Download Chapter 2. Exponents and Radicals

MODULE - 1
Exponents and Radicals
Algebra
2
Notes
EXPONENTS AND RADICALS
We have learnt about multiplication of two or more real numbers in the earlier lesson. You
can very easily write the following
4 × 4 × 4 = 64,11 × 11 × 11 × 11 = 14641 and
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
Think of the situation when 13 is to be multiplied 15 times. How difficult is it to write?
13 × 13 × 13 ×.................15 times?
This difficulty can be overcome by the introduction of exponential notation. In this lesson,
we shall explain the meaning of this notation, state and prove the laws of exponents and
learn to apply these. We shall also learn to express real numbers as product of powers of
prime numbers.
In the next part of this lesson, we shall give a meaning to the number a1/q as qth root of a.
We shall introduce you to radicals, index, radicand etc. Again, we shall learn the laws of
radicals and find the simplest form of a radical. We shall learn the meaning of the term
“rationalising factor’ and rationalise the denominators of given radicals.
OBJECTIVES
After studying this lesson, you will be able to
•
write a repeated multiplication in exponential notation and vice-versa;
•
identify the base and exponent of a number written in exponential notation;
•
express a natural number as a product of powers of prime numbers uniquely;
•
state the laws of exponents;
•
explain the meaning of a0, a–m and a q ;
•
simplify expressions involving exponents, using laws of exponents;
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Notes
•
identify radicals from a given set of irrational numbers;
•
identify index and radicand of a surd;
•
state the laws of radicals (or surds);
•
express a given surd in simplest form;
•
classify similar and non-similar surds;
•
reduce surds of different orders to those of the same order;
•
perform the four fundamental operations on surds;
•
arrange the given surds in ascending/descending order of magnitude;
•
find a rationalising factor of a given surd;
•
1
rationalise the denominator of a given surd of the form a + b x and
1
x+ y ,
where x and y are natural numbers and a and b are integers;
•
simplify expressions involving surds.
EXPECTED BACKGROUND KNOWLEDGE
•
Prime numbers
•
Four fundamental operations on numbers
•
Rational numbers
•
Order relation in numbers.
2.1 EXPONENTIAL NOTATION
Consider the following products:
(i) 7 × 7
(ii) 3 × 3 × 3
(iii) 6 × 6 × 6 × 6 × 6
In (i), 7 is multiplied twice and hence 7 × 7 is written as 72.
In (ii), 3 is multiplied three times and so 3 × 3 × 3 is written as 33.
In (iii), 6 is multiplied five times, so 6 × 6 × 6 × 6 × 6 is written as 65.
72 is read as “7 raised to the power 2” or “second power of 7”. Here, 7 is called base and
2 is called exponent (or index)
Similarly, 33 is read as “3 raised to the power 3”or “third power of 3”. Here, 3 is called the
base and 3 is called exponent.
Similarly, 65 is read as “6 raised to the power 5”or “Fifth power of 6”. Again 6 is base and
5 is the exponent (or index).
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From the above, we say that
The notation for writing the product of a number by itself several times is called the
Exponential Notation or Exponential Form.
Thus, 5 × 5 × .... 20 times = 520 and (–7) × (–7) × .... 10 times = (–7)10
Notes
In 520, 5 is the base and exponent is 20.
In (–7)10, base is –7 and exponent is 10.
Similarly, exponential notation can be used to write precisely the product of a ratioinal
number by itself a number of times.
16
Thus,
3 3
⎛3⎞
× × .........16 times = ⎜ ⎟
5 5
⎝5⎠
and
⎛ 1⎞ ⎛ 1⎞
⎛ 1⎞
⎜ − ⎟ × ⎜ − ⎟ × ..........10 times = ⎜ – ⎟
⎝ 3⎠ ⎝ 3⎠
⎝ 3⎠
10
In general, if a is a rational number, multiplied by itself m times, it is written as am.
Here again, a is called the base and m is called the exponent
Let us take some examples to illustrate the above discussion:
Example 2.1: Evaluate each of the following:
(i ) ⎛⎜ 2 ⎞⎟
⎝7⎠
3
⎛ 3⎞
(ii) ⎜ − ⎟
⎝ 5⎠
(i)
8
⎛ 2 ⎞ 2 2 2 (2 )
⎜ ⎟ = × × = 3=
⎝ 7 ⎠ 7 7 7 (7 ) 343
(ii)
81
⎛ 3 ⎞ ⎛ 3 ⎞⎛ 3 ⎞⎛ 3 ⎞⎛ 3 ⎞ (− 3)
=
⎜ − ⎟ = ⎜ − ⎟⎜ − ⎟⎜ − ⎟⎜ − ⎟ =
5
625
⎝ 5 ⎠ ⎝ 5 ⎠⎝ 5 ⎠⎝ 5 ⎠⎝ 5 ⎠ (5)
3
Solution:
4
3
4
4
Example 2.2: Write the following in exponential form:
(i) (–5) × (–5) × (–5) × (–5) × (–5) × (–5) × (–5)
⎛3⎞ ⎛3⎞ ⎛3⎞ ⎛3⎞
(ii) ⎜ ⎟ × ⎜ ⎟ × ⎜ ⎟ × ⎜ ⎟
⎝ 11 ⎠ ⎝ 11 ⎠ ⎝ 11 ⎠ ⎝ 11 ⎠
Solution:
(i) (–5) × (–5) × (–5) × (–5) × (–5) × (–5) × (–5) = (–5)7
⎛3⎞ ⎛3⎞ ⎛3⎞ ⎛3⎞ ⎛3⎞
(ii) ⎜ ⎟ × ⎜ ⎟ × ⎜ ⎟ × ⎜ ⎟ = ⎜ ⎟
⎝ 11 ⎠ ⎝ 11 ⎠ ⎝ 11 ⎠ ⎝ 11 ⎠ ⎝ 11 ⎠
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Example 2.3: Express each of the following in exponential notation and write the base
and exponent in each case.
Notes
(i) 4096
Solution:
(ii)
125
729
(iii) – 512
(i) 4096 = 4 × 4 × 4 × 4 × 4 × 4
= (4)6
Alternatively 4096 = (2)12
Base = 2, exponent =12
Here, base = 4 and exponent = 6
125
5 5 5 ⎛5⎞
(ii)
= × × =⎜ ⎟
729 9 9 9 ⎝ 9 ⎠
3
⎛5⎞
Here, base = ⎜ ⎟ and exponent = 3
⎝9⎠
(iii) 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
Here, base = 2 and exponent = 9
Example 2.4: Simplify the following:
3
⎛3⎞ ⎛4⎞
⎜ ⎟ ×⎜ ⎟
⎝2⎠ ⎝3⎠
4
3
3 3 3 33
⎛3⎞
⎜ ⎟ = × × = 3
2 2 2 2
⎝2⎠
Solution:
4
Similarly
44
⎛4⎞
⎜ ⎟ = 4
3
⎝3⎠
3
4
33 4 4
⎛3⎞ ⎛4⎞
×
⎜ ⎟ ⎜ ⎟ = 3× 4
2 3
⎝2⎠ ⎝3⎠
=
33 16 ×16 32
×
=
8
34
3
Example 2.5: Write the reciprocal of each of the following and express them in exponential
form:
5
(i) 3
42
⎛3⎞
(ii) ⎜ ⎟
⎝4⎠
2
⎛ 5⎞
(iii) ⎜ − ⎟
⎝ 6⎠
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Solution:
(i)
35 = 3 × 3 × 3 × 3 × 3
= 243
1
⎛1⎞
=⎜ ⎟
∴ Reciprocal of 3 =
243 ⎝ 3 ⎠
5
Notes
5
2
(ii)
32
⎛3⎞
⎜ ⎟ = 2
4
⎝4⎠
2
42 ⎛ 4 ⎞
⎛3⎞
∴ Reciprocal of ⎜ ⎟ = 2 = ⎜ ⎟
3
⎝4⎠
⎝3⎠
2
⎛ 5⎞
(− 5)9
⎜− ⎟ =
69
⎝ 6⎠
9
(iii)
9
9
⎛ 5⎞ −6 ⎛−6⎞
⎟
∴ Reciprocal of ⎜ − ⎟ = 9 = ⎜
5
⎝ 6⎠
⎝ 5 ⎠
From the above example, we can say that if
9
p
is any non-zero rational number and m is
q
m
m
⎛ p⎞
⎛q⎞
any positive integer, then the reciprocal of ⎜⎜ ⎟⎟ is ⎜⎜ ⎟⎟ .
⎝q⎠
⎝ p⎠
CHECK YOUR PROGRESS 2.1
1. Write the following in exponential form:
(i) (–7) × (–7) × (–7) × (–7)
⎛3⎞ ⎛3⎞
(ii) ⎜ ⎟ × ⎜ ⎟ × .... 10 times
⎝4⎠ ⎝4⎠
⎛ 5⎞ ⎛ 5⎞
(iii) ⎜ − ⎟ × ⎜ − ⎟ × .... 20 times
⎝ 7⎠ ⎝ 7⎠
2. Write the base and exponent in each of the following:
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(i) (–3)
Notes
5
⎛ 2⎞
(iii) ⎜ − ⎟
⎝ 11 ⎠
4
(ii) (7)
8
3. Evaluate each of the following
⎛3⎞
(i) ⎜ ⎟
⎝7⎠
4
⎛ –2⎞
(ii) ⎜
⎟
⎝ 9 ⎠
4
⎛ 3⎞
(iii) ⎜ – ⎟
⎝ 4⎠
3
4. Simplify the following:
5
⎛7⎞ ⎛3⎞
(i) ⎜ ⎟ × ⎜ ⎟
⎝3⎠ ⎝7⎠
2
6
⎛ 5⎞ ⎛ 3⎞
(ii) ⎜ − ⎟ ÷ ⎜ − ⎟
⎝ 6⎠ ⎝ 5⎠
2
5. Find the reciprocal of each of the following:
5
(i) 3
⎛ 3⎞
(iii) ⎜ − ⎟
⎝ 5⎠
4
(ii) (–7)
4
2.2 PRIME FACTORISATION
Recall that any composite number can be expressed as a product of prime numbers. Let
us take the composite numbers 72, 760 and 7623.
(i) 72 = 2 × 2 × 2 × 3 × 3
= 23 × 32
(ii) 760 = 2 × 2 × 2 × 5 × 19
2
2
2
3
72
36
18
9
3
2
2
2
5
760
380
190
95
19
3 7623
3 2541
7 847
(iii) 7623 = 3 × 3 × 7 × 11 ×11
11 121
= 32 × 71 × 112
11
We can see that any natural number, other than 1, can be expressed as a product of
powers of prime numbers in a unique manner, apart from the order of occurrence of
factors. Let us consider some examples
= 23 × 51 ×191
Example 2.6: Express 24300 in exponential form.
Solution: 24300 = 3 × 3 × 3 × 3 × 2 × 2 × 5 × 5 × 3
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∴ 24300 = 22 × 35 ×52
Example 2.7: Express 98784 in exponential form.
Solution:
2
2
2
2
2
3
3
7
7
98784
49392
24696
12348
6174
3087
1029
343
49
7
Notes
∴ 98784 = 25 × 32 × 73
CHECK YOUR PROGRESS 2.2
1. Express each of the following as a product of powers of primes, i.e, in exponential form:
(i) 429
(ii) 648
(iii) 1512
2. Express each of the following in exponential form:
(i) 729
(iv)
(ii) 512
1331
4096
(v) −
(iii) 2592
243
32
2.3 LAWS OF EXPONENTS
Consider the following
(i) 32 × 33
= (3 × 3) × (3 × 3 × 3) = (3 × 3 × 3 × 3 × 3)
= 35 = 32 + 3
(ii) (–7)2 × (–7)4 = [(–7) × (–7)] × [(–7) × (–7) × (–7) × (–7)]
= [ (–7) × (–7) × (–7) × (–7) × (–7) × (–7)]
= (–7)6 = (–7)2+4
3
4
⎛3⎞ ⎛3⎞ ⎛3 3 3⎞ ⎛3 3 3 3⎞
(iii) ⎜ ⎟ × ⎜ ⎟ = ⎜ × × ⎟ × ⎜ × × × ⎟
⎝4⎠ ⎝4⎠ ⎝4 4 4⎠ ⎝4 4 4 4⎠
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⎛3 3 3 3 3 3 3⎞
=⎜ × × × × × × ⎟
⎝4 4 4 4 4 4 4⎠
Notes
7
⎛3⎞ ⎛3⎞
= ⎜ ⎟ =⎜ ⎟
⎝4⎠ ⎝4⎠
3+ 4
(iv) a3 × a4 = (a × a × a) × (a × a × a × a) = a7 = a3+4
From the above examples, we observe that
Law 1: If a is any non-zero rational number and m and n are two positive integers, then
am × an = am+n
3
5
⎛ 3⎞ ⎛ 3⎞
Example 2.8: Evaluate ⎜ − ⎟ × ⎜ − ⎟ .
⎝ 2⎠ ⎝ 2⎠
3
Here a = − , m = 3 and n = 5.
2
Solution:
3
5
⎛ 3⎞ ⎛ 3⎞ ⎛ 3⎞
∴ ⎜− ⎟ ×⎜− ⎟ = ⎜− ⎟
⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠
3+ 5
8
⎛ 3 ⎞ 6561
= ⎜− ⎟ =
256
⎝ 2⎠
Example 2.9: Find the value of
2
⎛7⎞ ⎛7⎞
⎜ ⎟ ×⎜ ⎟
⎝4⎠ ⎝4⎠
Solution:
3
As before,
2
3
⎛7⎞ ⎛7⎞ ⎛7⎞
⎜ ⎟ ×⎜ ⎟ = ⎜ ⎟
⎝4⎠ ⎝4⎠ ⎝4⎠
2+3
5
⎛ 7 ⎞ 16807
=⎜ ⎟ =
1024
⎝4⎠
Now study the following:
75 7 × 7 × 7 × 7 × 7
= 7 × 7 = 7 2 = 7 5−3
(i) 7 ÷ 7 = 3 =
7
7×7×7
5
3
(ii) (–3) ÷ (–3) =
7
4
(− 3)7 = (− 3)× (− 3)× (− 3)× (− 3)× (− 3)× (− 3)× (− 3)
(− 3)× (− 3)× (− 3)× (− 3)
(− 3)4
= (− 3)(− 3)(− 3) = (− 3)3 = (–3)7– 4
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From the above, we can see that
Law 2: If a is any non-zero rational number and m and n are positive integers (m > n), then
am ÷ an = am–n
Notes
16
13
⎛ 35 ⎞
⎛ 35 ⎞
Example 2.10: Find the value of ⎜ ⎟ ÷ ⎜ ⎟ .
⎝ 25 ⎠
⎝ 25 ⎠
16
13
⎛ 35 ⎞
⎛ 35 ⎞
⎜ ⎟ ÷⎜ ⎟
⎝ 25 ⎠
⎝ 25 ⎠
Solution:
⎛ 35 ⎞
=⎜ ⎟
⎝ 25 ⎠
16 −13
3
3
343
⎛ 35 ⎞
⎛7⎞
=⎜ ⎟ =⎜ ⎟ =
125
⎝ 25 ⎠
⎝5⎠
In Law 2, m < n ⇒ n > m,
1
a m ÷ a n = a −(n − m ) =
then
a
m−n
Law 3: When n > m
am ÷ an =
1
a
m−n
6
⎛3⎞ ⎛3⎞
Example 2.11: Find the value of ⎜ ⎟ ÷ ⎜ ⎟
⎝7⎠ ⎝7⎠
Solution:
Here a =
9
3
, m = 6 and n = 9.
7
6
1
9
⎛ 3 ⎞ ⎛ 3 ⎞ ⎛ 3 ⎞ 9−6
∴ ⎜ ⎟ ÷⎜ ⎟ = ⎜ ⎟
⎝7⎠ ⎝7⎠ ⎝7⎠
=
7 3 343
=
33 27
Let us consider the following:
(i)
(3 )
3 2
= 33 × 33 = 33+3 = 36 = 33×2
5
⎡⎛ 3 ⎞ 2 ⎤ ⎛ 3 ⎞ 2 ⎛ 3 ⎞ 2 ⎛ 3 ⎞ 2 ⎛ 3 ⎞ 2 ⎛ 3 ⎞ 2
(ii) ⎢⎜ ⎟ ⎥ = ⎜ ⎟ × ⎜ ⎟ × ⎜ ⎟ × ⎜ ⎟ × ⎜ ⎟
⎢⎣⎝ 7 ⎠ ⎥⎦ ⎝ 7 ⎠ ⎝ 7 ⎠ ⎝ 7 ⎠ ⎝ 7 ⎠ ⎝ 7 ⎠
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⎛3⎞
⎜ ⎟
⎝7⎠
Notes
2+ 2+ 2+ 2+ 2
10
⎛3⎞
⎛3⎞
=⎜ ⎟ =⎜ ⎟
⎝7⎠
⎝7⎠
2×5
From the above two cases, we can infer the following:
Law 4: If a is any non-zero rational number and m and n are two positive integers, then
(a )
m n
= a mn
Let us consider an example.
⎡⎛ 2 ⎞ 2 ⎤
Example 2.12: Find the value of ⎢⎜ ⎟ ⎥
⎣⎢⎝ 5 ⎠ ⎦⎥
3
3
Solution:
⎡⎛ 2 ⎞ 2 ⎤ ⎡ 2 ⎤ 2×3 ⎛ 2 ⎞6
64
⎢⎜ ⎟ ⎥ = ⎢ ⎥ = ⎜ ⎟ =
5
⎢⎣⎝ ⎠ ⎥⎦ ⎣ 5 ⎦
⎝ 5 ⎠ 15625
2.3.1 Zero Exponent
Recall that a m ÷ a n = a m − n , if m > n
=
1
a
, if n > m
n−m
Let us consider the case, when m = n
∴ a m ÷ a m = a m−m
am
⇒ m = a0
a
⇒ 1 = a0
Thus, we have another important law of exponents,.
Law 5: If a is any rational number other than zero, then ao = 1.
Example 2.13:Find the value of
⎛2⎞
(i) ⎜ ⎟
⎝7⎠
0
⎛ −3⎞
⎟
(ii) ⎜
⎝ 4 ⎠
0
0
Solution:
48
⎛2⎞
(i) Using a = 1, we get ⎜ ⎟ = 1
⎝7⎠
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Mathematics Secondary Course
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Algebra
0
⎛ −3⎞
⎟ = 1.
(ii) Again using a = 1, we get ⎜
⎝ 4 ⎠
0
Notes
CHECK YOUR PROGRESS 2.3
1. Simplify and express the result in exponential form:
3
2
⎛3⎞ ⎛3⎞
(ii) ⎜ ⎟ × ⎜ ⎟
⎝4⎠ ⎝4⎠
3
(i) (7) ×(7)
2
1
2
⎛ 7⎞ ⎛ 7⎞ ⎛ 7⎞
(iii) ⎜ − ⎟ × ⎜ − ⎟ × ⎜ − ⎟
⎝ 8⎠ ⎝ 8⎠ ⎝ 8⎠
3
2. Simplify and express the result in exponential form:
8
⎛3⎞ ⎛3⎞
(ii) ⎜ ⎟ ÷ ⎜ ⎟
⎝4⎠ ⎝4⎠
(i) (− 7 ) ÷ (− 7 )
9
7
2
18
⎛−7⎞
⎛−7⎞
⎟ ÷⎜
⎟
(iii) ⎜
⎝ 3 ⎠
⎝ 3 ⎠
3
3. Simplify and express the result in exponential form:
⎡⎛ 3 ⎞3 ⎤
(ii) ⎢⎜ ⎟ ⎥
⎢⎣⎝ 4 ⎠ ⎥⎦
( )
(i) 26
3
5
0
2
⎡⎛ 5 ⎞3 ⎤
(iii) ⎢⎜ − ⎟ ⎥
⎢⎣⎝ 9 ⎠ ⎥⎦
0
⎛ 11 ⎞ ⎛ 15 ⎞
⎛ 7⎞ ⎛ 7⎞
(iv) ⎜ ⎟ × ⎜ ⎟ (v) ⎜ − ⎟ × ⎜ − ⎟
⎝3⎠ ⎝7⎠
⎝ 11 ⎠ ⎝ 11 ⎠
5
3
4. Which of the following statements are true?
5
3
3
(i) 7 × 7 = 7
4
7
2
⎡⎛ 4 ⎞5 ⎤ ⎛ 4 ⎞9
(iii) ⎢⎜ ⎟ ⎥ = ⎜ ⎟
⎢⎣⎝ 9 ⎠ ⎥⎦ ⎝ 9 ⎠
⎡⎛ 3 ⎞ 6 ⎤ ⎛ 3 ⎞8
(iv) ⎢⎜ ⎟ ⎥ = ⎜ ⎟
⎢⎣⎝ 19 ⎠ ⎥⎦ ⎝ 19 ⎠
0
2
⎛3⎞
(v) ⎜ ⎟ = 0
⎝ 11 ⎠
5
2
⎛3⎞ ⎛3⎞ ⎛3⎞
(ii) ⎜ ⎟ × ⎜ ⎟ = ⎜ ⎟
⎝ 11 ⎠ ⎝ 11 ⎠ ⎝ 11 ⎠
6
9
⎛ 3⎞
(vi) ⎜ − ⎟ = −
4
⎝ 2⎠
0
⎛ 8 ⎞ ⎛7⎞ ⎛ 8 ⎞
(vii) ⎜ ⎟ × ⎜ ⎟ = ⎜ ⎟
⎝ 15 ⎠ ⎝ 6 ⎠ ⎝ 15 ⎠
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2.4 NEGATIVE INTEGERS AS EXPONENTS
Notes
i) We know that the reciprocal of 5 is
1
. We write it as 5–1 and read it as 5 raised to
5
power –1.
1
ii) The reciprocal of (–7) is − . We write it as (–7)–1 and read it as (–7) raised to the
7
power –1.
iii) The reciprocal of 52 =
1
. We write it as 5–2 and read it as ‘5 raised to the power (–2)’.
52
From the above all, we get
If a is any non-zero rational number and m is any positive integer, then the reciprocal of am
1 ⎞
⎛
⎜ i.e. m ⎟ is written as a–m and is read as ‘a raised to the power (–m)’. Therefore,
⎝ a ⎠
1
= a −m
m
a
Let us consider an example.
Example 2.14: Rewrite each of the following with a positive exponent:
⎛3⎞
(i) ⎜ ⎟
⎝8⎠
−2
⎛ 4⎞
(ii) ⎜ − ⎟
⎝ 7⎠
−7
Solution:
−2
1
1 82 ⎛ 8 ⎞
⎛ 3⎞
=
= =⎜ ⎟
(i) ⎜ ⎟ =
2
32 32 ⎝ 3 ⎠
⎝8⎠
⎛ 3⎞
⎜ ⎟
82
⎝8⎠
−7
2
1
77
⎛ 7⎞
⎛ 4⎞
(ii) ⎜ − ⎟ =
=
= ⎜− ⎟
7
7
(− 4) ⎝ 4 ⎠
⎝ 7⎠
⎛ 4⎞
⎜− ⎟
⎝ 7⎠
7
From the above example, we get the following result:
If
p
is any non-zero rational number and m is any positive integer, then
q
⎛ p⎞
⎜⎜ ⎟⎟
⎝q⎠
50
−m
m
qm ⎛ q ⎞
= m = ⎜⎜ ⎟⎟ .
p
⎝ p⎠
Mathematics Secondary Course
MODULE - 1
Exponents and Radicals
Algebra
2.5 LAWS OF EXPONENTS FOR INTEGRAL EXPONENTS
After giving a meaning to negative integers as exponents of non-zero rational numbers, we
can see that laws of exponents hold good for negative exponents also.
Notes
For example.
−4
3
3
1
⎛3⎞ 3
⎛3⎞
⎛3⎞
×⎜ ⎟ =
(i) ⎜ ⎟ × ⎜ ⎟ =
4
⎝5⎠
⎝5⎠ ⎛3⎞ ⎝5⎠ 5
⎜ ⎟
⎝5⎠
−2
−3
−3
−7
3– 4
1
1
1
⎛ 2⎞
⎛ 2⎞
⎛ 2⎞
×
=
= ⎜− ⎟
(ii) ⎜ − ⎟ × ⎜ − ⎟ =
2
3
2+3
⎝ 3⎠
⎝ 3⎠
⎝ 3⎠
⎛ 2⎞ ⎛ 2⎞
⎛ 2⎞
⎜− ⎟ ⎜− ⎟
⎜− ⎟
⎝ 3⎠ ⎝ 3⎠
⎝ 3⎠
⎛ 3⎞
⎛ 3⎞
(iii) ⎜ − ⎟ ÷ ⎜ − ⎟
⎝ 4⎠
⎝ 4⎠
3
− 2 −3
7
⎛ 3⎞ ⎛ 3⎞
=
÷
=
×⎜− ⎟ = ⎜− ⎟
3
7
3
⎛ 3⎞ ⎛ 3⎞
⎛ 3⎞ ⎝ 4⎠ ⎝ 4⎠
⎜− ⎟ ⎜− ⎟
⎜− ⎟
⎝ 4⎠ ⎝ 4⎠
⎝ 4⎠
1
1
1
7 −3
3
⎛ ⎛ 2 ⎞ − 2 ⎞ ⎡⎛ 7 ⎞ 2 ⎤ ⎛ 7 ⎞ 6 ⎛ 2 ⎞ −6 ⎛ 2 ⎞ − 2×3
(iv) ⎜ ⎜ ⎟ ⎟ = ⎢⎜ ⎟ ⎥ = ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟
⎜ ⎝ 7 ⎠ ⎟ ⎢⎝ 2 ⎠ ⎥ ⎝ 2 ⎠ ⎝ 7 ⎠
⎝7⎠
⎝
⎠ ⎣
⎦
Thus, from the above results, we find that laws 1 to 5 hold good for negative exponents
also.
∴ For any non-zero rational numbers a and b and any integers m and n,
1. am × an
= am+n
2. am ÷ an
= am–n if m > n
= an–m if n > m
3. (am)n = amn
4. (a × b)m = am × bm
CHECK YOUR PROGRESS 2.4
−2
p
⎛ −3⎞
⎟ as a rational number of the form :
1. Express ⎜
q
⎝ 7 ⎠
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Algebra
2. Express as a power of rational number with positive exponent:
Notes
⎛3⎞
(i) ⎜ ⎟
⎝7⎠
⎡⎛ 3 ⎞ −3 ⎤
(iii) ⎢⎜ ⎟ ⎥
⎢⎣⎝ 13 ⎠ ⎥⎦
−4
(ii) 125 × 12 −3
4
3. Express as a power of a rational number with negative index:
⎛3⎞
(i) ⎜ ⎟
⎝7⎠
4
⎡⎛ 3 ⎞ 2 ⎤
(iii) ⎢⎜ − ⎟ ⎥
⎣⎢⎝ 4 ⎠ ⎦⎥
[ ]
(ii) (7 )2
5
5
4. Simplify:
−3
⎛3⎞
⎛3⎞
(i) ⎜ ⎟ × ⎜ ⎟
⎝2⎠
⎝2⎠
−3
7
⎛ 2⎞
⎛ 2⎞
(ii) ⎜ − ⎟ × ⎜ − ⎟
⎝ 3⎠
⎝ 3⎠
4
−4
⎛ 7⎞
⎛ 7⎞
(iii) ⎜ − ⎟ ÷ ⎜ − ⎟
⎝ 5⎠
⎝ 5⎠
−7
5. Which of the following statements are true?
(i) a–m × an = a–m–n
(ii) (a–m)n = a–mn
(iii) am × bm = (ab)m
⎛a⎞
(iv) a ÷ b = ⎜ ⎟
⎝b⎠
m
m
m
(v) a–m × ao = am
2.6 MEANING OF ap/q
You have seen that for all integral values of m and n,
am × an = am+n
What is the method of defining a1/q, if a is positive rational number and q is a natural
number.
Consider the multiplication
1
q
1
q
1
q
1
q
a × a × a ........× a = a
1 1 1
+ + + .....q times
q q q
q times
q
= aq = a
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Mathematics Secondary Course
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Algebra
1
In other words, the qth power of a q = a
or
1
in other words a q is the qth root of a and is written as q a .
Notes
For example,
1
1
1
1
1 1 1 1
+ + +
4 4 4
74 ×74 ×74 ×74 = 74
4
= 7 4 = 71 = 7
1
or 7 4 is the fourth root of 7 and is written as 4 7 ,
Let us now define rational powers of a
If a is a positive real number, p is an integer and q is a natural number, then
p
q
aq = ap
We can see that
p
q
p
q
p
q
p
q
a × a × a ........× a = a
p p p
+ + + .....q times
q q q
=a
p
.q
q
= ap
q times
p
q
q
∴a = a p
∴ ap/q is the qth root of ap
2
Consequently, 7 3 is the cube root of 72.
Let us now write the laws of exponents for rational exponents:
(i) am × an = am+n
(ii) am ÷ an = am–n
(iii) (am)n = amn
(iv) (ab)m = am bm
m
am
⎛a⎞
(v) ⎜ ⎟ = m
b
⎝b⎠
Let us consider some examples to verify the above laws:
Example 2.15: Find the value of
(i) (625)
1
4
(ii) (243)
Mathematics Secondary Course
2
5
⎛ 16 ⎞
(iii) ⎜ ⎟
⎝ 81 ⎠
−3 / 4
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MODULE - 1
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Algebra
Solution:
(i)
Notes
1
(625)4 = (5 × 5 × 5 × 5) 4 = (54 )4 = 5
1
1
2
2
−3
−3
( )
(ii) (243)5 = (3 × 3 × 3 × 3 × 3) 5 = 35
2
5
4×
1
4
=5
5×
=3
2
5
= 32 = 9
⎛ 16 ⎞ 4 ⎛ 2 × 2 × 2 × 2 ⎞ 4
(iii) ⎜ ⎟ = ⎜
⎟
⎝ 81 ⎠
⎝ 3× 3× 3× 3 ⎠
−3
⎛ −3 ⎞
⎡⎛ 2 ⎞ 4 ⎤ 4 ⎛ 2 ⎞ 4×⎜⎝ 4 ⎟⎠ ⎛ 2 ⎞ −3 ⎛ 3 ⎞ 3 27
=⎜ ⎟ =⎜ ⎟ =
= ⎢⎜ ⎟ ⎥ = ⎜ ⎟
8
⎝3⎠
⎝3⎠
⎝2⎠
⎣⎢⎝ 3 ⎠ ⎦⎥
CHECK YOUR PROGRESS 2.5
1. Simplify each of the following:
(i) (16)
3
⎛ 27 ⎞
(ii) ⎜
⎟
⎝ 125 ⎠
4
−
2
3
2. Simplify each of the following:
1
1
(i) (625)− 4 ÷ (25)− 2
−
1
1
3
⎛ 7 ⎞ 4 ⎛ 7 ⎞2 ⎛ 7 ⎞4
(ii) ⎜ ⎟ × ⎜ ⎟ × ⎜ ⎟
⎝8⎠
⎝8⎠ ⎝8⎠
−
3
1
3
⎛ 13 ⎞ 4 ⎛ 13 ⎞ 4 ⎛ 13 ⎞ 2
(iii) ⎜ ⎟ × ⎜ ⎟ × ⎜ ⎟
⎝ 16 ⎠
⎝ 16 ⎠ ⎝ 16 ⎠
2.7 SURDS
We have read in first lesson that numbers of the type 2 , 3 and 5 are all irrational
numbers. We shall now study irrational numbers of a particular type called radicals or
surds.
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Algebra
A surd is defined as a positive irrational number of the type n x , where it is not possible to
find exactly the nth root of x, where x is a positive rational number.
The number
x is a surd if and only if
n
Notes
(i) it is an irrational number
(ii) it is a root of the positive rational number
2.7.1 Some Terminology
In the surd n x , the symbol
is called a radical sign. The index ‘n’ is called the order
of the surd and x is called the radicand.
Note: i)
When order of the surd is not mentioned, it is taken as 2. For example, order
of
(
)
7 = 2 7 is 2.
ii)
3
iii)
2 + 2 , although an irrational number, is not a surd because it is the square
root of an irrational number.
8 is not a surd as its value can be determined as 2 which is a rational.
2.8 PURE AND MIXED SURD
i) A surd, with rational factor is 1 only, other factor being rrational is called a pure surd.
For example, 5 16 and 3 50 are pure surds.
ii) A surd, having rational factor other than 1 alongwith the irrational factor, is called a
mixed surd.
For example, 2 3 and 3 3 7 are mixed surds.
2.9 ORDER OF A SURD
In the surd 53 4 , 5 is called the co-efficient of the surd, 3 is the order of the surd and 4
is the radicand. Let us consider some examples:
Example 2.16: State which of the following are surds?
(i) 49
(ii) 96
Mathematics Secondary Course
(iii) 3 81
(iv) 3 256
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Algebra
Solution:
(i)
49 = 7, which is a rational number.
∴
49 is not a surd.
Notes
96 = 4 × 4 × 6 = 4 6
(ii)
∴ 96 is an irrational number.
⇒ 96 is a surd.
(iii)
3
81 = 3 3 × 3 × 3 × 3 = 33 3 , which is irrational
∴
(iv)
3
3
81 is a surd.
256 = 3 4 × 4 × 4 × 4 = 43 4
∴ 3 256 is irrational.
⇒
3
256 is a surd
∴ (ii), (iii) and (iv) are surds.
Example 2.17: Find “index” and “radicand” in each of the following:
(i) 5 117
(ii) 162
Solution:
(iii) 4 213
(i)
index is 5 and radicand is 117.
(ii)
index is 2 and radicand is 162.
(iii)
index is 4 and radicand is 213.
(iv)
index is 4 and radicand is 214.
(iv) 4 214
Example 2.18: Identify “pure” and “mixed” surds from the following:
(i) 42
(ii) 4 3 18
Solution:
(i)
(iii) 2 4 98
42 is a pure surd.
(ii) 4 3 18 is a mixed surd.
(iii) 2 4 98 is a mixed surd.
2.10 LAWS OF RADICALS
Given below are Laws of Radicals: (without proof):
(i)
56
[ a] = a
n
n
Mathematics Secondary Course
MODULE - 1
Exponents and Radicals
Algebra
(ii)
(iii)
n
a n b = n ab
n
a n a
=
b
b
n
Notes
where a and b are positive rational numbers and n is a positive integer.
Let us take some examples to illustrate.
Example 2.19: Which of the following are surds and which are not? Use laws of radicals
to ascertain.
(i) 5 × 80
(ii) 2 15 ÷ 4 10
(iii) 3 4 × 3 16
(iv) 32 ÷ 27
Solution:
5 × 80 = 5 × 80 = 400 = 20 .
(i)
which is a rational number.
∴ 5 × 80 is not a surd.
(ii)
2 15 ÷ 4 10 =
=
2 15
15
=
4 10 2 10
15
15
3
=
=
, which is irrational.
8
2 × 2 ×10
40
∴ 2 15 ÷ 4 10 is a surd.
(iii)
(iv)
3
4 × 3 16 = 3 64 = 4 ⇒ It is not a surd.
32 ÷ 27 =
32
32
=
, which is irrational
27
27
∴ 32 ÷ 27 is a surd.
CHECK YOUR PROGRESS 2.6
1. For each of the following, write index and the radicand:
(i) 4 64
(ii) 6 343
Mathematics Secondary Course
(iii) 119
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Algebra
2. State which of the following are surds:
Notes
(i) 3 64
(ii) 4 625
(iv) 5 × 45
(v) 3 2 × 5 6
(iii) 6 216
3. Identify pure and mixed surds out of the following:
(i) 32
(ii) 2 3 12
(iii) 13 3 91
(iv) 35
2.11 LAWS OF SURDS
Recall that the surds can be expressed as numbers with fractional exponents. Therefore,
laws of indices studied in this lesson before, are applicable to them also. Let us recall them
here:
(i)
1
n
1
n
x.
n
x
x
xn ⎛ x ⎞n
or 1 = ⎜⎜ ⎟⎟
=n
y
y
y
yn ⎝ ⎠
n
n
1
(ii)
n
1
y = xy or x .y = ( xy )n
n
1
1
1
x = mn x = n
1
⎛ 1 ⎞m
⎛ 1 ⎞n
x or ⎜⎜ x n ⎟⎟ = x mn = ⎜⎜ x m ⎟⎟
⎝ ⎠
⎝ ⎠
(iii)
m n
(iv)
n
x m = x n or x m
(v)
n
x p = mn x pn or x p
m
m
( )
1
n
m
= xn
( )
1
m
p
pn
( )
= x m = x mn = x pn
1
mn
Here, x and y are positive rational numbers and m, n and p are positive integers.
Let us illustrate these laws by examples:
(i)
3
(ii)
(5)3
1
(9)3
3
3
1
58
1
3
1
3
1
8 = 3 × 8 = (24)3 = 3 24 = 3 3 × 8
1
5
⎛ 5 ⎞3
=⎜ ⎟ =3
9
⎝9⎠
Mathematics Secondary Course
MODULE - 1
Exponents and Radicals
Algebra
1
(iii)
3 2
1
⎛ 1 ⎞3
7 = 7 = ⎜⎜ 7 2 ⎟⎟ = 7 6 = 6 7 = 2×3 7 = 2
⎝ ⎠
1
2
3
3
7
Notes
(iv)
5
3
5
( )
43 = 4
1
3 5
9
15
= 4 = 4 = 15 49 = 3×5 43×3
Thus, we see that the above laws of surds are verified.
An important point: The order of a surd can be changed by multiplying the index of the
surd and index of the radicand by the same positive number.
For example
and
3
2 = 6 22 = 6 4
4
3 = 8 32 = 8 9
2.12 SIMILAR (OR LIKE) SURDS
Two surds are said to be similar, if they can be reduced to the same irrational factor,
without consideration for co-efficient.
For example, 3 5 and 7 5 are similar surds. Again consider
12 = 2 3 . Now
similar surds.
75 and
75 = 5 3 and
12 are expressed as 5 3 and 2 3 . Thus, they are
2.13 SIMPLEST (LOWEST) FORM OF A SURD
A surd is said to be in its simplest form, if it has
a) smallest possible index of the sign
b) no fraction under radical sign
c) no factor of the form an, where a is a positive integer, under the radical sign of index n.
For example,
3
125 3 125 ×12 5 3
12
=
=
18
18 ×12 6
Let us take some examples.
Example 2.20: Express each of the following as pure surd in the simplest form:
(i) 2 7
(ii) 44 7
Mathematics Secondary Course
(iii)
3
32
4
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MODULE - 1
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Algebra
Solution:
(i) 2 7 = 2 2 × 7 = 4 × 7 = 28 , which is a pure surd.
Notes
(ii) 44 7 = 4 4 4 × 7 = 4 256 × 7 = 4 1792 , which is a pure surd.
(iii)
9
3
32 = 32 × = 18 , which is a pure surd.
16
4
Example 2.21: Express as a mixed surd in the simplest form:
(i) 128
(ii) 6 320
(iii) 3 250
Solution:
(i) 128 = 64 × 2 = 8 2 ,
which is a mixed surd.
(ii)
6
320 = 6 2 × 2 × 2 × 2 × 2 × 2 × 5
= 6 26 × 5 = 26 5 , which is a mixed surd.
(iii)
3
250 = 3 5 × 5 × 5 × 2 = 53 2 , which is a mixed surd.
CHECK YOUR PROGRESS 2.7
1. State which of the following are pairs of similar surds:
(i) 8 , 32
(ii) 5 3 ,6 18
(iii) 20 , 125
2. Express as a pure surd:
(i) 7 3
(ii) 3 3 16
(iii)
5
24
8
3. Express as a mixed surd in the simplest form:
(i) 3 250
(ii) 3 243
(iii) 4 512
2.14 FOUR FUNDAMENTAL OPERATIONS ON SURDS
2.14.1 Addition and Subtraction of Surds
As in rational numbers, surds are added and subtracted in the same way.
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Algebra
For example,
and
5 3 + 17 3 = (5 + 17 ) 3 = 22 3
12 5 − 7 5 = [12 − 7 ] 5 = 5 5
For adding and subtracting surds, we first change them to similar surds and then perform
the operations.
For example
Notes
i) 50 + 288
=
5 × 5 × 2 + 12 ×12 × 2
= 5 2 + 12 2 = 2 (5 + 12 ) = 17 2
ii) 98 − 18
=
7 × 7 × 2 − 3× 3× 2
= 7 2 − 3 2 = (7 − 3) 2 = 4 2
Example 2.22: Simplify each of the following:
(i) 4 6 + 2 54
(ii) 45 6 − 3 216
Solution:
(i) 4 6 + 2 54
= 4 6 + 2 3× 3× 6
= 4 6 + 6 6 = 10 6
(ii) 45 6 − 3 216
= 45 6 − 3 6 × 6 × 6
= 45 6 − 18 6
= 27 6
Example 2.23: Show that
24 45 − 16 20 + 245 − 47 5 = 0
Solution:
24 45 − 16 20 + 245 − 47 5
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Algebra
= 24 3 × 3 × 5 − 16 2 × 2 × 5 + 7 × 7 × 5 − 47 5
= 72 5 − 32 5 + 7 5 − 47 5
Notes
=
5 [72 − 32 + 7 − 47 ]
=
5 × 0 = 0 = RHS
Example 2.24: Simplify: 23 16000 + 83 128 − 33 54 + 4 32
Solution:
23 16000 = 23 10 ×10 ×10 × 8 × 2 = 2 ×10 × 23 2 = 403 2
83 128 = 83 4 × 4 × 4 × 2 = 323 2
33 54 = 33 3 × 3 × 3 × 2 = 93 2
4
32 = 24 2
∴ Required expression
= 403 2 + 323 2 − 93 2 + 24 2
= (40 + 32 − 9)3 2 + 24 2
= 633 2 + 24 2
CHECK YOUR PROGRESS 2.8
Simplify each of the following:
1.
175 + 112
2.
32 + 200 + 128
3. 3 50 + 4 18
108 − 75
4.
5.
3
24 + 3 81 − 83 3
6. 63 54 − 23 16 + 43 128
7. 12 18 + 6 20 − 6 147 + 3 50 + 8 45
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Algebra
2.14.2 Multiplication and Division in Surds
Two surds can be multiplied or divided if they are of the same order. We have read that the
order of a surd can be changed by multiplying or dividing the index of the surd and index
of the radicand by the same positive number. Before multiplying or dividing, we change
them to the surds of the same order.
Notes
Let us take some examples:
[ 3 and
3 × 2 = 3× 2 = 6
12 ÷ 2 =
2 are of same order
]
12
= 6
2
Let us multiply 3 and 3 2
3 = 6 33 = 6 27
3
2 =6 4
∴ 3 × 3 2 = 6 27 × 6 4 = 6 108
and
3
3 6 27 6 27
=
=
4
2 64
Let us consider an example:
Example 2.25:(i) Multiply 53 16 and 113 40 .
(ii) Divide 153 13 by 66 5 .
Solution:
(i) 53 16 × 113 40
= 5 ×11× 3 2 × 2 × 2 × 2 × 3 2 × 2 × 2 × 5
= 55 × 2 × 23 2
3
5
= 220 3 10
153 13 5 6 132 5 6 169
= . 6
=
(ii) 6
2
2 5
6 5
5
Example 2.26: Simplify and express the result in simplest form:
2 50 × 32 × 2 72
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Algebra
2 50 = 2 5 × 5 × 2 = 10 2
Solution:
32 = 2 × 2 × 2 × 2 × 2 = 4 2
Notes
2 72 = 2 × 6 2 = 12 2
∴ Given expression
= 10 2 × 4 2 × 12 2
= 960 2
2.15 COMPARISON OF SURDS
To compare two surds, we first change them to surds of the same order and then compare
their radicands along with their co-efficients. Let us take some examples:
Example 2.27: Which is greater
1
or
4
3
1
?
3
3
1 6 ⎛1⎞
1
= ⎜ ⎟ =6
64
4
⎝4⎠
Solution:
3
1 6 1
=
3
9
1 1
1
1
1
1
>
⇒6 >6
⇒3 >
9 64
9
64
3
4
Example 2.28: Arrange in ascending order: 3 2 ,
Solution:
3 and
6
5.
LCM of 2, 3, and 6 is 6.
∴ 3 2 = 6 22 = 6 4
3 = 6 33 = 6 27
6
Now
6
5=6 5
4 < 6 5 < 6 27
⇒3 2<6 5< 3
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Algebra
CHECK YOUR PROGRESS 2.9
1. Multipliy 3 32 and 53 4 .
Notes
2. Multipliy 3 and 3 5 .
3. Divide 3 135 by
3
5.
4. Divide 2 24 by
3
320 .
5. Which is greater 4 5 or 3 4 ?
6. Which in smaller: 5 10 or 4 9 ?
7. Arrange in ascending order:
3
2,
6
3,
3
4
8. Arrange in descending order:
3
2,
4
3,
3
4
2.16 RATIONALISATION OF SURDS
Consider the products:
1
1
(i) 3 2 × 3 2 = 3
7
4
(ii) 511 × 511 = 5
1
3
(iii) 7 4 × 7 4 = 7
In each of the above three multiplications, we see that on multiplying two surds, we get the
result as rational number. In such cases, each surd is called the rationalising factor of the
other surd.
(i) 3 is a rationalising factor of 3 and vice-versa.
(ii)
11
54 is a rationalising factor of 11 57 and vice-versa.
(iii)
4
7 is a rationalising factor of 4 73 and vice-versa.
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In other words, the process of converting surds to rational numbers is called rationalisation
and two numbers which on multiplication give the rational number is called the rationalisation
factor of the other.
Notes
For example, the rationalising factor of
x , of 3 + 2 is 3 − 2 .
x is
Note:
(i) The quantities x − y and x + y are called conjugate surds. Their sum and product
are always rational.
(ii) Rationalisation is usually done of the denominator of an expression involving irrational
surds.
Let us consider some examples.
Example 2.29: Find the rationalising factors of 18 and 12 .
Solution:
18 =
3× 3× 2 = 3 2
∴ Rationalising factor is 2 .
12 =
2× 2×3 = 2 3 .
∴ Rationalising factor is 3 .
Example 2.30: Rationalise the denominator of
2+ 5
=
2− 5
Solution:
=−
(
(
2+ 5
2− 5
)(
)(
) (
)
2+ 5
=
2+ 5
(
(
66
)(
)(
4+3 5 4+3 5
4+3 5
=
4−3 5 4+3 5
4−3 5
=
2+ 5
−3
)
2
7 + 2 10
7 2
=− −
10
3
3 3
Example 2.31: Rationalise the denominator of
Solution:
2+ 5
.
2− 5
4+3 5
.
4−3 5
)
)
16 + 45 + 24 5
61 24
=− −
5
16 − 45
29 29
Mathematics Secondary Course
MODULE - 1
Exponents and Radicals
Algebra
1
Example 2.32: Rationalise the denominator of
1
Solution:
3 − 2 +1
=
[(
=
(
(
Solution:
) ][(
)
Notes
) ]
3 − 2 −1
3 − 2 −1
2
.
3 − 2 −1
=
3 − 2 −1
4−2 6
3 − 2 −1 4 + 2 6
×
4−2 6
4+2 6
4 3 −4 2 −4+6 2 −4 3 −2 6
16 − 24
=−
Example 2.33: If
)
3 − 2 −1
3 − 2 +1
=
=
3 − 2 +1
2 −2− 6
6− 2+2
=
4
4
3+ 2 2
= a + b 2 , find the values of a and b.
3− 2
3+ 2 2 3+ 2 2 3+ 2 9+ 4+9 2
=
×
=
9−2
3− 2
3− 2 3+ 2
=
13 + 9 2 13 9
= +
2 =a+b 2
7
7 7
⇒a=
13
,
7
b=
9
7
CHECK YOUR PROGRESS 2.10
1. Find the rationalising factor of each of the following:
(i) 3 49
(ii) 2 + 1
(iii) 3 x 2 + 3 y 2 + 3 xy
2. Simplify by rationalising the denominator of each of the following:
(i)
12
5
(ii)
2 3
17
Mathematics Secondary Course
(iii)
11 − 5
11 + 5
(iv)
3 +1
3 −1
67
MODULE - 1
Exponents and Radicals
Algebra
3. Simplify:
2+ 3 2− 3
+
2− 3 2+ 3
Notes
4. Rationalise the denominator of
1
3 − 2 −1
1
5. If a = 3 + 2 2 . Find a + .
a
2+5 7
= x + 7 y , find x and y.
2−5 7
______________________________________________________________
6. If
LET US SUM UP
•
a × a × a × ..... m times = am is the exponential form, where a is the base and m is the
exponent.
•
Laws of exponent are:
m
m
n
m+n
(i) a × a = a
( )
(v) a m
p
q
= a mn
(iii) (ab) = a b
(vi) ao = 1
(vii) a
n
m–n
m
−m
=
m m
am
⎛a⎞
(iv) ⎜ ⎟ = m
b
⎝b⎠
1
am
•
a = ap
•
An irrational number n x is called a surd, if x is a rational number and nth root of x is
not a rational number.
•
In
•
A surd with rational co-efficient (other than 1) is called a mixed surd.
•
The order of the surd is the number that indicates the root.
•
The order of
•
Laws of radicals (a > 0, b > 0)
(i)
68
n
(ii) a ÷ a = a
m
q
n
x , n is called index and x is called radicand.
[ a] = a
n
n
n
x is n
n
(ii) a × b = ab
n
n
n
(iii)
n
a n a
=
b
b
Mathematics Secondary Course
MODULE - 1
Exponents and Radicals
Algebra
•
Operations on surds
⎛
x × y = ( xy ) ; ⎜⎜ x
⎝
1
n
(x )
1
m n
1
n
1
n
m
n
=x ;
m
x =
a
1
n
1
m
⎞
⎟ =x
⎟
⎠
mn
x
an
1
mn
1
1
n
1
⎛ x ⎞n
⎛ ⎞
=
= ⎜⎜ x ⎟⎟ ; 1 ⎜⎜ y ⎟⎟
yn ⎝ ⎠
⎝ ⎠
1
m
( )
or x
1
a m
xn
a
m
=x =x
an
mn
Notes
( )
= x an
1
mn
•
Surds are similar if they have the same irrational factor.
•
Similar surds can be added and subtracted.
•
Orders of surds can be changed by multiplying index of the surds and index of the
radicand by the same positive number.
•
Surds of the same order are multiplied and divided.
•
To compare surds, we change surds to surds of the same order. Then they are compared
by their radicands alongwith co-efficients.
•
If the product of two surds is rational, each is called the rationalising factor of the
other.
•
x + y is called rationalising factor of x − y and vice-versa.
TERMINAL EXERCISE
1. Express the following in exponential form:
(i) 5 × 3 × 5 × 3 × 7 × 7 × 7 × 9 × 9
⎛−7⎞ ⎛−7⎞ ⎛−7⎞ ⎛−7⎞
⎟×⎜
⎟×⎜
⎟×⎜
⎟
(ii) ⎜
⎝ 9 ⎠ ⎝ 9 ⎠ ⎝ 9 ⎠ ⎝ 9 ⎠
2. Simplify the following:
3
2
⎛ 5⎞ ⎛7⎞ ⎛3⎞
(i) ⎜ − ⎟ × ⎜ ⎟ × ⎜ ⎟
⎝ 6⎠ ⎝5⎠ ⎝7⎠
2
⎛ 3 ⎞ 35 ⎛ 1 ⎞
(ii) ⎜ ⎟ × × ⎜ − ⎟
⎝ 7 ⎠ 27 ⎝ 5 ⎠
3
2
3. Simplify and express the result in exponential form:
(i) (10 )2 × (6 )2 × (5)2
Mathematics Secondary Course
69
MODULE - 1
Exponents and Radicals
Algebra
20
⎛ 37 ⎞
⎛ 37 ⎞
(ii) ⎜ − ⎟ ÷ ⎜ − ⎟
⎝ 19 ⎠
⎝ 19 ⎠
Notes
⎡⎛ 3 ⎞3 ⎤
(iii) ⎢⎜ ⎟ ⎥
⎢⎣⎝ 13 ⎠ ⎥⎦
20
5
4. Simplify each of the following:
(i) 3o + 7o + 37o – 3
(ii) ( 7o + 3o) ( 7o – 3o)
5. Simplify the following:
−3
(i) (32 ) ÷ (32 )
12
(ii) (111) × (111)
−6
−5
6
−3
11
⎛3⎞
⎛3⎞
⎛3⎞
6. Find x so that ⎜ ⎟ × ⎜ ⎟ = ⎜ ⎟
⎝7⎠
⎝7⎠
⎝7⎠
−2
⎛ 2⎞
⎛ 2⎞
(iii) ⎜ − ⎟ × ⎜ − ⎟
⎝ 9⎠
⎝ 9⎠
5
x
−9
⎛3⎞
⎛3⎞
⎛3⎞
7. Find x so that ⎜ ⎟ × ⎜ ⎟ = ⎜ ⎟
⎝ 13 ⎠
⎝ 13 ⎠
⎝ 13 ⎠
2 x +1
8. Express as a product of primes and write the answers of each of the following in
exponential form:
(i) 6480000
(ii) 172872
(iii) 11863800
9. The star sirus is about 8.1 × 1013 km from the earth. Assuming that the light travels at
3.0 × 105 km per second, find how long light from sirus takes to reach earth.
10. State which of the following are surds:
(i)
36
289
(iii) 3
(ii) 9 729
5 +1
(iv) 4 3125
11. Express as a pure surd:
(i) 32 3
(ii) 53 4
(iii) 55 2
12. Express as a mixed surd in simplest form:
(i) 4 405
(ii) 5 320
(iii) 3 128
13. Which of the following are pairs of similar surds?
(i) 112 , 343
70
(ii) 3 625 , 3 3125 × 25
(iii) 6 216 , 250
Mathematics Secondary Course
MODULE - 1
Exponents and Radicals
Algebra
14.Simplify each of the following:
(i) 4 48 −
5 1
+6 3
2 3
Notes
(ii) 63 + 28 − 175
(iii)
8 + 128 − 50
15. Which is greater?
(i)
2 or
3
3
(ii) 3 6 or 4 8
16. Arrange in descending order:
(i) 3 , 3 4 , 4 5
(ii) 2 , 3 , 3 4
17. Arrange in ascending order:
3
16 , 12 , 6 320
18. Simplify by rationalising the denominator:
(i)
3
6− 7
(ii)
12
7− 3
(iii)
5 −2
5+2
19. Simplify each of the following by rationalising the denominator:
(i)
20. If
1
1+ 2 − 3
(ii)
1
7 + 5 − 12
5+ 2 3
= a + b 3, find the values of a and b, where a and b are rational numbers.
7+4 3
1
21. If x = 7 + 4 3 , find the value of x + .
x
ANSWERS TO CHECK YOUR PROGRESS
2.1
10
4
1. (i) (–7)
⎛3⎞
(ii) ⎜ ⎟
⎝4⎠
Mathematics Secondary Course
⎛ −5⎞
⎟
(iii) ⎜
⎝ 7 ⎠
20
71
MODULE - 1
Exponents and Radicals
Algebra
2.
Notes
Base
Exponent
(i) – 3
5
(ii) 7
4
2
11
(iii) −
8
3. (i)
81
2401
(ii)
16
6561
4. (i)
3
7
(ii)
625
324
⎛1⎞
5. (i) ⎜ ⎟
⎝3⎠
5
⎛ 1⎞
(ii) ⎜ − ⎟
⎝ 7⎠
(iii) −
4
27
64
4
⎛ 5⎞
(iii) ⎜ − ⎟
⎝ 3⎠
2.2
1. (i) 31 × 111 × 131
(ii) 23 × 34
(iii) 23 × 33 × 71
2. (i) 36
(ii) 29
(iii) 25 × 34
113
(iv) 12
2
(v)
(− 7 )3
25
2.3
5
1. (i) (7)
2
2. (i) (–7)
18
3. (i) 2
⎛ 11 ⎞
(iv) ⎜ ⎟
⎝3⎠
5
⎛3⎞
(ii) ⎜ ⎟
⎝4⎠
5
⎛ 7⎞
(iii) ⎜ − ⎟
⎝ 8⎠
⎛3⎞
(ii) ⎜ ⎟
⎝4⎠
6
⎛ 7⎞
(iii) ⎜ − ⎟
⎝ 8⎠
⎛3⎞
(ii) ⎜ ⎟
⎝4⎠
6
⎛ 5⎞
(iii) ⎜ − ⎟
⎝ 9⎠
⎛ 7⎞
(v) ⎜ − ⎟
⎝ 11 ⎠
6
15
15
3
4. True: (i), (ii), (vii)
False: (iii), (iv), (v), (vi)
72
Mathematics Secondary Course
MODULE - 1
Exponents and Radicals
Algebra
2.4
1.
49
9
Notes
⎛7⎞
2. (i) ⎜ ⎟
⎝3⎠
4
⎛7⎞
3. (i) ⎜ ⎟
⎝3⎠
−4
4. (i)
12
⎛ 13 ⎞
(iii) ⎜ ⎟
⎝3⎠
2
(ii) 12
81
16
⎛1⎞
(ii) ⎜ ⎟
⎝7⎠
(ii) −
−10
2
3
⎛ 4⎞
(iii) ⎜ − ⎟
⎝ 3⎠
(iii) −
−10
343
125
5. True: (ii), (iii), (iv)
2.5
1. (i) 8
(ii)
25
9
2. (i) 1
(ii)
7
8
(iii)
13
16
2.6
1. (i) 4, 64
(ii) 6, 343
(iii) 2, 119
2. (i) 147
(ii) 3 432
(iii)
3. (i) 53 2
(ii) 33 9
(iii) 44 2
2. 22 2
3. 27 2
2. (iii), (iv)
3. Pure: (i), (iv)
Mixed: (ii), (iii)
2.7
1. (i), (iii)
75
8
2.8
1. 9 7
Mathematics Secondary Course
4.
3
73
MODULE - 1
Exponents and Radicals
Algebra
5. − 3 3
7. 51 2 + 36 5 − 42 3
6. 303 2
2.9
Notes
1. 203 2
2. 33 5
3. 3
5.
6.
7.
3
4
4
9
6
3, 3 2 , 3 4
4. 6
216
25
8.
4 , 4 3, 3 2
3
2.10
12
5
5
2. (i)
(iii) 3 x − 3 y
(ii) 2 − 1
1. (i) 3 7
(ii)
2 51
17
(iii)
8
55
−
3
3
(iv) 2 + 3
3. 14
4. −
[
1
2+ 6 + 2
4
]
5. 6
6. −
179 20 7
−
171 171
ANSWERS TO TERMINAL EXERCISE
1. (i) 52 × 32 × 73 × 92
2. (i) −
5
56
⎛ 7⎞
(ii) ⎜ − ⎟
⎝ 9⎠
(ii)
4
1
105
15
4
2
3. (i) 2 × 3 × 5
4. (i) zero
(ii) 1
⎛3⎞
(iii) ⎜ ⎟
⎝ 13 ⎠
(ii) zero
18
5. (i) (32)
74
4
(ii) 111
⎛2⎞
(iii) ⎜ ⎟
⎝9⎠
2
Mathematics Secondary Course
MODULE - 1
Exponents and Radicals
Algebra
6. x = 8
7. x = – 6
8. 27 × 34 × 54
Notes
9. 33 × 107 seconds
10. (ii), (iii), (iv)
11. (i) 2 27
(ii) 3 500
(iii) 5 6250
12. (i) 34 5
(ii) 25 10
(iii) 43 2
(ii) zero
(iii) 5 2
13. (i), (ii)
14. (i)
127
3
6
15. (i) 3 3
(ii) 3 6
16. (i) 3 , 3 4 , 4 5
(ii) 3 , 3 4 , 2
17. 3 16 , 6 320 , 12
(
18. (i) − 3 6 + 7
19. (i)
2+ 2 + 6
4
)
(
(ii) 3 7 + 3
(ii)
)
(ii) 9 − 4 5
7 5 + 5 7 + 2 105
70
20. a = 11, b = –6
21. 14
Mathematics Secondary Course
75