The Better Quadratic Formula You Won't Be Taught

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/ @brithemathguy
Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
Corrections:
2:32 Make sure a=1 (divide by the leading term)
#math #brithemathguy #quadraticequation

[@BriTheMathGuy]

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[@That_One_Guy...]

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

[@atharvasinha3205]

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

[@wyattstevens8574]

​@@That_One_Guy...Exactly what I know it as!

[@SuperMtheory]

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]

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

[@magicbaboon6333]

It slow tho .original best

[@sanjayluthra3464]

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

[@somerandomuserfromootooob]

​@@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]

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

[@johnbutler4631]

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]

good luck find imaginary roots with this

[@marcusmelander8055]

@@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]

Yeah i would say, its about the same

[@thereeldeal7858]

@@marius4363 cykabled??

[@kbin7042]

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

[@yoyoezzijr]

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]

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]

Thanks for explaining it😭 Quadratic makes my life hell

[@onradioactivewaves]

The jokes not funny if you explain it 🙄

[@pauselab5569]

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]

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

[@vice214]

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]

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

[@alifsheikh4237]

​@@JeeAdvancerause this when b is even

[@bilkishchowdhury8318]

Midnight equation

[@emelieforsmark28]

This is what we learn in Sweden as well!

[@demonnauki]

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

[@nigelmansfield3011]

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.

[@acidishot2]

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)))

[@fireflyslight8155]

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.

[@mattotonton]

you cannot solve higher degree polynomials in any straight forward general way past degree 2, approximation methods are used past degree 2. The main reason we learn to solve polynomials is to solve linear differential equations and it turns out degree 2 is in fact the most important case, one way to think about why this is so is because the most of the important equations of Physics are degree 2. Hence the quadratic formula is in fact very important and not some afterthought.

[@the-boy-who-lived]

√(b²/4 - C) = U x = b/2 - U or b/2 + U x = b/2 ± √(b²/4 - C) x = [b ± √(b² - 4c)]/2 This is true only if a = 1 So if a ≠ 1, let's devide everything by a to make a = 1. So b becomes b/a and c becomes c/a and a becomes a/a and as we wanted equals to 1. x = [b/a ± √(b²/a² - 4c/a)]/2 x = [b/a ± √(b² - 4ac)/a²]/2 x = [1/a × b ± 1/a × √(b² - 4ac)]/2 x = 1/a[b ± √(b² - 4ac)]/2 x = [b ± √(b² - 4ac)]/2a So this is exactly the quadratic equation. So saying this a "better" way is not the right way to say. This just explains you what is really going on and gives you an idea of the concept.

[@joelpaddock5199]

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]

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.

[@kingsleys.1319]

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]

Dorfuchs?

[@toyb-chan7849]

die jute alte pq formel

[@jofx4051]

Simpliflied quad equation is just easier to remember

[@yama123numbercauseytdemand4]

@@toyb-chan7849 Absolut zuverläßig. :)

[@wernerviehhauser94]

@@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.

[@ianmathwiz7]

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

[@ASHOKSHARMA-so5rj]

Im so happy to see this verified by a reputable sources. I found it using the general form of the factored quadratic like 2 years ago at 13 and kept using it. I always used (a+b)/2 directly which led to some mistakes, and this really helped!

[@maxanimator9547]

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

[@superiontheknight963]

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]

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]

@@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]

@@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.

[@isaacnewton5364]

This is very similar to the “Completing the square” method I use despite teachers and classmates calling me crazy for doing so. All I can say is I beat everyone using the quad formula Math on bingo review days with it EVERY SINGLE TIME

[@NoGoatsNoGlory.]

-16x²+32x+2=7 solve using completing the square 16x²-32x-2=-7 16x²-32x=-5 16(x²-2x)=-5 16(x²-2x+1)=11 X²-2x+1=11/16 (X-1)²=11/16 √(x-1)²=±√(11/16) X-1=±√(11)/4 X=1±√(11)/4 Meanwhile quadratic formula got that answer 4 steps ago

[@nathanjiang100]

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.

[@anuragmondal]

This is life changing!

[@anudeepaghosh]

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.

[@islandsociete]

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).

[@mcalkis5771]

I had never heard of this before. Thanks Brian!

[@BriTheMathGuy]

You bet!

[@hubusgaming7718]

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

[@eagle32349]

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]

Isn't that difference of roots √Δ / a

[@Fire_Axus]

your feelings were irrational

[@RealJackBolt-NITJ]

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

[@asamuelmannareddy8988]

He literally said a=1​@@RealJackBolt-NITJ

[@genoric4094]

This is the formula we’re taught in Sweden. The quadratic is there to use and absolutely valid for anyone who wishes to do so but we’re never taught to do it. It doesn’t appear unless you choose to take more advanced mathematics/physics courses at university level.

[@kidkid]

Det är inte sant, många skolor/lärare går igenom det snabbt någon gång i mattematiken, och den "vanliga" formeln visades även på formelsamlingen på det nationella provet även om pq formeln är standarden här.

[@genoric4094]

@ Den finns på formelbladet ja, som jag sa så är den där för den som vill använda den. Men det är inget vi lär oss mer än att den finns och har typ samma funktion. Har svårt att se varför någon lärare skulle lägga mer tid på den andra mer än att kanske visa hur man löser med den en gång. pq-formeln är däremot central i matte 2 och 3 och återkommer om och om igen

[@Anmol_Sinha]

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!

[@reio4641]

But beware, the coefficient A must be equal to 1

[@Anmol_Sinha]

@@reio4641 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]

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]

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

[@danieltemelkovski9828]

@@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.)

[@lukeerikblue958]

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]

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

[@peg8354]

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

[@willabytes]

This version is actually primarily taught here in Sweden! Here it's called the "Pq-formula", where x = (-P/2) +- sqrt((P/2)^2 - q)

[@crustyoldfart]

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 < c/a then the roots are imaginary. 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]

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]

@@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.

[@mtaur4113]

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.

[@coolstar7819]

Thank You, that's a very interesting and intuitive approach!!✨

[@amos083]

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?)

[@BusterFortnite]

We get taught this in schools already (for GSCE's/A-Levels etc. at least) for quadratics we learn this method, quadratic formula, and completing the square

[@哲子仮免]

That thing do not put too much plugging in quantities.

[@Matthew_Klepadlo]

(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]

√-1 2^3 Σ π

[@UpdateFreak33]

This is what I've always been taught in school, I was always confused why others use the weird one that I still don't understand

[@yuuishigami8988]

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

[@adrianpilikser2195]

My man literally invented the quadratic formula a second time

[@Neko-san]

2:53 "Quickly solved for u"

[@hanstristanvirkus6814]

Yes?

[@Neko-san]

@@hanstristanvirkus6814 sorry, I already forgot what was the point of this comment

[@delta1404]

​@@Neko-sanepic comment

[@fr11dy]

​@@Neko-san Play on words maybe? "U" as "You", and "U" as a mentioned number

[@Neko-san]

@@fr11dy yeah that was it

[@Vi_prak]

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

[@statsy150]

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

[@msi6824]

we use this in sweden. x2+px+q=0 PQ: x = -p/2 +- Squareroot of: (p/2)^2 - q

[@invincible8714]

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

[@cowflick1180]

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

[@Dan-1031]

When we learned to graph parabolas this seems so obvious yet so clever. Thanks bae

[@Mayank81joshi]

Shridharacharya was a legend to deduce this formula centuries ago

[@dawon7750]

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.

[@panlomito]

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.

[@racool911]

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

[@joshuaalter4994]

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

[@3bitclash463]

bro this is such a good technique, really fast too!

[@nel385]

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.

[@loser-537]

After following this channel I started falling in love with Maths

[@Secret64462]

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.

[@Malidictus]

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.

[@kephalopod3054]

Beautiful!

[@per2632]

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

[@MathPhysicsEngineering]

To those who are interested in the derivation of formulas for polynomial equations, I have a playlist on my channel with the derivation of all the formulas from order 2 up to 4. Please check it out!

[@TastyMade3asy]

bro why is my brain just not braining rn

[@beuh0623]

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]

*linear algebra, phone keyboards suck lol

[@albondar]

There is no need for any form of quadratic formula when you can just complete the square.

[@rOceans-XbL]

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

[@AJ-et3vf]

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]

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]

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

[@MrEdrum]

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]

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]

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

[@yannik4966]

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

[@hackergaming6869]

We learned both and called it abc formula and pq formula

[@r2studion299]

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]

idt there is any advantage

[@uraharakisuke1329]

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

[@andrewkarsten5268]

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]

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]

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

[@giulimborges]

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

[@Sir_Drive_Safety]

Love You!!!

[@them_gamer8460]

This only works when a = 1 (the quadratic equation has its coefficient of x^2 as 1)

[@18yearoldcel]

So u divide everything by a and u get a =1

[@Musicrafter12]

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.

[@tonyboy_c3120]

Honestly if you don’t use the formula you should just factorise with like trinôme produit somme

[@justarandomdood]

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]

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

[@peterhofer8998]

That's valuable, thank you ☺

[@anshulbadhani]

In India this method is taught in school but explained in a different way this method is called Completing the square method.

[@nobody-fk1is]

Oh well Thanks you saved my time

[@bruh____784]

10th class Polynomials lmao

[@anonamos_]

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

[@rssl5500]

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

[@callumvanheerden1530]

What an amazing video!!

[@Apple-sq4wr]

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

[@hashtags_YT]

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.

[@BILGI_PASA]

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

[@DeJay7]

You can never convince me *ax² + bx + c = 0 x = [-b ± √(b² - 4ac)]/2a* is difficult to compute. For being 100% accurate every single time (with a≠0 obviously) at solving an equation otherwise challenging, it is so simple and actually some very simple arithmetic, it cannot be improved.

[@stratonikisporcia8630]

Yeah I mean if a = 0 it's even easier, x = -c/b linear done. No like seriously, who in the world ever struggled with the quadratic formula? Even my borderline-illiterate classmates who keep forgetting what (a-b)² is can remember and apply this...

[@yb3604]

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]

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

[@prasanthkumar1770]

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?

[@jeffreykalb9752]

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...

[@DriftinVr]

I remember I accidentally derived this before I knew of this formula and only knew how to complete the square so I tried to see if I could make a way to solve if a was always 1

[@aug3842]

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

[@angelmendez-rivera351]

Rather than learning _any_ of the variants of the quadratic formula (and let us all stop lying to ourselves: the Po Shen Loh method is just the quadratic formula in disguise, just presented differently), the best way to teach students to solve arbitrary quadratic equations to is to just be robust with teaching them completing the square. Rather than forcing them to memorize an algorithm, teaching them to complete the square helps set up for factoring situations, and it also just gives them a better understanding of algebraic manipulations in general. Having learn the formula as the primary solution method is pointless.

[@crustyoldfart]

Your comment is exactly what one would expect a teacher to say. In the real world, the solution to the quadratic is itself of importance, to be found as quickly as possible. Also, let's face it, the real world is seldom so obliging and convenient so as to render the traditional pedagogic methods possible.

[@angelmendez-rivera351]

@@crustyoldfart *In the real world, the solution to the quadratic is itself of importance, to be found as quickly as possible.* No, not really. There is no career out there in the world that demands the skill of solving quadratic equations as fast you can. Most careers never even encounter quadratic equations, so you are inflating its importance, but for those careers for which solving quadratic equations is valuable, speed of solution is not an important component. If anything, the usefulness behind solving the equation is derived from the analysis that comes in solving them. In this cases, the finding the solution set is not as important as the actual execution of the solving method. To be honest, if all you care about are the actual solutions, and the coefficients are all numerical, then there are calculators that take care of that for you, no quadratic formula needed. *Also, let's face it, the real world is seldom so obliging and convenient so as to render the traditional pedagogic methods possible.* I am literally not advocating for traditional pedagogic methods here. I know you said that I say things a teacher would say. I do not. In the education system today, most teachers would simply tell you to memorize the formula and get it done with. This is the opposite of what I am advocating, and it seems obvious to me you are trying to appeal to the "you're a teacher" to discredit my post without actually putting any thought into it. And with that being said, in many jobs, such a sin engineering or in physics, completing the square is a skill that actually is used very frequently.

[@crustyoldfart]

@@angelmendez-rivera351 I apologize if I gave he impression that I was criticizing you [ and/or other teachers on a personal level ]. I absolutely abhor personal criticism in a public forum. The point I was trying to make is that there is the pedagogic world and then there is the real world. As individuals we need the first to prepare ourselves for the second. As with anything there are good teachers and bad teachers. Looking back on my own continuous 17 years of sitting on hard wooden benches, listening to teachers of varying degrees of ability, I can honestly say that I found less than one in ten I would rate as good at their job. In the case of an estimated two out of ten I would estimate that I had to unlearn the rubbish they were purveying as truth.

[@ankurgupta9106]

That is a very good advice.

[@KingKevin108]

I think my entire brain fell out of my head about 45 seconds into this video

[@NotFiftyTwo-52]

Holy hell bro just changed the entire math game

[@砖递鷵橡孂㨩錘墭]

i love your vids keep it up

[@BriTheMathGuy]

Thanks for the support!

[@scottleung9587]

That's really cool!

[@BriTheMathGuy]

Glad you think so!

[@BitcoinMotorist]

I was taught the quadratic formula with a whole story to match each part of the formula in 8th grade. Anytime I needed to remember the formula, I just remembered the story.

[@animationcity8178]

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.

[@opoaotoroiocoko]

My high school math teacher asked me if I rode the short bus to school when I presented that to my class.

[@stuartl7761]

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]

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

[@rubensramos6458]

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).

[@kobethebeefinmathworld953]

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.

[@uraharakisuke1329]

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

[@thymenwaterman8672]

The reason why most schools learn the quadratic formula instead of the German version or the Midpoint way, is that you instantly learn how to solve for the discriminant and what you can do with it. Once you know how to calculate the discriminant, it is quite easy to apply to cases and it is actually confusing to learn other methods, however, ofcourse these methods work and all are up to preference.

[@b_hav_6365]

I know right. Try showing the graph of parabola to a 13 yo. Quadratic eq is tough as it is(back then, ofc). You only arrive at stuff like this after playing with maths for 4-5 years. And at that point you develop you own FAST method to solve quads. I like completing the sqaure method. Dont even need a damn formula.

[@lukeskywalker2255]

When I didn't know the quardatic formula I used something like this (about 5 month ago)

[@amritkumar7615]

A very great method to solve quadratics