"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.
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.
@@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: ua-cam.com/video/6lUFxO-aAMU/v-deo.html - my favorite approach.
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!
@@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)
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
@@vmin9848 First of all I didn't thought UA-cam 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 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!
@@_kreap 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😂😂
@@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😂😂😂
@@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...
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!
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.
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
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
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!
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.😅
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.
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.
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!
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.
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.
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
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.😂
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.
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❤
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.
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.
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
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.
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.
This is a whole different level of learning online. If all topics were taught like this I’d be Einstein! I’m an AE major so please do more engineering classes related!!
When you break down the quadratic equation into a product of binomials =0 it should be pointed out that the boxes are, most likely, for different values. I would have liked to see an extra 5 minutes on the video so that you could show proof, perhaps starting from the usual quadratic solution. This would allow some discussion of possible complex solutions. Having said all of this, I really liked the video and the thinking that went into it. A possible added topic would to look at the zero crossings of y=x^2 -60x+899. Keep up the good work!
This begs a meta question about teaching. He's very good and engaging. How do teachers learn to become engaging? It's always been a goal for me. Passion and skill aside, how does one become engaging within teaching or mentoring?
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]
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]
@@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!!
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
@@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.
The "never ask a math professor to multiply" is so real haha Im doing the toughest math for my year in high school and still struggle with elementary school math like dividing by decimals, multiplying large numbers, factorising, etc.
bro ... this is splitting the middle term.... I'm a math tutor.... Factorisation is a topic I have to teach my students everytime... and this splitting always surprises and excites my students interest in maths... btw... one of my students now only likes to study maths now ....
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.
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.
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.
Thank you for sharing this method. I think it’s a great method combining taking parts of factoring method and complete the square method to make quadratic equations easier to solve. From the two examples you showed, both examples had coefficient of x (b) as an even number (and negative), and the constant terms (c) as a positive value. I wonder if the method will still work just as well when b is odd (and positive) and when c is negative.
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. 😊
I'm not a math professor, but I like learning about Geometry. I have some very rudimentary proofs on my channel in the background of sermons. Just visual proofs, for why Sine and Cosine work and Radians and Fractions and one of them shows the difference between a circle and square. Another visually represents Quadratic Equations. Also, my guess with an odd number, is you divide it in half, too, it just turns out a fraction.
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.
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!
wonderful , a different way of solving quadratic equation . I tried doing the above equations in 3 different ways , for me personally shreedhacharya formula was way more faster and easy
When you have been doing math for so long, you will eventually find new ways of solving problems. He is the coach of the US International Math Olympiad team.
“In algebra, if you got an equation and there's an equal sign, you're allowed to beat up the equation as much as you want as long as the left and the right sides get beaten up equally badly” This is so quotable, I swear
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.
- this method is smooth... replacing a more difficult equation with a less difficult one... that's smooth... that's the idea behind integration-by-parts... - if i write the number sixty... in base sixty... it will look like 10 *no that's not a ten that's a one in the sixties place* 12:47
this has to be the singlemost helpful math trick EVER because i have to solve like 100-150 problems per day and quadrarics come up so, so often. ans the radical formula is way too clunky when factorisation isnt possible
saya blom pernah melihat ppenyelesaian aljabar kuadrat jadi menarik, seandianya ada banyak guru yg menarik seperti kamu di negeri fufufafa ini mungkin matematika akan jadi mainan yg indah😁😁😁😁😁😁
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. 😶
This was wonderful! Makes perfect sense and seems a much better way to factor quadratics than the way we teach with cooked up easily factorable quadratics so that students have a possibility of factoring.
When he said *we mathematicians are lazy* it hit my mind and reminded me of my matrix maths teacher who used to told us whenever he used a short trick in a solution 😔😂
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.
I am missing something! In the beginning, it should be pointed out that the problem is solvable e.g., that the solutions are Integers! And if so, I remember from my School-Algebra that every second order equation can be written as X^2 - Sx + P =0 S being the Sum and P the Product of the solutions. And, if the solutions are real and an integer, they can be easily found. Example: X^2 - 5x + 6 = 0 S=5 and P=6 It is easily seen that x1=2 and x2=3 In the given example the only numbers when multiplied gives a 9 are 1 and 9! Therefore to get the sum 60 we must only check 9/51, 19/41, 29/31, 39/21, 49/11 and 59/1 and see which two multiplied comes near 900 and conclude relatively quick to 29 and 31! Another possibility that works sometimes fine is to find the divisors of the product and check-up the sum! PS. When you have for example 12x^2 -x - 6 = 0 This approach will be more difficult. But I wonder, even in this case, it is easier to proceed like in the video, or find two numbers who's sum is 1/12 and their product -1/2! By the way, the answer is 3/4 and -2/3
Outstanding Professor! Wished I knew this method years ago, it would have made the difference between a B and an A for sure! (smile). More than ever, I shall subscribe to your channel. A statement that learning should always be continuous! Blessings Prof! Thanks for this method!
This guy literally had me smiling the entire time he talked. When people are passionate about something it is infective.
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.
It's infectious
how we will proceed if there is sum of roots is odd number???
@@krisna4455 I was thinking the same. Can someone answer us please. I want to use this method
Show Respect. Is this how you refer to educated humans around you?? “Guy”.. this “guy” seriously?
This is why YT has become the uni of maths. Thanks Professor Po.
This is the funnest math teacher I've ever seen.
he is the best i know stfu
Then you have never met me
Habibi come 2 Bangladesh
@@EramMahmood Fr Bhai xDD
@@EramMahmood who in their right mind wants to go to Bangladesh!
"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.
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.
@@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: ua-cam.com/video/6lUFxO-aAMU/v-deo.html - my favorite approach.
@@chocolateangel8743 thank you for that link! Much appreciated!
@@Big_Glizzy. You're welcome. 😁
Hey, what if the sum is odd or if the u isn't a perfect square?
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
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!
@@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)
Very good my guy these nonCreative suckers should learn the lesson@@ukulelevillain4170
I'm literally crying 😭 right now my biggest fear in maths is quadratic equations and professor just turned suspence into mist thanks outlier ❤️
Don't worry quadratic equations won't rouble you but other things will
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
@@vmin9848 First of all I didn't thought UA-cam 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!
Wait until you get to calculus 2
@@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!
Amazing! In my 77 years I've never seen this method...It's brilliant.
it's actually pretty basic
in india
they teach it in 8th grade
@@_kreap 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😂😂
@@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😂😂😂
@@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...
Better luck next time, try to be born in India or China in your next life
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!
Honestly... for me, this was most useful in actually understanding square roots. This method teaches itself and all it's friends. Well done.
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.
This is out of topic but isn't youtube banned in china? how are you using youtube?
@@K.Parth_Singh he is cheating the party by using the VPN
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.
@@johnnyli4993 >the politics of my country is none of my business
The absolute state of chinese people lmao
@@vinicius8644 OP literally said he is a foreigner it's not his country ret4rd
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
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
@@philipsamways562 no, i truly appreciate your input, thanks for the extra knowledge.
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!
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.😅
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.
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.
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!
@@kelliamaral6891 can u solve this quadratic xsquare -8√3x - 27 = 0 in this process
Let me know
@@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
@@death_nope ok
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.
This is absolutely gorgeous. I have been learning and doing quadratic equations all wrong my entire life.
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.
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.
Absolutely brilliant!!! The presentation was flawless. My favorite part was "don't ask mathematicians to multiply".
Mine too hahah
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
thanks for the method, i am going to use this for my entire life.
What a genuine guy ❤ I want to listen to him. That’s such a huge part of being a good teacher… being engaging.
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
This method was taught in my school even before the quadratic formula
What country are you from?
@@Queenbree469 India for sure
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.😂
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.
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❤
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.
When I saw this my mathematics search was over truely mathematics is a very cool think and more cool with outlier ❤️♥️❤️
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.
This is EXCEPTIONAL. I have No words to describe the beauty it holds. Never thought of the quadratic equation this way.
3:35 minutes in and you hooked me. This is amazing. Difference of squares is truly the work horse of algebra.
Brilliant
This is Called the Middle Factorisation Method.
It is Very Common in Countries like India,Nepal,Bangladesh,China etc.
Yes I know ithis method since 10th standard I don't get something special in this vedio@@bandanaacharjee-cf5ye
What a joyous and patient explanation. Brilliant!
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.
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
Mate you are Awesome, i wish there were more math professors like you
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.
It's surprising how much less advanced Western mathematics seems compared to the methods commonly used by school children in Asian countries.
Yea... This is the common method we learn in Jr high school for factorization. Still fluent in it even after 2.5 decades.
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.
@@souratanayroy3554 Yes, Teacher and Teaching style is amazing.
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)
This is just the start try INMO Paper and reply back
This is a whole different level of learning online. If all topics were taught like this I’d be Einstein! I’m an AE major so please do more engineering classes related!!
This is brilliant, we like to think we know more now than we did then, but I'm not convinced.
Great video needed this for my linear algebra exam where i couldnt recall the quadratic formula
Thanks! This is gonna help out so much since we use quad. equations everywhere
When you break down the quadratic equation into a product of binomials =0 it should be pointed out that the boxes are, most likely, for different values. I would have liked to see an extra 5 minutes on the video so that you could show proof, perhaps starting from the usual quadratic solution. This would allow some discussion of possible complex solutions. Having said all of this, I really liked the video and the thinking that went into it. A possible added topic would to look at the zero crossings of y=x^2 -60x+899. Keep up the good work!
This begs a meta question about teaching. He's very good and engaging. How do teachers learn to become engaging? It's always been a goal for me. Passion and skill aside, how does one become engaging within teaching or mentoring?
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]
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]
@@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!!
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
@@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.
@@krishnachoubey8648can u solve this quadratic xsquare -8√3x - 27 = 0 in this process
Let me know
Wow thanks for this method. This method is a life saver if you your stuck
The "never ask a math professor to multiply" is so real haha
Im doing the toughest math for my year in high school and still struggle with elementary school math like dividing by decimals, multiplying large numbers, factorising, etc.
bro ... this is splitting the middle term.... I'm a math tutor.... Factorisation is a topic I have to teach my students everytime... and this splitting always surprises and excites my students interest in maths... btw... one of my students now only likes to study maths now ....
Posted 2 years ago and less than 300k views. So many people in the world are missing out on this brilliant method.
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!
Glad i was able to guess his method and able to solve in my mind before the video started
I was taught this method in o level maths in an East Midlands comprehensive in 1975. We lived in a shoebox in the middle of the road.
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.
I still prefer completing the square method.
As in this method you can easily get the plus minus signs wrong.
Brilliant. And if I use this method with just the coefficients, we're back at the quadratic formula.
Thank you for explaining this so clearly! I struggle with math but you make it easy to understand and apply.
Ok. I like you. Factoring has always been the most difficult parts of early math for me.
You go a long way, then come back! Your method is just a combination of completing the square method and the factor theorem.
Woooow! I just applied this in many numbers, its like magic.
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.
Brilliant, Thank You. Had to test it X^2 + 7X - 450 convinced me. Thank You.
how did you divide the 7?(i need help)
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.
Thank you for sharing this method. I think it’s a great method combining taking parts of factoring method and complete the square method to make quadratic equations easier to solve. From the two examples you showed, both examples had coefficient of x (b) as an even number (and negative), and the constant terms (c) as a positive value. I wonder if the method will still work just as well when b is odd (and positive) and when c is negative.
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. 😊
I'm not a math professor, but I like learning about Geometry. I have some very rudimentary proofs on my channel in the background of sermons. Just visual proofs, for why Sine and Cosine work and Radians and Fractions and one of them shows the difference between a circle and square. Another visually represents Quadratic Equations.
Also, my guess with an odd number, is you divide it in half, too, it just turns out a fraction.
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.
Very cool! Never seen anybody solve the "unfactorable" kind without using the quadratic equation!
Thanks Professor Po it a really amazing math video i have ever seen
I much admire your teaching ability.
Thank you for the explicit and complete breakdown of this method and for the beautiful presentation.
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!
wonderful , a different way of solving quadratic equation . I tried doing the above equations in 3 different ways , for me personally shreedhacharya formula was way more faster and easy
When you have been doing math for so long, you will eventually find new ways of solving problems. He is the coach of the US International Math Olympiad team.
“In algebra, if you got an equation and there's an equal sign, you're allowed to beat up the equation as much as you want as long as the left and the right sides get beaten up equally badly”
This is so quotable, I swear
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.
I like the math history, someone had to discover/invent it. It’s cool how ancient ideas make it all the way to the present.
BRO I REMEMBER WATCHING HIS LIVESTREAMS BACK IN COVID19 CUZ MY MOM FORCED ME TO. OMGGGGG
- this method is smooth... replacing a more difficult equation with a less difficult one... that's smooth... that's the idea behind integration-by-parts...
- if i write the number sixty... in base sixty... it will look like 10 *no that's not a ten that's a one in the sixties place* 12:47
Dudeee.....u literally blew my brains out....
this has to be the singlemost helpful math trick EVER because i have to solve like 100-150 problems per day and quadrarics come up so, so often. ans the radical formula is way too clunky when factorisation isnt possible
This is awesome. Finally found something out of the box.
Thank you ! Bro for teaching me another approach 😀😊
At 1:49 he wasn't making a joke, he was double checking his memory.
Thank u so much ❤
My mother told me this when I was in 10th grade that we could find the square roots of a and c, any of those 2 numberr will satisfy the equation
This man is so great. I love this man for his outstanding work
saya blom pernah melihat ppenyelesaian aljabar kuadrat jadi menarik, seandianya ada banyak guru yg menarik seperti kamu di negeri fufufafa ini mungkin matematika akan jadi mainan yg indah😁😁😁😁😁😁
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. 😶
This was wonderful! Makes perfect sense and seems a much better way to factor quadratics than the way we teach with cooked up easily factorable quadratics so that students have a possibility of factoring.
Amazing trick to solve Quadratic Equations. I will use it from now on.
Yo you made my stress less , i thank u a lot for this gift you gave us , you are great, you created history
Mind boggling. Now I’m off to revision my maths 😆
When he said *we mathematicians are lazy* it hit my mind and reminded me of my matrix maths teacher who used to told us whenever he used a short trick in a solution 😔😂
THIS IS THE MOST USEFUL VIDEO I HAVE SEEN ON THIS APP
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.
I am missing something! In the beginning, it should be pointed out that the problem is solvable e.g., that the solutions are Integers!
And if so, I remember from my School-Algebra that every second order equation can be written as
X^2 - Sx + P =0
S being the Sum and P the Product of the solutions. And, if the solutions are real and an integer, they can be easily found.
Example:
X^2 - 5x + 6 = 0
S=5 and P=6
It is easily seen that x1=2 and x2=3
In the given example the only numbers when multiplied gives a 9 are 1 and 9! Therefore to get the sum 60 we must only check 9/51, 19/41, 29/31, 39/21, 49/11 and 59/1 and see which two multiplied comes near 900 and conclude relatively quick to 29 and 31!
Another possibility that works sometimes fine is to find the divisors of the product and check-up the sum!
PS. When you have for example
12x^2 -x - 6 = 0
This approach will be more difficult. But I wonder, even in this case, it is easier to proceed like in the video, or find two numbers who's sum is 1/12 and their product -1/2!
By the way, the answer is 3/4 and -2/3
Outstanding Professor! Wished I knew this method years ago, it would have made the difference between a B and an A for sure! (smile). More than ever, I shall subscribe to your channel. A statement that learning should always be continuous! Blessings Prof! Thanks for this method!
More people need to watch this
It woyld help a lot, if you would put the writing board upeight in the camera while explaining
This is going to be so handy on the GRE - thanks!
Well, too time consuming and does not work if there are no solns
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.
Why not went sideway and use fraction un-adding, I.e=(3/2-u)(3/2+u)?
I couldn't figure this out , as well😢😅
6:09 don't ask a professor to multiply: I thought I was the only one with a fear of multiplying on the spot.