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@That_One_Guy...

Жыл бұрын

Wait isn't it m +- sqrt(m^2 - p) instead of m +- sqrt(m^2 - c) ? that would make c = c/a

@atharvasinha3205

5 ай бұрын

Hello can we use this formula for solving the equation consist of iota i in the coefficient of b in the equation?

This is a great way to use the quadratic formula without "using" the quadratic formula. I've seen this method used on other channels.

@BriTheMathGuy

Жыл бұрын

It's true! It really sort of is the formula (with a bit of trickery)

@magicbaboon6333

Жыл бұрын

It slow tho .original best

@sanjayluthra3464

Жыл бұрын

@@magicbaboon6333 but if you practice it is very much faster than quadractic formula

@somerandomuserfromootooob

Жыл бұрын

​@@BriTheMathGuy but it doesn't work for some quadratic equations which dosent form a parabola in the graph. Example : x²-350x+660.,, .....(to be continued for eternities lol)

@RahulMaru3507

Жыл бұрын

@@somerandomuserfromootooob It does form a parabola (plot it on desmos and zoom out), every quadratic does

I don't know if "better" is the right word, but I think this is "as good." It highlights a different facet of quadratic equations and functions in a really cool and enlightening way. But the quadratic formula we were taught in school also highlights some important relationships, namely, that of the vertex and the discriminant.

@marius4363

Жыл бұрын

good luck find imaginary roots with this

@marcusmelander8055

Жыл бұрын

@@marius4363 I mean, it actually works just as well, and arguably the same. If U ends up as the square root of a negative number, congrats! You've just found the imaginary roots.

@Mrtamps

Жыл бұрын

Yeah i would say, its about the same

@thereeldeal7858

Жыл бұрын

@@marius4363 cykabled??

@kbin7042

Жыл бұрын

​@@marius4363 this method IS the quadratic formula, so it's completely possible to find complex solutions, u2 will be negative and that's it

m ± √(m² - c) is because we let m = -b/2, and so it is -b/2 ± √(b²/4 - c) which becomes -b/2 ± √(b² - 4c) / 2 which is the quadratic equation in the case of a = 1. And we can always ÷ the equation to get a = 1

@andrewkarsten5268

Жыл бұрын

Yeah that’s why it gives you the same answer, it is mathematically equivalent. The point is to show an intuitive way to “derive” the expression that isn’t completing the square

@vaishirv2237

Жыл бұрын

Thanks for explaining it😭 Quadratic makes my life hell

@onradioactivewaves

Жыл бұрын

The jokes not funny if you explain it 🙄

@pauselab5569

Жыл бұрын

Yes it’s basically x=-b/2a also graphically, x=-b/2a moves the vertex of the parabola to the y axis in the function f(x)=ax^2+bx+c which leaves us with only the constant and the square. This is kinda better than completing the square as moving special points of a graph is useful for simplifying any equation to ax^n+bx+c=0 (though there isn’t a simple formula way of solving this for n >=5

@elen1ap

6 ай бұрын

If you derive m=-b/2a, then you get (-b +- sqrt(b²- 4a²c))/2a. Why does this happen?

The formula we learn here in Germany is x1,x2=-p/2+-sqrt((p/2)^2-q) where in a quadratic ax^2+bx+c=0 p=b/a, q=c/a has the nice name of pq-Formel

@JeeAdvancera

2 ай бұрын

It's the same formula but simplified to a complicated equation 😅

@alifsheikh4237

Ай бұрын

​@@JeeAdvancerause this when b is even

@bilkishchowdhury8318

29 күн бұрын

Midnight equation

@emelieforsmark28

24 күн бұрын

This is what we learn in Sweden as well!

@demonnauki

23 күн бұрын

Ach ja, und natürlich auch das schöne "pq-Formel Lied" vom DorFuchs

In Germany in school we learned to solve quadratic equations by setting a to 1, i.e. to bring it into the form of x^2 + px + q = 0. To solve you simply use x = - p/2 +- sqr ( (p/2)^2 - q). I personally think thats easier to memorize than the other formula

@yama123numbercauseytdemand4

Жыл бұрын

Dorfuchs?

@toyb-chan7849

Жыл бұрын

die jute alte pq formel

@jofx4051

Жыл бұрын

Simpliflied quad equation is just easier to remember

@yama123numbercauseytdemand4

Жыл бұрын

@@toyb-chan7849 Absolut zuverläßig. :)

@wernerviehhauser94

Жыл бұрын

@@jofx4051 from first hand experience, that is not true. The pq formula is messed up more often by forgetting to divide b and c by a..... Most states in Germany prefer the "larger" formula for a reason, it also is easier to use when solving physics problems that usually come with crooked numbers.

Introducing the a factor into this equation is surprinsingly simple. Thank you for the guidance ; i hadnt realised what the b factor represented.

This is essentially equivalent to dividing the entire equation by a, and replacing the linear term bx with 2Mx.

It’s interesting. This approach is more similar to how financial professionals approach pricing securities. One can use Ito’s lemma to price non-linear products. The key result is that the one can deduce the width of the quadratic curve using: Sqrt(2*Θ/Γ) Where Θ = the absolute value of the minimum of the curve, or F(F’(X)=0). This is given by: B^2 * (1-2a)/4a^2 + c Where Γ = F’’(X), which for a quadratic is just a constant. Defining M as b/2a (from ax^2 + bx + c). The roots become: -M +- sqrt(abs(b^2 * (1-2a)/4a^3 + c/a)) That all simplifies to: -M +- sqrt(abs((M^2 -bM + c)/a)))

We learnt this extensively in school.. but in college the 'usual' formula was always used and i had forgotten the trick, ie, which was addition, which was multiplication. Thanks for helping in recalling it.

This was in our high school textbook when learning the quadratic formula. This shows the intuition of the quadratic formula. It is much easier to just remember the one quadratic formula, than remember and figure out these steps in the video. Less room for mistakes too. (I did university engineering and economics btw).

I think the reason we are taught to solve these problems the way we are is because students quickly move on to equations that aren't quadratic. They need more robust strategies to deal with more advanced polynomials. While this is a great trick and something that I can definitely see coming in handy or saving some time, learning to factor through some method is truly much more helpful in my experience. Past more introductory levels of algebra, solving quadratics was never a consistent part of any of my courses.

I had never heard of this before. Thanks Brian!

@BriTheMathGuy

Жыл бұрын

You bet!

@hubusgaming7718

2 ай бұрын

its actually called Viete's theorem or Viete's equations , they relate solutions for anypolynomial and are really useful, I wrote a paper and had like 20 usages of them in Olympic exercises

I'm going to start using this. Thanks Brian!

I think possibly the reason we get the 'clunky' form of the quadratic formula is just to follow certain conventions: "always combine fractions" and "always bring common factors out of radical expressions." So often you can see recent textbook presentations of the quadratic formula will actually derive this more intuitive expression, and then they apply the rules to get everything in one big fraction. I sort of wonder if older books went a different way because typesetting for math was fairly limited, and these rules had a real benefit in cases like that.

@SmallSpoonBrigade

6 ай бұрын

We don't generally use this method because it requires external calculations. If you're going to use a canned formula, you might as well use one that has everything necessary in it when available. All this is to hide details that you still have to remember and deal with in an opaque variable.

The content you deliver is awesome! Can you please let me know that from which textbook source did you study all the methodologies of this particular lecture?

Completing the square has other uses outside of just finding zeroes, such as when representing rational functions as partial fractions in a form suitable for integrating. But this does look shorter, you just have to simplify to a more specific form first.

So, basically we just manually remove the coefficient 'a' by dividing and substitute m=-b/2 to make it more simpler in looks but actually harder. While I will not call it the 'better quadratic formula', I did like the new perspective of the derivation of the formula brought in the video!

@Anmol_Sinha

Жыл бұрын

@Reio4 yeah, I said that we can remove it by dividing. (By dividing I mean dividing the polynomial by a). Thanks for highlighting it for the readers of this comment.(if that was what you actually meant to do)

@TamissonReis

Жыл бұрын

Actually it mixes the summ and product rule (factoring) in a way that you don't have to guess. To me it is good because you can use summ and product trying to guess and, if you can't, you use this method as an expansion.

@programmingpi314

Жыл бұрын

@@TamissonReis You don't have to guess with the quadratic formula (which this is just a poor imitation of) either.

@danieltemelkovski9828

Жыл бұрын

@@programmingpi314 This method puts you more 'in touch' with what is happening graphically, imo. The quadratic formula just has you blithely punching numbers into a calculator without lending much understanding of what you're doing. Although if you learn the derivation of the quadratic formula from 'completing the square' you will gain some insight into what you're doing. (All this presuming the person doing the calculations actually cares about understanding, rather than just 'getting the right answer' for a test or something.)

@user-pr6ed3ri2k

Жыл бұрын

@@danieltemelkovski9828 you don't have to know how the black box works to use the black box

while this formula looks shorter, when you sub -b/2a for M and do a little manipulation, you still get the same old quadratic formula, so I don’t know how effective it actually is in the real world.

I always noticed that the difference between x1 and x2 was always the discriminant and due to almost always working with a = 1, I always felt that the value between those two was somewhat special. Well, now I at least know I was onto something!

@stratonikisporcia8630

6 ай бұрын

Isn't that difference of roots √Δ / a

@Fire_Axus

3 ай бұрын

your feelings were irrational

@RealJackBolt

Ай бұрын

the difference of roots for a quadratic is : √∆/ |a|

So glad I'm not the only one using xtop and distance delta plus/minus from xtop. With this formula you will know xtop, the discriminator and delta, each important to understand a parabola.

Amazing formula I’ve never thought about this and I haven’t been teaches this trick Thanks a lot man ❤️🌹🙏

The midpoint approach fails geometrically when the roots of the polynomial are *complex.* Consider x² + x + 1 = 0. M = -1/2 and we discover that U² = -1/2. If U is interpreted as a distance and it's positive, then how do we interpret U = √(-3/2) as a distance between the midpoint and the roots, as U isn't a real number? Sure, the midpoint approach works algebraicly, but not geometrically when the roots are complex.

@angeldude101

Жыл бұрын

Well of course you're not going to find a ℝeal distance to the roots when the roots themselves aren't ℝeal. However I have actually seen a concept of an sphere with an imaginary radius in the past which could be thought of in a similar manner. The two roots of a quadratic can be thought of as a single 0-sphere with a centre of M and a radius of U. In this case U is imaginary.

@davidbrisbane7206

Жыл бұрын

@@angeldude101 Where did he say in words that he was restricting the solution to only the real numbers? He does only show examples where the solution are real. I'll stick to the "Clunky" formula, as its solution allow complex solutions also.

@davidbrisbane7206

Жыл бұрын

@@angeldude101 Of course, we could use the discriminant to determine if the solution has complex roots or not 🤣😂🤣😂.

@angeldude101

Жыл бұрын

@@davidbrisbane7206 Ultimately, this formula is algebraically equivalent to the other. M = -b/2a, U² = M² - c/a. = M ± √(M² - c/a) = -b/2a ± √((-b/2a)² - c/a) = -b/2a ± √(b²/4a² - c/a) = -b/2a ± √((b² - 4a²c/a)/4a²) = -b/2a ± (√(b² - 4ac))/2a = (-b ± √(b² - 4ac))/2a Any result the traditional formula will give, this alternate formula will also give. The only difference is that the traditional formula is "simplified" into a single fraction.

@davidbrisbane7206

Жыл бұрын

@@angeldude101 Except for the graph 😀

Beautiful!

This formula reminds me of a variant of the quadratic formula to use when b is even, if you set M=b/2 then the overall formula can simplify to (-M+-sq(MxM-ac))/a, removing the clunkiness of the x4 in the square root and the x2 in the denominator

Awesome video! Thank you! This very interesting from a computational standpoint. Seems much faster to implement compared to the usual formula. That being said, I wonder if this is still applicable for quadratic formula with complex roots. It seems that the graphical intuition won't be applicable then.

@andrewkarsten5268

Жыл бұрын

It does still work, the only difference is you get U^2 to be a negative term instead of positive. Then when taking the square root on both sides, i pops up

@kbin7042

Жыл бұрын

is this really faster for computers? because this is literally the quadratic formula

It makes so much more sense... I've discovered it myself the first time I tried to write a computer program to solve this. Later I've found out that this is the method taught in high schools in some countries (Germany? China?)

this method was explained only in special cases where the a = 1. but incase you want to know them for the general cases, for the sum of x1 + x2 = -b/a, and for the product of x1 . x2 = c/a

The Po Shen Lo Method! I usually teach my kids the formula x = (-B/2) +/- sqrt( (B/2)^2 - C) after we've divided out a first. Very similar to the related formula x = M +/- sqrt( M^2 - P). Granted I show them the actual formula too. Both is good.

@wyattstevens8574

7 ай бұрын

"M plus or minus square root of M^2-p!" -Tim Blais (he made a short tune for it, used in some 3Blue1Brown videos)

Factoring, completing the square, the quadratic formula are all easy to learn and use. Learning how to derive these is the key to enjoying this level of maths and sets the groundwork for harder things to come.

This really reminds me of how you find the eigenvectors of a 2x2 matrix where M is the mean of the diagonals and the C term is the product of the diagonals. 3b1b did a great video on it in their literary algebra series

@beuh0623

22 күн бұрын

*linear algebra, phone keyboards suck lol

That's really interesting! We had to factor for quite a while as we hadn't learnt the quadratic formula yet last year, so it got me thinking about an easier way to do it rather than just in my head. I attempted to use simultaneous equations but eventually it lead me right back to square 1.

Yes, this is what i do with the po-shen loh style. The most simple if you use the -b/2a , but a=1, so it becomes -b/2 only. Vertex formula to find the “h” value of the vertex: (h,k). That is actually the AOS or Axis of Symmetry. Po-Shen Loh is hiding this one.

That's valuable, thank you ☺

I was never taught the vertex formula my entire life, I had a test a couple months ago and used completing the square for finding the vertex. I know, such a waste of time, but now I actually know the vertex formula and proved it using calculus. Thanks.

I saw the Poh-shen-loh method before! It's really cool! It got me to experiment and see how much I could simplify it(like this video). I found a few fellas in a comment section to some other video(can't remember which, it was a while ago unfortunately) mention a few ideas that got really close to what I was looking for, but I still thought it could be slightly neater. So I played with it a little and found the following: Starting from a typical trinomial: ax² + bx + c = 0 -x² - (b/a)x - c/a = 0 B = -b/(2a) C = -c/a -x² + 2Bx + C = 0 x² - 2Bx + B² = B² + C (x - B)² = B² + C x = B ± √(B² + C) So what all that means if we ignore the two steps before the last, is that we can divide everything by "-a", and jump directly from this: -x² + 2Bx + C = 0 ...to this: x = B ± √(B² + C) Basically, this is just yet another way of looking at what was just shown in the video. Really awesome however you look at it, and I'm glad to see someone put out a video to make this more known! :)

@kevinkasp

Жыл бұрын

Can you (or someone) help me out? I tried this on a very simple problem that I can do in my head, and using this video's method my answer comes out with the wrong signs. When I apply this method to: x^2 - 5x -14 = 0 I get a mid point of 5/2 and a distance U of 9/2 My set of solutions for the zeros of the quadratic then become 5/2 - 9/2 = -2 and 5/2 + 9/2 = 7 But that's not correct. The correct solution is positive 2 and negative 7. What am I doing wrong?

@superiontheknight963

Жыл бұрын

@@kevinkasp The set of solutions you got from the video are actually the correct solutions. So I'll do this both the way the video did it, and the way I did it in my comment. :) We'll start with the videos approach: x² - 5x - 14 = 0 M = -(-5)/(2(1)) = 5/2 c = -14 U = √(M² - c) U = √((5/2)² - (-14)) = √(25/4 + 14) = √(25/4 + 56/4) = 9/2 x₁ = M - U x₁ = 5/2 - 9/2 = -4/2 = -2 x₂ = M + U x₂ = 5/2 + 9/2 = 14/2 = 7 Now if we do it the way shown in my comment: x² - 5x - 14 = 0 -x² + 5x + 14 = 0 B = 5/2 C = 14 x = B ± √(B² + C) x = 5/2 ± √((5/2)² + 14) = 5/2 ± √(25/4 + 56/4) = 5/2 ± 9/2 x₁ = 5/2 - 9/2 = -4/2 = -2 x₂ = 5/2 + 9/2 = 14/2 = 7 Either way you get to the correct solutions of x = -2 or x = 7. We can check this too and see that our answers are correct. (-2)² - 5(-2) - 14 = 0 4 + 10 - 14 = 0 0 = 0 True! (7)² - 5(7) - 14 = 0 49 - 35 - 14 = 0 0 = 0 True! We can also check the other two solutions you came up with, and see they don't work. (2)² - 5(2) - 14 = 0 4 - 10 - 14 = 0 -20 = 0 False! (-7)² - 5(-7) - 14 = 0 49 + 35 - 14 = 0 70 = 0 False! Perhaps you were confused when this is factored. Remember, that "x² - 5x - 14 = 0" can be factored as the following: (x - 7)(x + 2) = 0 Remember though, x is still "-2" or "7". This can be seen by plugging in for x. ((-2) - 7)((-2) + 2) = 0 (-9)(0) = 0 0 = 0 True! ((7) - 7)((7) + 2) = 0 (0)(9) = 0 0 = 0 True! Hope this helps! :)

@kevinkasp

Жыл бұрын

@@superiontheknight963 Got it. Brain fart on my part. It's been so many years since I had to do problems like this I did in fact make the exact mistake you suggested. Meaning, I got the correct solution, but "checked" my answer by looking at the factored form of the equation and did the brain fart of thinking ( x + 2) shows +2 is a factor, instead of solving x + 2 = 0, which would have proved that I had the correct solution. That's as bad as it gets. To solve a problem correctly and then discard the answer because you take the time to check it, but then do that incorrectly. Thank you for taking the time to set me straight. Also, your method is the way I would teach it to kids learning algebra. I would use the video's method of visually showing what is to be accomplished, and yours to do problems. Thanks again.

The midpoint M, a.k.a. the axis of symmetry, is just -b/(2a); the distance U from the midpoint to one of the solutions is then the square root of (M^2 - c), a.k.a. the square root of the quantity of the discriminant D divided by 4a^2.

normally completing the square (x+b/2)^2 + B is the go to for quadratics with even 'b' value and then factorising (x+A)(x+B) for prime 'b' values

personally i love factoring and will try to use it in any case unless necessary for the quadratic formula

A very great method to solve quadratics

admiring the simplicity of this, i am left wondering why this is not how everyone is taught this formula. thank you for opening my eyes to this wonder. god bless you, stay healthy.

@stratonikisporcia8630

6 ай бұрын

Simplicity ? This method is probably the hardest way to do it, why struggle with this graph-based algorithm when you can just remember one goddamn formula, that's even easy to find again if you forget since the proof is quite simple

I'm actually proud of myself rn because I already came up with this exact thought before

Love You!!!

This method much way faster than the original. Thank you so much!

What an amazing video!!

Dude this save me from the factoring quadratics exam thanks!

I'm in middle school and on top of the quadratic formula we were taught another method: make an x shape and put the product of a times c into the top. Put b into the bottom. Figure out what 2 numbers add to b but multiply to that other product(if you can't then there's no solutions).Then, put the original a into the top left of a 2x2 box and the c into the bottom right. Fill in the other 2 boxes with those factors from earlier but they're both multiplied by x. Find the greatest common factor in each row so one row was one factor and the the other row was the other.

Isn't this just like another way to think about/write completing the square? Because that IS something that they teach at least in my state.

This is literally the quadratic formula but simplifying the 2a from the denominator within the square root

Is this the same or similar to the one that 3b1b showed on his livestreams from about a year ago? This method is super cool, glad to see it again as a refresher :D

@BriTheMathGuy

Жыл бұрын

I didn't see the stream but it very well could have been!

Bravo ! Your approach is what I myself have been advocating for years. That is, it's based on the simple fact you point out - that the two roots are equidistant from the vertex value of the parabola.. A proof might run as follows : Let R1 and R2 be the roots of the equation so ( x - R1 ) * ( x - R2 ) = 0 ; Arbitrarily R2 > R1. x^2 - ( R1 + R2 ) * x + R1 * R2 = 0 which is equivalent to x^2 - b/a*x + c/a = 0 but m [ mean ] is ( R1 + R2 ) / 2 which is obvious from geometry. So m = - b / ( 2 * a ). We can define d as the difference between the value m and R1, R2 such that R1= m - d and R2 = m + d . But R1 * R2 = c / a -> ( m - d ) * ( m + d ) -> m^2 - d^2 From which we can assert that d = sqrt ( m^2 - c / a ) In summary then the two roots are : ( m - d ) and ( m + d ) where m = - b / ( 2* a ) and d = sqrt ( m^2 - c/a ) A quick check that if m^2 Many teachers will deplore this approach, since they feel that the various ways of solving the quadratic [ completing the square, factorization ...etc ] are important pedagogic tools. Schooldays behind them, those who have to solve quadratics in real life want the roots quickly and your method is logical and fast, and based on simple geometry. " When I was a child, I thought as a child ; now I am a man, and have put aside childish things. "

@asadmahmood2007

Жыл бұрын

I love that you're teaching so many people and getting them to reexamine the quadratic formulae ❤🙏👍 That being said, It's a prerequisite for High School level Math classes here in the Indian Sub-Continent. I'm back in Pakistan and yep, I make my students create their own sums. That's the best way to test their skills i.e. working backwards and forward. A good tip for teachers/tutors is to collect them and make a test using all the questions

@crustyoldfart

Жыл бұрын

@@asadmahmood2007 SO, if I understand you correctly, you are saying that as a professional teacher at High School level, your main task is to teach students how to meet the the pedagogic dictates of the examining authority. One wonders if there is any room in such a system to imbue students with a feeling of awe and wonderment regarding the nature and immensity of mathematics. Speaking for myself I endured the pedagogic dictates of public exams up to the age of 22 when I graduated in engineering. It is only since then that I have developed a reverence for the greater part of the mathematical world, but with the reservation that I am not only incapable of understanding anything but a small part of that world, but there is much of it which holds no appeal at all.

for basic quadratic equation i was taught in school to do: -b/a = Sum of both x and c/a = Product of both x so for instance in the example you showed -b/a= -(-4)/1=4 and c/a= 3 that means the x1 and x2 are equivalent to 1 and 3

@andrewkarsten5268

Жыл бұрын

That’s for factoring, but yea

After following this channel I started falling in love with Maths

I use the -p/2+-root of (p/2) squared-q. p and q are b and c. These points can also be used in the (x-x1)*(x-x2) formula ( sry for the writing dunno How to Write root of something on iPhone)

I wish i had seen this video 1 day before I had my test

This is super cool. I think this can be linked to Vieta's formula right??

I will hold on the completing the square method

Btw please do more videos like these since I’m going to 10th grade and I need fast methods to solve problems since I’m my country after year 12 we have a big exam in order to go to university I’d really appreciate it ❤️🌹

i love your vids keep it up

@BriTheMathGuy

Жыл бұрын

Thanks for the support!

Isnt there an easier way? In this case, the formula you mentioned can be sovled using the Fact that a+b+c=0 in which case one of the roots is 1 and the other is c/a I know this only works for this equation but equations are usually given in a form to make it easy for you to find the roots and this formula is really easy to specify the roots.

yeah i mean, this is kind of a derivation of the dharacharya formula, still pretty impressive, gives more visualisation of "roots" of equation

That's really cool!

@BriTheMathGuy

Жыл бұрын

Glad you think so!

This is such a cool way of visualizing how (a + b)(a - b) = a^2 - b^2 with two points b distance away from a

I LOVE this, and I am very sad that I have never seen this in my high school

In Turkey, we are learning these in high school starting from 11. grade.

I love how I have been taught this from the beginning (I live in Sweden)

I wish I had learned about this starting Algebra II

I started using this method a few years ago when Lo published it. Of course I'm way beyond math class being about solving quadratics, but sometimes they need to be solved incidentally in the process of solving a harder problem. I've confused more than a few professors by employing this method! It's so much faster than the full quadratic formula, and avoids the guess-and-check element of factoring.

In germany, we learn the pq formula, which is a bit simpler than the quadratic formula you showed at the beginning: -p/2 ± √(p/2)^2 - q) (p is the x¹ term and q the x⁰ term) You need to normalize the quadratic equation by deciding by the factor of the x² term and then plug p and q into the equation. I don't know if it makes it a lot easier to calculate, but I think it's more elegant than having the whole formula devided by 2a at the end

@SmallSpoonBrigade

6 ай бұрын

IIRC, we wouldn't use that in the US typically because the rational zeros theorem that that's based on isn't taught until later. It can be used, but there's little utility in using it on a 2nd degree polynomial when we've got both completing the square and the quadratic formula as well.

@filipbergman4232

6 ай бұрын

We are taught this in Sweden too. I feel like it is less clunky compared to the other formula.

@yannik4966

6 ай бұрын

I just checked it. If you puzzle the steps he did in the video together, this exact formula is the result

question, if the coefficient term A is negative, would it be safe to multiply the entire equation by -1? (-x^2+bx+c=0 would be the same as x^2-bx-c=0.) Or is this method no good in that case.

A very good method. Fairly obvious if you use to sketch the graphs though.

What I do is x^2 - 4x + 3=> x(x-4)=-3 And then I take -3’s multipliers (which is 1,-3; -1 and 3) and just try them to see which one is 4 smaller than the other.

been using the factoring method ever since i started learning about quadratic it is good but when things get complex the quadratic formula is better (in india it is normal to use both we were taught this in school )

Since M= -a/(2b), you'll often be stuck with simplifying the square root of a fraction, even when the coefficients are integers. The reason why the quadratic formula is written the way it is, is to eliminate the simplification necessary every time. Notice that the presenter only chooses equations with integral values of M...

in germany we learn the p-q Formel which is a formula where you need the x squared part to be 1x squared to apply it find the p which is the number with x and the q which is the number without any x then plug it into -p halves plus minus the square root of phalves squared minus q it is very easy for me to remember due to a song made by the youtuber dorfuchs

The discriminant comes an alternate formula of the quadratic function: f(x) = a(x + b)^2 + c. The three scalars here determine the precise location of the extremum point and the slope of the curve. a determines the slope, b determines the inverse of the abscissa offset of the extremum and c determines the ordinate offset. The discriminant comes from taking the f(x) = ax^2 + bx + c function and transforming it into the above. I'm not going to do the full derivation in plain text, but it comes out to f(x) = a(x + b/(2a))^2 + (-b^2 + 4ac)/4a.

I always knew this as pq-Formel and in Germany it is taught as the main method to solve quadratic equations

You are a freaking genius

This is quite interesting!

I discovered this myself when I didn’t understand the way it was being taught. I’m really intelligent and work for like 5 hours to come up with this

(x+1)^3:”I think I forgot something.” x^3+1:”If you forgot, it probably wasn’t that important.” (x+1)^3:”Yeah, you’re right.” 3x^2+3x: “-_-“

@Mono_Autophobic

Жыл бұрын

√-1 2^3 Σ π

if i cannot factor the equation i will usually just manually derive the quadratic formula by completing the square, shifting c-(b²/2a)² over to the right and then solving for x - it just feels nicer that way

I remember learning this in school, but just looked a little harder(we didn't use the formula just factoring ones), so I never tried it.

you can also use a simpler formula for quadratics with even B rather than the general one, cause it is little bit faster. k=b/2, x12=(-k+-sqrt(k^2-ac))/a

It's so cool and obvious when you see it. haha I love it!

To find an analytical solution for ax^2+bx+c = 0 is easy. However, what is the analytical solution for ax^(2+e)+bx+c=0 with ‘e’ being a real number? The solutions are x1=(b/(az))Wq(((-c/b)^z)(a/b)z)^(1/z), where z = (1+e) and q = 1-1/z. x2 = (-y(a/b)Wq((-1/y)(b/a)((-c/a)^(-1/y))))^(-1/(1+e)) where y = (2+e)/(1+e) and q = 1+y Wq is the Lambert-Tsallis Wq function (a generalization of the Lambert function). Sometimes the correct solution is x1, in other cases the correct one is x2 and there are cases where x1 = x2, depending on the values of a, b and c. For example the solution of x^(2.5)+x-1 = 0 is x1 = x2 = 0.6540 (up to 4 decimals).

In India we do this at the very beginning of the quadratic equation chapter

Yeah, this is my go to method to solve quads. It's pretty easy to quickly do in your head in most situations. They should definitely teach it in schools.

@vamsikrishna6101

Жыл бұрын

no I don't think so it's not possible to get roots when we have quadratic equation with complex roots

@Gotouchsomegrass-bs9nu

Жыл бұрын

I'd say that the original formula is much easier to do in the head, and it works in everything

I'm Swedish, and this is the formula I've always learned in school, we call it the P-q formula. I've always seen the other version on the internet however which always confused me since it never seemed to have an advantage and it's harder (more time-comsuming) to type it into a calculator. The only thing I've noticed is that the quadratic formula can be used on all quadratic equations without any "set-up" while the p-q formula require a=1 (correct me if I'm wrong, I never use the quadratic formula normally). If there is any real advantage to the quadratic formula I'm not aware of, please do inform me because I'm genuinely curious of why everyone I see always use a formula I see no value in.

@mohammedsaqibkalsekar1058

Жыл бұрын

idt there is any advantage

@uraharakisuke1329

Жыл бұрын

Same here, the pq Formula is just way easier to use

@andrewkarsten5268

Жыл бұрын

Yes, the only real “advantage” is that the quadratic formula doesn’t require any “set up.” However, as far as typing into a calculator goes, I wouldn’t know which is “faster” since I never use the calculators. I’m a senior in college majoring in math, and I haven’t used a calculator in about 7 years. I only used them in stats for distribution calculations. For quadratics, I always just completed the square and solved algebraically since I could always do that faster than I could type it into a calculator. Plus I hate computers

@justintroyka8855

Жыл бұрын

I agree the P-q formula is better in many cases, but I prefer the traditional quadratic formula in cases where b is odd or a is not 1.

@onradioactivewaves

Жыл бұрын

The advantage of the quadratic formula would be for any application that requires anything other than the roots (zeros).

we were taught the difference of squares method at my school. Do other schools not go over it?

With practice, this could be faster than the traditional quadratic formula when you're calculating with numbers. But I feel like it would get tricky to follow if a, b, c are themselves variable expressions. In that case a 1-step solution like the traditional quadratic formula seems more straightforward to me. Maybe that's just me, though....

this is literally just the formular we've been taught here in germany

My man literally invented the quadratic formula a second time

We actually learn that in Germany. It's called the p-q-formula, because if you put all steps into a single formula for x² + px + q = 0 the solution is x = -(p/2) ± √[ (p/2)^2 - q ]. Maybe it looks a bit complicated because of all the parentheses but you usualy write (p/2) as a fraction so it looks much tidier (like this www.gut-erklaert.de/images/mathematik/pq-formel-loesungsgleichung.jpg ). It's so convenient that at least in the part of Germany I live in (Lower Saxony) we don't even learn the whole quadratic formula because you can always just divide by a to get rid of the x²'s coefficient.

@prakharvermaT1nO

Жыл бұрын

germany always was ahead in STEM no surprise there, glad i learned this from the video it will come in handy sometime hopefully

@darki6339

Жыл бұрын

Was about to comment this, but you did a better job at explaining than I ever could have, thank you

nice!

Actually quadratic formula is best way to solve quadratic equations. If you need some analysis of graph for some advance problems, Vieta is good, also transformation to the form where it's easy to see, how much offset is for x and y

In which class did u learn this formula?(in school)(sophomore,frehmen,high school)

I actually had derived this on my own to save time 😭😭😭

we have been taught -p/2 + / - sqrt(p²/4-q) which practicalli is the solution here but without the geometry stuff ^^ And from -p/2 you simply have to square it to get the p²/4, so it's fast.

But does this work for all equations? What are the limitations of using such a method? I know that the Quadratic Equation works for every single Quadradic in existence and that Factoring is really only used for simple Quadratics that can be done in less than a minute or so. What if one of the answers is a complex number? What if the graph has only 1 zero or even no zeroes? I know I'm being pedantic but as someone who loves math, I care about the details. Details are important.

@DavoDovox

Жыл бұрын

I think it works for every quadratic since, as I read from a comment, it can be derived from the standard one and I don't do any "invalidating" operation as long as a ≠ 0 -b/2a ± √(b²-4ac)/2a Let a equals one, so in case it isn't divide the whole equation and bring in the root the 2 -b/2 ± √(b²/4 - c) b²/4 = (b/2)² = (-b/2)² Now let -b/2 = m m ± √(m² - c)

@creepstriod8147

Ай бұрын

No quadratics cannot have one real and one complex roots. I think if we check the discriminant first the verify that it has real roots then this will work fine.

@RealJackBolt

Ай бұрын

Well if you really want to minimize the effort of learning the quadratic formula, then I guess you can do the following: 1. Make the value of a = 1 by dividing the equation by a on both sides. 2. Differentiate the quadratic and equate it to zero to get a value of X (Call it B, where it stands for Bolt's Number) 3. Calculate J = (√∆)/2 (∆ = discriminant) (J stands for Jack's Number) 4. Now the roots of the equation is given by: X1,2= B ± J ( im sure you can remember this easily!😂) Note: if Jack's Number (J) has a real value, the root(s) with be real aswell given that a,b,c were real. If J is an imaginary number, roots will also be imaginary. You can call this The "BJ" Theorem.