Factor ANY Quadratic Equation Without Guessing | Outlier.org

[@borisdorofeev5602]

This guy literally had me smiling the entire time he talked. When people are passionate about something it is infective.

[@jarettthomas348]

This guy took over the US Math Olympiad team and we went from not winning an IMO since 1994 to getting 4 wins in 5 years. He is an incredible teacher.

[@ridhamsharma3857]

It's infectious

[@krisna4455]

how we will proceed if there is sum of roots is odd number???

[@backyardagric5191]

@@krisna4455 I was thinking the same. Can someone answer us please. I want to use this method

[@sao_rav]

Show Respect. Is this how you refer to educated humans around you?? “Guy”.. this “guy” seriously?

[@boredomgotmehere]

This is why YT has become the uni of maths. Thanks Professor Po.

[@Yojitt--r]

[@TheChillennial]

This is the funnest math teacher I've ever seen.

[@TSS4LYFE]

he is the best i know stfu

[@gamerzgang3264]

Then you have never met me

[@EramMahmood]

Habibi come 2 Bangladesh

[@nobodybutnobody]

@@EramMahmood Fr Bhai xDD

[@willthecat3861]

@@EramMahmood who in their right mind wants to go to Bangladesh!

[@chocolateangel8743]

"I think that's how you multiply." 🤣 That's especially funny coming from a math professor -- but I get it. Ever since I learned about the area model of multiplication (and variations of it), I no longer use the standard algorithm. . As for this method of solving Quadratics, it took me seeing it demonstrated, using a graph, before I was able to understand it. Now, it's easy, and I like it. I'm able to visualize a graph in my head now.

[@Big_Glizzy.]

I did it more sloppy, probably the an inefficient way, knowing the square root of 900 is 30, and it's only missing 1. I figured 1 would have to be added to 30 and subtracted. I multiplied 29 and 31 and I got 899.

[@chocolateangel8743]

@@Big_Glizzy. You used your understanding of number relationships to figure it out. That's great! I often use alternative approaches to solve problems and don't necessarily pay attention to how many steps it takes for me to solve something. Since learning about the Po-Shen Loh method, I've also discovered CTS by equating coefficients: kzbin.info/www/bejne/bJ24d6uFYsZ0g7c - my favorite approach.

[@Big_Glizzy.]

@@chocolateangel8743 thank you for that link! Much appreciated!

[@chocolateangel8743]

@@Big_Glizzy. You're welcome. 😁

[@raghavswindowseat5789]

Hey, what if the sum is odd or if the u isn't a perfect square?

[@alessio2980]

This dude did such an easy thing that none of my teacher never teach me… only 200k views? C’mon YT, give this dude a damn rewards

[@ukulelevillain4170]

what if I told you that you already were taught it? look closer the second problem for example. our final answer was what? 4 [+ or -] sqrt(3), right? but how did we find it? lets see, that 4 came from b/a, then divided by 2, and multiplied by negative one. this happened because he set it up as (x-[__])(x-[__]), effectively factoring out a -1. this gives us -b/2a. sounding familiar? okay put a pin in that. next, that sqrt(3) was the value of u. so what was u? we got u by taking the square root of u^2, and u^2 we got from taking (-b/2a)^2 (which is 4^2)and subtracting (c/a). 16-13=3, and 3 was u^2, right? lastly, bring that +/- along. remember, u is (+ or -) because we took the square root, but it doesn't matter in "his" method because we're putting it in once for (x+[_]) and once for (x-[_]). okay, so lets write that all out. our solutions to this quadratic was (you may want to write this out on paper) (-b/2a) [+ or -] sqrt( (-b/2a)^2 - (c/a) ) lets distribute that square inside the square root (-b/2a) [+ or -] sqrt( (b^2/4a^2) - (c/a) ) now multiply c/a by (4a/4a) so we can combine those fractions (-b/2a) [+ or -] sqrt( (b^2)-4ac / (4a^2) ) and lastly, we distribute the square root onto the fraction. the top stays the same, but the bottom gives us 2a. so we finally get: (-b/2a) [+ or -] sqrt( (b^2)-4ac ) / 2a combine the fractions and what do we get? (-b [+ or -] sqrt(b^2-4ac) )/2a The quadratic formula. You already have learned it, but instead of taking a bunch of intuitive steps, you get a neat formula to do it. tada!

[@alessio2980]

⁠@@ukulelevillain4170I’m dyslexic, so it has been difficult to follow to me. But you are right, the only difference as you said, is that we learned a formula, and not few intuitive steps. A formula can be forgot, because in our mind it’s just it, otherwise understanding the process with reasoning makes it all easier. Thank you for explanation (Sorry for bad English)

[@nirmalhadiya219]

Very good my guy these nonCreative suckers should learn the lesson​@@ukulelevillain4170

[@Priyanka-tl8nj]

Exactly :) More empowerment to us 😊 This is why I find subjects with formulas, ahm.. difficult, ​@@alessio2980

[@kenhaley4]

Amazing! In my 77 years I've never seen this method...It's brilliant.

[@Aamir_Maqbool]

it's actually pretty basic in india they teach it in 8th grade

[@AyyoGamer]

​​@@Aamir_Maqbool we are a lot advanced in math than rest of the world...even in terms of technological advancement in studies ,we r ahead of most western countries.. just show any online class video to any westerner, they can't believe we are using Digital Boards here so extensively , digital boards are rare in west😂😂

[@aa6eheia156]

​@@AyyoGamer advanced or wasting resources and talent in unnecessary stuffs? 😂😂😂 Zero innovation and creativity to create new technologies while using hundreds of small tricks to solve simple problems for entrance examinations😂😂😂

[@AyyoGamer]

@@aa6eheia156 you are right in that matter... but when we create advances in something ,we can use the same advances in different fields of intrest.. nowadays, its a trend to study for government exam ,the trend will diminish in the future , then we will use same technological advancements in other fields also.. it will take some time because a whole ecosystem has been established for Competitive exams , they just want us to study for Government exams...

[@divyeshkumarbalar7732]

Better luck next time, try to be born in India or China in your next life

[@lakshya5946]

I'm literally crying 😭 right now my biggest fear in maths is quadratic equations and professor just turned suspence into mist thanks outlier ❤️

[@parvgoyal7997]

Don't worry quadratic equations won't rouble you but other things will

[@vmin9848]

Quadratic equations are heaven imo, a lot of stuff boils down to factorizing a quadratic equation and they’re just so precious. Personifying quadratic equations is kinda weird but they are like that one cousin you adore

[@lakshya5946]

@@vmin9848 First of all I didn't thought KZbin will show me a reply of 2 year old comment that I've forgetten. I had learnt them and I have passed my class 12 now too! I haven't done enough practice of them in class 10. When they were Introduced to me properly and that was my mistake but everything i fine now so!

[@firemonkey1015]

Wait until you get to calculus 2

[@lakshya5946]

@@firemonkey1015 how you guys are suddenly commenting on this video, do anyone mentioned it anywhere? and I'm not in college right now I'm just making some personal projects and enjoying a gap year!

[@Tletna]

I liked the joke about how math professors forget how to do basic arithmetic operations. When, I was just an undergrad engineering student I already started to forget, haha. While I do feel like this was aimed at children, as an adult, I still liked this lesson. It was very simple. He showed clearly how and why it worked. He repeated it twice to show it worked with two different kinds of equations. He showed that we can divide away the a in ax^2 + bx + c. He showed that there will be a + and - version of the answer but that they will be equivalent anyway. His tone was cheerful and kept my attention even if it was a little patronizing sounding. But, he seemed so genuinely into it that I didn't take it as patronizing but just that he really loved what he was doing. I love math, but even so, sometimes it is hard to make math that is really tedious fun but he sort of made it fun. What I really liked too is that he was just using basic principles instead of memorization. His simple method would work on almost any quadratic equation. Great video, great teacher!

[@johnnyli4993]

As soon as I look at the title, I supposed this was about the famous Po-Shen Loh method. It's a big pleasure for me to meet through the internet this young talented man. Greetings from a 老外 in China.

[@K.Parth_Singh]

This is out of topic but isn't youtube banned in china? how are you using youtube?

[@ZelenoJabko]

@@K.Parth_Singh he is cheating the party by using the VPN

[@johnnyli4993]

As soon as you don't mess with politics (and it's none of my business), they just turn a blind eye to my VPN.

[@vinicius8644]

@@johnnyli4993 >the politics of my country is none of my business The absolute state of chinese people lmao

[@raven44400]

@@vinicius8644 OP literally said he is a foreigner it's not his country ret4rd

[@kindreddarkness]

Honestly... for me, this was most useful in actually understanding square roots. This method teaches itself and all it's friends. Well done.

[@Tetsujinfr]

You are my héro !!! Never found such a clear, generic, and powerful explaination on quadratic eq factorisation. Real neat, such a talent to teach here. You deserve a pedagogy price if that exists. Thanks a million times professor!

[@wren4077]

899 is one short of 900 which is 30 squared so I tried 29 and 31 and it worked something I did in this surprised me because I just kinda did it in my head and never really thought about it before, the square of a number will always be one more than the product of one less and one more than the number a^2 = (a-1)*(a+1) + 1 if you break this open it's obvious how. but I never thought about it this way. In working my way back to finding factors of a product. For instance, now I know what the product of 1599 can be shown as 39*41 Neat.

[@A.A.Iretomiwa]

💀 👁👄👁 Bruh I would have gone my whole life without know that formula Oh My God

[@wren4077]

@@A.A.Iretomiwa lol thanks for commenting causei forgot about it again.

[@deadeye_plays]

it’s a standard algebraic formula no? a^2-b^2 = (a+b)(a-b) you put the value for 1 for b and you get it

[@sigmarule-lb1sr]

the fact that this man can see any type of quadratic equation and know its factors right away is absolutely mind blowing. and this coming from a person who is not bad at maths for his age

[@philipsamways562]

Sorry to dampen your enthusiasm but the example is very carefully chosen. In an Instant it's clear that the ,899 is one less than 30 squared. In real life, real quadratic equations are likely to have the 899 replaced by cos (Pi/2)Sin(à). This method is of very limited use except to solve second-level maths exams

[@sigmarule-lb1sr]

@@philipsamways562 no, i truly appreciate your input, thanks for the extra knowledge.

[@lebesgue-integral]

We used to call this in Brazil "the sum and product method," but I never thought of why this works. Your explanation was great! Thanks for sharing

[@ericcartman2294]

Thank you, Master Po! Although, I forgot basic math more than 2 decades ago and currently have almost 0 math knowledge, this video was quite entertaining. 👏

[@peguisreal]

This is absolutely gorgeous. I have been learning and doing quadratic equations all wrong my entire life.

[@ora-knowledgepressultimate6386]

why this channel is so underrated 😓 found it today and loved it 🙌 😍 thanks to whoever bringing these quality content

[@krishnachoubey8648]

Once my maths teacher gave our class a puzzle like - a+b=11; a·b=28 Find a & b. He gave some hints & tricks to solve questions & puzzles of such kind, which do come a lot in competitive exams. I generalised it in an equation(not that much work rly) to get a nice equation with a very versatile application- (a,b) = {s±√(s² - 4p)}/2 Where, s = a+b And, p = ab For any number a & b. (we only know the value of s & p and not of a&b) From the above question, If we plug in the value of s=11 & p=28 Then value of a & b can be calculated through- (a,b) = {s±√(s² - 4p)}/2 = {11 ± √(11² - 4·28)}/2 = {11 ± √(121 - 112)}/2 = {11 ± √9}/2 = {11 ± 3}/2 (a,b) = (7, 4) Which by cross verification, is correct. [Derivation of the eqn. in reply section]

[@krishnachoubey8648]

Derivation of the eqn.- Let a&b be any given real numbers. Given:- s = a+b; & p = a·b Find:- a & b (a+b)² = a² + b² + 2ab s² = a² + b² + 2ab s² - 4p = a² + b² - 2ab s² - 4p = (a - b)² (a - b) = √(s² - 4p) a = {(a + b) + (a - b)}/2 = {s + √(s² - 4p)}/2 b = {(a + b) - (a - b)}/2 = {s - √(s² - 4p)}/2 Since a & b have a symmetric equation, (a,b) = {s ± √(s² - 4p)}/2 [And hence we have the not-so-hard-but-clever equation]

[@mathonize]

​@@krishnachoubey8648i always remembered this as (a+b)²-(a-b)² = 4ab Then i would go on to solve for (a-b) and use elimination for finding a and b. Never thought of making a formula out of it! Thanks man!!

[@angleth]

I’ve gone through competitive math books and seen questions like this. I hated memorizing the formula because I felt that it would weaken my actual understanding of the subject. Instead, I learned the why and the how

[@mathonize]

@@angleth when it comes to quadratics I think there are only 3 solid methods - middle term splitting, completing the square, and the quadratic formula . All the others are derived somehow from these. So yeah you're correct learning the right concept clears all the doubts.

[@Giebu2b]

​@@krishnachoubey8648can u solve this quadratic xsquare -8√3x - 27 = 0 in this process Let me know

[@juanitoviejo2121]

A glimmer of hope @3:07, then a miracle! @3:57. That's how I memorized enough algebra back in the old days to get a B on the final! Thanks Professor Po for your cheery and bright style of instruction.

[@mandolinic]

Interesting but leaves more questions than answers. For instance: What do you do when the b (middle) term isn't an even number? What do you do when there is no nice division by the a (first) term? What do you do when the b and c factors are both positive? Or negative? What do you do when the solutions aren't real? The beauty of the (-b ± √(b^2-4ac))/(2a) formula is that it works in every case, and IMHO it's easier to remember.

[@kelliamaral6891]

It works in all those cases! If the coefficient of the leading term is say 7 and the coefficient of the linear term is say 3, then you would use 3/7 as the sum you want and use half of that in your difference of squares!

[@Giebu2b]

​@@kelliamaral6891 can u solve this quadratic xsquare -8√3x - 27 = 0 in this process Let me know

[@death_nope]

@@Giebu2b you're right, using foil or the way pro teach is for some z numbers or ratio, not for thiis. i would rather using formula to calculate

[@Giebu2b]

@@death_nope ok

[@epsi]

For any equation of the form Ax² + Bx + C = 0 , you need to determine _u_ in [B/(2A) - u][B/(2A) + u] = C/A to obtain the result x = -B/(2A) ± u . Btw, if you try solving for _u_ in the equation involving C/A above, you will see that the term B²/(4A²) is used. That is equal to [±B/(2A)]², so you could actually use -B/(2A) and get the same result. That means you can do this: B²/(4A²) - C/A = u² q = -B/(2A) → q² = B²/(4A²) q² - C/A = u² √(q² - C/A) = u x = -B/(2A) ± u x = q ± √(q² - C/A) You can manipulate the quadratic formula to obtain the same form: x = [-B ± √(B² - 4AC)] / (2A) x = -B/(2A) ± √(B² - 4AC)/(2A) x = q ± √((B² - 4AC)/(4A²)) x = q ± √(B²/(4A²) - 4AC/(4A²)) x = q ± √(q² - C/A) This form works well when B/(2A) and C/A are both integers, especially when A=1. Ex: 3x² - 24x + 9 = 0 (3x² - 24x + 9)/3 = 0/3 x² - 8x + 3 = 0 q = -(-8)/(2•1) = 4 x = 4 ± √(4² - 3/1) x = 4 ± √(16 - 3) x = 4 ± √13 That's definitely better than the standard quadratic formula in my opinion: (3x² - 24x + 9)/3 = 0/3 x² - 8x + 3 = 0 x = [-(-8) ± √((-8)² - 4•1•3)] / (2•1) x = (8 ± √(64 - 12))/2 x = (8 ± √52)/2 x = (8 ± 2√13)/2 x = 4 ± √13 And if the equation wasn't divided by 3, a value >100 would be used as the input to the square root function with the standard quadratic equation, unlike the modified form that has division by A built into it.

[@L1ght214]

I never knew this till now, I could have just think outside the box, but instead stuck with the idea of only relying on the instructions of the teacher rather than thinking about radical solutions, which really brings out the beauty of math.

[@TheOriginalJohnDoe]

Posted 2 years ago and less than 300k views. So many people in the world are missing out on this brilliant method.

[@theoneobviousguy7328]

The ways he described are very unique and interesting, however, you can also just use the quadratic formula for a lot (it’s seriously not hard to memorize) and when there is a number in front of the x squared, you can use the slide and divide method where you take the number in front of the x squared and then drag it to the non x term, you then multiply it by the term then you factor that equation, BUT THEN you divide by the number you dragged to the non x term, you divide that number from each of the factored numbers, if its a pretty number, its that, but if you cant divide and get a pretty number, you just put that in fromt of that x term, you then get the zeros. The way this guy described it was good and all, but he did more work then needed.

[@mikeeisler6463]

3:35 minutes in and you hooked me. This is amazing. Difference of squares is truly the work horse of algebra. Brilliant

[@bandanaacharjee-cf5ye]

This is Called the Middle Factorisation Method. It is Very Common in Countries like India,Nepal,Bangladesh,China etc.

[@banshthappa3222]

Yes I know ithis method since 10th standard I don't get something special in this vedio​@@bandanaacharjee-cf5ye

[@janthran]

i've seen this taught before, but not in so much depth and with so much clarity and joy. felt like i was discovering a cheat code with you!

[@donbelisario8811]

I saw him with Lex fridman, I'd love to read a math book from such a humble and insightful person, the kind of math teacher I always wanted to have

[@mjs28s]

Um....so many math instructors in multiple schools and college and not ONE TIME did this come up. Oy! Glad I found this by accident! I love it.

[@mystyx7]

It's surprising how much less advanced Western mathematics seems compared to the methods commonly used by school children in Asian countries.

[@vzx]

Yea... This is the common method we learn in Jr high school for factorization. Still fluent in it even after 2.5 decades.

[@souratanayroy3554]

As a JEE aspirant in India, I agree. No hate to the teacher though, he seems amazing, and its really great that he discovered this by himself. Loved to know the philosophy part of this video.

[@mystyx7]

@@souratanayroy3554 Yes, Teacher and Teaching style is amazing.

[@chinmaychaligeri4654]

Like these things arent as hard as they seem and we do solve these kind of quadratics in the middle of a random question(as a part of the quenstion)

[@NirjharChakraborty-f9u]

This is just the start try INMO Paper and reply back

[@sarfarazsarkar-uz4gr]

this is so diff then i have known maths the way he shares his knowledgee is just amazingggg and makes u feel intrested and smile the whole time

[@omdesai6058]

At 1:30 claiming that the minus signs allows for us to see the subtrahend as the solution is correct but is an interesting leap over some algebra. Is it because the algebra behind it are a prerequisite to learning from this video or to avoid having to explain those algebraic steps? Anyways great video!

[@chibicretster]

I think it is due to the form, (x-s)(x-r)=0. X subtracting the unknown intercept will always lead to zero due to reflection of parabolas. For example x-3=0, x=3. The intercept is 3 and therefore subtracting the intercept to x should always get 0. TLDR: He gets it from factored forms formula

[@nzarambafelicien9617]

The best of best way without guessing. My whole university studies in applied mathematics, I never met a such easy way for quadratic equation. Surprisingly this method was used by Babylonians......Fun indeed.

[@Taric25]

His number u is just √(b²-4ac)/(2a), so his missing numbers in the square boxes are -b/(2a)±u, which is just the quadratic formula. This is just the quadratic formula broken down into smaller, longer steps.

[@amortalbeing]

This was amazingly done. loved it and enjoyed every second of it. 🥰

[@lakshya5946]

When I saw this my mathematics search was over truely mathematics is a very cool think and more cool with outlier ❤️♥️❤️

[@extraincomesuz]

I failed Algebra twice. So I was unable to advance in mathematics. And, it made my desire to go into scientific research impossible. However, watching this guy makes perfect sense. I wish he had been my teacher. Im 61 now.😅

[@anuragiyer4664]

This method was taught in my school even before the quadratic formula

[@Queenbree469]

What country are you from?

[@nobodybutnobody]

@@Queenbree469 India for sure

[@HeatherHolt]

What a genuine guy ❤ I want to listen to him. That’s such a huge part of being a good teacher… being engaging.

[@jrousselle7828]

Absolutely brilliant!!! The presentation was flawless. My favorite part was "don't ask mathematicians to multiply".

[@superclass.ad20]

Mine too hahah

[@paulbork7647]

2:55 Early morning for me. From here, I multiplied the 30 times 30 got 900. Close to 899. 29 and 31. Immediately, units place is 9, so potentially correct. Multiply to check these numbers. Change math to make easier (30-1)*(30+1). This is 900-30+30-1 or 899. QED. Thanks for an interesting mental path.

[@Nishtha_ns7]

THIS IS A MIRACLE. I'm seriously blown away. I didn't think that this was actually what you could do to figure out the middle term split but just wow. This was what I always do in my mind to figure out zeros for big and complex polynomials! Thank you for simplifying it more and making it so easy.

[@verysb]

That’s a good method for someone that don’t want to memorise the formula. But it would be better if u show another situations such as ax^2+bx+c=0 when c is negative, I think it should still work, but maybe need more process cause that it would bounce (x+alpha)(x-beta) where we need to find 2 numbers where 1 is positive and 1 is negative and since one is pos and one is neg, it would be a subtraction in our mind to get b another thing that I hope the video could add is when the solution is imaginary where b^2-4ac

[@torietorreano6613]

I really wish I understood math better. I'm actually pretty good at adding things mentally, multiplying, doing percentages. But I don't understand higher mathematics. At all. When mathematics are presented in such a format to where an engaging person explains it just makes me wish I had paid better attention in high school... Great video, because I never thought I'd want to watch anything about a quadratic equation. I was intrigued!

[@mukulshakya571]

This approach is nothing but a new way of using the quadratic formula. We have, ax² + bx + c = 0 => x² + (b/a)x + c/a = 0 To get the factors, => (b/(2a) + u)(b/(2a) - u) = c/a => b²/(4a²) - u² = c/a => u = ±(√(b²-4ac))/(2a) (x - x1)(x - x2) = 0 where x1 = -(b/(2a) + u) and, x2 = -(b/(2a) - u) => x = x1 or x = x2 => x = -b/(2a) ± (√(b²-4ac))/(2a) => x = (-b ± √(b²-4ac))/(2a)

[@ta_ogboy9998]

Yeah you can derive this by noticing that the first solution plus the second solution is -b/a, and the midpoint between the two solutions is therefore -b/2a. The distance between the midpoint and the two solutions is the unknown variable "u". So -b/2a + u and -b/2a - u are the two solutions. Then you can note that the first solution times the second solution is c/a, and then you solve for "u" to get the distance between the midpoint and the solutions (distance between midpoint and the solutions: sqrt(b^2-4ac)/2a) Note: midpoint is referring to the vertex, while solutions is referring to where the line passes the x-axis. The distance between the vertex (-b/2a) and the two solutions is denoted by "u", which you find to be the sqrt(b^2-4ac)/2a

[@mukulshakya571]

@@ta_ogboy9998 Yeah, you are right! This can also be another way to think this.

[@a3wassver47]

Can't you think out of the box? Everyone can remember the formulas dude

[@jackett_2122]

@a3wassver47 the very reason he can think out of the box is why he wrote this comment. he could derive the formula from this so called out of the box method, which as he says is just a simpler way to use the quadratic formula!

[@mukulshakya571]

@@a3wassver47 Sorry brother! I’m definitely not a genius! Consequently, I was simply able to derive this basic formula, which you can check out using the following link: drive.google.com/file/d/1opFvHgBMDmd2cl-QY-QsdQCkJOAqBln-/view?usp=drivesdk I'm not sure if this formula has already been discovered, but I derived it independently. This formula is not exactly a quadratic formula since it still works when a=0. In that case, it works as a formula for a linear equation, although we don’t really need a formula for that. Still, it provides the correct solutions for any quadratic equation. This formula shows that not only x = -b/(2a) gives the real and equal roots when D=0, but x = (-2c)/b also works if we put D=0 in the formula I derived.

[@malielanijoy]

Thank you! Honestly, I just enjoyed watching this video because you're just so enjoyable to listen to. You can just tell you love math, and as someone who also has loved math for as long as I can remember, it just feels so great to listen to someone with a common passion. 😊

[@pinkribbongigikoo]

thanks for the method, i am going to use this for my entire life.

[@JustWatchingVideo56]

Been doing this in school the whole time already. I always felt like it was very roundabout. But I’m so glad to see someone so excited about this.

[@LydellAaron]

6:09 don't ask a professor to multiply: I thought I was the only one with a fear of multiplying on the spot.

[@monsieurfrog]

this is actually really interesting and a really simple concept to understand. you do this sort of thing with (4-u, 4+u) in topology and it's called the ball of center 4 and radius u. it's used to prove the neighbourhoods of a number, never thought i'd stumble over it here

[@tonywmckinney]

I enjoyed this video a lot. Very interesting approach. Much simpler than, say, completing the square and I was able to back into the quadratic equation more easily using this logic.

[@N0Xa880iUL]

I still prefer completing the square method. As in this method you can easily get the plus minus signs wrong.

[@piotrcthlu]

Mate you are Awesome, i wish there were more math professors like you

[@flocci22n]

This is "completing the square", a method found in every algebra textbook.

[@moiratimmons8755]

It's related, (middle term over 2 is familiar) but not "completing the square" acc to nelson textbooks

[@arc8588]

Depending on the curriculum and grade in all general algebra textbooks yes but not all highschool textbooks but good for u on being a no it all

[@flocci22n]

@@arc8588Just stating fact. It’s not some secret factoring method. Lame clickbait. It’s been known for centuries and routinely taught to algebra students. Sorry if you are offended by truth.

[@Grizzly01-vr4pn]

@@arc8588’no it all’ 😂😂

[@shyam6468]

This is EXCEPTIONAL. I have No words to describe the beauty it holds. Never thought of the quadratic equation this way.

[@AkshaySinghJamwal]

At 1:49 he wasn't making a joke, he was double checking his memory.

[@SamSam-xz8rd]

Thank you very much. That is very helpful.

[@madhesh720]

may i know ow to un add if the number is odd?

[@JH-dj8hd]

Just use half of the sum number that is the x coefficient, which will be a number and a half...?

[@gwzeejir]

X - 2

[@oh-range]

This is basically the same as the pq-formula (a variant of the well-known abc-formula which depends on the coefficient in front of the x^2 to be 1) and we also learn it in parts of Germany. Still the Video explains it very well and it’s fun to watch.

[@vinayakmalhotra1184]

What if the middle term (x term) is an odd number instead of even number?

[@agwa.shorts]

Up

[@sokka90ml]

Multiple equation by 2 or just divide that odd number by 2 and find (x/2+u)(x/2-u), whichever convenient follow

[@epsi]

For any equation of the form Ax² + Bx + C = 0 , you need to determine _u_ in [B/(2A) - u][B/(2A) + u] = C/A to obtain the result x = -B/(2A) ± u . Btw, if you try solving for _u_ in the equation involving C/A above, you will see that the term B²/(4A²) is used. That is equal to [±B/(2A)]², so you could actually use -B/(2A) and get the same result. That means you can do this: B²/(4A²) - C/A = u² q = -B/(2A) → q² = B²/(4A²) q² - C/A = u² √(q² - C/A) = u x = -B/(2A) ± u x = q ± √(q² - C/A) You can manipulate the quadratic formula to obtain the same form: x = [-B ± √(B² - 4AC)] / (2A) x = -B/(2A) ± √(B² - 4AC)/(2A) x = q ± √((B² - 4AC)/(4A²)) x = q ± √(B²/(4A²) - 4AC/(4A²)) x = q ± √(q² - C/A) This form works well when B/(2A) and C/A are both integers, especially when A=1. Ex: 3x² - 24x + 9 = 0 (3x² - 24x + 9)/3 = 0/3 x² - 8x + 3 = 0 q = -(-8)/(2•1) = 4 x = 4 ± √(4² - 3/1) x = 4 ± √(16 - 3) x = 4 ± √13 That's definitely better than the standard quadratic formula in my opinion: (3x² - 24x + 9)/3 = 0/3 x² - 8x + 3 = 0 x = [-(-8) ± √((-8)² - 4•1•3)] / (2•1) x = (8 ± √(64 - 12))/2 x = (8 ± √52)/2 x = (8 ± 2√13)/2 x = 4 ± √13 And if the equation wasn't divided by 3, a value >100 would be used as the input to the square root function with the standard quadratic equation, unlike the modified form that has division by A built into it.

[@sokka90ml]

@@epsi great explanation and method solving, hope I didn't use gpt, even u understood the differences Great bro

[@kallewirsch2263]

that's actually the well known pq formula explained (or a possible derivation of it) X^2 + px + q = 0 with the solution x = - p/2 +- sqrt( (p/2)^2 - q ) You can match every term in the pq-solution 1:1 to your explanation. The term under the square root is just a formalized version of what you did to solve for the needed difference to your u. Nevertheless a nice explanation for why the pq formula works.

[@primefactor5417]

awesome technique .....

[@SebastianGomez-cb9vu]

Great teaching! Thank you!

[@Riya-o4v]

Sir the method you suggested does it works only if value of b is even . So what about when value of b turns to be odd ex x²-5x+6 and what about when both b and c's values are odd ex x²-7x+9. And what about when b is prime x²-5x+3 . Please response as soon as possible sir .

[@Riya-o4v]

Thanks sir I have finded it out that it works for all the situations given above

[@isaiahsams8790]

Yeah you just put the 5 over 2 so you get -5/2 +5/2

[@Riya-o4v]

@@isaiahsams8790 which one are you answering to

[@sindonophany7024]

Just woke up early as in 01.30 and came across this after looking at something about proof by contradiction which I found a bit difficult because of the speed the young man was speaking -but I think you tube is marvelous for many things and helping me get into maths again after my love of this subject was eroded by school 50 years ago. This gentleman is obviously in love with his subject. Fully understood .I'll now try to put it into practice as required.😂

[@irtza947]

What to do when the sum is a odd number??

[@KONAMAN100]

What a joyous and patient explanation. Brilliant!

[@richardjafrate5124]

Brilliant! I was tagged as "not being good at math" because I was not good at guessing factors and was just as bad at arithmetic as the professor. However, I excelled at calculus and engineering. 😶

[@callierobillard5809]

This method is so cool. I was halfway to figuring it out in HS, but then forgot about it. I'm now a teacher and would love to share this with students. Is there an example where b is odd? This is the type that I am unsure of. Thank you!

[@SITANSUSEKHARDas-i6n]

What if the coefficient of middle term is odd??

[@Roronoro--zoro-99]

Hiw

[@notanother7396]

Allergic to decimals?

[@muzaffarrahaman-v3h]

cause, otherwise just follow the same, but with decimals

[@MuhammadBilal-qg6ny]

How do I apply this technique to x^2 +3x -6 = 0? How do I solve for the unknown number when u^2 = -15/4? Please explain!

[@xenvector]

always calculate your discriminant before factorizing, cause if they're just asking for real solutions and u get a negative one u can just skip the exercise

[@jsurinderveygal561]

It was fun watching you. Cannot wait to teach my math class on Monday on how to do you what I just learnt. I teach math in a NZ school. THANK YOU!

[@agrocassiano]

(2x2 - 16x + 26 = 0) : 2 a b c *x² = 8x - 13* a b c Fórmula _ *X = b/2 + - ^[ (b/2)² + (c) ]* X = 8/2 + - ^[ (8/2)² + (- 13) X = 4 + - ^[ 16 - 13 ] *X = 4 + - ^[ 3 ]* Parabéns Outlier, sucesso sempre.

[@robwin0072]

Professor, I have questions about what you just covered on the board. First, if the middle term (or sum term) is odd, how does that affect the solution? In the second example, should we consider the coefficient of the first term (‘2’) when showing the final solution? I also wanted to share a general thought. Math feels more manageable when there are multiple ways to reach a solution. We’re often taught a specific method in class, but exam questions demand different approaches. I wonder if this is to assess our understanding of the core concept rather than just our ability to follow a set formula.

[@thatonedynamitecuber]

ok when did we all get recommended this

[@J9-di4ry]

Mind blowing, thank you to solve my 4 years problen

[@davidthomas6043]

I tried this out on the following equation; x (squared) +5x -39 = 0; . I got u squared =45.25.. so u = squ rt 45.25. This then gave me answers for x as (2.5 + this u-value) and (2.5 - this u-value) I don't think these answers are correct . Solving for x the usual way gives quite different answers ! Can someone comment please and show the proper workings.? I am not a youngster, actually 93. .

[@manhhao-p7w]

Solving by this method gives out that u squared is negative, which is still an answer, if you consider imaginary numbers. Using the quadratic formula, then b²-4ac is negative, which gives out imaginary answers too.

[@jackett_2122]

you got √45.25 because you put the values into the formula as such: (2.5)² - u² = -39 you dont need to equate it as -39 but 39 which would make u-number = √-32.75 or √(-131/4) hope this helps

[@davidthomas6043]

@@jackett_2122 Thanks for replying.. Doing it my way, I now get the same two answers using both methods, The answers I got were x equals 4.23 and 9.23. Of these only 4.23 satisfies the equation. I'm baffled why you treated -39 as just 39.. I have no idea why 9.23 is not acceptable as an answer.

[@foodsafari-rj3uq]

​​@@davidthomas6043 Using this trick we get (2.5)² - u² = -39 => u² = 45.25. Clearly, (2.5+√(45.25))×(2.5−√(45.25)) = -39. Or if you'd like to use rational approximations then, (2.5+6.23)(2.5-6.23) ≈ -39 (39.0429). The equation thus reduces to (x+9.23)(x-4.23) = 0. Meaning the solutions are 4.23 and -9.23.

[@blocky9158]

Quick question, what would happen if the b coefficient (ax^2 + bx + c) isn't an even number (Cannot be divided equally into 2).

[@RiverHeights]

This seems very confusing at first, but I'm going to learn it, and keep trying so I get good at it

[@Suhan_Jadhao]

if it seems confusing and all you wanna do is solve the quadratic without understanding the how, just learn the quadratic formula. It will be much faster, and will work in many more ways

[@RiverHeights]

@@Suhan_Jadhao That was originally the plan, but some exams do not allow you to use a scientific calculator. You only have an arithmetic one. No squareroots or anything.

[@annelizebronkhorst4932]

Thank you for explaining this so clearly! I struggle with math but you make it easy to understand and apply.

[@Hit-Knowledge-A]

We can also solve this without pen and paper in just 10 seconds, By use unit digit - .... Guess two numbers - + = 60 × = 899 .... Unit digit of is [899] 9, So prime Factor of 9 = 3×3 Or 9×1 But the unit digit in 60 is 0 So we Don't need 3×3 Because if we add 3,3 it will give 6 And if we add 9,1 it will give 0 So we need 9×1 factor .... And now use numbers like 19 × 41, 29 × 31 .... And we find the factor 29×31 =899 29+31 = 60 So the factor will be 29 ,31 .... If you do this in our mind it's so easy to find factor of Any quadratic equation❤

[@lanevalhalla1225]

You saved my life.. I don’t like using the clunky formula and plus for integration. You don’t find the value of X1 and X2 and so you just factorize it and I just hate factorizing using the formal method and you just save me a lot of time❤

[@richardslater677]

I’m not that good at maths so I’m probably wrong here but when you say you can use this method to solve “any” quadratic equation, then how does that work when the equation has no real roots, as most of them do. If b^2 is less than 4ac, there are no (real) roots.

[@shivamchouhan5077]

Agreed, factorisation only helps when we have natural numbers as factors otherwise ain't nobody solving it that way.but Quadratic equation always works!

[@Illuminous_]

I much admire your teaching ability. Thank you for the explicit and complete breakdown of this method and for the beautiful presentation.

[@tallguy3]

What if the sum isnt an even number?

[@ehtesham998]

Then, you can divide by 2

[@ThaoNguyen-ki9yb]

Thanks Professor Po it a really amazing math video i have ever seen

[@int_17]

10:50 we mathematicians are lazy.....welll...yeahhh i guess..😩😩

[@Simba.Luna.]

Absolutely amazing!! Thank You 🙏

[@extremeaviation5500]

Avg. IIT JEE aspirant watching and laughing in the corner

[@prabhakargautam4776]

I can tell you, THAT aspirant, always remained an aspirant.

[@prabhakargautam4776]

I can tell you, THAT aspirant, always remained an aspirant.

[@Rocketjake]

I always just fallback on the quadratic formula, I've already memorized it so it's what I use. This is a really interesting way to tackle the problem and I think it should be the standard way of teaching how to factor quadratics.

[@Cicyro4716]

Great video needed this for my linear algebra exam where i couldnt recall the quadratic formula

[@Bannerkim1]

Good and sound explanation.

[@grandcoder3482]

This thing saved me a lot time. Thanks

[@Flame-bv6wv]

How about quadratic equations (ax^2 + bx + c = 0) with: 1. b is not divisible by 2 (odd number). 2. a is not 1 and cannot be divided to both sides.

[@naqibzahid9628]

Why not went sideway and use fraction un-adding, I.e=(3/2-u)(3/2+u)?

[@JJVolfe]

I couldn't figure this out , as well😢😅

[@goelkiran-ms1yg]

Simply good, thanks for sharing

[@rkalle66]

The formula of difference of squares is a nice way to shorten many multiplications where the difference of factors is even. It can be done in your head. You just have to memorize squares. Example: 22*28 ... using the formula leads to (25+3)(25-3) = 25²-3² = 625-9 = 616 Thus 899 was not difficult to factorize to me.

[@CptUnique]

My friend's dad was trying to tell me this trick. Holy crap, I finally understand it.

[@MuikkuSami]

Thank you so so much for sharing this! You are explaining it so clearly and finally i have a way to do factoring that is actually fun problem solving, instead of rote memorizing.

[@neetavardani8709]

Again i found a fabulous person on my years of youtube journey😅

[@beomini555]

Thanks! This is gonna help out so much since we use quad. equations everywhere

[@expeditioner9322]

I often wonder about the thought process of those who are excellent problem solvers. Do they think about the algorithm first or do they figure it out as they solve the problem. I mean does a person like Lo first figure out how the problem can be solved and the steps that would lead to the solution or do they just start with an approach an fugure out the steps that would be required as they go along.

[@user-ohionguy]

Yo you made my stress less , i thank u a lot for this gift you gave us , you are great, you created history