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Algebra I: High School20 chapters | 168 lessons | 1 flashcard set

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Lesson Transcript

Instructor:
*Jennifer Beddoe*

Solving radical equations is not any more difficult than solving other algebraic equations. This lesson will show you how to solve equations containing a square root and give some real-world examples.

Sherlock Holmes is famous for his crime-solving abilities. No matter how small the clue, he could figure out 'who done it.' This was partly due to his detailed knowledge on a wide variety of topics.

Well, with some practice and knowledge of your own, you can become the Sherlock Holmes of algebra mysteries. You can solve for *x* no matter what the circumstances. This lesson will give you some of the knowledge that you need to become an algebra master.

What is a radical? You might be thinking (and rightly so) that a radical is someone who speaks out against injustices. They want to do things in a new and unconventional way. While this is true, it will not help at all with solving algebra problems.

For the purposes of this lesson, a **radical** is an algebraic term under a square root symbol. For example:

It does not have to be a square root; it can be a cubed root, or fourth root or any other number. The square root is the most common and is what is implied when the radical symbol is used alone. If another root is called for, there will be a small superscript number in the 'v' of the root symbol, like this:

The 3 indicates it is a **cube root**, which means that the term inside the radical equals a term that has been multiplied to itself three times. For example:

This is correct because 2 * 2 * 2 = 8.

Every mathematical operation has an **inverse operation**, an operation that is its opposite: for addition, it's subtraction; for multiplication, it's division; and for the square root, it's squared. Inverse operations are critical for solving algebraic equations.

For example, if you want to solve the following: *x* + 2 = 3. You need to get *x* alone. To do that, you perform the inverse operation, to move the 2 to the right side of the equation. That means, you subtract (the opposite of add) 2 from both sides of the equation. *x* + 2 = 3. Subtract 2 from both sides, and you get *x* = 1. And your problem is solved.

The method for solving equations containing radicals involves the same process. In order to isolate the *x* (or whatever the variable happens to be), you need to perform inverse operations to move all the numbers to the right side of the equation. It doesn't have to be the right side. That is just the most conventional way of doing things. You'll get the same answer either way.

Let's try an example. Solve for *x*:

The first step to solving is to remove the radical symbol by performing the inverse operation, which is to square both sides. (âˆš(x-1))^2 = 4^2. When you square the left side, the square and square root cancel each other out, so you are left with *x* - 1. On the right side, you simplify 4^2 = 16. Now, the problem is simple. *x* - 1 = 16. Just add one to both sides and the solution is *x* = 17.

With radicals, however, the problem does not end there. There is always the possibility that you will get an answer that is called an **extraneous root**. This is an answer that seems to work, but isn't right when you check your answers. This is why, with problems containing radicals, you always have to check your answer. It is not common, but there can be an answer that doesn't work. The only way to know is to check your answer by substituting it for *x* in the original problem.

In this case, the original problem was âˆš(*x* - 1) = 4. Substitute 17 in for *x*, and we get âˆš(17 - 1) = 4, or âˆš(16) = 4. For this problem, the number 17 works, so we have confirmed it as the answer.

Problems containing other roots are solved in the same manner. For example:

Since it is a cube root, the first thing we need to do is cube both sides.

Which gives us:

Then, subtract 4 from both sides to get *x* = 23. Again, because we have to check for extraneous roots, we will substitute the answer back into the original equation.

Because it checks out, we know that 23 is the correct answer.

There are many areas where radicals are used in real-life situations. One of the most common places is in the calculation of areas and also in using the Pythagorean Theorem. There are two other common equations that use radicals. The first is the visibility formula, which says that *v* = 1.225 * âˆš*a*, where *v* = visibility (in miles), and *a* = altitude (in feet). This formula will tell you how far away you can see on a clear day.

The other formula helps physicians in prescribing medicine. Certain medications require knowing the patient's body surface area (BSA), and the equation for determining BSA = âˆš(*wh* / 3600), where *w* = weight (in pounds), and *h* = height (in cm).

Let's try an example. A woman on a hang glider can see 29 miles to the horizon. Using the visibility formula, how far above the ground is she?

The visibility formula again is *v* = 1.225 * âˆš*a*, where *v* = visibility (in miles), and *a* = altitude (in feet). By substituting in what we know, we get this equation to solve: 29 = 1.225 * âˆš*a*.

To solve, first divide both sides by 1.225 to get 23.67 = âˆš*a*. The next step is to square both sides to get rid of the square root symbol: (23.67)^2 = (âˆš*a*)^2. So, 560 ft = *a*, which means the hang glider is 560 feet above the ground.

To solve any algebraic equation, you need to perform inverse operations to isolate the variable. If the equation contains a radical, the inverse operation to solve it is to square (or cube) it. That will eliminate the radical and allow you to solve the equation.

The process of studying this video lesson's topics could enable you to:

- Define 'radical' as it relates to mathematics
- Solve a radical equation using inverse operations
- Understand the use of radicals in real-life scenarios

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Algebra I: High School20 chapters | 168 lessons | 1 flashcard set

- How to Find the Square Root of a Number 5:42
- Estimating Square Roots 5:10
- Simplifying Square Roots When not a Perfect Square 4:45
- Simplifying Expressions Containing Square Roots 7:03
- Division and Reciprocals of Radical Expressions 5:53
- Radicands and Radical Expressions 4:29
- Evaluating Square Roots of Perfect Squares 5:12
- Factoring Radical Expressions 4:45
- Simplifying Square Roots of Powers in Radical Expressions 3:51
- Multiplying then Simplifying Radical Expressions 3:57
- Dividing Radical Expressions 7:07
- Simplify Square Roots of Quotients 4:49
- Rationalizing Denominators in Radical Expressions 7:01
- Addition and Subtraction Using Radical Notation 3:08
- Multiplying Radical Expressions with Two or More Terms 6:35
- Solving Radical Equations: Steps and Examples 6:48
- Go to High School Algebra: Radical Expressions

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