INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith

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• Some formulas are quadratic in nature if they contain a squared term.
• Remember we can use the principle of square roots to help isolate the squared variable.
• Some word problems lead to quadratic equations. These usually include problems with distance, right triangles, or area.
• The Pythagorean Theorem says a2 + b2 = c2, where a and b are the legs of the right triangle and c is the longest side.

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 0:09
• Quadratic Formulas and Applications 0:35
• Squared Variable
• Principle of Square Roots
• Example 1 1:09
• Example 2 2:04
• Quadratic Formulas and Applications Cont. 3:34
• Example 3 4:42
• Example 4 13:33
• Example 5 20:50

### Transcription: Quadratic Formulas & Applications

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at some quadratic formulas and some applications of quadratics.0003

Specifically we will look at how you can recognize some quadratic equations or some formulas that have a quadratics in them.0012

Think of that squared term.0020

Some situations that tend to be very common with quadratic equations are the Pythagorean Theorem and the distance formula.0022

Look for some of these when we get to our application problems.0031

When looking at various different formulas sometimes they might involve a squared variable in them.0038

If you go through the process of trying to solve for that variable, you might have to try and get it all by itself.0044

Since it is going to be squared we will have to use the principle square roots to get it completely by itself.0052

Remember that when you are using the principle of square roots, not only will we have one solution that is positive,0058

we will also have another one that is negative.0064

Let us see this quick small formula.0072

In this formula I have E = m × c2.0075

Let us go ahead and try and solve for c.0080

As I work and try to get it all by itself, the first thing I will do is divide both sides by m.0084

E ÷ m = c.0092

To get it completely all alone, this is where we will go ahead and take the square root of both sides.0096

From the principle of square roots I will have ± √e÷m.0104

You can think of this as two different ones.0111

That c = √e÷m and c = -√e÷m.0114

In this other formula you will see that we have pretty much the same situation.0127

We are looking to solve for the r and r2.0132

We have to do a little bit more work to get it all by itself, but it is not too bad.0136

Let us start this one off by multiplying both sides by 3.0142

We are doing that just to get rid of that 1/3 that I see out front.0145

1/3 pi r2h.0154

This will give me 3v = and then the 3 and 1/3 will cancel each other out, pi r2 h.0161

To get a little bit closer to getting it all by itself, let us divide both sides by pi and h.0171

3v pi × h = r2.0180

We have not used that principles of square roots yet but now it is going to play a large part0187

because in order to get rid of that square, we will square root both sides.0192

Remember that we need a + and -.0196

Our formula is ±√3v÷pi h.0207

In some applications of quadratics we will have to do a little bit more work and set it up ourselves.0215

Some things that will help with the setup process is to watch for things such as right triangles or maybe even problems involving area.0222

The reason why this is important is these situations often lead to things that are being squared.0230

Here is a quick example.0236

When looking at a right triangle, triangle with a right angle in it, the sides are related.0238

If I take the two legs of the triangle, they are related to the hypotenuse in this way right here at a2 + b2 =c2.0247

This is known as the Pythagorean Theorem.0258

If you have a problem that deals with a right triangle, you can be sure that this will probably show up along the way.0266

If we are going to end up solving it with all the squared terms in there, we are going to need many of our quadratic tools to tackle it.0275

Let us see that triangle as we look at this first application problem.0286

This one says that we have two ships and they are leaving port at the same time.0290

One is headed due south and the other one is headed due east.0294

Several hours after departure the ships are 195 miles apart.0298

The ship traveling south travel 105 miles farther than the other one, how many miles did they each travel individually.0304

Let us see if we can get a diagram of what is going on here.0313

Maybe here I have my port, we will make a nice illusion here and I have two ships.0317

One is traveling due south and the other one is going due east.0325

What I know so far is that after a few hours that they are exactly 195 miles apart.0336

You can see that what we have created here is one of those right triangles.0353

We cannot solve that yet, we need a little bit more information about the legs of the triangle.0358

The ship traveling south traveled 105 miles farther than the other one.0362

We have no idea how far the ship going east went, we better call that x.0368

x is the distance the east ship travelled.0375

The other one is exactly 105 miles more.0392

This gives me information about all sides of that triangle.0400

I can relate them using the Pythagorean Theorem.0404

That would be one of our legs square + the value of the other leg square = the value of the hypotenuse square.0409

Looking over, we can see that because of that squared terms we definitely have a quadratic equation.0430

We are going to use our quadratic tools to tackle it.0435

Let us go ahead and copy that formula down so we can look at it.0444

If we are going to solve this we need to foil things out and get everything over on one side.0460

That way we could do something like the quadratic formula or even factoring.0466

I'm going to foil this out and see what we get.0472

Our first terms x × x =x2.0480

Outside would be 105x and so would be the inside terms.0485

Our last terms would take 105 × 105.0491

On the other side we have 1952.0503

The numbers are quite large.0514

There are lots of things in here that we can go ahead and combine.0516

I have some x2 that will go ahead and put together.0519

I have some x’s and then we will move this 38,025 to the other side so we can combine it with our other constant.0522

Let us see what this gives us.0534

2x2 + 210x = -27,000.0538

Now it definitely looks more like a quadratic, we are getting closer.0557

Always check to see if you can pull out a common factor.0560

One thing I see here is everything is divisible by 2.0563

Let us divide by 2.0567

x2 + 105x - 13,500.0573

We have several options open to us.0588

We could try factoring but I’m a little hesitant because the numbers are quite large.0590

I will identify a, b, and c.0602

Negative the value of b ±√(b2 )-4 ×a ×c ÷ 2a.0607

Do not be intimidated by those large numbers just work carefully through the problem and try not to make a mistake.0628

Let us take care of that 105 being squared underneath the square root.0637

I get -105 ±√11025 and we have 4 × 13, 500.0647

All of that is being divided by 2.0669

We will go ahead and add these together and end up taking their square root.0678

I’m getting -105 ± and I borrowed down the square root in here so 255 ÷ 2.0692

I have two things to consider either -105 + 255 ÷ 2 or -105 – 255 ÷ 2.0705

That will be 150 ÷ 2 that will simply be 75.0724

The other one being -360 ÷ 2 =-180.0731

Remember when we first identified what x was, it was the distance that our east ship was traveling.0742

We have a bit of a problem with one of our solutions.0748

Notice how this one turned out to be -180.0752

We can not have our ship traveling a negative distance.0755

We are going to get rid of that as one of our possible solutions.0758

It just does not make sense in the context of the problem.0761

We will keep that the east ship went 75 miles.0763

That is only half of the problem.0768

The other ship went exactly 105 more so we can add that to this and now get the distance of both ships traveled.0771

We get 75 + 105 = 180 that would be our south ship 180 miles and the other one we will say as our east ship 75 miles of travel.0786

We have the distance for both of them.0809

Let us try and set up another one of these and see how it works.0816

Harold and his wife have recently installed a built-in rectangular swimming pool.0821

It measures 18 ft × 22 ft.0826

They want to take this swimming pool and add a nice decorative tile boarding, a nice uniform width all around the pool.0829

The question is how wide can they make the title border if they only purchased enough to cover 176 ft2?0836

I think in this one, we need to draw a picture as well.0845

What we have here is a nice rectangular pool and we know about its dimensions.0849

This pool is 22 × 18 ft.0858

What we are going to do with this pool is actually build a nice border that goes around it.0864

The thing is that we do not know how wide this border should be.0874

That is what we are trying to figure out, how wide is this border?0877

Let us set that as our unknown.0881

x is the width of the border.0885

I want to be able to figure out how the area of just the border is related to the 176.0897

Let us see what we can do in here.0905

Well, if I'm looking at the area of just this tiled part, I could look at that as the area of the entire rectangle0907

and then subtract out that area of the pool in between.0917

Let us try and do that.0921

What is the total length of the large rectangle and the total length and width of the rectangle?0924

Since we are adding these little border pieces on the outside, this would be 22 +.it has 1, 2 widths of x.0932

That would be 18 + 2x.0944

The large rectangles area would be (22 + 2x) × (18 + 2x), that is the area of the large rectangle.0949

We will subtract out the area of the other one.0964

The difference of those two rectangles should give us just the area of the border, 176.0972

Now we have our quadratic equation that we can go ahead and solve from here.0981

Let us copy that down.0990

At first it does not look like a quadratic but notice I will end up multiplying these x’s here1010

and will get that x2 term, so this is going to be quadratic after all.1017

We have some multiplication to do, let us go ahead and start foiling things out so we can eventually get everything over on one side.1022

Our first term is 22 × 18 = 396.1034

Our outside terms 44x, inside terms 36x and our last terms 4x2.1040

I have 18 × 22= 396, equals 176.1057

There are no squared terms that need and there is a few other things that we can combine.1066

Here we have 396 and -396, those will go away.1072

We can go ahead and put these two together.1077

The largest term in here is 4x2 that takes care of that one.1083

Then I will combine together my x’s 44 + 36 = 80x.1090

It takes care of both of those since that 396 is cancel out and1100

we will get the 176 on the same side as everything else by subtracting it from both sides of the equation.1106

Okay, now it is looking much better.1113

Definitely you will it is quadratic.1116

There is my 4x2 at the very beginning.1117

I need to just work on solving it.1120

The numbers are getting large, but I think everything in here is divisible by 4.1123

Let us go ahead and divide everything by 4.1129

It should make things much smaller.1135

Definitely looking much better now .1147

This one, we can solve using our quadratic techniques, quadratic formula, factoring.1149

In this one I think I see a way that this could factor.1155

Let us sit down a couple of parentheses and try reverse foil.1160

Two things that will multiply to give me x2 would have to be x and x.1165

Some possibilities for the second term would be 1 and 44, 2 and 22.1171

I'm pretty sure it is going to be those so that I can get the 20 in the middle, 22 and -2.1178

Now that we have both of our factors we will take each of the factors and set them equal to 0.1193

Solving each of these separately I have x = -22 and x = 2.1205

If you think back when we are first deciding what x was, x represents the width of the border.1213

We can not have a negative width for our border.1220

It simply does not make sense.1224

We can have a positive border that will work out just fine.1227

Let us say that the border is 2 ft wide.1231

It does take a little bit of work to set these equations up but definitely watch for quadratic to show up in there somewhere.1243

One last problem.1252

In this one we have a rectangular piece of metal and it is 15 inches longer than it is wide.1254

There are going to be little squares that are 3 inches long cut from the four corners and we are going to fill up the flaps to form an open box.1260

The volume of this box is supposed to be a 1092 in3.1269

What were the original dimensions of the piece of metal?1275

We have a very interesting situation going on here.1279

We are starting off with a nice rectangular piece of metal, but then what is going to happen1282

when little squares are cut out from each of the corners.1289

Our metal is starting to look more like this.1295

Then it is folded up into a type of open box.1299

We want to look at some of the variables present and see what we can set up now.1317

Also what do we know about the original piece of metal?1323

It was 15 inches longer than it is wide.1327

I have no idea how wide it is but the length is going to be 15 inches wider.1330

W is the width of the metal.1336

We are going to go along and cut out these little 3 inch corners.1348

Consider when we do that, these new sides here represent the new lengths on the box.1357

How long are those pieces going to be?1367

This first one started off as W but we cut a couple of 3 inches off of it.1372

This length is W – 6, 2 of those 3.1378

We did the same thing for the other length so W + 15 – 6 = W + 9.1384

Look at our correspondence to the folded up box, it means that this length down here is W – 6.1393

This one over here is our W + 9.1400

That is all the information we need.1406

What about the height of this box?1408

That comes from our corners and we know that they are 3 inches.1412

We know the height is 3.1417

Once we have all of our dimensions we can finally relate it to the volume.1421

The volume comes from multiplying the length × the width × the height of this box.1426

Volume = length × width × height or 1092 = a length of W + 9, a width of W - 6 and the height of 3.1433

Let us go ahead and borrow that equation and see what we can do.1453

1092 = W + 9W - 6 and 3.1464

Let us go ahead and start off by maybe multiplying things out using foil.1484

W × W = W2 I have my outside terms and my inside terms, and -54.1495

Things are looking a little bit better but I still have that 3 out there.1510

Unfortunately 1092 is divisible by 3, let us go ahead and divide both sides by 3 so we can get rid of it.1515

It is looking much better.1534

Combine our middle terms here W2 + 3W1535

I’m going to go ahead and subtract this 364 from both sides and we will have that term on there as well.1548

-418 looking good.1568

I just have to simply try and solve this quadratic equation.1574

What do we got?1581

I could try and use something like the AC method.1584

I could try and factor this.1587

We have lots of different options to actually try and break this down.1589

My first leading term here is just got a 1W2, let us see what possibilities we have to multiply and get that 418.1598

As soon as we find them we can say that this is factored.1607

It could be 1418, could be 2209, we got a lot of different things.1613

Packing up a lot of things go in for 18, it looks like 19 does.1626

19 and 22 do, which is good because I'm trying to that middle term 3.1642

Let us go ahead and borrow 22 – 19.1648

Remember you can also use the quadratic formula on this now we will also find what we need.1655

We did two things here, either that W + 22 = 0 or that W -19 = 0.1662

I know that -22 could be W or that 19 could be W.1671

What is W exactly?1678

Remember it represents the width of our original piece of metal.1680

We do not want to deal with a negative width because that does not make sense in the context of the problem, let us get rid of that.1685

But if it is perfectly okay to have the width at 19.1692

At least we know about one of the dimensions of our piece of metal.1699

If you want the other dimension, remember that it is just a little bit larger.1707

In fact, it is exactly 15 larger.1713

You take 19 add 15 and see what we get.1716

It looks like the other length will be 34 inches.1723

When dealing with applications of quadratics, it does take a little bit of work to set them up1730

but when you do get to that quadratic part use all tools available to you for solving those quadratics.1736

Thank you for watching www.educator.com.1742