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### Quadratic Formulas & Applications

- 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 a
^{2}+ b^{2}= c^{2}, where a and b are the legs of the right triangle and c is the longest side.

### Quadratic Formulas & Applications

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

### Algebra 1 Online Course

### 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 × c ^{2}.*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 r ^{2}.*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 r ^{2}h.*0154

*This will give me 3v = and then the 3 and 1/3 will cancel each other out, pi r ^{2} 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 = r ^{2}.*0180

*We have not used that principles of square roots yet but now it is going to play a large part*0187

*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 a ^{2} + b^{2} =c^{2}.*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

*We would be talking about this distance right here.*0344

*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 =x ^{2}.*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 195 ^{2}.*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 x ^{2} 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

*2x ^{2} + 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

*x ^{2} + 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’m going to go ahead and jump to something like a quadratic formula and go more direct root.*0596

*I will identify a, b, and c.*0602

*Negative the value of b ±√(b ^{2} )-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 ft ^{2}?*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 rectangle*0907

*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 here*1010

*and will get that x ^{2} 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 4x ^{2}.*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 4x ^{2} 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 and*1100

*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 4x ^{2} 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 x ^{2} 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 in ^{3}.*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 happen*1282

*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 = W ^{2} 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 W ^{2} + 3W*1535

*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 1W ^{2}, 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 up*1730

*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

1 answer

Last reply by: Professor Eric Smith

Sat Dec 30, 2017 6:55 PM

Post by Juan Collado on December 27, 2017

SORRY

NEVER MIND

SQUARE ROOT OF R^2 IS R

I FEEL LOST IN THIS LESSON AND THE LAST ONE.

0 answers

Post by Juan Collado on December 27, 2017

In Example 2, you went from

_+ square root of 3V/pieH = Square root of R^2

to +- square root of 3V/pieH = R

How do you just get rid of the square root on one side without doing the same to the other.

I always have problems with square roots