02 - Solve Perfect Square Quadratic Equations Part 1 - By Math and Science
Transcript
00:00 | Hello . Welcome back to algebra . The title of | |
00:02 | this lesson is solving perfect square quadratic equations . This | |
00:06 | is part one . So the road map here is | |
00:08 | we're going to be learning different ways to solve quadratic | |
00:11 | equations . Remember quadratic just means the highest power of | |
00:15 | X is a two . So it's an X squared | |
00:17 | is the highest exponent in the equation . You already | |
00:20 | know how to solve some quadratic equations . We've actually | |
00:22 | done it quite a bit remember we factor under those | |
00:25 | quadratic under the quadratic . We try to factor into | |
00:27 | two . Binomial setting it equal to zero setting each | |
00:30 | of those binomial is equal to zero . And solving | |
00:33 | the only problem is we can't factor a whole lot | |
00:36 | of quadratic . You can't factor them most of the | |
00:38 | time . So when you can't factor those things great | |
00:40 | . But if you can't factor then you're gonna have | |
00:42 | to use some of these methods we're gonna be learning | |
00:43 | in these lessons here . So these types of quadratic | |
00:46 | . So we're going to have here are called perfect | |
00:48 | square quadratic . They're very special . And you'll see | |
00:51 | very obviously why they're special . So the first one | |
00:55 | is very , very simple . So it's X squared | |
00:57 | is equal to three . This is a perfect square | |
01:00 | quadratic . Because you'll see as we get into more | |
01:03 | problems , the power on the X here is very | |
01:06 | easy to eliminate by taking the square root of both | |
01:09 | sides . And it will be again a little more | |
01:11 | easy to understand exactly why it's called a perfect square | |
01:14 | when I can get a few more problems on the | |
01:15 | board . But now that we've conquered the idea of | |
01:18 | radicals , you know how to solve an equation like | |
01:20 | this . So you didn't know how to solve this | |
01:21 | before . He talked about radicals . But now that | |
01:23 | we have , you know , you can take a | |
01:25 | square root of both sides , we take the square | |
01:28 | on the left and it cancels with the X square | |
01:31 | just leaving the eggs . When we apply a square | |
01:33 | root ourselves , we must put the plus minus . | |
01:35 | We talked about that in great detail in the past | |
01:38 | and we'll take the square root of the three . | |
01:39 | So we take the square root of both sides . | |
01:41 | But we just have to uh put a plus minus | |
01:43 | in front . So because we have a plus and | |
01:45 | a minus , there's actually two solutions to this . | |
01:48 | The first solution is positive square root of three and | |
01:51 | the other solution is negative square root of three . | |
01:55 | So you can circle this the plus minus square root | |
01:57 | of three if you like . Or you can circle | |
01:58 | the two solutions written out plus or minus if you | |
02:01 | like . Now we talked about this in the last | |
02:04 | lesson , but what does it actually mean when you | |
02:05 | get to two solutions to roots of a quadratic like | |
02:09 | this ? So we talked about it extensively in the | |
02:11 | last lesson , I encourage you to go watch that | |
02:13 | last lesson if you haven't already done so . So | |
02:15 | what it means is you have these quadrants , they | |
02:17 | all look like parabolas , either smiley faces like this | |
02:20 | or upside down frowny faces . You know uh Parabolas | |
02:24 | and the fact that we get to answers means these | |
02:27 | are the values where the the actual parabola crosses the | |
02:30 | X axis , the intersection points . Because those are | |
02:33 | the places where if we were to subtract this , | |
02:36 | X squared minus three , that would be where that | |
02:38 | function is equal to zero and it's X is equal | |
02:40 | to positive square 23 and X is equal to negative | |
02:43 | square 23 So just to drive it home , we | |
02:46 | talked about it before , but if you were to | |
02:47 | graph this thing , it might look something like this | |
02:51 | , right ? So the function your if you were | |
02:53 | to rewrite it F of X is equal to the | |
02:56 | X squared and subtract the three to put everything on | |
02:59 | one side . So this is the function x squared | |
03:01 | minus three . If we set this thing equal to | |
03:04 | zero , then we're finding the intersection points when the | |
03:09 | function is equal to zero and one of them is | |
03:11 | that positive square root of three right there ? And | |
03:13 | the other one is negative square root of three , | |
03:15 | which is right there . So that's just a little | |
03:17 | sketch . But in general when you're solving equations , | |
03:19 | you can always move everything to one side . This | |
03:21 | is the function and you're setting that function equal to | |
03:23 | zero there , which is the intersection points . All | |
03:28 | right . So let's just crank through a bunch of | |
03:30 | these guys . Perfect square quadratic . Another one might | |
03:33 | be like this , X minus one squared is equal | |
03:36 | to three . How would you solve that ? We | |
03:39 | want to get X by itself , but not only | |
03:41 | X , but x minus one is wrapped up in | |
03:44 | a parenthetical all squared , but we can get rid | |
03:47 | of the square . We can release the variable for | |
03:49 | lack of a better word by taking the square to | |
03:51 | both sides . Taking the square root of the left | |
03:54 | is just gonna leave what's under here by itself because | |
03:57 | the square in the square ruble than cancel . And | |
03:59 | then when we apply a square in ourselves , we | |
04:01 | have to put a plus minus square root of three | |
04:04 | . You see how these two solutions look very similar | |
04:07 | . We just took the square root of the left | |
04:09 | and the squared of the right square of the left | |
04:10 | and squared of the right . But now we have | |
04:12 | this one here , so then we need to move | |
04:14 | it to the right hand side , we'll add it | |
04:16 | . So then we'll have one plus or minus the | |
04:19 | square root of three . So again , when you | |
04:21 | have these pluses and minuses , you can leave it | |
04:23 | like this . This means there's two distinct solutions . | |
04:25 | The first solution is one plus the square root of | |
04:28 | three , that's one answer , and the other solution | |
04:31 | is that it's one minus the square root of three | |
04:34 | . So these are the two answers . So you | |
04:36 | can circle it either as the two answers separately written | |
04:39 | out , or you can circle it with the plus | |
04:41 | minus . This means there's two different answers , and | |
04:44 | again , I'm not gonna draw another sketch . But | |
04:45 | what it basically means is if you were to take | |
04:47 | and jag this three over And make the function f | |
04:50 | of X is equal to this -3 . That's the | |
04:53 | function . You can plot it on a graph . | |
04:55 | And when you set that function equal to zero , | |
04:57 | you're finding the crossing points where the function is equal | |
05:00 | to zero . So when you solve this equation it's | |
05:03 | exactly the same thing . I guess I will write | |
05:04 | it down . It's exactly the same thing as saying | |
05:07 | the function F of X being equal to x minus | |
05:11 | one quantity squared minus three . Because I just move | |
05:13 | it over here . That function , I'm gonna set | |
05:16 | this function equal to zero . So x minus one | |
05:19 | quantity squared minus three is equal to zero . This | |
05:21 | equation that you're solving here is exactly the same as | |
05:23 | what you have . Because notice if you're going to | |
05:26 | solve this , you'd move the three over first , | |
05:28 | which is what we were given , then we take | |
05:30 | the square root . And we would solve so pretty | |
05:32 | much any equation you ever solve , even if there's | |
05:34 | numbers on the right or not is exactly the same | |
05:38 | thing as moving everything on one side . Calling it | |
05:41 | a function and setting that function equal to zero . | |
05:44 | That's why we spend so much time learning how to | |
05:47 | find the roots of polynomial is because all of these | |
05:49 | equations can be thought of as just talking about where | |
05:52 | the function crosses the X axis like this and there's | |
05:55 | two crossing points because there's just like there's two crossing | |
05:58 | points here , we have two crossing points in this | |
06:00 | case . Now we're going to stop drawing graphs and | |
06:04 | things and just continue writing down equations . So here | |
06:07 | we had X squared is equal to three . Here | |
06:08 | we have quantity X minus one squared is equal to | |
06:11 | three . And now we'll change it just a little | |
06:13 | bit further and we'll say two X -1 , quantity | |
06:16 | squared is equal to three . Now you can see | |
06:18 | the pattern . How do you solve this equation ? | |
06:20 | Well we have to get rid of the square , | |
06:21 | so we're gonna take the square root of both sides | |
06:23 | . That will cancel and leave this on the right | |
06:26 | . It will be plus or minus the square root | |
06:27 | of three . But now I have to solve for | |
06:29 | X . So I move the 1/2 , X is | |
06:32 | one plus or minus the square root of three . | |
06:35 | And now I want to divide by the two . | |
06:37 | So it's gonna be one plus or minus the square | |
06:39 | root of three , all divided by quantity of two | |
06:42 | . Now there's various ways you can ride it , | |
06:44 | I can break this up into two fractions if I | |
06:46 | want . But I'm just gonna leave it like this | |
06:48 | because you can see there's two answers . One plus | |
06:50 | the square root of three divided by two , or | |
06:53 | one minus the square root of three divided by two | |
06:56 | . So now that we can see the general pattern | |
06:59 | here , you can kind of see why it's called | |
07:00 | a perfect square quadratic . Because what's happening here is | |
07:04 | the term that involves the variable is just a term | |
07:07 | squared C . You have the entire term , the | |
07:10 | entire parentheses is quantity squared uh here . So it's | |
07:15 | a perfect square , right ? This entire term that | |
07:18 | involves the variable , The quantity is squared . This | |
07:21 | entire term that involves the variable is also square , | |
07:23 | but it's just a simple case of just X being | |
07:25 | squared . So it's not so obvious . You know | |
07:27 | , in general , when you look at these polynomial | |
07:29 | in general , a quadratic polynomial looks like something like | |
07:32 | this X squared plus three x minus four . So | |
07:36 | this is not a perfect square quadratic . It's not | |
07:38 | a perfect square quadratic because you have an X squared | |
07:42 | here and you have an X . Here . So | |
07:43 | it's difficult if I were going to set this thing | |
07:45 | equal to zero , I mean you might be able | |
07:46 | to factor it . Okay . But if you don't | |
07:48 | know how to factor it , until we learn the | |
07:49 | methods , you don't know how to solve this because | |
07:52 | the only way we really learned how to solve polynomial | |
07:54 | , like that is to factor them and set them | |
07:56 | equal to zero . But what if this thing is | |
07:58 | not factory able then you're kind of stuck and that's | |
08:00 | what we're gonna be doing in this in this lesson | |
08:02 | . But this is not a perfect square because you | |
08:04 | have the variable in two places and one is squared | |
08:06 | and one is not . But in this equation , | |
08:09 | the only place where the variable exists is a term | |
08:12 | that is squared on one side of the equal sign | |
08:14 | . The only place where the variable exists is a | |
08:16 | quantity that is squared on one side of the equal | |
08:18 | side . That makes this class of problems easy to | |
08:21 | solve because we can just take the square to both | |
08:23 | sides . That's why we learn at first , it's | |
08:25 | the easiest kind of equation to learn how to solve | |
08:28 | . That's a quadratic equation . So another example of | |
08:31 | that would be why plus seven , quantity squared is | |
08:34 | equal to 16 ? How do you solve this ? | |
08:36 | The variable is wrapped up on one side , nice | |
08:39 | and neat like a bow with a square . So | |
08:41 | we can take the square root of the left . | |
08:42 | Leaving why plus seven left over the square root of | |
08:45 | the right will be the square root of 16 . | |
08:47 | So we now know that the square root of 16 | |
08:50 | is four , so it's plus minus four and then | |
08:53 | we can move that seven to the other side . | |
08:55 | So what we'll have is when we subtract seven , | |
08:57 | what we're going to get is negative seven on the | |
08:59 | other side plus or minus four . We just subtract | |
09:01 | seven subtract seven . Now you see in these other | |
09:04 | cases we lifted this plus minus square root of three | |
09:06 | , we left this plus minus square +23 That's because | |
09:09 | the square root of three is an irrational number . | |
09:11 | And if you want to be exact , you have | |
09:13 | to leave it as the square root of three . | |
09:15 | But in this case it's plus or minus four . | |
09:17 | So it's actually very nice just to be able to | |
09:20 | ride it out because it's easier to see negative seven | |
09:23 | plus four . And then why will be negative seven | |
09:26 | minus four ? You can all do addition and subtraction | |
09:28 | . So what you have is negative seven plus four | |
09:30 | is going to give you negative three and this will | |
09:32 | be negative seven minus four . Which would be negative | |
09:34 | 11 . So you have two solutions negative three and | |
09:38 | negative 11 . All right . So the word , | |
09:40 | the rule of thumb is if you get down to | |
09:42 | the plus minus step and you can actually do the | |
09:44 | math and get numbers then do it . If you | |
09:46 | have radicals involved or weird fractions involved in just leave | |
09:49 | it alone . All right . So now that we | |
09:54 | know the idea of what a perfect square quadratic is | |
09:56 | . Now we can just crank through a bunch of | |
09:58 | problems to get practice and they're all going to be | |
10:00 | the same sort of thing . So what if we | |
10:02 | have three ? Y plus seven quantity squared is equal | |
10:05 | to 16 . It's almost exactly the same thing as | |
10:08 | we had here . But now we just put a | |
10:09 | three in front of the y . Here . All | |
10:12 | that's going to happen here is we're gonna take the | |
10:13 | square root of both sides . Three y plus seven | |
10:16 | is leftover plus or minus four on the right then | |
10:20 | we have uh not sorry not plus or minus . | |
10:22 | Uh Let's leave it like this plus or minus square | |
10:25 | root of 16 . We take the square root of | |
10:27 | both sides and on the right we have plus or | |
10:29 | minus four . Now we move the 7/3 . Y | |
10:33 | is equal to negative seven plus or minus four . | |
10:37 | And then why is negative seven plus or minus four | |
10:41 | all divided by three ? So it's almost exactly the | |
10:43 | same problem . We're just dividing by three at the | |
10:45 | end essentially here . And so what you're gonna end | |
10:49 | up having is you'll have two possible scenarios negative seven | |
10:53 | plus 4/3 or negative seven minus 4/3 . So what | |
11:00 | you're going to have in this case negative seven plus | |
11:02 | four is negative 3/3 . So that why will be | |
11:06 | negative one circle that as an answer or you'll have | |
11:10 | negative seven minus four is negative 11/3 . Now , | |
11:12 | I can't really do much with that . I can't | |
11:14 | simplify it any further . So this is just one | |
11:17 | of the answers and this is the other negative one | |
11:19 | And -11/3 . All right . Just have a couple | |
11:23 | of additional problems to go here . I have one | |
11:25 | that I want to draw . A quick plot of | |
11:26 | . What if we have ? Let me give myself | |
11:29 | a little space here . What if we have the | |
11:30 | equation as follows , X squared is equal to negative | |
11:33 | four . Now this equation looks almost identical to the | |
11:37 | first equation . We solved this one here . We | |
11:39 | took the square root of both sides and we got | |
11:42 | to real numbers one was a negative square root of | |
11:44 | 31 was a positive square 23 And we said that | |
11:47 | that corresponded to a graph where it below the axis | |
11:51 | and one of the roots was negative and one of | |
11:52 | the roots was positive . That's what it actually means | |
11:55 | to have two routes like that here . The form | |
11:57 | of the equation is the same . We take the | |
11:59 | square root of both sides , revealing X is plus | |
12:01 | or minus the square root of negative four . But | |
12:03 | now you all know how to take the square root | |
12:05 | of negative numbers . We have to use the imaginary | |
12:07 | numbers . So the square root of the four is | |
12:09 | the two and the squares of the negative one is | |
12:11 | the eye . So we get plus or minus two | |
12:13 | . I So we get to answers just like before | |
12:16 | positive to I and -2 . I so I'm gonna | |
12:18 | go in and circle this and let's do a quick | |
12:22 | plot of what that might look like . So as | |
12:25 | we've looked at earlier in the demo , we suspect | |
12:29 | that there's not gonna be any crossing values on the | |
12:31 | real access because of what we learned before . And | |
12:34 | what this graph actually does look like is it looks | |
12:37 | like this , the base graph is X squared , | |
12:40 | but we have this modification of minus four here . | |
12:43 | So it looks like this 1234 The graph actually goes | |
12:47 | down like this because if you're going to plot this | |
12:50 | function , what you really have here , you have | |
12:52 | to solve and get everything on one side . F | |
12:54 | of X is equal to X squared plus . For | |
12:59 | all you do is you move forward to the other | |
13:01 | side . So it's X squared plus four . That's | |
13:03 | the function X squared plus four . And then you're | |
13:05 | setting it equal to zero . Of course there are | |
13:08 | no values where it's equal to zero because if you | |
13:10 | plot this function , you're taking the X squared function | |
13:13 | , the regular parabola and you're adding forward to it | |
13:15 | . The regular parabola goes all the way down to | |
13:17 | the axis . When you add for two , it | |
13:20 | shifts the whole thing up . So this thing never | |
13:22 | intersects the axis anymore . So because of that you | |
13:25 | don't have real solutions , You have imaginary solutions . | |
13:28 | And we showed that in the demo that we did | |
13:30 | before when you have these any kind of graph that's | |
13:33 | not crossing the access that's up above or down below | |
13:35 | . It's not there's no crossing values and you don't | |
13:37 | have any real solutions . So but we do have | |
13:39 | two solutions . They're just imaginary numbers like this . | |
13:43 | All right now , while we have this on our | |
13:46 | mind , let's go and do one more . That's | |
13:48 | very similar to it . What if we change the | |
13:51 | problem a little bit and we say X plus seven | |
13:54 | quantity squared equals negative for . So this is almost | |
13:58 | exactly the same problem instead of X squared is negative | |
14:01 | for we're just changing a little bit to be this | |
14:04 | . But we saw that the same way . Take | |
14:06 | the square root of both sides giving you X plus | |
14:08 | seven is plus or minus the square root of negative | |
14:10 | four , X plus seven is plus or minus . | |
14:13 | This becomes too I just like it did last time | |
14:16 | and then we have to take the seven over . | |
14:18 | So we're gonna have negative seven plus or minus two | |
14:20 | . I so you see there's still two answers negative | |
14:23 | seven plus two i in negative seven minus two I | |
14:25 | two answers . But they're complex numbers . So because | |
14:28 | the solutions that we got for both of the roots | |
14:31 | of this polynomial or this quadratic are both uh complex | |
14:35 | , right ? Because of that , then we expect | |
14:37 | that if you were to graph the function of this | |
14:39 | thing , there would be no crossing points on the | |
14:42 | X axis . So this is the function right now | |
14:45 | and we haven't set equal to negative four . But | |
14:47 | if we move the four over , we can say | |
14:49 | that this is equivalent to a function which would be | |
14:51 | F of X is equal to X plus seven , | |
14:55 | quantity squared plus four . Why A plus sign ? | |
14:59 | Because what we have here is this equals to this | |
15:01 | ? So if we move this to one side , | |
15:03 | thereby making it zero over here , then the function | |
15:06 | itself is x squared plus seven squared . I'm sorry | |
15:08 | , X plus seven squared plus the four . That's | |
15:11 | the function . And then we're looking for these crossing | |
15:15 | points . If you move it over here , you're | |
15:16 | going to get uh X plus seven quantity squared plus | |
15:20 | four equals zero . If you move the four back | |
15:23 | over here , you're gonna get a zero . So | |
15:25 | this is the function , we're setting the function equal | |
15:27 | to zero , that's the crossing points . But what | |
15:30 | you're going to have happen or c in this case | |
15:33 | is you're gonna have the 1234 that's an important tick | |
15:36 | mark and the 1234567 This function actually is going to | |
15:41 | look like I have a problem that looks something like | |
15:44 | this , it's going to be shifted four units up | |
15:48 | because of the plus four here . But because of | |
15:50 | the seven in here , the parable of the basic | |
15:53 | parable of X squared is going to be shifted seven | |
15:55 | units to the left . So the parabola again is | |
15:58 | sitting out in empty space with no crossings on the | |
16:01 | X axis down here . So because of that , | |
16:03 | we know it doesn't have any real roots , it | |
16:05 | has just the two complex routes that we see there | |
16:08 | now . Don't worry so much if you don't understand | |
16:10 | why this thing is shifted where it is actually , | |
16:12 | we're gonna have complete lessons on that later down the | |
16:15 | road . We talk about taking a function and shifting | |
16:18 | it around and what the graph of it and what | |
16:19 | the equation looks uh looks like when you start shifting | |
16:22 | the function around . But I'll just give you a | |
16:24 | little bit of a secret punch line . The basic | |
16:27 | shape of this graph is just F . Of X | |
16:31 | . Is X squared , It's just an X squared | |
16:33 | parabola . But because we have a plus seven , | |
16:36 | that shifts the problem to the left . And because | |
16:38 | we have a plus for it shifts the problem up | |
16:40 | by four . And so this is the basic shape | |
16:43 | . But these numbers just turn out to shift it | |
16:46 | around into a different location . We're gonna talk about | |
16:48 | exactly why it moves it here . As I know | |
16:50 | it is a little confusing negative versus positive here . | |
16:53 | It looks a little confusion . We'll get to that | |
16:54 | a little bit later . Mostly what I want you | |
16:56 | to know here is that the graph of this function | |
16:59 | here , the graph of this function does not intersect | |
17:01 | the X axis . So because of that we get | |
17:03 | the two uh imaginary or complex roots rather than real | |
17:09 | roots . And so our last problem is going to | |
17:12 | be very similar to this . We're just gonna take | |
17:14 | two X plus seven . We're gonna square it and | |
17:17 | say negative for it . We've done it so much | |
17:18 | , we'll just fly through it real quick . We | |
17:20 | take the square root of both sides uh here and | |
17:23 | then when we take the square root of the right | |
17:25 | hand side we're gonna get the plus or minus two | |
17:27 | . I when we take the square root of the | |
17:28 | negative four and then we're gonna move the seven over | |
17:31 | . So we're gonna have negative seven plus or minus | |
17:33 | two . I and then we're gonna divide by two | |
17:35 | . So we're gonna have a negative seven plus or | |
17:37 | minus two . I divided by two . Now we | |
17:41 | could leave it like this , I could I would | |
17:42 | probably accept this but since there's no radicals anywhere it's | |
17:45 | better to break it up . You could say that | |
17:47 | this is gonna be negative seven plus two . I | |
17:50 | All divided by two And you could say that this | |
17:53 | could also be equal to negative 7 -2 . I | |
17:55 | all divided by two . Like this . So there's | |
17:58 | two possible routes . You can see they're going to | |
18:00 | be complex because we have this imaginary number in there | |
18:03 | . Mm and then we're going to simplify further . | |
18:06 | You could leave it like this but we could also | |
18:08 | break it up as follows . You could say that | |
18:10 | this is negative seven halves plus two . I over | |
18:15 | to and that this one over here will be negative | |
18:17 | seven halves minus two . I over to you know | |
18:21 | breaking fractions up like this should it make sense ? | |
18:24 | If you go backwards there's a common denominator here , | |
18:26 | so if you have the negative seven plus the two | |
18:29 | , I if you were to add these back together | |
18:30 | you get back exactly what you had and the same | |
18:33 | thing here , negative seven minus two . I would | |
18:34 | give you exactly what you have here . If you | |
18:36 | were to go backwards . So we're just taking the | |
18:38 | fraction of breaking it up instead of going backwards . | |
18:41 | So why do we do that ? Because we have | |
18:43 | a cancellation of two ? So have negative 7/2 plus | |
18:47 | Just I . And then we have negative 7/2 -1 | |
18:53 | . So there's one complex number and there's another one | |
18:56 | . The real part is here , the imaginary part | |
18:58 | is here , the real part is here , the | |
19:00 | imaginary part is here , so is negative seven half | |
19:02 | plus , I negative seven halves minus side . And | |
19:05 | again if we were to graph this function , we | |
19:07 | would find out that this parable was hanging out in | |
19:10 | space somewhere where it never crosses the X axis . | |
19:13 | So therefore there's no real roots , there's no crossings | |
19:16 | . So we have just these purely complex congregates . | |
19:19 | Notice one is a plus and one is a minus | |
19:21 | . And by the way , it's a good time | |
19:22 | to tell you that when you get complex conjugate complex | |
19:26 | numbers as roots of polynomial , they're always going to | |
19:29 | pop up in pairs and they're always going to pop | |
19:31 | up its congregants . So you will never just see | |
19:33 | one of these by themselves . You'll always have them | |
19:36 | in pairs and you'll always have them as candidates plus | |
19:39 | and minus . That's what candidates are when the imaginary | |
19:41 | part has a plus and a minus kind of sister | |
19:44 | term . So that's what all I wanted to talk | |
19:46 | to you about here , we're gonna solve more equations | |
19:48 | . But the general idea is that when you have | |
19:50 | these perfect square quadratic , so you can just take | |
19:54 | the square root of both sides to reveal the variables | |
19:57 | . So by using radicals we can solve these very | |
19:59 | easily because of that . And you're always going to | |
20:02 | get to answers because taking the radical of both sides | |
20:05 | of the square root of both sides introduces a plus | |
20:08 | or minus term which is consistent with . We know | |
20:11 | that we're gonna always get to solutions for all of | |
20:13 | these quadratic . So follow me on to the next | |
20:15 | lesson , will solve more of these perfect square quadratic | |
20:18 | will increase the problem complexity and build your skills from | |
20:21 | there . |
Summarizer
DESCRIPTION:
Quality Math And Science Videos that feature step-by-step example problems!
OVERVIEW:
02 - Solve Perfect Square Quadratic Equations Part 1 is a free educational video by Math and Science.
This page not only allows students and teachers view 02 - Solve Perfect Square Quadratic Equations Part 1 videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.