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