## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### 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

- ( √{4x − 16} )
^{2}= ( 3√2 )^{2} - 4x − 16 = 9 ×2
- 4x − 16 = 18
- 4x = 34

- 10x + 12 = 64 ×3
- 10x + 12 = 192
- 10x = 180

- 16 ×5 = 6y + 8
- 80 = 6y + 8
- 72 = 6y

- x
^{2}= ( √{x + 30} )^{2} - x
^{2}= x + 30 - x
^{2}− x − 30 = 0 - ( x + 5 )( x − 6 ) = 0

- x
^{2}= ( √{x + 56} )^{2} - x
^{2}= x + 56 - x
^{2}− x − 56 = 0 - ( x + 7 )( x − 8 ) = 0

- x
^{2}= ( √{13x + 30} )^{2} - x
^{2}= 13x + 30 - x
^{2}− 13x − 30 = 0 - ( x + 2 )( x − 15 ) = 0

- √{x + 4} = x + 4 + 2
- √{x + 4} = x + 6
- ( √{x + 4} )
^{2}= ( x + 6 )^{2} - x + 4 = x
^{2}+ 12x + 36 - 4 = x
^{2}+ 11x + 36 - 0 = x
^{2}+ 11x + 28 - x
^{2}+ 11x + 28 = 0 - ( x + 4 )( x + 7 ) = 0

- √{4x − 11} = x − 1 − 1
- √{4x − 11} = x − 2
- ( √{4x − 11} )
^{2}= ( x − 2 )^{2} - 4x − 11 = x
^{2}− 4x + 4 - − 11 = x
^{2}− 8x + 4 - 0 = x
^{2}− 8x + 15 - x
^{2}− 8x + 15 = 0 - ( x − 5 )( x − 3 )

- − √{3 − 11x} = 3 − x
- ( − √{3 − 11x} )
^{2}= ( 3 − x )^{2} - 3 − 11x = ( − x + 3 )
^{2} - 3 − 11x = x
^{2}− 6x + 9 - − 11x = x
^{2}− 6x + 6 - 0 = x
^{2}+ 5x + 6 - 0 = ( x + 3 )( x + 2 )

- − √{5 − 7x} = 7 − x
- ( − √{5 − 7x} )
^{2}= ( − x + 7 )^{2} - 5 − 7x = x
^{2}− 14x + 49 - − 7x = x
^{2}− 14x + 44 - 0 = x
^{2}− 7x + 44 - 0 = ( x − 11 )( x + 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.

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
- Radical Equations
- 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

### Algebra 1 Online Course

### 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 √x-5 - √x-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 I’m talking about with √3x + 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

*√3x+ 1 = 4*0092

*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 = 4 ^{2} which is 16*0109

*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

*I’m going to write this out here -4 does that equals 0 or not.*0146

*Let us put in 5, 3 × 5 would be 15 and all of that is underneath the root and then 15 + 1 would be 16.*0156

*√16=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 + √x+ 7 = x.*0191

*Isolate our radical by subtracting 5 from both sides of the equation √ 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 × x – 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 x ^{2}, outside terms - 5x, inside terms - 5x, and our last terms 25.*0256

*We can go ahead and combine to see what we will get x ^{2} - 10x + 25.*0270

*You can see that this one is turning into a quadratic equation because we have x ^{2} 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 = x ^{2} - 11x + 18*0301

*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 × x = x ^{2} then 2 × 9 = 18*0322

*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 + √ 2 + 7 does it equal 2?*0347

*Let us find out, 2 + √7 = √ 9, √9 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 √16 =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 √x - 3 + √x + 5 = 4*0477

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

*I’m going to work to isolate it and get all by itself x + 5 = 4 – and I have subtracted the other radical to the other side.*0490

*Since √x+ 5 is the one I’m trying to get rid of, I’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 × 4 = 16, our outside terms we are taking 4 × -√x – 3, -4 × √x - 3.*0545

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

*negative × negative = positive and then we have √x -3 × √x -3.*0570

*That will give us √x - 3 ^{2} 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 √x -3*0614

*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 - 8√x -3.*0650

*Let us go ahead and subtract a 13 from both sides, -13 – 13, -8 = -8√x -3.*0665

*It looks like we can divide both sides by -8, 1 =√x -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 + √4 + 5 we are checking does that equal 4.*0741

*Let us see what we have.*0756

*4 - 3 would give us √1 , 4 + 5 would be 9, so√4 =1 and √9 = 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 + 2*0810

*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 × putting that -10 - 8 and 8 × -10 + 2*0845

*Let us see if they are equal.*0866

*This time I’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

1 answer

Last reply by: Professor Eric Smith

Wed Jul 30, 2014 2:45 PM

Post by Peter Spicer on July 22, 2014

Wouldn't you add 3 to both sides since we have an x-3?