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Hawkes Learning Systems:
College Algebra
Section 1.4b: Properties of Radicals
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Objectives
o Combining radical expressions.
o Rational number exponents.
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Combining Radical Expressions
Often, a sum of two or more radical expressions can be
combined into one. This can be done if the radical
expressions are like radicals, meaning that they have
the same index and the same radicand. It is frequently
necessary to simplify the radical expressions before it
can be determined if they are like or not.
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Like and Unlike Radicals
Like
The expression below can be
combined because both the
index and radicand are the
same, therefore making the
radical expressions like
radicals.
2 3x  4 3x
 (2  4 ) 3 x
 6 3x
Unlike
The expression below cannot be
combined because the radicand is
not the same, therefore making
this radical expression an unlike
radical.
2 4x  4 5x
Likewise, the below expression
also cannot be combined because
the index is not the same,
therefore making this radical
expression an unlike radical.
23 4x  4 4x
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Example 1: Combining Radical Expressions
Combine the radical expressions, if possible.
a.  5 12 x  27 x
5
 5 3  2  x  x 
2
  10 x
4
3x  3 3x
2
   10 x  3  3 x
2
3 3 x
2
We begin by simplifying the radicals.
Note: the two radicals have the same
index and the same radicand and can
be combined.
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Example 1: Combining Radical Expressions (cont.)
Combine the radical expressions, if possible.
b. 3 432 x 3  32 x 2

3
3 2 2 x 
3
3
 6x 3 2  4 x
3
2
4 2 x
2
2
We begin by simplifying the radicals.
Note: the two radicals have the same
radicand but not the same index.
Therefore, they cannot be combined.
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Example 2: Combining Radical Expressions
Combine the radical expression, if possible.
1
20

16

1
2 5

2
45
 1

2 5


3 5
65
5 5
30


 5
6
4
3 5
2
 3 5   4


 3 5   3 5
8 5
65
  2 5  Here, we rationalize
 the denominator.

 2 5 
Note: the two
radicals have the
same index and the
same radicand and
can be combined.
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Rational Number Exponents
1
Meaning of a n : If n is a natural number and if n a is a
1
real number, then a n 
x
n
a.
1
5

5
x
and
x
1
3

3
x
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Rational Number Exponents
m
Meaning of a n : If m and n are natural numbers, then
m
an 
Either n a m o r
 
n
a
m
n
a

m
 a
n
can be used to evaluate a n , as
m
n
.
m
they are equal.
Note: a
m
is defined to be
1
m
a
n
.
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Example 3: Rational Number Exponents
Simplify the following expression, writing your answer
using the same notation as the original expression.
3
a.
32
5
 5
  32 


1
2

1
2

3
1
8
3
3
To simplify the expression, we begin by noting
3
that 32
5
 15 
  32 


expression.
3
, then simplify the
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Example 3: Rational Number Exponents (cont.)
Simplify the following expression, writing your answer
using the same notation as the original expression.
b.
10
16 x
2

10
4
2 x
4
2
2  16
4
2
 2 10 x 10
2
1
 25 x5

5
4x
Simplify.
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Example 3: Rational Number Exponents (cont.)
Simplify the following expression, writing your answer
using the same notation as the original expression.
9
c.
7 x  65 7 x  6
2
2
 7 x  6
2
 7 x  6
2
 7 x  6
2
9 4
   
5 5
5
 
5
4
5
n
m
nm
Recall, ( a )( a )  a .
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Example 4: Rational Number Exponents
Simplify each of the following expressions.
a.
7 3
x
4

( 7 )( 3 )

21
x
4
x
4
Since
m n
a 
mn
a
Because 4 and 21 have no common factors, the
radical is in simplest form.
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Example 4: Rational Number Exponents (cont.)
Simplify each of the following expressions.
b.
7x  2y
1
7 x  2 y  5
c.
6
x
3
 7 x  2 y 
 1 
1 

 5 
6
 7 x  2 y 5
 x
3
1
 x2

x

1
6
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Example 5: Rational Number Exponents
Write the expression as a single radical.
5
243
1
1
 25 34
4
First make the exponents equal. We do so by finding the
1
1
least common denominator of and and writing
5
4
both fractions with this common denominator.
5
 2 20  3 20
 2
1
4
  3 
5
20
1
1
Use the property a m n   a n  .
m
20
1
 1 6  2 0   2 4 3  2 0
 3888 

20
3888
1
20
Use the property a b   ab  .
n
n
n
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Heron’s Formula
Heron’s formula applied to an
equilateral triangle of length d is
3
A

 3d    3d 
. Expressing this


d
 2  2 

in terms of d, each triangle has area
A
 3d   d 

 
 2  2 
3
or
A
3d
4
16
.
d
Simplifying this radical, we obtain
A
d
2
4
3
.
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Example 6: Rational Number Exponents
Find the area of a regular hexagon (pictured on the right).
Since the hexagon is made up
of six of these triangles,
the total area A of the
hexagon is A 
3d
2
2
3
.
d