Download Solution

8.2 Multiplying, Dividing, and Simplifying Radicals
1
Multiply square root radicals.
2
Simplify radicals by using the product rule.
3
Simplify radicals by using the quotient rule.
4
Simplify radicals involving variables.
5
Simplify other roots.
Multiply square root radicals.
Product Rule for Radicals
For nonnegative real numbers a and b,
a  b  a b
and
a b 
a  b.
That is, the product of two square roots is the square root of the product,
and the square root of a product is the product of the square roots.
It is important to note that the radicands not be negative
numbers in the product rule. Also, in general, x  y  x  y .
EXAMPLE 1 Using the Product Rule to Multiply Radicals
Find each product. Assume that
x  0.
Solution:
3 5
 3 5
 15
6  11
 6 11
 66
13  x
 13  x
 13x
10  10
 10 10
 100
 10
Simplify radicals by using the product rule.
A square root radical is simplified when no perfect square factor
remains under the radical sign.
This can be accomplished by using the product rule:
a b  a  b
EXAMPLE 2 Using the Product Rule to Simplify Radicals
Simplify each radical.
Solution:
60
 4  15
 2 15
500
 100  5
 10 5
17
It cannot be simplified further.
EXAMPLE 3 Multiplying and Simplifying Radicals
Find each product and simplify.
Solution:
10  50
6 2
3 5  4 10
 10  50
 62
 500  100  5
 12
 10 5
2 3
 3  4 5 10  12 50  12 25  2  12 25  2
 12  5  2  60 2
Simplify radicals by using the quotient rule.
The quotient rule for radicals is similar to the product rule.
EXAMPLE 4 Using the Quotient Rule to Simplify Radicals
Simplify each radical.
Solution:
4
49

4
49

48
3

48
3
 16
5
36

5
36

2
7
5
6
4
EXAMPLE 5 Using the Quotient Rule to Divide Radicals
Simplify.
Solution:
8 50
4 5
8 50
 
4
5
50
 2
5
 2  10
 2 10
EXAMPLE 6 Using Both the Product and Quotient Rules
Simplify.
Solution:
3 7

8 2
3 7


8 2
21

16
21

16
21

4
Simplify radicals involving variables.
Radicals can also involve variables.
The square root of a squared number is always nonnegative. The
absolute value is used to express this.
The product and quotient rules apply when variables appear under
the radical sign, as long as the variables represent only nonnegative
real numbers
For any real number a,
a2  a .
x  0, x  x.
EXAMPLE 7 Simplifying Radicals Involving Variables
Simplify each radical. Assume that all variables represent positive
real numbers.
Solution:
x
6
100 p 8
7
y4
x
Since  x
3
 100  p 8

7
y4
 10 p 4
7
 2
y

3 2
 x6
Find cube, fourth, and other roots.
Finding the square root of a number is the inverse of squaring a
number. In a similar way, there are inverses to finding the cube of a
number or to finding the fourth or greater power of a number.
The nth root of a is written
n
In
n
a.
a , the number n is the index or order of the radical.
Index
Radical
sign
n
Radicand
a
It can be helpful to complete and keep a list to refer to of third and fourth
powers from 1-10.
Simplify other roots.
To simplify cube roots, look for factors that are perfect cubes. A
perfect cube is a number with a rational cube root.
3
For example, 64  4 , and because 4 is a rational number, 64 is a
perfect cube.
Properties of Radicals
For all real number for which the indicated roots exist,
n
n
a na
a  b  ab and n 
b  0 .
b
b
n
n
EXAMPLE 9 Finding Cube Roots
Find each cube root.
Solution:
3
64
4
3
27
 3
3
512
8
EXAMPLE 8 Simplifying Other Roots
Simplify each radical.
Solution:
3
108
 3 27  3 4
 33 4
4
160
 4 16 10
 4 16  4 10
4
16
625
4
16
4
625
2

5
 2 4 10
EXAMPLE 10 Finding Other Roots
Find each root.
Solution:
4
81
 4 81
3
 3
4
81
Not a real number.
5
243
3
5
243
 3
EXAMPLE 9 Simplifying Cube Roots Involving Variables
Simplify each radical.
Solution:
 z3
3
z
3
8x 6
 3 8  3 x6
3
54t 5
 3 27t 3  2t 2
3
a15
64
9
3
15
a
 3
64
 2x 2
 3 27t 3  3 2t 2
a5

4
 3t 3 2t 2
8.3 Adding and Subtracting Radicals
1
Add and subtract radicals.
2
Simplify radical sums and differences.
3
Simplify more complicated radical expressions.
Add and subtract radicals.
We add or subtract radicals by using the distributive property. For
example,
8 36 3
 86 3
 14 3.
Only like radicals — those which are multiples of the same root of
the same number — can be combined this way. The preceding
example shows like radicals. By contrast, examples of unlike radicals
are
2 5 and 2 3,
as well as 2 3 and 2 3 3.
Note that
5 + 3 5 cannot be simplified.
Radicands are
different
Indexes are different
EXAMPLE 1 Adding and Subtracting Like Radicals
Add or subtract, as indicated.
8 52 5
3 11 12 11
7  10
Solution:
 8  2 5
  3  12  11
 10 5
 9 11
It cannot be
added by the
distributive
property.
Simplify radical sums and differences.
Sometimes, one or more radical expressions in a sum or difference
must be simplified. Then, any like radicals that result can be added or
subtracted.
EXAMPLE 2 Simplifying Radicals to Add or Subtract
Add or subtract, as indicated.
27  12
2 3 54  4 3 2
5 200  6 18
Solution:
3 32 3
5

100  2  6
 
5 3
5

100  2  6
92
 

9 2
2


3

27  3 2  4 3 2


 2 3 3 2  4 3 2
 50 2 18 2
 63 2  43 2
 32 2
 10 3 2
Simplify more complicated radical expressions.
When simplifying more complicated radical expressions, recall the
rules for order of operations.
A sum or difference of radicals can be simplified only if the radicals
are like radicals. Thus,
5  3 5  4 5, but 5  5 3
cannot be simplified further.
EXAMPLE 3 Simplifying Radical Expressions
Simplify each radical expression. Assume that all variables represent
nonnegative real numbers.
7  21  2 27
6  3r  8r
Solution:
 7  21  2 27
 7 3  2 27
 147  2 27
 6  r  2 2r
 3 2r  2 2r
 7 3  2 3 3
 6  3r  2 2r
 5 2r
 49  3  2 27
7 36 3
 18r  2 2r
 49  3  2 27
 13 3
 9  2r  2 2 r
 
EXAMPLE 3 Simplifying Radical Expressions (cont’d)
Simplify each radical expression. Assume that all variables represent
nonnegative real numbers.
y 72  18 y 2
3
81x4  5 3 24 x4
Solution:
 9  8  9  2y 
 y 3 8   3 2 y 
 y 3   2 2   3 2 y 
y
2
2
2

 
y 6 2  3
2  y2


 
 6 2y 3 2y

 3 2y
 3x  3 3x  5  2 x   3 3x
 3y 2
 3x  3 3x  10 x  3 3x
3
27 x3  3 3x  5

 13x 3 3x
 
3
8 x 3  3 3x


8.4 Rationalizing the Denominator
1
Rationalize denominators with square roots.
2
Write radicals in simplified form.
3
Rationalize denominators with cube roots.
Rationalize denominators with square roots.
It is easier to work with a radical expression if the denominators do not
contain any radicals.
1
1 2
2


2
2 2 2
This process of changing the denominator from a radical, or irrational
number, to a rational number is called rationalizing the denominator.
The value of the radical expression is not changed; only the form
is changed, because the expression has been multiplied by 1 in
the form of
2
.
2
EXAMPLE 1 Rationalizing Denominators
Rationalize each denominator.
Solution:
18
24
18 6


2 6 6
16
8

16
2

2 2 2
18 6

26

16 2

22
16 2

4
18 6
12
3 6

2
4 2
Write radicals in simplified form.
Conditions for Simplified Form of a Radical
1. The radicand contains no factor (except 1) that is a perfect square
(in dealing with square roots), a perfect cube (in dealing with cube
roots), and so on.
2. The radicand has no fractions.
3. No denominator contains a radical.
EXAMPLE 2 Simplifying a Radical
5
.
18
Simplify
Solution:
5

18

3 5  2

18
5 18

18 18
3 10

18

5  18
18
10

6

5  92
18

5 9 2
18
EXAMPLE 3 Simplifying a Product of Radicals
Simplify
1 5
 .
2 6
Solution:

1 5

2 6

5
12

5
12

5
3

2 3 3

5 3
6

15
6
EXAMPLE 4 Simplifying Quotients Involving Radicals
Simplify. Assume that p and q are positive numbers.
5 p2q2
7
5p
q
Solution:

5p
q

q
q
5 pq

q

5 p2q2
7
35 p 2 q 2

7
pq 35

7


5 p2q2
7
7

7
p 2 q 2 35
7
EXAMPLE 5 Rationalizing Denominators with Cube Roots
Rationalize each denominator.
Solution:
3
3
5
6
2
3
3
5 3 62
3 
6 3 62
3
2 3 32
 3 
3 3 32
3


3
5  62
3
3
63
2  32
3

3
33
180
6
3
18

3
3
3
3
2
3
3
3 3 42 x 2
3 16 x 2 3 3  2  8 x 2
8

6
x
,x  0 




3
3
3
2
2
3
3
3
4x
4x
4x
4x 4 x
4 x
3
3
6 x2

2x
8.5 More Simplifying and Operations with Radicals
1
Simplify products of radical expressions.
2
Use conjugates to rationalize denominators of radical expressions.
3
Write radical expressions with quotients in lowest terms.
EXAMPLE 1 Multiplying Radical Expressions
Find each product and simplify.
2

8  20



2 5 3 
32 2

Solution:

 2 2 2  45

 2 2 2


4 5
 2 2 2 2 5
 42 5 2
 4  2 10


 2

 3   2  2 2   5 3  3   5 3  2 2 
 6  11  10 6
 11  9 6
EXAMPLE 1 Multiplying Radical Expressions (cont’d)
Find each product and simplify.


2 5 
10  2

Solution:
 2
 10   2  2   5  10   5  2 
 20  2  50  10
 2 5  2  5 2  10
EXAMPLE 2 Using Special Products with Radicals
Find each product. Assume that x ≥ 0.

5 3

2
4
2 5



2  x 
2
2
Solution:

 5
2
2
 5   3  3
2
 4 2
2


 2 4 2  5  5
 56 5 9
 32  40 2  25
 14  6 5
 57  40 2
Remember only like radicals can be combined!
2
 2  2  2
2
 x x
 44 x  x
2
Using a Special Product with Radicals.
Example 3 uses the rule for the product of the sum and difference of
two terms,
2
2
x

y
x

y

x

y
.



EXAMPLE 3 Using a Special Product with Radicals
Find each product. Assume that

32

32

y  0.

y 4


 y
y 4

Solution:

 3
 3 4
 1
2
  2
2
2
 y  16
  4
2
Use conjugates to rationalize denominators of radical
expressions.
The results in the previous example do not contain radicals. The
pairs being multiplied are called conjugates of each other.
Conjugates can be used to rationalize the denominators in more
complicated quotients, such as
2
.
4 3
Using Conjugates to Rationalize a Binomial
Denominator
To rationalize a binomial denominator, where at least one of those
terms is a square root radical, multiply numerator and denominator
by the conjugate of the denominator.
EXAMPLE 4 Using Conjugates to Rationalize Denominators
Simplify by rationalizing each denominator. Assume that
3
2 5
t  0.
5+3
2 5
Solution:


3
2 5

2 5 2 5

2


 5
3 2 5
2
2

3 2 5
45



3 2 5



1

 3 2  5


  2  5 
5 2  5 
5 3
2
2 5 563 5
2 2  52
5 5  11

45

5 5  11
1


  5 5  11
 11  5 5
EXAMPLE 4 Using Conjugates to Rationalize Denominators (cont’d)
Simplify by rationalizing each denominator. Assume that
3
2 t
Solution:

3
2 t

2 t 2 t



2  t 
3 2  t 

3 2 t
2
2
4t
t  0.
EXAMPLE 5 Writing a Radical Quotient in Lowest Terms
Write
5 3  15
10
Solution:

5

3 3
10
3 3

2

in lowest terms.
8.7 Using Rational Numbers as Exponents
1
Define and use expressions of the form a1/n.
2
Define and use expressions of the form am/n.
3
Apply the rules for exponents using rational exponents.
4
Use rational exponents to simplify radicals.
Define and use expressions of the form a1/n.
Now consider how an expression such as 51/2 should be defined, so
that all the rules for exponents developed earlier still apply. We define
51/2 so that
51/2 · 51/2 = 51/2 + 1/2 = 51 = 5.
This agrees with the product rule for exponents from Section 5.1. By
definition,
 5  5   5.
Since both 51/2 · 51/2 and 5  5 equal 5,
this would seem to suggest that 51/2 should equal 5.
Similarly, then 51/3 should equal 3 5.
Review the basic rules for exponents:
a m a n  a mn
a 
m n
 a mn
am
mn

a
an
Slide 8.7-4
Define and use expressions of the form a1/n.
a1/n
If a is a nonnegative number and n is a positive integer, then
a1/ n  n a .
Notice that the denominator of the rational exponent is the index
of the radical.
Slide 8.7-5
EXAMPLE 1 Using the Definition of a1/n
Simplify.
Solution:
491/2
1000
811/4
1/3
 49
7
 3 1000
 10
 4 81
3
Slide 8.7-6
Define and use expressions of the form am/n.
Now we can define a more general exponential expression, such as
163/4. By the power rule, (am)n = amn, so
 
 16
1/ 4 3
3/ 4
16
4

3
16
 23  8.
However, 163/4 can also be written as
3/ 4
16
 16

3 1/ 4
  4096 
1/ 4
 4 4096  8.
Either way, the answer is the same. Taking the root first involves
smaller numbers and is often easier. This example suggests the
following definition for a m/n.
am/n
If a is a nonnegative number and m and n are integers with n > 0, then
a
m/ n
 a
  a
1/ n m
n
m
.
Slide 8.7-8
EXAMPLE 2 Using the Definition of am/n
Evaluate.
Solution:
9
5/2
 9
8
5/3
 8
–27

 35
 243

 25
 32
1/ 2 5
1/ 3 5
2/3
   27

1/ 3 2
   3  9
2
Slide 8.7-9
Using the definition of am/n.
Earlier, a–n was defined as
an 
1
an
for nonzero numbers a and integers n. This same result applies to
negative rational exponents.
a−m/n
If a is a positive number and m and n are integers, with n > 0, then
a
m / n

1
a
m/n
.
A common mistake is to write 27–4/3 as –273/4. This is incorrect. The
negative exponent does not indicate a negative number. Also, the negative
exponent indicates to use the reciprocal of the base, not the reciprocal of
the exponent.
Slide 8.7-10
EXAMPLE 3 Using the Definition of a−m/n
Evaluate.
Solution:
36–3/2
1
 3/ 2
36
81–3/4
1
 3/ 4
81


1
 36 
1/ 2 3
1
81 
1/ 4 3
1
 3
6
1

216
1
 3
3
1

27
Slide 8.7-11
Apply the rules for exponents using rational exponents.
All the rules for exponents given earlier still hold when the exponents
are fractions.
Slide 8.7-13
EXAMPLE 4 Using the Rules for Exponents with Fractional Exponents
Simplify. Write each answer in exponential form with only positive
exponents.
Solution:
71/ 3  7 2 / 3
 71/ 3 2 / 3
7
92 / 3
9 1/ 3
 92 / 31/ 3
9
 27 
 
 8 
5/ 3
31/ 2  32
35/ 2
5/ 3
27
 5/ 3
8
1/ 2 4/ 25/ 2
3
27 


8 
35
 5
2
3
3
1/ 3 5
1/ 3 5
2/ 2
Slide 8.7-14
EXAMPLE 5 Using Fractional Exponents with Variables
Simplify. Write each answer in exponential form with only positive
exponents. Assume that all variables represent positive numbers.
Solution:
a

2 / 3 1/ 3 2 6
b c
r 2 / 3  r1/ 3
r 1
a 
 1/ 4 
b 
2/3
3
 a
 b  c 
2/ 3 6
1/ 3 6
2 6
 a12 / 3b 6 / 3c12
 a 4b 2 c12
 r 2/ 31/ 33/ 3
 r6/3
 r2
a 


b 
a6 / 3
 3/ 4
b
a2
 3/ 4
b
2/3 3
1/ 4 3
Slide 8.7-15
Use rational exponents to simplify radicals.
Sometimes it is easier to simplify a radical by first writing it in
exponential form.
Slide 8.7-17
EXAMPLE 6 Simplifying Radicals by Using Rational Exponents
Simplify each radical by first writing it in exponential form.
Solution:
4
12
2
 
6
x
3
 12
2 1/ 4

 x

3 1/ 6
 121/ 2
 12
 x1/ 2
 x  x  0
2 3
Slide 8.7-18
8.6 Solving Equations with Radicals
1
Solve radical equations having square root radicals.
2
Identify equations with no solutions.
3
Solve equations by squaring a binomial.
4
Solve radical equations having cube root radicals.
Solving Equations with Radicals.
A radical equation is an equation having a variable in the radicand,
such as
x 1  3
or
3 x  8x  9
Solve radical equations having square root radicals.
To solve radical equations having square root radicals, we need a new
property, called the squaring property of equality.
Squaring Property of Equality
If each side of a given equation is squared, then all solutions of the
original equation are among the solutions of the squared equation.
Be very careful with the squaring property: Using this property can give a new
equation with more solutions than the original equation has. Because of this
possibility, checking is an essential part of the process. All proposed solutions
from the squared equation must be checked in the original equation.
EXAMPLE 1 Using the Squaring Property of Equality
Solve.
9 x  4
Solution:

9 x

2
4
2
9  x  16
9  x  9  16  9
x  7
x  7
7
It is important to note that even though the algebraic work may be done
perfectly, the answer produced may not make the original equation true.
EXAMPLE 2 Using the Squaring Property with a Radical on Each Side
Solve.
3x  9  2 x
Solution:

3x  9
  2 x 
2
2
3x  9  4 x
3x  9  3x  4x  3x
x 9
9
EXAMPLE 3 Using the Squaring Property When One Side Is Negative
Solve.
x  4
Solution:
 x
2
  4 
Check:
x  4
16  4
4  4
2
x  16
False

Because
x represents the principal or nonnegative square root of x in
Example 3, we might have seen immediately that there is no solution.
Solving a Radical Equation.
Solving a Radical Equation
Step 1 Isolate a radical. Arrange the terms so that a radical is
isolated on one side of the equation.
Step 2 Square both sides.
Step 3 Combine like terms.
Step 4 Repeat Steps 1-3 if there is still a term with a radical.
Step 5 Solve the equation. Find all proposed solutions.
Step 6 Check all proposed solutions in the original equation.
EXAMPLE 4 Using the Squaring Property with a Quadratic Expression
Solve x  x2  4x 16.
Solution:
x2 

x 2  4 x  16

2
x 2  x 2  x 2  4 x  16  x 2
0  4x  4x 16  4x
4 x 16

4
4
x  4
Since x must be a positive number the solution set is Ø.
EXAMPLE 5 Using the Squaring Property when One Side Has Two Terms
Solve
2 x  1  10 x  9.
 2 x  1   10 x  9 
2
4 x  4 x  1  10 x  9  10 x  9  10 x  9
Solution:
2
2
4 x 2  14 x  8  0
 2x  1 2 x  8  0
2x 1  0
1
x
2
or
2x  8  0
x4
Since 2x-1 must be positive the solution set is {4}.
EXAMPLE 6 Rewriting an Equation before Using the Squaring Property
Solve.
25 x  6  x
Solution:
25 x  6  6  x  6

25 x

2
  x  6
2
25 x  25 x  x 2  12 x  36  25 x
0  x 2  13x  36
0   x  4 x  9
or
0  x4
0  x 9
x4
x 9
The solution set is {4,9}.
Solve equations by squaring a binomial.
Errors often occur when both sides of an equation are squared. For
instance, when both sides of
9x  2x 1
are squared, the entire binomial 2x + 1 must be squared to get 4x2 + 4x
+ 1. It is incorrect to square the 2x and the 1 separately to get 4x2 + 1.
EXAMPLE 7 Using the Squaring Property Twice
Solve.
x 1  x  4  1
Solution:

x 1  1 x  4
x 1
  1 
2
x4
x  1  1  2 x  4   x  4

4  2 x4
2
16  4x 16
32 4x

4
4
x 8
The solution set is {8}.

2

2
Solve radical equations having cube root radicals.
We can extend the concept of raising both sides of an equation to
a power in order to solve radical equations with cube roots.
EXAMPLE 8 Solving Equations with Cube Root Radicals
Solve each equation.
3
Solution:

3
7 x  3 4x  2
 
3
7x
3
4x  2
7 x  4x  2
3x 2

3 3
2
x
3
2
 
3
3

3
x2  3 26 x  27
 x  
3
2
3
3
26 x  27

3
x 2  26 x  27
0  x 2  26  27
0   x  27  x 1
or
0  x  27
0  x 1
x  27
x 1
27,1