@@RegardedBee185 Many people do need to understand how algorithms work in order to successfully execute them. It's actually how they learn. I'm one such person.
In Taiwan, we learn the “Factoring Method”(not sure what it’s called in English) by finding two factors of -12 which are -6&2, -6 and 2 can be add up as the constant 4 in x term. Times them (x-6)(x+2)=0, and there’s the easy answer. If the constant term isn’t dividable, I’ll use the quadratic formula x= (-b±√(b²-4ac))/(2a). Pretty simple imo. The way you use is quite non-intuitive for me. Still pretty nice to learn a new way.
The reason you’d do this is to make it easier to do algebraic proof for increasing/decreasing functions You could solve the above equation by (x-2)^2-16 = 0 (X-2)^2 = 16 x-2=+/- 4 x = 6 or -2 The factoring method could be quicker for numbers that could fully factorise but I think it’s probably quicker than if you’d use the quadratic formula on numbers that couldn’t fully factorise
I also find it non-intuitive, also aren't you supposed to have everything at grade 1? (I mean without exponents, I don't know how you say it in english) In italy we call the first method you used "trinomio notevole" or "trinomio speciale", (special thrinomial) where you find the 2 numbers that when multiplied are equal to c (so -12) and when added are equal to b (so -4) those 2 numbers being -6 and 2, so you can rewrite the entire thrinomial as (x-6)(x+2). I think you have to have everything without exponents, so is (x-2)²-16 a complete answer? And obviously after (x-6)(x+2)=0 the solutions are 6 and -2
@@giovanji7883 yeah, I think learning each method is totally applicable in simple math. But when it comes to advanced math, like Laplace transform or Fourier transform, the exponent 2,3,4… will become really annoying. Using first method will make it easier to calculate. Btw, what do you mean by having everything in grade 1? I didn’t get it.
Completing the square is used to make an "unfactorable" problem easily solvable using square roots. If you can factor it's probably easier to just factor.
That's all well and good in this example where you end up with "a^2 - b^2" where a=x-2 and b = 4. This factors out to (a+b)(a-b). Substituting, we get ((x-2)+4)((x-2)-4), which becomes (x+2)(x-6) with roots x=6, x=-2. If c had been (say) 11, you wouldn't have been so lucky.
A far faster method to solve for X, take -12. Look for two numbers that multiply to -12, but also add up to -4. Those numbers would be +2 and -6. So you would have (x+2) (x-6). You can then just set each of them equal to 0. X+2=0 and X-6=0. This is also known as the Master Product method. Usually for me I start here. If this doesn't work I just go straight to quadratic formula. Usually works out better for me that way.
Oh yea I NEVER solve quadratic equations this way. I do the exact same process as you. Try to factor first and if that doesn’t work then plug into the quadratic formula. Completing the square is just beneficial for finding the vertex of a parabola (which we can now do in this problem) as well as finding the centers of conic sections.
@@datboy038 bro literally if the x^2 coefficient is anything other than 1 it won't work. Edit technically it can work, but in those scenarios it is far more likely to get errors than if you just go straight to quadratic. Or you can master product by putting factors of "a" in front of the 2 different x equations
divide b by 2 you get -2 right. then the last number has to be -2*-2 which is 4 (foil method is what i usde to determine that (a+b)^2 = a^2 +2ab + b^2 so it would be 4 at the end to make it wrk. so all u do is add 16 to each side. and get (x-2)^2=16 which you could also get with his way by adding 16 to both sides. then u do the sqrt get x-2=4 then x=6
This is more efficient but it's still important to understand the other method because you actually learn why it works not just that it works. It's like how we have to use limits to take derivatives in calculus before we are allowed to use power rule and the other shortcuts.
I agree, except i think this method can be learned initially instead of the other one. It just needs to be explained why it actually works like you’re saying.
@@SkullPrince8 I agree. For now, this video will at least get kids completing the square. If they want a video explaining why it works, I’ll make that video.
@@LudusYTbeing able to do it is useless. The reason why is of 100% importance. If kids are able to do it by memorisation only. Then they cannot do it at all. Like completing the square in everyday life is useless. But understanding algebra and how to manipulate things properly at least teaches how the relationships between variables can be taken advantage of.
I love the quadratic formula, and it certainly is faster for solving quadratic equations, but completing the square has other uses like getting a quadratic into vertex form which is what I’m doing here. You can’t use quadratic formula for that.
I have never completed the square in my entire life and I have a degree in engineering. Maybe I had to relearn it for differential equations or something but that was for a specific thing and it was quickly forgotten. Not an important skill, just understand concepts.
That's because you live in an age when you have access to a computer and the internet in the professional world. Had you been an engineer 50 years ago, you'd likely need to use this all the time. Even if it is unlikely one will need to use this skill, it's still important to use so you understand what a formula is doing, rather than just blindly using it like a machine.
If you throw a ball in the air, its motion will trace out a parabola. Completing the square allows us to find the vertex of the parabola which can give us how high the ball goes. It also allows us to find the zeros which is where the ball is thrown and where it lands.
Why are people talking about A, coefficient of X² here? You dont need a solution for the A coefficient, just divide the rest of the terms by A if it's bothering you... 0 divided by anything is still 0 anyways
If you were going to teach completing the square, but you write the equation with the letter A as the coefficient of the quadratic term, then it is incumbent upon you to explain what to do with it. Otherwise you should say that you are assuming that A equals 1.
With a quadratic this easy, why not just factorise. x²-4x-12=0⇒(x-6)(x+2)=0⇒x=-2 or 6. For a method that does not require x² on its own or adding the magic number (square and subtract stage) see: th-cam.com/video/Imc4ViiD1xY/w-d-xo.html
It’s very simple why. (x+a)^2 = x^2 + 2ax + a^2 . So, (x+a/2)^2 = x^2 + ax + (a/2)^2 . So, x^2 + ax + b = (x+a/2)^2 - (a/2)^2 + b It’s more important students learn WHY it works, more so than how it works
Yea i hear ya. Obviously I couldn’t explain both the steps and why they work in this video which was limited to 60s. I can make another video on why it works if people want it. Mostly, people just want to know how to do it.
But this is a tutorial for, you know, completing the square lol. CTS has other uses like finding the vertex of a parabola which you can’t do as easily by factoring.
Hi, I see the equation is factorable, so why are you trying to complete the square. There's nothing wrong in completing the square when it's factorable, it will still work. but the main use-cases people use the completing the square method is when a quadratic equation isn't factorable. There's 4 methods for quadratic equations; factoring, extraction of roots, completing the square, and the quadratic formula. Decending from easiest to most complex, If your showing a simple problem with a complex process, it doesn't show the importance of this method. What I would advise is taking a problem that is "perfect" for completing the square.
I get your point but I don’t consider solving to be a main use-case of completing the square (unless you’re plugging directly into the quadratic formula). The main things we use CTS for is finding the vertex of a parabola, finding the center of a circle from an equation, etc. That’s why my example is fine here, I have no intention of solving this, so I picked a nice and easy quadratic and got it in vertex form.
This video assumes you already know what completing the square is. I only have 60 seconds, but I agree that I should make a video on why you complete the square and also tie in how completing the square is where the quadratic formula comes from! Thanks for the suggestion!
Or you test quickly the solutions, you need to get +12 so X will be even, you need to have -4X to be positive so -2 works Then you try 4 and -4, easy to see it doesn’t work, then you see quickly that -6 works That’s the end
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Don’t learn by memorization, understand how it work and you can apply to more cases.
memorize it if you are using it way too many times and traditional method taking time.
The geometrical meaning of completing the square
U don’t need to understand how it works.
Can someone explain to me why we do this because I don’t understand it but I can do the method?
@@RegardedBee185 Many people do need to understand how algorithms work in order to successfully execute them. It's actually how they learn. I'm one such person.
This works great if the coefficient on x^2 is 1. For other cases you need to factor first, not just use half the value of b.
Yes, I should have specified that in this video. Thanks for clarifying!!
In Taiwan, we learn the “Factoring Method”(not sure what it’s called in English) by finding two factors of -12 which are -6&2, -6 and 2 can be add up as the constant 4 in x term.
Times them (x-6)(x+2)=0, and there’s the easy answer.
If the constant term isn’t dividable, I’ll use the quadratic formula x= (-b±√(b²-4ac))/(2a). Pretty simple imo.
The way you use is quite non-intuitive for me. Still pretty nice to learn a new way.
The reason you’d do this is to make it easier to do algebraic proof for increasing/decreasing functions
You could solve the above equation by
(x-2)^2-16 = 0
(X-2)^2 = 16
x-2=+/- 4
x = 6 or -2
The factoring method could be quicker for numbers that could fully factorise but I think it’s probably quicker than if you’d use the quadratic formula on numbers that couldn’t fully factorise
I also find it non-intuitive, also aren't you supposed to have everything at grade 1? (I mean without exponents, I don't know how you say it in english)
In italy we call the first method you used "trinomio notevole" or "trinomio speciale", (special thrinomial) where you find the 2 numbers that when multiplied are equal to c (so -12) and when added are equal to b (so -4) those 2 numbers being -6 and 2, so you can rewrite the entire thrinomial as (x-6)(x+2).
I think you have to have everything without exponents, so is (x-2)²-16 a complete answer?
And obviously after (x-6)(x+2)=0 the solutions are 6 and -2
@@giovanji7883 yeah, I think learning each method is totally applicable in simple math. But when it comes to advanced math, like Laplace transform or Fourier transform, the exponent 2,3,4… will become really annoying. Using first method will make it easier to calculate.
Btw, what do you mean by having everything in grade 1? I didn’t get it.
Completing the square is used to make an "unfactorable" problem easily solvable using square roots. If you can factor it's probably easier to just factor.
That is actually the method many(mostly asians and by asians I mean the whole asia, not the one in memes) use.
I’ve been done with math for a while now but thanks for helping get a 750 on the math section of my SAT 3 years ago
Hey!!! I’m so glad I could help ya do that!!! What are ya studying now?
@@LudusYT right now I’m getting my paramedic license but I did an AA in Bio and I’m gonna go back to university after a couple of years for Microbio.
@@sean5364 that’s so cool!!! Totally keep me posted with how that’s going and lmk if I can help in anyway with anymore math/physics videos!
That's all well and good in this example where you end up with "a^2 - b^2" where a=x-2 and b = 4. This factors out to (a+b)(a-b). Substituting, we get ((x-2)+4)((x-2)-4), which becomes (x+2)(x-6) with roots x=6, x=-2. If c had been (say) 11, you wouldn't have been so lucky.
A far faster method to solve for X, take -12. Look for two numbers that multiply to -12, but also add up to -4. Those numbers would be +2 and -6. So you would have (x+2) (x-6). You can then just set each of them equal to 0. X+2=0 and X-6=0. This is also known as the Master Product method. Usually for me I start here. If this doesn't work I just go straight to quadratic formula. Usually works out better for me that way.
Oh yea I NEVER solve quadratic equations this way. I do the exact same process as you. Try to factor first and if that doesn’t work then plug into the quadratic formula.
Completing the square is just beneficial for finding the vertex of a parabola (which we can now do in this problem) as well as finding the centers of conic sections.
Yeah but this isn’t always reliable so why risk it
@@datboy038 neither is completing the square. Quadratic is the only thing that can always solve it.
@@walterp1028 what drugs are you on? Completing the square will always solve it’s literally just a simpler form of the quadratic formula
@@datboy038 bro literally if the x^2 coefficient is anything other than 1 it won't work.
Edit technically it can work, but in those scenarios it is far more likely to get errors than if you just go straight to quadratic. Or you can master product by putting factors of "a" in front of the 2 different x equations
It doesn't matter what sign is in the bracket, just subtract the squared number
divide b by 2 you get -2 right. then the last number has to be -2*-2 which is 4 (foil method is what i usde to determine that (a+b)^2 = a^2 +2ab + b^2 so it would be 4 at the end to make it wrk. so all u do is add 16 to each side. and get (x-2)^2=16 which you could also get with his way by adding 16 to both sides. then u do the sqrt get x-2=4 then x=6
It's a simple rule: sum of factors is b: product of factors is c - holds for all qudratics
Me literally in calc II having to watch this cuz I forgot xD
too real, one of the only things I never fully understood
cam you please make a detailed explaination on this. I dont want to just memorize and want to actually understand ahat we r doing
This is more efficient but it's still important to understand the other method because you actually learn why it works not just that it works.
It's like how we have to use limits to take derivatives in calculus before we are allowed to use power rule and the other shortcuts.
I agree, except i think this method can be learned initially instead of the other one. It just needs to be explained why it actually works like you’re saying.
@@LudusYT just showing this method doesn't explain why it works. Maybe you can start with this method after showing a proof that it works.
@@SkullPrince8 I agree. For now, this video will at least get kids completing the square. If they want a video explaining why it works, I’ll make that video.
@@LudusYTbeing able to do it is useless. The reason why is of 100% importance. If kids are able to do it by memorisation only. Then they cannot do it at all. Like completing the square in everyday life is useless. But understanding algebra and how to manipulate things properly at least teaches how the relationships between variables can be taken advantage of.
@@LudusYTnah bro I’m in University and my prof taught the same method you did. You don’t need to know how it works
Hi, I am a Mathematics Teacher, so can I download your videos and share them with my students?
Thanks for revision man, i forgot this trick.
X^2 - 4x - 12
(X - 6) ( x+2)
Thank you ! 😊
In India, we call it
middle term factorisation
This is really easier than the quadratic formula, which has already completed the square for you? No way.
I love the quadratic formula, and it certainly is faster for solving quadratic equations, but completing the square has other uses like getting a quadratic into vertex form which is what I’m doing here. You can’t use quadratic formula for that.
I have never completed the square in my entire life and I have a degree in engineering. Maybe I had to relearn it for differential equations or something but that was for a specific thing and it was quickly forgotten. Not an important skill, just understand concepts.
It is fundamentally important in so many different ways if u study in a math degree and more advanced
That's because you live in an age when you have access to a computer and the internet in the professional world. Had you been an engineer 50 years ago, you'd likely need to use this all the time.
Even if it is unlikely one will need to use this skill, it's still important to use so you understand what a formula is doing, rather than just blindly using it like a machine.
Factoring is a different goal than completing the square.
Much easier just to apply Vieta's formula
This is the common method we use in finding conic equations in india😂
What if B is odd
In our country, we just factor it.
Thanks. Please can we have a real world example.
of what, a quadratic equation? lol
If you throw a ball in the air, its motion will trace out a parabola. Completing the square allows us to find the vertex of the parabola which can give us how high the ball goes. It also allows us to find the zeros which is where the ball is thrown and where it lands.
How many people use this in the real world. Why do we learn this maths?
Why are people talking about A, coefficient of X² here? You dont need a solution for the A coefficient, just divide the rest of the terms by A if it's bothering you... 0 divided by anything is still 0 anyways
Uh.... How abou just factor it? Just look, you need two numbers that sum is -4 and product is -12. So 2 and -6 seems good.
If you were going to teach completing the square, but you write the equation with the letter A as the coefficient of the quadratic term, then it is incumbent upon you to explain what to do with it. Otherwise you should say that you are assuming that A equals 1.
Yea, I agree I should’ve said we need a = 1 to do this.
With a quadratic this easy, why not just factorise. x²-4x-12=0⇒(x-6)(x+2)=0⇒x=-2 or 6. For a method that does not require x² on its own or adding the magic number (square and subtract stage) see:
th-cam.com/video/Imc4ViiD1xY/w-d-xo.html
On the one hand, quick and you get the answer….
On the other hand, you don’t understand the process and lose marks in an evaluation…..
how about just differentiate it with power rule and solve for the vertex coordinate
Bruh this is an algebra tutorial… lolol.
+16 both sides and x=±4-2
You need to include the caveat of this only working on monics. Kids would not know.
It’s very simple why.
(x+a)^2 = x^2 + 2ax + a^2 . So,
(x+a/2)^2 = x^2 + ax + (a/2)^2 . So,
x^2 + ax + b = (x+a/2)^2 - (a/2)^2 + b
It’s more important students learn WHY it works, more so than how it works
Yea i hear ya. Obviously I couldn’t explain both the steps and why they work in this video which was limited to 60s.
I can make another video on why it works if people want it. Mostly, people just want to know how to do it.
Or you could, you know, factor it into x-6 and x+2…
But this is a tutorial for, you know, completing the square lol. CTS has other uses like finding the vertex of a parabola which you can’t do as easily by factoring.
It is much faster to solve through Vieta's theorem, the roots are primitive
Yea I don’t consider CTS to be a fast solving method. CTS is more for getting quadratics in vertex form.
High school 2 in Turkey
just do it how you were taught. no need to complicate things you already know
You can if you want 🤷🏻♂️. I just like this way the best.
He made it a lot simpler, r u slow?
Hi, I see the equation is factorable, so why are you trying to complete the square.
There's nothing wrong in completing the square when it's factorable, it will still work. but the main use-cases people use the completing the square method is when a quadratic equation isn't factorable. There's 4 methods for quadratic equations; factoring, extraction of roots, completing the square, and the quadratic formula. Decending from easiest to most complex, If your showing a simple problem with a complex process, it doesn't show the importance of this method. What I would advise is taking a problem that is "perfect" for completing the square.
I get your point but I don’t consider solving to be a main use-case of completing the square (unless you’re plugging directly into the quadratic formula). The main things we use CTS for is finding the vertex of a parabola, finding the center of a circle from an equation, etc.
That’s why my example is fine here, I have no intention of solving this, so I picked a nice and easy quadratic and got it in vertex form.
Might have helped if you had started by defining completing the square
This video assumes you already know what completing the square is. I only have 60 seconds, but I agree that I should make a video on why you complete the square and also tie in how completing the square is where the quadratic formula comes from! Thanks for the suggestion!
Or you test quickly the solutions, you need to get +12 so X will be even, you need to have -4X to be positive so -2 works
Then you try 4 and -4, easy to see it doesn’t work, then you see quickly that -6 works
That’s the end
I’m not trying to solve the quadratic formula. I wanted it in vertex form.
@@LudusYT I didn’t see the full name of the referenced video my bad
I didn’t know that vertex form, that’s a great finding, thanks a lot 🙏
(x - 2)^2 - 16 = 0
(x - 2 - 4) * (x - 2 + 4) = 0
(x - 6) * (x + 2) = 0
x^2 - 6x + 2x - 12 = 0
x^2 - 4x - 12 = 0
Therefore, the method can also be used to find that x^2 - 4x - 12 = (x - 6) * (x + 2).
-6*2 = -12
-6+2 = -4
Where is the fkin mystery.. ?
Only a Dumas would complete square here
Divide by 2a...
I always factor out the a if it’s not equal to 1.
Is this not how everyone does it? Anything else looks weird
answers 2+-4i
The roots are real. A monic quadratic with a negative constant term at the end, always has real solutions.
The solutions are x = +6 and x= -2.
X=6?
X=6
It’s b/2a, not b/2
If you have an a, you would want to factor it out first! That way it’s still b/2. I can upload a sequel video eventually
thanks
X=18,-14?