Download Section 1.2 Radicals and Irrational numbers

Section 1.2
Radicals and Irrational numbers
What are irrational numbers and why do they exist?
length (l) = 4.5 inches
Width = 2 inches
You remember how to calculate the area of a rectangle or square, don’t you? The area
of a rectangle is the length times the width: A = l  w. For a square, the area is the
length of a side times itself: A = s  s = s2. Measurements of area are in square units, so
that if the dimensions are in inches, the area will be in square inches and if the
dimensions are in meters the area will be in square meters.
Area = l  w
= 4.5 inches  2 inches
= 9 square inches
side (s) =
5 meters
side (s) =
5 meters
Area = s2
= (5 meters)2
= 25 square meters
Calculating the area of a rectangle given the sides is simple enough. Calculating the
sides of a square to give a certain area is a little harder. Suppose you wanted to plant a
square field of wheat that had an area of 36 square meters, and you wanted to compute
the length of one side of the field. You would need to use the square root to compute
the side, as shown:
Area  s 2
36  s 2
Area =
36 m2
solve by taking the square root of both sides:
36  s 2
Side length = s
6s
Since we are assured that s will be positive, the square root of s squared is simply s.
Algebraically, s 2  s . The answer to our question is that a field 6 meters on a side will
have an area of 36 square meters.
What if we wanted to plant half as much wheat (meaning half the area)? Now what side
length would you use? If you answered 3 meters, then you are incorrect. The area of a
field 3 meters on a side is 3 meters  3 meters = 9 square meters, which is not half of 36
square meters. If we want to have a square with an area of 18 square meters, then we
follow the same procedure we did for the 36 square meter field:
Area  s 2
18  s 2
solve by taking the square root of both sides:
18  s 2
18  s
The result is a side length of 18 meters. It may seem strange for you to say it this
way, but it is correct. 18 Does not have an exact rational equivalent, as 36 did.
18 is approximately equal to 4.2426, the first five digits of a non-terminating, nonrepeating decimal. If you want to express an exact answer, only 18 will do (or
preferably 3 2 , which we’ll talk more about later). The figure below illustrates a similar
concept using 1-foot square floor tiles.
Floor A
Area = 36 Square feet
uses 36 1-foot by 1-foot tiles
Each side length is 6 feet
Floor B
Side length is half of Floor A
Area = 9 Square feet
uses 9 1-foot by 1-foot tiles
Each side length is 3 feet
Floor C
Area is half of floor A
Area = 18 Square feet
uses 18 1-foot by 1-foot tiles (but not all are whole tiles)
Each side length is 18 feet (approximately 4.2426 feet)
The number 18 is important to us as the exact side length of a square with an area of
18 square units. It is as useful a number as 6 was when it represented the exact side
length of a square with an area of 36 square units. However, as mentioned above it has
no rational equivalent (meaning that it cannot be expressed as the ratio of two integers
or as a terminating or repeating decimal). Therefore, it is an irrational number.
Now suppose we have a circular field of wheat, with a radius of 8 meters. The area of
this field is
A   r2
Radius =
8 meters
A   (8) 2
A  64 m 2
A  201.06 m 2
The area of the circular field is exactly 64 square meters. Again, it may seem strange
for you to say it this way, but the value 64 is the only value that is exactly correct. The
area can be rounded to approximately 201.06 square meters. The number 64 is also
an irrational number. The product of an irrational number and a real number is
irrational.
Now, suppose we have a very small farm, with an 18 square meter area wheat field next
to a square shed that is 5 meters on each side. Away from the square field and shed is
a circular wheat field with an 8 meter radius.
Irrational Farm
5 meters
Square
Shed
3 2 meters
Square
Wheat Field
Area =
18 m2
Circular
Wheat Field
Area =
64 m2
Brick Walkway
Length of Brick Walkway = 5  3 2 meters
Total area of both wheat fields =
18 + 64 square meters
The total area of our wheat plantings would be 18  64 square meters. There is no
simpler way to give an exact answer. The number 18  64 is an irrational number,
since you cannot express it as a ratio of two integers. Whenever you add a rational
number and an irrational number, the result is an irrational number. You can see from
the diagram that there is a brick walkway along the side of the shed and square wheat
field. The length of this walkway is 5  3 2 meters. Once again, the number may look
complicated, but there is no simpler way to give an exact
answer. Irrational numbers such as these have two
Examples of
terms. Terms are parts of an algebraic expression that
Irrational Numbers
are separated by + or – signs. If we wanted an
11
approximate answer for the length of the walkway, we
2
would use a calculator to evaluate the irrational number:
5  3 2  9.243 meters. As the table to the right shows,
irrational numbers can take many forms
4 21
3  5
1 1

5
Most of the irrational numbers we will be using in this
2 3
book will involve radicals. We will be doing all kinds of
arithmetic operations with irrational numbers and
radicals. That’s why operations and simplification with radicals is the topic of the
remainder of this section.
Check what you’ve learned so far
Example 1
Classify each of the following as rational or irrational, and give its exact
decimal value or approximate its value to the nearest thousandth.
A)
9
16
B)
5
11
C)
5 6
D)

4
Click Here for the Answer
Radicals in simplest form
You have certainly used radicals in some of the mathematics you have studied so far.
Expressions like 2 , 10 , and x probably look familiar to you. You may even be
3
. If you are not
4
comfortable with these, you will be by the end of this section. The following box shows
some properties of radicals:
comfortable handling radicals such as 3 15 ,
x 3 ,
9d , and
Properties of Radicals
The number beneath (or inside) the radical is called the radicand. For example, for
the radicand is 7.
7
Radicals can be added if the radicands are the same. They add in the same way that
variables do. You know that 2x + 5x = 7x. Applying this same idea to radicals
gives 2 11  5 11  7 11 . Furthermore, 5 3  4 15 cannot be simplified, since the
radicands are not the same. This is comparable to saying that 5y – 4z cannot be
simplified (which is true).
For positive real numbers a and b,
ab  a  b
a
a

b
b
 a
2
(The Product Property of radicals)
(The Quotient Property of radicals)
a
 a  a
2
The last four properties are useful when working to express radicals in simplest form.
Just as fractions have an accepted simplest form, so do radicals. The rules for
expressing radicals in simplest form are shown in the following box:
Radicals in Simplest Form
For a radical to be considered in simplest form, it must meet the following criteria:
1. The radicand must not have any factors that are squares of integers
2. The radicand must not be a fraction
3. There must not be radicals in the denominator of any fraction
While it’s pretty simple to detect if a radical is not in simplest form, simplifying it to
comply with the rules takes a little bit of work.
Simplifying a radical when the radicand has a factor that is a square of an
integer
When we were talking about the square wheat field with an area of 18 m2, we briefly
mentioned that it is preferable to express 18 as 3 2 . How do we figure out that those
two numbers are equivalent? Let’s start by examining the radicands. A radicand of 18
has 9 as a factor (18 = 9  2). Since 9 is a perfect square (9 = 32), 18 is not in simplest
form. We begin the simplification by factoring the 18 under the radical:
18  9  2
Next ,use the product property of radicals to write the expression as two radicals:
 9 2
Use the fact that 9 = 32 to change the radicand of the first radical:
 32  2
Now use the property
 a   a to complete the simplification:
2
 3 2  3 2
Let’s try it again with
75 :
75  25  3
 25  3
 52  3
5 3
Sometimes a radicand has more than one perfect square factor. Consider 72 , for
example. We could factor the radicand as 4  18 , 9  8 , or 36  2 . Let’s see what
happens in each of those situations:
72  4  18
72  9  8
72  36  2
 4 18
 9 8
 36 2
 2 18
3 8
6 2
These results are somewhat disturbing. After all, shouldn’t there be one simplest form
for a radical? In fact, there is only one simplest form. You see, after you have
simplified a radical, you must examine the new radicand to make sure there are no
remaining perfect square factors. Of the three simplifications for 72 , only one, 6 2 ,
is fully simplified. The other two have radicands with perfect square factors, and must
be simplified further.
72  2 18
72  3 8
 2 9 2
 3 4 2
2 9 2
3 4 2
 23 2
 3 2 2
6 2
6 2
As you can see, the result is 6 2 regardless of how you start, as long as you keep
checking the radicand to make sure the radical is completely simplified. As you may
have noticed, if you find the largest possible perfect square factor, then the
simplification will be completed in one step.
The main task of radical simplification is recognizing
the perfect square factor in the radicand. It will be
very helpful to memorize the squares of the first twelve
to fifteen integers (see the box below).
Squares of the First Fifteen Integers
12  1
22  4
32  9
42  16
52  25
6 2  36
7 2  49
82  64
9 2  81
102  100
112  121
12 2  144
132  169
142  196
152  225
If you have to simplify a radical, say 300 , it is best to start with larger numbers from
the perfect square list (shown in the box above) and work your way down. You don’t
need to check any numbers greater than half of the radicand, since it’s impossible to
have a factor that’s greater than half of a number (do you understand why?) For 300 ,
we first check 144 (which is not a factor), then 121 (which is not), then 100, which is a
factor of 300. Now, the simplification will go like this:
300  100  3
 100 3
 10 3
Check what you’ve learned so far
Example 2
Simplify each of the following expressions.
A)
56
C)
108
B)
5 24
Click Here for the Answer
One of the reasons for simplifying all radicals is so that we can add them and express
the result in simplest form. As we learned, you can add or subtract radicals only if they
have the same radicand, such as:
4 5  18 5  13 5
At first, you may think that this means you can’t add 10 7  3 28 , but you can’t make
that decision until all the radicals are simplified. 3 28 in simplest form is 6 7 , so
10 7  3 28
 10 7  6 7
 16 7
Let’s get back to our farm. If you had to separately fence a square tomato patch with an
area of 20 square meters and a square pumpkin patch with an area of 125 square
meters, how many meters of fence do you need?
Pumpkin
Patch
125 m2
Tomato
Patch
20 m2
Side of tomato patch = 20 meters
Fence for tomato patch = 4 20 meters
Side of pumpkin patch = 125 meters
Fence for pumpkin patch = 4 125 meters
Total fence length = 4 20  4 125 meters
 4 4 5  4 25 5
 8 5  20 5
 28 5 meters
Check what you’ve learned so far
Example 3
Perform the indicated operation. Express your answer in simplest form.
A)
C)
13 5  22 5
B)
8 2  6 3  15 2
18  20  50
Click Here for the Answer
Multiplying Radicals
Earlier, we learned the product property of radicals: ab  a  b . We can use the
converse of this property when we need to multiply radicals:
11  5  11 5  55
Of course, we need to express the result in simplest radical form, as shown in the
following:
6  21  6  21  126
 9  14  3 14
If we are multiplying a couple of more complicated irrational numbers
that have two terms, we need to recall how we multiply binomial
expressions, such as  x  4  2 x  3 . By applying the distributive
property, the resulting product is the sum of each term in the first factor
times each term in the second factor.You may have learned the
acronym FOIL to remind you to add the products of the First, Outer,
Inner, and Last terms.
 x  4  2 x  3   x  2 x    x  3   4  2 x    4  3
I remember FOIL!
 2 x 2  3x  8 x  12
 2 x 2  5 x  12
Let’s apply this concept to multiplying 4  3 5 times 2  5 :
 4  3 5  2  5    4  2    4   5    3 5   2    3 5  5 
 8  4 5  6 5  3 5 
2
 8  2 5  3 5
 7  2 5
Check what you’ve learned so far
Example 4
Perform the indicated operation. Express your answer in simplest form.
A)
3 6  7 3
C)
3

B)
2 7 62 2

D)
9 5 
 6  3  6  3 
2
Click Here for the Answer
As shown in the solutions to Example 4, the answer to part D) is 33. This answer has
no radicals! This will always happen when the two numbers being multiplied have the
same rational part and opposite irrational parts. A pair of numbers such as these is
called a conjugate pair, or simply conjugates. Conjugates should
remind you of multiplying binomial expressions that represent the sum
I remember
2
that!
and difference of the same two terms, such as  x  7  x  7   x  49 .
You will need to create and work with conjugates when we work on
dividing radical expressions.
Dividing Radical Expressions
Suppose you need to construct the framing of the house roof
shown in the figure. You want to know the length c of the
center ridge pole. We recall from the properties of special
right triangles (there’s that Geometry knowledge again) that
c
1

20
3
c
30°
20 feet
c 3  20
c
20
feet
3
20
is not in simplest form because it has a radical in the denominator,
3
so we have to simplify it.
The expression
The steps in simplifying a fraction with a radical in the denominator are similar to the
steps in changing a fraction to an equivalent fraction with a different denominator. If you
want to change 23 to an equivalent fraction that has 12 as its denominator, you multiply
by 1, the multiplicative identity. You should choose 44 as the preferred version of 1 for
the fraction multiplication:
2 4 8
 
3 4 12
Simplifying radicals is similar: you multiply by 1 by choosing an appropriate fraction
whose numerator and denominator are the same. When there is a single term radical in
the denominator, you simply use that radical as the numerator and denominator. Let’s
see how to simplify the expression from the roofing example:
20 3 20 3
3


multiply by 1 in the form of
2
3 3
3
3
 

20 3
the result has no radical in the denominator
3
Sometimes, it may be easier to simplify the radical in the denominator before
multiplying, such as when completing the following example:
12 3 12 3

50
5 2

12 3 2

5 2
2

12 6
5
 2

12 6
10

6 6
5
2
If there is a 2-term radical in the denominator, you need to multiply by 1 in the form of
the conjugate of the denominator over itself, as shown in this example:
7
5  2 11
35  14 11


5  2 11 5  2 11 25  (2 11)(2 11)


35  14 11
25  4 
 11
2
35  14 11
Woo hoo! The denominator has no radical!
25  44

35  14 11
19
As you can see, multiplying by 1 in the form of the conjugate of the denominator over
itself resulted in a denominator with no radical. This is what we were seeking.
Reminder: Simplifying an expression means changing it to a generally accepted form
without changing its value. This means that we can perform indicated multiplications,
add similar terms, and use identity elements. Specifically, we can add zero and multiply
by one in whatever interesting forms are useful to us. However,
we must be careful not to do any operations that will change the
value of the expression. Students often are so concerned with
figuring out what to use to multiply the denominator that they forget
to multiply the numerator by the same thing. This means they
have not multiplied by the identity, and they have changed the
value. This kind of infraction will get you arrested by the math
police.
Check what you’ve learned so far
Example 5
Perform the indicated operation. Express your answer in simplest form.
A)
49
16
B)
56
14
C)
22
2
D)
17
12
E)
4
2 6
F)
3 2 3
25 3
Click Here for the Answer
Section 1-2 Student Problems
GUIDED PRACTICE
1)
Simplify each of the following expressions.
A)
2)
3)
4)
B)
63
C)
108
7 120
D)
11 72
Perform the indicated operation. Express your answer in simplest form.
A)
8 23  17 23
B)
46 5  7 5
C)
6 29  3 14  8 29
D)
7 15  9 60
E)
54  150
F)
11 45  9 5  12 20
Perform the indicated operation. Express your answer in simplest form.
A)
6  30
D)
10 13  5 2


B)
3 7  2 21
E)
3  9 5  2  4 5 
C)
 4 13 
2
Perform the indicated operation. Express your answer in simplest form.
A)
25
81
B)
30
6
C)
D)
9
20
E)
6
4 7
F)
3
6
11  3
45 3
INDEPENDENT PRACTICE
Simplify each of the following expressions.
1)
76
2)
80
3)
168
4)
180
5)
3 28
6)
7 32
7)
19 24
8)
23 82
Perform the indicated operation. Express your answer in simplest form.
9)
4 17  8 17
12)
24  54
13)
15) 8 44  99
98  8  100
18)
11) 5 13  7 11  3 11  2 13
10) 13 29  28 29
20  45
14)
27  18
16) 7 196  3 72
17) 6 52  5 48
19) 3 48  4 75  7 150
20) 4 63  2 147  7 80
Perform the indicated operation. Express your answer in simplest form.
2 6
21)
22)
24) 2 24  9 18
27)

30)
 18 
56

2
39)
4
 3
4 5 
29)
7 3 


8  7 
26)
2
31) 3 2  4 3
33) 2 5 6  9 5
36)
25) 7 28  5 40
28)
4
23) 3 6  5 15
3 8

2

3 5 63 2



32)
2  11 3

34)
6
37)
2
6 4 59 6
40)
8
3 7 6


2

2
2

3 45 2
35)
3  2  4  2 
38)
6  3 5 
41)
2
2
1
11
43)
45)
4
5 12
46)
1
8 3
9
25
44)
47)

7  3 2 5 6  3 14
Perform the indicated operation. Express your answer in simplest form.
42)

1
8
27
49

48)
28
8
49)
16
7
50)
32
2
51)
98
18
52)
75
9
53)
12
5
54)
1
1 2
55)
5
3  14
56)
6
4  3 22
57)
4 5
7 5
58)
3 7 6
11  2 7
59)
8 3
5 36
60)
2  4 11
4 11  2
61)
2 37
3 5
62)
5 6 2 3
4 6 7 3
Evaluate the expression
b  b2  4ac
for each of the following problems. Express your
2a
answer in simplest form.
63) a = 4, b = 13, c = 9
64) a = 2, b = –7, c =5
65) a = 1, b = –8, c = 13
66) a = 4, b = 4, c = –19
Determine whether the following statements are True or False.
67) 2 3  3 2  5 6
68)
69)
ab  a  b
70)
71)
4x 2 x

y
y
72)
3  7  21
3 6 
2
 54


x a 3  y 2   a  y  5x
CONNECTION AND EXTENSION
73) Determine whether the problem shown below is correct or incorrect. If it is incorrect,
identify the mistake in the work.
3

8 5 7 2 3 9 2

=
6

2 5 7 2 3 9 2

= 12 6  54 4  10 21  45 14
= 12 6  108 2  10 21  45 14
74) Determine whether the problem shown below is correct or incorrect. If it is incorrect,
identify the mistake in the work.
18
18
6
6 3



15
5 27 15 3 5 3
75) Which of the following is not equivalent to
A)
B)
2 20
4 10
76) How many different ways can
80 ?
C)
160
2
D)
4 15
3
648 be expressed in radical form? List all of them.
77) Recall that the reciprocal of the number a may be written as 1a , where a ≠ 0. Find the
reciprocal of each of the following numbers. Simplify each expression.
A)
3 4 2
B)
58 7
MIXED REVIEW
78) List the following numbers in order from least to greatest. Express each number is
simplest form.
2 2  3 3 2  12 2 6
79) Classify each of the following numbers using all the terms that apply: natural (counting),
whole, integer, rational, irrational, and real.
A)
7.1923
B)
65
C)
D)
197
80) Simplify each of the following expressions.
A)
2

 7
2
3
 4  8  6  1

B)
81) If a  b = 3a2 + 2b, then evaluate 5 6  3.
6
4
3
 11 5
2 49
26
11
Section 1-2 Additional Material
Section 1-2 Solutions to Sample Problems
Example 1
Answer:
Classify each of the following as rational or irrational, and give its exact
decimal value or approximate its value to the nearest thousandth.
A)
9
16
B)
A)
9
 0.75
16
rational
5
11
B)
5
 0.455
11
rational
C)
5 6  12.247
irrational
D)

6
 0.524
irrational
Back to text
C)
5 6
D)

6
Example 2
Simplify each of the following expressions.
A)
B)
56
4 14
4  14
2 14
C)
108
5 24
5 4  6
5  4  6
5  2 6
10 6
4  27
9 12
2 27
2 93
3 12
3 43
6 3
6 3
Yahoo!! They’re all the same!!
Back to text
36  3
6 3
6 3
6 3
Example 3
Perform the indicated operation. Express your answer in simplest form.
A)
13 5  22 5
B)
7 2 6 3
9 5
C)
8 2  6 3  15 2
18  20  50
3 2  2 5 5 2
2 2  2 5
Back to text
Example 4
Perform the indicated operation. Express your answer in simplest form.
A)
3 6  7 3
B)
3  7  6  3
21 18
9 5 
9 5 9 5 
2
or 92 
 5
9  9  5  5 or 81 5
81 25 or 405
815 or 405
405 or 405
63 2
Need to use FOIL for C) and D):
C)
3

2 7 62 2

D)
 6  3  6  3 
= 18 2  6 4  42 14 2
= 36  6 3  6 3 
= 18 2 12  42 14 2
= 30  4 2
= 36  3
= 33
Back to text
 3
2
2
Back to text
Example 5
Perform the indicated operation. Express your answer in simplest form.
A)
C)
49
=
16
7
49
=
4
16
22 2
22
=

2
2
2
17
17
=
2 3
12
D)
22 2
4
22 2
=
2
=
=
17
3

2 3 3
17 3
2 9
17 3
=
6
=
= 11 2
E)
56
=
14
56
=
14
B)


4 2 6
4
4
2 6
=
=

2 6
2 6 2 6
2 6 2 6



84 6
42 6 2 6 6
84 6
=
10
8 4 6

=
10 10
4 2 6
42 6
= 
or
5
5
5
=
F)




3 2 3 25 3
3 2 3 2 5 3
3 2 3
=
=


2 5 3 2 5 3
25 3
25 3 25 3
6  15 3  4 3  10 9
4  10 3  10 3  25 9
6  11 3  30
=
4  75
24  11 3
=
71
24 11 3
24  11 3

=
or
71
71
71
=
Back to text


4 = 2