction
of two integers
2
2
6
1
π 1
2 4 10

11. Think!
Which of the following is NOT a irrational number?
2
x
 1 
3 
0011 0010 1010 1101 0001 0100 1011
25
x  27  1 2 1
( ) =
2
3 9
5=
5
26 1
4
1
Irrational number
1
27
1
x
Irrational number
36
11

12. Surd Rules
1 1 1 1 1 1 1
+
0011 0010 1010 1101 0001 0100 1011
a ×a = a
m n m n (a ) = a
m n mn
1
2
1 1 1 1 1 1 1
−
a ÷a = a
m n m n a × b = (ab)
m m m
4
1 1 1
a ÷ b = ( a ÷ b)
m m m
We can use the above rules to:
• simplify two or more surds or
•combining them into one single surd
12

13. Simplify each of the following surds
27 + 243
0011Ques.11010 1101 0001 0100 1011
0010
2
= 9 × 3 + 81× 3 1 1 1
(ab) = a × b
m m m
( ) ( )
1
4
= 9× 3 + 81 × 3
=3 3 +9 3
Common Factor : 3
= 3 (9 + 3)
= 12 3 13

14. Simplify each of the following surds
Question 2
175 + 112 − 28
0011 0010 1010 1101 0001 0100 1011
= 25 × 7 + 16 × 7 − 4 × 7 1 1 1
2
(ab) = a × b
m m m
= ( ) ( ) (
25 × 7 + 16 × 7 − 4× 7
1
)
4
=5 7 +4 7 −2 7
Common Factor : 7
= 7 (5 + 4 − 2)
=7 7
14

15. Simplify each of the following surds
4 3 × 12 ÷ 27
0011 0010 1010 1101 0001 0100 1011
Question 3
2
= 4 3 × 4 × 3 ÷ 3× 9
= 4 3×2 3 ÷3 3
1
4
= 3 (4 × 2) ÷ 3
x
8 3 2 3
= =2
3 3
15

16. Simplify each of the following surds
Question 4
2 0001 0100 27 ÷ 343
0011 0010 1010 1101
21 × 1011
3
21× 27
=2
2
343 49
3 × 27
1
4
=2
49
81
= 2( )
49
2 81 2× 9 4
= = =2
49 7 7 16

17. Simplify each of the following surds
Question 5 8 3 ÷ 2 48 × 243
0011 0010 1010 1101 0001 0100 1011
2
= 8 3 ÷ 2 16 × 3 × 81× 3
= 8 3 ÷ 2( 16 ) 3 × ( ) ( 81) 3
1
4
( )
= 8 3 ÷8 3 ×9 3
=9 3
17

18. Simplify each of the following surds
Question 6
(2 3 +3 2 ) 2
0011 0010 1010 1101 0001 0100 1011
(2 a) 2 + 2ab3 ) (3 2 ) +(3 b) 2
= ( 2 a3 + 3b2 ) = 3 (2 2 2
2
= ( 2) ( 3) 2
( )
+ 2( 2 × 3) 3 × 2 + (3) 2 ( 2 ) 2
1
2
= ( 4 )( 3) + 2( 2 × 3) ( 6 ) + (9)(2)
4
= 12 + 12( 6 ) + 18
= 6(5 + 2 6 )
18

19. Simplify each of the following surds
Question 7
( 2 + 3 ) − (3 − 3 )
0011 0010 1010 1101 0001 0100 1011 2
2
1
2
=( 2 + 3 )
a 2
−(3 − 3 )
b 2
= (2 +a 3 + 3 −b 3 )( 2 + 3 −(3 −b 3 ) )
a
4
(
= ( 5) 2 − 3 + 3 + 3 )
= ( 5) ( − 1 + 2 3 )
= 5(2 3 − 1)
19

20. Simplify each of the following surds
Question 8
(2 3
) (
3 −1 + 1− 4 3
0011 0010 1010 1101 0001 0100 1011 ) 2
( )( )
1 − 2(1) 4 3 + 4 3 2 
2
( ) ( )
2
= 2 3 −1 2 3 −1 +  
 
1
( )( 2
= 2 3 −1 2 3 −1 + 

)
1 − 8 3 + 4 3 2 


( )
4
( )( ) (
 2 3 2 − 4 3 + 1 + 1 − 8 3 + (16 × 3)
= 2 3 −1 



)
( )(
= 2 3 − 1 13 − 4 3 + 49 − 8 3 ) ( )
= 26 3 − 24 − 13 + 4 3 + 49 − 8 3
= 22 3 + 12 = 2(11 3 + 6) 20

21. Rationalization of the Denominator
•Process of removing a surd from the denominator
0011 0010 1010 1101 0001 0100 1011
1
•Example:
2
1
2
1
= ×1 =
2
4
2 2
1 2
= ×
Multiply together!
2 2
2 2
= =
( 2 )( 2 ) 2 21

22. 2 2
•Example: = ×1 3 −2
3+2 =
3+2 3 −2
2 3−2 Note!
0011 0010 1010 1101 0001 0100 1011 ×
=
3+2 3−2 Multiply together!
( )
2
Objective is 2 3−2
=
to remove
( )( ) Conjugate Surds
1
surds from 3+2 3−2
denominator
( )
4
2 3−2
=
( 3 ) − ( 2)
2 2 (a + b)(a − b) = a 2 − b 2
2( 3 − 2 ) 2 3−4
= =
( 3 − 4) ( − 1)
= −2 3 + 4 22

23. Re-look into the question: 2
3+2
0011 0010 1010 1101 0001 0100 1011 2 3−2
Multiply conjugate = ×
surds 3+2 3−2
2
(
2 3−2 )
1
=
Multiply denominator
( 3+2 3−2 )( )
4
and numerator
=
(
2 3−2 )
(a + b)(a − b) = a − b
2 2 ( 3) 2
− ( 2)
2
2 3−4
= = 4−2 3
−1 23

24. Question 2
Rationalize the denominators: 7 −2
0011 0010 1010 1101 0001 0100 1011 2 7 +2
Multiply conjugate = ×
surds 7 −2 7 +2
2
(
2 7 +2 )
1
=
Multiply denominator
( )(
7 −2 7 +2 )
4
and numerator
=
(
2 7 +2 )
(a + b)(a − b) = a − b
2 2 ( 7) 2
− ( 2)
2
=
2 7 +2(=
2 7 +2 ) ( )
7−4 3 24

25. Question 3 −3
Rationalize the denominators: 3 −1
0011 0010 1010 1101 0001 0100 1011 3 −3 3 +1
Multiply conjugate = ×
surds 3 −1 3 +1
2
( 3 −3 )( 3 +1 )
1
=
Multiply denominator
( 3 −1 )( 3 +1 )
4
and numerator
=
( 3) 2
+ 3 −3 3 −3
( 3) 2
− (1)
2
(a + b)(a − b) = a 2 − b 2
=
3− 2 3 −3 − 2 3
= =− 3
( )
3 −1 2 25

26. Question 2 2 +1
Rationalize the denominators: 4 2 −3
2 2 +1 4 2 +3
= ×
Multiply conjugate
0011 0010 1010 1101 0001 0100 1011
4 2 −3 4 2 +3
surds
2
2 2 +1 4 2 +3
= ×
1
Multiply denominator 4 2 −3 4 2 +3
and numerator
4
=
(2 )(
2 +1 4 2 + 3 )
(4 2 ) − ( 3)
2 2
(a + b)(a − b) = a 2 − b 2
=
(16 + 6 2 +4 2 +3 )
32 − 9
19 + 10 2 26
=
23

27. Question 3 2 −2 3
Rationalize the denominators: 3 2 + 2 3
3 2 −2 3 3 2 −2 3
= ×
Multiply conjugate
0011 0010 1010 1101 0001 0100 1011
3 2+2 3 3 2 −2 3
surds
(3 2 −2 3 ⋅ 3 2 −2 3 )( )
2
=
(3 2 +2 3 ⋅ 3 2 −2 3 )( )
1
Multiply denominator
and numerator (3 ) 2
4
2 −2 3
=
(3 2 ) − ( 2 3 )
2 2
(a + b)(a − b) = a 2 − b 2
=
(3 2 ) 2
( )( ) (
−23 2 2 3 + 2 3 ) 2
(3 2 ) − ( 2 3 )
2 2
1 1 1
a × b = (ab)
m m m
=
( 9 × 2) − 12 6 + ( 4 × 3)
( 9 × 2 ) − ( 4 × 3)
30 − 12 6
= = 5−2 6 27
6

28. Question 7 3+2 5
Rationalize the denominators: 3 5 −2 3
0011 0010 1010 1101 0001 0100 = 7 3 + 2
Multiply conjugate 1011 5 3 5+2 3
×
3 5−2 3 3 5+2 3
surds
(7 )( )
2
3+2 5 3 5+2 3
=
(3 )(
5−2 3 3 5+2 3 )
1
(21 15 + 42 + 30 + 4 15 )
Multiply denominator =
( 45 − 12)
4
and numerator
25 15 + 72
=
33
(a + b)(a − b) = a 2 − b 2
28

29. Question Rationalize the denominators:
1 1
+
Determine the LCM
0011 0010 1010 1101 0001 0100 1011 3+ 7 3− 7
( 3+ 7 )( 3− 7 )
2
1 ( 3− 7 ) 1 ( 3+ 7 )
= ( ) + ( )
1
Multiply denominator 3+ 7
3− 7
3− 7
3+ 7
and numerator
4
3− 7 3+ 7
= +
( )(
3+ 7 3− 7 ) ( )(
3+ 7 3− 7 )
3− 7+ 3+ 7 2 3
=
(
3+ 7 3− 7 )( ) =
3−7
2 3 3
= =−
−4 2 29

30. Question
3− 5 3+ 5
Rationalize the denominators: +
( 5 + 3) ( 3 − 5)2
2
Expand the denominators =
0011 0010 1010 1101 0001 0100 1011
3− 5
+
3+ 5
( 5 )2 + 2 5 3 + ( 3)2 ( 3)2 − 2 3 5 + ( 5 )2
3− 5 3+ 5
2
= +
=
3− 5
+
3+ 5 5 + 2 15 + 3 3 − 2 15 + 5
2(4 + 15 ) 2(4 − 15 )
1
3− 5 3+ 5
= +
8 + 2 15 8 − 2 15
4
Find the LCM 24( 4 − ) 15 )(+ + 15 )+
3− 5 ( − 15 3
4
5 (4 15 )
= × + ×
2(4 + 15 ) (4 − 15 ) 2(4 − 15 ) (4 + 15 )
(12 − 3 15 − 4 5 + 75 ) + (12 + 3 15 + 4 5 + 75 )
=
2(16 − 15)
2(12 + 75 )
= = 12 + 5 3 30
2(1)

31. Question 5+3 2 3− 2
+
Rationalize the denominators: 4 − 2 5 4 − 2 5
0011 0010 1010 1101 0001 0100 1011 5+3 2 +3− 2
=
4−2 5
2
Same denominator 8+ 2 2 4+ 2 5
= ×
1
4−2 5 4+2 5
4
(8 + 2 2 )(4 + 2 5 )
=
(4 − 2 5 )(4 + 2 5 )
Rationalization
32 + 16 5 + 8 2 + 4 10
=
16 − 20
32 + 16 5 + 8 2 + 4 10
= = −8 − 4 5 − 2 2 − 10
−4 31

32.  5+ 3 1
Find k + 2
2
k = 
Question : given  5− 3
  k
1
 5+ 3
2 k + 2
2
k =
2
 k
 5− 3
 
0011 0010 1010 1101 0001 0100 1011 8 + 2 15 8 − 2 15
= +
k2 =
( 5+ 3 ) 2
8 − 2 15 8 + 2 15
2
( 5− 3)
2
1 8 − 2 15
= (8 + 2 15 ) ( 8 + 2 15 )+ (8 − 2 15 ) (8 − 2 15 )
=
(8 − 2 15 ) ( (8 + 2 15 )
k 2 8 + 2 15
1
× ×
8 + 2 15 ) (8 − 2 15 )
(5 + 2 15 + 3 )
4
k 2
=
(5 − 2 15 + 3 ) =
(8 + 2 15 ) + (8 − 2 15 )
2 2
(8 − 2 15 )(8 + 2 15 )
k =
8 + 2 15
2 =
(64 + 32
15 + 60 + 64 − 32 15 + 60 ) ( )
8 − 2 15 ( 64 − 60)
1 1 =
(124 + 124)
=
k 2 8 + 2 15 ( 4)
8 − 2 15 248
= = 62 32
4

33. 1 1
OR =
Rationalize k 2 31 + 8 15
 5+ 3 k First
2
1 31 − 8 15
k2 =   = ×
 5− 3 31 + 8 15 31 − 8 15
 
0011 0010 1010 1101 0001 0100 1011
k2 = [
(
5+ 3
2
]2 ) =
31 − 8 15
( )(
5− 3 5+ 3 ) ( 961 − 64 × 15) 1
2
k + 2
2
(5 + 2 15 + 3) 31 − 8 15 k
1
k2 = [ ]2 =
( 5 − 3) ( 961 − 960)
4
= 31 + 8 15 + 31 − 8 15
8 + 2 15
k2 = [ ]2 = 31− 8 15 = 62
2
k 2 = (4 + 15 ) 2 = 16 + 2 × 4 × 15 + 15
= 31+ 8 15
33

34. Homework
0011 0010 1010 1101 0001 0100 1011
2
Find an
1
attached
4
worksheet
Remember:
Success is 99%perspiration and 1% inspiration
34