09 - The Quadratic Formula Explained, Part 1 (Practice Problems & Solutions) - By Math and Science
Transcript
00:00 | Hello , welcome back to algebra . The title of | |
00:02 | this lesson is called the quadratic formula Part one . | |
00:05 | Now I have two objectives in this lesson . Both | |
00:07 | of them are very important . One objective is that | |
00:10 | I want you to be able to use this quadratic | |
00:11 | formula to find the solution to any quadratic equation to | |
00:14 | be able to actually use the quadratic formula that's in | |
00:17 | every algebra text book ever written . And the second | |
00:20 | thing I think actually maybe even more important than that | |
00:23 | is I want you to understand where the quadratic formula | |
00:26 | actually comes from because most books almost every book I've | |
00:30 | ever seen just kind of throws it in your lap | |
00:31 | and says here's the quadratic formula and then you use | |
00:35 | it , you learn how to use it but you | |
00:36 | don't really know where it comes from . So it | |
00:37 | just seems like this magical thing that really smart people | |
00:40 | figured out a long time ago . The truth is | |
00:42 | you have the knowledge now to understand yourself where the | |
00:45 | quadratic formula comes from . So what I'm gonna do | |
00:47 | in the first half of this lesson is derive it | |
00:49 | and show you where it comes from , I encourage | |
00:52 | you even if you don't care where it comes from | |
00:53 | to watch me go through it because it will give | |
00:56 | you skills and practice and doing a nice derivation and | |
00:59 | also at the end of it you'll be really proud | |
01:00 | of yourself for understanding one of the most famous formulas | |
01:03 | and all of math . And then at the end | |
01:05 | of that we'll use the quadratic formula to solve some | |
01:07 | problems in this lesson . And in the remaining lessons | |
01:10 | . So we all know that we can take any | |
01:13 | quadratic equation and we can use the completing the square | |
01:16 | method to solve any quadratic equation . It turns out | |
01:20 | that if you apply that , completing the square operation | |
01:24 | to a general quadratic equation with general values , then | |
01:28 | you can derive this thing that we call the quadratic | |
01:30 | formula , which then we normally just use the quadratic | |
01:32 | formula to find all of the solutions going forward from | |
01:35 | here on out . And so the quadratic formula , | |
01:39 | what we're talking about , So quad radic formula . | |
01:46 | So I think it's really important when I can to | |
01:49 | show you where some of these things come from Now | |
01:51 | . The quadratic formula is one of the most famous | |
01:53 | formulas . You know , everybody learns this guy and | |
01:56 | so what it is is for any quadratic equation this | |
02:02 | means quadratic equation that looks like this A times X | |
02:06 | squared plus B , times X plus C is equal | |
02:10 | to zero . In other words , we've been solving | |
02:12 | lots of quadratic equations with completing the square . Maybe | |
02:15 | it was two x squared plus three X plus four | |
02:18 | . Maybe it was negative five X squared minus three | |
02:21 | plus one equals zero . In other words A and | |
02:24 | B . And C . Or just numbers . And | |
02:26 | they can be any numbers . And if you choose | |
02:28 | any random numbers for A and B and C . | |
02:30 | Then you have different quadratic equations and we know how | |
02:33 | to solve all of them because completing the square operation | |
02:36 | makes it so that I can factor the left hand | |
02:38 | side of any of those things and solve them . | |
02:40 | We've done tons of examples using completing the square . | |
02:43 | If you haven't watched my lessons on completing the square | |
02:45 | , I recommend you go do that because that's what | |
02:47 | we're gonna do basically here . But the quadratic formula | |
02:50 | says as you will see it displayed in most of | |
02:53 | your books that the solutions to this uh is as | |
02:57 | follows , the solution is something like that like this | |
03:00 | X . Is equal to negative B plus or minus | |
03:04 | the square root of B squared minus four times a | |
03:09 | times C . And this whole monster numerator is over | |
03:12 | the denominator of two times a . Now , one | |
03:14 | thing I'm gonna tell you right away is that you're | |
03:16 | going to get used to using the quadratic formula because | |
03:18 | you use it all the time . A lot of | |
03:20 | students will say , well we already learned completing the | |
03:22 | square . Why do we need this quadratic formula thing | |
03:24 | ? Well , the truth is completing the square is | |
03:26 | a really useful thing that we learn to learn how | |
03:28 | to use algebra . But once you learn the quadratic | |
03:31 | formula and where it comes from , there really isn't | |
03:33 | much of a need to ever really use completing the | |
03:35 | square too much because we can just plug the values | |
03:38 | into this equation and get the solutions directly instead of | |
03:41 | going through the whole process of completing the square . | |
03:43 | In other words , if I know any a any | |
03:45 | be in any seat for any quadratic , I just | |
03:47 | put them in these locations and then calculate the answer | |
03:51 | and I get always two solutions because you have the | |
03:53 | plus and the minus . And if you remember back | |
03:55 | when we did the completing the square , all of | |
03:57 | those solutions had a plus and a minus there as | |
03:59 | well . And that's because of the way we did | |
04:02 | the solution technique when you take the square root of | |
04:04 | both sides of something , you have to insert the | |
04:05 | plus and minus and we get all that . We've | |
04:08 | done all that before and you can see that for | |
04:10 | completing the square , we always got those two answers | |
04:12 | . So you can already see the similarity here because | |
04:14 | we have a plus and a minus and here that | |
04:16 | is ultimately going to come from completing the square . | |
04:19 | So this is the famous quadratic formula , so I'll | |
04:21 | go ahead and box it . It's the most important | |
04:23 | formula that would probably have had to pick one . | |
04:25 | I would say the quadratic formula is probably the most | |
04:26 | important formula in algebra . Most students don't know where | |
04:29 | it comes from . So let's figure this out . | |
04:31 | Where does it come from ? You have the knowledge | |
04:33 | , let's take five minutes and figure out where one | |
04:34 | of the most famous formulas and math actually come from | |
04:37 | . So here we have a general quadratic equation , | |
04:39 | X squared plus bx plus C is equal to zero | |
04:44 | . If I told you , how do you solve | |
04:46 | this equation using completing the square , what would you | |
04:48 | do ? A lot of students would just freeze up | |
04:50 | because they have all these letters everywhere . They don't | |
04:52 | know what to do . But in your mind just | |
04:53 | pretend that it was like two x squared plus three | |
04:56 | X plus four is equal to zero . Well , | |
04:58 | the first thing you would do is you would take | |
05:00 | the constant term . This term here is this is | |
05:02 | the square term , this is the X term . | |
05:04 | This is the constant term . The first thing you | |
05:06 | do is take that square term and you would move | |
05:08 | it to the other side of the equal sign . | |
05:09 | That's what we always do when we start to complete | |
05:11 | the square . So what we would get from this | |
05:13 | step is the following . All we do is subtract | |
05:17 | see from the left so it disappears and we subtract | |
05:20 | it from the right . So we get a negative | |
05:21 | . See , the next thing we do is we | |
05:23 | take a look at the coefficient in front of the | |
05:25 | X square term . When we did the completing the | |
05:27 | square , we said we have to have this coefficient | |
05:29 | equal to one . If it's not equal to one | |
05:32 | , which in this case it's a it's not equal | |
05:34 | to one . We have to divide both sides of | |
05:36 | the equation by a to get rid of it . | |
05:38 | So what we'll do is we'll say , well we | |
05:39 | have a X squared plus B . X equals negative | |
05:43 | . See I've just written the whole thing down again | |
05:45 | , we'll divide the left side by a , well | |
05:47 | divide the right side by because we can do anything | |
05:49 | . We want to both sides to be legal . | |
05:52 | So then what we do is we look at this | |
05:54 | , we want to do the cancellation , we're gonna | |
05:55 | break this up into two fractions like this , we're | |
05:58 | gonna break it up into a X squared over a | |
06:02 | . The plus sign comes from here and then be | |
06:04 | X over A . And on the right hand side | |
06:06 | it's gonna be negative C over A . So all | |
06:08 | I've done here is break it up . Now , | |
06:10 | if this looks unfamiliar to you , if I ask | |
06:12 | you , how do I add these fractions together ? | |
06:14 | Because there are fractions I want to add them . | |
06:16 | How would you go backwards ? Well I have a | |
06:18 | common denominator , so the common denominator of my answer | |
06:21 | should be A . Then I would just take a | |
06:23 | X squared plus bx for the numerator of the answer | |
06:25 | . So if I add these fractions , I get | |
06:28 | this , therefore I can break this up and go | |
06:29 | backwards and get this . And the reason I'm doing | |
06:32 | that is because here the A . Is going to | |
06:35 | cancel with the A . Leaving me only with X | |
06:38 | squared here , I still have A B over A | |
06:41 | . And an X . There . So I'm gonna | |
06:43 | write this like this , I'm gonna write it is | |
06:44 | B over A , times X . It's just the | |
06:48 | coefficient of X . For lack of a better word | |
06:50 | is the fraction B over A . And on the | |
06:52 | right hand side is negative . See over it . | |
06:55 | Now , I'm not gonna lie to you , this | |
06:56 | is an ugly looking equation , but it is no | |
06:58 | different than any of the other equations . We've done | |
07:00 | by completing the square . You know , we've done | |
07:03 | exactly the same steps , we move the constant term | |
07:05 | to the right , then we divide through and make | |
07:07 | this coefficient of X square , just equal to one | |
07:10 | . In the course of doing that , we introduced | |
07:12 | these fractions , which is not nice , I agree | |
07:14 | with you , it's not fun . But the next | |
07:16 | step of the process is very straightforward . What we | |
07:18 | have to do is take the coefficient in front of | |
07:21 | X . For completing the square . We take the | |
07:23 | coefficient whatever it is , we divide it by two | |
07:26 | and then we square it and then we added to | |
07:28 | both sides . That's what we've done for every one | |
07:30 | of these , completing the square problems . So what | |
07:33 | we do , as we say , okay , I'm | |
07:36 | going to let me see what I want to do | |
07:38 | this . Actually , I think I want to do | |
07:39 | it on the next board . Let me rewrite this | |
07:41 | as it sits , I'm gonna rewrite it on the | |
07:43 | next board . It's gonna be X squared plus the | |
07:47 | fraction B over a times X . That's the coefficient | |
07:50 | of X . Is equal to negative C over a | |
07:53 | . Now , for the next step , what I | |
07:54 | need to do is take one half of this uh | |
07:59 | and square it . So what I want to do | |
08:01 | is say , well , what is that equal to | |
08:03 | ? So be over a that's a fraction , right | |
08:06 | ? And I'm going to take that and divide by | |
08:08 | two . And I need to square it . This | |
08:10 | is what I'm gonna have to actually add to both | |
08:12 | sides of the equation . Right ? So what is | |
08:15 | this equal to ? Well , when I divide by | |
08:16 | two , I'm really divided by 2/1 because any number | |
08:19 | is just over one , so I can change this | |
08:21 | fraction division into multiplication and then the 2/1 . When | |
08:26 | I flip it over to multiply becomes one half . | |
08:29 | This whole thing is still squared right ? I have | |
08:32 | to take one half of this coefficient and then square | |
08:34 | . So by dividing it by two , I flip | |
08:37 | it over and multiply by a half . And now | |
08:39 | I can see that this is going to be B | |
08:40 | over two . A . You can already see the | |
08:44 | quadratic formula starting to pop out here . The quadratic | |
08:47 | formula is negative B plus or minus B squared minus | |
08:50 | four . A . C over two . A . | |
08:52 | There's a to A here . And you can already | |
08:54 | see a to a starting to formulate in the answer | |
08:57 | . Of course we're not there yet , but you | |
08:58 | can see the seeds of it is kind of starting | |
09:01 | to happen right here . So this thing d over | |
09:04 | to a quantity squared , come on , is what | |
09:07 | I need to add to both sides of this equation | |
09:11 | up here . So let me switch colors and we're | |
09:13 | gonna add that quantity to both sides . So what | |
09:15 | it's going to be as X squared plus quantity B | |
09:19 | over A , times X plus this thing , which | |
09:22 | is one half of that coefficient squared , So B | |
09:25 | over two A quantity squared is equal to negative C | |
09:29 | over A plus B over two , A quantity square | |
09:34 | . So I just added that to both sides of | |
09:36 | the equation , which is what I have to do | |
09:39 | . Okay , now , the next thing I want | |
09:40 | to do is I want to work on the right | |
09:42 | hand side because ultimately I'm gonna have to add this | |
09:45 | stuff together and combine it into one term . But | |
09:48 | it is ugly the way it sits . So I'm | |
09:49 | going to draw a little arrow here , telling you | |
09:52 | that I'm only now working on this stuff , so | |
09:54 | let's see what it would come out to be negative | |
09:56 | C . Over A . Plus . Now this square | |
09:59 | is gonna apply to the B . And also to | |
10:00 | the to A . So it's going to be B | |
10:02 | . Squared and then it's going to apply to the | |
10:04 | to A . As a unit , which means it's | |
10:07 | going to be four times a square . It's gonna | |
10:09 | apply to the to making it two squared into the | |
10:12 | A . Making it a . Squared . So for | |
10:13 | a squared . All right , now we have to | |
10:16 | add these fractions together . I know it's ugly , | |
10:18 | but we have to do it we have to combine | |
10:20 | them . So what we're gonna have is this is | |
10:22 | gonna be negative C . Over A . And we'll | |
10:24 | be adding it to be squared over for a . | |
10:27 | But I have different denominators here . What can I | |
10:30 | do to make these denominators the same ? Well I | |
10:32 | can first of all realize that in the course of | |
10:35 | this it's actually for a squared on the bottom . | |
10:37 | Sorry about that . It's just for a squared on | |
10:39 | the bottom of that denominator . So in order to | |
10:42 | get a common denominator I need to multiply this one | |
10:44 | by four A . Over four A . Because then | |
10:47 | I'll have four A squared in the denominator . So | |
10:50 | when I do that I'm gonna have negative C . | |
10:53 | Times for a . So negative for a C . | |
10:56 | Notice that this by itself is starting to look reminiscent | |
10:59 | I have a four A . C . In there | |
11:00 | . So I'm kind of feeling like this is looking | |
11:02 | a little bit like the quadratic formula negative four A | |
11:05 | . C on the bottom is four A squared . | |
11:09 | All right . And then I'm adding it to be | |
11:11 | squared over for a squared . So now I have | |
11:15 | a common denominator and so what I'm going to have | |
11:17 | is if you think carefully , the denominator of this | |
11:20 | answer is going to be for a squared and it's | |
11:22 | gonna be negative for a C plus B square . | |
11:25 | But I can flip the whole thing around and say | |
11:27 | this is B squared minus four A . C . | |
11:30 | And that she looked really familiar because the B squared | |
11:32 | minus four A . C . Is exactly what's under | |
11:34 | the radical in the quadratic formula anyway , so I'm | |
11:37 | adding the negative four plus this but I'm just writing | |
11:40 | it backwards as B squared minus four ac . So | |
11:42 | this whole thing is what the right hand side of | |
11:45 | the equal sign is . Now , we've got to | |
11:47 | look and see what is the left hand side . | |
11:49 | Normally , when you complete the square , you get | |
11:51 | to this point and you try to factor you should | |
11:53 | have a perfect square . Um I think I need | |
11:56 | a little more room actually . Um you should have | |
11:58 | a perfect square here . Uh and so we know | |
12:01 | that we're gonna factor , that's the right hand side | |
12:03 | is going to be b squared minus four A . | |
12:06 | C . Over for a squared . And I'm really | |
12:11 | really , really want to be careful that we don't | |
12:13 | get into trouble thinking that this stuff runs together . | |
12:15 | So this is kind of a border here . Now | |
12:17 | on the left hand side , we want to factor | |
12:19 | this and I know you're looking at that thinking that's | |
12:21 | crazy . How do I factor it ? Well let's | |
12:23 | just look at it one step at a time . | |
12:24 | We need the next time , the next to give | |
12:25 | us an X squared the last term of the the | |
12:29 | tri no meal , there is a square term . | |
12:31 | So we need two things to multiply together to give | |
12:34 | us this square . So the only thing that really | |
12:36 | works is B over two , A Times B over | |
12:40 | two a think about it . If you multiply these | |
12:42 | together , you're going to get this to be squared | |
12:45 | the to a squared and everything is a positive signs | |
12:48 | that we need a positive and a positive . So | |
12:50 | we think we factored it correctly , but you should | |
12:53 | always check yourself X times X gives us X squared | |
12:57 | the last terms multiplied together to give us the last | |
12:59 | terms . The inside terms is going to be B | |
13:03 | over two A times X and the outside terms are | |
13:08 | also going to be B over two A times X | |
13:10 | . And we have to add them together . So | |
13:12 | we're gonna have a baby over to a . We | |
13:16 | know that the X . Is going to be here | |
13:17 | . So I'm going to add B over two A | |
13:18 | plus B over two A . And check it . | |
13:20 | We have a common denominator , so we'll get to | |
13:22 | be over to A . Which is B over A | |
13:25 | . So the two is canceled . So be over | |
13:27 | A . X . That's what we got for the | |
13:29 | middle term . I'm doing this to show you that | |
13:31 | this factored form that we have is exactly giving you | |
13:33 | the middle term that you need , right ? So | |
13:36 | then we're getting very close to the punch line . | |
13:38 | These are identical terms . So just like we always | |
13:41 | do in completing the square , it's X plus B | |
13:44 | over two A . We're gonna make it quantity squared | |
13:47 | because they're multiplied times each other on the right . | |
13:50 | We're gonna have B squared minus four A . C | |
13:54 | . Over for a squared . And it doesn't look | |
13:58 | like we've made much progress but actually the only variable | |
14:01 | which is what we're trying to solve the for is | |
14:02 | in here . So to get him by himself , | |
14:04 | we have to take the square root of both sides | |
14:06 | . And this is how the famous radical is going | |
14:08 | to pop up in the quadratic formula here when we | |
14:11 | take the square root , so when we take the | |
14:13 | square root of the left we're gonna have X . | |
14:14 | Plus B over two . A . Because we take | |
14:18 | the square , it's gonna cancel with this on the | |
14:20 | right . We're gonna have plus or minus a giant | |
14:22 | radical around all this stuff B squared minus four A | |
14:27 | . C over for a squared like this , we | |
14:31 | had to insert the plus and minus . When we | |
14:33 | take the square root of both sides . This is | |
14:35 | where the famous plus minus is coming from in the | |
14:37 | quadratic formula . And then what do we do ? | |
14:40 | Let's go ahead and move this now over . So | |
14:43 | X . Is going to then be equal negative B | |
14:46 | over two A . Because we move it over to | |
14:48 | the other side of the equal sign plus or minus | |
14:50 | . And for this radical , I'm gonna make it | |
14:52 | a radical of the top in a radical of the | |
14:54 | bottom . So it's gonna be square root of B | |
14:57 | squared minus four Ac . Now you can see it's | |
15:00 | starting to happen on the bottom , it's going to | |
15:02 | be the square root of four A squared , right | |
15:05 | ? So what I want to do here is I | |
15:08 | think probably the easiest thing to do is just make | |
15:09 | an equal sign because we're getting really close to the | |
15:11 | answer here , so we have a negative B over | |
15:13 | two A . We have negative B over two A | |
15:18 | plus or minus B squared minus four A . C | |
15:22 | . All with the radical . And then when we | |
15:24 | take the square root of four a square , the | |
15:26 | square root of fourth to the square where they squared | |
15:29 | is A . So on the bottom all you get | |
15:31 | to to a . Now notice what you have here | |
15:33 | , you have this fraction plus or minus this fraction | |
15:37 | but the denominators are the same , the denominators are | |
15:40 | the same . So what do you get to see | |
15:42 | if I can squeeze it at the bottom here ? | |
15:43 | You get the following X . Is equal to um | |
15:47 | The denominators are the same . So we can just | |
15:49 | add the numerator . It'll be negative B plus or | |
15:52 | minus B squared minus four times a times C . | |
15:56 | With a radical around all of that over the common | |
15:59 | denominator of two times a . All I do is | |
16:02 | say well I'm adding these together like fractions . So | |
16:03 | we have the same denominator in the answer and I | |
16:06 | add or subtract these enumerators . The plus minus must | |
16:09 | come along for the ride negative B plus or minus | |
16:11 | B squared minus four A C squared of all that | |
16:14 | over to a . That's the famous quadratic formula . | |
16:17 | It's one of the most famous formulas in all of | |
16:19 | algebra . The most useful formulas in all of algebra | |
16:22 | because what it says is if I know any quadratic | |
16:26 | with any A . In any be in any , | |
16:28 | see all I need to do is put the values | |
16:30 | in here and calculate the answers and I will always | |
16:33 | get two answers because they have a plus and minus | |
16:35 | here and so now you can see where it comes | |
16:37 | from because you know it's kind of a lot to | |
16:39 | slog through in the beginning . But now that we've | |
16:41 | done so much completing the square , you can see | |
16:44 | that the quadratic formula comes from taking a general quadratic | |
16:48 | and just completing the square with it . And that | |
16:50 | is why we learn completing the square . But then | |
16:53 | oftentimes students forget about it because once you understand the | |
16:55 | quadratic formula , you really don't have a need to | |
16:58 | do completing the square very much . Because if I | |
17:01 | have a quadratic , I'm just gonna stick it into | |
17:02 | this thing . It's a lot easier to do . | |
17:05 | So in order to show you that let's do a | |
17:08 | couple of quick examples here with how to use this | |
17:10 | famous quadratic formula , for instance , three x squared | |
17:15 | plus x minus one equals zero . Now , if | |
17:19 | I was going to complete the square with this , | |
17:20 | I would have to move the one over and then | |
17:22 | I have divide by three to get all coefficients right | |
17:25 | and I have to take one half of this square | |
17:27 | , it added to both sides . And then I | |
17:29 | have to factor the left hand side and then I'd | |
17:31 | have to take the square to both sides and then | |
17:33 | have to move it around and I'd finally get the | |
17:34 | answer . But now that I understand and have derived | |
17:37 | the quadratic formula , I know that in this equation | |
17:39 | A is equal to three , B is equal to | |
17:41 | the one here , and C is equal to the | |
17:44 | negative one , which is the constant . And that | |
17:46 | the solutions of this is just negative B plus or | |
17:49 | minus B squared minus four times a times C . | |
17:53 | Uh And the entire thing is divided by two A | |
17:56 | . And so your equation is going to be negative | |
17:58 | B which is a one . So negative one here | |
18:00 | , plus or minus on the inside , you're still | |
18:03 | going to have a one squared B squared minus four | |
18:07 | times A . Which is three times C . Which | |
18:10 | is negative one . The square root applies to this | |
18:13 | whole thing right here and on the bottom it's gonna | |
18:15 | be two times a two times three like this . | |
18:19 | And so what you're going to get on the inside | |
18:21 | , you just crank through it one step at a | |
18:22 | time , you're gonna get negative one plus or minus | |
18:25 | on the inside . What do you have ? You | |
18:26 | have one minus four times three is 12 Right ? | |
18:30 | But negative times negative is positive . So it's going | |
18:32 | to actually be positive 12 like this and on the | |
18:36 | bottom you're gonna have a six . So what you'll | |
18:38 | have over here then is negative one plus or minus | |
18:42 | one plus 12 is 13 over six like this . | |
18:48 | Now usually you can just leave the answer like that | |
18:50 | . But since it's our first problem , I'll just | |
18:52 | write it explicitly out what this means is you have | |
18:54 | , the answer is negative one negative one plus square | |
18:58 | root of 13 divided by six . That's one answer | |
19:01 | . And the other answer is negative one minus the | |
19:04 | square root of 13/6 . This is exactly the same | |
19:08 | answer , negative one plus 13 . Route 13/6 , | |
19:11 | negative one minus Route 13/6 . This is exactly the | |
19:14 | same set of answers you would get if you just | |
19:15 | completed the square with the original thing , but because | |
19:18 | we now have the quadratic formula that comes from completing | |
19:21 | the square , there's no reason to complete the square | |
19:23 | in this problem . We'll just take the coefficients and | |
19:25 | stick it in there and get the answers . But | |
19:27 | now , you know where it comes from and it | |
19:28 | doesn't seem like so much black magic because it it | |
19:32 | comes from something you already understand . All right , | |
19:36 | let's take a look at a couple more real quick | |
19:40 | . What if it was the polynomial X squared plus | |
19:45 | six X plus four Equals zero . I want to | |
19:48 | solve the quadratic equation . Well , the first thing | |
19:51 | you identify A . B . And C . And | |
19:53 | from this equation A . Is gonna equal to the | |
19:55 | coefficient of X , which is one . B is | |
19:57 | going to be equal to the six and C is | |
19:59 | going to be equal to the four . Those are | |
20:00 | the three coefficients . That's basically all you have to | |
20:02 | do there . And I always encourage you to write | |
20:05 | the quadratic formula down every time . So you don't | |
20:07 | make any mistakes , negative B plus or minus B | |
20:10 | squared minus four times . Hmc The radical has to | |
20:14 | apply that whole thing over to A And then you | |
20:18 | just stick it in here negative B means negative six | |
20:21 | plus or minus . We have be square which means | |
20:23 | six squared minus four times a time . See . | |
20:29 | And then the whole thing don't forget this radical . | |
20:31 | It has to sit outside the whole thing over two | |
20:32 | times a two times one . So you put everything | |
20:36 | in place and then in the following steps you just | |
20:38 | crank through and get the answer . So what we | |
20:39 | have is X . Is equal to negative six plus | |
20:42 | or minus under this radical . It'll be 36 minus | |
20:47 | the 16 and that radical has to apply to the | |
20:50 | whole thing over to so it'll be negative six plus | |
20:54 | or minus what is 36 minutes 16 . Let me | |
20:56 | double check my math . Here we have a 36 | |
20:58 | minus 16 is 20 . It's gonna be the square | |
21:01 | root of 20 and you'll have a tube so now | |
21:04 | it's a little bit harder . You shouldn't say harder | |
21:06 | but because you have a square root of 20 you | |
21:08 | need to simplify that as much as you can . | |
21:10 | So 10 times two is 20 and five times two | |
21:13 | is 10 and I have a pair of twos here | |
21:15 | . So if I was going to write this final | |
21:17 | answer down it be negative six plus or minus , | |
21:19 | but the square root of 20 is two times the | |
21:21 | square root of five , two times the square root | |
21:24 | of five over to . Now you could circle that | |
21:27 | and say that you're done . You can however you | |
21:30 | see that this is divisible by two . This is | |
21:32 | divisible by two and this is divisible by two . | |
21:34 | So it's better to break it up and cancel everything | |
21:37 | . So let's write , it is negative 6/2 plus | |
21:39 | or minus two . Route 5/2 and all I'm doing | |
21:44 | is breaking it apart as much as I showed you | |
21:47 | in the past . So if you were to add | |
21:49 | these together , you have a common denominator and then | |
21:51 | this plus minus this . So all I've done is | |
21:52 | broken it apart . And the reason I did that | |
21:55 | is because the answer then is more easily read as | |
21:58 | negative 6/2 is negative three plus or minus . The | |
22:02 | two is now cancel meaning to square +25 is left | |
22:05 | over . So all you have is negative three plus | |
22:07 | or minus the square root of five . Yeah , | |
22:09 | two answers negative three plus the square to five . | |
22:12 | Negative three minus the square root of five . And | |
22:14 | that's the final answer . All right now , for | |
22:18 | our last one , since we've done a little bit | |
22:20 | of this stuff for our last one , we'll crank | |
22:23 | through it . Just a touch faster . What if | |
22:25 | we have the polynomial or the quadratic Y squared minus | |
22:28 | four times ? Y whoops . Not X . Y | |
22:33 | Plus 13 is equal to zero . So a . | |
22:37 | is one b is negative for and see us 13 | |
22:40 | . It's always a good idea in the beginning here | |
22:41 | to write that down . So a . is one | |
22:44 | B is negative for c . is 13 like this | |
22:48 | . And now you have to plug it into the | |
22:49 | quadratic formula . So you have negative B plus or | |
22:52 | minus and all that stuff . I do recommend in | |
22:54 | the beginning you write it down every time you use | |
22:55 | it negative B plus or minus B squared minus four | |
22:59 | . A C square root goes around all of that | |
23:01 | over to a . Mhm . And so you just | |
23:05 | plug things in . Now the trick here is you | |
23:07 | have a negative B . But B itself is negative | |
23:09 | . So in order to avoid any problems , the | |
23:11 | negative comes from here . You write that down in | |
23:14 | parentheses , you put the negative four that you're substituting | |
23:16 | in for me . And that way you never make | |
23:18 | any mistakes with with the signs , then you have | |
23:21 | plus or minus . And on the inside B is | |
23:23 | squared but B is negative . So you have to | |
23:25 | write it as negative four squared minus four times A | |
23:29 | , which is one times C , which is 13 | |
23:32 | . And the square root is wrapped around this whole | |
23:34 | thing . It's very important that you write it down | |
23:36 | like this so you don't make any problems with the | |
23:38 | signs on the bottom is two times A , which | |
23:40 | is two times one . So then X is going | |
23:43 | to be equal to the negative of negative four is | |
23:45 | positive for plus or minus on the inside , four | |
23:48 | squared negative four squared is positive . 16 . Then | |
23:51 | you have a minus four times 13 , Four times | |
23:54 | 13 is 52 . So you're gonna have 16 -52 | |
23:59 | , you have a square root that lives around that | |
24:01 | whole thing . And on the bottom it's gonna be | |
24:02 | over to So then continuing on you'll have four plus | |
24:07 | or minus what is 16 -52 is negative 36 but | |
24:11 | we still have a square root wrapped around that over | |
24:13 | two . And then you have four plus or minus | |
24:16 | . Now we know the square root of 36 to | |
24:17 | 6 . We know the square root of negative ones | |
24:19 | . I so you get six I and on the | |
24:22 | bottom you're gonna have this to the same kind of | |
24:24 | thing . I could probably leave it like this if | |
24:26 | I want but I would rather cancel everything . And | |
24:29 | so what I'm gonna do is break this thing apart | |
24:30 | so I can cancel it . And say what I'm | |
24:32 | gonna have is it's gonna be four over to the | |
24:34 | plus minus comes down from here and then six I | |
24:37 | over two and that's going to be equal to four | |
24:40 | divided by two is two and then I have plus | |
24:43 | or minus and then I have the 6/2 is going | |
24:45 | to be three . I So this is gonna be | |
24:47 | what X . is going to be equal to . | |
24:49 | So here we have the final answer here to plus | |
24:51 | or -3 I is equal to X . But again | |
24:53 | we have two solutions but both of them are imaginary | |
24:56 | complex I should say . So in this lesson I | |
24:59 | had two objectives . I want you to be able | |
25:03 | to use the quadratic formula . So we've done that | |
25:05 | here and I want you to go on to the | |
25:07 | next lesson in the next several because we're gonna be | |
25:08 | doing and using the quadratic formula more and more and | |
25:11 | more just to get you really comfortable with it . | |
25:13 | But equally important with that , I wanted you to | |
25:15 | know where it comes from . It comes from completing | |
25:17 | the square which we've learned and learned and used and | |
25:20 | used . Now . Once we have this quadratic formula | |
25:23 | in our tool bag usually going forward , you'll just | |
25:25 | solve most problems that are quadratic using this quadratic formula | |
25:28 | . But that doesn't mean that completing the square was | |
25:31 | useless because everything that we learned in algebra builds and | |
25:33 | we'll find out that when we solve more kind of | |
25:35 | complicated equations , some of the ideas that we used | |
25:38 | in completing the square will still be useful . So | |
25:40 | try not to forget it . But the truth is | |
25:42 | that most quadratic that you'll solve going forward would probably | |
25:46 | use the quadratic formula if you can't factor it very | |
25:49 | easily . So follow me on to the next lesson | |
25:51 | and we'll continue getting practice with the quadratic formula right | |
25:54 | now . |
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