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2.18 Algebraic Surds:
We have studied the surds and their representation on
number line in section 1.7.
In this section, the variables we are going to use(x, a,
b, n.) are all natural numbers.
In the surd represented by
we call m as the ‘order’ and n as the ‘radicand’
Definition: ‘Like surds’ are
Group of surds which have same order and same radicand after simplification (in their
simplest form)
A group of surds of different order or different radicand
in their simplest form
are called ‘unlike surds’
Example: Let us observe the
following surds
1.![]()
2.![]()
If we had not simplified the above two surds, we would
have classified them as unlike surds. This is because their radicands (48, 12)
are not same, though the order is same. We need to compare surds only after
converting them to their simplest form.
Thus the above 2 surds are like surds as their order=2 and
radicand=3 in their simplest form
3.
--à
order = 3, radicand=2
4.
--à
order = 4, radicand=5
The above surds do not have same radicand and order and
hence they are called unlike surds.
Observe the way we do the following operations
1. 5a+3a =(5+3)a =8a
2. 7a-2a =(7-2)a= 5a
We do additions/subtractions of surds in a similar manner.
1. Sum
or difference of like surds in the simplest form is obtained by adding or
subtracting their co-efficients
2.18
Problem 1:
Simplify ![]()
Solution:
![]()
2.18
Problem 2.
Simplify
Solution:
=
=
= 2x (1+1/2)
= 2x3/2
2.18
Problem 3.
Subtract
from ![]()
Solution:
Result = (
) –(
)
= ![]()
=![]()
Observe following:
![]()
2. We
followed the rule
similar to the rule (ab)
n= an *bn
2.18
Problem 4.
Multiply
by ![]()
Solution:
We know that
=51/2= 52/4= (52)1/4=
(25)1/4
![]()
=(25)1/4*
31/4= 751/4=![]()
What did we do?
Steps followed for multiplication of surds:
Step 1 : Write the surds in
index form.
Step 2 : Find LCM of orders of
given surds
Step 3 : Convert surds to have
equal orders
Step 4 : Multiply radicands by
following the rule![]()
2.18
Problem 5.
Multiply
by ![]()
Solution:
|
No |
Step |
Explanation |
|
1 |
|
Write
the surds in index form. |
|
2 |
The
orders of the surds are 3 and 4. Their LCM
is 12 |
Find
LCM of orders |
|
3 |
|
Change
indices of surds |
|
4 |
|
Change
indices of surds |
|
5 |
( |
|
We know that
is an irrational
number. How do we convert
to a rational number?
Let us multiply
by
then we have
*
=
= 5. Note 5 is a rational number
Definition: The procedure of multiplying a surd by another
surd to get a rational number is called ‘Rationalisation’
The operands are called rationalizing
factor (RF) of the other.
In the above example
is RF of ![]()
2.18
Problem 6 What is
the RF of
?
Solution:
Note that in the surd only
is irrational .It’s
co- efficient 6 is rational number. Therefore we need to find RF only of
.
The RF of
is
because
*
=
= (a-b)
Now Multiply
by ![]()
Result= 6(a-b)1/3*((a-b)2)1/3
= 6(a-b)1/3*(a-b)2/3
= 6(a-b)(1+2)/3 =6(a-b) which is
a rational number
Definition: A binomial
surd is an algebraic sum (sum or difference) of 2 terms both of
which could be surds or one could be a rational number and another a surd
Examples of Binomial surds are
,
,![]()
RF or ‘Conjugate’ of
a binomial surd is the term which when multiplied by the binomial surd, results
in a rational number.
(Conjugate of binomial surd* Binomial surd = Rational
number)
2.18
Problem 7 : Find the conjugate of ![]()
Solution:
Note
= 2(
)
We need to find a term such that the result has x and y
with rational co-efficients
We also know (a+b)(a-b) = a2-b2 and hence
appears to be the conjugate of ![]()
Therefore
* ![]()
= 2(
)*(
)
= 2{(
)2-(
)2}
= 2{22*(
)2-(
)2`}
= 2(4x-y) =8x-2y which is a
rational number
For rationalization of the surd
in the denominator, we follow the following steps:
1) Find the RF of denominator
2) Multiply both numerator and denominator of surd by RF
of denominator
2.18
Problem 8:
Rationalise denominator and simplify 2/(
)
Solution:
Since (
)*(
) =(x-y) (
(a-b)(a+b)
= a2-b2 with a =
b =
)
We note that
is conjugate of the
denominator. We multiply numerator and
denominator by this conjugate
2/(
)
={2/(
)}*{(
)/(
)}
=2 (
)/(x-y)
2.18
Problem 9:
Rationalise denominator and simplify (
)/(
)
Solution:
As in the above example
is conjugate of ![]()
(
)/(
)
= {(
)/(
)}*{(
)/(
)}
= (
)*(
)/(9*2-5) (
(
)2=9*2 and (
)2=5)
= (
)/13
=(
)/13
= (
)/13
2.18
Problem 10 : Rationalise denominator and simplify 7
/ (
) -
/ (
)
Solution:
Let us rationalise the terms
separately
1. Multiply both numerator and denominator of the first
term by
which is conjugate of ![]()
Note (
)*(
)= 10-3 =7
7
/ (
)
= 7
*(
)/((
)*(
))
= 7(
)/7 (![]()
*
= 3)
= 3+![]()
2. Multiply both numerator
and denominator of the second term by
which is conjugate of ![]()
![]()
/ (
)
=
(
)/(6-2)
=
/4
=
/2
7
/ (
) -
/ (
)
= (3+
) -
/2
= (6+ 2
-
+
)/2
= (6+
+
)/2
2.18 Summary of learning
|
No |
Points to remember |
|
1 |
|
|
2 |
Rationsalisation
is a process of finding a term such
that the product of this term and the surd is
a rational number |