A radical equation is an equation that includes a square root or cubed root. The easiest way to solve an equation with a square or cubed root is to square or cube it. But beware that this technique doesn’t always lead to the right answer, so its best to check your answer by substituting it back into the equation if you use this method. The Advanced Algebra: Strategies for Success offers you step by step tutorials on how to solve advanced algebra equations. This blog is based on one of the lessons from this course and it will show you how easy it is to solve advanced algebraic equations if you understand the steps.

Let’s examine the steps we needed for solving radical equations:

## Steps Required to Solve Radical Equations

There are really three steps for solving radical equations. The first step is that we need to isolate the radical on one side of the equation. Step two, is you can eliminate a radical by squaring both sides if it’s a square root, or by raising both sides to a power equal to an index number if it is other than a square root. The third step is to solve whatever equation results from that process. Let me show you what I mean here.

## Example 1: A Simple Square Root Example

In example one, we have a square root, and then it’s being added to 3 and is set equal to 4.

Well, we said step one is to isolate the radical. If we subtract 3 from both sides, this will eliminate the 3s on the left, and we’ll be left with square root of x is equal to 1:

The radical is all alone; that isolated the radical.

Step two is to raise both sides of the equation to the same power as our index number. In this case, the index number is a 2, a square root, so we will square both sides of the equation. Square root of x squared, when it gets raised to the second power, the squared and the square root eliminate, simply leaving us the radicand x. And 1 to the second power is still 1:

It’s important that you always go back and check your solutions on radical equations because we do have domain restrictions, and that we cannot have the square root of negative numbers, and we just need to make sure the values we get are correct. Occasionally, you can wind up with extraneous answers – meaning answers that come out when we solve the process, but when we substitute them back into the equation, they do not work. So, let’s check our answer of 1 by substituting it back into the equation for the variable. Square root of 1 is 1 and 1 + 3 = 4, and that equals the 4 that we were given so this one checked out. x = 1 is a good answer.

This is one of the step-by-step tutorials offered by the Advanced Algebra: Strategies for Success course. The course also includes over 60 video lectures that are approximately 10 minutes in length each that will teach you everything you need to know to solve advanced algebra equations.

## Example 2: A binomial Square Root

Let’s look at another example.

This time we have a binomial under the radical but it works in the same way. We can add 3 to both sides of the equation. In this case, we’re left with the square root of 2x + 5 – let me clean that up a little bit – of 2x + 5 is equal to 4 + 3 = 7.

Now, we can square both sides of the equation. That’ll leave us simply with 2x + 5 = 7 squared, which is 49. To solve this equation, we can subtract 5 from both sides, and that leaves us with 2x = 44. And dividing both sides by 2, we wind up with x = 22.

We can check our answer by substitution. So we have the square root of 2 times our answer, 22 plus 5, is subtracting 3 to give us the answer of 4. Order of operations says lets simplify our square root first, so this is going to be the square root of 2 x 22 = 44, and 44 + 5 = 49. We know the square root of 49 is really 7, and 7 – 3, sure enough, that gives us 4 = 4. So this is a correct answer, x = 22.

## Example 3: Try it Yourself

I have one here for you to try. Go ahead and work through this using that three-step process. Stop here and take a minute to solve this. When you’re ready, read on and we’ll see how you did.

Our first step is to isolate the radical, and we’ll do this by subtracting 2 from both sides. That leaves us the square root of 5x – 1 = 9. Now we can square both sides of the equation. That’ll leave us 5x – 1 = 81.

By adding 1 to both sides, we’ll isolate the variable term, so we have 5x is equal to, looks like, 82. And now dividing both sides by 5, looks like this time we get a fractional expression that x = 82/5.

You can get a decimal approximation for this, if you’d like. So 82 divided by 5 is approximately– oh no it works out exactly, that gives us exactly 16.4. Using our substitution process, a 16.4 is a little bit easier to work with than 82/5.

So let’s come back in and let’s substitute to check our answer. We have the square root of– we have 5 times, in this case 16.4, minus 1, and then we have to add 2 and then this is equal to 11. Simplifying the square root first, 16.4 x 5 = 82, and 82 – 1 leaves us the square root of 81. The square root of 81, we know is 9, and 9 + 2, sure enough, leaves us 11 = 11. This is a true statement and so our answer is correct. You could use 16.4 or you could use 82/5; both are correct answers.

Solving radical equations doesn’t have to be hard work. In fact, if you know the steps and what you need to do at each step of the process, then you will be solving radical equations quickly and painlessly. In no time at all, you’ll be able to graph, work with polynomials, radical equations and more!

For complete video lessons on the steps you need to understand how to solve advanced algebra equations and radical equations, sign up for the Advanced Algebra: Strategies for Success course today.