18 - Writing Quadratic Equations when Given its Roots - Part 1 - By Math and Science
Transcript
| 00:00 | Hello . Welcome back . I'm Jason with math and | |
| 00:02 | science dot com . The title of this lesson is | |
| 00:04 | called writing quadratic equations when we're given their roots and | |
| 00:09 | this is part one . We're actually gonna have four | |
| 00:11 | parts to this topic here . In the beginning . | |
| 00:13 | I'll explain the concept with a couple of quick examples | |
| 00:16 | to get you comfortable with what it is . Uh | |
| 00:18 | and then the following sections will increase the complexity so | |
| 00:21 | that you can get a lot more practice with more | |
| 00:23 | complex problems , but they're all gonna be utilizing the | |
| 00:25 | basic skills that we learn here . I'm actually really | |
| 00:28 | excited to teach this lesson because what we're gonna do | |
| 00:30 | is learn about the theorem in algebra that usually is | |
| 00:33 | thrown at you in a book , usually as a | |
| 00:36 | box around it or something and you have to use | |
| 00:38 | it , but most students have no idea where it | |
| 00:39 | comes from and so just like everything else that just | |
| 00:42 | thrown at you , you think that you're supposed to | |
| 00:44 | use it and you end up forgetting about it and | |
| 00:45 | that's how a lot of students think algebra is just | |
| 00:48 | a bunch of black magic because they don't really know | |
| 00:50 | where things come from . So what we're gonna do | |
| 00:52 | is we're gonna learn how to use this theorem , | |
| 00:54 | but we're also going to learn where it comes from | |
| 00:57 | and here's what we're going to learn in a nutshell | |
| 00:59 | . You all know that these quadratic equations , they | |
| 01:03 | can oftentimes cut through the X axis . We call | |
| 01:06 | those the roots of the equation where why is equal | |
| 01:09 | to zero . Those are the places along the X | |
| 01:11 | axis where the function cuts through . We call that | |
| 01:13 | the roots and we know how to solve quadratic to | |
| 01:16 | find the roots . We can factor them , you | |
| 01:18 | know , and find the roots of the equation . | |
| 01:20 | We can use the quadratic formula and find the roots | |
| 01:22 | of the equation . In this lesson . We're gonna | |
| 01:24 | go backwards we're gonna go the exact opposite direction . | |
| 01:27 | What I'm gonna do is I'm gonna give you two | |
| 01:29 | routes . All right . Two routes . I might | |
| 01:32 | tell you that Route number one is three and route | |
| 01:34 | number two is negative five and from the roots and | |
| 01:38 | only from the roots . What we want to do | |
| 01:40 | is reconstruct the parent quadratic equation that has those routes | |
| 01:44 | . So , let me say that one more time | |
| 01:45 | . I'm gonna give you two routes . And then | |
| 01:47 | your job is to come up with the quadratic function | |
| 01:49 | that has those routes . All right . So , | |
| 01:51 | in order to understand how that works , what I | |
| 01:53 | want to do is go the other direction , first | |
| 01:56 | , the direction that we have done a lot . | |
| 01:58 | So let me go ahead and review a couple of | |
| 01:59 | things , and then we'll show you where it comes | |
| 02:01 | from , and then we'll generalize the theorem and do | |
| 02:03 | an example . All right . So let's start with | |
| 02:06 | what everybody knows how to do . Everybody knows how | |
| 02:09 | it is because we covered it so many times before | |
| 02:11 | . So we already know how to solve this type | |
| 02:14 | of quadratic . Let's say we have X squared minus | |
| 02:18 | six X plus five and that equals zero . So | |
| 02:22 | right now we're just solving this quadratic . And by | |
| 02:24 | the way , when we saw that , what we're | |
| 02:25 | saying is this quadratic is a parable of some sort | |
| 02:28 | . We know it opens up because this is a | |
| 02:29 | positive one . It's a quadratic . And um if | |
| 02:34 | there are any crossing points of this quadratic to the | |
| 02:37 | X axis , we set it equal to zero . | |
| 02:38 | We set the equation where we set the y value | |
| 02:41 | equal to zero . So when we factor and solve | |
| 02:43 | this guy , we're finding the crossing points where this | |
| 02:46 | equation crosses . So typically the way that we do | |
| 02:49 | that is we factor the thing . All right . | |
| 02:52 | And so we say or we try to factor it's | |
| 02:54 | not always fact , herbal . It's not always fashionable | |
| 02:57 | , but we're going to try to factor it initially | |
| 02:59 | . So we have an X square term . So | |
| 03:01 | we know we have to have an X . Times | |
| 03:02 | and X . And the only way to make five | |
| 03:04 | with multiplication is one times five . And the only | |
| 03:07 | way to get the plus here in the minus here | |
| 03:09 | is to have a minus and a minus . Make | |
| 03:10 | sure you agree with me , X times X is | |
| 03:12 | X squared . This is negative X . This is | |
| 03:15 | negative five X . That adds together to give you | |
| 03:18 | the negative six X . And then these multiply together | |
| 03:20 | to give you a positive five . So that's the | |
| 03:22 | factored form . And so from this factor form , | |
| 03:25 | we've done this a million times , We know that | |
| 03:28 | X -1 . This term can be equal to zero | |
| 03:30 | , which means that one of the roots is that | |
| 03:32 | X is equal to one . And we also know | |
| 03:35 | This term can be equal to zero , which means | |
| 03:38 | that this route is five . So what it means | |
| 03:41 | is that if we were to plot this function , | |
| 03:43 | you know , now we take away the zero F | |
| 03:44 | of X is equal to this , it's going to | |
| 03:46 | create some parabola . These two numbers are the crossing | |
| 03:50 | points . It's where the problem cuts into the X | |
| 03:52 | axis . In those two locations , that X is | |
| 03:54 | equal to one and X is equal to five . | |
| 03:56 | We call those numbers the roots of the equation . | |
| 03:59 | Also , you can see them referred to as the | |
| 04:01 | zeros of the equation . It means the same thing | |
| 04:04 | . All right . So what we have done is | |
| 04:06 | we've gone the forward direction because we have already done | |
| 04:08 | this kind of thing . We've we've gone this direction | |
| 04:11 | where we have given the equation to figure out what | |
| 04:13 | the roots are . Now , we want to talk | |
| 04:15 | about that . We can go backwards and by backwards | |
| 04:24 | . What I mean is if I give you in | |
| 04:27 | a problem , so I want you to kind of | |
| 04:29 | forget that we've done this now , but I'm gonna | |
| 04:30 | give you a problem given that the roots of some | |
| 04:34 | quadratic , some parabola is X is equal to one | |
| 04:38 | and X is equal to five . I'm choosing the | |
| 04:40 | exact same thing we've done here because we're familiar with | |
| 04:42 | it . Then find a quadratic with these routes . | |
| 04:56 | All right . So how do you think you would | |
| 04:57 | do it ? Obviously we know the answer . We | |
| 04:59 | know that this is the answer . But how do | |
| 05:01 | we if all we are given is this that these | |
| 05:03 | two routes exist ? How do we figure out what | |
| 05:05 | that quadratic is ? And here is the secret sauce | |
| 05:08 | of what you do to generalize what you do . | |
| 05:10 | If you know that the routes exist right then what | |
| 05:14 | you do is you say that the quadratic is going | |
| 05:18 | to basically look like this , it's gonna look like | |
| 05:20 | x minus one times x minus five and that's got | |
| 05:23 | to be equal to zero . How do I know | |
| 05:25 | that ? I'm not copying this from what I've already | |
| 05:28 | done up here . What I'm saying is that if | |
| 05:30 | you know the route is one and you know that | |
| 05:32 | when you solve quadratic you always get them in this | |
| 05:35 | form like x minus one , x minus five . | |
| 05:38 | Then you know that if the route is one this | |
| 05:40 | term has to be a minus one because that's what | |
| 05:42 | you would set equal to zero and that's how you | |
| 05:45 | would get a one as a route . And then | |
| 05:47 | you know that the other term multiplied by , it | |
| 05:48 | must be x minus five because that's the only way | |
| 05:51 | I could get a root of five . So going | |
| 05:53 | backwards is a simple matter of taking the routes that | |
| 05:56 | I have and writing the factor form directly from the | |
| 05:59 | roots and you can always do that because every quadratic | |
| 06:02 | that can be factored is always going to be some | |
| 06:05 | parentheses times some parentheses . And if you know what | |
| 06:08 | the roots are , you can always fill in what | |
| 06:10 | those parentheses are because those are the roots . That's | |
| 06:12 | what you would calculate as the roots . So It | |
| 06:16 | has to be X -1 , that would give me | |
| 06:17 | a root of one , it has to be X | |
| 06:19 | -5 and they have to be multiplied together . Now | |
| 06:21 | this is not the answer you want to circle . | |
| 06:23 | This is just the starting point . Now , what | |
| 06:25 | we do is we multiply all this stuff out , | |
| 06:27 | which we of course have all done before . Up | |
| 06:28 | above . But let's multiply X times X is X | |
| 06:31 | squared . This term gives me negative X . This | |
| 06:34 | outside term is negative five X . And this term | |
| 06:37 | is positive five . Of course , it's still equal | |
| 06:39 | to zero and then I'm gonna go and finish it | |
| 06:43 | off here . This is gonna be negative six , | |
| 06:44 | X Plus five is equal to zero . So this | |
| 06:48 | is the quadratic . So this is the answer for | |
| 06:51 | 10 . We had never done this first part of | |
| 06:52 | the problem before . I'm only showing you this so | |
| 06:56 | that you obviously are working with something in your mind | |
| 06:59 | and you can see that it makes sense . And | |
| 07:00 | it's real . But the real problem would be everything | |
| 07:03 | you would never have known about this above . You | |
| 07:05 | would say , well I only know these two routes | |
| 07:06 | are here , so I'm gonna have to put this | |
| 07:07 | in . Multiply it all out . This is the | |
| 07:09 | quadratic . So I would know that this quadratic , | |
| 07:14 | that's what the quad means . This quadratic has roots | |
| 07:19 | of one and five , X is equal to one | |
| 07:22 | and X is equal to five . And this is | |
| 07:24 | the answer that you would circle right ? Or you | |
| 07:26 | could uh take away the zero . But anyhow , | |
| 07:29 | when you set it equal to zero , that's what | |
| 07:30 | the roots are basically going to to be . Now | |
| 07:35 | here , that part was I think pretty easy to | |
| 07:37 | understand , right ? Because you can take any routes | |
| 07:39 | , you can write the factored form directly from the | |
| 07:41 | roots , multiply the whole thing out and that is | |
| 07:43 | always going to give you a quadratic , that has | |
| 07:46 | those roots . However , there's more than one quadratic | |
| 07:50 | that can have the same routes . In fact there's | |
| 07:53 | an infinite number of quadratic functions that can have roots | |
| 07:56 | one in five at just as an example . So | |
| 07:58 | I can pick any two routes , I want any | |
| 08:01 | two routes and I can always find a quadratic by | |
| 08:03 | this method . But then there's always an infinite number | |
| 08:06 | of other ones that are related to that one that | |
| 08:08 | have the same routes . And let's just see how | |
| 08:10 | that is the case . So we know that this | |
| 08:12 | is a quadratic that has roots one and five . | |
| 08:15 | But think about it , this thing is equal to | |
| 08:17 | zero . So what I could do is I can | |
| 08:19 | take this guy and I can say okay x squared | |
| 08:22 | minus six X Plus five equals 0 . Right . | |
| 08:26 | And remember this is an equation equals zero . I | |
| 08:29 | can take and multiply anything I want to this equation | |
| 08:32 | let's multiply by two . But if I multiply it | |
| 08:35 | by two on the left and I have to multiply | |
| 08:37 | it by two on the right . And when I | |
| 08:39 | do that , what am I gonna get ? I'm | |
| 08:40 | gonna get two X squared -2 times six is 12 | |
| 08:44 | x Plus two times 5 is 10 . And on | |
| 08:48 | the right hand side it's still equal to zero . | |
| 08:50 | Now this quadratic is different than this one . I | |
| 08:53 | mean it's completely different . It's multiplied by two . | |
| 08:55 | All the numbers are larger , right ? But this | |
| 08:58 | quadratic still has roots one and five . So you | |
| 09:06 | see you can take the original one that you find | |
| 09:08 | and you can multiply it by a number . In | |
| 09:09 | this case we just pick two and you will get | |
| 09:12 | another quadratic that has exactly the same route . And | |
| 09:15 | just to bring it home let's go back let me | |
| 09:18 | grab another color here . Let's go back to the | |
| 09:19 | original quadratic that we had . Let's say we start | |
| 09:22 | out with X squared minus six . X plus five | |
| 09:25 | equals zero . I can multiply let me give myself | |
| 09:29 | some room . I can multiply this quadratic by let's | |
| 09:33 | try three . So if I multiply by three on | |
| 09:36 | the left I got to multiply by three on the | |
| 09:37 | right . So if I do that , what am | |
| 09:39 | I gonna get ? I'll get three X squared -3 | |
| 09:42 | times six is 18 x three times five is 15 | |
| 09:46 | plus 15 . Again equals zero . This quadratic if | |
| 09:50 | you graph it looks different than this one which looks | |
| 09:53 | different than this one . But they all have roots | |
| 09:55 | one and five . So this quadratic is another one | |
| 09:58 | that has roots one and five . So you might | |
| 10:03 | ask yourself , let me actually catch up and make | |
| 10:06 | sure I've done the math correct . So we have | |
| 10:07 | two X squared minus 12 X plus 10 . We | |
| 10:09 | have three X squared minus 18 plus 15 . That's | |
| 10:11 | all correct . Okay so we have three quadratic that | |
| 10:15 | . I have found that I'm claiming all have the | |
| 10:17 | same routes one in five and I found them just | |
| 10:19 | by multiplying by the number two and number three . | |
| 10:21 | So you might imagine I can take that original quadratic | |
| 10:24 | . I found I can multiply by four . I | |
| 10:26 | can multiply by five . I can also multiply by | |
| 10:28 | negative numbers . I can multiply the whole thing by | |
| 10:30 | negative two or negative 10 or negative 100 . I | |
| 10:33 | could multiply by one half on both sides and I'm | |
| 10:36 | still going to have the roots the same exact routes | |
| 10:39 | Because if you think about it , if you were | |
| 10:41 | to solve this guy and figure out what the roots | |
| 10:43 | are , what are you going to do ? Well | |
| 10:44 | you're gonna first factor out of three which is this | |
| 10:47 | step and then you're gonna factor this expression which is | |
| 10:50 | gonna factor to this , but you're going to have | |
| 10:52 | a three out in front . So I would divide | |
| 10:54 | away the three because I will divide by three and | |
| 10:56 | divide by three , so it would disappear and I | |
| 10:58 | would have exactly the same route . So for any | |
| 11:00 | of these equations , when I factor out the number | |
| 11:02 | , I'm left with the same kind of like child | |
| 11:05 | polynomial behind , which will have the exact same roots | |
| 11:08 | of one in five . That's why there's an infinite | |
| 11:11 | family of routes , because that can take of infinite | |
| 11:14 | family of quadratic that have the same routes because I | |
| 11:16 | can take the original one and I can multiply by | |
| 11:19 | anything . I want all the way up into its | |
| 11:21 | close to infinity is I want to get and I'll | |
| 11:23 | get another quadratic . It has the same routes so | |
| 11:26 | I want to take a second . And the reason | |
| 11:28 | I'm going over this is because the theorem that we're | |
| 11:30 | going to talk about in just a minute when I | |
| 11:31 | get there , we'll talk a little bit about this | |
| 11:34 | multiplication stuff . And it's much easier to understand in | |
| 11:36 | terms of numbers than it is in terms of the | |
| 11:38 | theorem like that . So let's go and see how | |
| 11:41 | is it possible that we can have quadratic That looks | |
| 11:44 | so different but have the same routes . All right | |
| 11:47 | . So we said that this quadratic , the original | |
| 11:50 | one had roots of one and five and the original | |
| 11:52 | quadratic was x squared minus six . X plus five | |
| 11:56 | . That was the original one . So let me | |
| 11:57 | go over here and say this is X . Is | |
| 11:59 | equal to one and this is X is equal to | |
| 12:02 | five . So the original quadratic , I'm just gonna | |
| 12:05 | do my best to sketch it okay , it's not | |
| 12:07 | gonna be perfect , so forgive me , but it's | |
| 12:08 | basically going to look , something like this is gonna | |
| 12:10 | go through this point , it's going to reach a | |
| 12:11 | minimum , it's gonna go up through five , something | |
| 12:14 | like that . It's not exact . I never claimed | |
| 12:16 | to be an artist , but this is basically it | |
| 12:18 | and that equation is the very first one we found | |
| 12:21 | . So this is basically um how do I want | |
| 12:24 | to , how do I want to annotate it ? | |
| 12:25 | Let's go , I'll do it like this . I'll | |
| 12:27 | say I want to stay away from everything , You'll | |
| 12:29 | see why in a second I'll do this , I'll | |
| 12:31 | say f of X , the function is X squared | |
| 12:36 | minus six , X Plus five . So red goes | |
| 12:39 | with red , right ? This is the equation . | |
| 12:42 | Now , the second one that we found was this | |
| 12:44 | 12 X squared minus 12 X plus 10 . We | |
| 12:47 | said that when that was equal to zero , that | |
| 12:49 | was a quadratic also . So notice what is the | |
| 12:51 | difference between these two ? Obviously the numbers are different | |
| 12:54 | but we know it's a quadratic , we know it | |
| 12:57 | opens up because they're both positive in front of the | |
| 12:59 | X squared term . And remember the bigger than number | |
| 13:02 | in front of X squared means it closes up more | |
| 13:05 | . Like if you get larger and larger and larger | |
| 13:08 | numbers of that first coefficient it gets more and more | |
| 13:10 | narrow so that second one has exactly the same routes | |
| 13:14 | . It's going to look something like this , it's | |
| 13:15 | gonna go down like this and something like this , | |
| 13:23 | right ? And let me just double check my where | |
| 13:26 | am I gonna write it ? I guess I'll do | |
| 13:27 | it here and the second one will be F of | |
| 13:31 | X . Is equal to Um two x squared minus | |
| 13:38 | 12 X Plus 10 . So blue goes with blue | |
| 13:42 | and as you might guess the third one , what | |
| 13:43 | was the third one ? It was three X squared | |
| 13:45 | minus 18 X plus 15 . So there's a three | |
| 13:47 | so it's basically positive opens up but it's even more | |
| 13:50 | narrow than the previous one and it has to have | |
| 13:53 | the same rights as well . So of course I | |
| 13:55 | can't draw it exactly but it's going to go through | |
| 13:56 | and have the exact same routes and come up something | |
| 13:59 | like this . Of course the shape of it is | |
| 14:01 | not right but you get the idea . So F | |
| 14:03 | . Of X . For the third one is three | |
| 14:06 | X squared minus 18 X Plus 15 . And we | |
| 14:10 | take this away plus 15 . So green goes with | |
| 14:13 | green . So now you can see how this idea | |
| 14:17 | of finding a polynomial or finding a quadratic with the | |
| 14:21 | roots one in five is really kind of a weird | |
| 14:23 | question because there is not just one quadratic that has | |
| 14:26 | these routes , there's always a family of quadratic . | |
| 14:29 | So I'm gonna use the word family and infinitely related | |
| 14:33 | family . These guys are basically the same quadratic , | |
| 14:35 | but because we multiplied by a number to generate the | |
| 14:38 | family , they have basically the same characteristics . They | |
| 14:41 | go through the same routes with a factor any one | |
| 14:45 | of these guys . I'm just gonna pull out the | |
| 14:46 | number , I'm still gonna have the factored form in | |
| 14:48 | there . I'm gonna have the same exact routes for | |
| 14:50 | for all of them . But obviously they're getting steeper | |
| 14:52 | and steeper And if I go if I have another | |
| 14:55 | one where it's 100 and as the first coefficient , | |
| 14:57 | that thing is gonna be really , really , really | |
| 14:58 | steep and go off of my chart here . But | |
| 15:01 | they're always going to go through the same exact routes | |
| 15:03 | . So when a problem says find the quadratic that | |
| 15:06 | goes through these routes , it's really kind of not | |
| 15:08 | true . There's an infinite number of them . So | |
| 15:11 | if you find one of them , you've satisfied the | |
| 15:13 | problem . But there's always more that you can find | |
| 15:14 | just by multiplying the equation through . So just to | |
| 15:18 | bring it home , we could say something like in | |
| 15:22 | general , so we can generalize this in general . | |
| 15:26 | In general , we can take a number A and | |
| 15:29 | we can multiply by that base equation X squared minus | |
| 15:33 | six X plus five . That's the one that we | |
| 15:35 | found the first time and any time I take this | |
| 15:38 | guy multiplied by any number A . It always has | |
| 15:44 | roots one and five . And the number A . | |
| 15:49 | Here that you multiply by determines the shape of whatever | |
| 15:52 | particular parabola that you have . So most books are | |
| 15:56 | not going to go through a talk like that with | |
| 15:58 | a specific example like this . Most books are just | |
| 16:01 | going to show you the theorem and the theory is | |
| 16:03 | gonna have a lot of letters and it makes people | |
| 16:05 | dizzy because they don't understand what it means . They | |
| 16:07 | don't have the big picture before they look at the | |
| 16:09 | theory , but now that you have the big picture | |
| 16:11 | and you'll understand what it's trying to say . Now | |
| 16:13 | , I'm going to present the theorem as you might | |
| 16:15 | see it in a textbook and you'll understand a lot | |
| 16:17 | more how it works and why it works . All | |
| 16:20 | right . So this is not going to be so | |
| 16:21 | scary . So we're going to generalize this to the | |
| 16:24 | following thing that you might see in a book given | |
| 16:27 | roots of a polynomial , the roots of a quadratic | |
| 16:30 | . However you want to say it are number one | |
| 16:32 | and our number two . So it's in terms of | |
| 16:34 | R . One and R . Two because I'm generalizing | |
| 16:36 | it any two routes in that example , it was | |
| 16:39 | routes one in five , but R . One and | |
| 16:40 | R . Two could be any two routes negative three | |
| 16:42 | , positive seven , whatever . It doesn't matter what | |
| 16:44 | the roots are there were going to call them R | |
| 16:46 | . One and R . Two . If I knew | |
| 16:48 | that I had this guy two routes one and two | |
| 16:50 | , how would I find the polynomial ? Well what | |
| 16:52 | did I do ? I took these two routes , | |
| 16:54 | I wrote them in the factored form and I multiplied | |
| 16:56 | it out . That's how I got this equation . | |
| 16:58 | And then once I got this equation , I knew | |
| 17:00 | that I can multiply by these whole numbers to generate | |
| 17:04 | the family of equations . So I'm gonna do exactly | |
| 17:07 | the same thing here . I'm gonna write the factored | |
| 17:09 | form . I'm going to do it in a different | |
| 17:11 | color . If I have two routes are number one | |
| 17:13 | or number two , then I could say that the | |
| 17:14 | equation x minus R . One times X minus R | |
| 17:18 | . Two has to equal zero . This will be | |
| 17:20 | the polynomial . How do I know ? Because if | |
| 17:24 | I were to get a polynomial factored into this form | |
| 17:26 | then this one would correspond . Moving this over would | |
| 17:29 | correspond to our one route . This one would correspond | |
| 17:31 | to the R . To route exactly is the problem | |
| 17:33 | that we have already done and now we can multiply | |
| 17:36 | this out . X times X is going to give | |
| 17:38 | you X squared this time , this is negative . | |
| 17:41 | Are one X . So negative are one X . | |
| 17:45 | Right ? This one here , don't forget the negative | |
| 17:48 | sign is negative are two X . And this guy | |
| 17:52 | here negative times negative positive are one times are too | |
| 17:57 | so that looks ugly . See why this gets kind | |
| 17:59 | of confusing when you look at it in a textbook | |
| 18:01 | because they've got all the R . One and R | |
| 18:02 | . Two's everywhere . But bear with me we're going | |
| 18:04 | to simplify it a little bit . This middle term | |
| 18:07 | has a common factor of X . So let's just | |
| 18:09 | factor out that X squared minus . I'm gonna pull | |
| 18:13 | out the X . And actually do I want to | |
| 18:15 | pull yeah I want to pull it out like this | |
| 18:17 | . Then on the inside I'm gonna have our one | |
| 18:19 | plus R . Two and this last term is gonna | |
| 18:23 | stay the same . Make sure you understand that ? | |
| 18:25 | Why did I change it to A plus ? Well | |
| 18:26 | because I pulled out a basically a minus X . | |
| 18:29 | And so when I multiply it in this times this | |
| 18:31 | gives me negative are one X . Multiply ? This | |
| 18:34 | times this gives me negative are two times X . | |
| 18:36 | So you have to always go backwards and make sure | |
| 18:39 | that's why I needed a plus sign . There was | |
| 18:40 | because I actually pulled out of minus X to begin | |
| 18:42 | with . And now one other small change is this | |
| 18:47 | term . I just want to flip around how I've | |
| 18:49 | written it here . So let me switch colors to | |
| 18:52 | something else and I want to write it like this | |
| 18:56 | , I'm gonna put the minus sign but this time | |
| 18:57 | I'm gonna flip it around I'm gonna write it is | |
| 18:58 | R . One R . Whoops the wrong thing are | |
| 19:01 | one plus R . Two times X plus R . | |
| 19:05 | One R two equals zero . So all I did | |
| 19:08 | was since they're multiplied three times four is 12 and | |
| 19:11 | also four times three is 12 flipping it around doesn't | |
| 19:14 | matter . So these are multiplied . So I just | |
| 19:16 | flip the order of them . That's all I've done | |
| 19:18 | . Now . You might look at this and say | |
| 19:19 | oh my gosh , this is crazy . What does | |
| 19:21 | this mean ? All right . What this means is | |
| 19:25 | that if I give you roots R . One and | |
| 19:27 | R . Two , Right ? R . one and | |
| 19:29 | R . two . Then all you have to do | |
| 19:31 | to figure out the polynomial or the quadratic that has | |
| 19:34 | those routes is sticking into this equation . It's already | |
| 19:37 | built . You stick are one in and are too | |
| 19:40 | and you add them together and that's gonna be in | |
| 19:42 | front of the X . Term . And then these | |
| 19:43 | two R . One and R two are multiplied together | |
| 19:46 | and then you have a polynomial have X squared minus | |
| 19:49 | something times X . Plus some number . That is | |
| 19:52 | a quadratic . That is a polynomial that will have | |
| 19:54 | roots R . One and R . Two . Okay | |
| 19:58 | , so I'll put this on the side . This | |
| 20:01 | is a quadratic has roots R . one and R | |
| 20:11 | . two . R . one and R . two | |
| 20:13 | . So this is exactly the same . I'm not | |
| 20:15 | actually gonna do it . I'm gonna go ahead and | |
| 20:17 | do it with the example that we've had . But | |
| 20:19 | this is exactly what we have already done but totally | |
| 20:22 | generalized . We took the roots , we wrote them | |
| 20:24 | in the factor form . We multiplied it out and | |
| 20:26 | we collected terms . That's all that we have done | |
| 20:28 | here . So if you know R . One and | |
| 20:30 | R . Two just stick it in here and this | |
| 20:31 | will give you the polynomial . But you also have | |
| 20:34 | the idea that this is actually a family , a | |
| 20:38 | family here . So this is the base one I | |
| 20:40 | guess you could say or we know that we can | |
| 20:44 | take this polynomial of X squared minus R one plus | |
| 20:49 | R two times X plus R one . R two | |
| 20:53 | equals zero . We can take that polynomial . Let | |
| 20:57 | me go and stick this a little bit further over | |
| 21:00 | and we can multiply the whole thing by some number | |
| 21:04 | A and that will generate the family of polynomial . | |
| 21:07 | Why did I not multiplying on the right hand side | |
| 21:09 | ? Well I did , but it's zero . So | |
| 21:10 | if I write eight times zero is still zero on | |
| 21:12 | the right , just like it was over here when | |
| 21:15 | I multiplied by two , then I multiply two times | |
| 21:18 | zero is still zero , multiplied by three . It's | |
| 21:20 | still three times zero and zero . So you see | |
| 21:23 | really you only have to multiply the left hand side | |
| 21:25 | because the right hand side is going to be zero | |
| 21:26 | , no matter what . So the idea is this | |
| 21:29 | is the family with roots R . one and R | |
| 21:38 | . two . So really this is the answer to | |
| 21:42 | this is the theorem that you'll see in a textbook | |
| 21:44 | . The textbook might say something like this . They | |
| 21:46 | might say given a polynomial has roots R . One | |
| 21:48 | and R . Two . This generates a quadratic function | |
| 21:51 | , a quadratic function with those roots . And this | |
| 21:55 | other term here can generate a family of of quadratic | |
| 21:58 | that all have the same routes . So that's what | |
| 22:00 | it is . But the first time I looked at | |
| 22:01 | it a long time ago , I was very confused | |
| 22:03 | like why is there an A here ? Why isn't | |
| 22:05 | there only one answer ? Why is there two answers | |
| 22:07 | ? And so I had to do the other example | |
| 22:09 | to show you that the base when it comes from | |
| 22:11 | just putting the roots in there and multiplying them out | |
| 22:14 | . But I can multiply that quadratic that I get | |
| 22:16 | buy anything I want and still have the same routes | |
| 22:19 | because of what I've drawn here . Now there's kind | |
| 22:23 | of an easier way to remember this . This is | |
| 22:25 | an ugly equation . I'm not gonna lie to you | |
| 22:26 | . So easy to remember . I'm gonna write that | |
| 22:29 | down easy to remember . Or easier to remember . | |
| 22:36 | Remember it like this a times X squared minus instead | |
| 22:41 | of all this stuff . Remember this as the sum | |
| 22:44 | of the routes that you're given and you have to | |
| 22:47 | multiply by X . And then plus the , I'm | |
| 22:50 | gonna right here , product of the roots given right | |
| 23:01 | equals zero . So this is a little bit easier | |
| 23:03 | to remember if you think of it in terms of | |
| 23:04 | some of roots and product of roots , some of | |
| 23:06 | roots and product of roots , then really what you | |
| 23:09 | have is you have X squared minus to some of | |
| 23:11 | the routes times X plus the product of roots and | |
| 23:14 | that's their base equation . And then I can multiply | |
| 23:16 | that basic equation by any number . I want to | |
| 23:18 | generate the whole family of other equations because oftentimes what's | |
| 23:23 | going to happen that we're gonna find as we solve | |
| 23:25 | problems is when we take the roots and we put | |
| 23:27 | them in there , we multiply them out . We're | |
| 23:28 | gonna we might oftentimes have fractional coefficients for the polynomial | |
| 23:32 | that's going to be a correct polynomial , but oftentimes | |
| 23:35 | we don't want fractional coefficients . So just multiply through | |
| 23:38 | by any number , you want to clear all the | |
| 23:40 | fractions away and you'll get rid of them . So | |
| 23:42 | this theorem is telling you , you can generate the | |
| 23:44 | base the base quadratic and then you can multiply that | |
| 23:47 | by anything you want to basically clear any fractions away | |
| 23:50 | or get rid of terms that you don't want . | |
| 23:52 | As long as you multiply every term by some number | |
| 23:55 | A it's still gonna have exactly the same routes . | |
| 23:57 | So now what I wanna do right underneath this and | |
| 24:00 | I'm going to uh solve a separate problem on the | |
| 24:03 | other board but I want to do it right underneath | |
| 24:05 | . This was uh fresh in our mind . I | |
| 24:08 | want to apply this guy To our example when we | |
| 24:11 | have roots one in 5 . So let's say I | |
| 24:13 | give you and I tell you that route number one | |
| 24:16 | is one and route number two is five . And | |
| 24:20 | I say uh give me a polynomial that has a | |
| 24:23 | quadratic function that has both of those as the roots | |
| 24:26 | . And then you would come back to your theory | |
| 24:28 | and say well yeah I can multiply by a but | |
| 24:30 | I'm just gonna take a equal to one right now | |
| 24:32 | and just use the basic one . So I'm going | |
| 24:33 | to say that it's going to be X squared minus | |
| 24:36 | the sum of these routes . That some of these | |
| 24:38 | routes is one plus five . So let's leave it | |
| 24:40 | as one plus five now times X . And then | |
| 24:42 | plus the product of these routes , the product is | |
| 24:45 | one times five , so one times five . Right | |
| 24:48 | ? And I'm hopeful that this is gonna work . | |
| 24:50 | So I have X squared here , I have minus | |
| 24:52 | six X plus five Equals zero . And that is | |
| 24:58 | exactly what we have is exactly what we did . | |
| 25:00 | Right ? So there's two ways to do all of | |
| 25:02 | these problems . You can take the routes , you | |
| 25:04 | can put them there and you can multiply it all | |
| 25:06 | out and you can get the answers or you can | |
| 25:09 | just use what the theorem is telling you because it | |
| 25:11 | comes from the same place . Some of roots product | |
| 25:14 | of routes . Follow this guy . All right , | |
| 25:17 | now we're going to close the lesson out with one | |
| 25:21 | more quick example and then we're going to move on | |
| 25:24 | to the next lesson where we have , you know | |
| 25:25 | , a lot more examples where the problems are a | |
| 25:28 | little more complex , more fractions to deal with other | |
| 25:30 | little things you have to worry with . But ultimately | |
| 25:32 | we're just going to be doing using the same theorem | |
| 25:34 | over and over again . All right . So let's | |
| 25:37 | say we want to find a quadratic with what we | |
| 25:40 | call integral coefficients , meaning , you know , whole | |
| 25:43 | number coefficients where the roots of this guy , R | |
| 25:48 | X is equal to two and X is equal to | |
| 25:50 | five . Those are the roots of some guy . | |
| 25:52 | And we want to figure out what the what the | |
| 25:56 | quadratic or what a quadratic with integral coefficients are . | |
| 25:58 | So , the first thing you should do is you | |
| 26:00 | want to find the some of the routes in the | |
| 26:01 | product of the roots because that's what's in the equation | |
| 26:04 | that we have there . So if you want to | |
| 26:07 | figure out what the some of the routes are , | |
| 26:10 | It's just two plus 5 , Which is obviously seven | |
| 26:14 | . And then the product of the roots is just | |
| 26:16 | two times five , which is 10 . So now | |
| 26:19 | that you have the some of the routes in the | |
| 26:20 | product of the roots , you just have to put | |
| 26:22 | your equation together , X squared minus to some of | |
| 26:25 | the routes times X plus the product of the roots | |
| 26:27 | equals zero . And then if you want you can | |
| 26:29 | multiply that by some number a and generate a different | |
| 26:32 | one . But you know , you don't often have | |
| 26:34 | to but you could yeah , so the quadratic function | |
| 26:38 | is going to be this X squared minus the sum | |
| 26:41 | of the roots are too times X plus the product | |
| 26:46 | of the roots equals zero . So we have x | |
| 26:49 | squared minus the sum of the roots . We already | |
| 26:51 | figured out with seven , so it's going to be | |
| 26:52 | seven X plus the product of the roots , which | |
| 26:55 | was 10 , which was equal to zero . So | |
| 26:57 | x squared minus seven X plus 10 equals zero . | |
| 27:00 | So this equation already had integral coefficients . The problem | |
| 27:04 | statement here is find a quadratic function with integral coefficients | |
| 27:08 | . But what if the roots were weird ? What | |
| 27:10 | if the roots were one half and 3/10 ? Well | |
| 27:13 | then when I do all of this sum and product | |
| 27:16 | , I'm going to get fractions for all of these | |
| 27:18 | coefficients . If I get fractions for the coefficients , | |
| 27:20 | this theorem tells me that I'm free to multiply this | |
| 27:25 | equation by any number I want because multiplying both sides | |
| 27:28 | leaves the right hand side unchanged . I can multiply | |
| 27:30 | by anything I want and still have the same routes | |
| 27:32 | . So if I did this different problem like I | |
| 27:35 | will here in the future and I got you know | |
| 27:36 | like a one half in the front , like one | |
| 27:38 | half X squared when I did all this math and | |
| 27:40 | I got a one half here , I don't want | |
| 27:42 | a Quadratic with 1/2 as a coefficient , I want | |
| 27:45 | integral coefficients . But if I did get a one | |
| 27:48 | half here , then the theorem tells me I could | |
| 27:50 | just multiply the whole thing by two , which would | |
| 27:52 | clear the denominator and clear the one half fraction . | |
| 27:55 | And we have to multiply by two , multiply this | |
| 27:57 | by to multiply this by to multiply this by two | |
| 27:59 | and then I would have a new quadratic that would | |
| 28:01 | have exactly the same routes and it would have integral | |
| 28:04 | coefficients . Now this is using the theorem . Okay | |
| 28:07 | . Which I want you to do , that's why | |
| 28:09 | it's here , but there's another way to do it | |
| 28:12 | . I want you to I don't want to ever | |
| 28:13 | forget that another way . Actually , it ends up | |
| 28:16 | being the same way . But there's another way to | |
| 28:18 | do it . If I know the routes are two | |
| 28:19 | and five , and I can just build the thing | |
| 28:21 | like this , X minus two , X minus five | |
| 28:24 | equals zero because these would be the routes two and | |
| 28:26 | five , and I can multiply this and get X | |
| 28:28 | squared minus two , X minus five X . And | |
| 28:34 | then the last guys plus 10 . And you can | |
| 28:37 | see right away this would be X squared minus seven | |
| 28:39 | , X plus 10 is equal to zero , which | |
| 28:41 | is exactly the same thing . So when you do | |
| 28:43 | these things and of course that could multiply this by | |
| 28:46 | any whole number . I want to get another another | |
| 28:48 | quadratic that has exactly the same routes . So when | |
| 28:51 | you do these problems , you're gonna do them the | |
| 28:52 | same way . You can either build them like this | |
| 28:55 | and do the multiplication out or you can just remember | |
| 28:57 | the theorem and remember what it tells you and just | |
| 28:59 | plug in the values . I'll probably do a little | |
| 29:01 | bit of a mixture as I solve most of the | |
| 29:03 | problems . But if you ever get a quadratic back | |
| 29:05 | as an answer that has like a fraction in it | |
| 29:07 | , just multiply the whole thing through by whatever the | |
| 29:10 | denominator of that fraction is to clear it out . | |
| 29:13 | And you will be guaranteed to have a resulting polynomial | |
| 29:16 | with exactly the same routes as the parent because they're | |
| 29:19 | all form of family as we've drawn on the board | |
| 29:21 | here . So make sure you can solve all these | |
| 29:23 | problems and make sure you understand the concept most of | |
| 29:25 | all . I want you to want to make sure | |
| 29:27 | you understand this theorem given roots R . One and | |
| 29:29 | R . Two . This is how you generate the | |
| 29:31 | family of quadratic that has those roots , and then | |
| 29:34 | follow me on to the next lesson where we will | |
| 29:37 | get practice solving more and more complicated problems of this | |
| 29:40 | nature . |
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