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Simplifying Basic
Radical Expressions 2
         By L.D.




                    S
Table of Contents

S   Slide 3: Mini Lesson

S   Slide 4: Instructions

S   Slide 5: Problem 1

S   Slide 7: Example Problems

S   Slide 9: Instructions

S   Slide 10: Problem 2

S   Slide 13: Example Problems

S   Slide 15: Instructions

S   Slide 16: Problem 3

S   Slide 19: Problem 4

S   Slide 22: Example Problems
Mini Lesson


S Another word for the square root of a number is
  saying radical _______(insert number).

S 5 = radical 5

S 4 = radical 4
Instructions


S Rationalize the denominator (make it so the denominator
  has no square roots)
Problem 1


7

6
Problem 1


7

6

The first thing we need to do is get rid of the 6. As explained
in the previous presentation, another way to completely get
rid of something squared is to multiply it by itself. But to do
this we also have to multiply the rest of the problem by
radical 6.
   7       6         7 6        7 6      This is the final answer
                                         since the instructions were
    6   x   6
                =     36   =     6       to get rid of the
                                         denominator’s square
Example Problems


1. 3

  5
               3. 3

2. 3             5x

  3
Example Problems


1. 3           5                 3 5           3 5
       x               =                   =
  5            5                 25             5
                                               3. 3           5x       15x
                                               15x        x        =          =
2. 3       3               3 3       3 3
       x           =             =                   5x       5x       25x2
  3        3               9         3         5x
Instructions


S Find the sum or difference
Problem 2


7 14 + 21 – 4 14
Problem 2


7 14 + 21 – 4 14

The first step is to see if any of the square roots can be
simplified.

Unfortunately, they can’t.
Problem 2


7 14 + 21 – 4 14

The next step is to combine the like square roots (remember
combing like terms).

Combining the terms:

7 14 + - 4 14 = 3 14

Now the problem looks like 3 14 + 21
Example Problems


1. 2 7 + 2 7




2. 2 7 + 9 7
Example Problems


1. 2 7 + 2 7 = 4 7




2. 2 7 + 9 7 = 11 7
Instructions


S Find the product
Problem 3


3(2 + 12)
Problem 3


3(2 + 12)

In this problem we simply need to use distributive.

3(2 + 12)

2 3 + 36
Problem 3


2 3 + 36

Now we just need to simplify to get our answer.



2 3+6
Problem 4


( 2 + 5 )( 3 – 3 5 )
Problem 4


( 2 + 5 )( 3 – 3 5 )

To solve the problem we need to use F.O.I.L.

( 2 + 5 )( 3 – 3 5 )

6 – 3 10 + 15 – 3 25
Problem 4


6 – 3 10 + 15 – 3 25

Now we need to simplify everything

6 – 3 10 + 15 – 15

After this you would need to add the like terms if there were
any.
Example Problems


1. 5 (4 - 20)




2. ( 7 + 2)( 7 – 3 2)
Example Problems


1. 5 (4 - 20)

 4 5 - 100

 4 5 - 10

2. ( 7 + 2)( 7 – 3 2)

  49 – 3 14 + 14 – 3 4

   7 – 2 14 – 6

   1 – 2 14
Simplifying basic radical expressions 2

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Simplifying basic radical expressions 2

  • 2. Table of Contents S Slide 3: Mini Lesson S Slide 4: Instructions S Slide 5: Problem 1 S Slide 7: Example Problems S Slide 9: Instructions S Slide 10: Problem 2 S Slide 13: Example Problems S Slide 15: Instructions S Slide 16: Problem 3 S Slide 19: Problem 4 S Slide 22: Example Problems
  • 3. Mini Lesson S Another word for the square root of a number is saying radical _______(insert number). S 5 = radical 5 S 4 = radical 4
  • 4. Instructions S Rationalize the denominator (make it so the denominator has no square roots)
  • 6. Problem 1 7 6 The first thing we need to do is get rid of the 6. As explained in the previous presentation, another way to completely get rid of something squared is to multiply it by itself. But to do this we also have to multiply the rest of the problem by radical 6. 7 6 7 6 7 6 This is the final answer since the instructions were 6 x 6 = 36 = 6 to get rid of the denominator’s square
  • 7. Example Problems 1. 3 5 3. 3 2. 3 5x 3
  • 8. Example Problems 1. 3 5 3 5 3 5 x = = 5 5 25 5 3. 3 5x 15x 15x x = = 2. 3 3 3 3 3 3 x = = 5x 5x 25x2 3 3 9 3 5x
  • 9. Instructions S Find the sum or difference
  • 10. Problem 2 7 14 + 21 – 4 14
  • 11. Problem 2 7 14 + 21 – 4 14 The first step is to see if any of the square roots can be simplified. Unfortunately, they can’t.
  • 12. Problem 2 7 14 + 21 – 4 14 The next step is to combine the like square roots (remember combing like terms). Combining the terms: 7 14 + - 4 14 = 3 14 Now the problem looks like 3 14 + 21
  • 13. Example Problems 1. 2 7 + 2 7 2. 2 7 + 9 7
  • 14. Example Problems 1. 2 7 + 2 7 = 4 7 2. 2 7 + 9 7 = 11 7
  • 17. Problem 3 3(2 + 12) In this problem we simply need to use distributive. 3(2 + 12) 2 3 + 36
  • 18. Problem 3 2 3 + 36 Now we just need to simplify to get our answer. 2 3+6
  • 19. Problem 4 ( 2 + 5 )( 3 – 3 5 )
  • 20. Problem 4 ( 2 + 5 )( 3 – 3 5 ) To solve the problem we need to use F.O.I.L. ( 2 + 5 )( 3 – 3 5 ) 6 – 3 10 + 15 – 3 25
  • 21. Problem 4 6 – 3 10 + 15 – 3 25 Now we need to simplify everything 6 – 3 10 + 15 – 15 After this you would need to add the like terms if there were any.
  • 22. Example Problems 1. 5 (4 - 20) 2. ( 7 + 2)( 7 – 3 2)
  • 23. Example Problems 1. 5 (4 - 20) 4 5 - 100 4 5 - 10 2. ( 7 + 2)( 7 – 3 2) 49 – 3 14 + 14 – 3 4 7 – 2 14 – 6 1 – 2 14