INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith

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For more information, please see full course syllabus of Algebra 1

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 1 answerLast reply by: Professor Eric SmithWed Jul 30, 2014 2:45 PMPost by Peter Spicer on July 22, 2014Wouldn't you add 3 to both sides since we have an x-3?

### Solving Radical Equations

• A radical equation is an equation in which variables are present in the radical.
• To solve a radical equation, isolate the radical and raise each side of the equation to an appropriate power.
• You must check your solutions with these types of problems to ensure that they work in the original.
• If there is more than one radical expression in the equation, isolate them one at a time. Also be very careful with raising each side to a power. Some instances require you to FOIL.

### Solving Radical Equations

âˆš{4x âˆ’ 16} = 3âˆš2
• ( âˆš{4x âˆ’ 16} )2 = ( 3âˆš2 )2
• 4x âˆ’ 16 = 9 Ã—2
• 4x âˆ’ 16 = 18
• 4x = 34
x = 8[1/2]
âˆš{10x + 12} = 8âˆš3
• 10x + 12 = 64 Ã—3
• 10x + 12 = 192
• 10x = 180
x = 18
4âˆš5 = âˆš{6y + 8}
• 16 Ã—5 = 6y + 8
• 80 = 6y + 8
• 72 = 6y
12 = y
x = âˆš{x + 30}
• x2 = ( âˆš{x + 30} )2
• x2 = x + 30
• x2 âˆ’ x âˆ’ 30 = 0
• ( x + 5 )( x âˆ’ 6 ) = 0
x = âˆ’ 5,6
x = âˆš{x + 56}
• x2 = ( âˆš{x + 56} )2
• x2 = x + 56
• x2 âˆ’ x âˆ’ 56 = 0
• ( x + 7 )( x âˆ’ 8 ) = 0
x = âˆ’ 7,8
x = âˆš{13x + 30}
• x2 = ( âˆš{13x + 30} )2
• x2 = 13x + 30
• x2 âˆ’ 13x âˆ’ 30 = 0
• ( x + 2 )( x âˆ’ 15 ) = 0
x = âˆ’ 2,15
âˆ’ 2 + âˆš{x + 4} = x + 4
• âˆš{x + 4} = x + 4 + 2
• âˆš{x + 4} = x + 6
• ( âˆš{x + 4} )2 = ( x + 6 )2
• x + 4 = x2 + 12x + 36
• 4 = x2 + 11x + 36
• 0 = x2 + 11x + 28
• x2 + 11x + 28 = 0
• ( x + 4 )( x + 7 ) = 0
x = âˆ’ 4, âˆ’ 7
1 + âˆš{4x âˆ’ 11} = x âˆ’ 1
• âˆš{4x âˆ’ 11} = x âˆ’ 1 âˆ’ 1
• âˆš{4x âˆ’ 11} = x âˆ’ 2
• ( âˆš{4x âˆ’ 11} )2 = ( x âˆ’ 2 )2
• 4x âˆ’ 11 = x2 âˆ’ 4x + 4
• âˆ’ 11 = x2 âˆ’ 8x + 4
• 0 = x2 âˆ’ 8x + 15
• x2 âˆ’ 8x + 15 = 0
• ( x âˆ’ 5 )( x âˆ’ 3 )
x = 5,3
x âˆ’ âˆš{3 âˆ’ 11x} = 3
• âˆ’ âˆš{3 âˆ’ 11x} = 3 âˆ’ x
• ( âˆ’ âˆš{3 âˆ’ 11x} )2 = ( 3 âˆ’ x )2
• 3 âˆ’ 11x = ( âˆ’ x + 3 )2
• 3 âˆ’ 11x = x2 âˆ’ 6x + 9
• âˆ’ 11x = x2 âˆ’ 6x + 6
• 0 = x2 + 5x + 6
• 0 = ( x + 3 )( x + 2 )
x = âˆ’ 3, âˆ’ 2
x âˆ’ âˆš{5 âˆ’ 7x} = 7
• âˆ’ âˆš{5 âˆ’ 7x} = 7 âˆ’ x
• ( âˆ’ âˆš{5 âˆ’ 7x} )2 = ( âˆ’ x + 7 )2
• 5 âˆ’ 7x = x2 âˆ’ 14x + 49
• âˆ’ 7x = x2 âˆ’ 14x + 44
• 0 = x2 âˆ’ 7x + 44
• 0 = ( x âˆ’ 11 )( x + 4 )
x = 11, âˆ’ 4

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Solving Radical Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Objectives 0:07
• Solving Radical Equations 0:17
• Isolate the Roots and Raise to Power
• Example 1 1:13
• Example 2 3:09
• Solving Radical Equations Cont. 7:04
• Solving Radical Equations with More than One Radical
• Example 3 7:54
• Example 4 13:07

### Transcription: Solving Radical Equations

Welcome back to www.educator.com.0000

In this lesson we are going to go ahead and solve some radical equations.0003

We only have one thing to pickup in this lesson it is just all the nuts and bolts on how you solve on these radical equations.0009

A radical equation is an equation in which we have variables present in our radical.0018

That will be something like this, I have âˆšx-5 - âˆšx-3 = 1.0026

To solve these we essentially want to end up isolating these roots then raise each side of our equation to a power in order to get rid of them.0033

It has to be the appropriate power.0044

For dealing with square roots we will square both sides.0045

If we have say 4th root then we will raise both sides to the 4th power.0048

With these ones it is extremely important that you go ahead and you check your solutions when you are done.0054

In the solving process and you raise both sides to a power we may introduce solutions or possible solutions that do not work in the original.0060

Always check your solutions for these types of equations.0069

Let us get an idea of what Iâ€™m talking about with âˆš3x + 1 - 4 = 0.0075

The very first thing you want to do when solving something like this is get that root all by itself.0082

Let us go to both sides of the equation so we can do that.0088

âˆš3x+ 1 = 40092

We have isolated the root we are going to get rid of it by raising both sides to the power of two.0097

The reason why we are doing this is because we have a square root.0104

3x + 1 = 42 which is 160109

Now they got rid of your roots, we simply solve directly.0117

The type of equation that we might end up with could be quadratic, could be linear, but we use all of our other tools from here on out.0122

-1 from both sides we will get 15 and divide both sides by 3 will give us 5.0129

You will realize that this looks like a possible solution always go back to your original for these ones and check to see if it does work.0136

Iâ€™m going to write this out here -4 does that equals 0 or not.0146

Let us put in 5, 3 Ã— 5 would be 15 and all of that is underneath the root and then 15 + 1 would be 16.0156

âˆš16=4, 4 -4 does equal 0, I know that this checks out and x equals 5 is our solution.0172

This one, we will go ahead and try and solve it is 5 + âˆšx+ 7 = x.0191

Isolate our radical by subtracting 5 from both sides of the equation âˆš x + 7 = x - 5.0200

To get rid of that radical let us go ahead and raise both sides to the power of 2.0213

On the left side we will just have x + 7.0222

Be very careful on what is going on over here, keep your eyes peeled because notice how we have a binomial and we are squaring it.0226

In fact, you should look at it like this x - 5 Ã— x â€“ 5.0238

That way you can remember that this is how it should be foiled instead of trying to do some weird distribution with the two.0244

It does not work like that.0250

Let us see what we have when we foil.0253

Our first terms are x2, outside terms - 5x, inside terms - 5x, and our last terms 25.0256

We can go ahead and combine to see what we will get x2 - 10x + 25.0270

You can see that this one is turning into a quadratic equation because we have x2 term.0285

Since it is quadratic we will get all of our x on to the same side and see if we can use some of our quadratic tools in order to solve it.0291

0 = x2 - 11x + 180301

How shall we solve this quadratic?0312

It is not too bad, it looks like we can use reverse foil to go ahead and break it up and see what our parts are.0315

x Ã— x = x2 then 2 Ã— 9 = 180322

I know that 2 and 9 are good candidate because If I add 2 and 9 I will get that 11.0330

Let us make these both negative.0335

Have it x = 2 or x = 9.0337

Two possible solutions and they are possible because they might not work.0342

Let us check in the original 5 + âˆš 2 + 7 does it equal 2?0347

Let us find out, 2 + âˆš7 = âˆš 9, âˆš9 is 3 so I get 8.0359

Unfortunately that looks like it is not the same as the other side.0372

This one does not work and we can mark it off our list.0377

Let us try the other one 5 + the square roots and we are testing out a 9 so we will put that in 9 and 9.0383

Underneath the square root, 9 + 7 would give us a 16 and âˆš16 =4, I will get 9 which is the same thing as the other side.0396

That one looks good.0410

You will keep that one as your solution.0413

Be very careful and always check these types of ones back in the original, you will know which ones you should throw out.0416

If there happens to be more than one radical in your equation then we can solve these by isolating the radicals one at a time.0427

Do not try and take care of them both at once.0434

Just focus on one, get rid of that radical and focus on the other one and then get rid of that radical.0436

It does not matter which radical you choose first.0442

If you have a bunch of them just go ahead choose one and go after it.0446

Be very careful as you raise both sides to a power since you have more than one radical in there,0450

when you raise both sides to power it will often end having to get foiled.0456

Let us see exactly what I'm talking about with this step, but will have to be extremely careful to make sure we multiply correctly.0460

Always make sure you check your solutions to see that they work in the original.0468

But this one has two radicals in it.0474

I have âˆšx - 3 + âˆšx + 5 = 40477

It does not matter which radical you choose at the very beginning and I'm going to choose this one.0483

Iâ€™m going to work to isolate it and get all by itself x + 5 = 4 â€“ and I have subtracted the other radical to the other side.0490

Since âˆšx+ 5 is the one Iâ€™m trying to get rid of, Iâ€™m going to square both sides to get rid of it.0505

That will leave me with just x + 5 on that left side.0514

Be very careful what happens over here on the other side, it is tempting to try and distribute the 2 but that is not how those work.0520

In fact if we are going to take look at this as two binomial so we can go ahead and distribute.0528

Let us go ahead and take care of this very carefully.0542

Our first terms 4 Ã— 4 = 16, our outside terms we are taking 4 Ã— -âˆšx â€“ 3, -4 Ã— âˆšx - 3.0545

Inside terms would be the same thing -4 âˆšx - 3 and now we have our last terms.0560

negative Ã— negative = positive and then we have âˆšx -3 Ã— âˆšx -3.0570

That will give us âˆšx - 32 since it will be multiplied by themselves.0581

When you do that step the first time, it usually looks like you have made things way more complicated.0590

I mean, we are trying to get rid of our root but now it looks like I have 3 of them running around the page.0596

It is okay we will be able to simplify and in the end we will end up with just one root,0600

which is good because originally we started with two and if I get it down to one we are moving forward with this problem.0605

Let us see what we can do.0612

On that left side I have 16 these radicals are exactly the same, I will just put their coefficients together -8 âˆšx -30614

The last one square of square root would be x -3.0626

Now you can start to combine a few other things.0634

I will drop these parentheses here, if I subtract x from both sides that will take care of both of those x.0639

Let us see I can go ahead and subtract my 3 from the 16, 5 =13 - 8âˆšx -3.0650

Let us go ahead and subtract a 13 from both sides, -13 â€“ 13, -8 = -8âˆšx -3.0665

It looks like we can divide both sides by -8, 1 =âˆšx -3.0682

That is quite a bit of work, but all of that work was just to get rid of one of those radicals and we have accomplished that.0689

We got rid of that one.0695

But we still have the other radical here to get rid of.0696

We go through the same process of getting it all alone on one side, isolating it, squaring both sides get rid of it, and solving the resulting equation.0700

This one is already isolated we are good there.0709

I will simply move forward by squaring both sides.0711

1 = x -3 let us get some space.0716

If I add 3 both sides this will be 4 = x.0724

Even after going through all of that work and just when you think you have found a solution, we got to check these things.0729

We have to make sure that it will work in the original.0735

4 - 3 + âˆš4 + 5 we are checking does that equal 4.0741

Let us see what we have.0756

4 - 3 would give us âˆš1 , 4 + 5 would be 9, soâˆš4 =1 and âˆš9 = 3 and fortunately 4 does equal 4.0758

We know that this solution checks out, 4 =x.0774

If you have more than one of those radicals just try and get rid of them one at a time.0781

You might be faced with some higher roots and that is okay.0788

You will end up simply using a higher power on both sides of your equation in order to get rid of them.0791

In this one I have the 3rd root of 7x - 8 = 3rd root of 8x+ 2.0796

If I'm going to get rid of these cube roots I will raise both sides of it to the power of 3.0802

Leaving me with 7x - 8 = 8x + 20810

I can work on just giving my x together.0818

-7x from both sides, -8 = x + 2.0822

We will subtract 2 from both sides and this will give us -10 = x.0833

Let us quickly jump back up here to the top and see if that checks out.0841

3rd root of 7 Ã— putting that -10 - 8 and 8 Ã— -10 + 20845

Let us see if they are equal.0866

This time Iâ€™m dealing with 3rd root of (-70-8) to be the 3rd root of -78.0868

The other side I have -80 + 2 which is the 3rd root of -78.0880

It looks like the two agree.0887

I know that the -10 is my solution.0890

With those radicals try and isolate them, and then get rid of them by raising them to a power.0894

Always check your solutions with these ones to make sure they work in the original.0899

Thanks for watching www.educator.com.0904