If you pay attention this method is basically completing the square and you could use it as a a proof for the quadratic equation. But it is a good explanation of why completing the square actually works. I like that it incorporates difference of squares.
Everyone is saying there is nothing new...Of course there is nothing ..but the algorithm he has used is simple! You know..what ever method you use to solve any question...ultimately all of them are related...so of course use any method to solve equation ..it will ultimately give the "quadratic formula" result..because that's what we are supposed to find!!!!
I recall my teacher using this to let us solve quadratic equations in a more guided way given that the quadratic formula was supposed to be taught a year after. He just didn't generalize it to solving any such equations. I don't think the quadratic formula is that complicated, though, and as other pointed out, it's necessary anyway when solving for complex numbers.
PRO Shen we solve with this methods last 5 years its many many methods u can used in algebra i hope every one work in this methods to make the math easy
Too long? Here are some jump spots 0:00 Me being excited 1:48 Start here if you want all the math 4:50 Start here if you're in Algebra and want to see the full explanation including background Algebra 7:12 Start here if you're past Algebra, still remember it, and just want to see the new method 16:17 TL;DR Start here if you're good at math and can get it just from a summary
I mean.. it's literally just the quadratic formula. -b/2 plus or minus that z term which us (b/2)^2-c a was set to 1 anyway and sqrt(x)/2 is (sqrt(x/4)) as everyone knows. this is l i t e r a l l y the exact same calculation as the quadratic formula for a=1, which always reduces to: -b/2 ± sqrt(b^2/4-c)
I had been using this graphical approach in my high school exams, simply because I was good with calculus based solving. But this method is the same in terms of computers if not more inefficient. This approach would take a computer more time to solve than just using the quadratic formula
I see a lot of criticism on how this is not necessarily "new" or that it involves more steps than the quadratic formula. I agree, but not so harshly. I think it's one more intuitive way to think about solving quadratic equations.. but also think it may be most helpful to those who already intuitively understand it. For other students, quadratic formula is fine, if clunky.
New or not, If it demystifies the concept clearly and helps us understand how a system works and how their concepts interact with each-other it's a win in my book. ...Actually In a sense it is new: not because it's an invention but rather an new way of explanation I guess.
If m not wrong then there's this mistake i found, at 17:25 it's shown the formula, -b/2 +,- z ; it should be (-b +,- z)/2 because if you don't do that, uve to take lcm, which makes the ans wrong
No,it's correct in the video as we have to take (avg+z)&(avg-z) where avg is b/2 not (b+z)/2..by taking (b+,-z)/2 it would make the avg. of b&z..check point 2 at 17:25
Perhaps this is a new method in the US, but this is how we were taught to solve quadratic equations in South Africa over two decades ago. The only thing that's new to me here is the Quadratic Formula shown at 2:03, which we were never taught & it's obvious why... it's far more complicated & more difficult to remember.
@Quasar Supernova the method works with complex roots, and it is geometric. the idea is the same. rather than having two roots that are symmetric about a real number a, in the form a+b, a-b where both a and b are real, you have the two roots are complex conjugates, ie. c+di, c-di, which are also geometrically symmetric about c in the complex plane. the method works perfectly for complex numbers too: Suppose x^2-ax+b = (x-p)(x-q) = 0. Then we have p+q=a, pq=b. We know that p,q are symmetric about a either across a vertical (imaginary) line or the real axis. So we can write p = a/2 + c, q = a/2 - c. (here c can be nonreal). Then we do the same thing, and plug this into pq = b to get (a/2 + c)(a/2 - c) = a^2/4 - c^2 = b, so c = +- sqrt(a^2/4 - b). Then you plug back in to find p and q. For example: x^2-4x+5 = (x-p)(x-q) = 0. So, p+q = 4, pq = 5. Thus, p = 2+c, q = 2-c, and pq = (2+c)(2-c) = 4-c^2=5, so c^2 = -1, meaning c = +- i. Plugging back in gives p and q are 2+i, 2-i.
This is quite a good exposition of the technique. You also make the video fun and easy to listen to. You should consider teaching math more generally on RUclips.
Don't show it to students though. He's not consistent, saying something is the average when it's actually the sum at one point and repeatedly saying equation instead of formula.
This is a very well known formula. If you practice quadratics equations a lot , you would obviously recognise this pattern. This is nothing new. I guess nobody bothered to write an actual proof so the guy takes all the credit
This method isn't better at all, because if you'll try to write at as a formula you'll get normal quadratic equation formuła, there is no difference, seriously, formuła for roots of quadratic equation is so intuitive, when it is well explained, just see the proof, with geometrical illustration and you'll know, how simple it is.
I believe the key to learning this math in general lies in understanding the relationship between the numbers and the lines and the points of a line-being able to better see the graph with your minds eye so to speak. Often, too little emphasis is placed on concepts while too much emphasis is placed on finding the answer to problems, i.e. what is the answer to this one, x^2 + x + 1 = 0?, now what about this one, x^2 + 2x + 3 = 0?, now here are ten more problems for your homework. The advantage of Dr. Loh's discovery is that it removes the need to recall the quadratic equation, used by high school students studying algebra. The discovery employs principles of geometry, such as that every segment has a midpoint, which frees the student to focus more on basic principles and concepts and the relationship between numbers and lines and the points of a line, rather than the quadratic formula itself. Lets look at some of the underlying assumptions. As for the assumption that every segment has a midpoint, that every segment can be bisected was Euclid’s Proposition 10. However, note that Euclid made a tacit assumption in his proof of Proposition 10, which means that there is an overlooked assumption. It is one that caused Wikipedia to wrongly state that Euclidean and non-Euclidean geometry share as many as 28 of Euclid’s elementary geometry propositions, when the number is fewer than 10. See the Facebook Note, Wikipedia Contradicted by Euclid's Proposition 10, Youngsters with Ruler and Compass, facebook.com/notes/reid-barnes/wikipedia-contradicted-by-euclids-proposition-10-youngsters-with-ruler-and-compa/577085739010671/. Also assumed is the basics of the coordinate system. Along these lines, the following is from the article by Caroline Delbert: "Since a line crosses just once through any particular latitude or longitude, its solution is just one value." This statement depends on Hilbert's Axiom I. 2, that two such lines cannot share the same pair of points. When David Hilbert added a coordinate line, the line with the features to comprise a number line, to Euclid’s geometry, the very earliest axioms required subtle modifications. From Euclid's to draw a line from one point to any other, and extend it in a straight line, Hilbert first produced, two points determine a line and added, they determine it completely. But this eventually became every pair of points is in some line (Axiom I. 1) and two different lines cannot contain the same pair of points (Axiom I. 2, paraphrased). This 'line' is what became a coordinate line. The term "line" in Axiom I. 2 is an elementary term, which means it has no definition that is used in a proof. Non-Euclidean geometry depends on the stipulation that its term for "line" is an elementary term and therefore has no definition that is used in a proof. So this opens the door to interpreting the meaning of what is meant by the elementary terms, "line" or "plane," and then applying the logic of the geometry axioms. One type of non-Euclidean geometry says, there are no parallel lines. Well, if the “lines” are the great circles on the surface of a sphere, and the surface is their “plane,” then there are no parallel “lines” because great circles on the same sphere always intersect. (Parallel lines are defined as “lines” in the same “plane” that do not intersect.) Euclidean geometry says, through a point not on a “line” there is only one parallel to the line. When you interpret the “line” as a straight line, this seems right. So given an undefined line, the Euclidean geometry and non-Euclidean were seen as both logically consistent (just not logically consistent with each other). But what has been forgotten is that the non-Euclidean geometry with no parallels (called Riemannian geometry) is not logically compatible with the elementary axioms necessary for including coordinates in the geometry, such as Hilbert's Axiom I. 2. Given this inclusion, the non-Euclidean geometry then becomes self-contradicting because you can prove there are parallel lines, which contradicts the assumption that there are no parallel lines. This is described in a brief Facebook Note: Self-Contradicting Non-Euclidean Geometry, facebook.com/notes/reid-barnes/self-contradicting-non-euclidean-geometry/766736476712262/.
What!?!? People find the Quadratic Formula hard to remember? And this would be easier? I don't think so. Consider roots to be A and B A+B =-b/a AB = c/a Solve this and you will arrive at the Quadratic formula. Your explanation and this method would not work fast for complex roots. And the Quadratic formula is much easier and a more robust way.
Does this still apply to equations where the a coefficient ≠ 1? Edit: Read the paper, from what i understand you bring the a coefficient down to 1 by either multiplying/dividing/adding/subtracting it from the rest of the equation
This feels like I've done this and things similar to this but still product of roots thing and after that was brilliant and it may not be new but haven't seen this in any book either
Are you kidding me? This method is super old, is incredible that is being published as “a new disruptive method”. What is next? 2+2 is equal to 2*2? Wooow
@@SaurabhSingh-hk6qj Quadratic formula works for every quadratic equation. But this one, only for small and easy numbers. So I guess you could say that this method is a subset of the quadratic formula.
And he is just using the quadratic formula anyway! The difference of two squares bit is basically just the discriminant! Halving b is what you do when completing the square as well - you find the max/min which is always on the line of symmetry, directly between the roots! It's just been manipulated a bit to relate it to the graph but it won't be good for massive numbers or complex roots. I'm a mathS teacher and this is more confusing to students than the formula or completing the square.
Why on earth is every body excited about this "new method"? It is totally equivalent to completing the square or in other words, pq-formula. Somekind of mathematical clickbait.
Please tell me this is a joke. I cannot buy that a university prof. is wasting time restating a known formula by breaking down into two. Basically you are assuming that a is 1 so the quadratic formula is left to -b/2 (+-) root(b^2-4c)/2. And that is EXACTLY what the "new" method is... calling the first part (-b/2) as average, and the other part z! You got to be kidding me.!! BTW, the correct average is -b/2a but you got away with -b/2 by assuming a as 1, which makes the whole thing even more pathetic.
All you geniuses who came up with the idea yourselves but never wrote it up should now turn your attention to quantum entanglement. All you geniuses who say this is the same as the quadratic formula should realize you can also compute the answer with perturbation theory but you're not going to teach that to your high school algebra class.
Hmmm, I'd rather go into algebraic topology since I think it's a more interesting field than quantum entanglement. I am no genius, but you don't need a genius to see that this neither new nor groundbreaking. The "scientific news" has claimed that this worldly genius has found a better method of solving equations that it's gonna make everything change. Seriously? If he actually said that this is a new teaching method and that it's approach is only to make people in high school understand easier, then I'd be fine with it. But he claimed that it has slipped the minds of people for quite a long time Yada Yada Yada...
Way too much intro and unneeded explanation. Should be a 6-7 minutes ... max. Too many math teachers confuse or turn off students by over explaining concepts as they introduce them. They will figure out the ancillary information as they continue they progression in mathematics.
Is there a mathematical way to find the mean of the roots, or the speaker is using visual que on the graph to find it? It it is the latter then one could very well find the roots visually, which may not be precise.
Hi. If suppose x1 and x2 are the roots, then by mutliplying out a(x-x1)(x-x2) and equating it to ax^2 + bx + c, you can find that the sum of the two roots equals -b/a. Therefore the average of the roots (x1+x2)/2 = -b/(2a)
There's NOTHING new here. This method is EXACTLY the well-known derivation of the quadratic formula, just done numerically. The derivation proceeds by assuming WLOG that the two roots are of the form r = u ± v, multiplying out, and equating coefficients. (x-u-v)(x-u+v) = x^2 + (-u-v)x + (-u+v)x + (u-v)(u+v) = x^2 -2ux + u^2 - v^2 = x^2 + bx + c => u = -b/2 ; u^2-v^2 = c => v^2 = u^2-c Taking sqrt for v yields quadratic formula for a=1. THERE IS NOTHING NEW HERE!
It is a different approach, of course it is not a new formula because there cannot exists a 2 formula for 1 same problem so it MUST be derived from same formula
Lol... I learned this method like 18years ago while on middle school. Its called sum-product solve for quadratic equations. This method its usable only when the roots are easy to find and are integers. Also a = 1. If you think that finding roots of quadratic equations are hard its better recheck your career objectives..
Salut Bong c est Kim comment sais-tu si ils existent des racines réelles ? supposons que P(x)=x*x+6x+100 le minimum est atteint à x0 avec P'(x0)=0 c-a-d x=-3 or P(-3) > 0 donc pour tout x P > 0 mais tu peux penser travailler dans C et chercher des solutions complexes avec cette méthode..
I taught this lesson to two of my high school classes TODAY: ruclips.net/video/Db-8OAz9pYM/видео.html I was very pleased with the comprehension and my students' confidence to now be able to solve ANY quadratic (even those with imaginary or irrational solutions). I was also inspired by Po-Shen Loh's method, but I decided to have my students briefly investigate their factoring solutions (and look at the visual connections) to bridge the gap to the new method. Thanks!
Was he kidding? I mean, we’ve been using it in parabola for ages! Maybe it is considered as new. Because i’ve never seen a formula like this in any textbook. Rather than having formula, we’ve always had steps. First step draw the parabola, second get the average and goes on. Maybe because he made all these steps to fit in just one formula, it is considered as new? Am i right? Am o wrong? I’m geninunely asking.
It's definitely not new and basically breaks the quadratic formula into small steps which for some students is more to remember. Also, finding the point in the middle is exactly the same as what completing the square does anyway.
There is nothing new about this method. As the author recognises, the method is at least hundreds of years old. Furthermore, after trying it on same examples, you will see why it never caught up. There's nothing here that can't be solved quicker with other current methods.
What I have said still stands for solving easier. The video gives the impression that the only other option I have to solve the examples shown is to use the quadratic formula, when in fact I don't need it at all. I would call this long forgotten method as an early primitive form of the "completing the square" method.
Ricardo Reis I would say it’s a less confusing approach to completing the square; which was never my preferred or go to method. I really like Loh’s Method. I will continue my practicing, and coding in C++, FORTRAN, PASCAL, and my TI-nspire CAS CX.
@@robwin0072 It really doesn't make any difference. It's the same quadratic formula rearranged into multiple steps. I guess writing it out into more number of steps will actually take more computer time. A simple quadratic formula would've been very efficient in any language
The "ingenious" new method in a german song (from 2013): ruclips.net/video/tRblwTsX6hQ/видео.html btw: The paper is about a simple proof of the quadratic formula and not about a better method to solve quadratic equations.
when you use classic method, you get something like (x+something1(x+something2)=0, where something1+something2 equals B But for (x+something1(x+something2) to = 0, either (x+something1) or (x+something2) = 0 for them to equal 0, x has to equal -something1 and -something2 so, the sum of the roots is (-something1)+(-something2), factor out -1, -(something1+something2), since something1+something2=b, the sum of the roots is -B. is your case, B=6, therefore the sum is -6
The reason is because the "roots" add up to -6. When you solve this: x^2 + 6x + 5 = 0 the factors are (x + 1)(x + 5) and the factors add up to +6, but the "roots" add up to -6. The roots are x = -1 and x = -5. The solution being demonstrated in NOT finding the factors. It is finding the roots. I hope my answer if helpful.
x²+6x+10=0 We know that x²+6x+9=(x+3)² So x²+6x+10=x²+6x+9+1=(x+3)²+1=0 (x+3)²=-1 x=-3±i Much easier method and you're able to find even complex roots and irrational roots
good luck explaining to the young students where -b/2a comes from, need to teach derivation or other ways to find the mid/turning point of the quadratic equation curve.
For factoring into (x+A)(x+B), we know the sum (of the factoring constants A, B) is b. So the average is b/2a. And the roots are the opposite of the factoring constants (roots are -A and -B), so their average is -b/2a. No calculus.
@@vudomath I guess young children learn to stop asking why and accept the explanation. Well, i'll be the evil guy who keeps asking why. Now the question would be average of what? avg(A,B)=(A+B)/2 =b/2, a=1. well, why is the average the midpoint. This explanation already assumes the student already accepts that the midpoint is smack middle between A and B.
For all the assholes critics, please go read Dr. Loh's paper. These are not new concepts, but a new way of proving them. From my understanding, it will be useful in other number systems like the complex numbers. "Perhaps the reason is because it is actually mathematically nontrivial to make the reverse implication: that x^2 + Bx + C = 0 always has two roots (counting multiplicity), and that those roots have sum −B and product C. Early mathematicians did not know how to reason with a full (algebraically closed) system of numbers. Indeed, al-Khwarizmi did not even use negative numbers, nor did Vi`ete, not to mention the complex numbers that might arise in general.'
I agree completely with your assessment of Professor Loh's method. This should be adopted by every high school math teacher immediately, and that is not a guess.
To be fair, that method isn’t even new. That method has been around for 4000 years to solve for the roots in quadratics, no wonder why this method is so forgotten over time w/ teachers not teaching students this method. I should tell certain algebra teachers that don’t know this method on how to do that method, so they can teach the students how to do that kind of method instead of always the quadratic formula & completing the square. I did that method today, & it was fun to use that method instead of always QF, & completing the square. There are not 4 methods, there are 5 methods to solve quadratics. There were always 5 methods to solve quadratics solutions for many years. 1 - Square root (If no “bx” term) 2 - Factor 3 - Complete the square 4 - Quadratic formula 5 - Sum & products of roots There may be more methods, but all I know is that there are 5 methods to solve quadratics functions
3Blue1Brown did this method on his series Quarantine Math two months ago, and then now some professor claims to have “discovered” a new method? I call bullshit!
If you follow this method using the generic ax^2+bx+c=0, you end up deriving the formula for the general solution. Obvious maybe, but fascinating nevertheless.
If you pay attention this method is basically completing the square and you could use it as a a proof for the quadratic equation. But it is a good explanation of why completing the square actually works. I like that it incorporates difference of squares.
Everyone is saying there is nothing new...Of course there is nothing ..but the algorithm he has used is simple! You know..what ever method you use to solve any question...ultimately all of them are related...so of course use any method to solve equation ..it will ultimately give the "quadratic formula" result..because that's what we are supposed to find!!!!
Correct.
I recall my teacher using this to let us solve quadratic equations in a more guided way given that the quadratic formula was supposed to be taught a year after. He just didn't generalize it to solving any such equations.
I don't think the quadratic formula is that complicated, though, and as other pointed out, it's necessary anyway when solving for complex numbers.
Thanks for sharing your impression of this method!
Thanks for stopping by! And thanks for figuring this out!
U should do maths videos to help out students like us
PRO Shen we solve with this methods last 5 years its many many methods u can used in algebra
i hope every one work in this methods to make the math easy
Too long? Here are some jump spots
0:00 Me being excited
1:48 Start here if you want all the math
4:50 Start here if you're in Algebra
and want to see the full explanation including background Algebra
7:12 Start here if you're past Algebra, still remember it, and just want to see the new method
16:17 TL;DR Start here if you're good at math and can get it just from a summary
Thank you, this guy takes a million years to get to the point
@@Laaaaaaurenism You just responded to "that guy that takes a million years to get to the point " 😂😂
@@hunterlivie7988 xD
@@Laaaaaaurenism no wonder his students are afraid of quadratic formula. They probably fell asleep by the time he got to it.
I mean.. it's literally just the quadratic formula. -b/2 plus or minus that z term which us (b/2)^2-c
a was set to 1 anyway and sqrt(x)/2 is (sqrt(x/4)) as everyone knows.
this is l i t e r a l l y the exact same calculation as the quadratic formula for a=1, which always reduces to:
-b/2 ± sqrt(b^2/4-c)
Old wine in new bottle, I was aware of this method since I was in class 9th. Days have come where miniscule discoveries like these are hyped so big
Just kidding with the viewers , just manipulated the quadretic formula 😂😂LOL
This is so amazing! Now I can finally conquer those quadratics on my SAT without spending 5 minutes on one question!
I had been using this graphical approach in my high school exams, simply because I was good with calculus based solving. But this method is the same in terms of computers if not more inefficient. This approach would take a computer more time to solve than just using the quadratic formula
It's actually way more effective.
For x^2+6x+7=0 we can immediately write 9-z^2=7 and it's almost done.
Thanks.
Pretty much haha
Always to the tune of "Pop goes the weasel"! x equals negative b, plus or minus the square root, b squared minus 4ac, ALL over 2a!
It is just finding D=b^2-4ac the different way, I did this several times to solve some specific type of questions. Lol
I this x=-b+-√b^2-4ac/2a is the easier way
@@diptirindani8190 It depends what questions you get, sometimes the derivation is easier
Thank you sir , you method is far more simple and better than the convectional one , thanks again
I see a lot of criticism on how this is not necessarily "new" or that it involves more steps than the quadratic formula.
I agree, but not so harshly. I think it's one more intuitive way to think about solving quadratic equations.. but also think it may be most helpful to those who already intuitively understand it.
For other students, quadratic formula is fine, if clunky.
thumb ups for ya
New or not, If it demystifies the concept clearly and helps us understand how a system works and how their concepts interact with each-other it's a win in my book.
...Actually In a sense it is new: not because it's an invention but rather an new way of explanation I guess.
If m not wrong then there's this mistake i found, at 17:25 it's shown the formula, -b/2 +,- z ; it should be (-b +,- z)/2 because if you don't do that, uve to take lcm, which makes the ans wrong
No,it's correct in the video as we have to take (avg+z)&(avg-z) where avg is b/2 not (b+z)/2..by taking (b+,-z)/2 it would make the avg. of b&z..check point 2 at 17:25
Perhaps this is a new method in the US, but this is how we were taught to solve quadratic equations in South Africa over two decades ago. The only thing that's new to me here is the Quadratic Formula shown at 2:03, which we were never taught & it's obvious why... it's far more complicated & more difficult to remember.
This is one of those "I should have thought of that!" moments.
Wake me up when you are able to find complex roots in a geometric way.
@twistedblktrekie I prefer this way ruclips.net/video/d1YBv2mWll0/видео.html
@Quasar Supernova the method works with complex roots, and it is geometric.
the idea is the same. rather than having two roots that are symmetric about a real number a, in the form a+b, a-b where both a and b are real, you have the two roots are complex conjugates, ie. c+di, c-di, which are also geometrically symmetric about c in the complex plane.
the method works perfectly for complex numbers too:
Suppose x^2-ax+b = (x-p)(x-q) = 0. Then we have p+q=a, pq=b. We know that p,q are symmetric about a either across a vertical (imaginary) line or the real axis. So we can write p = a/2 + c, q = a/2 - c. (here c can be nonreal). Then we do the same thing, and plug this into pq = b to get (a/2 + c)(a/2 - c) = a^2/4 - c^2 = b, so c = +- sqrt(a^2/4 - b). Then you plug back in to find p and q.
For example: x^2-4x+5 = (x-p)(x-q) = 0. So, p+q = 4, pq = 5. Thus, p = 2+c, q = 2-c, and pq = (2+c)(2-c) = 4-c^2=5, so c^2 = -1,
meaning c = +- i. Plugging back in gives p and q are 2+i, 2-i.
Excellent explanation!!! Tks for sharing a video explaining Professor Loh!
This is quite a good exposition of the technique. You also make the video fun and easy to listen to. You should consider teaching math more generally on RUclips.
Don't show it to students though. He's not consistent, saying something is the average when it's actually the sum at one point and repeatedly saying equation instead of formula.
This thing really needed 19 minute video!!
xD
The reason why this method is "better" is that it's intuitive, whereas the quadratic formula is rote memorization with very little intuition.
This is a very well known formula. If you practice quadratics equations a lot , you would obviously recognise this pattern. This is nothing new. I guess nobody bothered to write an actual proof so the guy takes all the credit
This method isn't better at all, because if you'll try to write at as a formula you'll get normal quadratic equation formuła, there is no difference, seriously, formuła for roots of quadratic equation is so intuitive, when it is well explained, just see the proof, with geometrical illustration and you'll know, how simple it is.
I believe the key to learning this math in general lies in understanding the relationship between the numbers and the lines and the points of a line-being able to better see the graph with your minds eye so to speak. Often, too little emphasis is placed on concepts while too much emphasis is placed on finding the answer to problems, i.e. what is the answer to this one, x^2 + x + 1 = 0?, now what about this one, x^2 + 2x + 3 = 0?, now here are ten more problems for your homework.
The advantage of Dr. Loh's discovery is that it removes the need to recall the quadratic equation, used by high school students studying algebra. The discovery employs principles of geometry, such as that every segment has a midpoint, which frees the student to focus more on basic principles and concepts and the relationship between numbers and lines and the points of a line, rather than the quadratic formula itself.
Lets look at some of the underlying assumptions. As for the assumption that every segment has a midpoint, that every segment can be bisected was Euclid’s Proposition 10. However, note that Euclid made a tacit assumption in his proof of Proposition 10, which means that there is an overlooked assumption. It is one that caused Wikipedia to wrongly state that Euclidean and non-Euclidean geometry share as many as 28 of Euclid’s elementary geometry propositions, when the number is fewer than 10. See the Facebook Note, Wikipedia Contradicted by Euclid's Proposition 10, Youngsters with Ruler and Compass, facebook.com/notes/reid-barnes/wikipedia-contradicted-by-euclids-proposition-10-youngsters-with-ruler-and-compa/577085739010671/.
Also assumed is the basics of the coordinate system. Along these lines, the following is from the article by Caroline Delbert: "Since a line crosses just once through any particular latitude or longitude, its solution is just one value."
This statement depends on Hilbert's Axiom I. 2, that two such lines cannot share the same pair of points.
When David Hilbert added a coordinate line, the line with the features to comprise a number line, to Euclid’s geometry, the very earliest axioms required subtle modifications. From Euclid's to draw a line from one point to any other, and extend it in a straight line, Hilbert first produced, two points determine a line and added, they determine it completely. But this eventually became every pair of points is in some line (Axiom I. 1) and two different lines cannot contain the same pair of points (Axiom I. 2, paraphrased). This 'line' is what became a coordinate line.
The term "line" in Axiom I. 2 is an elementary term, which means it has no definition that is used in a proof. Non-Euclidean geometry depends on the stipulation that its term for "line" is an elementary term and therefore has no definition that is used in a proof. So this opens the door to interpreting the meaning of what is meant by the elementary terms, "line" or "plane," and then applying the logic of the geometry axioms.
One type of non-Euclidean geometry says, there are no parallel lines. Well, if the “lines” are the great circles on the surface of a sphere, and the surface is their “plane,” then there are no parallel “lines” because great circles on the same sphere always intersect. (Parallel lines are defined as “lines” in the same “plane” that do not intersect.) Euclidean geometry says, through a point not on a “line” there is only one parallel to the line. When you interpret the “line” as a straight line, this seems right.
So given an undefined line, the Euclidean geometry and non-Euclidean were seen as both logically consistent (just not logically consistent with each other). But what has been forgotten is that the non-Euclidean geometry with no parallels (called Riemannian geometry) is not logically compatible with the elementary axioms necessary for including coordinates in the geometry, such as Hilbert's Axiom I. 2.
Given this inclusion, the non-Euclidean geometry then becomes self-contradicting because you can prove there are parallel lines, which contradicts the assumption that there are no parallel lines. This is described in a brief Facebook Note: Self-Contradicting Non-Euclidean Geometry, facebook.com/notes/reid-barnes/self-contradicting-non-euclidean-geometry/766736476712262/.
Literally It is a god gift to me
Now I do have to do that guessing work between my exam.
Woohoo!!! Sharing with everyone I know!
What!?!?
People find the Quadratic Formula hard to remember?
And this would be easier?
I don't think so.
Consider roots to be A and B
A+B =-b/a
AB = c/a
Solve this and you will arrive at the Quadratic formula.
Your explanation and this method would not work fast for complex roots.
And the Quadratic formula is much easier and a more robust way.
All of us math people find the quadratic formula really simple until we start teaching it to 30-40 students who can't figure out how to use it.
My mans just shown the derivation of the quadratic formula and is acting like he’s had a breakthrough lmaooooooo
Does this still apply to equations where the a coefficient ≠ 1?
Edit: Read the paper, from what i understand you bring the a coefficient down to 1 by either multiplying/dividing/adding/subtracting it from the rest of the equation
Yes. Just divide both sides by a first.
Bruh our teacher has been teaching this to us and her ex students from at least 10 years
This was discovered long back
This was discovered about 1400 years ago to be exact
This feels like I've done this and things similar to this but still product of roots thing and after that was brilliant and it may not be new but haven't seen this in any book either
In india we have been using this trick since our childhood. I dont fjnd anything special about it
In Russia we learn this method when we are fetuses. I was laughing when this so-called professor reinvent "new" method.
Yeah man I used to thought chinese were geniuses but after this oh man😱
I still have my notes from high school where I figured this out if you want to see them
Totally cool
Are you kidding me? This method is super old, is incredible that is being published as “a new disruptive method”.
What is next? 2+2 is equal to 2*2? Wooow
Ikr?!?! I thought it'd be something easy...omfg!!
Yeah, also this method confuses me sometimes, quadratic formula is pretty straightforward
@@SaurabhSingh-hk6qj Quadratic formula works for every quadratic equation. But this one, only for small and easy numbers. So I guess you could say that this method is a subset of the quadratic formula.
Someone needs to tell this man that he's late to the party. Waay too late.
And he is just using the quadratic formula anyway! The difference of two squares bit is basically just the discriminant! Halving b is what you do when completing the square as well - you find the max/min which is always on the line of symmetry, directly between the roots!
It's just been manipulated a bit to relate it to the graph but it won't be good for massive numbers or complex roots. I'm a mathS teacher and this is more confusing to students than the formula or completing the square.
Why on earth is every body excited about this "new method"?
It is totally equivalent to completing the square or in other words, pq-formula.
Somekind of mathematical clickbait.
*Thanks for the method* 💙 💖 💙
I always struggled with physics problems that had quadratic equation for time, image distance etc
Great explanation
Great video. Glad it blew up ;)
Even though you can't visualise the answer on a graph, this method works with complex roots.
Please tell me this is a joke. I cannot buy that a university prof. is wasting time restating a known formula by breaking down into two. Basically you are assuming that a is 1 so the quadratic formula is left to -b/2 (+-) root(b^2-4c)/2. And that is EXACTLY what the "new" method is... calling the first part (-b/2) as average, and the other part z! You got to be kidding me.!! BTW, the correct average is -b/2a but you got away with -b/2 by assuming a as 1, which makes the whole thing even more pathetic.
Thank you very much for your video. Thanks to it now I know somenthing new, it was very helpful for me.
Cheers man!!!
Will it still work when "a" is other than 1.
yeah but you would have to take c/a instead of c and -b/2a instead of -b/2
Literally just divide by 'a', redefine what b and c are, and do it.
@@dogukanbas8402 and a≠1 makes this method worse than quadratic formula
@@SaurabhSingh-hk6qj nah I think the opposite
All you geniuses who came up with the idea yourselves but never wrote it up should now turn your attention to quantum entanglement. All you geniuses who say this is the same as the quadratic formula should realize you can also compute the answer with perturbation theory but you're not going to teach that to your high school algebra class.
This guy = HUGE PP
Hmmm, I'd rather go into algebraic topology since I think it's a more interesting field than quantum entanglement. I am no genius, but you don't need a genius to see that this neither new nor groundbreaking. The "scientific news" has claimed that this worldly genius has found a better method of solving equations that it's gonna make everything change. Seriously? If he actually said that this is a new teaching method and that it's approach is only to make people in high school understand easier, then I'd be fine with it. But he claimed that it has slipped the minds of people for quite a long time Yada Yada Yada...
Vou escrever em português mesmo pois não sou obrigado a nada.
RELAÇÕES DE GIRARD
Isso aí não é mesma cosia que soma e produto? kkkkkkkkkkkkk
@@matheusvitti3330 sim, e tudo isso é derivado de que toda equação do 2°Grau pode ser escrita como a(x-r1)(x-r2)
Why the average is negative 3 not positive 3?
6:28
HaoNhien Vu b in the equation is +6 though. So 6/2 = 3, not - 6/2 = -3.
@@georgecaplin9075 Yes, but the roots add to -b not b
Way too much intro and unneeded explanation. Should be a 6-7 minutes ... max. Too many math teachers confuse or turn off students by over explaining concepts as they introduce them. They will figure out the ancillary information as they continue they progression in mathematics.
This idea is very old it is the basic understanding of quadratic equation. It's not at all new.
This helped so much, thanks!
I have always used this method, I figured everyone did. This is not new at all
We used this method in school 10 years ago, so not sure why it's branded as being new
And if we pay attention its just another form of the formula used....
Boy this is what we are doing in high school now
Exactly
@@RiceMan31 aynen öyle canım çıkıyor şu an bu konudan
bir de profesör yeni keşfetmiş te Allahım ya
Why can't we take (average + z)+(average - z)=6 ???
Also because the z's would cancel out and you still don't know what z is.
I like it but when it comes to find the roots I prefer the quadratic formula over this
Screw haters... I'm teaching this as of tomorrow
But obviously a new approach to quadratic
Which makes it exciting at first sight
Is there a mathematical way to find the mean of the roots, or the speaker is using visual que on the graph to find it? It it is the latter then one could very well find the roots visually, which may not be precise.
Hi. If suppose x1 and x2 are the roots, then by mutliplying out a(x-x1)(x-x2) and equating it to ax^2 + bx + c, you can find that the sum of the two roots equals -b/a. Therefore the average of the roots (x1+x2)/2 = -b/(2a)
There's NOTHING new here. This method is EXACTLY the well-known derivation of the quadratic formula, just done numerically.
The derivation proceeds by assuming WLOG that the two roots are of the form r = u ± v, multiplying out, and equating coefficients.
(x-u-v)(x-u+v) = x^2 + (-u-v)x + (-u+v)x + (u-v)(u+v)
= x^2 -2ux + u^2 - v^2 = x^2 + bx + c
=> u = -b/2 ; u^2-v^2 = c => v^2 = u^2-c
Taking sqrt for v yields quadratic formula for a=1.
THERE IS NOTHING NEW HERE!
It is a different approach, of course it is not a new formula because there cannot exists a 2 formula for 1 same problem so it MUST be derived from same formula
Please can you solve more complex questions ...
I have more here: ruclips.net/video/JlmOINYW0ZM/видео.html
Lol... I learned this method like 18years ago while on middle school. Its called sum-product solve for quadratic equations. This method its usable only when the roots are easy to find and are integers. Also a = 1.
If you think that finding roots of quadratic equations are hard its better recheck your career objectives..
To read Professor Po-Shen Loh's paper: arxiv.org/abs/1910.06709v1
Salut Bong c est Kim comment sais-tu si ils existent des racines réelles ? supposons que P(x)=x*x+6x+100 le minimum est atteint à x0 avec P'(x0)=0 c-a-d x=-3 or P(-3) > 0 donc pour tout x P > 0 mais tu peux penser travailler dans C et chercher des solutions complexes avec cette méthode..
282K views with mathematics that is great
Salut Kim. Le methode marche avec les racines complexes et non-reelles aussi. J'ai un autre video pour ca.
ruclips.net/video/JlmOINYW0ZM/видео.html
What if root's are unreal then how to solbe
I taught this lesson to two of my high school classes TODAY: ruclips.net/video/Db-8OAz9pYM/видео.html I was very pleased with the comprehension and my students' confidence to now be able to solve ANY quadratic (even those with imaginary or irrational solutions). I was also inspired by Po-Shen Loh's method, but I decided to have my students briefly investigate their factoring solutions (and look at the visual connections) to bridge the gap to the new method. Thanks!
Thanks for the feedback.
Was he kidding? I mean, we’ve been using it in parabola for ages! Maybe it is considered as new. Because i’ve never seen a formula like this in any textbook. Rather than having formula, we’ve always had steps. First step draw the parabola, second get the average and goes on. Maybe because he made all these steps to fit in just one formula, it is considered as new? Am i right? Am o wrong? I’m geninunely asking.
It's definitely not new and basically breaks the quadratic formula into small steps which for some students is more to remember. Also, finding the point in the middle is exactly the same as what completing the square does anyway.
There is nothing new about this method. As the author recognises, the method is at least hundreds of years old. Furthermore, after trying it on same examples, you will see why it never caught up. There's nothing here that can't be solved quicker with other current methods.
It not about solving quicker. It is about solving fairly easily even if you haven't memorized the quadratic formula.
What I have said still stands for solving easier. The video gives the impression that the only other option I have to solve the examples shown is to use the quadratic formula, when in fact I don't need it at all. I would call this long forgotten method as an early primitive form of the "completing the square" method.
Ricardo Reis I would say it’s a less confusing approach to completing the square; which was never my preferred or go to method. I really like Loh’s Method. I will continue my practicing, and coding in C++, FORTRAN, PASCAL, and my TI-nspire CAS CX.
@@robwin0072 wtf?
@@robwin0072 It really doesn't make any difference. It's the same quadratic formula rearranged into multiple steps. I guess writing it out into more number of steps will actually take more computer time. A simple quadratic formula would've been very efficient in any language
Where did he get -6
The -b since he showed that the sum of the roots is -b
In india have a different approach to solve quadratic equation
U should visit
The "ingenious" new method in a german song (from 2013): ruclips.net/video/tRblwTsX6hQ/видео.html
btw: The paper is about a simple proof of the quadratic formula and not about a better method to solve quadratic equations.
It was always in front of the eyes but we didn't get it to be a new way
Why do you keep saying negative 6 when the term in the middle is a positive 6. They need to add up to positive 6, not negative 6
when you use classic method, you get something like (x+something1(x+something2)=0, where something1+something2 equals B
But for (x+something1(x+something2) to = 0, either (x+something1) or (x+something2) = 0
for them to equal 0, x has to equal -something1 and -something2
so, the sum of the roots is (-something1)+(-something2), factor out -1, -(something1+something2), since something1+something2=b, the sum of the roots is -B.
is your case, B=6, therefore the sum is -6
The reason is because the "roots" add up to -6. When you solve this:
x^2 + 6x + 5 = 0
the factors are (x + 1)(x + 5) and the factors add up to +6, but the "roots" add up to -6. The roots are x = -1 and x = -5.
The solution being demonstrated in NOT finding the factors. It is finding the roots. I hope my answer if helpful.
6:28
So how did you know -3 was the average without drawing a graph?
Vincent Aquila simple. you gotta do -b/2. B was 6 in that question. -b= -6 -b/2= -3
This might be new to you, but we learned also this way in addition to ABC and PQ method.
This shit appeared in Science Alert as a new genius discovery... LOL!!! XD
Kis class Ka h
Wtf man i learned this before I learned about x=-b±√b²-4ac/2a
x²+6x+10=0
We know that x²+6x+9=(x+3)²
So
x²+6x+10=x²+6x+9+1=(x+3)²+1=0
(x+3)²=-1
x=-3±i
Much easier method and you're able to find even complex roots and irrational roots
good luck explaining to the young students where -b/2a comes from, need to teach derivation or other ways to find the mid/turning point of the quadratic equation curve.
For factoring into (x+A)(x+B), we know the sum (of the factoring constants A, B) is b. So the average is b/2a. And the roots are the opposite of the factoring constants (roots are -A and -B), so their average is -b/2a. No calculus.
@@vudomath I guess young children learn to stop asking why and accept the explanation. Well, i'll be the evil guy who keeps asking why. Now the question would be average of what? avg(A,B)=(A+B)/2 =b/2, a=1. well, why is the average the midpoint. This explanation already assumes the student already accepts that the midpoint is smack middle between A and B.
But anyway, great vid tho. Really simplifies the computing part of the math for those who understand why it works.
U done here mistake A^2-B^ ,z is A ans -3 is B
OMG You are amazing
@Coherent Mango
Mind your business
Sir your graph does not show -3 ...kindly check your graph
-3 is the value of the x coordinate, so the graph is correct
Preciso de tradução =(
For all the assholes critics, please go read Dr. Loh's paper. These are not new concepts, but a new way of proving them. From my understanding, it will be useful in other number systems like the complex numbers.
"Perhaps the reason is because it is actually mathematically nontrivial to make the reverse
implication: that x^2 + Bx + C = 0 always has two roots (counting multiplicity), and that those
roots have sum −B and product C. Early mathematicians did not know how to reason with a full
(algebraically closed) system of numbers. Indeed, al-Khwarizmi did not even use negative numbers,
nor did Vi`ete, not to mention the complex numbers that might arise in general.'
It's just one of Vieta formulas or Vieta theorem, whatever you name it. I've been taught it in school
You don't actually compute the roots in Vieta's theorem
Yes it is a little new method I think, but it is more harder than the old way.
This is beautiful.
I agree completely with your assessment of Professor Loh's method. This should be adopted by every high school math teacher immediately, and that is not a guess.
When you substitute the root in the equation you must get 0 but you get 12√22
I hate maths it is so confusing, but I can't resist b/c I still love maths.
That's neat!
if i have a non factorable coefficient in the first term, I find that it is alot easier to use the quadratic formula instead of this method.
Looks like he has rediscovered en.wikipedia.org/wiki/Quadratic_formula#By_Lagrange_resolvents ?
To be fair, that method isn’t even new. That method has been around for 4000 years to solve for the roots in quadratics, no wonder why this method is so forgotten over time w/ teachers not teaching students this method. I should tell certain algebra teachers that don’t know this method on how to do that method, so they can teach the students how to do that kind of method instead of always the quadratic formula & completing the square. I did that method today, & it was fun to use that method instead of always QF, & completing the square. There are not 4 methods, there are 5 methods to solve quadratics. There were always 5 methods to solve quadratics solutions for many years.
1 - Square root (If no “bx” term)
2 - Factor
3 - Complete the square
4 - Quadratic formula
5 - Sum & products of roots
There may be more methods, but all I know is that there are 5 methods to solve quadratics functions
this eq was commonly used for solving complex numbers nothing new
I don’t see it. It’s the same as completing the square.
3Blue1Brown did this method on his series Quarantine Math two months ago, and then now some professor claims to have “discovered” a new method? I call bullshit!
How the hell does it get 1k likes
I mean it's the formula I studied in 9th grade
Same as quadratic formula
It's nothing but another way of adding up elements and proove Sri Dharacharya Niyama
That's literally just a longer roundabout way of the Quadratic Formula. :/ ???????????
If you follow this method using the generic ax^2+bx+c=0, you end up deriving the formula for the general solution. Obvious maybe, but fascinating nevertheless.
Change the name of "Quadratic formula" to " shreedharacharya formula" in your published paper.. I'll help you to get 1M views 😂
This is great, but it doesnt help me solve complex roots any faster. Quadratic equation for the win.
x²+6x+10=0
We know that x²+6x+9=(x+3)²
So
x²+6x+10=x²+6x+9+1=(x+3)²+1=0
(x+3)²=-1
x=3±i
Much easier method and you're able to find even complex roots
Eg. M totally agreed