Algebra 1 Equations in Quadratic Form

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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 ax² + 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 x⁴ is not quadratic.0060

This one has x²/3, 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 x’s now you just have a whole bunch of u’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 x⁴ - 5x² + 4 = 0.0132

This is not a quadratic because I’m dealing with x⁴ out front.0136

What I'm going to do is I’m going to turn it into a quadratic form.0141

I’m going to let u = x².0146

The reason why I’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 u’s running around.0163

Let us take that little equation that we created u = x².0166

Let us square both sides of that.0172

That would give us that u² = x⁴ and that gives us a way that we can also swap out this other x over here.0176

x⁴ is the same as u².0186

I can take our equation here and write it as u² - 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 x’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 x² = 4.0289

This one is x² = 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

I’m going to use this to help me swap out my x's for u’s.0361

Let u =x¹/3.0367

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

u² = (x¹/3)² which is the same as x²/3.0379

I can swap out both of these.0388

u² – 2u -15 = 0.0392

Everything else is the same, but now I have those u’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

x¹/3 x¹/3 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

5³ would be 125, -3³ = -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 - 1² + 2 × 3x - 1 = 8.0521

One way that you could handle this is simply to multiply everything out0529

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 – 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 u² + 2 × 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 – 1.0620

We are going to put that in for both of our u’s.0624

3x – 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 -6¹ - 1 = 6 × y – 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

I’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 – 6^-1.0739

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

We have 6u² + u – 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 ±√4, (±√b² – 4) × (a × c) ÷ 2a.0787

I think this will simplify nicely.0812

I have 4 × 6 = 24 × 2 = 48 ÷ 12 or -1 ± √49/12 or -1 ± 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 ½ .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 y’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 = ½ 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

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