how to solve radical math problems

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Roots and Radicals

Solve Radical Equations

Learning Objectives

By the end of this section, you will be able to:

Before you get started, take this readiness quiz.

${\left(y-3\right)}^{2}.$

In this section we will solve equations that have a variable in the radicand of a radical expression. An equation of this type is called a radical equation .

An equation in which a variable is in the radicand of a radical expression is called a radical equation .

As usual, when solving these equations, what we do to one side of an equation we must do to the other side as well. Once we isolate the radical, our strategy will be to raise both sides of the equation to the power of the index. This will eliminate the radical.

Solving radical equations containing an even index by raising both sides to the power of the index may introduce an algebraic solution that would not be a solution to the original radical equation. Again, we call this an extraneous solution as we did when we solved rational equations.

In the next example, we will see how to solve a radical equation. Our strategy is based on raising a radical with index n to the n th power. This will eliminate the radical.

$\text{For}\phantom{\rule{0.2em}{0ex}}a\ge 0,\phantom{\rule{0.2em}{0ex}}{\left(\sqrt[n]{a}\right)}^{n}=a.$

When we use a radical sign, it indicates the principal or positive root. If an equation has a radical with an even index equal to a negative number, that equation will have no solution.

$\sqrt{9k-2}+1=0.$

Because the square root is equal to a negative number, the equation has no solution.

$\sqrt{2r-3}+5=0.$

If one side of an equation with a square root is a binomial, we use the Product of Binomial Squares Pattern when we square it.

$\begin{array}{}\\ \\ {\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}\\ {\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}\end{array}$

Don’t forget the middle term!

$\sqrt{p-1}+1=p.$

When the index of the radical is 3, we cube both sides to remove the radical.

${\left(\sqrt[3]{a}\right)}^{3}=a$

Sometimes the solution of a radical equation results in two algebraic solutions, but one of them may be an extraneous solution !

$\sqrt{r+4}-r+2=0.$

When there is a coefficient in front of the radical, we must raise it to the power of the index, too.

$\text{3}\phantom{\rule{0.2em}{0ex}}\sqrt{3x-5}-8=4.$

Solve Radical Equations with Two Radicals

If the radical equation has two radicals, we start out by isolating one of them. It often works out easiest to isolate the more complicated radical first.

In the next example, when one radical is isolated, the second radical is also isolated.

$\sqrt[3]{4x-3}=\sqrt[3]{3x+2}.$

Sometimes after raising both sides of an equation to a power, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and raise both sides of the equation to the power of the index again.

$\sqrt{m}+1=\sqrt{m+9}.$

We summarize the steps here. We have adjusted our previous steps to include more than one radical in the equation This procedure will now work for any radical equations.

Isolate one of the radical terms on one side of the equation.

If yes, repeat Step 1 and Step 2 again.

${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$

Use Radicals in Applications

As you progress through your college courses, you’ll encounter formulas that include radicals in many disciplines. We will modify our Problem Solving Strategy for Geometry Applications slightly to give us a plan for solving applications with formulas from any discipline.

One application of radicals has to do with the effect of gravity on falling objects. The formula allows us to determine how long it will take a fallen object to hit the gound.

On Earth, if an object is dropped from a height of h feet, the time in seconds it will take to reach the ground is found by using the formula

$t=\frac{\sqrt{h}}{4}.$

It would take 2 seconds for an object dropped from a height of 64 feet to reach the ground.

$t=\frac{\sqrt{h}}{4}$

Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed , in miles per hour, a car was going before applying the brakes.

If the length of the skid marks is d feet, then the speed, s , of the car before the brakes were applied can be found by using the formula

$s=\sqrt{24d}$

Access these online resources for additional instruction and practice with solving radical equations.

Key Concepts

$\begin{array}{c}{\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}\\ {\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}\end{array}$

Practice Makes Perfect

In the following exercises, solve.

$\sqrt{5x-6}=8$

$\sqrt{3y-4}=-2$

In the following exercises, solve. Round approximations to one decimal place.

$s=\sqrt{A}$

Writing Exercises

$\sqrt{x}+1=0$

Answers will vary.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

The table has 4 columns and 4 rows. The first row is a header row with the headers “I can…”, “Confidently”, “With some help.”, and “No – I don’t get it!”. The first column contains the phrases “Solve radical equations”, “solve radical equations with two radicals”, and “use radicals in applications”. The other columns are left blank so the learner can indicate their level of understanding.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Intermediate Algebra by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Solving Radical Equations

How to solve equations with square roots, cube roots, etc.

Radical Equations

We can get rid of a square root by squaring (or cube roots by cubing, etc).

Warning: this can sometimes create "solutions" which don't actually work when we put them into the original equation. So we need to Check!

Follow these steps:

Then continue with our solution!

Example: solve √(2x+9) − 5 = 0

Now it should be easier to solve!

Check: √(2·8+9) − 5 = √(25) − 5 = 5 − 5 = 0

That one worked perfectly.

More Than One Square Root

What if there are two or more square roots? Easy! Just repeat the process for each one.

It will take longer (lots more steps) ... but nothing too hard.

Example: solve √(2x−5) − √(x−1) = 1

We have removed one square root.

Now do the "square root" thing again:

We have now successfully removed both square roots.

Let us continue on with the solution.

It is a Quadratic Equation! So let us put it in standard form.

Using the Quadratic Formula (a=1, b=−14, c=29) gives the solutions:

2.53 and 11.47 (to 2 decimal places)

Let us check the solutions:

There is really only one solution :

Answer: 11.47 (to 2 decimal places)

See? This method can sometimes produce solutions that don't really work!

The root that seemed to work, but wasn't right when we checked it, is called an "Extraneous Root"

So checking is important.

A radical equation is an equation in which a variable is under a radical. To solve a radical equation:

Isolate the radical expression involving the variable. If more than one radical expression involves the variable, then isolate one of them.

Raise both sides of the equation to the index of the radical.

If there is still a radical equation, repeat steps 1 and 2; otherwise, solve the resulting equation and check the answer in the original equation.

By raising both sides of an equation to a power, some solutions may have been introduced that do not make the original equation true. These solutions are called extraneous solutions.

how to solve radical math problems

Isolate the radical expression.

how to solve radical math problems

Raise both sides to the index of the radical; in this case, square both sides.

how to solve radical math problems

This quadratic equation now can be solved either by factoring or by applying the quadratic formula.

how to solve radical math problems

Now, check the results.

how to solve radical math problems

Isolate one of the radical expressions.

how to solve radical math problems

This is still a radical equation. Isolate the radical expression.

how to solve radical math problems

This can be solved either by factoring or by applying the quadratic formula.

how to solve radical math problems

Check the solutions.

how to solve radical math problems

So x = 10 is not a solution.

how to solve radical math problems

The only solution is x = 2.

how to solve radical math problems

Isolate the radical involving the variable.

how to solve radical math problems

Since radicals with odd indexes can have negative answers, this problem does have solutions. Raise both sides of the equation to the index of the radical; in this case, cube both sides.

how to solve radical math problems

The check of the solution x = –15 is left to you.

Previous Quiz: Solving Equations in Quadratic Form

Next Quiz: Solving Radical Equations

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Section 2.10 : Equations with Radicals

Solve each of the following equations.

Solving Radical Equations

Learning how to solve radical equations requires a lot of practice and familiarity of the different types of problems. In this lesson, the goal is to show you detailed worked solutions of some problems with varying levels of difficulty.

What is a Radical Equation?

An equation wherein the variable is contained inside a radical symbol or has a rational exponent. In particular, we will deal with the square root which is the consequence of having an exponent of \Large{1 \over 2} .

1) Isolate the radical symbol on one side of the equation

2) Square both sides of the equation to eliminate the radical symbol

3) Solve the equation that comes out after the squaring process

4) Check your answers with the original equation to avoid extraneous values

Examples of How to Solve Radical Equations

Example 1 : Solve the radical equation

The radical is by itself on one side so it is fine to square both sides of the equations to get rid of the radical symbol. Then proceed with the usual steps in solving linear equations.

You must ALWAYS check your answers to verify if they are “truly” the solutions. Some answers from your calculations may be extraneous. Substitute x = 16 back into the original radical equation to see whether it yields a true statement.

Yes, it checks, so x = 16 is a solution.

Example 2 : Solve the radical equation

The setup looks good because the radical is again isolated on one side. So I can square both sides to eliminate that square root symbol. Be careful dealing with the right side when you square the binomial (x−1). You must apply the FOIL method correctly.

We move all the terms to the right side of the equation and then proceed with factoring out the trinomial. Applying the Zero-Product Property, we obtain the values of x = 1 and x = 3 .

Caution: Always check your calculated values from the original radical equation to make sure that they are true answers and not extraneous or “false” answers.

Looks good for both of our solved values of x after checking, so our solutions are x = 1 and x = 3 .

Example 3 : Solve the radical equation

We need to recognize the radical symbol is not isolated just yet on the left side. It means we have to get rid of that −1 before squaring both sides of the equation. A simple step of adding both sides by 1 should take care of that problem. After doing so, the “new” equation is similar to the ones we have gone over so far.

Our possible solutions are x = −2 and x = 5 . Notice I use the word “possible” because it is not final until we perform our verification process of checking our values against the original radical equation.

Since we arrive at a false statement when x = -2, therefore that value of x is considered to be extraneous so we disregard it! Leaving us with one true answer, x = 5 .

Example 4 : Solve the radical equation

The left side looks a little messy because there are two radical symbols. But it is not that bad! Always remember the key steps suggested above. Since both of the square roots are on one side that means it’s definitely ready for the entire radical equation to be squared.

So for our first step, let’s square both sides and see what happens.

It is perfectly normal for this type of problem to see another radical symbol after the first application of squaring. The good news coming out from this is that there’s only one left. From this point, try to isolate again the single radical on the left side, which should force us to relocate the rest to the opposite side.

As you can see, that simplified radical equation is definitely familiar . Proceed with the usual way of solving it and make sure that you always verify the solved values of x against the original radical equation.

I will leave it to you to check that indeed x = 4 is a solution.

Example 5 : Solve the radical equation

This problem is very similar to example 4. The only difference is that this time around both of the radicals has binomial expressions. The approach is also to square both sides since the radicals are on one side, and simplify. But we need to perform the second application of squaring to fully get rid of the square root symbol.

The solution is x = 2 . You may verify it by substituting the value back into the original radical equation and see that it yields a true statement.

Example 6 : Solve the radical equation

It looks like our first step is to square both sides and observe what comes out afterward. Don’t forget to combine like terms every time you square the sides. If it happens that another radical symbol is generated after the first application of squaring process, then it makes sense to do it one more time. Remember, our goal is to get rid of the radical symbols to free up the variable we are trying to solve or isolate.

Well, it looks like we will need to square both sides again because of the newly generated radical symbol. But we must isolate the radical first on one side of the equation before doing so. I will keep the square root on the left, and that forces me to move everything to the right.

Looking good so far! Now it’s time to square both sides again to finally eliminate the radical.

Be careful though in squaring the left side of the equation. You must also square that −2 to the left of the radical.

What we have now is a quadratic equation in the standard form. The best way to solve for x is to use the Quadratic Formula where a = 7, b = 8, and c = −44.

So the possible solutions are x = 2 , and x = {{ - 22} \over 7} .

I will leave it to you to check those two values of “x” back into the original radical equation. I hope you agree that x = 2 is the only solution while the other value is an extraneous solution, so disregard it!

Example 7 : Solve the radical equation

There are two ways to approach this problem. I could immediately square both sides to get rid of the radicals or multiply the two radicals first then square. Both procedures should arrive at the same answers when properly done. For this, I will use the second approach.

Next, move everything to the left side and solve the resulting Quadratic equation. You can use the Quadratic formula to solve it, but since it is easily factorable I will just factor it out.

The possible solutions then are x = {{ - 5} \over 2} and x = 3 .

I will leave it to you to check the answers. The only answer should be x = 3 which makes the other one an extraneous solution.

You might also be interested in:

Simplifying Radical Expressions Adding and Subtracting Radical Expressions Multiplying Radical Expressions Rationalizing the Denominator

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Unit 10: Lesson 2

Solving square-root equations: one solution

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Video transcript

Solving Radical Equations

Learning Objective(s)

· Solve equations containing radicals.

· Recognize extraneous solutions.

· Solve application problems that involve radical equations as part of the solution.

An equation that contains a radical expression is called a radical equation . Solving radical equations requires applying the rules of exponents and following some basic algebraic principles. In some cases, it also requires looking out for errors generated by raising unknown quantities to an even power.

Squaring Both Sides

A basic strategy for solving radical equations is to isolate the radical term first, and then raise both sides of the equation to a power to remove the radical. (The reason for using powers will become clear in a moment.) This is the same type of strategy you used to solve other, non-radical equations—rearrange the expression to isolate the variable you want to know, and then solve the resulting equation.

Notice how you combined like terms and then squared both sides of the equation in this problem. This is a standard method for removing a radical from an equation. It is important to isolate a radical on one side of the equation and simplify as much as possible before squaring. The fewer terms there are before squaring, the fewer additional terms will be generated by the process of squaring.

In the example above, only the variable x was underneath the radical. Sometimes you will need to solve an equation that contains multiple terms underneath a radical. Follow the same steps to solve these, but pay attention to a critical point—square both sides of an equation, not individual terms . Watch how the next two problems are solved.

Extraneous Solutions

Following rules is important, but so is paying attention to the math in front of you—especially when solving radical equations. Take a look at this next problem that demonstrates a potential pitfall of squaring both sides to remove the radical.

Look at that—the answer a = 9 does not produce a true statement when substituted back into the original equation. What happened?

Incorrect values of the variable, such as those that are introduced as a result of the squaring process are called extraneous solutions . Extraneous solutions may look like the real solution, but you can identify them because they will not create a true statement when substituted back into the original equation. This is one of the reasons why checking your work is so important—if you do not check your answers by substituting them back into the original equation, you may be introducing extraneous solutions into the problem.

It may be difficult to understand why extraneous solutions exist at all. Thinking about extraneous solutions by graphing the equation may help you make sense of what is going on.

Although x = −1 is shown as a solution in both graphs, squaring both sides of the equation had the effect of adding an extraneous solution, x = −6. Again, this is why it is so important to check your answers when solving radical equations!

Solving Application Problems with Radical Equations

Radical equations play a significant role in science, engineering, and even music. Sometimes you may need to use what you know about radical equations to solve for different variables in these types of problems.

A common method for solving radical equations is to raise both sides of an equation to whatever power will eliminate the radical sign from the equation. But be careful—when both sides of an equation are raised to an even power, the possibility exists that extraneous solutions will be introduced. When solving a radical equation, it is important to always check your answer by substituting the value back into the original equation. If you get a true statement, then that value is a solution; if you get a false statement, then that value is not a solution.

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