01 - Solving Equations in Quadratic Form - Part 1 (Learn to Solve Equations in Algebra) - Free Educational videos for Students in K-12 | Lumos Learning

## 01 - Solving Equations in Quadratic Form - Part 1 (Learn to Solve Equations in Algebra) - Free Educational videos for Students in k-12

#### 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
00:19 equations . Why do we continue to continue to work
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
00:55 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
01:22 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
02:35 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
05:07 . 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
07:13 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
09:40 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
14:29 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
15:24 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
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
18:19 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
19:41 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
21:35 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
23:28 problems left in this section , but it is important
23:30 for you to understand that the main concept of what
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
24:23 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
25:08 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
33:03 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
33:35 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
34:06 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
35:29 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
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