01 - Solving Equations in Quadratic Form - Part 1 (Learn to Solve Equations in Algebra) - By Math and Science
Transcript
00:00 | Hello , welcome back to algebra . The title of | |
00:02 | this lesson is called solving equations that are in quadratic | |
00:06 | form . This is part one of several lessons . | |
00:08 | Now , if you've been working with me up to | |
00:10 | this point in algebra , we've been working with quadratic | |
00:12 | equations , otherwise known as parabolas for quite some time | |
00:16 | . And you might be saying , well don't we | |
00:17 | already know everything there is to know about these quadratic | |
00:19 | equations . Why do we continue to continue to work | |
00:22 | on this ? Well , the answer is , quadratic | |
00:24 | equations are just that important . They are so very | |
00:27 | important . There's a lot to it . So what | |
00:29 | we're gonna be doing in this batch of lessons is | |
00:31 | kind of finishing up the discussion with quadratic this lesson | |
00:34 | . We're talking about equations , what we call that | |
00:36 | are in quadratic form and I'll show you what that | |
00:39 | means in a second . But I want to give | |
00:40 | you a little bit of road map for this batch | |
00:42 | of lessons in the sequence pretty soon . In the | |
00:45 | next few lessons after this topic will be talking about | |
00:47 | how to take these quadratic functions , these parabolas and | |
00:50 | how to shift them around the xy plane . In | |
00:53 | other words , you already know about f of X | |
00:55 | is equal to x squared , right ? It's that | |
00:57 | nice parabolas centered at the origin . Well , we | |
01:00 | can change that equation slightly and we can move it | |
01:02 | to the left , we can move it to the | |
01:03 | right , we can move it up , we can | |
01:04 | move it down . So we're going to be learning | |
01:06 | how to shift these equations around and we'll have some | |
01:09 | computer demos as we as we get to that point | |
01:11 | to show you graphically what's really going on . Uh | |
01:13 | And then we'll actually be wrapping up the class with | |
01:16 | this batch of lessons with , learning how to uh | |
01:20 | take roots of a quadratic equation . We already have | |
01:22 | the roots and how to go backwards and reconstruct a | |
01:26 | quadratic function that uh is a parent or that the | |
01:29 | roots come from . In other words , if I | |
01:30 | give you the answers the roots , how do you | |
01:32 | go backwards and construct the quadratic equation from that and | |
01:36 | in between here and there , there's a lot of | |
01:37 | little details and theorems that we have to cover . | |
01:40 | Uh and so I want to just kind of give | |
01:42 | you that overall idea that these quadratic functions , these | |
01:45 | quadratic equations are so very important . One parting thought | |
01:49 | as we as I want you to keep in the | |
01:50 | back of your mind as we do these bachelor lessons | |
01:53 | dispatch lessons is just one little example that occurred to | |
01:56 | me , I want to bring up to you , | |
01:57 | A lot of students say , why did we spend | |
01:59 | so much time on quadratic ? Well , as I | |
02:02 | said already before they're ubiquitous , they're everywhere in Science | |
02:05 | and math and I'll give you one example of that | |
02:07 | . I'm not gonna write anything on the board . | |
02:08 | I just want you to kind of work with me | |
02:09 | . This is kind of advanced , but I want | |
02:11 | you to understand how useful they are when you get | |
02:14 | passed algebra and you go on into calculus and you | |
02:16 | get past calculus , you're going to take a class | |
02:18 | called differential equations , which is an advanced calculus class | |
02:22 | . That's all it means . But differential equations is | |
02:24 | really how the world works . It's all about using | |
02:26 | calculus to predict how complex systems are gonna move . | |
02:29 | And probably one of the most famous differential equations in | |
02:33 | advanced calculus you probably even have already heard of , | |
02:35 | and it's the famous Newton's second law of motion F | |
02:38 | equals M . A . If you haven't taken a | |
02:40 | physics class , maybe you haven't heard of that . | |
02:42 | But certainly it's one of the first things you learn | |
02:44 | in a basic physics class , force is equal to | |
02:47 | mass times , acceleration . It turns out when you | |
02:50 | get past basic physics and you kind of decompose that | |
02:53 | famous equation . It turns out to be a differential | |
02:56 | equation , which means a calculus equation . And the | |
02:59 | reason I'm bringing all this up because it seems like | |
03:01 | it's coming from nowhere is because that famous equation ethical | |
03:04 | , when you break it out into the calculus notation | |
03:07 | is actually what we call a second order differential equation | |
03:11 | , which means it has a second derivative or a | |
03:14 | second , a term that has a calculus derivative taken | |
03:17 | two times in it . Why do I explain all | |
03:20 | of this to you now in algebra class ? Well | |
03:21 | , the reason is because when you learn how to | |
03:23 | solve those equations , those second order differential equations , | |
03:27 | The solution method that we all learn when we take | |
03:30 | those classes down the road . As part of that | |
03:33 | solution method , you have to solve a quadratic equation | |
03:37 | because it's a second order differential equation , it ends | |
03:40 | up requiring us to solve a 2nd 2nd degree polynomial | |
03:45 | , which is a quadratic equation . In other words | |
03:48 | , one of the most famous equations of all time | |
03:50 | . That's used to calculate trajectories to the moon . | |
03:53 | That's used to calculate anything , you know , moving | |
03:56 | around on the surface of the earth in terms of | |
03:58 | physics , which means F is equal to force is | |
04:00 | equal to mass times acceleration when you grow up a | |
04:03 | little bit , so to speak , and really deal | |
04:05 | with that thing . In terms of calculus , you | |
04:06 | end up having to solve these quadratic equations in the | |
04:10 | process of solving that advanced calculus equation . So we're | |
04:13 | building the bedrock here of things that will be used | |
04:16 | for many , many years if you're going to go | |
04:18 | into any branch of science and engineering . So now | |
04:21 | that I had to get that off my chest because | |
04:22 | I want people to understand why we're learning things . | |
04:25 | Let's turn our attention to what we call solving equations | |
04:27 | and quadratic form . So I'm gonna go and start | |
04:31 | with the beginning uh here and show you a quadratic | |
04:34 | equation that I know should look pretty familiar to . | |
04:36 | What if you had the equation X squared plus four | |
04:39 | times X minus three is equal to zero . You | |
04:42 | all know how to solve this many , many different | |
04:44 | ways . We can try to factor it and solve | |
04:45 | it . We can try to use the quadratic formula | |
04:48 | and of course we'll be able to solve it . | |
04:49 | We also know that we can complete the square on | |
04:51 | this thing and solvent . So we have now basically | |
04:53 | three main methods of trying to solve this thing , | |
04:56 | but it's a quadratic equation . But more specifically , | |
04:59 | I want to write something down here . This equation | |
05:01 | is what we call quadratic quadratic in the variable X | |
05:07 | . I'm putting X and quotations . You'll see why | |
05:09 | in a second . Why is it quadratic and X | |
05:11 | ? Because X is the thing that squared and then | |
05:14 | plus a number four in this case times X . | |
05:17 | And then plus or minus a number . And then | |
05:19 | really there's an invisible X to the zero power because | |
05:22 | remember anything to the zero is just one . So | |
05:25 | really you have X squared and X . Two the | |
05:27 | first and then X to the zero , which is | |
05:29 | invisible because anything to zero is one . So it | |
05:32 | just disappears . That's the general form of these things | |
05:34 | we call quadratic equations . The thing that is being | |
05:37 | squared is just the variable X . So we say | |
05:39 | it's quadratic in the variable X . Now let's use | |
05:43 | the same thing as a skeleton and introduce a similar | |
05:47 | equation . What if I give you three X plus | |
05:49 | the number one . Right ? I'm gonna square that | |
05:52 | . And then I'm gonna add to that the term | |
05:55 | four times three X plus one . And then I'm | |
05:58 | gonna start track three and I'm gonna have this equal | |
06:00 | to zero notice how this is exactly the same thing | |
06:03 | . Except for one huge difference . The X squared | |
06:05 | is replaced by three X plus one . This whole | |
06:08 | quantity is what is now squared the number in front | |
06:11 | of the X . Is one , this is one | |
06:13 | here and there's a one in front here , there's | |
06:15 | a four in front of here and then a three | |
06:17 | and the negative three is the same as well . | |
06:18 | So you see how it matches the pattern exactly something | |
06:21 | squared plus four times something minus something as a constant | |
06:25 | in all the numbers are the same . The only | |
06:27 | thing is we swapped this this guy out so what | |
06:29 | does that mean ? It turns out this is a | |
06:31 | quadratic equation also it looks more complicated than this one | |
06:35 | because of the more complicated thing that's being squared and | |
06:38 | that's carried through into the other terms . But this | |
06:40 | one's not quadratic and X we say this one is | |
06:43 | quadratic . Can you guess what in the quantity ? | |
06:49 | Three ? X plus one ? So this one's quadratic | |
06:52 | and X . Because that's the thing that's being squared | |
06:55 | , the second one is quadratic in a totally different | |
06:57 | term . Right ? So let me give you one | |
07:00 | more example and we're gonna start to solve a problem | |
07:02 | . But this is the goal of this this lesson | |
07:04 | is to teach you how to solve equations like this | |
07:06 | . They're quadratic but they're quadratic . Uh their equations | |
07:10 | in quadratic form because this is the quadratic form but | |
07:13 | it's more complicated than what we've dealt with in the | |
07:15 | past . So here's another example . What if I | |
07:18 | give you 1/3 X quantity squared ? What's not to | |
07:22 | the third power ? We don't know how to solve | |
07:23 | those yet . To the second power plus four times | |
07:27 | the quantity 1/3 eggs uh minus three . And this | |
07:31 | is equal to zero . Right ? So same sort | |
07:34 | of thing . It matches the form exactly something squared | |
07:37 | plus four times something minus three . But this one | |
07:40 | is quadratic in the quantity 1/3 X . So these | |
07:46 | are all quadratic equations , right ? It's just that | |
07:50 | this one is quadratic in the simple variable X . | |
07:52 | And the other ones are quadratic and more complicated terms | |
07:55 | . So what we're gonna do now is learn how | |
07:57 | to solve these . Now the first thing you can | |
07:59 | do of course when you're giving any equation like this | |
08:02 | is you can just brute force through it . In | |
08:04 | fact if you wanted to solve this when you already | |
08:05 | know how to do it Because if I cover up | |
08:08 | the rest of the problem , you know how to | |
08:10 | expand this thing , right ? This is a binomial | |
08:12 | times itself , so you can multiply that out with | |
08:15 | foil , right ? With this one , you can | |
08:17 | just take the four and distribute in . That's gonna | |
08:19 | give you a bunch of terms and then the -3 | |
08:21 | . So when you expand this whole thing out you're | |
08:23 | gonna get lots and lots and lots of terms and | |
08:26 | then you collect like terms . And so you'll have | |
08:28 | a new quadratic equation which you can then try to | |
08:31 | solve . Of course you can always do that , | |
08:32 | that's what we call the brute force way of doing | |
08:34 | it , you just expand that sucker out and try | |
08:37 | to collect terms and then you're gonna get a quadratic | |
08:39 | and you solve it using quadratic formula or other methods | |
08:43 | that we have learned . But what we're gonna do | |
08:45 | in this class or in this lesson is we're gonna | |
08:47 | learn how to solve these things in a much more | |
08:49 | clever way . We're going to use the concept of | |
08:51 | what we call substitution and you're gonna find that it's | |
08:54 | actually really cool . It makes it much , much | |
08:56 | easier to solve than blowing this whole thing out , | |
08:59 | which is , you know , most people would probably | |
09:01 | try that first , including me if I wasn't thinking | |
09:04 | clearly about it . But when we use the concept | |
09:07 | of substitution is going to make it much , much | |
09:09 | simpler to solve this . Also in the back of | |
09:11 | your mind , I want you to keep in mind | |
09:12 | the idea of substitution because when we get to calculus | |
09:16 | later down the road , probably half of the class | |
09:19 | and I do mean it literally about half of the | |
09:21 | calculus class is using a concept called substitution to solve | |
09:26 | to solve problems in calculus . So this idea of | |
09:28 | substitution actually goes way beyond algebra . And so we're | |
09:32 | kind of inching our way into what this means . | |
09:34 | But you'll be using substitution for many , many years | |
09:37 | to come . So let's just talk about this first | |
09:40 | problem right here . Now , I don't want to | |
09:41 | solve this particular one . I want to solve one | |
09:43 | . That's a little bit easier first . So let | |
09:44 | me give you another problem . That will be all | |
09:46 | . Call this problem one A What if I give | |
09:49 | you x plus three quantity squared minus five times X | |
09:54 | plus three plus four equals zero . Now this is | |
09:58 | a quadratic in the quantity X plus three because that | |
10:01 | is what is squared , that's what's to the first | |
10:03 | power . And that's what's to the zero power there | |
10:05 | . So it's quadratic equation . So what we're gonna | |
10:07 | do is we're gonna make a substitution and you're free | |
10:11 | to make any substitution with equations that you want . | |
10:15 | The goal with the substitution is to make the equation | |
10:17 | simpler to solve . If your substitution makes it harder | |
10:20 | to solve , it's the wrong move . But if | |
10:23 | your substitution makes it easier to solve , it's the | |
10:25 | right move and you'll see how that happens right here | |
10:27 | . This guy is the obvious thing that we want | |
10:30 | to to uh substitute . So we're gonna say we're | |
10:32 | gonna let some new variable which isn't even in this | |
10:36 | problem . We're gonna call it Z . You can | |
10:37 | call it what you want . But I'm gonna say | |
10:39 | let's let Z . Is equal to X plus three | |
10:41 | . Why did I choose X . Plus three ? | |
10:43 | Because we know that this thing squared minus five times | |
10:47 | this thing and so on is in the quadratic form | |
10:50 | . So we're going to basically take this thing and | |
10:51 | represented as a new variable . Because then when we | |
10:54 | do that we can make this substitution since Z is | |
10:57 | now equal to this , we can say that I | |
10:59 | have a new equation called Z squared -5 times e | |
11:03 | plus four . Make sure you understand before we go | |
11:06 | on that these two equations are the same exactly the | |
11:10 | same . Because now that we've let Z equal to | |
11:12 | this , if I take this and stick it in | |
11:14 | here , I'm gonna get the first term back if | |
11:16 | I take it and stick it in here , parentheses | |
11:18 | don't forget it's equal to this whole thing . So | |
11:19 | you have to put a parentheses that would give me | |
11:21 | this term . And then the four comes along for | |
11:23 | the ride . So by making a substitution , a | |
11:26 | clever substitution , right then I can transform this ugly | |
11:29 | equation into a much simpler equation because now I pretend | |
11:33 | that the rest of this problem above doesn't even exist | |
11:36 | if I give you this equation straight out of the | |
11:38 | gate and I say solve it . What do you | |
11:40 | do ? Well , there's lots of choices . You | |
11:41 | can try to factor it and solve it . You | |
11:44 | can use the quadratic formula , you can do completing | |
11:46 | the square . You have freedom to do whatever you | |
11:48 | want . You know how to solve polynomial are quadratic | |
11:51 | when you set them equal to zero . So now | |
11:53 | that we have this in place , let's try to | |
11:56 | solve it by factoring So we're gonna open up a | |
11:59 | couple of parentheses and we have a Z square . | |
12:02 | So this will be easy and this will be easy | |
12:05 | multiply to give us this . And then of course | |
12:07 | for four , I could do two times too , | |
12:08 | but that's not going to add to give me five | |
12:10 | . I'm going to try one times four because I | |
12:12 | think I may have a shot of giving this . | |
12:13 | And the only way it's gonna work is within minus | |
12:15 | and a minus . And you should always check yourself | |
12:18 | Z times Z Z squared inside terms gives me negative | |
12:22 | Z . Outside terms gives me negative four Z . | |
12:24 | Those add together to give me the negative five Z | |
12:26 | . And then the last term's multiply to give me | |
12:29 | the positive force . So this is the factored form | |
12:31 | Of this . And so then I can say now | |
12:34 | I can solve for Z , Z -1 is equal | |
12:36 | to zero . So that means Z is equal to | |
12:38 | the number one . Over here , I have Z | |
12:41 | -4 is equal to zero and Z is equal to | |
12:44 | four . So so far we've done exactly what we | |
12:46 | would do if I just gave you this equation and | |
12:48 | told you to solve it , you factor it , | |
12:50 | set each thing zero and get the things back . | |
12:53 | The only problem is my original equation doesn't have any | |
12:56 | disease in there at all . It has excess . | |
12:59 | So now you have to take the answers that you | |
13:01 | got and use them somehow to get the answers to | |
13:05 | the equation to the original equation that you were given | |
13:07 | . Luckily you made a nice substitution , you wrote | |
13:10 | it down in your paper and you should always write | |
13:12 | it on your paper , you let Z equals this | |
13:14 | . So then what I can do since Z was | |
13:16 | equal to this to begin with , as I go | |
13:19 | , put it in here , Z can be substituted | |
13:22 | back into these answers because that's what Z is equal | |
13:24 | to its equal to this . So then if I | |
13:26 | go and take this guy and stick them back in | |
13:29 | here , then I'm gonna have X plus three is | |
13:32 | equal to one because I just stick this on the | |
13:35 | left hand side , it's equal to one . Now | |
13:37 | I move the three over by subtraction , X is | |
13:39 | equal to negative two and I do the same substitution | |
13:43 | here . X plus three goes on . The left | |
13:45 | hand side is equal to four . And when I | |
13:47 | subtract this four minus three means I'm gonna have a | |
13:50 | one . So the two answers I get is X | |
13:52 | is equal to negative two , and X is equal | |
13:54 | to positive one . So that is the concept of | |
13:57 | substitution right ? You're given an equation that looks very | |
14:01 | complicated to solve . Remember the only other alternative to | |
14:03 | solving this equation is to to literally do foil and | |
14:06 | blow this term out and then to multiply and distribute | |
14:09 | this out and then to collect all the like terms | |
14:11 | and you would have a polynomial there that you could | |
14:13 | then solve and you would get these answers . But | |
14:16 | this is much faster because since you recognize it's a | |
14:18 | quadratic form if you do the substitution , you get | |
14:20 | a nice quadratic equation which is very very simple for | |
14:23 | you to solve . And so that's the method that | |
14:26 | we're going to use going forward when you get down | |
14:27 | and you get your answers in terms of Z . | |
14:29 | Then you substitute back in to get the answers in | |
14:32 | terms of X . So now we're gonna go through | |
14:35 | the rest of the lessons solving several other problems . | |
14:38 | Some similar to this , some a little more complicated | |
14:40 | but they will all have the same thing in common | |
14:42 | . You have to make a substitution solve and then | |
14:45 | do another substitution and to get to the final answer | |
14:47 | . And by the way , this whole idea of | |
14:49 | substitution is basically how it goes in calculus two , | |
14:51 | you have a complicated calculus equation and you make a | |
14:54 | substitution to make it simpler and then you go on | |
14:56 | with the solution from there . So let's move on | |
14:58 | and continue working some of these problems in algebra . | |
15:02 | Alright for our next problem , I'm going to call | |
15:05 | this one B . And you'll understand why I'm calling | |
15:07 | it one A and one B in the second . | |
15:08 | Let's take a different equation . Let's say it's two | |
15:11 | x minus one quantity squared minus five times two X | |
15:17 | minus one Plus four is equal to zero . Now | |
15:21 | this is a quadratic equation and quadratic form because you | |
15:24 | have some term squared a constant some term and then | |
15:28 | a constant with just with no term there in that | |
15:31 | quantity . The this equation here is quadratic in the | |
15:35 | quantity two X -1 . So it's the same sort | |
15:37 | of deal . So let's make a substitution to make | |
15:40 | this guy simpler to solve the obvious one is whatever | |
15:42 | is being squared here , we're gonna let Uh some | |
15:46 | variables that we're gonna make it up equal to two | |
15:48 | X -1 . Now , when we do that we | |
15:52 | get a new quadratic equation back . We just put | |
15:54 | it in here . We're gonna have at that point | |
15:55 | , Z squared minus five times E plus four equals | |
15:59 | zero because we just take this and everything is right | |
16:01 | here . So it's Z squared and Z . Here | |
16:03 | and there's no Z there at all . Now we | |
16:04 | know how to solve this guy . We're gonna try | |
16:06 | anyway . First thing you always try is to factor | |
16:09 | , it may not always work out if it if | |
16:10 | it isn't factory , well then just use the quadratic | |
16:12 | formula . So you have a Z . Here in | |
16:14 | the Z . Here two times two is four . | |
16:16 | But wait a minute . I'm thinking , hey , | |
16:17 | this looks pretty familiar . What's going on ? If | |
16:20 | I go back once I did this substitution here , | |
16:23 | I got Z squared minus five , Z plus four | |
16:25 | . Here I did a different substitution and I got | |
16:28 | Z squared minus five Z plus four . It's exactly | |
16:31 | the same equation in Z . Why is that ? | |
16:33 | It's because this equation Is very similar to the other | |
16:36 | one . This one is uh this thing squared minus | |
16:40 | five times this thing plus four . This thing squared | |
16:43 | minus five times this thing plus four . You see | |
16:45 | all the coefficients of the equation . Both of them | |
16:48 | are the same . It's just the thing that's it's | |
16:50 | in quadratic in different functions in different terms . So | |
16:54 | the form of the equation is the same . So | |
16:55 | I want you to your substitution , you actually get | |
16:57 | the same thing back so I don't really have to | |
16:59 | go through this but I'm just gonna do it anyway | |
17:01 | one times four and then minus minus . That's what | |
17:04 | we did last time . So if we set this | |
17:05 | equal to zero and we set this equal to zero | |
17:07 | and solve , we're gonna get Z is equal to | |
17:09 | one here and we're gonna get Z is equal to | |
17:11 | four from here . Exactly the same thing as before | |
17:14 | . Now here is where it diverges because now to | |
17:17 | get the actual answer we have to take this value | |
17:19 | of Z and put it into the left hand side | |
17:21 | . So what we'll have Is two X -1 is | |
17:24 | equal to one here . And when we put that | |
17:27 | in over here we'll have to x minus one is | |
17:30 | equal to four . Now I can move the one | |
17:32 | over giving me two , X is equal to two | |
17:35 | because I add one to both sides and then I | |
17:37 | can divide by two , so I'll get to over | |
17:39 | two , so X is equal to one . And | |
17:42 | then over here , when I move this over I'll | |
17:43 | get a 52 X is equal to five because four | |
17:47 | plus one is five . And then when I divide | |
17:49 | by two , I'll get five house . So let | |
17:51 | me just double check myself . I have an answer | |
17:53 | of one and five halves and these are the correct | |
17:55 | answers . So notice we get two answers because we | |
17:58 | had an equation that was quadratic in nature , right | |
18:01 | ? We had the highest power was a square . | |
18:03 | Even if you foil this thing out , the highest | |
18:05 | power you're going to get us a square term . | |
18:07 | So because the highest power is a square , we | |
18:09 | expect to solutions , we get the two solutions here | |
18:12 | . If on your test , you circle these two | |
18:14 | values of Z . You're going to get the wrong | |
18:16 | answer . If you on your test you circle these | |
18:19 | two values of Z here , the one in the | |
18:21 | forum on this problem , Same thing . You're gonna | |
18:23 | get the wrong answer why ? Because this equation is | |
18:25 | not an equation and Z at all , we just | |
18:27 | use it as a tool . It's like whenever you | |
18:29 | fix the sink , you got to get a wrench | |
18:31 | out . You know , you gotta get some putty | |
18:33 | or maybe some screwdriver . Those are just your tools | |
18:35 | . So the substitution thing is just a tool . | |
18:37 | We use it . We pull it out of the | |
18:39 | bag to make this equation simple . Then when we | |
18:41 | solve it , this is just an intermediate answer . | |
18:44 | It's not the answer to the problem , It's just | |
18:46 | intermediate to help us get to the actual answer . | |
18:49 | Again , this is many steps shorter than blowing this | |
18:52 | whole thing out and going from there to find the | |
18:55 | solution . So we're gonna do one more and I'm | |
18:57 | gonna call this problem one C . And you'll see | |
18:59 | why I'm calling it one . See in a second | |
19:01 | , what if I give you X to the fourth | |
19:03 | power -5 times x squared Plus four is equal to | |
19:08 | zero . Now you might look at this and say | |
19:10 | wait a minute . This is completely different than anything | |
19:12 | we've learned because you have 1/4 power to this equation | |
19:15 | and it's true . We've we've not really learned how | |
19:18 | to solve for the fourth power polynomial . But what | |
19:21 | you do know is that you should expect for answers | |
19:24 | because the highest power of this guy is gonna always | |
19:26 | dictate how many solutions you get how many crossing points | |
19:29 | you get now . Again , sometimes they could be | |
19:31 | real . Sometimes it could be imaginary , but you | |
19:33 | should always expect for answers because of the highest power | |
19:36 | for here . So what you should do is start | |
19:39 | thinking about this . Okay , I'm gonna show you | |
19:41 | uh kind of one thing here and then you may | |
19:43 | or may not do it on your own paper , | |
19:45 | but you can rewrite this as follows this . X | |
19:48 | to the fourth power . When you think about it | |
19:49 | , you can write it as X squared Quantity squared | |
19:54 | . Would you agree that this is the same thing | |
19:56 | as X to the 4th power ? Yes . Because | |
19:58 | when you have a exponent raised to another exponent , | |
20:00 | you multiply the exponents . So this is X to | |
20:02 | the fourth power . Okay , then , for this | |
20:05 | , next thing you say , well , I have | |
20:06 | a five and then I'm gonna put in princes just | |
20:09 | to make it clearer . Five X . Where would | |
20:11 | you agree that this term is exactly the same as | |
20:13 | this one ? Yes , of course . Because you | |
20:15 | just take the parentheses away and you have what you | |
20:16 | have and then I have plus four is equal to | |
20:19 | zero . So the one thing I want to point | |
20:21 | out to you , I'm just writing it this way | |
20:22 | to show you that this equation is quadratic in the | |
20:26 | quantity X squared . Why ? Because when you go | |
20:31 | back here , you know this one's quadratic in two | |
20:33 | X minus one . Because you have two x minus | |
20:35 | one squared minus 52 X minus one . And then | |
20:38 | nothing here . This guy is quadratic in the quantity | |
20:41 | X squared . Because X squared is what squared ? | |
20:44 | And then you have X squared to the first power | |
20:46 | . So you have second power . First power . | |
20:48 | Zero power . It's exactly the same form of a | |
20:51 | quadratic equation . You have to have something squared , | |
20:54 | something to the first , something to the zero , | |
20:56 | which means it disappears , these coefficients are the same | |
20:59 | . So because of the way this is laid out | |
21:01 | , you obviously want to let you want to let | |
21:05 | the quantity Z . To solve this thing equal to | |
21:10 | X squared ? Why ? Because when you substitute it | |
21:13 | in , you're gonna have Z Squared -5 times z | |
21:19 | plus four . Yes , is equal to zero . | |
21:22 | Make sure you understand that . So we've transformed this | |
21:24 | thing . I could have skipped this step and just | |
21:26 | told you to let Z is equal to this . | |
21:28 | But I think it's really instructive to write it out | |
21:30 | so that you can see it . But you think | |
21:31 | about this if Z is going into here X squared | |
21:35 | square , that gives me the fourth power . This | |
21:37 | going in here five X square , That's what's here | |
21:39 | . And then you have the four right there . | |
21:41 | So now we have to solve this guy uh again | |
21:44 | the same as we always have . But notice that | |
21:46 | this equation is z squared minus five , Z plus | |
21:49 | four . It's exactly the same thing as we have | |
21:51 | here . Five Z squared minus five Z plus four | |
21:54 | . Again , the reason it's the same is just | |
21:56 | because these coefficients are the same . That's why I | |
21:58 | have them , one , A one B and one | |
22:00 | C . Because all of these equations look different . | |
22:02 | But they're all have the same basic form because they | |
22:05 | all have the same coefficients . So I don't even | |
22:07 | have to factor this thing , I've solved it so | |
22:09 | many times . I already know that . Let me | |
22:12 | go what I want to do this , let's go | |
22:14 | over here . I already know that Z can have | |
22:17 | two values , Z can be equal to one and | |
22:21 | Z can be equal to four . Those are the | |
22:23 | two values . Those are the two values I got | |
22:25 | before . But I have this uh substitution meaning uh | |
22:30 | that Z Z is equal to X square . So | |
22:35 | I can substitute this guy back in here . Let | |
22:37 | me switch colors over here . So what I'm gonna | |
22:39 | do is put it over here and say X squared | |
22:41 | is equal to one . Now you already know how | |
22:43 | to solve this . I take the square root of | |
22:45 | both sides . When I take the square to both | |
22:47 | sides , I'm gonna have an X on the left | |
22:48 | . I got to insert my plus or minus and | |
22:51 | the squared of one . So X is going to | |
22:53 | be equal to squared of one is one . So | |
22:54 | you get plus or minus one , that's two values | |
22:56 | plus one and minus one . Here . When I | |
22:59 | make the same substitution , Z being x squared , | |
23:02 | I'm gonna have X squared is equal to four . | |
23:05 | So the same thing , I'm gonna take the square | |
23:06 | root and it's gonna be plus or minus the square | |
23:08 | root of four and I know what this is , | |
23:11 | X is going to be equal to plus or minus | |
23:12 | square to four is the number two . This is | |
23:14 | two values here . So the answer is that you | |
23:16 | get is plus one in minus one and plus two | |
23:19 | and minus +24 values for 1/4 order polynomial . That's | |
23:22 | exactly what you expect plus or minus to plus or | |
23:25 | minus one for your answers . So we have more | |
23:28 | problems left in this section , but it is important | |
23:30 | for you to understand that the main concept of what | |
23:33 | we're doing is now in your hands , you already | |
23:35 | know now the main idea . All of the other | |
23:37 | problems that we're going to do are just gonna be | |
23:39 | slightly more complicated . Maybe a few more steps . | |
23:42 | But ultimately you have to identify something to be substituted | |
23:45 | to transform the parent equation into a child equation that's | |
23:49 | much simpler to solve . And then you might need | |
23:52 | to do another substitution to bring it back in to | |
23:56 | the variable that you have to soft forward to begin | |
23:58 | with . So let's do one more . We have | |
24:02 | , I should say two more X to the fourth | |
24:04 | power minus three times X squared minus four equals zero | |
24:09 | . Again , I have 1/4 power , a second | |
24:12 | power and then a nothing power , which is very | |
24:14 | similar to what I had before . 1/4 power , | |
24:17 | a second power and then a zeroth power . So | |
24:20 | I can write this or I can think about this | |
24:23 | as I can think about it as X squared quantity | |
24:27 | squared minus three , X squared minus four . That's | |
24:32 | just the way I think about it . It's something | |
24:33 | squared , something to the first , something to the | |
24:35 | zero . So I'm gonna make this substitution the same | |
24:39 | one that I just did . I'm gonna let Z | |
24:41 | is equal to X squared , I'm gonna let Z | |
24:45 | is equal to X square . And when I do | |
24:47 | that I'm going to transform this into this equation is | |
24:51 | going to be Z squared minus three times E minus | |
24:55 | four is equal to zero . Now I need to | |
24:57 | try to solve this . I can do the quadratic | |
24:59 | formula but I always try to factor first Abbas scene | |
25:03 | . I have a Z . I could do two | |
25:04 | times too , but that's not gonna be able to | |
25:06 | address of track to give me three . So instead | |
25:08 | I'm gonna do one times four and I'm gonna try | |
25:10 | to get the signs right . So I have a | |
25:12 | negative sign here . So the only way it's gonna | |
25:13 | work is like this and you should always check yourself | |
25:16 | by the way , the interior terms are gonna give | |
25:18 | me positive Z . The exterior terms negative four times | |
25:22 | the positive Z will give me negative four . Z | |
25:24 | . When I add these I'll get the negative three | |
25:26 | . Z . So I have Z times Z . | |
25:28 | A Z square the interior and exterior terms give me | |
25:31 | this . And then these multiply to give me negative | |
25:33 | four . So that's the correct factor form . So | |
25:36 | then I have Z plus one equals zero or z | |
25:41 | minus four . Can also be equal to zero . | |
25:43 | So I just solve and say that Z can be | |
25:45 | equal to negative one and Z can be equal to | |
25:47 | four . Now a lot of students will stop here | |
25:49 | in circle but that's wrong because the answer is not | |
25:51 | Z equals anything . The answer is you want X | |
25:53 | . That's what your problem was given to you in | |
25:55 | . So I'm gonna take and substitute this in . | |
25:57 | Z . Is equal to X squared . So I'm | |
25:59 | gonna switch colors to do that . And then what | |
26:02 | I'm going to have is X squared is negative one | |
26:05 | . And then over here I'm gonna have because Z | |
26:07 | is equal to X squared X squared is four . | |
26:10 | Now let's work on this one first . Let's work | |
26:12 | on this one first . To solve this , I'm | |
26:14 | gonna have to take the square to both sides , | |
26:17 | right . I'm gonna have to take a square to | |
26:18 | both sides . So I'm gonna take the square root | |
26:20 | of the left and the right . I'm gonna have | |
26:22 | to insert my plus or minus square root before the | |
26:25 | square in the square root cancel giving me X . | |
26:27 | Here . So over on the right hand side have | |
26:29 | plus or minus two . Now on this side I | |
26:33 | have to take the square root of both sides . | |
26:34 | So that's going to reveal the X . Here plus | |
26:36 | or minus the square root of negative one . So | |
26:39 | we've done complex and imaginary numbers by now . So | |
26:41 | you know the square root of negative one is just | |
26:43 | I . So that's plus or minus I . So | |
26:46 | the two answers I get here are plus or minus | |
26:48 | I . And the two answers I get here are | |
26:49 | plus or minus two . So altogether actually have four | |
26:52 | answers , which is exactly what I expect . So | |
26:56 | if you were to graph this thing , what you | |
26:57 | would find is that you have two crossing points on | |
26:59 | the axis Which are at plus or -2 . These | |
27:03 | other two are imaginary and we have already explained . | |
27:06 | I've already done another lesson and what imaginary roots mean | |
27:09 | . It just means that in the complex plane , | |
27:11 | the crossing point , so to speak , on the | |
27:13 | imaginary side of things . So the real crossings are | |
27:15 | here , the imaginary crossings are here . And if | |
27:18 | you're more curious about why sometimes you get imaginary roots | |
27:20 | , go look at my lesson on on what imaginary | |
27:23 | roots mean . And I go through a lot of | |
27:25 | 45 minutes of teaching you what that means . All | |
27:28 | right . But the bottom line is you get four | |
27:29 | answers for 1/4 order equation . So notice that the | |
27:32 | form of this was something to the 43 something to | |
27:35 | the squared . And then this guy uh constant . | |
27:38 | And now I want to do one last equation which | |
27:41 | looks very , very different . Um but actually turns | |
27:45 | out to be similar . So this last problem is | |
27:47 | going to be x minus three times the square root | |
27:50 | of x minus four equals zero . Now this equation | |
27:54 | has no squares in it , anywhere . There's a | |
27:57 | radical in it . Uh it looks very different than | |
28:00 | this equation , totally different than any of these equations | |
28:02 | right now . Most people , including me , honestly | |
28:06 | , if I looked at it , I would probably | |
28:07 | not know what to do . But the truth is | |
28:10 | you can try different substitution , you can try any | |
28:13 | substitution you want , you can make Z equals two | |
28:16 | squared of X plus two plus five minus seven , | |
28:19 | anything you want . But if it doesn't make the | |
28:21 | problem simpler , you're wasting your time . So my | |
28:24 | advice is if you're given a weird equation and you | |
28:26 | want to try substitution , just try something . If | |
28:28 | it falls out to be a nice quadratic , you've | |
28:30 | made the right choice . This one is not clear | |
28:33 | , it's not it's not setting off alarm bells . | |
28:36 | But what I want you to do is let Z | |
28:40 | equal to the square root of X . Y squared | |
28:42 | of X . Because it's the only weird thing in | |
28:44 | here and we know if we have a square root | |
28:46 | running around in this polynomial , it's gonna be very | |
28:49 | hard to solve . So by doing a substitution , | |
28:51 | we're going to try to remove the radical from our | |
28:54 | simpler equation down below by setting it equal to Z | |
28:57 | . Let's just try , we may not get what | |
28:58 | we want to go , but we're gonna try now | |
29:01 | when you set Z equal to X squared . This | |
29:04 | also means if you square both sides of this equation | |
29:07 | , that Z squared is equal to X , right | |
29:09 | ? Because if you let this happen and you square | |
29:11 | both sides , Z squared squared of X squared gives | |
29:14 | you this . This is exactly equivalent and comes from | |
29:18 | this substitution that you have just made . So why | |
29:20 | do I do that ? Because I have an X | |
29:21 | . Here . But I know that X Z squared | |
29:24 | because of what I've just done . So then by | |
29:26 | putting in next here , I'm going to actually have | |
29:28 | a first term which is Z squared . And then | |
29:31 | it's gonna be -3 times the square root of X | |
29:33 | . But square root of X was just see And | |
29:36 | then I'm gonna have -4 is equal to zero . | |
29:38 | Now this should blow your mind because the original equation | |
29:41 | is very ugly . No , I don't know how | |
29:42 | to solve that . But by doing a clever substitution | |
29:46 | it actually turns out to be a very nicely behaved | |
29:48 | quadratic Z squared minus three Z minus four . Which | |
29:52 | actually turns out to be exactly the one we got | |
29:54 | here , Z squared minus three Z minus four . | |
29:57 | So I guess I'm doing this to show you that | |
29:59 | looks can be deceiving sometimes sometimes you can do these | |
30:01 | things and it's very obvious what to substitute sometimes , | |
30:04 | like in this one you don't have any idea . | |
30:06 | But when you when you try it and you get | |
30:08 | a simpler equation then you just grow rock and roll | |
30:12 | with it because you know that's what you do . | |
30:15 | So here we're gonna try to factor and we know | |
30:18 | it . It's fact trouble because we just did it | |
30:20 | . So we have Z . We have Z . | |
30:21 | We have one , we have four and we have | |
30:24 | plus and we have minus . And we've already verified | |
30:26 | this is right . So we know that Z is | |
30:28 | equal to negative one . That comes from sub setting | |
30:31 | that equal to zero . And we know that Z | |
30:32 | is equal to positive four . Because these are exact | |
30:35 | same answers we got before now , where it gets | |
30:37 | a little bit tricky is going from here to the | |
30:39 | answer Z is -1 , Z is square root of | |
30:44 | X . So I can substitute that squared of X | |
30:46 | into here , square root of X is equal to | |
30:48 | negative one . Now , before I solve anything further | |
30:51 | , let's go over here , Z squared of X | |
30:53 | . I put it into here , square root of | |
30:54 | X is equal to four . Now , what I'm | |
30:59 | gonna do is work on this one first and you'll | |
31:01 | see why in just a second , square root of | |
31:03 | X is equal to four . I mean you can | |
31:05 | think of it several ways . I mean it's nice | |
31:07 | even number . So you already know the square root | |
31:09 | of some number has to be four . So we | |
31:13 | know the square root of 16 is four . So | |
31:14 | you already know that X is 16 . But if | |
31:16 | you didn't know that , what the way you would | |
31:18 | do it is you would square both sides of this | |
31:20 | equation . right ? So you would say square root | |
31:23 | of X , quantity squared Is for square , because | |
31:27 | I'm doing the same thing to both sides , so | |
31:29 | I'm gonna have X is equal to 16 . And | |
31:30 | you know that that's right , because when you put | |
31:32 | 16 in here , that's what you get . So | |
31:34 | you get an answer of 16 , I'm gonna circle | |
31:36 | that answer , right ? But then I go over | |
31:39 | here and I try to apply the same logic . | |
31:43 | So your first thought is going to be , well | |
31:45 | , I'm gonna solve this guy , I'm gonna square | |
31:46 | this and square this . The problem is you're gonna | |
31:49 | here's where you gonna run into problems you see here | |
31:52 | , I'm getting the square root of some number equals | |
31:55 | a positive for and I know that that exists . | |
31:57 | I know that I can take square roots of numbers | |
31:59 | and get positive answers , but here , there is | |
32:02 | no value of X . Where I can take the | |
32:04 | square root of it and get a negative number back | |
32:06 | . Square roots never give you negative numbers back , | |
32:10 | Right , the square root of 16 is four , | |
32:12 | but the square root of -16 doesn't give you negative | |
32:15 | for it gives you four I which is imaginary . | |
32:18 | So there is no way to take the square root | |
32:20 | of anything and get a negative number out . So | |
32:22 | this equation , even though you want to solve it | |
32:25 | to get the other half of the solution , there's | |
32:26 | nothing else to do . This is this is basically | |
32:29 | impossible . Um This is uh I'm just gonna put | |
32:32 | no solution to the side of it . There is | |
32:35 | no value of X that you can put in here | |
32:38 | to make it work and sometimes you just have to | |
32:40 | look at what you're doing to recognize that's the case | |
32:42 | . But even if you didn't do that , let's | |
32:45 | say you just ignored that warning or whatever and you | |
32:47 | just decided to try it anyway , then what you | |
32:50 | would get here is you'd square both sides . So | |
32:51 | you would get rid of the square root and you | |
32:53 | have an X . And you would have a negative | |
32:55 | one squared because you'd square this , you square this | |
32:57 | . So you would say X is equal to one | |
32:59 | , and you would probably circle this answer and then | |
33:01 | you'd also circle this answer and you get two answers | |
33:03 | and you think you're done . But here's the problem | |
33:06 | . If you go back up to the original problem | |
33:08 | and put it one into this equation , does it | |
33:11 | work ? One minus three times a squared of one | |
33:14 | is just gonna give you one . So one minus | |
33:16 | three is negative to negative , two minus four is | |
33:19 | negative six . Do that again , one minus 23 | |
33:22 | is negative to negative two minus 24 is negative six | |
33:25 | . So when I put this in here it doesn't | |
33:27 | work . So even if you don't see it here | |
33:30 | , when you get down to the end , you | |
33:31 | know technically you're always supposed to be substituting your answers | |
33:35 | back in most of the time , we don't substitute | |
33:37 | the answers that we get back in because for the | |
33:39 | regular quadratic equations , everything is well behaved . But | |
33:42 | when you have radicals in here and you start squaring | |
33:44 | things , then you can introduce other solutions . And | |
33:47 | so this basically is not true and you can see | |
33:50 | it by looking at this equation realizing there's no solutions | |
33:53 | of this equation . So this equation actually only has | |
33:56 | one root . Now , you might also look at | |
33:59 | it and say , well wait a minute . I | |
34:00 | thought all these quadratic equations have two routes because they're | |
34:03 | quadratic right ? So I got curious about this one | |
34:06 | and I graphed it if you look at the original | |
34:08 | equation , which is this equation , this this weird | |
34:10 | one up here . This is what it basically looks | |
34:12 | like . So you have and access like this , | |
34:15 | here's X . Here's why so f of X is | |
34:18 | equal to this minus this minus four . That's what | |
34:20 | you're solving , why is equal to x minus three | |
34:24 | . Route X minus four . This is the equation | |
34:26 | , your graphic , what it looks like is this | |
34:30 | it starts here and it kind of goes up Like | |
34:34 | this and this crossing point is X is equal to | |
34:36 | 16 , which is exactly what we calculated . That's | |
34:39 | the crossing point , the route . But the graph | |
34:41 | doesn't even go past this axis . Uh here . | |
34:44 | Why ? Because this is all positive values of X | |
34:46 | going in . But here at the negative values of | |
34:49 | X , there's no there's nothing the function doesn't exist | |
34:52 | for real numbers because if you put a negative number | |
34:54 | in here squared of negative number gives you an imaginary | |
34:57 | answer . So when you're plotting the thing on a | |
34:59 | real access , the function stops here . It doesn't | |
35:01 | even continue to to another crossing point because it literally | |
35:05 | stops dead in its tracks . Because for any negative | |
35:08 | value of X that you put in the function becomes | |
35:11 | imaginary , it doesn't exist for as a real number | |
35:14 | . So You get two answers technically quote unquote because | |
35:20 | of the quadratic because of the substitution that you did | |
35:23 | created a quadratic equation that had two answers . These | |
35:26 | two answers but these two answers didn't correspond to two | |
35:29 | answers . In the original problem . I guess my | |
35:31 | point is when you do substitution right ? Technically you | |
35:35 | should always take the two answers you get and plug | |
35:37 | them both back in and verify . They're both correct | |
35:39 | . But the real story of it is for all | |
35:41 | of these other problems where everything is well behaved , | |
35:43 | they're always going to work fine because the process doesn't | |
35:46 | involve any squaring or any radicals . It's really only | |
35:49 | when you have squaring and radicals going on that you | |
35:52 | can introduce additional solutions . Extraneous solutions That aren't really | |
35:56 | a solution of the parent equation and that's why I | |
35:59 | wanted to do this problem . So you get the | |
36:01 | two values of Z , you get the two solutions | |
36:03 | technically you would plug them back in . But here | |
36:05 | we even realize this can't have anything so we can | |
36:07 | throw this away . But even if you did , | |
36:09 | you'd realize it doesn't work . If you stick to | |
36:11 | 16 in there , you'll see it works fine because | |
36:13 | you have 16 minus three . This scored a 16 | |
36:16 | 4 , so 16 minus three times four is 12 | |
36:20 | , so 16 minus four . I'm sorry 16 minus | |
36:22 | 12 is 44 minus four is zero . So this | |
36:25 | value works , this value doesn't and I'm trying to | |
36:27 | show you what the equation . So here is the | |
36:30 | idea of solving equations in quadratic forms . Do a | |
36:33 | substitution . If it makes the equation look like a | |
36:36 | nice , well behaved quadratic , solve it and then | |
36:39 | you have to do another back substitution at the end | |
36:41 | to get the final answers . And then as a | |
36:43 | check , you really should be putting everything back in | |
36:45 | to make sure . But I'm just telling you that | |
36:47 | if the equations are well behaved , they're always going | |
36:49 | to work fine . You have radicals are squaring involved | |
36:52 | . Uh then I would always check both solutions to | |
36:54 | make sure . So make sure you understand this . | |
36:56 | Follow me on to the next lesson . We're not | |
36:58 | done yet . We're going to do more complicated problems | |
37:00 | to get you very comfortable with solving these equations in | |
37:03 | quadratic form . |
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