Become a math whiz with AI Tutoring, Practice Questions & more.

Join for Free
HotmathMath Homework. Do It Faster, Learn It Better.

# Solving Radical Equations

During our journey through the world of math, we may be asked to solve radical equations. If you've never encountered one of these before, the name alone can be somewhat daunting. But rest assured that there's nothing exceedingly "radical" about radical equations. In fact, they follow a number of predictable rules. We can use these rules to our advantage, equipping ourselves with plenty of tips and tricks to solve radical equations. But what exactly is a radical equation, anyway? How do we solve them? Let's find out:

## What is a radical equation?

A radical equation includes a variable underneath a square root sign. As we may recall, this square root symbol is also called a "radical symbol" or simply a "radix." This is why we call them "radical equations."

These are all examples of radical equations:

$\sqrt{x+2}=3$

$\sqrt{8-x}=4$

$\sqrt{4+x}=16$

## The technique for solving radical equations

Now that we know what radical equations are, we can solve them using a pretty straightforward technique:

All we need to do is follow a few easy steps:

1. Isolate the square root symbol and its contents on one side of the equation.

2. Square both sides of the equation

3. Solve the equation normally (it should be easy to do so by this point)

## Solving radical expressions: Practicing our skills

As we all know, reading about how to solve rational equations is very different compared to actually solving rational expressions yourself.

Let's go through a few examples of these problems:

$\sqrt{x+2}=3$

Note that the radical symbol is already isolated on one side of the equation. This means that we can move directly to the second step, which is to square both sides:

${\left(\sqrt{x+2}\right)}^{2}={3}^{2}$

We are left with the following once we simplify:

$x+2=9$

In other words, $x=7$ .

As a general rule, we need to check our solutions when solving radical equations. Let's see if this answer actually makes sense:

$\sqrt{7+2}$

Does this equal 3?

We can simplify this as:

$\sqrt{9}=3$

$3=3$

In this case, we know for sure that $x=7$ .

Now let's try something a little bit more difficult:

$\sqrt{12x+4}+5=x$

Remember, the first step is to isolate the square root on one side of the equation. Fortunately, we can do this by simply subtracting 5 from both sides:

$\sqrt{12x+4}=x-5$

Our next step is to square both sides of the equation:

${\left(\sqrt{12x+4}\right)}^{2}={\left(x-5\right)}^{2}$

If we simplify, we are left with the following:

$12x+4={x}^{2}-10x+25$

We can simplify this even further:

${x}^{2}-22x+21=0$

Do you notice anything special about this equation? That's right -- it's now in the form of a quadratic equation -- which means that we can solve it by factoring.

$\left(x-21\right)\left(x-1\right)=0$

Now we can apply the zero product property, which states that if $ab=0$ , then either $a=0$ , $b=0$ , or both. This must mean that x either equals 21 or 1, since both of these values would cause the result to become zero.

You may have noticed that when we square both sides of the equation, we are potentially left with a few "extraneous solutions."

As you might recall, an extraneous solution is the root of a transformed equation that is not a root of the original equation. We get these values because they were excluded from the domain of the original equation.

If this is the case, then we must check the validity of our solutions after finding our x values.

Let's start by substituting x for 21:

$\sqrt{\left(12×21\right)+4}+5$

Does this equal 21?

$\sqrt{256}+5=21$

$21=21$

Now we know that this answer is valid.

But what if we substitute x for our second possible answer, which is 1?

$\sqrt{\left(12×1\right)+4}+5$

Does this equal 1?

$\sqrt{16}+5=9$

Since 9 clearly does not equal 1, we know that this potential answer is not valid.

Therefore, we are left with only one possible solution, which is $x=21$ .

## Flashcards covering the Solving Radical Equations

Algebra II Flashcards

## Pair your student with a tutor who can explain how to solve radical equations

Solving radical equations can be tricky -- especially when students don't have the chance to ask questions or practice adequately. As classroom teachers need to move ahead quickly to cover all the topics in the curriculum, students can often feel left behind. Fortunately, a little extra help from a tutor can give them the opportunity to have those burning questions answered in a patient, clear manner. Students can also take all the time they need to practice operations like solving radical equations until they feel totally confident in their abilities. Varsity Tutors always matches your student with a tutor who has been carefully vetted and interviewed. Speak with our Educational Directors today to get started.

;
Download our free learning tools apps and test prep books