S

Simplifying Basic Radical Expressions

2By L.D.

Table of Contents

Slide 3: Mini Lesson

Slide 4: Instructions

Slide 5: Problem 1

Slide 7: Example Problems

Slide 9: Instructions

Slide 10: Problem 2

Slide 13: Example Problems

Slide 15: Instructions

Slide 16: Problem 3

Slide 19: Problem 4

Slide 22: Example Problems

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

5 = radical 5

4 = radical 4

Mini Lesson

Instructions

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 x6

6=

7 6

36=

7 6

6

This is the final answer since the instructions were to get rid of the denominator’s square roots.

Example Problems

1. 3

5

2. 3

3

3. 3

5x

Example Problems

1. 3 5 3 5 3 5

5 5 25 5

2. 3 3 3 3 3 3

3 3 9 3

x = =

3. 3 5x 15x 15x

5x 5x 25x2

5x

x = =

x = =

Instructions

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

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

Visit

myratatemyhomework.blogsp

ot.com fo

r more m

ath

tutoria

ls!