Al-khowarizmi - Develop Schools

7
Types of surds - Like and
unlike surds
Surds
This unit facilitates you in,
Binomial surds
identifying like surds and unlike surds.
Operation on surds addition,
subtraction and multiplication
defining binomial surd.
Multiplication of binomial
surds.
Rationalisation of surds
Rationalising the denominator
adding and subtracting like surds.
finding the product of given surds.
solving problems related to addition,
subtraction and multiplication of surds.
finding the product of binomial surds.
explaining the process of rationalisation of
surds.
rationalising the given surds.
simplifying the surds by rationalising the
denominator.
Mathematics is concerned only with the
Al-khowarizmi
(about 780 – 850 AD)
Al-khowarizmi spoke of rational
numbers as "audible" and of
surds as "inaudible" and it is the
later that gave rise to the word
"surd" (deaf, mute). The term
was also used by Fibonacci, but
to represent a number that has
no root.
enumeration and comparision of relations.
-Carl Friedrich Gauss
154
UNIT-7
Like and unlike surds
We have already learnt about surds. Observe the given surds, their order and radicand in
the table.
Sl. No
Surd
Simplest form
Order
Radicand
1.
3
3
2
3
2.
12
2 3
2
3
3.
75
5 3
2
3
4.
3 27
9 3
2
3
5.
2 3x 2
2x 3
2
3
Compare the orders and radicands of the surds in their simplest form.
All the above surds, in their simplest form have the same order and same radicand.
Such surds are called like surds.
A group of surds having same order and same radicand in their simplest form are
called Like surds.
Now, observe the given groups of surds and complete the table.
Group
Surds
1.
8, 12, 20, 54
2.
50, 3 54, 4 32
3.
18, 3 24, 4 64, 5 192
Simplest
form
same
Order
Radicand
different same
different
In the above table, we observe that,
in group 1 : the orders are same, radicands are different.
in group 2 : the orders are different, radicands are same.
in group 3 : the orders are different, radicands are different.
Such group of surds are called unlike surds.
Groups of surds having different orders or different radicands or both in their
simplest form are called unlike surds.
Binomial Surds
Study the following examples :
1.
3
5
Sum of two surds
Surds
155
2.
3
Difference of two surds
3. 6 3
5
Sum of a surd and a rational number.
4. 6 x
5 y
Difference of two surds.
7
Observe that, each example is a sum or difference of two surds or surd and a rational
number. Such surds are called binomial surds.
Addition and subtraction of surds:
Surds can be classified as like surds and unlike surds. Can we perform addition
and subtraction on these two types of surds?
To answer this question, recall the addition and subtraction of like and unlike
algebraic terms.
We know that only like terms can be added or subtracted. Similarly, only like surds
can be added or subtracted.
That is, surds having same order and same radicand in their simplest form can be
added or subtracted.
Hence, to add or subtract surds, we have to use the following steps. First reduce
them to their simplest form and then add their co-efficients by using distributive law.
ILLUSTRATIVE EXAMPLES
Example 1 : Find the value of
Sol:
2
3 2
2
3 2
(surds are in their simplest form
They are like surds).
5 2
= (1 3
5 2
(By using distributive property)
5) 2
= 9 2
 2 3 2 5 2 =9 2
Example 2 : Simplify 4 63
Sol:
4 63
5 7
= 4 9
7
5 7
8 28 .
8 28
5 7
= (12 + 5 – 16)
8 4
7 =
7 = 12 7
16 7
(By using distributive property)
7
 4 63 5 7 8 28
5 7
7
Example 3 : Simplify: 2 3 16 3 81 3 128 3 192
Sol: Reduce the surds to their simplest form
23 16 = 2 3 8
3
128 =
3
64
23 2
2
2
2
43 2
43 2
3
81 =
3
192 =
3
27
3
64
3
3
33 3
43 3
156
UNIT-7
 23 16
3
3
81
3
128
192 = 4 3 2
33 3
= 432
43 2
432
43 3
33 3
4 3 3 (by re-arranging the like terms)
= (3 + 4) 3 3 = 7 3 3
 23 16
3
3
81
3
128
192 = 7 3 3
Example 4 : Find the sum of 4 x
Sol:
4 x
6 y
3 4 x
= 4 x
6 y
= 4 x
12 x
= (4 12) x
= 16 x
 (4 x
6 y and 3 4 x
3 y
3 y
12 x
9 y
(By removing the brackets)
6 y
9 y
(By re-grouping the like surds)
(6
(adding and subtracting of like surds)
9) y
3 y
6 5) 3(4 x
3 y ) 16 x
3 y
EXERCISE 7.1
I. Simplify the following surds :
1.
75
108
3.
45
3 20
5. 3x x
3 x3
7. 4 63
5 7
9. 4 7
192
3 5
2 9x 3
50
4. 2 2a
3 8a
6.
50
10.
5 343
II. Find sum of the following surds.
1. x y ,2x y , 4x y
2. 5 3 p, 33 p, 23 p
3. x x , y y , 3 x 3 , 4 y 3
4. ( 12
20), (3 3
5. ( 3
2), (2 2
6. ( x
2 y ), (2 x
12
8. 2 3 32
8 28
3 252
2. 4 12
2 5), ( 45
3 3), (4 2
3 y ), (3 x
90)
3 3)
y)
1
50
8
33 4
7 48
2a
5 3
3
1
75
6
147
32
500
1
18
8
1
3
3
Surds
III.
157
1. Subtract 5 x from 9 x and express the result in index form.
2. Subtract 3 p from 10 p .
3. Subtract 3 a from the sum of 4 a and 2 a
4. Subtract 2 x
3 y from 5 x
y
Multiplication of surds
We have learnt addition and subtraction of surds. The condition is that, the given surds
should have same order and same radicand in their simplified form, to find their sum or
difference. Does the same condition apply for mutliplication of surds?
We have learnt that surds can be expressed in index form. Let us recall the
multiplication of two exponential forms using the first law of exponents, am × an = am + n .
This law holds good only when the bases are equal. How to find the product, when the
bases are not equal?
1
1
For example, a n × b n can be rewritten in the form of surds as
equal to
n
n
a×b
n
n
a
b which is
ab by IV law of exponents.
We can conclude that the essential condition for multiplication of surds is that,
their orders should be same, but the radicands can either be same or different.
Now, let us study some cases of multiplication of surds.
Case 1: Multiplication of surds having same order
Observe the following examples.
Example 1 : Simplify
Sol:
3 =
5
5
5 3 =
Example 3 : Simplify
Sol:
3
5
6 =
=
 3( 5
3
3
3
Example 2 : Multiply
15
Sol:
5
6
5
3
3 5
3
= 15
18
6)
3 2
15
23 3
33 4
(by using
6
15
9
33 4
2
35
15
n
a
n
b
n
ab )
n
b
n
c
3 2
(same order)
3
= 24 3 24
 23 3 3 3 4 43 2
7 5 =
43 2
43 2
= 2
5 =
5
(using distributive law)
6
Example 4 : Find the value of 2 3 3
Sol:
7
7 by
4
3
3
24 3 3
4
23
(by using
2
24
n
a
n
abc )
2 3 3 = 48 3 3
48 3 3
Observe that, in all the above examples, the surds are having same order.
158
UNIT-7
Case 2 : Multiplication of surds having different orders
If the surds are not of the same order, we cannot follow the process of multiplication
directly as discussed in the previous section.
Then, how do we multiply surds having different orders?
Study the following example.
Multiply
5 and
3
2
Since the essential condition for mutliplication of surds is that the surds should be
of same order, we should convert these surds into same order.
How to convert them to same order?
Observe these steps of converting surds of different orders into same order :
5  3 2  5½  21/3
Step 1 : Express the given surds in index form
Observe that the surds written in index form have the exponents in rational form.
Note that in these exponents the denominators are representing the orders of the
surds. Hence, if we have to convert the surds to same order, the rational forms
should be converted to common denominator.
How to convert them into common denominator? Recall that we find LCM to convert
them into common denominator.
Therefore, we have to find LCM of the orders of the surds to convert them into same
order.
Step 2: Find the LCM of 2 and 3
 LCM of 2 and 3 is 6
Step 3: Multiply and divide each

index by LCM and simplify
Step 4: Rewrite in surd form
Step 5: Use the rule
 5
3
2
n
a
6
n
b
n
ab
5
3
1
2
 53
6
6
1
6
2
2
2

6
53

6
125
6
1
3
2
6
6
1
6
22 = 6 125
6
4
500
500
Steps to be followed for multiplying surds of different orders :
1.
Find the LCM of orders of the given surds.
2.
Convert each surd into surds of same order.
3.
Multiply the surds by using the rule
n
a
n
b
n
ab
6
4
Surds
159
ILLUSTRATIVE EXAMPLES
Example 1 : Find the product of
Sol: Orders are 3 and 4.
3
2
4
1
=
2
1
3
4
3
3.
(index form of the surds)
4
12
12
4
4
LCM of 3 and 4 is 12.
34
= 212
32
2 and
1
= 23
3
3
3
1
4
3
12
12
3
1
12
4
312 = 2
=
12
24
12
=
12
24
33 =
=
12
432
12
432
(each index is multiplied and divided by the LCM)
3
1
12
3
(rewrite the index form in surd form)
33
12
16
(using the rule
27
n
a
n
b
n
ab )
From the above two examples, we can conclude that when two surds of different
orders are converted to same order, the order is equal to LCM of the orders of the two
surds, i.e., given surds are raised to the LCM of the two surds.
Example 2 : Find the product of
3 and
3
5
Sol: Alternate method :
LCM of orders of the surds = 6
 Raise the order of 3 and 3 5 to order 6.
i.e. We have to multiply the order of the surd
3 which is 2, by 3.
Observe what happens if we do that,
3
2 3
3
6
3 But,
3
6
3
We observe that the value of the surd changes. What should we do to retain the
original value? Observe the following,
3
2 3
33
3
3
1 1
2 3
1
32
3
In this process, the original value of the surd is retained.
This implies, while converting the order of the surd, to retain its original value,
raise the radicand to the same power as we are mutliplying the original order of the
given surd.
160
UNIT-7
Similarly,
3
5 =
 3
3 2
5=
3
5
2 3
=
 3
2
52
6
1 1
3 2
33
3 2
27
6
5
1
3
3
5
52
25 =
6
27
25
6
675 =
6
675
5 = 6 675
3
Case 3 : Multiplication of Binomial surds.
6
Example 1 : Multiply
6
2
=
6
2
=
6
Sol:
2
6
2 by
6
2
2
2
It is in the form of (a+b)2
2 6
2
2
2
 (a+b)2 = a2 + 2ab + b2
= 6 + 2 12 + 2
= 8 + 2 12
Example 2 : Multiply x
Sol:
x
2 3
= x2
2 3
x
2 3 by x
3 3
3 3
3 3 x
It is in the form of (x+a) (x+b)
2 3
3 3
 (x+a) (x+b) = x2 + (a + b) x + ab
= x2 + 5x 3 + 18
Example 3 : Multiply 3 18
Sol:
3 18
= 9 2
= 9 2
2 12
4 3
5 2
50
5 2
2 12 by
3 3
3 3 +4 3 5 2
= 54 – 7 6
36
27
27
each surd is reduced to its simplest form.
3 3
= 45 × 2 – 27 6 + 20 6 – 12 × 3
= 90 – 7 6
50
Surds
161
EXERCISE 7.2
I. Simplify:
1.
5.
3
6
6
2
7
2.
3
4
3
5
3.
5
6.
n
x
n
y
7. 2 3 7
4
4
4
6
4.
33 4
8.
5
10
5
18
11
27
128
II. Find the product of the following surds.
1.
2 and
5.
5 and
3
3
4
3
2.
3
3 and
4
2
3.
3
2 and
4
3
4.
6.
3
4 and
5
2
7.
3
5 and
4
4
8.
3
3 and
4
5
2 and
6
5
III. Simplify
1.
3 2
2 3
2 3
4 2
2.
75
45
20
12
3.
3 x
2 y
3 y
2 x
4.
6 a
5 b
6 a
5 b
5.
6 2
7 3
6 2
7 3
6.
3 27
5
9 3
7
Rationalising factor (R.F) and Rationalisation of surds.
In the table below, surds, their product and the result are given. study them.
Sl.No.
Surds
1.
7,
2.
5 x,
3.
x
4.
ab ,
ab
5.
6 3
5
6.
8 x
7
Result
7× 7=7
Rational number
x = 5x
5 x
x
y,
x
y
x
y
ab
6 3
y
Product
5
8 x
6 3
y
8 x
x
2
x
y
– 52 =108 – 25 = 83
–
y
2
Rational number
Rational number
ab
ab
2
y
Rational number
64x
y
Rational number
Rational number
In all the above examples, product of two surds is a rational number.
When the product of two surds is rational, then each surd is called rationalising factor
(R.F) of the other.
Hence,
7 is the R.F. of
x
6 3
7
y is the R.F. of
x
y
5 is the R.F. of ( 6 3
5)
162
UNIT-7
This process of muliplying a surd by another surd to get a rational number is
called rationalisation.
The rationalising factor of binomial surd is also called conjugate of binomial
surd.
If the product of two binomial surds is a rational number, then each surd is
called the conjugate of the other.
ILLUSTRATIVE EXAMPLES
Example 1 : Verify that the conjugate of
Sol: Consider ( 3
2)( 3
= ( 3)2
( 2)2 = 3  2 = 1
= 5 x
2
3 y
2
[using the identity (a + b) (a  b) = a2  b2]
2) is ( 3
Example 2 : Rationalise the surd (5 x
3 y) 5 x
3
2)
 The conjugate of ( 3
Sol: (5 x
2 is
3
2) .
3 y)
3 y
2
{ (a – b)(a + b) = a2 – b2}
= 25x – 9y
 5 x
3 y
is rationalised by using its conjugate 5 x
Example 3 : The conjugate of 3
5
x is 3
Sol: If the the conjugate of 3
5
x
is 3
5
x
5
3 y .
x . Verify.
5
x
then the product must be a
rational number
3
5
x
3
5
x
= 32
2
= 9 – (5 + x) = 9 – 5 – x = 4 – x
(4 - x) is a rational number.
 3
5
x
is the conjugate of 3
1
Example 4 : Find the R.F of 3 3  3
Sol: Let a = 3
1
3
and b = 3
3
x .
5
1
3
1
3
1 3
3
then,
a3 = 313

a3 – b3 = 3
but
a3 – b3 = (a – b)(a2 + ab + b2)
= 3 and b3 = 3
1
3
9 1
3
8
3
= 3
1
1
3
Surds
163
1

33
1 2
1
3
3
1
33
3
33
1
3
3
1
3
3
1
33
Since
1
3
3
1
3
3
2
3
3
1
3
1
3
2
33
3
2
3
3
8
3
=
8
3
8
1 = 3
2
3
8
is a rational number, R.F of
3
=
2
1
3
3
1
33
3
2
1
3
is 3 3
3
2
3
1
EXERCISE 7.3
I. Write the rationalising factor for the following surds.
(a)
(b) 2 x
a
(c) 7 y
(d)
(e)
xy
1
p
(h) a ab
(i) x mn
2
II. Write the conjugates of the following binomial surds.
(f) 8 x
(a)
a
y
(g)
(b)
b
(e) 10 2
x
(f) 5
3 5
2 y
(c) 3 p
3
(g)
8
(d)
x
(h)
3 7
7 3
1
1
1
2
x
y
(j)
(k) x a y b
(l) xy z
2
2
2
III. Find the rationalising factor of the following binomial surds.
1
1
1
1
1 y
b
3 y
(i)
(a) 2 3 2 3
(b) 5 3 5 3
(c)
1 y
Simplification of surds by rationalising the denominator
q
(j) 5p a
2 q
5
4 p
yz x
1
(d) x 2
x
1
2
While rationalising the denominator, both the numerator and the denominator must be
multiplied by the rationalising factor of the denominator.
ILLUSTRATIVE EXAMPLES
Example 1 : Rationalise the denominator and simplify
Sol:
3
3
=
5

R.F of
5
3
=
5
3
5
5
5

3
5
5
=
5
5 is
15
5
5

3
=
5
Example 2 : Rationalise the denominator and simplify
Sol:
6
8
6 8
=
8
8
8
6 4 2
8
3
12 2
82
3
.
5
3 2
2
15
5
6

8
6
8
.
3 2
2
164
UNIT-7
bc
a
Example 3 : Rationalise the denominator and simplify
bc
Sol:
bc
=
a
a
bc
abc

a
a
3
Example 4 : Rationalise the denominator and simplify
5
3
Sol:
×
a
bc
=
a
3
5

3
5
3
=
3
5
3
3
=

3
5
=
3
5
3
5
3
5
3
3
2
5
=
abc
a

3 is the denominator.
5
3
a
a
2
3
5
2
=
3
5
3
5 3
3
R.F of
5
2
6
3
6
3
=
R.F of
6
3
6
3
6
3
6
3
6
=
6
=
=

6
3
2
6
3
6
3
Example 6 : Simplify
Sol:
5
5

2 6
3
9
6 2
3
5
5
5
5
5
3 3
2 2
3
2
5
6
3
6
3
= 3
2 2
2
R.F of
3
2
3
3
2
5
6
2 2
3
2
3
 (a + b)2 = a2 + 2ab + b2
3
2 18
3
3
 (a – b) (a + b) = a2 – b2
3
6
3 is
6
2
2
3
6
3
5
2
3
2
3 is
3
Example 5 : Simplify by rationalsing the denomiantor
Sol:
5
2
=
5
5
2
5
2
5
2
2 is
5
3
5
2
5
2 and R.F of
5
2
5
2
5
2 is
5
2
Surds
165
5
=
5
5
=
5
5
= 5
5

5
3
2
5
2
Sol:
4
4
1
3
1
48
2
15
5
2
22
2 3
4
15
2 5
3
2 3
15
2 3
4
4 3
2
3
4 3
4 3
=
2
3
3
8 3
 4
5
2 5
4
48
2
4 3
4
=
2
3
3
2
5
2
1
48
2
4
1
48 =
2
3
1
3

= 5
3
2
2 5
1
3
Example 7 : Simplify 4
2
2
R.F of
3 is
3
4 3
2
10
20
3
12 3
10 3
=
=
63
6
3
10 3
3
EXERCISE 7.4
I. Rationalise the denominator and simplify.
A.
(1)
B.
(1)
C.
(1)
8
(2)
3
3
(3)
2 x
5
2y
x
2
3
2
3
2
3
2
(2)
(2)
x
(3)
y
5 2
3
3 2
5
(3)
(4)
10
5
3
4 3
2
3
2
1 2a
2 5
3 5
(4)
(4)
(5)
6
3
3
(5)
3 5
6
ab
a
b
6
3
6
II. Simplify each of the following:
(1)
(4)
2
3
5
2
3
7 3
6
(7) If y =
2
2 5
3
8
2
5
(2)
(5)
x
a
x
a
x
a
x
a
5
3
3
21
3
5
3
2 5
7
show that y
21
5
1
y
2x
a
(3)
6
5
2
5
(6) If x = 2 6
2
5 find x
1
x
166
UNIT-7
Surds
Meaning of
Surd
n
Radicand
a
a
Addition of surds
Order of
surd
n
Types of Surds
Rationalisation of surds
Operations on Surds
Rationalising factor
Subtraction of surds
Like Surds
Conjugate of
binomial surd
Multiplication of surds
Unlike Surds
with same
order
Binomial Surds
ANSWERS
EXERCISE 7.1
I.
1] 3 3
2
10]
4
5] 3
EXERCISE 7.2
II.
1]
6
2] 5 2 20 3 3] 0 4] 7 2a 5] 0 6] 2 7] 7 8] 6 3 4 9] 21 7
3
II. 1] 7x y 2] 10 3 p 3] 4x x 5y y
4] 5 3 7 5 3 10
2
1
2] 7 p 3] 3 a 4] 3 x 4 y
7 2 6] 6 x III. 1] 4 x 2
I. 1] 21
128 2]
12
2]
3
20 3]
648 3]
12
4
24 4]
1] –12 – 2 6 2] 4 15 3] 5 xy
EXERCISE 7.3
(a)
(j) a
(f) 5
a (b)
b II. (a)
2 3 (g) 8
2
3
III. (a) 2
b (b)
1
7 3 (i) 2
5 (h) 3 7
2
23
p
2
6
1,125 6]
(f) x
q
2 y (c) 3 p
x
5] 6 10 6]
110
15
n
xy 7] 6 3 28
8,192 7]
12
8] 144 3
40,000 8]
6
20
6y 4] 36a – 25b 5] 219 – 84 6 6] 278 + 108 3
6x
x (c) y (d) xy (e)
a
5
432 4] 4 45 5]
III.
I.
Simplication by
rationalising the
denominator
with different
orders
2 (j)
y (g)
2 q (d) x
1
x
2
p (h)
ab (i) mn
3 y (e) 10 x
1
y (k) x a
2
3 5
y b (l) xy z
yz x
2
1 (b) 5 3
1 5 3 (c) 1 y
1 y
1
(d) x 2
x
1
2
1
or x 2
1
2
x
EXERCISE 7.4
I. A. 1]
3] 10
5
10y
3 x
4 3
2]
3]
3
2x
3 4]
5
2y
6
3 5]
2
3]
4]
10
21
3 6
2
2 1
3
3
4]
6
2
2 6
11
10
5]
14 3
10a
5]
10
4]
6 7
8
ab
a
a
b
b
II.1] 2
105
30
2
6
5
B. 1] 2
C.1] 5
15
6] 10
3
2
x
2]
x
x
y
30 5 10 3 6
2 6 2]
13
10 2] 4 3]
3 30
10
15
y
15
5
2