Download CHAPTER 4 RADICAL EXPRESSIONS 4.1 The nth Root of a Real

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CHAPTER 4 RADICAL EXPRESSIONS
th
4.1 The n Root of a Real Number
A real number a is called the nth root of a real number b if a n = b .
Thus, for example:
2 is a square root of 4 since 22 = 4 .
2
−2 is also a square root of 4 since ( −2 ) = 4 .
2 is a cube root of 8 since 23 = 8 .
Note that 8 has no other real cube roots.
3
−2 is a cube root of −8 since ( −2 ) = −8 .
Note that −8 has no other real cube roots.
2 is a fourth root of 16 since 24 = 16 .
4
−2 is also a fourth root of 16 since ( −2 ) = 16 .
2 is a fifth root of 32 since 25 = 32 .
Note that 32 has no other real fifth roots.
5
−2 is a fifth root of −32 since ( −2 ) = −32 .
Note that −32 has no other real fifth roots.
Note:
1.
2.
3.
4.
A real number has two nth roots when n is even and
only one nth root when n is odd.
th
The odd n root of a positive number is positive and
the odd nth root of a negative number is negative.
We use the notation n b to denote the principal nth root of a real number b.
The above notation is called a radical expression, b is called the radicand, n
is the index of the radical, and the symbol
is called the radical sign. The
index n is omitted when the index is 2.
The principal root is chosen to be the positive root for the case when the
index is even; it is chosen to be the unique root when n is odd. For
example,
3
4=2
3
Note:
even.
n
8=2
−8 = −2
4
5
16 = 2
5
32 = 2
−32 = −2
b is undefined in the set of real numbers when b is negative and n is
Thus, we say that
−36 is undefined over the reals or is not a real number.
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4.2 Simplifying Radical Expressions
There are three conditions for a radical expression to be in simplest form:
1. There are no perfect n power factors under a radical of index n.
2. There can be no fractions under a radical sign OR there can be
no radicals in the denominator.
3. The index should be the smallest possible.
In this section we will explain how to accomplish the first condition. We will use
the properties
n
ab = n a ⋅ n b for n ≥ 2
a na
=
for n ≥ 2
b nb
where we assume that a and b are such that the expressions do not become
undefined.
n
Let us first consider numerical radicands.
Examples
1. 98
Solution: 98 is not in simplest form there is a perfect square factor 49 under
the radical sign:
98 = 49 ⋅ 2
2.
3
72 ⋅ 2
OR
= 49 ⋅ 2
= 72 ⋅ 2
=7 2
=
7 2
16
Solution: 3 16 is not in simplest form since there is a perfect cube factor 8
under the radical sign:
3
16 = 3 8 ⋅ 2
= 23 ⋅ 3 2
3
=2 2
4
23 ⋅ 2
3
= 38⋅32
3.
3
OR
2⋅3 2
=
80
Solution: 4 80 is not in simplest form since there is a perfect 4th power factor
16 under the radical sign:
4
80 = 4 16 ⋅ 5
= 4 16 ⋅ 4 5
=
4
2 5
4
OR
24 ⋅ 5
= 4 24 ⋅ 4 5
=
24 5
Next, let us consider variable radicands. Let us also first consider the case
when the index n is odd. We observed in Section 4.1 that odd nth roots follow
the sign of the radicand, thus
63
n
x n = x for odd n and any real number x .
For example,
3
x 3 = x,
5
y 5 = y,
7
m7 = m ,
and so on.
Next, let us consider the case when the index n is even. We defined in
Section 4.1 the principal nth root to be the positive root when n is even.
Consider the following:
22 = 2 and also
4
4
2 = 2 and also
( −2 )
4
2
( −2 )
4
=2
=2
From the above we conclude that for any real number x and even index n,
n
n
xn = x
n
x = −x
if x is positive, and
if x is negative
which can be abbreviated to
n
xn = x
for any real number x .
Thus, for example:
x2 = x
4
w4 = w
6
c6 = c
If we want to remove the cumbersome absolute value notation we need to
make the assumption that all variables represent positive real numbers.
Thus, with this assumption we will have
x2 = x
4
w4 = w
6
c6 = c
Let’s now simplify radicals with variable radicands. We will assume that the
variables can be any real number.
Examples
x4
4. Simplify:
4
x = x2 = x2
Solution:
5. Simplify:
Solution:
y6
y6 = y3
64
a12
6. Simplify:
12
Solution:
= a 6 = a6
a
n 22
7. Simplify:
Solution:
n
8. Simplify:
Solution:
3
3
3
4
3
4
5
Solution:
m2
p8
4
w 36
w 36 = w 9
5
n12
n12 = 5 n10 ⋅ n 2 = 5 n10 ⋅ 5 n 2 = n2
13. Simplify:
6
4
3
p8 = p2 = p2
12. Simplify:
Solution:
m 23
m 23 = 3 m21 ⋅ m 2 = 3 m21 ⋅ 3 m 2 = m7
11. Simplify:
Solution:
x4
x = 3 x3 ⋅ x = 3 x3 ⋅ 3 x = x 3 x
10. Simplify:
Solution:
= n11
5
9. Simplify:
Solution:
22
x
6
18
5
n2
x18
= x3
Now let us consider a combination of numbers and variable factors for
radicands. We will continue to assume that the variables can represent any
real number.
32x 4 y 6
14. Simplify:
32 x 4 y 6 = 16x 4 y6 ⋅ 2 = 16x 4 y6 ⋅ 2 = 4 x 2 y3
Solution:
128a10 b16c 2 = 64a10b16c2 ⋅ 2 = 64a10b16c2 ⋅ 2 = 8 a5b 8c
Solution:
3
2
128a10 b16c 2
15. Simplify:
Solution:
16. Simplify:
2 = 4 x2 y 3
3
2 = 8 a5c b8 2
54n 5 p10
54n 5 p10 = 3 27n3p9 ⋅ 2n 2 p = 3 27n3p9 ⋅ 3 2n 2 p = 3np3 3 2n 2 p
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17. Simplify:
3
375x 32 y 46 z 21
Solution:
3
375 x 32 y 46 z 21 = 3 125x 30 y 45 z21 ⋅ 3 x 2 y = 3 125x 30 y 45 z 21 ⋅ 3 3 x 2 y = 5x 10 y15 z 7 3 3 x 2 y
18. Simplify:
4
162a 8 m12e 26
Solution: 4 162a 8 m12e 26 = 4 81a 8m12 e 24 ⋅ 2e 2 = 4 81a 8m12 e 24 ⋅ 4 2e 2
= 3 a2m3 e6
19. Simplify:
4
4
2e 2 = 3a2 m 6 m3
4
2e2
16y 30 n 34
Solution:
4
16 y 30 n 34 = 4 16y28n32 ⋅ 2y 2 n 2 = 4 16y28n32 ⋅ 4 2y 2 n 2 = 2 y7n8
4
2y 2 n 2 = 2 y 7 n 8 4 2y 2 n 2
Next, we have examples showing what to do when there are factors outside
the radical expression.
20. Simplify: 3ab 2 5 64a6 b13
Solution: 3ab2 5 64a 6 b13 = 3ab2 5 32a5b10 ⋅ 2ab3 = 3ab 2 5 32a 5b10 ⋅ 5 2ab3
= 3ab2 ⋅ 2ab2 ⋅ 5 2ab3 = 6a 2b 4
5
2ab3
21. Simplify: −4w 2 6 64 y 18w 36
Solution: −4w 2 6 64 y 18w 36 = −4w 2 6 64y18 w 36 = −4w 2 ⋅ 2 y3 w 6 = −8w 8 y 3
Finally, let us look at examples involving rational radical expressions. For
these we will assume that the variables represent nonzero real numbers.
63a16 b8
25c 22
22. Simplify:
63a16 b8
=
25c 22
Solution:
23. Simplify:
Solution:
3
3
7 =
3 7 a8 b 4
5 c 11
40m7
27n15
40m 7
=
27n15
24. Simplify:
9a16b 8 ⋅ 7 3 a8b 4
=
25c22
5 c11
3
8m6 ⋅ 5m 2m2 3
=
5m
27n15
3n5
2 4 48x18 y 42
xy 4 z
z4
Solution:
2 4 48 x18 y 42
2 4 16x16 y 40 ⋅ 3 x 2 y 2
2
2x 4 y10 4 2 2 4 x 3 y 6 4 2 2
=
=
⋅
3x y =
3x y
4
4
4
4
4
xy z
z
xy z
z
xy z
z
zz
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4.3 Adding or Subtracting Radical Expressions
We can add or subtract only like radicals, i.e., those radical expressions that
have the same index and the same radicand.
Examples Perform the operations.
1. 2 3 + 3 3 + 7 3 − 4 3 3
Solution:
3
2 3 + 3 3 + 7 3 − 43 3 = 2
3 +7 3 +
3−
4 3
3 =9 3 − 33 3
2. x y − y 4 x − 4 x y + 4 y 4 x
Solution:
x y − y 4 x − 4 x y + 4 y 4 x = x y − 4 x y − y 4 x + 4 y 4 x = −3 x y + 3 y 4 x
Recall that sometimes the radicals need to be simplified first. In the following
assume that the variables represent positive real numbers.
8 y 3 − 4 y 18 y
3.
Solution:
8 y 3 − 4 y 18 y = 4y2 ⋅ 2y − 4 y 9 ⋅ 2 y = 2y 2 y − 4 y ⋅ 3 2y
= 2y 2 y − 12 y 2 y = −10 y 2y
6 3 48a 5 + a 3 6a 2
4.
Solution: 6 3 48a 5 + a 3 6a 2 = 6 3 8a 3 ⋅6a 2 + a 3 6a 2 = 6 ⋅ 2a 3 6a 2 + a 3 6a 2
= 12a
3
6a 2 + a 3 6a 2 = 13a 3 6a 2
3 4 48 x 5 y − 6 x 4 243 xy
5.
Solution: 3 4 48 x 5 y − 6 x 4 243 xy = 3 4 16x 4 ⋅ 3 xy − 6 x 4 81 ⋅ 3 xy
= 3 ⋅ 2x 4 3 xy − 6 x ⋅ 3 4 3 xy = 6 x 4 3 xy − 18 x 4 3 xy = −12 x 4 3 xy
5
6.
−32a 7 b11 − 4ab 5 a 2b 6
Solution:
=
5
( −2 )
5
5
−32a 7 b11 − 4ab 5 a 2b 6 =
5
( −2 )
5
a 5b10 ⋅ a 2 b − 4ab 5 a 2b5 b
a 5b10 ⋅ 5 a 2 b − 4ab 5 b5 ⋅ 5 a 2 b = −2ab 2 5 a 2b − 4ab 2 5 a 2b = −6ab 2 5 a 2 b
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4.4 Multiplying Radical Expressions
to multiply radicals with the same index n ≥ 2 we use the following:
n
Examples
1. Multiply:
Solution:
a ⋅ n b = n ab .
12 ⋅ 30
12 ⋅ 30 = 12 ⋅ 30 = 360 = 36 ⋅ 10 = 6 10
Sometimes when the numbers are large it is better to use prime factorization
rather than carry out the multiplication.
3
2. Multiply:
Solution:
3
20 ⋅ 3 54
20 ⋅ 3 54 = 3 20 ⋅ 54 = 3 2
⋅2⋅5⋅2
⋅ 3
⋅ 3
⋅ 3 = 3 23 ⋅ 33 ⋅ 5 = 2 ⋅ 3 3 5 = 6 3 5
20
54
Let us review multiplication involving variable radicands. We will assume all
variables represent positive real numbers.
x 5 y ⋅ xy 6
3. Multiply:
x 5 y ⋅ xy 6 =
Solution:
x 5 y ⋅ xy 6 =
x6y 7 =
x 6 y6 ⋅ y = x 3 y3 y
Let us now look at the case where we have both numerical and variable
factors.
4. Multiply:
Solution:
3
3
4ab 4 ⋅ 3 −2a 3 b 2
4ab 4 ⋅ 3 −2a 3 b 2 =
3
(
)
4ab 4 ⋅ −2a 3 b 2 =
3
−8a 4 b 6 =
3
−8a 3b6 ⋅ a = −2ab 2
3
a
There could be factors sitting outside the radicals.
(
5. Multiply: 2y
(
Solution: 2y
4
4
)(
8x5y ⋅ 3x
)(
8x 5 y ⋅ 3x
= 6 xy
4
4
4
4x2y 3
)
)
4 x 2 y 3 = 2y ⋅ 3 x ⋅ 4 8 x 5 y ⋅ 4 x 2 y 3 = 6 xy
16x 4 y 4 ⋅ 2 x 3 = 6 xy ⋅ 2xy 4 2 x 3 = 12 x 2 y 2
4
2x 3
One or more of the factors could consist of two or more terms.
(
6. Multiply: 2 3 4 2 + 7 6
(
)
)
Solution: 2 3 4 2 + 7 6 = 2 3 ⋅ 4 2 + 2 3 ⋅ 7 6 = 8 6 + 14 18
= 8 6 + 14 9 ⋅ 2 = 8 6 + 14 ⋅ 3 2 = 8 6 + 42 2
4
32 x 7 y 4
68
(
)(
7. Multiply: 3 5 + 8 −2 10 − 3 2
)
Solution:
(3
)(
5 + 8 −2 10 − 3 2
(
)
)
(
)
(
)
(
= 3 5 ⋅ −2 10 + 3 5 ⋅ −3 2 + 8 ⋅ −2 10 + 8 ⋅ −3 2
= − 6 50
− 9 10
− 2 80
− 3 16
= − 6 25 ⋅ 2
− 9 10
− 2 16 ⋅ 5
−3⋅4
= − 6⋅5 2
− 9 10
− 2⋅4 5
− 12
− 30 2
− 9 10
−8 5
− 12
=
(
8. Multiply: 6 3 + 2 5
)
)
2
Solution: One way to do this is to rewrite the problem as
(6
)(
)
3 + 2 5 6 3 + 2 5 and perform the multiplication as in Example 7.
2
Another way is to apply the special product formula ( a + b ) = a 2 + 2ab + b 2
which we will do:
(6
3 +2 5
2
) = (6 3 )
2
(
)(
) (
+2 6 3 2 5 + 2 5
= 36 9
+ 24 15
+ 4 25
= 36 ⋅ 3
+ 24 15
+ 4⋅5
= 108
+ 24 15
+ 20
)
2
= 128 + 24 15
(
)(
9. Multiply: 4 8 − 5 6 4 8 + 5 6
)
Solution: We can multiply as in Example 7 but then again we notice that we
can also use the special product formula ( a − b )( a + b ) = a 2 − b 2 as follows:
(4
)(
) (
8 −5 6 4 8 +5 6 = 4 8
10. Multiply:
(
2+ 6
)
2
) − (5 6 )
2
= 16 ⋅ 8 − 25 ⋅ 6 = 128 − 150 = −22
3
3
Solution: Using the special product formula ( a + b ) = a 3 + 3a 2b + 3ab 2 + b 3 ,
we get
(
2+ 6
3
) = ( 2)
= 23
3
+3
2
2
( 2 ) ( 6 ) + 3 ( 2 )( 6 ) + ( 6 )
+ 3⋅2⋅ 6
+ 3 2⋅6
+ 63
= 22 ⋅ 2 + 6 6
+ 18 2
+ 62 ⋅ 6
=2 2
+ 18 2
+6 6
+6 6
= 20 2 + 12 6
3
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There could be variables in the product. We again assume that all variables
represent positive real numbers.
(
11. Multiply: x 2 x + 2 y
)(
3 x − y 2y
)
Solution: We apply FOIL.
(x
)(
2x + 2 y
3 x − y 2y
)
= x 2 x ⋅ 3 x − xy 2 x ⋅ 2y + 2 y ⋅ 3 x − 2y y ⋅ 2y
= x 6 x 2 − xy 4 xy + 2 3 xy − 2y 2y 2
= x 2 6 − 2 xy xy + 2 3 xy − 2y 2 2
The index could be larger than 2.
(
12. Multiply: 4 3 6 2 3 4 + 3 3 9
)
Solution:
(
43 6 23 4 + 33 9
)
3
= 4 ⋅ 2 6 ⋅ 4 + 4 ⋅3 3 6 ⋅ 9
= 8 3 3 ⋅ 2 ⋅ 2 ⋅ 2 + 12 3 2 ⋅ 3 ⋅ 3 ⋅ 3 OR 8 3 3 ⋅ 23 + 12 3 2 ⋅ 33
= 8 ⋅ 2 3 3 + 12 ⋅ 3 3 2
= 16 3 3 + 36 3 2
13. Multiply: 3 −2 x 2 y
Solution:
3
2
−2 x y
14. Multiply:
(
3
(
3
(
3
4 xy 2 − 3 −32 x 4 y 5
)
)=
3
2
3
4
4 xy − −32x y
7−35
)(
3
5
49 + 3 35 + 3 25
−8 x 3 y 3 − 3 64 x 6 y 6 = −2 xy − 4 x 2 y 2
)
Solution: Note that the problem looks like the special product
( a − b ) ( a 2 + ab + b2 ) = a 3 − b3 .
(
)( 49 + 35 + 25 ) = (
= ( 7) − ( 5) = 7 − 5 = 2
3
7−35
3
3
3
3
3
3
3
3
7 − 3 5 

) ( 7)
3
2
+37⋅35+
( 5 ) 
3
2
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4.5 Dividing Radical Expressions; Rationalizing
Denominators
To divide radical expressions with the same index n ≥ 2 , we have the
following rule:
n
n
a
b
=
n
a
, b≠0
b
This means that we simply divide the radicands.
Examples
98
1. Divide:
2
98
Solution:
2
3
2. Divide:
Solution:
5
40
3
3. Divide:
40
3
3
98
= 49 = 7
2
=
=
5
3
21 4 64
35 4 4
21 4 64
Solution:
35 4 4
4. Divide:
40 3
= 8=2
5
=
21 4 64
7 ⋅3 4
3
6
16 = ⋅ 2 =
=
35 4
5
5
7 ⋅5
4 x 2 3 48 y 5
8 xy 3 6 y
Solution:
4 x 2 3 48y 5
=
8 xy 3 6 y
4x 2
8 xy
3
48y 5
x
=
6y
2y
3
8y 4 =
x
2y
3
8y3 ⋅ y =
x
⋅ 2y 3 y = x 3 y
2y
Multiplication and division can be combined:
3
5.
3ab 2 ⋅ 3 45a 3 b 6
3
5ab 2
Solution:
3
3ab 2 ⋅ 3 45a 3 b 6
3
5ab
2
=
3
3ab 2 ⋅ 45a 3 b 6
3 ⋅ 3 ⋅ 3 ⋅ 5 a 4b8 3 3 3 6
=3
= 3 a b = 3ab 2
5ab 2
5 ab 2
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There could be two or more terms in the dividend:
6.
4 8 x 3 y + 30 50 xy 3
2 2xy
Solution:
4 8 x 3 y + 30 50 xy 3
2 2 xy
=
4 8x3y
+
2 2 xy
30 50 xy 3
2 2 xy
= 2 4 x 2 + 15 25 y 2
= 2 ⋅ 2 x + 15 ⋅ 5 y = 4 x + 75 y
Note that radicals are NOT ALLOWED IN THE DENOMINATOR if a radical
expression is to be considered in simplest form. Let us next consider how the
radicals can be removed from the denominator.
Rationalizing the Denominator
Rationalizing the denominator means making the denominator a rational
number or removing radicals like 2, 5, 3 3, ... which are irrational numbers.
Let us consider the case where there is only one radical and only one term in
the divisor. If the radical is of index n, the idea is to multiply and divide the
expression by a radical that will make the radicand a perfect n power.
Examples Rationalize the denominator.
6
7. Simplify:
3
Solution: 6 = 6 ⋅ 3 = 6 3 = 6 3 = 2 3
2
3
3
3
3
3
There could be variable factors as well (assume these are positive real
numbers):
8
8. Simplify:
2x
8
Solution:
=
2x
8
2x
⋅
2x
2x
=
8 2x
4x 2
=
8 2x 4 2x
=
2x
x
The index could be larger than 2:
7
9. Simplify: 3
5
Solution:
7
7 3 52 7 3 25 7 3 25
= 3 ⋅
=
=
3 3
5
5
5 3 52
5
3
10. Simplify:
Solution:
6
4
3
6
6 4 33 6 4 27 6 4 27
=
⋅
=
=
= 2 4 27
4
4 4
3
3 4 3 4 33
3
72
3a
11. Simplify:
Solution:
5
3a
4a 3
5
4a 3
5
3a
=
5
⋅
22 a 3
5
23 a2
23 a2
=
5
3a
5
23 a2
25 a 5
=
3a 5 23 a2 3
=
2a
2
5
8a 2
Having a fraction or rational expression under a radical sign is the same as
having a radical in the denominator:
4
11
12. Simplify:
4
4
4
11
=
=
⋅
=
11
11
11 11
Solution:
4
4 11
=
⋅
=
11
11 11
OR
13. Simplify:
Solution:
3
3
4 ⋅ 11 2 11
=
11
11
4 ⋅ 11 2 11
=
11
11
5
2xy 2
5
=
2xy 2
3
3
3
5
2xy
2
=
3
3
5
2xy
2
⋅
3
22 x 2 y
2
2
3
=
2 x y
20 x 2 y
2 xy
Now let us consider the case where there are two or more terms in the
denominator. For the case of index 2 radicals, we can use the fact that
( a − b )( a + b ) = a 2 − b 2 to remove the radicals from the denominator.
4
14. Simplify:
3− 7
(
)
( ) ( )
4 3+ 7
4
4
3+ 7
=
⋅
=
2
3− 7
3− 7 3+ 7
3 − 7
Solution:
4
=
3+ 7
3−7
)= 4(
3+ 7
−4
)=−
3− 7
10
15. Simplify:
Solution:
(
2
2 3 +5 2
(
)
)
10
10
2 3 − 5 2 10 2 3 − 5 2
=
⋅
=
2
2
2 3 +5 2 2 3 +5 2 2 3 −5 2
2 3 − 5 2
(
10 2 3 − 5 2
=
4 ⋅ 3 − 25 ⋅ 2
(
5 2 3 −5 2
=
−19
) = 10 ( 2
) = 10
3 −5
12 − 50
( ) (
2 ) 10 ( 2 3 − 5 2 )
=
−38
3 − 25 2
−10 3 + 25 2
or
−19
19
There could also be two terms in the numerator:
73
16. Simplify:
5 3+6
2 5 −9
5 3 + 6 5 3 + 6 2 5 + 9 5 3 ⋅2 5 + 5 3 ⋅9 + 6⋅2 5 + 6⋅9
Solution:
=
⋅
=
2
2 5 −9 2 5 −9 2 5 +9
2 5 − 92
(
)
=
10 15 + 45 3 + 12 5 + 54 10 15 + 45 3 + 12 5 + 54
=
4 ⋅ 5 − 81
20 − 81
=
10 15 + 45 3 + 12 5 + 54
−61
There could be variables involved:
17. Simplify:
2x − 3 x
3x + 2 x
2 x − 3 x 2x − 3 x 3 x − 2 x
Solution:
=
⋅
3x + 2 x 3 x + 2 x 3 x − 2 x
=
6 x 2 − 4x x − 9 x x + 6x
(3x )
2
(
− 2 x
)
2
6 x 2 − 13 x x + 6 x
9 x 2 − 4x
=
There could be radicals of index 3:
1
18. Simplify: 3
2+37
Solution: As in the case for index 2 radicals we can use a special product
formula, namely ( a + b ) a 2 − ab + b 2 = a 3 + b 3 , which we actually learned as a
(
)
factoring formula.
1
3
2+37
3
1
=
3
2+37
⋅
3
19. Simplify:
3
3
22 − 3 2 ⋅ 7 + 3 7 2
22 − 3 2 ⋅ 7 + 3 7 2
3
4 − 3 14 + 3 49 3 4 − 3 14 + 3 49
=
2+7
9
=
x
x2 + 23 x + 4
( x)
3
Solution: The denominator looks like
2
+ 3 x ⋅ 2 + 22 = a 2 + ab + b 2 so if we
multiply the numerator and denominator by a − b = 3 x − 2 and apply the
(
)
formula ( a − b ) a 2 + ab + b 2 = a 3 − b 3 , we will get
3
3
2
x
3
x +2 x +4
3
⋅
3
x −2
=
x −2
3
x
(
3
( x)
3
x −2
3
3
−2
) = x(
3
3
x −2
x −8
).
74
4.6 Rational Exponents
In this section we will learn how to manipulate expressions involving rational or
fractional exponents.
1
We define a n where a is a real number and n is an integer with n ≥ 2 as
1
an = n a .
The following will show how to use the above definition to simplify expressions
1
that look like a n .
Examples
1
1. Evaluate: 4 2
1
Solution: 4 2 =
4= 2
( )
Note that 41 2
2
1
⋅2
= 4 2 = 4 and
( 4)
2
= 4 , thus the definition makes sense.
1
2. Evaluate: ( −8 ) 3
1
Solution: ( −8 ) 3 =
3
−8 = −2
1
 16  4
3. Evaluate:  
 81 
1
16 4 16 2
 16  4
Solution:   = 4
=
=
81 4 81 3
 81 
Now let us consider the case when the numerator in the rational exponent is
not 1. Let us look at how we can handle this case:
m
an = a
m⋅
1
n
1
( )
= am
n
= n am
OR
m
1
a n = an
m
⋅m
 1
=  a n  =
 
( a)
n
m
Let us look at how these will work in the following examples.
75
2
4. Evaluate: 8 3
2
2
Solution: 8 3 = 3 82 = 3 64 = 4 OR 8 3 =
( 8)
3
2
2
= ( 2) = 4
Note that the second way seems “better” in the sense that we deal with smaller
numbers.
5
5. Evaluate: 16 4
5
(
Solution: 16 4 =
4
16
)
5
= 25 = 32
5
 1 3
6. Evaluate:  
 27 
5
5
5
1
 1 3  1 
 1
Solution:   =  3
=  =

 27 
3
243
 27 




4
7. Evaluate: ( −32 ) 5
4
(
Solution: ( −32 ) 5 =
5
−32
)
4
4
= ( −2 ) = 16
3
8. Evaluate: −16 4
3
Solution: −16 4 = −
(
4
16
)
3
= −23 = −8
5
9. Evaluate: 2 2
Solution: For this problem it would make more sense to use the first way of
interpreting the rational exponent.
5
2 2 = 25 = 24 ⋅ 2 = 22 2 .
The exponent could be negative:
10. Evaluate: 25
Solution: 25
3
−
2
−
3
2
1
=
3
=
25 2
11. Evaluate: ( −8 )
Solution:
( −8 )
−
−
2
3
=
1
(
25
2
3
1
2
=
( −8 ) 3
12. Evaluate: −81
Solution: −81
3
−
4
−
)
1
1
=
53 125
=
3
1
(
3
(
4
−8
)
2
=
1
( −2 )
2
=
1
4
3
4
=−
1
3
814
=−
1
81
)
3
=−
1
1
=−
33
27
76
 8 
13. Evaluate:  
 27 
 8 
Solution:  
 27 
14. Evaluate: 2
Solution: 2
−
3
2
5
3
−
−
3
+
5
3
5
5
 27 
243
3
=  3
 =   =
32
2
 8 
3
2
2
−
4
3
1
=
3
2
 1
15. Evaluate:  − 
 3
 1
Solution:  − 
 3
5
 27 
=

 8 
1
=
−
−
1
=
23
22 ⋅ 2
=
1
2 2
4
3
= ( −3 )
+
4
3
=
3
( −3 )
4
=
3
3
( −3 ) ⋅ ( −3 ) = −3 3 −3 = 3 3 3
Let us next consider expressions involving variable factors. Again let us
assume all the variables represent positive real numbers.
1
16. Simplify: ( 27 x 3 y 6 ) 3
1
Solution: ( 27 x 3 y 6 ) 3 = 3 27 x 3 y 6 = 3 xy 2
3
17. Simplify: (16a8b16 ) 4
3
(
Solution: 16a8 b16
)
4
=
(
4
16a8 b16
)
3
(
= 2a2 b 4
)
3
= 23 a 6 b12 = 8a 6 b12
2
 27m12n 9  3
18. Simplify:  −

8 p6 

2
 27m12n 9  3 
27m12n 9
Solution:  −
 =  3 −
6
8p
8p6



2
2

 3m 4 n 3 
9m 8 n 6
 = −
 =
2

4p4
 2p 

4.7 Multiplying or Dividing Radicals with Different Indices
Now that we know about rational exponents we can talk about multiplying or
dividing radicals whose indices are not the same.
The idea is to write each radical using rational exponents. The goal is to first
make the indices the same. Since the index corresponds to the denominator of
the rational exponent, the idea is similar to finding the LCD for the exponents
and then converting each exponent to its equivalent fraction.
77
Examples
1. Multiply:
2⋅32
1
1
1⋅ 3
1⋅ 2
3
2
2 ⋅ 3 2 = 2 2 ⋅ 23 = 2 2⋅3 ⋅ 2 3 ⋅2 = 2 6 ⋅ 2 6 = 6 23 ⋅ 6 22 = 6 25 = 6 32
Solution:
x ⋅ 4 x3 ⋅ 6 x5
2. Multiply:
1
5
3
1⋅ 6
3⋅ 3
5 ⋅2
6
9
10
x ⋅ 4 x 3 ⋅ 6 x 5 = x 2 ⋅ x 4 ⋅ x 6 = x 2⋅ 6 ⋅ x 4 ⋅3 ⋅ x 6 ⋅2 = x 12 ⋅ x 12 ⋅ x 12
Solution:
=x
6 + 9 +10
12
25
= x 12 = 12 x 25 = 12 x 24 ⋅ x = x 2 12 x
2
3. Divide:
3
2
1
Solution:
2
3
4. Divide:
=
2
1
22
= 22
1
−
3−2
1
3
1
= 26 = 6 2
6
=2
23
3
y
4
y
1
3
Solution:
4
y
y
5. Simplify:
1
y3
=
= y3
1
−
4−3
1
4
=y
1
= y 12 = 12 y
12
y4
2x ⋅ 3 2x
4
2x
Solution:
1
2x ⋅ 3 2x
4
2x
=
1
( 2x ) 2 ( 2 x ) 3
1
( 2x )
1
= ( 2x ) 2
+
1
3
−
6+4−3
1
= ( 2x )
4
7
7
= ( 2 x )12 = 12 ( 2x ) = 12 128 x 7
12
4
4.8 Reducing the Index
The notion of rational exponents we can explain why an answer like
NOT considered to be in simplest form. To see why not, we have
2
6
1
6
x 2 is
3
x , where we assume x is positive.
Another way to arrive at the same answer is to use the following:
m n
a =
mn
a
which one can easily prove using rational exponents. Thus,
x2 =
3
x 2 is
x2 = x 6 = x 3 = 3 x
and thus the simplest form of
6
6
x 2 = 3 x . On the other hand,
6
x3 =
3
x3 = x .
78
Examples
24. Simplify:
6
8
1
Solution:
6
25. Simplify:
12
12
6
1
= 2 6 = 2 2 = 2 OR
6
3
8=
8 =
2
x4
1
4
Solution:
3
1
( )
8 = 8 6 = 23
x 4 = x 12 = x 3 =
3
OR
x
12
x4 =
3 4
x4 =
3
x
26. Simplify: 15 32a10 b5
1
Solution:
OR
15
15
(
32a10 b 5 = 32a10 b 5
32a10 b 5 =
3 5
)
15
1
1
2
1
= ( 25 a10b 5 )15 = 2 3 a 3 b 3 = 3 2a 2 b
32a10 b 5 = 3 2a 2 b
4.9 Complex Numbers
In our previous work whenever we encountered an expression like
−3, −16, ... we concluded that the expression is undefined (over the reals)
or is not a real number. When we were solving equations in elementary
algebra, whenever we reached a step, say, x = −9 , we concluded that the
equation has no real solution. In this section we will enlarge the set of
numbers that we are considering to include numbers such as these numbers.
We define a complex number to be any number of the form a + bi , where a
−1 . a is called the real part of the complex
1 3
number and b is called the imaginary part. −3 + 2i , − i , − 0.3i − 4.5i are
2 5
examples of complex numbers. All the real numbers are complex numbers –
the imaginary part is 0 for a real number.
When the real part is 0, the complex number is a pure imaginary number.
Two complex numbers a + bi and c + di are said to be equal if and only if the
real parts are equal and the imaginary parts are equal, i.e.
and b are real numbers and i =
a + bi = c + di ⇔ a = c and b = d .
Adding or Subtracting Complex Numbers
To add or subtract two complex numbers a + bi and c + di , we simply
add/subtract the real part of one to/from the real part of the other, the imaginary
part of one to/from the imaginary part of the other, i.e.
( a + bi ) + ( c + di ) = ( a + c ) + ( b + d ) i
( a + bi ) − ( c + di ) = ( a − c ) + ( b − d ) i
79
Examples
1. Add: 30 − 28i and − 42 + 15i
Solution: ( 30 − 28i ) + ( −42 + 15i ) = 30 − 28i − 42 + 15i
= 30 − 42 − 28i + 15i = −12 − 13i
2. Subtract 25 − 16i from − 13 + 12i .
Solution: ( −13 + 12i ) − ( 25 − 16i ) = −13 + 12i − 25 + 16i
= −13 − 25 + 12i + 16i = −38 + 28i
3. Simplify: 6 − −4 + −9 − 5
Solution: 6 − −4 + −9 − 5 = 6 − 4 ⋅ ( −1) + 9 ⋅ ( −1) − 5
= 6 − 4 ⋅ −1 + 9 ⋅ −1 − 5 = 6 − 2i + 3i − 5 = 1 + i
Multiplying Complex Numbers
To multiply two complex numbers a + bi and c + di we simply apply the
distributive property and remember that i =
−1 so that i 2 = −1 .
( a + bi )( c + di ) = ac + adi + bci + bdi 2 = ac + ( ad + bc ) i − bd
= ( ac − bd ) + ( ad + bc ) i
It is not necessary to memorize the formula; one can simply perform the
multiplication. It might be interesting to note some pattern to make the
multiplication faster:
a
+
bi
× c
di
+
( ac − bd ) + ( ad + bc ) i
Examples
4. Multiply: ( 2 − 5i )( −4 + 7i )
Solution: ( 2 − 5i )( −4 + 7i ) = −8 + 14i + 20i − 35i 2 = −8 + 34i + 35 = 27 + 34i
5. Expand: ( −3 − 6i )
Solution:
( −3 − 6i )
2
2
2
2
= ( −3 ) − 2 ( −3 )( 6i ) + ( 6i ) = 9 + 36i + 36i 2
= 9 + 36i − 36 = −27 + 36i
(
)(
−5 )( 3 −
6. Multiply: 2 − −5 3 − −6
Solution:
(2 −
)
) (
)(
)
−6 = 2 − i 5 3 − i 6 = 6 − 2i 6 − 3i 5 + i 2 30
(
) (
)
= 6 − 2i 6 − 3i 5 − 30 = 6 − 30 − 2 6 + 3 5 i
Check what happens if you do not do the second step and simply multiplied
directly. Can you explain why you got a slightly different answer?
80
Dividing Complex Numbers
Suppose we want to divide the complex number 2 + 3i by the complex
number 1 − 2i . We can write the problem as
2 + 3i
.
1 − 2i
What do we do with the above expression? Note that since i = −1 if we
rewrite the problem it will look like an expression with a radical in the
denominator and following what we discussed in a previous section we can
remove i = −1 from the denominator by following a similar procedure as in
rationalizing the denominator. More precisely, we would need the following:
The conjugate of a complex number a + bi is the complex number a − bi . Let
us give a few examples:
Complex Number
Conjugate
1 − 2i
1 + 2i
−3 + 5i
−3 − 5i
12i
−12i
24i − 9
−24i − 9
Note that the product of a complex number and its conjugate is a real number:
( a + bi )( a − bi ) = a 2 − ( bi )
2
= a 2 − b 2i 2 = a2 + b2
To divide a complex number by another complex number, multiply both the
dividend and the divisor by the conjugate of the divisor:
a + bi a + bi c − di
=
⋅
c + di c + di c − di
Examples
7. Divide:
2 + 3i
1 − 2i
Solution:
2 + 3i 2 + 3i 1 + 2i 2 + 4i + 3i + 6i 2 2 + 7i − 6 −4 + 7i −4 + 7i
4 7
=
⋅
=
=
=
=
=− + i
2
1 − 2i
1 − 2i 1 + 2i
1 − 4i 2
1+ 4
5
5 5
12 − ( 2i )
8. Divide:
Solution:
OR
−8 + 2i
5i
2 8
−8 + 2i −8 + 2i −5i 40i − 10i 2 40i + 10 10 40
=
⋅
=
=
=
+
i = + i
5i
5i
−5i
−25i 2
25
25 25
5 5
−8 + 2i −8 + 2i i −8i + 2i 2 −8i − 2 2 8
=
⋅ =
=
= + i
5i
5i
i
5i 2
−5
5 5
81
Powers of i
We have the following first few powers of i:
i =
i5 = i
−1
2
i 10 = −1
7
i = −1
i = −1
i9 = i
6
i = −i
i = −i
i 11 = −i
i4 = 1
i8 = 1
i 12 = 1
3
…
Do you notice any pattern? Let us try the following:
9. Evaluate: i 21
Solution: We note above that i 4 n = 1 , n a positive integer, i.e. i raised to a
positive multiple of 4 always gives the answer 1. Thus, one way of figuring out
what i 21 is:
i 21 = i 20 ⋅ i = 15 ⋅ i = i
There is a faster way but let us look at another example first:
10. Evaluate: i 35
Solution: i 35 = i 32 ⋅ i 3 = 1⋅ i 3 = −i
Did you observe that all we really need is the remainder when the exponent is
divided by 4 and then we can look at the first column of powers of I given
above to find the answer? Let’s try the following:
i 46 = i 2 = −1
i
55
3
= i = −i
i 96 = i 0 = 1
i 201 = i 1 = i
4.10 Chapter Review
A real number b is called the nth root of a real number a if a n = b .
For example, the 5th root of 32 is 2 since 25 = 32 . We use the notation
n
b to denote the principal nth root of a real number b. The above notation
is called a radical expression, b is called the radicand, n is the index of the
radical, and the symbol
when the index is 2.
is called the radical sign. The index n is omitted
The principal root is chosen to be the positive root for the case when the index
is even; it is chosen to be the unique root when n is odd. For example,
6
64 = 2 and 7 −128 = −2 .
82
Simplifying Radical Expressions
A radical expression to be in simplest form if the following are satisfied (in the
examples we will assume all variables represent positive real numbers):
1. There are no perfect n power factors under a radical of index n.
8x 5 y 2 =
3
4x 4 y2 ⋅ 2 x = 2x 2 y 2 x
27a6 b10 = 3 27a6b9 ⋅ b = 3a2b3 3 b
2. There can be no fractions under a radical sign or there can be no radicals in
the denominator.
2
=
x
4
=
9a
3
y
2 5y
2 x
⋅ =
x x
3
=
2x
=
x2
4 3a 2
⋅
=
9a 3a 2
y
2 5y
3
5y
⋅
5y
2x
x
3
12a 2
12a 2
=
3
27a
3a
=
y 5y
2 25 y
2
=
y 5y
y 5y
5y
=
=
2 ⋅ 5y
10
10 y
x
x
2 x + 3 2x x + 3 x 2x x + 3 x
=
⋅
=
=
4x − 9
2 x −3 2 x −3 2 x +3
4 x2 − 9
3
3 2
3 2
1
1
a − 3 ab + 3 b 2
a − 3 ab + 3 b 2
= 3
⋅
=
3
3
3 2
3
2
a+b
a+ b
a+ b
a − 3 ab + b
3. The index should be the smallest possible.
4
4a 2b 6 =
6
64 x 3 y 15 z 21 =
4a2 b 6 = 2ab3 = b 2ab
3
64 x 3 y 15 z 21 =
4 xy 5 z 7 = 2y 2 z 3 xyz
Adding or Subtracting Radical Expressions
We can add or subtract only like radicals, i.e., those radical expressions that
have the same index and the same radicand.
Examples
1. 5 x − 4 5 x − 8 x + 6 5 x = 5 x − 8 x − 4 6 x + 6 5 x = −3 x + 2 5 x
2. −3b 20a 3 b + 8a 45ab 3 = −3b 4a 2 ⋅ 5ab + 8a 9b2 ⋅ 5ab
= −3b ⋅ 2a 5ab + 8a ⋅ 3b 5ab = −6ab 5ab + 24ab 5ab = 18ab 5ab
83
Multiplying Radical Expressions
A. For radical expressions with the same index n ≥ 2 we use the following:
n
Example
a ⋅ n b = n ab .
4 xy 2 ⋅ 3 2 x 2 y 5 = 3 8 x 3 y 7 = 2 xy 2 3 y
3
B. For radical expressions with different indices n ≥ 2 use rational
exponents.
1
(
xy ⋅ 3 xy 2 = ( xy ) 2 xy 2
Example
1
3
3
6
2
2 6
) = ( xy ) ( xy )
=
6
( xy )
3
(
⋅ 6 xy 2
)
2
= 6 x 3y 3 ⋅ 6 x 2y 4 = 6 x 5y 7 = y 6 x 5y
Apply the distributive property and/or special product formulas when multiplying
radicals involving two or more terms.
Example
(2
3a + b
2
) = (2
3a
)
2
(
)
+ 2 2 3a ⋅ b + b2 = 4 ⋅ 3a + 4b 3a + b2 = 12a + 4b 3a + b2
(
One can also perform the expansion above by writing 2 3a + b
(2
)(
)
2
as
)
3a + b 2 3a + b and applying FOIL.
Dividing Radical Expressions
A. For radical expressions with the same index n ≥ 2 use:
4
Example
80a 7 b17c
4
3
5a bc
5
n
a
n
b
=
=
4
n
a
, b≠0
b
80a 7 b17c 4 16a 4 b16
2ab 4
=
=
3
5
4
5a bc
c
c
B. For radical expressions with different indices n ≥ 2 use rational
exponents.
Example
3
4x2
4
3
8x
1
2 3
( 4x )
=
(8x )
1
3 4
4
2 12
(4x )
=
(8x )
3
3 12
= 12
44 x 8
28 x 8
1 211 x 11
= 12 9 9 = 12
⋅
=
3 9
8 x
2 x
2 x 211 x 11
Rationalizing Denominators
See condition 2 for simplified radicals.
12
211 x11
2x
84
Rational Exponents
1
For a real number a and n an integer with n ≥ 2 , we have: a n = n a .
Also, for an integer m, we have:
m
a n = n a m or
( a)
n
m
.
.
1
3
Examples 16 4 = 4 16 = 2 ;
5
−
7
2
=
1
5
=
7
2
1
57
( −32 ) 5
=
1
=
56 ⋅ 5
(
=
5
−32
)
3
3
= ( −2 ) = −8
1
1
=
53 5 125 5
Complex Numbers
We define a complex number to be any number of the form a + bi ,
where a and b are real numbers and i = −1 . We call a the real part of the
complex number and we call b the imaginary part. −4 + 8i is a complex
number; the real part is −4 and the imaginary part is 8.
Operations on Complex Numbers
Examples
1. Add: ( 3 − 5i ) + ( −4 − 2i ) = ( 3 − 4 ) + ( −5i − 2i ) = −1 − 7i
2. Subtract: ( −6 − 4i ) − ( 2 − 6i ) = −6 − 4i − 2 + 6i = −8 + 2i
3. Multiply: ( 4 − 2i )( −3 + i ) = −12 + 4i + 6i − 2i 2 = −12 + 10i + 2 = −10 + 10i
2
2
4. Expand: ( 3 + 6i ) = 3 2 + 2 ( 3 )( 6i ) + ( 6i ) = 9 + 36i + 36i 2
= 9 + 36i − 36 = −27 + 36i
5. Divide:
4 + 6i 4 + 6i
=
3 − 2i 3 − 2i
12 + 26i − 12
=
9+4
3 + 2i 12 + 8i + 18i + 12i 2
=
3 + 2i
9 − 4i 2
26i
=
= 2i
13
⋅
12
6. Evaluate: i 49 = i 48 ⋅ i = ( i 4 ) ⋅ i = i