Factoring Quadratics WITHOUT Guessing Product & Sum

[@saturday9032]

Okay, wow. This method should've been taught in schools. Incredible! Edit: I am actually lost for words. Sounds like glazing but it isn't

[@wernerviehhauser94]

It was. Do I use it? No. I'm faster with the quadratic formula. This method provides some insight to the attentive student, but no gain in speed over the quadratic formula for real-world numbers

[@farhanrejwan]

this proves how we try to overcomplicate things in attempts to solve them, while all we needed was a simple thinking process.

[@saftheartist6137]

@@wernerviehhauser94I agree the quadratic formula technically works for all these problems and provides even more. The need to guess and check just gets eliminated in all it’s entirety.

[@kevinstreeter6943]

@@wernerviehhauser94 I have taught math. I recommend using the quadratic formula as a last resort. They are too many ways you can make a mistake.

[@kevinstreeter6943]

@@saftheartist6137 I have taught math. I recommend using the quadratic formula as a last resort. They are too many ways you can make a mistake.

[@claude77573]

My only suggestion is that "c" is used to define two different things: The third "element" of the quadradic equation and the "constant value" added to or subtracted from the second element of the quadratic, b. To help prevent confusion, I would suggest calling the second constant value "k", so we don't have two c's with different meanings.

[@thomasdemilio6164]

Right, mathematical rigor must be provided

[@shabirmir9597]

k= sqrt{(b/2)-(c)}; m = (b/2)+(k); n =(b/2)-(k). Final factors: (x +/- m)(x +/- n). k must be within bracket to take care of sign e.g. x**2 + 5 x - 14 provides k = 4.5, m=2.5+4.5=7, n=2.5-4.5= (-2). Factors are (x+7)(x-2)

[@geoffstrickler]

Simpler to remember than the quadratic formula, arguably easier to calculate. But as others pointed out, it’s essentially the same method.

[@KRYPTOS_K5]

The same suggestion of gpt4 who doesn't know this universal method to my amazement!!! Brasil

[@fgvcosmic6752]

This is pretty much just how the quadratic formula works. This is how the quadratic formula should be taught, in my opinion; by starting with this as motivation

[@gregorymorse8423]

The quadratic formula is derived from the completing the square method. Rookie mistake.

[@notcraig3204]

@@gregorymorse8423 You trolling? This method is the same steps as completing the square.

[@DarIsANoob]

@@gregorymorse8423formulas can be derived from a multiple of methods. Rookie mistake.

[@prefabrication]

​@@gregorymorse8423 This is literally completing the square. He's doing so by finding c in 16² - c² = 192.

[@mikechrissan]

That is nothing like completing the square

[@danielnegussie4894]

It is called the Poh Shen Loh method. A math professor named Poh Shen Loh discovered this method. The math professor himself said he found this method buried in a very very old Math book. Here is the link kzbin.info/www/bejne/jnyliWOoaK9gb7ssi=LiskkjE6oaHBAPZY

[@wernerviehhauser94]

i've heard it called Hindi Method or Sridhara method. We used to learn it as an application of Vieta. Nonetheless, all I use is the quadratic formula - it's faster than everything else for real-world numbers in physics problems.

[@ramunasstulga8264]

​@wernerviehhauser94 this is why I don't give a shit about how a formula is named because it's only about politics

[@maanasdongre1581]

Well it is poh Shen loh method but it is explained better

[@glennschexnayder3720]

It’s the ancient Babylonians’ method. They were very advanced in math and astronomy. If NJ Wildberger ever gets his channel restored, you’ll be able to see his demonstration of a real, Babylonian worked problem using this method.

[@chocolateangel8743]

@@glennschexnayder3720 Professor Loh basically condensed the method and made it easier to understand. This is especially true when explained with a graph.

[@angelviloria4966]

I came here to refresh my memory on factoring so I can begin teaching my girls (8th & 6th grades). I love this method because I hate ever having to say “guess and test”. I always look for methods/process or creat one. This is excellent. I can’t wait to teach it.

[@angelviloria4966]

I bet they aren’t teaching this in their school.

[@biswanathbera1260]

For small numbers this method is sheer time consuming. But for bigger numbers it is helpful.

[@dmitrmax]

Dude, teach them the generic way to solve quadratic equations, not this magic shit

[@ranjittyagi9354]

​@@angelviloria4966I would assume the same. However, just as others advised, please don't skip the formula taught in school. Remember the roots and (-b +- sqrt (b²-4ac)) / 2a

[@bjornfeuerbacher5514]

You can also do it in an another way, which in the end is essentially equivalent to the one in the video: first complete the square, x² + 32x + 192 = x² + 32x + 16² - 16² + 192 = (x + 16)² - 64. Then write this as a difference of squares: (x + 16)² - 64 = (x + 16)² - 8² = ( (x + 16) + 8 ) ( (x + 16) - 8 ) = (x + 24) (x + 8).

[@Reginald425]

It always comes back to completing the square

[@50Steaks68]

You lost me at completing the square lmao I hate doing that. Guess and check with calculator is by far the fastest and easiest way to solve these

[@lookiii1]

Do it with with x² + 4x + 10

[@samhardy6030]

​@@lookiii1It works well. You complete the square, then add and subtract the square root of the completing adjustment (X+2+√6)(X+2-√6)

[@ranjittyagi9354]

Your equation furnishes complex roots. Here they are: ​-2 + 2.4494897427832i and -2 - 2.4494897427832i Please check.

[@rainerzufall42]

This is just a manual execution of the abc or pq formula (there are many names for it). Still impressive. For simplification I use a = 1 => pq formula for x² + px + q = 0 => x_{1/2} = - p/2 +/- sqrt( (p/2)² - q ) (x - x_1)(x - x_2) = 0 => m = -x_1 and n = -x_2. That's p/2 -/+ sqrt( (p/2)² - q ).

[@alemswazzu]

So completing the square is also called PQ method. Extra bonus is giving a result in vertex form.

[@isooo8175]

Congradulations, you reinvented the well known solution of quadratic equation. B/2+/- sqrt(b*b-4ac)/2 What you labeled as c is the sqrt part.

[@epsi]

You can also reverse the quadratic formula so that you don't need to work with numbers of such significant magnitude: x = (-b ± √(b² - 4ac))/2a x = -b/2a ± √(b² - 4ac)/2a x = -b/2a ± √((b² - 4ac)/4a²) x = -b/2a ± √(b²/4a² - c/a) (-b)² = b² x = -b/2a ± √((-b)²/(2a)² - c/a) x = -b/2a ± √((-b/2a)² - c/a) q = -b/2a x = q ± √(q² - c/a) Another nice thing about this variant is the fact that you only need to calculate _q_ once. As an example of these conveniences, consider 8x² + 14x - 15: q = -14/2(8) = -7/8 x = -7/8 ± √((-7/8)² - (-15)/8) x = -7/8 ± √(49/64 + 120/64) x = -7/8 ± √(169/64) x = -7/8 ± 13/8 x = 6/8 , -20/8 x = 3/4 , -5/2 8x² + 14x - 15 = (x - 3/4)(x + 5/2) = (4x-3)(2x+5) = 4x(2x) + 4x(5) + (-3)(2x) + (-3)(5) = 8x² + 20x - 6x - 15 = 8x² + 14x - 15 compared to: x = [-14 ± √((-14)² - 4(8)(-15))]/2(8) x = [-14 ± √(196 + 480)]/16 x = (-14 ± √676)/16 x = [-14 ± √(4•169)]/16 x = [-14 ± 2(13)]/16 x = (-7 ± 13)/8 x = 6/8 , -20/8 x = 3/4 , -5/2 The reason you can multiply things is because you're using the roots of the expression, meaning when you set the expression equal to 0, you are effectively multiplying 0 by whatever factors you choose, which doesn't affect the truth of the equation. While the use of _m_ and _n_ are important to understand the underlying concepts shown in the video, it's not truly necessary once you can use an algorithm like this one, and i definitely prefer this algorithm over the standard quadratic formula.

[@RexxSchneider]

Isn't the method shown no less effort than factorising ax^2 + bx + c by calculating D = b^2 - 4ac and then writing the factorisation as a.( x + (b + √D)/2a) ).( x + (b - √D)/2a )? That has the advantage of using a formula that's either already familiar, or at least will have further applications later on. If you are worried by big numbers, just calculate d = (b/2)^2 - ac and write the factorisation as a.( x + (b/2 + √d)/a ).( x + (b/2 - √d)/a) which scales thing down by a factor of 4 (like the video does).

[@fahrenheit2101]

Yes. It's just an alternate form for the quadratic formula. This is the famous Po Shen Loh method, and 3B1B has a good video covering it in a little more depth

[@soilsurvivor]

Great method, and VERY well-presented! Just wordy enough without feeling like you're dragging out minor details. Concise, but not sparse. Thank you!

[@viasciencemath]

Another way to think of this is that the solutions to x^2 + b x + c = 0 are x = - b/2 +/- sqrt( (b/2)^2 - c ) which is just a rewriting of the quadratic formula.

[@DaneBrooke]

Wonderful video. Since this amounts to using the quadratic formula for writing the factors of the quadratic, you could have just gone for "How to write the factors of a quadratic by using the quadratic formula". And *that* would have been a wonderful video. And nobody would have guessed or checked anything.

[@Sarutolity]

Sending this to every math teacher I’ve ever had 🔥

[@mr.cabbage4428]

This is a fun video, I found the general form for this method by not plugging anything in where x^2+bx+c=(x+((b/2)+sqrt(((b/2)^2)-c))*(x+((b/2)-sqrt(((b/2)^2)-c)) If there is a coefficient on x^2 divide it by every term and put it on the outside of the formula a(x^2+(b/a)x+(c/a)) We can treat b/a as “b” and c/a as “c” for the formula

[@matthewmanzanares6798]

This method is what I figured out in deriving the quadratic formula which I thought was that's how you actually derive it instead of completing the square. Although this ensures that you get its factors, I don't think it's the method I'm going to use mainly in my factoring, because to factor quickly, a method that allows you to being able to do it mentally or at least minimal writing would be my bet. But this method would require me a lot of thinking, writing, and/or storing numbers in my head. This method actually gave me an idea. instead of just using the quadratic formula, just use the formula derived from this method directly because comparing it to the quadratic formula, it is easier to compute the square root. when a = 1 m = b/2 + √( (b/2)² - c ) n = b/2 - √( (b/2)² - c ) it is essentially the same as the method, but for me using the formula is faster than doing all the work

[@kdaviper]

Or perhaps solving one equation for a variable and plugging it into the other equation

[@raizen_u]

Ive been looking for a way to solve factoring quadratics without guessing for months now so thank you !

[@Chemest_a]

Yall this is literally the quadratic formula broken down into steps. Fire video tho!

[@matthiasklein9608]

And this is exactly what is called the p-q formula (at least here in Germany) for solving quadratic equations: x^2 + px + q = 0 x = -p/2 +/- sqrt((p/2)^2 -q). It’s just not displayed as a single formula but broken down into two steps.

[@starlogonova]

Brilliant. I am sure my 8th grade math teacher in my school will find this interesting.

[@edobolo-h2q]

i've had thoughts of this but never found a way to make them useful to my math classes, well until i found this video i could actually do something with it. THanks!

[@megumarutakashishi3003]

I've seen OutlierOrg but different approach but same principle. Thank you so much. You earn a subscribe

[@jakubjiricek7806]

I don't know about the other parts of the world, but here in the Czech republic, some 35 years ago, we were not taught to guess, but we learned the standard formula for the roots, to which this "new" approach is equivalent.

[@unknwonboy-fm5ei]

we are literally solving a quadratic , to solve another quadratic . Although the approach is nice.

[@airking2883]

Well it's always a single term quadratic which is much easier than multi term quadratic equation

[@fahrenheit2101]

​@@airking2883yes, but completing the square is not only already familiar, but is equivalent.

[@neitoxotien2258]

Hmmm... I love this method, since back in the day I love Factoring than Quadratic formula. This method gives me a new ways. Thank you.

[@zulfiqarali4888]

Love this. So much of boring guessing work of stone age in school. m = sum/2 + sqrt((sum/2)^2 - product) n = sum/2 - sqrt((sum/2)^2 - product)

[@user-gr5tx6rd4h]

In fact I invented this method (or very similar) some decades ago (now retired from high school). But often with much less work. Take x^2 - 6x - 91 = 0. Only integer factoring of 91 is 7 x 13, so I write (x - 7) (x - 13), yes - both with a minus. Now it is easy to see if any - must be made into a + (which is "easier" than the opposite). Since 7 - 13 = -6, we must have + on the 7. Thus: (x + 7) (x - 13).

[@IRVINMILLER]

Very clever! x^2+bx+c=0 x=-b/2(+/-)(b^2-4c)^.5/2 Someone noticed that the roots where b/2 +/- the same number. Then it becomes obvious how you developed your approach. Telling the students your thought processes is the way math was supposed to be taught. It was not intuitively obvious to me until I looked at the quadratic formula.

[@شعرکوتاه-ع7ظ]

Mathematics is the basis of all sciences and you explained it well, thank you

[@rgmolpus]

FANTASTIC! I've suspected there was a formula or algorithm for finding the factors, but never found one. Thank you for the explanation!

[@gendo1123]

What i have done for similar problems is right m=b-n then n(b-n)=c so n^2-bn+c=o then use x=(-b±squareroot(b^2-4ac))/(2a) for n^2-bn+c as it would allow me to right ax^2+bx+c=0 in form of (x+n)(x+m)=0

[@simonekentish7491]

In my experience, most students who struggle with multiplication struggle with fractions even more. It’s probably more efficient to stick to finding factors of ac first, and for more challenging problems, make an organized list of the factor pairs. What does help from this method is to recognize that larger b values (compared to ac) mean that the numbers are further apart. This understanding helps with guess and check.

[@Adityasharma-00]

In this question 15:37 2x²-7x-4=0 it will written as *X²-7x-8* =0 after finding Root (x-8)(x+1)=0 then divide 2 both side (x-4)(x+1/2)=0 This also easy by their method *Maine isse aur aasan bna diya*

[@catburner1896]

This actually really cool and helpful, I might use this next time instead of the quadratic equation

[@diwakardas1807]

In the thumbnail it's clearly written a new amazing "factoring method". That is to find the factors..... Not a new method of finding the value of x in the quadratic equation Which we all know the best is the quadratic formula 😊 hope it explains the confusion

[@notnyw]

Always wondered if there was a mathematical way of determining the factors and was told I had to guess the two numbers. Never knew about this method. Very well presented. Many thanks Jensen.

[@Neekalos]

Why not just use the quadratic formula?

[@parinose6163]

I just wanted to let you know that I never learned that before! Many thanks. It's another pain in the back removed with a quadratic polynomial!

[@Jabba1973]

Nice explanation of Po Shen Loh's method, publish about 5 years ago. Po does say that the steps to doing this method have been know for a very long time, but the combination used here wasn't well recorded elsewhere, so he is standing on the shoulders of others.

[@karikosonen3047]

I havent seen this before. Absolutely brilliant! I wish i knew this in the 90´s !

[@VuNam_MCVN]

6:26 m= 32 - n Replace m in the product Check if n and m is negative or not

[@ahmadafgan6591]

I am very grateful for your method. It helps me a lot, during my exams and worksheets. I think this should be known further, you are an inventor.

[@krissyai]

I generalized it with the roots adding up to -b/a and their product being c/a, I did the same thing where one root is essentially the x value for the vertex plus h, the other is the x value for the vertex minus h m = -b/2a + h n = -b/2a - h Their product should be c/a m . n = (-b/2a + h)(-b/2a - h) m . n = b²/4a² - h² b²/4a² - h² = c/a At this point you may have realized that it's building up the main formula to solve for quadratics h² = b²/4a² - c/a We multiply c/a times 4a/4a h² = b²/4a² - 4ac/4a² h² = (b² - 4ac)/4a² h = √(b² - 4ac)/2a Yes, the discriminant shows up here, being b² - 4ac Let's denote the discriminant by using ∆ h = √∆/2a The main formula is x = (-b ± √∆)/2a So if we just replace √∆/2a for h x = -b/2a ± h I should point out that -b/2a is the x value for the vertex in every quadratic, where f(-b/2a) is the y value for this vertex of this same quadratic f(x) This in a way proves that each of the roots is the same distance away from the x value of the vertex, I believe this applies for every polynomial that has an even degree

[@pietergeerkens6324]

This is fundamentally Poh Shen Loh factoring, derived differently; but you've missed an opportunity to combine with the Australian (i.e. "down under", referring to the placement of a in the denominator) technique for factoring quadratics with a 1. When b is odd, we encounter an unusual circumstance where the decimal fraction 4.5 may be more convenient than the common fraction 9/2, using this shortcut for squaring a half integer: 1.5² = 0.25 + 100 * (1 * 2) = 2.25 2.5² = 0.25 + 100 * (2 * 3) = 6.25 3.5² = 0.25 + 100 * (3 * 4) = 12.25 4.5² = 0.25 + 100 * (4 * 5) = 20.25 5.5² = 0.25 + 100 * (5 * 6) = 30.25 : 9.5² = 0.25 + 100 * (9 *10) = 90.25 : The Australian method for handling a 1 proceeds by initially ignoring that a 1 as, using your example: 3.5² - c² = (2)(-4) = - 8 becomes c² = 12.25 + 8 = 20.25 = 4.5² and then factoring as [ (2x - 3.5 - 4.5) (2x - 3.5 + 4.5) ] / 2 = [ (2x - 8) (2x + 1) ] / 2 = [ (2x - 8) / 2 ] (2x + 1) = (x - 4) (2x + 1). Notice how the a value of 2 was initially placed in both factors, as well as "down under" in the denominator. When a is composite and must be factored across both final factors, this trick delays having to determine that factoring and distribution until a final step, not interfering with the rest. The factoring of a reduces to identifying the common factor in each numerator factor - much simpler than previous methods.

[@trien30]

Po Shen Loh, not Poh Shen Loh.

[@saeidabbasi7152]

Hi there 🎉good good job If you in put as a(x-m)(x-n)=0 and your factor form as ax^2+bx+c=0 and find the ROOTS as m&n then plug into first equation a(x-m)(x-n)=0 it's much more easier 😊

[@iikitsunex-x]

Mind blowing method yet so intuitive and simple! I wish this was taught in my classes!! Thank you 😊

[@BirhaneGselasie]

Amazing, this method should include in our curriculum .

[@NyteRazor]

You can also factor by grouping 2x^2-7x-4 = 2x^2 -8x+1x -4 = (2x^2-8x) + (1x-4) = 2x*(x-4) + (x-4) -> factor (x-4) from each group making -> (2x+1)(x-4)

[@biscotty6669]

Very nice video. Small suggestion: use c' in the m/n formulas or another name to avoid any confusion with c in the quadratic equation.

[@baileyradel3104]

Compelling method. Reminds me of "completing the square" since you're taking half the b value and factoring out an a>1. A "benefit" of using the c value and listing factors is that students can generate a table of y = c/x in their graphing calculators and look for integer coordinate pairs. I love the link to multiplying conjugate pairs in this method though! I'll stick with the AC method for a>1 for now.

[@naveenpokala6442]

Excellent and simple method, thanks for bringing it up

[@kurtmager1626]

Quite interesting. I have my own method here. Consider that x=0.5((x+y)+(x-y)) and that y=0.5((x+y)-(x-y)) So then x and y can be found easily as well if we can find (x-y). Well (x-y)²=(x+y)²-4xy So x-y=√((x+y)²-4xy)

[@sphereguanzon2217]

This is scarily accurate. Even with irrational radicals

[@JeanMarieGalliot]

Nice algorithm and very clear explanation!

[@seanleysinday9120]

I have a different method in solving this. I just take the factors of 192, then find the factors that adds to 32, which is the 8 and 24. It might not be applicable for all the problems involving this, but I believe it will solve most of the problem.

[@Sesquipedalian3]

Okay I'm a little disappointed because the thumbnail said this was new, but for everyone saying to use the quadratic formula instead... that's like adding five three times when asked to do five times three. It might be faster when you first learn, but oh my god is it slow. when I looked at the first problem, my first instinct was that it was eight, and that's not some weird boast like "oh I can do factors slightly faster than you," that's the bare minimum. It's honestly shocking to me that so many people are plugging these numbers into the quadratic formula thinking it's faster. Now that you've watched this video, do this until you have it down perfectly easily. memorizing the quadratic formula won't help you. Learn to factor.

[@Orrinn123]

This is not entirely correct. Memorizing the quadratic formula WILL help you. Factoring won’t help you when the factors don’t turn out to be nice whole numbers. In those cases (some ugly fraction, an irrational number or a complex one) the quadratic formula is very nice to have memorized. However I do agree that mindlessly memorizing the quadratic formula without learning how to factor is a bad idea. People should learn both, factoring first and then the formula.

[@SuchRyanmuchgaming]

This is amazing and when I become a teacher I am so teaching my students this

[@notcraig3204]

Played this on x2.6 speed and quadratic formula got the answer first with plenty to spare. While this certainly gives a cool perspective that seems to speak to students, we should probably spend more time pondering what to do about the fact that the formula (by definition the easiest way to do something) is so intimidating to students.

[@riseofgaming9429]

If you look very very very closely at the procedure, Its just the quadratic formula but in step by step method of solving the determinat to get C and adding and subtracting that to b/2a which will give the two roots

[@alanthayer8797]

NIIIIIICE WRK MY BOY! Simplicity is a Gift

[@jogeshjoshi2535]

Excellent Explanation Sir Ji Thankyou Sir

[@lastchance8142]

My daughter is a high school math teacher. She likes this method, but can't change the approved curriculum. This is why things don't get better in US schools.

[@jamesharmon4994]

IMO, factoring is a waste of time.

[@davidwright8432]

Surely there's nothing against teaching it after teaching the 'approved' method? It could be done easily, in a 40 min class time!

[@scottekoontz]

I taught HS math, and always found time for additional niceties like this. Doesn't matter if it is approved, although she may want to avoid quizzing students on it if it really is a backward administration. But for the test that wraps up all methods, i.e., solve using any method, students can use this method if they wish.

[@JavedAlam24]

​@@jamesharmon4994 you've clearly never had to integrate anything requiring partial fraction deconposition

[@pianoman5259]

​@@jamesharmon4994respectfully, your opinion is wrong unless you are implying there is a faster way

[@lusifargamingyt8965]

Thanks bro for giving this method bring more methods like this it appreciated😊😊❤❤

[@DJCray8472]

Nice visualization of the solutions of quadratic formula for a=1. And you showed it: all quadratic equations can be rewrite to the pq formula. Your solutions m and n are the negative solutions of this formula x = -b/2 ± sqrt[(-b/2)² - c]. so m/n = b/2 ± sqrt[(b/2)² - c]. I removed the minus under the square, as it is not important, if you square it anyways... but it is your solution, you showing for different b and c.

[@ramonalejano671]

Since I was not taught this, I was using the quadratic formula if I can't guess the factors. Thanks.

[@Steven-v6l]

x^2 + 32x + 192 = 0 :: average of roots is r. roots are (r+p) and (r-p); sum of roots = 32, product of roots = 192. Sum:: (r+p)+(r-p) = 2r = 32 so r = 16. Product:: (r+p)*(r-p) = 192 r^2 - p^2 = 192 256 - p^2 = 192 p^2 = 256-192 p^2 = 64 so p = √64 = 8 the roots are r ± p or 16 ± 8 or 24 and 8

[@cerisaac8149]

Wish I knew this wayy earlier. Thanks to this video I just realized that you can actually also use the Quadratic Formula for factoring! (by replacing -b to positive)

[@cerisaac8149]

Oh and you can derive the process/formula used in the video from the Quadratic Formula(where -b is positive) and vice versa, which makes sense since both ultimately gives the same answer

[@ralphbecket]

Thank you! I have been writing a maths "book" for my teenage son and my father who are both doing their secondary school maths (my father as a hobby project) and neither would say they have a natural feel for this kind of thing. My book is an attempt to provide the clearest explanations for everything they encounter in the curriculum. And, as is the invariable way with maths, when you think you have something down pat, someone like you comes along with a splendid video that makes me think, "why the heck didn't I think of that?". I hope you don't mind if I steal your explanation for my book.

[@ammaradil3796]

This method should go viral, we have to make children's lives easier. My school years would've been so not depressing if this method was around my time.

[@ranjittyagi9354]

Most methods were there in books outside of the ones we were told to follow. However, the teachers who could think of explaining how to approach a given problem from different angles weren't around in some respectable number. They followed a particular book and stopped at that. There has been a very asked cha ge in how children learn these days. So many follow good channels on KZbin as well. No wonder then, the depressing days had to be endured by us.

[@malluzenitsu]

x = [-b±√(b²-4ac)]/2a

[@scottekoontz]

This is introduced in many math classes, but alas too many parents will claim this is more difficult or "I never learned it that way" and blame this on Common Core.

[@Clint52279]

Watching this ticks me off because not only does the more efficient methid make sense, but the OG method makes more sense than what I was taught. My math education sucked noodles. I took calculous in HS, and was fumbling in the dark the whole time.

[@rlouisw]

Good method, and well presented. (The rest of this is not for you, but more for the existence of this method in the first place) Ridiculous though, factoring is not a difficult process, and this method is not only completely unnecessary but much more difficult than just factoring it. I speak to you as a math tutor, and when the kids come to me being confused by yet another factoring method to make things "easier", and I show them how very easy it is to just skip to the end and solve it directly, each one is more shocked than the last at how very easy this all is. And they are truly shocked when I show them how easy it is when a doesn't equal 1. I mean... look at what happened with x^2 + 9x + 20, fractions were introduced to what should have been HORRIBLY SIMPLE! Stop babying these kids with all these "factoring methods", they can handle the guessing and checking (not to mention, it is actually the far easier method). *frustrated rant over, I had a day...*

[@mastermichael0751]

I’ve been taught this in school since seventh grade. Seeing other comments, it appears to me that I was apparently very lucky to have known this method from as early as I have

[@JulietNovember9]

Mind blown. Thank you so much!!!!

[@yvg2009]

This method already taught to me 🔥🔥🔥

[@adityaroy5300]

Very thanks sir i got 5/5 in section of quadratic equations in my bank manager exam. I'm from India and a aspirant of govt bank exams

[@adityaroy5300]

I have to do 35 questions in 20 min which also include 5 questions from quadratic equations

[@simonlee962]

So the general case for this is: (b/2 + n), (b/2 -n) And the general case for n is: √((b/2)^2 - C) Putting all that together we get a simple and elegant single equation: b/2 +- √((b/2)^2 - C) Then we just take the negative to get the value of X, and we get: - b/2 +- √((b/2)^2 - C) ... Also known as, the (reduced) quadratic formula

[@francoislanctot2423]

Best math video ever!

[@TeaCupToast]

I have in fact seen that method and I am happy I did

[@zenersmytok3619]

This method is brilliant. I love it.

[@alphalunamare]

I double checked this method and it certainly works. It also leads to the standard equation taught in schools sqrt(b^2 - 4ac) etc. I am puzzled for why it works though. Who would have thought in the first place? It is quite magical in its own way. Well it was magical until You explained it so well :-) You are a good teacher. There is always the How and Why. Feynman says shut up and calculate. So faced with these equations just apply the standard formula and sattisfy the 'How.' You have explained The 'Why' however. That is quite meritorious :-) Physics is in the doldrums, brain dead from not thinking and just calculating. Maybe expalin Math to them :--)

[@alemswazzu]

To get to the quadratic equation (which you have most of there), is done by completing the square of the general quadratic equation "ax^2+bx+c" and simplifying.

[@alphalunamare]

@@alemswazzu Oh I get that. I am not good at grammar. My puzzle was with this new method working in the first place :-)

[@KAhmedsir]

I will be teaching quadratic today. Will add this concept as well.

[@SrisailamNavuluri]

It is Indian old method. If m+n=k then both m,n are equal,k/2. If m,n are unequal then one is more than k/2,other is less than k/2 m=k/2+u,n=k/2-u so that sum is k. mn=k^2/4-u^2 u^2=k^2/4 -mn This is reduced form of quadratic formula.

[@dendaGulliLapoch]

Didnt quite understand it

[@Deevick2017J]

It's the same thing taught in the video just written in very compact from ​@@dendaGulliLapoch

[@basheer9098]

Too nicely explained thanks a lot

[@adamzoltan1685]

Or this c is all about expressing n and m witj only 1 variable. You could also just say that if m + n = v then n = v - m and boom you got it. Also another thing you could do is if m + n = i, m*n = j, then (m+n )squared is m2 + 2mn a n2 and you know 2mn so you get the value of m2 and n2, then calculate m-n squared, get the root, and boom you got it

[@diwakardas1807]

Why people hear in the comment section are confused between quadratic formula and an factorising method In quadratic formula we find the exact value of 'x' Whereas this is a factorising method where we find the "factors" of a quadratic equation

[@deveshgarg7974]

Great method! You can also use shri dharacharya formula.

[@radhakrishnamohanty3807]

In India, this method is taught to 12-13 year student (8th level). My grandfather taught me this in my childhood who had learned it in 1920s.

[@NichaelCramer]

Other commenters have mentioned that this technique is basically (just) how the quadratic equation works. While, strictly speaking, this may be true I still think that this video is extremely valuable in the way that it views the QE from rather different direction. In short, you can never too many tools in your toolbox. ;-) (More to the point I think that in some cases this technique could be quicker/handier/less cumbersome than using the QE. For example, if your goal were to factor the equation rather than simply obtaining its roots -say as a sub-step in dealing with a larger problem.)

[@ronaldburke2857]

Who invented this method? It's great. I'm 76. I got a math degree many years ago. I never worked in math but did use simple math and algebra on occasion. But never saw this method.

[@trien30]

A professor named Po Shen Loh.

[@fahrenheit2101]

​@@trien30to give him credit for the invention is generous. Popularization, sure.

[@Steve_Stowers]

The presentation in this video is clear and interesting, but the method described is essentially the same as what you do when you solve a quadratic equation by Completing The Square.

[@mysteriousyoutuber2517]

Isn't that just the quadratic formula...

[@Yadav24Anurag]

Same ! I was also thinkin about it. 🧐

[@Grecks75]

Yes, of course, it is. It's ridiculous to call this a new method, or a method with a certain name attached. This is common knowledge probably since Bronce Age. 😂 Edit: Just using the memorized quadratic formula works even faster for me when doing the calculation in the head.

[@dawon7750]

Absolutely, you are right! But in a simpler way. In fact i can mentally do it, the simpler ones.

[@CliffSedge-nu5fv]

Yes.

[@bjornfeuerbacher5514]

No, this is a method which works _without_ using the quadratic formula explicitly. Or which can be used to _derive_ the quadratic formula.

[@tambuwalmathsclass]

Thanks to Poh shen Loh for making our work easier.

[@trien30]

His name is Po Shen Loh, not Poh Shen Loh.

[@scottekoontz]

I taught HS math, and always found time for additional niceties like this. For the test that wraps up all methods, i.e., solve using any method, students can use this method if they wish. If you are math-minded (you did view this video after all) then you are applauding this method and suggesting it be taught in schools. First, it is taught in many schools already, although probably not enough. Second, the problems occur when math aliterate parents begin their YT videos with a bad example and explain that this is Common Core and "why can't they just do it the way I learned?" Such parents would construct problems that may take longer with this method, and mock math teachers for CC when they have zero clue what CC is. I always introduced additional methods in my classes, and when we were caught up used Friday classes to go over an assortment of computer science and math topics. I must admit I had a lot of autonomy, but it is not hard for teachers to introduce something like this even if you do not test them on it.

[@turnandfacethedragon]

Don't factor out the a. Multiply a and c and do this process as if a=1 and c=ac. When you have m and n, fill in (ax+m)(ax+n) and toss out anything common for a and m or a and n, the product of which is a. (It is necessary to factor out a common from the quadratic first.) This is fantastic. I've always hated the standard instructions to just find the numbers that work. I devised a way that uses the factor pairs of AC. It worked well for my students, but I had to admit that it could be time consuming. This is better. Thanks for the excellent video.

[@aldy_0x278]

Isn't it just the same as Quadratic Formula? And probably more faster using Quadratic Formula For equation: ax^2 + bx + c = 0; the values of x that satisfy: x = ( -b +- \sqrt{ b^2 - 4ac } )/( 2a ) Tweak a little bit inside the root: \sqrt{ b^2 - 4ac } = \sqrt{ ( ( ( b^2 )/4 ) - ( ( 4ac )/4) ) * 4 } = \sqrt{ ( b/2 )^2 - ac } * \sqrt{ 4 } = 2 * \sqrt{ ( b/2 )^2 - ac } Formula rn: x = ( -b +- 2* \sqrt{ ( b/2 )^2 - ac } )/( 2a ); tweak more become x = ( -b )/( 2a ) +- ( \sqrt{ (b/2)^2 - ac } )/( a ) And if you make it become factor, just multiply by -1 Because: ( x+m ) * ( x+n ) = 0; x+m = 0 or x+n = 0; x =--m or x = -n x^2 + bx + c = 0; ( x + ( ( b/ ( 2a ) ) + ( \sqrt{ (b/2)^2 - ac } )/( a ) ) ) * ( x + ( ( b/ ( 2a ) ) - ( \sqrt{ (b/2)^2 - ac } )/( a ) ) ) *Note: +- mean plus or minus \sqrt{ } mean square root of ...

[@GhostyOcean]

After seeing this method presented a few ways, I realize it's about parametrizing using the sum equation instead of the product equation.

[@Mr.Nichan]

In summary, ax^2 + bx + c = (x+r)(x+s) iff. r = b/(2a) ± [b^2/(4a^2) - c]^0.5 and s = b/(2a) ∓ [b^2/(4a^2) - c]^0.5 .

[@nika_yt2303]

tysm this is a lifesaver

[@Arima_Kousei0_0]

1:50 he did not considered 8 and 24 because that's the answer

[@johncoster558]

Wow, amazing method

[@GermanyServer-b7o]

Or simply you can take lcm of c value in quadratic equation and add or subtract those factors and get b value simple