Download 4 3 4 3[ 4 3] where a is the coefficient where b is the radica

PRE CALCULUS MATH 11
Substantive Assignment Resource Material
ac b
where a is the coefficient
where b is the radicand
where c is the index [root ]
4 3 would be read as 4 root 3 [it indicates to multiple 4 times the square root of 3]
When there is no index number , it is a square root
In order to simplify radicals we can use one of the following two methods.
Method One: Choosing a factor from the radicand that has a perfect square root
Simplify
20
In this case, the coeficient is 1, and the radicand is 20
[This is called an entire radical when the coefficient is 1]
We can rewrite 20 as
 4  5 
4  2, so we have 2 times the square root of 5, or 2 5
[This is called a mixed radical when the coefficient is a number other than 1]
Method Two: Expressing the radicand as a product of prime numbers
The prime numbers we generally use are 2, 3, 5, 7, 11, 13, 17 …
Simplify 20
In this case, the coeficient is 1, and the radicand is 20
[This is called an entire radical when the coefficient is 1]
We can exp ress 20 as a product of prime numbers
20 
2  2  5 so we have the square root of 2 times 2 times 5
Notice that we have 2  2  5 or one group of twos and the index is also a two
We take the the pair of twos and multiply two by the coeffice int in front
of the radical , so we will have 2 5
[This is called a mixed radical when the coefficient is a number other than 1]
1
Example 1
Simplify the following radical using either method shown previously 162
Using Method Two
Simplify 162
In this case, the coeficient is 1, and the radicand is 162
We can exp ress 162 as a product of prime numbers
162 
3  3 3  3  2
Notice that we have 3  3  3  3  2 or two groups of threes and the index is also a two
We take the the 2 pairs of threes and multiply (3)(3) by the coeffice int in front
of the radical , so we will have 3  3 2 which is 9 2
Using Method One
Simplify 162
In this case, the coeficient is 1, and the radicand is 162
We can exp ress 162 as

81
 2 
The square root of 81 is 9, so we have  9 
 2  9
2
Example 2
What is the simplest mixed radical form of
80m5n3 , m  0, n  0
Using Method Two
Simplify 80m5 n3
In this case, the coeficient is 1, and the radicand is 80m5 n3
We can exp ress 80m5 n3 as a product of prime numbers and var iables
80m5 n3

2  2 2  2 5 m  m  m  m  m  n  n  n
Notice that we have two groups of twos, two groups of m ' s, and one group of n ' s
We take the the 2 pairs of twos, two pairs of m ' s, and the one pair of n ' s and multiply
(2)(2)  m  m  (n) by the coeffice int of 1 which is in front of the radical
So we will have 2  2  m  m  n 5mn , which is 4m 2 n 5mn
2
Using Method One
Simplify 80m5 n3
In this case, the coeficient is 1, and the radicand is 80m5 n3
We can exp ress 80m5 n3
80m5 n3

16  5  m 4  m  n 2  n
So we will have the
 16  
m4
 n  
2

5mn , which is 4m 2 n 5mn
Example 4
What is the entire radical form of 7 5
7 5 is a mixed radical. In order to change to an entire radical we need to move
the 7 [coefficient ] back inside the radical so the coefficient is 1.
7 5 is the same as 5  7  7 or
5  49  245
Now we have our entire radical 245
Example 5
Without the use of a calculator arrange in order from least to greatest. 4 3, 3 6, 2 7, 5 2
4 3 , 3 6 , 2 7 , and 5 2 , are all mixed radicals. In order to
arrange them in order we need to change them to entire radicals.
4 3 is the same as 3  4  4 or
3  16  48
3 6 is the same as
6  3  3 or
6  9  54
2 7 is the same as
7  2  2 or
7  4  28
5 2 is the same as 2  5  5 or 2  25  50
So in order from least to greatest we have :
2 7 [ 48], 5 2 [ 50], 4 3 [ 48],
3 6 [ 54],
3
Example 6 – Cube Roots
Simplify:
3
16b5
Using Method One
3
16b5 is a mixed radical. Notice now the index is a three. In order to
simplify we can write the radicand as factors, one of the factors should
have a perfect cube root.
 8   b   2  b 
  2  b   2  b   2b
3
16b5 
3
16b5
3
3
3
3
3
3
3
3
2b
Using Method Two
3
16b5 is a mixed radical. Notice now the index is a three.
3
16b5 

3
2 2 2 2bbbbb

Notice that we have one group of three 2 ' s, and one group of three b ' s
We need groups of three[3] prime numbers or var iables to be the same because it
is a cube root , not a square root as in the previous examples
3
16b5 

3

2  2  2  2  b  b  b  b  b  2b 3 2b 2
Example 7
Simplify: 3 5  2 5  5
Now we are adding radicals. The rule is that when the radicands are the same
we can add their coefficients. We have 3 5  2 5  5 , notice the radicands
are all 5. In this form we can add their coefficients.
Notice that if their is no coefficient , there is a 1.  3 5  2 5  1 5
We can add the 3, 2, and the 1 to give us 6 3 for a final answer
4
Example 8
Simplify:
3 27  12  2 45
Now we are adding radicals. The rule is that the radicands of the different
terms must be the same. We have 3 27  12  2 45 , notice the radicands
are 27, 12, and 45. In this form we could not add their coefficients but if
we simplified the radicands we may be able to add and subtract.
3  9  3  4  3  2  9  5  3  3 3  2  3  2  3  5
9 3 2 3 6 5
We can add the 9 and  2 to give us 7 3  6 5 for a final answer
Example 9
Simplify: 33 81  33 135  2 3 40
In this form we cannot add their coefficients but if we simplified
the radicands we may be able to add and subtract.
3  3 27  3 3  3  3 27  3 5  3  3 8  3 5
 3 3  3 3 
 5    3 2  5 
9  3   9  5    6  5 
  3 3
3
3
3
3
3
We can subtract the 9 and the 6 because they have the same radicand
93 3  93 5
Example 10
Simplify: 2 5

5 
6

We are multiplying radicals in this example. The rules for multiplying
and dividing are the same. You multiply or divide the coefficients with
coefficients and radicands with radicands. You do not mix them.
In the example  2 5


5  6 we are multipling  2 5 by both 5
and 6. It is sometimes best to write the products out as prime factors
in the radicands rather than having to simplify later.
2 5  5 
 2 56 
 2  5
1   2 


30   10  2 30
5
Example 11
Simplify:
4

6  2 3 2 10  2




 4 6  2 10    4 6   2    2 3  2 10    2 3   2 
8 3  2  2  5    4 3  2  2    2 3  2  5    2 3  2 
 2  8 3  5    2  4 3    2 30    2 6 
16 15    8 3    2 30    2 6 
In the example 4 6  2 3 2 10  2 we are foiling so we have
16 15  8 3  2 30  2 6
We cannot add any of the coefficients as the radicands are different
Example 12
Simplify:
4 10
6 5
4 10
we are dividing so we have reduce the coefficients
6 5
and the radicands
In the example
4 10
2 2

3
6 5
6
Example 13
Simplify:
12 10
6 3
12 10
we are dividing so we have reduce the coefficients
6 3
and the radicands
In the example
12 10
2 10

After reducing we still have a radical in the deno min ator , it
6 3
6 3
is an irrational number and we want to rationalize it
2 30
2 30
2 30
2 10
3
2 253






63
18
6 9
6 3
3
6 3 3
30
9
Example 14
Simplify:
12 10
6 3
12 10
we are dividing , this time we have a binomial in the
6 3
deno min ator , in order to rationalize we will multiply by it ' s conjugate
In the example
12 10  6  3 
 6  3  6  3 
Notice the conjugate is the same except the sign between the terms is opposite
12 10  6  3 
 6  3  6  3 

12  6  10  12
 6  6    6    3    6  
253
  3   3 
3 
72 10  12 30
72 10  12 30
72 10  12 30


36  3
33
36  6 3  6 3  9
Since all three of the coefficients can be reduced by the same number , we must reduce them
72 10  12 30
24 10  4 30

33
11
7
Example 15
Simplify:
12  10
6 3
12  10
we are dividing , this time we have a binomial in the
6 3
deno min ator , in order to rationalize we will multiply by it ' s conjugate
In the example
12  10  6  3 
 6  3  6  3 
12  10  6  3 
 6  3  6  3 

12  6   12    3    6   
 6  6    6    3    6  
   
3    3   3 
10   10  3
72 10  12 3  6 10  30
72 10  12 3  6 10  30
72 10  12 3  6 10  30


36  3
33
36  6 3  6 3  9
Since not all of the coefficients can be reduced by the same number , we cannot reduce them
72 10  12 3  6 10  30
33
Example 15
8
Example 16
Simplify
√
√
First, think of the perfect Cubes
Now think of what times 5 will give me one of the perfect cubes.
In this case it is 25 x 5 = 125
√
√5
√
√25
√5 √25
√
√25
√125
√25
5
9
Example 17
Simplify
3
16
Think of the number that you can times with 16 to get one of the perfect cubes
16 x 4 = 64
3
16
3
3
16
3 4
16 4
3
3
12
64
√12
4
10
Example 18
Simplify
5
√4
2 √3
Again, think of the perfect cubes
What number that you can multiply with 3 to give one of the perfect cubes?
And that would be 3 x 9 = 27
5
√4
2 √3
5
√4
√9
2 √3 √9
5 √9
√4 9
2 √3 9
5 √9
√36
2 √27
5 √9 √36
2 3
√36
5 √9
6
Click on this Youtube link for more lessons on Radicals.
11
Practice
Make sure you show all your work and check your answers.
When you are done, proceeded to the Substantive Assignment.
1.
2.
3.
4.
5.
12
Answers
1.
2.
3.
4.
5.
13