This document provides instructions and examples for simplifying radical expressions. It defines a radical as a square root expression. It then provides 5 problems with step-by-step explanations and solutions for simplifying radical expressions by finding perfect squares under the radical signs. The problems cover simplifying radicals of variables, combining like radicals, and simplifying fractional radicals.

1. Simplifying Basic
Radical Expressions
By L.D.
S

2. Table of Contents
S Slide 3: What is a Radical?
S Slide 4: Instructions For The Problems
S Slide 5: Problem 1
S Slide 11: Problem 1 Example Problems
S Slide 13: Problem 2
S Slide 16: Problem 3
S Slide 18: Problem 4
S Slide 21: Problem 5

3. What is a Radical?
“Radical” is a fancy name for a square root. A radical
expression is an expression with a square root.

4. Instructions For The Problems
S Simplify the problems until there are no perfect squares
under the square roots.

5. Problem 1
√24

6. Problem 1
√24
Your first thought when seeing this problem is probably that
24 is not a perfect square. This can easily be fixed.

7. Problem 1
√24
The first step to solve this problem is to find the factors to
24. I personally like the cake method (if you do not know
how to use this method, check the post on my blog titled
“GCF & LCM” since it has a how-to).
Factors: 3, 2, 2, 2

8. Problem 1
√24
Factors: 3, 2, 2, 2
Now we have the factors we need to choose two of them
that are the same number, we have two 2s. They can be
multiplied to make 4 and the leftover numbers 3 & 2 can be
multiplied for 6. We will put these numbers with a
multiplication sign under a square root sign

9. Problem 1
√6 x 4 can also be seen as √6 x √4 . We will simplify the
square root we can now simplify(√4 ).
The problem will now look like 4 x √6 or 4√6 .

10. Problem 1
Our final answer is 4√6 as there are no more factors that
are hidden in the numbers under the square roots.

11. Problem 1 Example Problems
1. √50
2. √48

12. Problem 1 Example Problems
1. √50 = √25 x 2 = 5√2
2. √48
In this problem we will get the factors 2, 2, 2, 2 & 3. Since
we have two sets of two we will make the 2s combine to be
what is below.
√48 = √16 x 3 = 4√3

13. Problem 2
√25g5

14. Problem 2
√25g5
To find the answer to this we will as usual find the factors.
Factors: 5, 5, g, g, g, g & g
We will do as usual and multiply all the doubles together.

15. Problem 2
√25g5
Factors: 5, 5, g, g, g, g & g
√25g5 = √25g4 x g = 5g2√g

16. Problem 3
√7 x √7

17. Problem 3
√7 x √7
Remember how we can split numbers under separate
square roots, well we can also put them back together.
√7 x √7 = √7 x 7 = √49 = 7

18. Problem 4
3√k x √2k3

19. Problem 4
3√k x √2k3
First we will combine the things under the square roots.
3√k x √2k3 = 3√2k4
Now we will find the factors of the thing under the square
root.
Factors: 2, k, k, k & k

20. Problem 4
Factors: 2, k, k, k & k
We have two sets of k so our problem will be shown below.
3√2 x k4 = 3 x k2 √2 = 3k2√2

21. Problem 5
√5/49

22. Problem 5
√5/49
To solve this problem we must first split it, we will put square
root signs over both and then solve.
√5/√49 = √5/7