02 - Solve Perfect Square Quadratic Equations Part 1 - Free Educational videos for Students in K-12

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 .

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