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Simplifying Basic Radical Expressions

By L.D.

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

What is a Radical?

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

Instructions For The Problems

Simplify the problems until there are no perfect squares under the square roots.

Problem 1

√24

Problem 1

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

Problem 1

√24The 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

Problem 1

√24Factors: 3, 2, 2, 2Now 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

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 .

Problem 1

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

Problem 1 Example Problems

1. √50

2. √48

Problem 1 Example Problems

1. √50 = √25 x 2 = 5√2

2. √48In 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

Problem 2

√25g5

Problem 2

√25g5

To find the answer to this we will as usual find the factors.Factors: 5, 5, g, g, g, g & gWe will do as usual and multiply all the doubles together.

Problem 2

√25g5

Factors: 5, 5, g, g, g, g & g√25g5 = √25g4 x g = 5g2√g

Problem 3

√7 x √7

Problem 3

√7 x √7Remember 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

Problem 4

3√k x √2k3

Problem 43√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

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

Problem 5

√5/49

Problem 5

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

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