### Equations in Quadratic Form

- Some equations are similar to quadratic equations, but involve different powers other than two. For these we can try substituting a new variable in, and see if it becomes a quadratic.
- When using a u-substitution, look for one term that has a power exactly twice as much as another power. This will give you a clue on what u-should replace in the original.
- Donâ€™t forget to return to the original variable, once you are done solving the quadratic. To do this, re-substitute in what u is equal to.

### Equations in Quadratic Form

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 0:08
- Equations in Quadratic Form 0:24
- Using a Substitution
- U-Substitution
- Example 1 2:07
- Example 2 5:36
- Example 3 8:31
- Example 4 11:14

### Algebra 1 Online Course

### Transcription: Equations in Quadratic Form

*Welcome back to www.educator.com.*0000

*In this lesson we are going to go ahead and take a look at equations that are quadratic in form.*0002

*This will be some special types of problems and that they are not quite quadratic,*0009

*but we can manipulate them so that they at least look like quadratic.*0013

*The reason why that is important is because then we can use our quadratic solving techniques.*0017

*Let us go ahead and tackle them.*0021

*We call that an equation is a quadratic equation if it is on the form ax ^{2} + bx + c =0.*0027

*As long as a, that first value is not 0. *0036

*If we can take an equation and write it in this form, then it is said to be in standard form.*0041

*We will see many different examples of quadratic equations and how to solve them.*0047

*There are a few different types of equations that are definitely not quadratic. *0055

*For example, this one is x ^{4} is not quadratic.*0060

*This one has x ^{2}/3, the interesting thing about some of these equations is that you can treat them as quadratics during the solving process.*0064

*The way you do this is you use a substitution process to go ahead and turn it into a much more manageable form.*0076

*The way the substitution process works is that we choose part of the equation and we let that equal u.*0088

*We will go ahead and we rewrite the entire equation using this new u.*0095

*What you will find is that you should replace all of the original variables in the problem. *0100

*If you had xâ€™s now you just have a whole bunch of uâ€™s.*0105

*When you are done solving the equation though, we will make sure that we go back to the original variables. *0110

*It is kind of tricky what I'm describing here, I want you to see it in action.*0116

*You will see that it does make things much easier for solving these types of equations.*0121

*Let us use one like this example.*0129

*I want to solve x ^{4} - 5x^{2} + 4 = 0. *0132

*This is not a quadratic because Iâ€™m dealing with x ^{4} out front.*0136

*What I'm going to do is Iâ€™m going to turn it into a quadratic form.*0141

*Iâ€™m going to let u = x ^{2}.*0146

*The reason why Iâ€™m doing that is I'm hoping to replace this entire variable right there with u.*0153

*If that was the only thing I replaced I would still be in trouble.*0160

*I still have x's and uâ€™s running around. *0163

*Let us take that little equation that we created u = x ^{2}. *0166

*Let us square both sides of that. *0172

*That would give us that u ^{2} = x^{4} and that gives us a way that we can also swap out this other x over here.*0176

*x ^{4} is the same as u^{2}.*0186

*I can take our equation here and write it as u ^{2} - 5u + 4 = 0.*0192

*The reason why that is just so important is because this new equation is quadratic in u.*0204

*I have open many different solving techniques that I can use on this thing. *0212

*I could try reverse foil to simply factor and use the 0 factor property. *0216

*I could try our quadratic formula and get the solution directly, but I have lots of things open to me.*0221

*This one is not too bad, so I'm just going to go ahead and factor it and then use that 0 factor property.*0228

*My first terms must be a u and u.*0237

* I need two things that multiply to get 4, but they add to give me -5.*0241

*I only have one option for that, just -4 and -1.*0249

*I know that u -4 could be 0 and u -1 = 0.*0255

*Solving each of the separately would give me u = 4 and u = 1.*0261

*Once we switch into a quadratic it is much more easier to manage.*0268

*The problem is that we solve it for u in and our original problem had xâ€™s.*0272

*We are not exactly done with this problem yet. *0277

*At this stage we want to borrow what we called u and swap out for our original x's.*0280

*This says that x ^{2} = 4.*0289

*This one is x ^{2} = 1. *0295

*I'm left with two smaller quadratic equations and I will solve each of these directly.*0301

*I will simply take the square root of both sides, so + - square root of 4 and x = + - the square root of 1.*0307

*Good thing both of these can be simplified.*0315

*+ -2 and + -1, so I have four different solutions for this particular problem.*0317

*What got our foot in the door was being able to write all of these variables using a new variable and reducing those powers.*0327

*You can do this for a variety of types of equations to put them in a quadratic form.*0338

*What we want to recognize is what you should swap out for that u value.*0344

*You can usually use this technique if you recognize that one of the powers is twice as large as the other power.*0350

*2/3 is exactly twice as large as 1/3.*0357

*Iâ€™m going to use this to help me swap out my x's for uâ€™s.*0361

*Let u =x ^{1}/3.*0367

*I remember that you can often figure out what the other one means to be by squaring both sides of this little equation.*0373

*u ^{2} = (x^{1}/3)^{2} which is the same as x^{2}/3.*0379

*I can swap out both of these.*0388

*u ^{2} â€“ 2u -15 = 0.*0392

*Everything else is the same, but now I have those uâ€™s in there and now it is quadratic in form.*0402

*I can use a lot of other techniques to go ahead and solve this one.*0406

*What shall we do with this one?*0411

*I think this is another one we can simply factor without too much problem. *0413

*My first terms better be u and now I need two things that will multiply and give me -15 but somehow add to get -2.*0418

*Let us use 5 and 3.*0428

*-5 and 3.*0432

*This will give me that u -5 = 0, and u + 3 = 0.*0435

*That comes from the 0 factor property.*0443

*Solving each of the separately I have u = 5 and u = - 3.*0447

*Each of these are looking pretty good.*0454

*We want to go back to our original variable.*0457

*I know exactly what u is, let us go ahead and put that in for both of these spots on here.*0460

*x ^{1}/3 x^{1}/3 and this is equal to 5 and equal to -3.*0467

*To solve this directly from here, I think I can take both sides and cube it.*0478

*5 ^{3} would be 125, -3^{3} = -27.*0491

*Just like the foil, now we have our solutions to the original problem. *0501

*We work all the way back to x, which was the original variable. *0507

*Now some might actually be quadratic but you can still use this technique to put it in a much nicer form.*0513

*In this next one, we are dealing with a 3x - 1 ^{2} + 2 Ã— 3x - 1 = 8.*0521

*One way that you could handle this is simply to multiply everything out *0529

*and then get everything on one side sand set it equal to 0 and use the quadratic formula.*0534

* I'm not going to do that because I noticed that I actually have this common piece of 3x â€“ 1.*0539

*I'm going to swap out that common piece and call it something else.*0547

*You will say let u = 3x - 1.*0551

*Now that we have that, this will become much easier to solve.*0558

*We will have a u ^{2} + 2 Ã— u = 8.*0563

*We can work to get everything on one side and now it is set equal to 0.*0576

*For this u, let us go ahead and reverse foil it. *0585

*u and u, need two numbers that would multiply to be -8 but add to be a positive 2, 4 and -2.*0593

*This gives us two solutions u = -4 or u = 2.*0603

*We must go back to that original variable.*0616

* We know what u is, u = 3x â€“ 1.*0620

*We are going to put that in for both of our uâ€™s.*0624

*3x â€“ 1 = -4 and 3x - 1 = 2.*0628

*We can solve each of these separately.*0641

*Adding 1 to both sides would be -3 and divide both sides by 3 would give us -1.*0646

*There is one of our solutions right there.*0654

*For the other one, we will add 1 to both sides and get 3, then divide both sides by 3 and get 1.*0660

*We have our solutions for this problem.*0669

*Let us do one more and this would involve some negative exponents.*0676

*This one has 2 - y -6 ^{1} - 1 = 6 Ã— y â€“ 6^{-2}.*0681

*This one is a little jumbled up, it is difficult to figure out what we should swap out. *0689

*It is good to know that we do have a y -6 that seems to be a common.*0693

*That will be a part of what we end up exchanging with u.*0698

*Let us go ahead and get everything onto one side.*0704

*Iâ€™m going to move everything over to the right side 6y - 6 ^{-2}. *0707

*I will add y -6 ^{-1} -2.*0717

* What I recognize here is that this -2 is exactly twice as large as that -1.*0724

*It is a pretty good indication that I will end up swapping on my u for this piece right here.*0731

*Let u = let us call this y â€“ 6 ^{-1}.*0739

*Let us go ahead and rewrite our equation using this.*0750

*We have 6u ^{2} + u â€“ 2.*0755

*What I can see here is that it is definitely quadratic and the numbers are much smaller, much nicer to deal with.*0767

* What do I do from here?*0774

*I still have to be able to figure out what the solutions of this quadratic are.*0776

*We got lots of tools available to us, let us go ahead and try the quadratic formula.*0780

*-b Â±âˆš4, (Â±âˆšb ^{2} â€“ 4) Ã— (a Ã— c) Ã· 2a.*0787

*I think this will simplify nicely.*0812

*I have 4 Ã— 6 = 24 Ã— 2 = 48 Ã· 12 or -1 Â± âˆš49/12 or -1 Â± 7/12.*0816

*u = -1 + 7/12 and I have u = -1- 7/12.*0841

*-1 + 7 = 6/12 would be -8/12.*0856

*Both of those reduce, this one to Â½ .*0863

*The other one 4 goes into the top and 4 goes in the bottom, -3/2.*0869

*I know little bit more about what u is equal to. *0876

*Of course, we can not stop there, we must go back to our original variable.*0880

*We must work back all the way to those yâ€™s.*0886

*Let us put those back in for u and see if we can solve this from here.*0890

*Let us see y - 6 ^{-1} = Â½ and y - 6^{-1} = -2/3.*0897

*To solve this from here I think I will raise both sides to that -1 power.*0912

*That will give me y - 6 and then with that -1 exponent, that will change the location of the 1 and 2, 2/1.*0920

*Let us do the same thing with our other equation.*0931

*Raise both sides to -1, this is y - 6 = -3/2, since it changes the location of the -2 and the 3.*0933

*These are almost done.*0945

*With the one on the left here, let us add 6 to both sides. *0948

*This will give us y = 8 and with this other one let us add 6 to both sides, -3/2 + 6.*0952

*It looks like we do have to get a common denominator, but this one is not so bad.*0965

*It looks like it is just 9/2.*0975

*We have worked all the way back to our original variable, we know what the solutions are.*0978

*It can take quite a bit of work to end up swapping out the u and solving from there.*0984

*The important part is that if we do not swap out the u these we do not have a lot of other techniques to solve equations like this.*0989

*Make sure you properly identify what u you need to swap out and put it back in so you can get back to your original variable.*0997

*Thank you for watching www.educator.com.*1005

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