problem solving radical equations

Radical 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 [latex]\Large{1 \over 2}[/latex].

Key Steps to Solve Radical Equations:

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 or solutions

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 [latex]x = 2[/latex], and [latex]x = {{ – 22} \over 7}[/latex].

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 [latex]x = {{ – 5} \over 2}[/latex] and [latex]x = 3[/latex] .

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

You may also be interested in these related math lessons or tutorials:

Radical Equations Practice Problems with Answers

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

8.6 Solve Radical Equations

Learning objectives.

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

• Solve radical equations
• Solve radical equations with two radicals
• Use radicals in applications

Be Prepared 8.16

Before you get started, take this readiness quiz.

Simplify: ( y − 3 ) 2 . ( y − 3 ) 2 . If you missed this problem, review Example 5.31 .

Be Prepared 8.17

Solve: 2 x − 5 = 0 . 2 x − 5 = 0 . If you missed this problem, review Example 2.2 .

Be Prepared 8.18

Solve n 2 − 6 n + 8 = 0 . n 2 − 6 n + 8 = 0 . If you missed this problem, review Example 6.45 .

Solve Radical Equations

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 .

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.

Example 8.56

How to solve a radical equation.

Solve: 5 n − 4 − 9 = 0 . 5 n − 4 − 9 = 0 .

Try It 8.111

Solve: 3 m + 2 − 5 = 0 . 3 m + 2 − 5 = 0 .

Try It 8.112

Solve: 10 z + 1 − 2 = 0 . 10 z + 1 − 2 = 0 .

Solve a radical equation with one radical.

• Step 1. Isolate the radical on one side of the equation.
• Step 2. Raise both sides of the equation to the power of the index.
• Step 3. Solve the new equation.
• Step 4. Check the answer in the original equation.

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.

Example 8.57

Solve: 9 k − 2 + 1 = 0 . 9 k − 2 + 1 = 0 .

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

Try It 8.113

Solve: 2 r − 3 + 5 = 0 . 2 r − 3 + 5 = 0 .

Try It 8.114

Solve: 7 s − 3 + 2 = 0 . 7 s − 3 + 2 = 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.

Binomial Squares

Don’t forget the middle term!

Example 8.58

Solve: p − 1 + 1 = p . p − 1 + 1 = p .

Try It 8.115

Solve: x − 2 + 2 = x . x − 2 + 2 = x .

Try It 8.116

Solve: y − 5 + 5 = y . y − 5 + 5 = y .

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

Example 8.59

Solve: 5 x + 1 3 + 8 = 4 . 5 x + 1 3 + 8 = 4 .

Try It 8.117

Solve: 4 x − 3 3 + 8 = 5 4 x − 3 3 + 8 = 5

Try It 8.118

Solve: 6 x − 10 3 + 1 = −3 6 x − 10 3 + 1 = −3

Sometimes an equation will contain rational exponents instead of a radical. We use the same techniques to solve the equation as when we have a radical. We raise each side of the equation to the power of the denominator of the rational exponent. Since ( a m ) n = a m · n , ( a m ) n = a m · n , we have for example,

Remember, x 1 2 = x x 1 2 = x and x 1 3 = x 3 . x 1 3 = x 3 .

Example 8.60

Solve: ( 3 x − 2 ) 1 4 + 3 = 5 . ( 3 x − 2 ) 1 4 + 3 = 5 .

Try It 8.119

Solve: ( 9 x + 9 ) 1 4 − 2 = 1 . ( 9 x + 9 ) 1 4 − 2 = 1 .

Try It 8.120

Solve: ( 4 x − 8 ) 1 4 + 5 = 7 . ( 4 x − 8 ) 1 4 + 5 = 7 .

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

Example 8.61

Solve: r + 4 − r + 2 = 0 . r + 4 − r + 2 = 0 .

Try It 8.121

Solve: m + 9 − m + 3 = 0 . m + 9 − m + 3 = 0 .

Try It 8.122

Solve: n + 1 − n + 1 = 0 . n + 1 − n + 1 = 0 .

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

Example 8.62

Solve: 3 3 x − 5 − 8 = 4 . 3 3 x − 5 − 8 = 4 .

Try It 8.123

Solve: 2 4 a + 4 − 16 = 16 . 2 4 a + 4 − 16 = 16 .

Try It 8.124

Solve: 3 2 b + 3 − 25 = 50 . 3 2 b + 3 − 25 = 50 .

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.

Example 8.63

Solve: 4 x − 3 3 = 3 x + 2 3 . 4 x − 3 3 = 3 x + 2 3 .

Try It 8.125

Solve: 5 x − 4 3 = 2 x + 5 3 . 5 x − 4 3 = 2 x + 5 3 .

Try It 8.126

Solve: 7 x + 1 3 = 2 x − 5 3 . 7 x + 1 3 = 2 x − 5 3 .

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.

Example 8.64

Solve: m + 1 = m + 9 . m + 1 = m + 9 .

Try It 8.127

Solve: 3 − x = x − 3 . 3 − x = x − 3 .

Try It 8.128

Solve: x + 2 = x + 16 . x + 2 = x + 16 .

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.

Solve a radical equation.

• Step 1. Isolate one of the radical terms on one side of the equation.
• Step 3. Are there any more radicals? If yes, repeat Step 1 and Step 2 again. If no, solve the new equation.

Be careful as you square binomials in the next example. Remember the pattern is ( a + b ) 2 = a 2 + 2 a b + b 2 ( a + b ) 2 = a 2 + 2 a b + b 2 or ( a − b ) 2 = a 2 − 2 a b + b 2 . ( a − b ) 2 = a 2 − 2 a b + b 2 .

Example 8.65

Solve: q − 2 + 3 = 4 q + 1 . q − 2 + 3 = 4 q + 1 .

Try It 8.129

Solve: x − 1 + 2 = 2 x + 6 x − 1 + 2 = 2 x + 6

Try It 8.130

Solve: x + 2 = 3 x + 4 x + 2 = 3 x + 4

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.

Use a problem solving strategy for applications with formulas.

• Step 1. Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
• Step 2. Identify what we are looking for.
• Step 3. Name what we are looking for by choosing a variable to represent it.
• Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Step 5. Solve the equation using good algebra techniques.
• Step 6. Check the answer in the problem and make sure it makes sense.
• Step 7. Answer the question with a complete sentence.

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.

Falling Objects

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

For example, if an object is dropped from a height of 64 feet, we can find the time it takes to reach the ground by substituting h = 64 h = 64 into the formula.

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

Example 8.66

Marissa dropped her sunglasses from a bridge 400 feet above a river. Use the formula t = h 4 t = h 4 to find how many seconds it took for the sunglasses to reach the river.

Try It 8.131

A helicopter dropped a rescue package from a height of 1,296 feet. Use the formula t = h 4 t = h 4 to find how many seconds it took for the package to reach the ground.

Try It 8.132

A window washer dropped a squeegee from a platform 196 feet above the sidewalk Use the formula t = h 4 t = h 4 to find how many seconds it took for the squeegee to reach the sidewalk.

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.

Skid Marks and Speed of a Car

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

Example 8.67

After a car accident, the skid marks for one car measured 190 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Try It 8.133

An accident investigator measured the skid marks of the car. The length of the skid marks was 76 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Try It 8.134

The skid marks of a vehicle involved in an accident were 122 feet long. Use the formula s = 24 d s = 24 d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

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

• Solving an Equation Involving a Single Radical
• Solving Equations with Radicals and Rational Exponents
• Solving Radical Equations
• Radical Equation Application

Section 8.6 Exercises

Practice makes perfect.

In the following exercises, solve.

5 x − 6 = 8 5 x − 6 = 8

4 x − 3 = 7 4 x − 3 = 7

5 x + 1 = −3 5 x + 1 = −3

3 y − 4 = −2 3 y − 4 = −2

2 x 3 = −2 2 x 3 = −2

4 x − 1 3 = 3 4 x − 1 3 = 3

2 m − 3 − 5 = 0 2 m − 3 − 5 = 0

2 n − 1 − 3 = 0 2 n − 1 − 3 = 0

6 v − 2 − 10 = 0 6 v − 2 − 10 = 0

12 u + 1 − 11 = 0 12 u + 1 − 11 = 0

4 m + 2 + 2 = 6 4 m + 2 + 2 = 6

6 n + 1 + 4 = 8 6 n + 1 + 4 = 8

2 u − 3 + 2 = 0 2 u − 3 + 2 = 0

5 v − 2 + 5 = 0 5 v − 2 + 5 = 0

u − 3 + 3 = u u − 3 + 3 = u

v − 10 + 10 = v v − 10 + 10 = v

r − 1 = r − 1 r − 1 = r − 1

s − 8 = s − 8 s − 8 = s − 8

6 x + 4 3 = 4 6 x + 4 3 = 4

11 x + 4 3 = 5 11 x + 4 3 = 5

4 x + 5 3 − 2 = −5 4 x + 5 3 − 2 = −5

9 x − 1 3 − 1 = −5 9 x − 1 3 − 1 = −5

( 6 x + 1 ) 1 2 − 3 = 4 ( 6 x + 1 ) 1 2 − 3 = 4

( 3 x − 2 ) 1 2 + 1 = 6 ( 3 x − 2 ) 1 2 + 1 = 6

( 8 x + 5 ) 1 3 + 2 = −1 ( 8 x + 5 ) 1 3 + 2 = −1

( 12 x − 5 ) 1 3 + 8 = 3 ( 12 x − 5 ) 1 3 + 8 = 3

( 12 x − 3 ) 1 4 − 5 = −2 ( 12 x − 3 ) 1 4 − 5 = −2

( 5 x − 4 ) 1 4 + 7 = 9 ( 5 x − 4 ) 1 4 + 7 = 9

x + 1 − x + 1 = 0 x + 1 − x + 1 = 0

y + 4 − y + 2 = 0 y + 4 − y + 2 = 0

z + 100 − z = −10 z + 100 − z = −10

w + 25 − w = −5 w + 25 − w = −5

3 2 x − 3 − 20 = 7 3 2 x − 3 − 20 = 7

2 5 x + 1 − 8 = 0 2 5 x + 1 − 8 = 0

2 8 r + 1 − 8 = 2 2 8 r + 1 − 8 = 2

3 7 y + 1 − 10 = 8 3 7 y + 1 − 10 = 8

3 u + 7 = 5 u + 1 3 u + 7 = 5 u + 1

4 v + 1 = 3 v + 3 4 v + 1 = 3 v + 3

8 + 2 r = 3 r + 10 8 + 2 r = 3 r + 10

10 + 2 c = 4 c + 16 10 + 2 c = 4 c + 16

5 x − 1 3 = x + 3 3 5 x − 1 3 = x + 3 3

8 x − 5 3 = 3 x + 5 3 8 x − 5 3 = 3 x + 5 3

2 x 2 + 9 x − 18 3 = x 2 + 3 x − 2 3 2 x 2 + 9 x − 18 3 = x 2 + 3 x − 2 3

x 2 − x + 18 3 = 2 x 2 − 3 x − 6 3 x 2 − x + 18 3 = 2 x 2 − 3 x − 6 3

a + 2 = a + 4 a + 2 = a + 4

r + 6 = r + 8 r + 6 = r + 8

u + 1 = u + 4 u + 1 = u + 4

x + 1 = x + 2 x + 1 = x + 2

a + 5 − a = 1 a + 5 − a = 1

−2 = d − 20 − d −2 = d − 20 − d

2 x + 1 = 1 + x 2 x + 1 = 1 + x

3 x + 1 = 1 + 2 x − 1 3 x + 1 = 1 + 2 x − 1

2 x − 1 − x − 1 = 1 2 x − 1 − x − 1 = 1

x + 1 − x − 2 = 1 x + 1 − x − 2 = 1

x + 7 − x − 5 = 2 x + 7 − x − 5 = 2

x + 5 − x − 3 = 2 x + 5 − x − 3 = 2

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

Landscaping Reed wants to have a square garden plot in his backyard. He has enough compost to cover an area of 75 square feet. Use the formula s = A s = A to find the length of each side of his garden. Round your answer to the nearest tenth of a foot.

Landscaping Vince wants to make a square patio in his yard. He has enough concrete to pave an area of 130 square feet. Use the formula s = A s = A to find the length of each side of his patio. Round your answer to the nearest tenth of a foot.

Gravity A hang glider dropped his cell phone from a height of 350 feet. Use the formula t = h 4 t = h 4 to find how many seconds it took for the cell phone to reach the ground.

Gravity A construction worker dropped a hammer while building the Grand Canyon skywalk, 4000 feet above the Colorado River. Use the formula t = h 4 t = h 4 to find how many seconds it took for the hammer to reach the river.

Accident investigation The skid marks for a car involved in an accident measured 216 feet. Use the formula s = 24 d s = 24 d to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 175 feet. Use the formula s = 24 d s = 24 d to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Writing Exercises

Explain why an equation of the form x + 1 = 0 x + 1 = 0 has no solution.

ⓐ Solve the equation r + 4 − r + 2 = 0 . r + 4 − r + 2 = 0 . ⓑ Explain why one of the “solutions” that was found was not actually a solution to the equation.

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

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

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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:

• isolate the square root on one side of the equation
• square both sides of the equation

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.

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8.6: Solving Radical Equations

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Learning Objectives

• Solve equations involving square roots.
• Solve equations involving cube roots

Radical Equations

A radical equation is any equation that contains one or more radicals with a variable in the radicand. Following are some examples of radical equations, all of which will be solved in this section:

\(\begin{array}{c}{\sqrt{x-1}=5} \\ {\sqrt{2 x-5}+4=x} \\ {\sqrt[3]{x^{2}+4}-2=0}\end{array}\)

We begin with the squaring property of equality; given real numbers a and b , we have the following:

If \(a=b\), then \(a^{2}=b^{2}\)

In other words, equality is retained if we square both sides of an equation.

\(\begin{array}{rlrl}{-3=-3} & {\Rightarrow} & {(-3)^{2}} & {=(-3)^{2}} \\ {} & {} & {9} & {=9} \:\:\color{Cerulean}{\checkmark} \end{array}\)

The converse, on the other hand, is not necessarily true:

This is important because we will use this property to solve radical equations. Consider a very simple radical equation that can be solved by inspection:

\(\sqrt{x}=3\)

Here we can see that \(x=9\) is a solution. To solve this equation algebraically, make use of the squaring property of equality and the fact that ((\sqrt{a})^{2}=\sqrt{a^{2}}=a\) when a is positive. Eliminate the square root by squaring both sides of the equation as follows:

\(\begin{aligned}\color{Cerulean}{(}\color{black}{\sqrt{x}}\color{Cerulean}{)^{2}} &\color{black}{=}\color{Cerulean}{(}\color{black}{3}\color{Cerulean}{)^{2}} \\ x &=9 \end{aligned}\)

As a check, we can see that \(\sqrt{9}=3\) as expected. Because the converse of the squaring property of equality is not necessarily true, solutions to the squared equation may not be solutions to the original. Hence squaring both sides of an equation introduces the possibility of extraneous solutions, or solutions that do not solve the original equation. For this reason, we must check the answers that result from squaring both sides of an equation.

Example \(\PageIndex{1}\)

\(\sqrt{x-1}=5\)

We can eliminate the square root by applying the squaring property of equality.

Next, we must check.

\(\begin{aligned}\color{black}{ \sqrt{\color{OliveGreen}{26}-1}} &=5 \\ \sqrt{25} &=5 \\ 5 &=5\:\:\color{Cerulean}{\checkmark} \end{aligned}\)

The solution is 26.

Example \(\PageIndex{2}\)

Begin by squaring both sides of the equation.

You are left with a quadratic equation that can be solved by factoring.

\(\begin{array}{cc}{x+5=0} & {\text { or } \quad x-1=0} \\ {x=-5} & {x=1}\end{array}\)

Since you squared both sides, you must check your solutions.

After checking, you can see that \(x=−5\) was extraneous; it did not solve the original radical equation. Disregard that answer. This leaves \(x=1\) as the only solution.

The solution is \(x=1\).

In the previous two examples, notice that the radical is isolated on one side of the equation. Typically, this is not the case. The steps for solving radical equations involving square roots are outlined in the following example.

Example \(\PageIndex{3}\)

\(\sqrt{2 x-5}+4=x\)

Step 1: Isolate the square root. Begin by subtracting 4 from both sides of the equation.

Step 2: Square both sides. Squaring both sides eliminates the square root.

Step 3: Solve the resulting equation. Here you are left with a quadratic equation that can be solved by factoring.

\(\begin{array}{cc}{x-3=0} & {\text { or } \quad x-7=0} \\ {x=3} & {x=7}\end{array}\)

Step 4: Check the solutions in the original equation. Squaring both sides introduces the possibility of extraneous solutions; hence the check is required.

\(\begin{array}{r|r}{\text {Check } x=3} & {\text { Check } x=7} \\ {\sqrt{2 x-5}+4=x} & {\sqrt{2 x-5}+4=x}\\{\sqrt{2(\color{OliveGreen}{3}\color{black}{)}-5}+4=\color{OliveGreen}{3}}&{\sqrt{2(\color{OliveGreen}{7}\color{black}{)}-5}+4=\color{OliveGreen}{7}}\\{\sqrt{6-5}+4=3}&{\sqrt{14-5}+4=7}\\{\sqrt{1}+4=3}&{\sqrt{9}+4=7}\\{1+4=3}&{3+4=7}\\{5=3\:\:\color{red}{x}}&{7=7\:\:\color{Cerulean}{\checkmark}} \end{array}\)

After checking, we can see that \(x=3\) is an extraneous root; it does not solve the original radical equation. This leaves \(x=7\) as the only solution.

The solution is \(x=7\).

Example \(\PageIndex{4}\)

Begin by isolating the term with the radical

Despite the fact that the term on the left side has a coefficient, it is still considered isolated. Recall that terms are separated by addition or subtraction operators.

\(\begin{aligned} 3 \sqrt{x+1} &=2 x \\(3 \sqrt{x+1})^{2} &=(2 x)^{2}\qquad\color{Cerulean}{Square\:both\:sides.} \\ 9(x+1) &=4 x^{2} \end{aligned}\)

Solve the resulting quadratic equation.

\(\begin{array}{rlrl}{4 x+3} & {=0} & {\text { or }} & {x-3=0} \\ {4 x} & {=-3} && {x=3} \\ {x} & {=-\frac{3}{4}}\end{array}\)

Since we squared both sides, we must check our solutions.

After checking, we can see that \(x=−\frac{3}{4}\) was extraneous.

The solution is 3.

Sometimes both of the possible solutions are extraneous.

Example \(\PageIndex{5}\)

Begin by isolating the radical.

Since both possible solutions are extraneous, the equation has no solution.

No solution, Ø

The squaring property of equality extends to any positive integer power n . Given real numbers a and b , we have the following:

If \(a=b\), then \(a^{n}=b^{n}\)

This is often referred to as the power property of equality. Use this property, along with the fact that \((\sqrt[n]{a})^{n} = \sqrt[n]{a^{n}}=a\), when a is positive, to solve radical equations with indices greater than 2.

Example \(\PageIndex{6}\)

\(\sqrt[3]{x^{2}+4}-2=0\)

Isolate the radical and then cube both sides of the equation.

\(\begin{array}{rlrl}{x+2} & {=0} & {\text { or }} & {x-2=0} \\ {x} & {=-2} & {} & {x=2}\end{array}\)

\(\begin{array}{r|r}{\text {Check } x=-2} & {\text { Check } x=2} \\ {\sqrt[3]{x^{2}+4}-2=0} & {\sqrt[3]{x^{2}+4}-2=0}\\{\sqrt[3]{(\color{OliveGreen}{-2}\color{black}{)}^{2}+4}-2=0}&{\sqrt[3]{(\color{OliveGreen}{2}\color{black}{)}^{2}+4}-2=0}\\{\sqrt[3]{4+4}-2=0}&{\sqrt[3]{4+4}-2=0}\\{\sqrt[3]{8}-2=0}&{\sqrt[3]{8}-2=0}\\{2-2=0}&{2-2=0}\\{0=0\:\:\color{Cerulean}{\checkmark}}&{0=0\:\:\color{Cerulean}{\checkmark}} \end{array}\)

Exercise \(\PageIndex{1}\)

\(\sqrt{2 x-1}+2=x\)

\(x=5\) (\(x=1\) is extraneous)

It may be the case that the equation has two radical expressions.

Example \(\PageIndex{7}\)

\(\sqrt{3 x-4}=\sqrt{2 x+9}\)

Both radicals are considered isolated on separate sides of the equation.

Check \(x=13\).

\(\begin{aligned} \sqrt{3 x-4} &=\sqrt{2 x+9} \\ \sqrt{3(\color{OliveGreen}{13}\color{black}{)}-4} &=\sqrt{2(\color{OliveGreen}{13}\color{black}{)}+9} \\ \sqrt{39-4} &=\sqrt{26+9} \\ \sqrt{35} &=\sqrt{35} \quad \color{Cerulean}{\checkmark} \end{aligned}\)

The solution is 13.

Example \(\PageIndex{8}\)

\(\sqrt[3]{x^{2}+x-14}=\sqrt[3]{x+50}\)

Eliminate the radicals by cubing both sides.

\(\begin{array}{rlrl}{x+8} & {=0} & {\text { or }} & {x-8=0} \\ {x} & {=-8} && {x=8}\end{array}\)

\(\begin{array}{r|r}{\text { Check } x=-8}&{\text{Check }x=8} \\ {\sqrt[3]{x^{2}+x-14}=\sqrt[3]{x+50}} & {\sqrt[3]{x^{2}+x-14}=\sqrt[3]{x+50}}\\{\sqrt[3]{(\color{OliveGreen}{-8}\color{black}{)}^{2}+(\color{OliveGreen}{-8}\color{black}{)}-14}=\sqrt[3]{(\color{OliveGreen}{-8}\color{black}{)}+50}}&{\sqrt[3]{(\color{OliveGreen}{8}\color{black}{)}^{2}+(\color{OliveGreen}{8}\color{black}{)}-14}=\sqrt[3]{(\color{OliveGreen}{8}\color{black}{)}+50}}\\{\sqrt[3]{64-8-14}=\sqrt[3]{42}}&{\sqrt[3]{64+8-14}=\sqrt[3]{58}}\\{\sqrt[3]{42}=\sqrt[3]{42}\:\:\color{Cerulean}{\checkmark}}&{\sqrt[3]{58}=\sqrt[3]{58}\:\:\color{Cerulean}{\checkmark}} \end{array}\)

We will learn how to solve some of the more advanced radical equations in the next course, Intermediate Algebra.

Exercise \(\PageIndex{2}\)

\(\sqrt{3 x+1}=\sqrt{2 x-3}\)

No solution for x

Key Takeaways

• Solve equations involving square roots by first isolating the radical and then squaring both sides. Squaring a square root eliminates the radical, leaving us with an equation that can be solved using the techniques learned earlier in our study of algebra. However, squaring both sides of an equation introduces the possibility of extraneous solutions, so check your answers in the original equation.
• Solve equations involving cube roots by first isolating the radical and then cubing both sides. This eliminates the radical and results in an equation that may be solved with techniques you have already mastered.

Exercise \(\PageIndex{3}\) Solving Radical Equations

• \(\sqrt{x}=2\)
• \(\sqrt{x}=7\)
• \(\sqrt{x}+7=8\)
• \(\sqrt{x}+4=9\)
• \(\sqrt{x}+6=3\)
• \(\sqrt{x}+2=1\)
• \(5\sqrt{x}−1=0\)
• \(3\sqrt{x}−2=0\)
• \(\sqrt{x−3}=3\)
• \(\sqrt{x+5}=6\)
• \(\sqrt{3x+1}=2\)
• \(\sqrt{5x−4}=4\)
• \(\sqrt{7x+4}+6=11\)
• \(\sqrt{3x−5}+9=14\)
• \(\sqrt{2x−1}−3=0\)
• \(\sqrt{3x+1}−2=0\)
• \(\sqrt[3]{x}=2\)
• \(\sqrt[3]{x}=5\)
• \(\sqrt[3]{2x+9}=3\)
• \(\sqrt[3]{4x−11}=1\)
• \(\sqrt[3]{5x+7}+3=1\)
• \(\sqrt[3]{3x−6}+5=2\)
• \(2\sqrt[3]{x+2}−1=0\)
• \(2\sqrt[3]{2x−3}−1=0\)
• \(\sqrt{8x+11}=\sqrt{3x+1}\)
• \(2 \sqrt{3 x-4}=\sqrt{2(3 x+1)}\)
• \(\sqrt{2(x+10)}=\sqrt{7 x-15}\)
• \(\sqrt{5(x−4)}=\sqrt{x+4}\)
• \(\sqrt[3]{5 x-2}=\sqrt[3]{4 x}\)
• \(\sqrt[3]{9(x−1)}=\sqrt[3]{3(x+7)}\)
• \(\sqrt[3]{3 x+1}=\sqrt[3]{2(x-1)}\)
• \(\sqrt[3]{9x}=\sqrt[3]{3(x−6)}\)
• \(\sqrt{4 x+21}=x\)
• \(\sqrt{8x+9}=x\)
• \(\sqrt{4(2x−3)}=x\)
• \(\sqrt{3(4x−9)}=x\)
• \(2 \sqrt{x-1}=x\)
• \(3\sqrt{2x−9}=x\)
• \(\sqrt{9 x+9}=x+1\)
• \(\sqrt{3x+10}=x+4\)
• \(\sqrt{x−1}=x−3\)
• \(\sqrt{2x−5}=x−4\)
• \(\sqrt{16−3x}=x−6\)
• \(\sqrt{7−3x}=x−3\)
• \(3 \sqrt{2 x+10}=x+9\)
• \(2\sqrt{2x+5}=x+4\)
• \(3\sqrt{x−1}-1=x\)
• \(2\sqrt{2x+2}−1=x\)
• \(\sqrt{10x+41}−5=x\)
• \(\sqrt{6(x+3)}−3=x\)
• \(\sqrt{8x^{2}−4x+1}=2x\)
• \(\sqrt{18x^{2}−6x+1}=3x\)
• \(5\sqrt{x+2}=x+8\)
• \(4 \sqrt{2(x+1)}=x+7\)
• \(\sqrt{x^{2}−25}=x\)
• \(\sqrt{x^{2}+9}=x\)
• \(3+\sqrt{6x−11}=x\)
• \(2+\sqrt{9x−8}=x\)
• \(\sqrt{4x+25}-x=7\)
• \(\sqrt{8x+73}−x=10\)
• \(2\sqrt{4x+3}−3=2x\)
• \(2\sqrt{6x+3}−3=3x\)
• \(2x−4=\sqrt{14−10x}\)
• \(3x−6=3\sqrt{3−24x}\)
• \(\sqrt[3]{x^{2}−24}=1\)
• \(\sqrt[3]{x^{2}−54}=3\)
• \(\sqrt[3]{x^{2}+6x}+1=4\)
• \(\sqrt[3]{x^{2}+2x}+5=7\)
• \(\sqrt[3]{25x^{2}−10x−7}=−2\)
• \(\sqrt[3]{9x^{2}−12x−23}=−3\)
• \(\sqrt{2 x^{2}-15 x+25}=\sqrt{(x+5)(x-5)}\)
• \(\sqrt{x^{2}−4x+4}=\sqrt{x(5−x)}\)
• \(\sqrt[3]{2\left(x^{2}+3 x-20\right)}=\sqrt[3]{(x+3)^{2}}\)
• \(\sqrt[3]{3x^{2}+3x+40}=\sqrt[3]{(x−5)^{2}}\)
• \(x^{1/2}−10=0\)
• \(x^{1/2}−6=0\)
• \(x^{1/3}+2=0\)
• \(x^{1/3}+4=0\)
• \((x−1)^{1/2}−3=0\)
• \((x+2)^{1/2}−6=0\)
• \((2x−1)^{1/3}+3=0\)
• \((3x−1)^{1/3}−2=0\)
• \((4x+15)^{1/2}−2x=0\)
• \((3x+2)^{1/2}−3x=0\)
• \((2x+12)^{1/2}−x=6\)
• \((4x+36)^{1/2}−x=9\)
• \(2(5x+26)^{1/2}=x+10\)
• \(3(x−1)^{1/2}=x+1\)
• The square root of 1 less than twice a number is equal to 2 less than the number. Find the number.
• The square root of 4 less than twice a number is equal to 6 less than the number. Find the number.
• The square root of twice a number is equal to one-half of that number. Find the number.
• The square root of twice a number is equal to one-third of that number. Find the number.
• The distance, d , measured in miles, a person can see an object is given by the formula \(d=\sqrt{\frac{3h}{2}\) where h represents the person’s height above sea level, measured in feet. How high must a person be to see an object 5 miles away?
• The current, I , measured in amperes, is given by the formula \(I=\sqrt{\frac{P}{R}}\) where P is the power usage, measured in watts, and R is the resistance, measured in ohms. If a light bulb requires 1/2 ampere of current and uses 60 watts of power, then what is the resistance of the bulb?

5. \(Ø\)

7. \(\frac{1}{25}\)

21. \(−3\)

23. \(−\frac{15}{8}\)

25. \(Ø\)

31. \(−3\)

35. \(2, 6\)

39. \(−1, 8\)

43. \(Ø\)

45. \(−3, 3\)

47. \(2, 5\)

49. \(4, −4\)

51. \(\frac{1}{2}\)

53. \(2, 7\)

55. \(Ø\)

59. \(−6, −4\)

61. \(−\frac{1}{2},\frac{3}{2}\)

63. \(Ø\)

65. \(−5, 5\)

67. \(−9, 3\)

69. \(\frac{1}{5}\)

71. \(5, 10\)

73. \(−7, 7\)

75. \(100\)

77. \(−8\)

81. \(−13\)

83. \(\frac{5}{2}\)

85. \(−6, −4\)

87. \(−2, 2\)

93. \(16 \frac{2}{3}\) feet

Exercise \(\PageIndex{4}\) Solving Radical Equations

The period, T , of a pendulum in seconds is given by the formula

\(T=2π\sqrt{L/32}\)

where L represents the length in feet. For each problem below, calculate the length of a pendulum, given the period. Give the exact value and the approximate value rounded off to the nearest tenth of a foot.

• \(1\) second
• \(2\) seconds
• \(\frac{1}{2}\) second
• \(\frac{1}{3}\) second

1. \(\frac{8}{\pi^{2}} ≈0.8\) foot

3. \(\frac{2}{\pi^{2}} ≈0.2\) foot

Exercise \(\PageIndex{5}\) Solving Radical Equations

The time, t , in seconds an object is in free fall is given by the formula

\(s=16\cdot t^{2}\)

where s represents the distance in feet the object has fallen. For each problem below, calculate the distance an object falls, given the amount of time.

• \(\frac{1}{4}\) second

Exercise \(\PageIndex{6}\) Solving Radical Equations

The x -intercepts for any graph have the form \((x, 0)\), where x is a real number. Therefore, to find x -intercepts, set \(y = 0\) and solve for x . Find the x -intercepts for each of the following.

• \(y=\sqrt{x−3}−1\)
• \(y=\sqrt{x+2}−3\)
• \(y=\sqrt[3]{x−1}+2\)
• \(y=\sqrt[3]{x+1}−3\)

1. \((4, 0)\)

3. \((−7, 0)\)

Exercise \(\PageIndex{7}\) Discussion Board

• Discuss reasons why we sometimes obtain extraneous solutions when solving radical equations. Are there ever any conditions where we do not need to check for extraneous solutions? Why?

1. Answers may vary

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Solving Radical Equations

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A radical equation is an equation that contains at least one radical sign that includes a variable. For an example you can consider the following equation:

\(\qquad \qquad \qquad \qquad \sqrt{x+2}=x-3\).

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 .

Introduction

Problem solving.

A radical is an exponential number where the exponent is a fraction . For an example, consider \( \sqrt[n]{x^m} = x ^ { m/n } \). A radical equation contains at least one radical sign that includes a variable.

A square root \(\left( \sqrt{x} \right)\) is simplified, when the radicand (value inside the square root sign, which is \(x\) in this case) has no square factors. This can be done effectively by factorizing the radicand first, to know what terms to pull out.

Here is a problem based on the basics of radicals.

True or False:

For all real numbers \(y \),

\[\Large y = \sqrt{y^2}\]

The leading step of solving a radical equation is to eliminate the radical exponent to obtain a polynomial equation and check whether the obtained roots satisfy the original equation to escape from extraneous roots.

Following are the steps to solve a radical equation:

Eliminate the radicals or rational exponents and obtain a polynomial equation. The key step is to raise each side of the equation to the same power. If \(a = b,\) then \(a^n = b^n\). [Powers property of equality] Then solve the new equation using standard procedures. Before raising each side of an equation to the same power, you should isolate the radical expression on one side of the equation. However, this is not it. When solving radical equations we must always be cautious of extraneous roots . An extraneous root is a root that satisfies the polynomial obtained by squaring, but does not satisfy the original radical equation.

Question: Why do extraneous roots appear?

Answer: Suppose we are solving the radical equation \(A=\sqrt{B},\) where \(A\) and \(B\) are both polynomials. Squaring both sides gives \(A^2=B,\) solving which would give solutions to both \(A=\sqrt{B}\) and \(A=-\sqrt{B}.\) The roots that satisfy \(A=-\sqrt{B}\) are not what we were looking for. Therefore we must never forget to check if the roots we obtained satisfy the original radical equation.

However, we will most often encounter radical equations that contain square roots. The way to solve such equations is simple. Just put the square root on one side of the equation and the non-radicals on the other. Then square both sides and we get a polynomial we can solve!

For demonstration we will take the following example and solve it using the above stated steps:

Solve the equation \(\sqrt{x+3}=x+1.\) Step 1: Squaring both sides (which will get rid of the square root) gives \(x+3=x^2+2x+1.\) Step 2: Rearranging terms, we have \[x^2+x-2=(x+2)(x-1)=0 \Rightarrow x=-2 \text{ or } x=1.\] Step 3: Observe that \(x=-2\) is an extraneous root because \[\text{LHS}=\sqrt{(-2)+3}=1 \ne (-2)+1=\text{ RHS}.\] Then since \[\text{LHS}=\sqrt{(1)+3}=2=(1)+1=\text{ RHS},\] the solution to the equation \(\sqrt{x+3}=x+1\) is \(x=1.\) \( _\square \)

Try to solve the following problems using the above steps.

No problem found with slug "real-ly"

\[ \sqrt{x+1} - \sqrt{x-1} = \sqrt{4x-1}, \ \ x = ? \]

This section is dedicated to enhance the problem solving skills through several examples and problems to try.

Solve the equation \(x-2=\sqrt{2x+31}.\) Squaring both sides gives \(x^2-4x+4=2x+31.\) Rearranging terms, we have \[x^2-6x-27=(x+3)(x-9)=0 \Rightarrow x=-3 \text{ or } x=9.\] Observe that \(x=-3\) is an extraneous root because \[\text{LHS}=(-3)-2 = -5 \ne \sqrt{2\cdot (-3)+31}=\text{ RHS}.\] Then since \[\text{LHS}=(9)-2=7= \sqrt{2\cdot (9)+31}=\text{ RHS},\] the solution to the equation \(x-2=\sqrt{2x+31}\) is \(x=9.\) \( _\square \)

Solve the equation \(\sqrt{2x^2+7}+1=2x.\) The equation \(\sqrt{2x^2+7}+1=2x\) is equivalent to \(\sqrt{2x^2+7}=2x-1.\) Squaring both sides gives \(2x^2+7=4x^2-4x+1.\) Rearranging terms, we have \[2x^2-4x-6=2(x^2-2x-3)=2(x+1)(x-3)=0 \Rightarrow x=-1 \text{ or } x=3.\] Observe that \(x=-1\) is an extraneous root since \[\text{LHS}=\sqrt{2\cdot(-1)^2+7}=3 \ne 2\cdot (-1)-1=\text{ RHS}.\] Then since \[\text{LHS}=\sqrt{2\cdot (3)^2+7}=5=2\cdot (3)-1=\text{ RHS},\] the solution to the equation \(\sqrt{2x^2+7}+1=2x\) is \(x=3.\) \( _\square \)

Here are the problems for you try.

Find the only integral value of \(x\) in the equation

\[\sqrt[3]{x-9}+\sqrt[4]{x+8}=7.\]

\[ \dfrac{ \sqrt{3x} + \sqrt{2x+1} }{\sqrt{3x} - \sqrt{2x+1}} = 5 \]

Find the value of \(x\) satisfying the equation above.

\[\] Clarification: For this question, we are working over the complex numbers.

\(\big(\sqrt{x+1}+1\big)\big(\sqrt{x+16}+4\big)-\big(\sqrt{x+4}+2\big)\big(\sqrt{x+9}+3\big)=0\)

How many real value(s) of \(x\) satisfy the above equation?

If \[\Large {\color{blue}{x}}^{\color{blue}{x}}= 2015,\] then which answer is equivalent to \[\Huge \color{blue}{x^{x^{x^x}}}?\]

Find the integer value of \(x\) satisfying the equation below.

\[ \sqrt{10 + \sqrt{x^3 + 100}} = 10 - \sqrt{x^3 + 100} \]

Find the domain of all real numbers \(x\) such that

\[ \sqrt{ x + 6 - 4 \sqrt{ x+2} } + \sqrt{ x+11 - 6 \sqrt{x+2}} = 1. \]

Systems of Equations

Quadratic Equations

Functional Equations

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Course: Algebra 2   >   Unit 10

• Extraneous solutions
• Equation that has a specific extraneous solution

Extraneous solutions of radical equations

• Extraneous solutions of equations

Introduction

Practice question 1.

• (Choice A)   Yes A Yes
• (Choice B)   No B No

Practice question 2

Practice question 3.

• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  

Practice question 4

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Roots and Radicals

Solve Radical Equations

Learning Objectives

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

• Solve radical equations
• Solve radical equations with two radicals
• Use radicals in applications

Before you get started, take this readiness quiz.

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.

• Isolate the radical on one side of the equation.
• Raise both sides of the equation to the power of the index.
• Solve the new equation.
• Check the answer in the original equation.

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.

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

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.

Don’t forget the middle term!

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

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

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

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.

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.

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.

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.

• Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
• Identify what we are looking for.
• Name what we are looking for by choosing a variable to represent it.
• Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Solve the equation using good algebra techniques.
• Check the answer in the problem and make sure it makes sense.
• Answer the question with a complete sentence.

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

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

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

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

• Solving an Equation Involving a Single Radical
• Solving Equations with Radicals and Rational Exponents
• Solving Radical Equations
• Radical Equation Application

Key Concepts

Practice Makes Perfect

In the following exercises, solve.

no solution

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

Writing Exercises

Answers will vary.

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

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

Intermediate Algebra Copyright © 2017 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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