Transcript Operations on Radical Expressions

Operations
on
Radicals
Module 14 Topic 2
Table of Contents
 Slides 3-5: Adding and Subtracting
 Slide 5: Simplifying
 Slides 7-10: Examples
 Slides 11-14: Multiplying
 Slides 15-16: Conjugates
 Slides 17-19: Dividing
 Slides 20-22: Rationalizing the Denominator
 Slides 23-28: Practice Problems
Audio/Video and Interactive Sites
 Slide 7: Gizmo
 Slide 25: Interactive
Addition and Subtraction
Adding and Subtracting radicals is similar to
adding and subtracting polynomials.
Just as you cannot combine 3x and 6y, since they are
not like terms, you cannot combine radicals unless they
are like radicals.
• If asked to simplify the expression 2x + 3x, we recognize that
they each share a common variable part x that makes them like
terms, hence we add the coefficients and keep the variable part
the same.
3x  2 x  5 x
• Likewise, if asked to simplify radicals. If they have “like”
radicals then we add or subtract the coefficients and keep the
like radical. Two radical expressions are said to be like radicals
if they have the same index and the same radicand.
2 5 3 5  5 5
• If the radicals in your problem are different, be sure to check
to see if the radicals can be simplified. Often times, when
the radicals are simplified, they become the same radical and
can then be added or subtracted. Always simplify first, if
possible.
4 5  3 125
4 5  35 5
4 5  15 5
19 5
Recall: Simplifying Radicals
Divide the number
under the radical.
If all numbers are not
prime, continue
dividing.
Find pairs, for a square
root, under the radical
and pull them out.
Multiply the items you
pulled out by anything
in front of the radical
sign.
 8 99x5 y3z 2
 8 9 11 x  x  x  x  x  y  y  y  z  z
 8 3  3 11 x  x  x  x  x  y  y  y  z  z
 8 3  3 11  x  x  x  x  x  y  y  y  z  z
3
Multiply anything left
under the radical .
x
x
y
 8  3  x  x  y  z 11xy
It is done!
 24 x 2 yz 11xy
z
Simplify:
3  12
*These are not like terms, however the 12 can be simplified.
12  2  2  3  4  3  2 3
Now you can simplify by using like terms.
3  12
 32 3
3 3
Gizmo:
Operations
with Radicals
Example:
6 2 6
3 6
Example: 4 7  8 63  4 7  8(3 7 )
 4 7  24 7
 28 7
63  3 7
DO NOT ADD THE NUMBERS UNDER THE RADICAL!
You may not always be able to simplify radicals.
Since the radicals are not the same, and both are in their
simplest form, there is no way to combine them. The
answer is the same as the problem:
Simplify4 7  2 3
Answer:
4 7 2 3
Example:
2 4 2x  6 4 2x  84 2 x
Example:
4 3 3x  5 3 10x  4 3 3x  5 3 10x
*Neither radical can be simplified. The expression is already in simplest form.
Example: A garden has a widt h of 13 and a lengt hof 7 13.
Whatis t heperimet erof t hegarden in simplest radical
form?
*(Hint: P = 2L + 2W)
P  2( 13)  2(7 13)
P  2 13  14 13
P  16 13
Multiplying Radicals
• To multiply radicals, consider the following:
9  4  3 2  6
3
 8  3 64  2  4  8
and
3
 8  64  3  512  8
and
9  4  36  6
Multiplying Radical Expression
n
a  n b  n ab
Multiplying Radicals
You can simplify the product of two radicals by writing
them as one radical then simplifying.
Here is an example of how you can use this
Multiplication property to simplify a radical expression.
Example:
3
7 x  6x
7
3
8
3
 7  6  x7  x8
 3 42x15
3
 7  2  3  x5  x5  x5
 x5 3 42
Example:
15 x 3 y  15 xy
 225 x 4 y 2
 15 15  x 2  x 2  y 2
 15x 2 y
Example:
Multiply
7x ( x  7 7 )
*Recall that to multiply polynomials
you need to use the Distributive
Property.
7x ( x  7 7 )
 7 x2  7 49x
 x 7  77 x
 x 7  49 x
(a + b)(c – d)
=a(c – d) + b(c – d)
=ac – ad + bc - bd
You should recognize these from Topic 1 Notes
x
1
x2
3
x
1
x3
4
x
1
x4
7
x
1
x7
You should also remember the rules of exponents, when you multiply terms with the
same base, you add exponents. When you divide, you subtract exponents.
Ex.
x5  x7  x5 7  x12
Putting the above information together, you should see that these rules can be
combined to solve problems such as those below.
x 7 x 
1)
5
2)
4
x
4
x

4
x5
1
x4
1
x2


1 1
  
x 2 7 
4 1
  
x 5 4 
11
x 20
1
 x7


9
x14
Example:
Conjugates
( 6  2 )( 6  2 )
 6( 6  2)  2( 6  2)
 36  12  12  4
 36  4
Distribute
(2 ways to do so)
 6( 6)  6( 2)  2( 6)  2( 2)
 36  12  12  4
 36  4
 62
 62
4
4
Notice that just like a difference of two squares, the middle terms cancel out.
We call these conjugate pairs.
The conjugate of
a  b is a  b .
They are a conjugate pair. When we multiply a conjugate pair, the radical cancels out and we
obtain a rational number.
Simplified Radical Expressions are recognized by…
 No radicands have perfect square factors other than 1.
 No radicands contain fractions.
 No radicals appear in the denominator of a fraction.
Dividing Radicals
• To divide radicals, consider the following:
3
64 4
 2
3
2
8
and
16 4
 2
4 2
and
3
16
16

 42
4
4
64 3 64 3

 82
3
8
8
Dividing Radical Expressions
n
n
a n a

b
b
Dividing Radicals
You can simplify the quotient of two radicals by
writing them as one radical then simplifying.
Example:
90x18
2x

90 x18
2x
 45x17
 3  3  5  x8  x8  x
 3x8 5x
Dividing Radicals
Example:
6 x8 y 9
5x2 y 4
6 x 6 y 5 x3 y 2 6 y


5
5
Notice that after simplifying the radical, we still have a
square root in the denominator.
We have to find a way to get rid of the radical.
This is called rationalizing the denominator.
Rationalizing the Denominator.
 Case I: There is ONE TERM in the denominator
and it is a SQUARE ROOT.
When the denominator is a monomial (one term), multiply
both the numerator and the denominator by whatever makes
the denominator an expression that can be simplified so that it
no longer contains a radical. In this case it happens to be
exactly the same as the denominator.
7
2
7
2
14
14



2
2
2
4
 Case II: There is ONE TERM in the denominator,
however, THE INDEX IS GREATER THAN TWO.
Sometimes you need to multiply by whatever makes the
denominator a perfect cube or any other power greater than 2
that can be simplified.
3
9
3
11
9 3 1111 3 1089
3

3
11
11 1111
3
 Case III: There are TWO TERMS in the
denominator. We also use conjugate pairs to
rationalize denominators.
3 6
3 6

3 6
3 6

3 6
3 6
9  18  18  36

9  18  18  36
3  2 18  6

9  36
3  2(3 3 )
36
36 3

 1 2 3
3

Be sure to enclose
expressions with
multiple terms in ( ).
This will help you to
remember to FOIL
these expressions.
Always reduce the root
index (numbers
outside radical) to the
simplest form (lowest)
for the final answer.
Application/Critical Thinking
A. Find a radical expression for the perimeter
and area of a right triangle with side lengths
4 3, 12,8 3.
Perimeter … (P = a + b + c)
4 3  12  8 3  12  12 3
Hint: In a right
triangle, the
longest side is the
hypotenuse of the
triangle.
1
Area  bh
2
8 3  13.86
A  (4 3 )(12)
This is longer than
12, so 12 is a leg.
A  48 3
Application/Critical Thinking
B. The areas of two circles are 15 square cm and
20 square cm. Find the exact ratio of the
radius of the smaller circle to the radius of the
larger circle.
Find the radius of each first . . . A = π r2
Smaller Circle
15   r 2
r
15
11
Larger Circle
20   r 2
r
20
11
15
11  15  3  3
20
4
2
20
11
Practice Problems
9  9
Not real

3x 3  2 5x

4
4
46 x12  14 x9
 3 x  2 3x  5 x
 46 x12  14 x9
 3 x  2 15 x
 644 x 21
 3 x  2 x 15
2
4
4
4
 4  161  x16  x5
4
 2  2  7  23  x 4  x 4  x 4  x 4  x5
4
 x 4 644 x5
Practice Problems and Answers
16
81

16
81
4

9

48
4
8
4
48
8
 6



4
64
4
128
100
3
27
222222
4
2255
2 4
100
24 4
4
100
2222222
3
24 8

3
4
4

4
4
4
2 4 400

100

22 5
10

4 5
10

2 5
5
100
100
 9  12 11  12 
4 7
3 3
 99  9 12  11 12  12

 111  20 12
4 7
3 3


3 3
3 3
(4  7 )(3  3 )
(3  3 )(3  3 )
 111  20 4  3
 111  40 3
16  8

12  4 3  3 7  21

93 3 3 3  9
12  4 3  3 7  21
6
5 8
16  8

5 8


5 8
5 8
(16  8 )(5  8 )
(5  8 )(5  8 )


 80  16 8  5 8  8
25  5 8  5 8  8
 72  11 8
17
w10
36

w5  w5
66
w5

6
t

125
3
3

2
x3
1
 x6
1 1 1
   t
5 5 5

2 1
  
x 3 6 
3

5
x6
t
5
6
 x
5
1
m5
1
m8

1 1
  
x 5 8 

 8 5 
  
x  40 40 

3
x 40

40 3
x