61

CHAPTER 4 RADICAL EXPRESSIONS

4.1 The nth

Root of a Real Number

A real number a is called the nth

root of a real number b if na b= .

Thus, for example:

2 is a square root of 4 since 22 4= .

2− is also a square root of 4 since ( )2

2 4− = .

2 is a cube root of 8 since 32 8= .

Note that 8 has no other real cube roots.

2− is a cube root of 8− since ( )3

2 8− = − .

Note that 8− has no other real cube roots.

2 is a fourth root of 16 since 42 16= .

2− is also a fourth root of 16 since ( )4

2 16− = .

2 is a fifth root of 32 since 52 32= .

Note that 32 has no other real fifth roots.

2− is a fifth root of 32− since ( )5

2 32− = − .

Note that 32− has no other real fifth roots.

Note:

1. A real number has two nth

roots when n is even and 2. only one n

th root when n is odd.

3. The odd nth root of a positive number is positive and

4. 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 54

3 5

4 2 8 2 16 2 32 2

8 2 32 2

= = = =

− = − − = −

Note: n b is undefined in the set of real numbers when b is negative and n is

even.

Thus, we say that 36− is undefined over the reals or is not a real number.

62

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 n nab a b= ⋅ for 2n ≥

n

nn

a a

b b= for 2n ≥

where we assume that a and b are such that the expressions do not become undefined. 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 2 OR 2

2 2

2 2

= ⋅ ⋅

= ⋅ = ⋅

= =

2

2

49 7

49 7

7 7

2. 3 16

Solution: 3 16 is not in simplest form since there is a perfect cube factor 8

under the radical sign:

33 3

33 33

3 3

16 2 OR 2

2 2

2 2

= ⋅ ⋅

= ⋅ = ⋅

= = ⋅

3

3

8 2

8 2

2 2

3. 4 80

Solution: 4 80 is not in simplest form since there is a perfect 4th power factor

16 under the radical sign:

44 4

44 4 4

4 4

80 5 OR 5

5 5

5 5

= ⋅ ⋅

= ⋅ = ⋅

= =

4

4

16 2

16 2

2 2

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 n

th roots follow

the sign of the radicand, thus

63

for odd and any real number n nx x n x= .

For example,

3 5 73 75, , x x y y m m= = = ,

and so on. Next, let us consider the case when the index n is even. We defined in Section 4.1 the principal n

th root to be the positive root when n is even.

Consider the following:

( )

( )

22

44 4 4

2 2 and also 2 2

2 2 and also 2 2

= − =

= − =

From the above we conclude that for any real number x and even index n,

if is positive, and

if is negative

n n

n n

x x x

x x x

=

= −

which can be abbreviated to

for any real number n nx x x= .

Thus, for example:

2

4 4

6 6

x x

w w

c 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

2

4 4

6 6

x x

w w

c c

=

=

=

Let’s now simplify radicals with variable radicands. We will assume that the variables can be any real number. Examples

4. Simplify: 4x

Solution: 4 2 2x x x= =

5. Simplify: 6y

Solution: 6 3y y=

64

6. Simplify: 12a

Solution: 12 6 6a a a= =

7. Simplify: 22n

Solution: 22 11n n=

8. Simplify: 3 4x

Solution: 3 3 35 3 3x x x x= ⋅ = ⋅ =3 3x x x

9. Simplify: 3 23m

Solution: 3 3 3 3 323 2 2 2m m m m= ⋅ = ⋅ =

21 21 7m m m

10. Simplify: 84 p

Solution: 8 2 24 p p p= =

11. Simplify: 4 36w

Solution: 4 36 9w w=

12. Simplify: 5 12n

Solution: 5 5 5 5 512 2 2 2n n n n= ⋅ = ⋅ =

10 10 2n n n

13. Simplify: 6 18x

Solution: 6 18 3 x x=

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.

14. Simplify: 4 632x y

Solution: 4 6 2 332 2 2 2 4 2x y x y= ⋅ = ⋅ = =4 6 4 6 2 316x y 16x y 4 x y

15. Simplify: 10 16 2128a b c

Solution: 10 16 2 5 8128 2 2 2 8 2a b c a c b= ⋅ = ⋅ = =

10 16 2 10 16 2 5 864a b c 64a b c 8 a b c

16. Simplify: 5 103 54n p

Solution: 5 10 2 2 23 3 3 3 354 2 2 2n p n p n p n p= ⋅ = ⋅ =3 9 3 9 327n p 27n p 3np

65

17. Simplify: 32 46 213 375x y z

Solution: 32 46 21 2 2 23 3 33 3375 3 3 3x y z x y x y x y= ⋅ = ⋅ =

30 45 21 30 45 21 10 15 7125x y z 125x y z 5x y z

18. Simplify: 4 8 12 26162a m e

Solution: 8 12 26 2 24 4 4 4162 2 2a m e e e= ⋅ = ⋅8 12 24 8 12 2481a m e 81a m e

2 2 6 3 24 42 3 2e a m m e= =2 3 63 a m e

19. Simplify: 30 344 16y n

Solution: 30 34 2 2 2 2 2 2 7 8 2 24 4 4 44 416 2 2 2 2 2 y n y n y n y n y n y n= ⋅ = ⋅ = =

28 32 28 32 7 816y n 16y n 2 y n

Next, we have examples showing what to do when there are factors outside the radical expression.

20. Simplify: 52 6 133 64ab a b

Solution: 2 6 13 2 3 2 35 5 5 53 64 3 2 3 2ab a b ab ab ab ab= ⋅ = ⋅5 10 5 1032a b 32a b

2 3 2 4 35 53 2 6 2ab ab a b ab= ⋅ ⋅ =22ab

21. Simplify: 2 18 3664 64w y w−

Solution: 2 18 36 2 2 8 36 64 64 4 4 8w y w w w w y− = − = − ⋅ = −18 36 3 664y w 2 y w

Finally, let us look at examples involving rational radical expressions. For these we will assume that the variables represent nonzero real numbers.

22. Simplify: 16 8

22

63

25

a b

c

Solution:16 8 8 4

22 11

63 7 3 3 77

25 5 5

a b a b

c c

⋅= = =

16 8 8 4

22 11

9a b a b

25c c

23. Simplify: 7

315

40

27

m

n

Solution: 7

33 315

40 55

27

m mm

n

⋅= =

6 2

15 5

8m 2m

27n 3n

24. Simplify: 18 42

44 4

2 48x y

xy z z

Solution:

18 42 40 2 2 3 62 2 2 24 44 4

4 4 4 4

2 48 2 3 2 43 3

x y x y x yx y x y

xy z z xy z xy z z z

⋅= = ⋅ =

16 4 10

4

16x y 2x y

z z

66

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. 3 32 3 3 7 3 4 3+ + −

Solution: 3 3 3 3 32 3 3 7 3 4 3 2 3 7 3 3 4 3 9 3 3 3+ + − = + + − = −����� �����

2. 4 44 4x y y x x y y x− − +

Solution: 4 4 4 4 44 4 4 4 3 3x y y x x y y x x y x y y x y x x y y x− − + = − − + = − +

Recall that sometimes the radicals need to be simplified first. In the following assume that the variables represent positive real numbers.

3. 38 4 18y y y−

Solution: 38 4 18 2 4 2 2 4 2y y y y y y y y y− = ⋅ − ⋅ = − ⋅24y 9 2y 3

2 2 12 2 10 2y y y y y y= − = −

4. 3 35 26 48 6a a a+

Solution: 5 2 2 2 2 23 3 3 3 3 36 48 6 6 6 6 6 6 6a a a a a a a a a+ = ⋅ + = ⋅ +38a 2a

2 2 23 3 312 6 6 13 6a a a a a a= + =

5. 54 43 48 6 243x y x xy−

Solution: 54 44 43 48 6 243 3 3 6 3x y x xy xy x xy− = ⋅ − ⋅416x 81

4 4 4 4 43 3 6 3 6 3 18 3 12 3xy x xy x xy x xy x xy= ⋅ − ⋅ = − = −2x 3

6. 7 11 2 65 532 4a b ab a b− −

Solution: ( )7 11 2 6 2 25 5 5532 4 4a b ab a b a b ab a b− − = − ⋅ −5 5 10 52 a b b

( ) 2 2 2 2 2 2 2 25 5 5 5 5 55 4 2 4 6a b ab a b ab a b ab a b ab a b= − ⋅ − ⋅ = − − = −5 5 10 52 a b b

67

4.4 Multiplying Radical Expressions

to multiply radicals with the same index 2n ≥ we use the following:

n n na b ab⋅ = .

Examples

1. Multiply: 12 30⋅

Solution: 12 30 12 30 360 10 10⋅ = ⋅ = = ⋅ =36 6

Sometimes when the numbers are large it is better to use prime factorization rather than carry out the multiplication.

2. Multiply: 3 320 54⋅

Solution: 33 3 3 3 3

3

20 54

20 54 20 54 2 2 5 2 3 3 3 5 5 6 5⋅ = ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ = ⋅ =3 32 3 2 3

��� �����

Let us review multiplication involving variable radicands. We will assume all variables represent positive real numbers.

3. Multiply: 5 6x y xy⋅

Solution: 5 6 5 6 6 7x y xy x y xy x y y y⋅ = ⋅ = = ⋅ =6 6 3 3x y x y

Let us now look at the case where we have both numerical and variable factors.

4. Multiply: 3 34 3 24 2ab a b⋅ −

Solution:

( )3 3 3 34 3 2 4 3 2 4 6 334 2 4 2 8ab a b ab a b a b a a⋅ − = ⋅ − = − = − ⋅ = −3 6 28a b 2ab

There could be factors sitting outside the radicals.

5. Multiply: ( ) ( )5 2 34 42 8 3 4y x y x x y⋅

Solution: ( ) ( )5 2 3 5 2 3 7 44 4 4 42 8 3 4 2 3 8 4 6 32y x y x x y y x x y x y xy x y⋅ = ⋅ ⋅ ⋅ =

3 3 2 2 34 446 2 6 2 12 2xy x xy x x y x= ⋅ = ⋅ =4 416x y 2xy

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 2 8 6 14 2 8 6 42 2= + ⋅ = + ⋅ = +9 3

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 2 9 10 2 5 3 4

6 2 9 10 2 5 12

30 2 9 10 8 5 12

+ − −

= ⋅ − + ⋅ − + ⋅ − + ⋅ −

= − − − −

= − ⋅ − − ⋅ − ⋅

= − ⋅ − − ⋅ −

= − − − −

25 16

5 4

8. Multiply: ( )2

6 3 2 5+

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.

Another way is to apply the special product formula ( )2 2 22a b a ab b+ = + +

which we will do:

( ) ( ) ( )( ) ( )2 2 2

6 3 2 5 6 3 2 6 3 2 5 2 5

36 9 24 15 4 25

36 3 24 15 4 5

108 24 15 20

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 ( )( ) 2 2a b a b a b− + = − as follows:

( )( ) ( ) ( )2 2

4 8 5 6 4 8 5 6 4 8 5 6 16 8 25 6 128 150 22− + = − = ⋅ − ⋅ = − = −

10. Multiply: ( )3

2 6+

Solution: Using the special product formula ( )3 3 2 2 33 3a b a a b ab b+ = + + + ,

we get

( ) ( ) ( ) ( ) ( )( ) ( )3 3 2 2 3

3 3

2 6 2 3 2 6 3 2 6 6

2 3 2 6 3 2 6 6

2 6 6 18 2 6

2 6 6 18 2 6

20 2 12 6

+ = + + +

= + ⋅ ⋅ + ⋅ +

= ⋅ + + + ⋅

= + + +

= +

2 22 6

2 6

69

There could be variables in the product. We again assume that all variables represent positive real numbers.

11. Multiply: ( )( )2 2 3 2x x y x y y+ −

Solution: We apply FOIL.

( )( )

2 2

2 2

2 2 3 2

2 3 2 2 2 3 2 2

6 4 2 3 2 2

6 2 2 3 2 2

x x y x y y

x x x xy x y y x y y y

x x xy xy xy y y

x xy xy xy y

+ −

= ⋅ − ⋅ + ⋅ − ⋅

= − + −

= − + −

The index could be larger than 2.

12. Multiply: ( )33 34 6 2 4 3 9+

Solution:

( )33 3

3 3

33 3 3

33

33

4 6 2 4 3 9

4 2 6 4 4 3 6 9

8 3 12 2 OR 8 3 12 2

8 3 12 2

16 3 36 2

+

= ⋅ ⋅ + ⋅ ⋅

= ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ ⋅ + ⋅

= ⋅ + ⋅

= +

3 32 2 2 3 3 3 2 3

2 3

13. Multiply: ( )2 2 4 53 3 32 4 32x y xy x y− − −

Solution: ( )2 2 4 5 3 3 6 6 2 23 3 3 3 32 4 32 8 64 2 4x y xy x y x y x y xy x y− − − = − − = − −

14. Multiply: ( ) ( )3 3 3 3 37 5 49 35 25− + +

Solution: Note that the problem looks like the special product

( )( )2 2 3 3a b a ab b a b− + + = − .

( )( ) ( ) ( ) ( )

( ) ( )

2 23 3 3 33 3 3 3 3 3 3

3 33 3

7 5 49 35 25 7 5 7 7 5 5

7 5 7 5 2

− + + = − + ⋅ +

= − = − =

70

4.5 Dividing Radical Expressions; Rationalizing Denominators

To divide radical expressions with the same index 2n ≥ , we have the following rule:

, 0n

nn

a ab

bb= ≠

This means that we simply divide the radicands. Examples

1. Divide: 98

2

Solution: 98 98

49 722

= = =

2. Divide:3

3

40

5

Solution: 3

333

40 408 2

55= = =

3. Divide:4

4

21 64

35 4

Solution: 4

44

21 64 21 64 7

35 435 4= =

3

7

⋅ 4 3 616 2

5 55= ⋅ =

⋅

4. Divide:2 53

3

4 48

8 6

x y

xy y

Solution:

2 5 2 5343 33

3

4 48 4 488

8 6 2 2 28 6

x y x y x x xy y

xy y y y yxy y= = = ⋅ =

38y ⋅ 2y 3 3y x y=

Multiplication and division can be combined:

5. 3 32 3 6

3 2

3 45

5

ab a b

ab

⋅

Solution:

2 3 6 2 3 63 3

3223

3 45 3 45 3 3 3 5

55

ab a b ab a b

abab

⋅ ⋅ ⋅ ⋅ ⋅= =

4 8

5

a b 3 3 6 233

23 3a b ab

ab= =

71

There could be two or more terms in the dividend:

6. 3 34 8 30 50

2 2

x y xy

xy

+

Solution: 3 3 3 3

2 24 8 30 50 4 8 30 502 4 15 25

2 2 2 2 2 2

2 2 15 5 4 75

x y xy x y xyx y

xy xy xy

x y x 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 32, 5, 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.

7. Simplify: 6

3

Solution: 2

6 6 3 6 3 6 32 3

33 3 3 3= ⋅ = = =

There could be variable factors as well (assume these are positive real numbers):

8. Simplify:8

2x

Solution: 2

8 8 2 8 2 8 2 4 2

22 2 2 4

x x x x

x xx x x x= ⋅ = = =

The index could be larger than 2:

9. Simplify:3

7

5

Solution: 3 3 3

3 3 333

7 7 5 7 25 7 25

55 5 55= ⋅ = =

2

2

10. Simplify:4

6

3

Solution: 4 4 4

4

4 4 444

6 6 3 6 27 6 272 27

33 3 33= ⋅ = = =

3

3

72

11. Simplify:5 3

3

4

a

a

Solution: 5 5 5

25

3 2 3 5 55 5 55

3 3 2 3 2 3 2 38

2 24 2 22

a a a a a a aa

aa a aa= ⋅ = = =

3 2 3 2 3 2

3 2

Having a fraction or rational expression under a radical sign is the same as having a radical in the denominator:

12. Simplify:4

11

Solution: 4 4 4 11 4 11 2 11

11 11 1111 11 11

⋅= = ⋅ = =

OR 4 4 11 4 11 2 11

11 11 11 11 11

⋅= ⋅ = =

13. Simplify: 32

5

2xy

Solution: 2 2 23 3 3 3

32 2 2 2 23 3 3

2 205 5 5

2 22 2 2

x y x y

xy xyxy xy x y= = ⋅ =

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

( )( ) 2 2a b a b a b− + = − to remove the radicals from the denominator.

14. Simplify:4

3 7−

Solution: ( )

( ) ( )2 2

4 3 74 4 3 7

3 7 3 7 3 7 3 7

++= ⋅ =

− − + −

( )4 3 7 4

3 7

+= =

−

( )3 7

4

+

−3 7= − −

15. Simplify: 10

2 3 5 2+

Solution: ( )

( ) ( )2 2

10 2 3 5 210 10 2 3 5 2

2 3 5 2 2 3 5 2 2 3 5 2 2 3 5 2

−−= ⋅ =

+ + − −

( ) ( ) ( )

( )

10 2 3 5 2 10 2 3 5 2 10 2 3 5 2

4 3 25 2 12 50 38

5 2 3 5 2 10 3 25 2 10 3 25 2 or

19 19 19

− − −= = =

⋅ − ⋅ − −

− − − += =

− −

There could also be two terms in the numerator:

73

16. Simplify: 5 3 6

2 5 9

+

−

Solution:

( )2

2

5 3 6 5 3 6 2 5 9 5 3 2 5 5 3 9 6 2 5 6 9

2 5 9 2 5 9 2 5 9 2 5 9

+ + + ⋅ + ⋅ + ⋅ + ⋅= ⋅ =

− − + −

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: 2 3

3 2

x x

x x

−

+

Solution: 2 3 2 3 3 2

3 2 3 2 3 2

x x x x x x

x x x x x x

− − −= ⋅

+ + −

( ) ( )

2 2

2 22

6 4 9 6 6 13 6

9 43 2

x x x x x x x x x x

x xx x

− − + − += =

−−

There could be radicals of index 3:

18. Simplify: 3 3

1

2 7+

Solution: As in the case for index 2 radicals we can use a special product

formula, namely ( )( )2 2 3 3a b a ab b a b+ − + = + , which we actually learned as a

factoring formula.

3 32 23 3 3 3 33 3

3 3 3 3 3 32 23

1 1 2 2 7 7 4 14 49 4 14 49

2 7 92 7 2 7 2 2 7 7

− ⋅ + − + − += ⋅ = =

++ + − ⋅ +

19. Simplify: 3

23 32 4

x

x x+ +

Solution: The denominator looks like ( )2

2 2 23 3 2 2x x a ab b+ ⋅ + = + + so if we

multiply the numerator and denominator by 3 2a b x− = − and apply the

formula ( )( )2 2 3 3a b a ab b a b− + + = − , we will get

( )

( )

( )3 3 3 33 3

3323 3 33

2 22

822 4 2

x x x xx x

xxx x x

− −−⋅ = =

−−+ + −

.

74

4.6 Rational Exponents In this section we will learn how to manipulate expressions involving rational or fractional exponents.

We define 1

na where a is a real number and n is an integer with 2n ≥ as

1

nna a= .

The following will show how to use the above definition to simplify expressions

that look like 1

na .

Examples

1. Evaluate: 1

24

Solution:1

24 4 2= =

Note that ( )1

221 2 24 4 4

⋅

= = and ( )2

4 4= , thus the definition makes sense.

2. Evaluate: ( )1

38−

Solution: ( )1

338 8 2− = − = −

3. Evaluate:

1

416

81

Solution:

14

44

4

16 16 16 2

81 81 381

= = =

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:

( )1 1m

mnm mn n na a a a

⋅

= = =

OR

( )1 1

mm

mmnn n na a a a

⋅ = = =

Let us look at how these will work in the following examples.

75

4. Evaluate:2

38

Solution: 2

3 2 338 8 64 4= = = OR ( ) ( )2

2 2338 8 2 4= = =

Note that the second way seems “better” in the sense that we deal with smaller numbers.

5. Evaluate:5

416

Solution: ( )5

554416 16 2 32= = =

6. Evaluate:

5

31

27

Solution:

5 5 53

31 1 1 1

27 27 3 243

= = =

7. Evaluate: ( )4

532−

Solution: ( ) ( ) ( )4 4 45532 32 2 16− = − = − =

8. Evaluate:3

416−

Solution: ( )3

334416 16 2 8− = − = − = −

9. Evaluate: 5

22 Solution: For this problem it would make more sense to use the first way of interpreting the rational exponent.

5

5 4 222 2 2 2 2 2= = ⋅ = .

The exponent could be negative:

10. Evaluate: 3

225−

Solution:

( )

3

23 3 3

2

1 1 1 125

5 1252525

−

= = = =

11. Evaluate: ( )2

38−

−

Solution: ( )( ) ( ) ( )

2

32 2 2

33

1 1 1 18

4288

−

− = = = =−−−

12. Evaluate:3

481−

−

Solution:

( )

3

43 3 3

44

1 1 1 181

3 278181

−

− = − = − = − = −

76

13. Evaluate:

5

38

27

−

Solution:

5 5 5 53 3

38 27 27 3 243

27 8 8 2 32

− + = = = =

14. Evaluate:3

22−

Solution: 3

23 3 22

1 1 1 12

2 22 2 22

−

= = = =⋅

15. Evaluate:

4

31

3

−

−

Solution: ( ) ( ) ( ) ( )

443 4 3 3 33 33

13 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.

16. Simplify: ( )1

3 6 327x y

Solution: ( )1

3 6 3 6 23327 27 3x y x y xy= =

17. Simplify: ( )3

8 16 416a b

Solution: ( ) ( ) ( )3 3 348 16 8 16 2 4 3 6 12 6 12416 16 2 2 8a b a b a b a b a b= = = =

18. Simplify:

212 9 3

6

27

8

m n

p

−

Solution:

2 2 212 9 12 9 4 3 8 63

36 6 2 4

27 27 3 9

8 8 2 4

m n m n m n m n

p p p p

− = − = − =

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: 32 2⋅

Solution: 1 1 1 3 21

6 6 63 2 53 63 2 3 6 622 2 2 2 2 2 2 2 2 2 2 32⋅ ⋅

⋅ ⋅⋅ = ⋅ = ⋅ = ⋅ = ⋅ = =

3 2

3 2

2. Multiply: 64 3 5x x x⋅ ⋅

Solution: 5 1 3 51 3 6 9 10

3 564 6 2 4 62 4 12 12 12x x x x x x x x x x x x⋅ ⋅ ⋅

⋅ ⋅ ⋅⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅

6 3 2

6 3 2

6 9 10 25

25 24 212 12 1212 12x x x x x x x+ +

= = = = ⋅ =

3. Divide:3

2

2

Solution:

13 21 1 12

62 3 6 613

3

2 22 2 2 2

22

−−

= = = = =

4. Divide: 3

4

y

y

Solution:

14 31 1 13 3

123 4 12 12144

y yy y y y

yy

−−

= = = = =

5. Simplify: 3

4

2 2

2

x x

x

⋅

Solution:

( ) ( )

( )( ) ( ) ( ) ( )

1 13 6 4 31 1 1 72 3 7 12 7122 3 4 12 12

14

4

2 22 22 2 2 2 128

2 2

x xx xx x x x x

x x

+ −+ −⋅

= = = = = =

4.8 Reducing the Index

The notion of rational exponents we can explain why an answer like 26 x is

NOT considered to be in simplest form. To see why not, we have

2 1

26 36 3x x x x= = =

and thus the simplest form of 26 x is 3 x , where we assume x is positive.

Another way to arrive at the same answer is to use the following:

m n mna a=

which one can easily prove using rational exponents. Thus,

32 26 3x x x= = . On the other hand, 3 36 3x x x= = .

78

Examples

24. Simplify: 6 8

Solution: ( )1 3 11

36 6 6 268 8 2 2 2 2= = = = = OR 6 38 8 2= =

25. Simplify: 12 4x

Solution: 14

12 4 3312x x x x= = = OR 312 44 4 3x x x= =

26. Simplify:15 10 532a b

Solution: ( ) ( )1 2 11 1

15 310 5 10 5 5 10 5 23 3 315 1532 32 2 2 2a b a b a b a b a b= = = =

OR 315 5 310 5 10 5 232 32 2a b a b a 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, 9x = − , 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

and b are real numbers and 1i = − . a is called the real part of the complex

number and b is called the imaginary part. 1 3

3 2 , , 0.3 4.52 5

i i i i− + − − − are

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 a bi c di a c b d+ = + ⇔ = = .

Adding or Subtracting Complex Numbers

To add or subtract two complex numbers and a bi 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 28 and 42 15i i− − +

Solution: ( ) ( )30 28 42 15 30 28 42 15i i i i− + − + = − − +

30 42 28 15 12 13i i i= − − + = − −

2. Subtract 25 16 from 13 12i i− − + .

Solution: ( ) ( )13 12 25 16 13 12 25 16i i i i− + − − = − + − +

13 25 12 16 38 28i i i= − − + + = − +

3. Simplify: 6 4 9 5− − + − −

Solution: ( ) ( )6 4 9 5 6 4 1 9 1 5− − + − − = − ⋅ − + ⋅ − −

6 4 1 9 1 5 6 2 3 5 1i i i= − ⋅ − + ⋅ − − = − + − = +

Multiplying Complex Numbers

To multiply two complex numbers a bi+ and c di+ we simply apply the

distributive property and remember that 1i = − so that 2 1i = − .

( )( ) ( )

( ) ( )

2a bi c di ac adi bci bdi 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 5 4 7i i− − +

Solution: ( )( ) 22 5 4 7 8 14 20 35 8 34 35 27 34i i i i i i i− − + = − + + − = − + + = +

5. Expand: ( )2

3 6i− −

Solution: ( ) ( ) ( )( ) ( )2 2 2 23 6 3 2 3 6 6 9 36 36i i i i i− − = − − − + = + +

9 36 36 27 36i i= + − = − +

6. Multiply: ( )( )2 5 3 6− − − −

Solution: ( )( ) ( )( ) 22 5 3 6 2 5 3 6 6 2 6 3 5 30i i i i i− − − − = − − = − − +

( ) ( )6 2 6 3 5 30 6 30 2 6 3 5i i 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

number1 2i− . We can write the problem as

2 3

1 2

i

i

+

−.

What do we do with the above expression? Note that since 1i = − 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 1i = − 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: Note that the product of a complex number and its conjugate is a real number:

( )( ) ( )22 2 2 2 2 2a bi a bi a bi a b i a b+ − = − = − = +

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 3

1 2

i

i

+

−

Solution:

( )

2

2 22

2 3 2 3 1 2 2 4 3 6 2 7 6 4 7 4 7 4 7

1 2 1 2 1 2 1 4 1 4 5 5 51 2

i i i i i i i i ii

i i i ii

+ + + + + + + − − + − += ⋅ = = = = = − +

− − + − +−

8. Divide: 8 2

5

i

i

− +

Solution: 2

2

8 2 8 2 5 40 10 40 10 10 40 2 8

5 5 5 25 25 25 25 5 5

i i i i i ii i

i i i i

− + − + − − += ⋅ = = = + = +

− −

OR 2

2

8 2 8 2 8 2 8 2 2 8

5 5 5 5 5 5

i i i i i ii

i i i i

− + − + − + − −= ⋅ = = = +

−

Complex Number Conjugate

1 2i− 1 2i+

3 5i− + 3 5i− −

12i 12i−

24 9i − 24 9i− −

81

Powers of i We have the following first few powers of i:

2

3

4

1

1

1

i

i

i i

i

= −

= −

= −

=

5

6

7

8

1

1

i i

i

i i

i

=

= −

= −

=

9

10

11

12

1

1

i i

i

i i

i

=

= −

= −

=

…

Do you notice any pattern? Let us try the following:

9. Evaluate: 21i

Solution: We note above that 4 1ni = , 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 21i is:

21 20 51i i i i i= ⋅ = ⋅ = There is a faster way but let us look at another example first:

10. Evaluate: 35i

Solution: 35 32 3 31i i i i 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:

46 2

55 3

1i i

i i i

= = −

= = −

96 0

201 1

1i i

i i i

= =

= =

4.10 Chapter Review

A real number b is called the nth

root of a real number a if na b= .

For example, the 5th root of 32 is 2 since 52 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 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, 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.

5 28 2 2x y x x= ⋅ =4 2 24x y 2x y

3 36 10 327a b b b= ⋅ =

6 9 2 327a b 3a b

2. There can be no fractions under a radical sign or there can be no radicals in the denominator.

2

2 2 2 2x x

x x x x= ⋅ = =

x

x

32 2

3 333

4 4 12 12

9 9 27 3

a a

a a a a= ⋅ = =

2

2

3a

3a

2

5 5 5

2 52 5 2 5 5 2 25

yy y y y yy y

yy y y y= ⋅ = = =

⋅

5

10

y

y

5

10

y=

2

2 3 2 3 2 3

4 92 3 2 3 2 3 4 9

x x x x x x x x x

xx x x x

+ + += ⋅ = =

−− − + −

2 2 2 23 3 3 33 3

3 3 3 3 2 23 33

1 1 a ab b a ab b

a ba b a b a ab b

− + − += ⋅ =

++ + − +

3. The index should be the smallest possible.

4 2 6 2 6 34 4 2 2a b a b ab b ab= = =

3 15 21 3 15 21 5 7 2 36 364 64 4 2x y z x y z xy z y z 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 4 5 8 6 5 5 8 4 6 6 5 3 2 5x x x x x x x x x x− − + = − − + = − +

2. 3 33 20 8 45 3 5 8 5b a b a ab b ab a ab− + = − ⋅ + ⋅2 24a 9b

3 5 8 5 6 5 24 5 18 5b ab a ab ab ab ab ab ab ab= − ⋅ + ⋅ = − + =2a 3b

83

Multiplying Radical Expressions

A. For radical expressions with the same index 2n ≥ we use the following:

n n na b ab⋅ = .

Example 2 2 5 3 7 23 3 3 34 2 8 2xy x y x y xy y⋅ = =

B. For radical expressions with different indices 2n ≥ use rational exponents.

Example ( ) ( ) ( ) ( ) ( ) ( )1 21 3 232 2 2 23 63 6 62 6xy xy xy xy xy xy xy xy⋅ = = = ⋅

3 3 2 4 5 7 56 6 6 6x y x y x y y x y= ⋅ = =

Apply the distributive property and/or special product formulas when multiplying radicals involving two or more terms. Example

( ) ( ) ( )2 2

2 2 22 3 2 3 2 2 3 4 3 4 3 12 4 3a b a a b b a b a b a b a b+ = + ⋅ + = ⋅ + + = + +

One can also perform the expansion above by writing ( )2

2 3a b+ as

( )( )2 3 2 3a b a b+ + and applying FOIL.

Dividing Radical Expressions

A. For radical expressions with the same index 2n ≥ use:

, 0n

nn

a ab

bb= ≠

Example 4 7 17 7 17 4 16 4

4 43 5 44 3 5

80 80 16 2

55

a b c a b c a b ab

a bc c ca bc= = =

B. For radical expressions with different indices 2n ≥ use rational exponents. Example

( )

( )

( )

( )

1 42 23 123 122 4 8 8 8 11 11 11 11

12 12121 3 3 9 9 9 11 114 3

3 34 12

4 44 4 2 1 2 2

8 2 2 2 288 8

x xx x x x x

x x x x xxx x

= = = = = ⋅ =

Rationalizing Denominators See condition 2 for simplified radicals.

84

Rational Exponents

For a real number a and n an integer with 2n ≥ , we have: 1

nna a= .

Also, for an integer m, we have:

( ) or m

mmn nna a a= .

.

Examples 1

4416 16 2= = ; ( ) ( ) ( )3 3 35532 32 2 8− = − = − = −

7

27 37 62

1 1 1 1 15

5 5 125 55 5 55

−

= = = = =⋅

Complex Numbers

We define a complex number to be any number of the form a bi+ ,

where a and b are real numbers and 1i = − . 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 5 4 2 3 4 5 2 1 7i i i i i− + − − = − + − − = − −

2. Subtract: ( ) ( )6 4 2 6 6 4 2 6 8 2i i i i i− − − − = − − − + = − +

3. Multiply: ( )( ) 24 2 3 12 4 6 2 12 10 2 10 10i i i i i i i− − + = − + + − = − + + = − +

4. Expand: ( ) ( )( ) ( )2 22 23 6 3 2 3 6 6 9 36 36i i i i i+ = + + = + +

9 36 36 27 36i i= + − = − +

5. Divide: 2

2

4 6 4 6 3 2 12 8 18 12

3 2 3 2 3 2 9 4

i i i i i i

i i i i

+ + + + + += ⋅ =

− − + −

12 26 12 262

9 4 13

i ii

+ −= = =

+

6. Evaluate: ( )12

49 48 4i i i i i i= ⋅ = ⋅ =