It was. Do I use it? No. I'm faster with the quadratic formula. This method provides some insight to the attentive student, but no gain in speed over the quadratic formula for real-world numbers
@@wernerviehhauser94I agree the quadratic formula technically works for all these problems and provides even more. The need to guess and check just gets eliminated in all it’s entirety.
My only suggestion is that "c" is used to define two different things: The third "element" of the quadradic equation and the "constant value" added to or subtracted from the second element of the quadratic, b. To help prevent confusion, I would suggest calling the second constant value "k", so we don't have two c's with different meanings.
k= sqrt{(b/2)-(c)}; m = (b/2)+(k); n =(b/2)-(k). Final factors: (x +/- m)(x +/- n). k must be within bracket to take care of sign e.g. x**2 + 5 x - 14 provides k = 4.5, m=2.5+4.5=7, n=2.5-4.5= (-2). Factors are (x+7)(x-2)
This is pretty much just how the quadratic formula works. This is how the quadratic formula should be taught, in my opinion; by starting with this as motivation
It is called the Poh Shen Loh method. A math professor named Poh Shen Loh discovered this method. The math professor himself said he found this method buried in a very very old Math book. Here is the link ua-cam.com/video/XKBX0r3J-9Y/v-deo.htmlsi=LiskkjE6oaHBAPZY
i've heard it called Hindi Method or Sridhara method. We used to learn it as an application of Vieta. Nonetheless, all I use is the quadratic formula - it's faster than everything else for real-world numbers in physics problems.
It’s the ancient Babylonians’ method. They were very advanced in math and astronomy. If NJ Wildberger ever gets his channel restored, you’ll be able to see his demonstration of a real, Babylonian worked problem using this method.
@@glennschexnayder3720 Professor Loh basically condensed the method and made it easier to understand. This is especially true when explained with a graph.
I came here to refresh my memory on factoring so I can begin teaching my girls (8th & 6th grades). I love this method because I hate ever having to say “guess and test”. I always look for methods/process or creat one. This is excellent. I can’t wait to teach it.
@@angelviloria4966I would assume the same. However, just as others advised, please don't skip the formula taught in school. Remember the roots and (-b +- sqrt (b²-4ac)) / 2a
You can also do it in an another way, which in the end is essentially equivalent to the one in the video: first complete the square, x² + 32x + 192 = x² + 32x + 16² - 16² + 192 = (x + 16)² - 64. Then write this as a difference of squares: (x + 16)² - 64 = (x + 16)² - 8² = ( (x + 16) + 8 ) ( (x + 16) - 8 ) = (x + 24) (x + 8).
This is just a manual execution of the abc or pq formula (there are many names for it). Still impressive. For simplification I use a = 1 => pq formula for x² + px + q = 0 => x_{1/2} = - p/2 +/- sqrt( (p/2)² - q ) (x - x_1)(x - x_2) = 0 => m = -x_1 and n = -x_2. That's p/2 -/+ sqrt( (p/2)² - q ).
In the thumbnail it's clearly written a new amazing "factoring method". That is to find the factors..... Not a new method of finding the value of x in the quadratic equation Which we all know the best is the quadratic formula 😊 hope it explains the confusion
Maths teaches- if something's similar doesn't mean it's congruent. If something approaches a place doesn't mean it's at the place- Calculus If this seems like the quadratic formula it doesn't mean it actually is it or completing the square method Clearly visualise: It is a good method of finding the factors while the quadratic formula is for finding the value of x. ❤❤❤love your finding efforts
i've had thoughts of this but never found a way to make them useful to my math classes, well until i found this video i could actually do something with it. THanks!
You can also reverse the quadratic formula so that you don't need to work with numbers of such significant magnitude: x = (-b ± √(b² - 4ac))/2a x = -b/2a ± √(b² - 4ac)/2a x = -b/2a ± √((b² - 4ac)/4a²) x = -b/2a ± √(b²/4a² - c/a) (-b)² = b² x = -b/2a ± √((-b)²/(2a)² - c/a) x = -b/2a ± √((-b/2a)² - c/a) q = -b/2a x = q ± √(q² - c/a) Another nice thing about this variant is the fact that you only need to calculate _q_ once. As an example of these conveniences, consider 8x² + 14x - 15: q = -14/2(8) = -7/8 x = -7/8 ± √((-7/8)² - (-15)/8) x = -7/8 ± √(49/64 + 120/64) x = -7/8 ± √(169/64) x = -7/8 ± 13/8 x = 6/8 , -20/8 x = 3/4 , -5/2 8x² + 14x - 15 = (x - 3/4)(x + 5/2) = (4x-3)(2x+5) = 4x(2x) + 4x(5) + (-3)(2x) + (-3)(5) = 8x² + 20x - 6x - 15 = 8x² + 14x - 15 compared to: x = [-14 ± √((-14)² - 4(8)(-15))]/2(8) x = [-14 ± √(196 + 480)]/16 x = (-14 ± √676)/16 x = [-14 ± √(4•169)]/16 x = [-14 ± 2(13)]/16 x = (-7 ± 13)/8 x = 6/8 , -20/8 x = 3/4 , -5/2 The reason you can multiply things is because you're using the roots of the expression, meaning when you set the expression equal to 0, you are effectively multiplying 0 by whatever factors you choose, which doesn't affect the truth of the equation. While the use of _m_ and _n_ are important to understand the underlying concepts shown in the video, it's not truly necessary once you can use an algorithm like this one, and i definitely prefer this algorithm over the standard quadratic formula.
Okay I'm a little disappointed because the thumbnail said this was new, but for everyone saying to use the quadratic formula instead... that's like adding five three times when asked to do five times three. It might be faster when you first learn, but oh my god is it slow. when I looked at the first problem, my first instinct was that it was eight, and that's not some weird boast like "oh I can do factors slightly faster than you," that's the bare minimum. It's honestly shocking to me that so many people are plugging these numbers into the quadratic formula thinking it's faster. Now that you've watched this video, do this until you have it down perfectly easily. memorizing the quadratic formula won't help you. Learn to factor.
Wonderful video. Since this amounts to using the quadratic formula for writing the factors of the quadratic, you could have just gone for "How to write the factors of a quadratic by using the quadratic formula". And *that* would have been a wonderful video. And nobody would have guessed or checked anything.
I don't know about the other parts of the world, but here in the Czech republic, some 35 years ago, we were not taught to guess, but we learned the standard formula for the roots, to which this "new" approach is equivalent.
This is a fun video, I found the general form for this method by not plugging anything in where x^2+bx+c=(x+((b/2)+sqrt(((b/2)^2)-c))*(x+((b/2)-sqrt(((b/2)^2)-c)) If there is a coefficient on x^2 divide it by every term and put it on the outside of the formula a(x^2+(b/a)x+(c/a)) We can treat b/a as “b” and c/a as “c” for the formula
Very clever! x^2+bx+c=0 x=-b/2(+/-)(b^2-4c)^.5/2 Someone noticed that the roots where b/2 +/- the same number. Then it becomes obvious how you developed your approach. Telling the students your thought processes is the way math was supposed to be taught. It was not intuitively obvious to me until I looked at the quadratic formula.
And this is exactly what is called the p-q formula (at least here in Germany) for solving quadratic equations: x^2 + px + q = 0 x = -p/2 +/- sqrt((p/2)^2 -q). It’s just not displayed as a single formula but broken down into two steps.
This method should go viral, we have to make children's lives easier. My school years would've been so not depressing if this method was around my time.
Most methods were there in books outside of the ones we were told to follow. However, the teachers who could think of explaining how to approach a given problem from different angles weren't around in some respectable number. They followed a particular book and stopped at that. There has been a very asked cha ge in how children learn these days. So many follow good channels on UA-cam as well. No wonder then, the depressing days had to be endured by us.
This is introduced in many math classes, but alas too many parents will claim this is more difficult or "I never learned it that way" and blame this on Common Core.
Isn't the method shown no less effort than factorising ax^2 + bx + c by calculating D = b^2 - 4ac and then writing the factorisation as a.( x + (b + √D)/2a) ).( x + (b - √D)/2a )? That has the advantage of using a formula that's either already familiar, or at least will have further applications later on. If you are worried by big numbers, just calculate d = (b/2)^2 - ac and write the factorisation as a.( x + (b/2 + √d)/a ).( x + (b/2 - √d)/a) which scales thing down by a factor of 4 (like the video does).
Yes. It's just an alternate form for the quadratic formula. This is the famous Po Shen Loh method, and 3B1B has a good video covering it in a little more depth
Watching this ticks me off because not only does the more efficient methid make sense, but the OG method makes more sense than what I was taught. My math education sucked noodles. I took calculous in HS, and was fumbling in the dark the whole time.
In fact I invented this method (or very similar) some decades ago (now retired from high school). But often with much less work. Take x^2 - 6x - 91 = 0. Only integer factoring of 91 is 7 x 13, so I write (x - 7) (x - 13), yes - both with a minus. Now it is easy to see if any - must be made into a + (which is "easier" than the opposite). Since 7 - 13 = -6, we must have + on the 7. Thus: (x + 7) (x - 13).
Nice explanation of Po Shen Loh's method, publish about 5 years ago. Po does say that the steps to doing this method have been know for a very long time, but the combination used here wasn't well recorded elsewhere, so he is standing on the shoulders of others.
Another way to think of this is that the solutions to x^2 + b x + c = 0 are x = - b/2 +/- sqrt( (b/2)^2 - c ) which is just a rewriting of the quadratic formula.
Played this on x2.6 speed and quadratic formula got the answer first with plenty to spare. While this certainly gives a cool perspective that seems to speak to students, we should probably spend more time pondering what to do about the fact that the formula (by definition the easiest way to do something) is so intimidating to students.
My daughter is a high school math teacher. She likes this method, but can't change the approved curriculum. This is why things don't get better in US schools.
I taught HS math, and always found time for additional niceties like this. Doesn't matter if it is approved, although she may want to avoid quizzing students on it if it really is a backward administration. But for the test that wraps up all methods, i.e., solve using any method, students can use this method if they wish.
You can also factor by grouping 2x^2-7x-4 = 2x^2 -8x+1x -4 = (2x^2-8x) + (1x-4) = 2x*(x-4) + (x-4) -> factor (x-4) from each group making -> (2x+1)(x-4)
I double checked this method and it certainly works. It also leads to the standard equation taught in schools sqrt(b^2 - 4ac) etc. I am puzzled for why it works though. Who would have thought in the first place? It is quite magical in its own way. Well it was magical until You explained it so well :-) You are a good teacher. There is always the How and Why. Feynman says shut up and calculate. So faced with these equations just apply the standard formula and sattisfy the 'How.' You have explained The 'Why' however. That is quite meritorious :-) Physics is in the doldrums, brain dead from not thinking and just calculating. Maybe expalin Math to them :--)
To get to the quadratic equation (which you have most of there), is done by completing the square of the general quadratic equation "ax^2+bx+c" and simplifying.
Wish I knew this wayy earlier. Thanks to this video I just realized that you can actually also use the Quadratic Formula for factoring! (by replacing -b to positive)
Oh and you can derive the process/formula used in the video from the Quadratic Formula(where -b is positive) and vice versa, which makes sense since both ultimately gives the same answer
Compelling method. Reminds me of "completing the square" since you're taking half the b value and factoring out an a>1. A "benefit" of using the c value and listing factors is that students can generate a table of y = c/x in their graphing calculators and look for integer coordinate pairs. I love the link to multiplying conjugate pairs in this method though! I'll stick with the AC method for a>1 for now.
Thank you! I have been writing a maths "book" for my teenage son and my father who are both doing their secondary school maths (my father as a hobby project) and neither would say they have a natural feel for this kind of thing. My book is an attempt to provide the clearest explanations for everything they encounter in the curriculum. And, as is the invariable way with maths, when you think you have something down pat, someone like you comes along with a splendid video that makes me think, "why the heck didn't I think of that?". I hope you don't mind if I steal your explanation for my book.
This is fundamentally Poh Shen Loh factoring, derived differently; but you've missed an opportunity to combine with the Australian (i.e. "down under", referring to the placement of a in the denominator) technique for factoring quadratics with a 1. When b is odd, we encounter an unusual circumstance where the decimal fraction 4.5 may be more convenient than the common fraction 9/2, using this shortcut for squaring a half integer: 1.5² = 0.25 + 100 * (1 * 2) = 2.25 2.5² = 0.25 + 100 * (2 * 3) = 6.25 3.5² = 0.25 + 100 * (3 * 4) = 12.25 4.5² = 0.25 + 100 * (4 * 5) = 20.25 5.5² = 0.25 + 100 * (5 * 6) = 30.25 : 9.5² = 0.25 + 100 * (9 *10) = 90.25 : The Australian method for handling a 1 proceeds by initially ignoring that a 1 as, using your example: 3.5² - c² = (2)(-4) = - 8 becomes c² = 12.25 + 8 = 20.25 = 4.5² and then factoring as [ (2x - 3.5 - 4.5) (2x - 3.5 + 4.5) ] / 2 = [ (2x - 8) (2x + 1) ] / 2 = [ (2x - 8) / 2 ] (2x + 1) = (x - 4) (2x + 1). Notice how the a value of 2 was initially placed in both factors, as well as "down under" in the denominator. When a is composite and must be factored across both final factors, this trick delays having to determine that factoring and distribution until a final step, not interfering with the rest. The factoring of a reduces to identifying the common factor in each numerator factor - much simpler than previous methods.
In this question 15:37 2x²-7x-4=0 it will written as *X²-7x-8* =0 after finding Root (x-8)(x+1)=0 then divide 2 both side (x-4)(x+1/2)=0 This also easy by their method *Maine isse aur aasan bna diya*
Hi there 🎉good good job If you in put as a(x-m)(x-n)=0 and your factor form as ax^2+bx+c=0 and find the ROOTS as m&n then plug into first equation a(x-m)(x-n)=0 it's much more easier 😊
Good method, and well presented. (The rest of this is not for you, but more for the existence of this method in the first place) Ridiculous though, factoring is not a difficult process, and this method is not only completely unnecessary but much more difficult than just factoring it. I speak to you as a math tutor, and when the kids come to me being confused by yet another factoring method to make things "easier", and I show them how very easy it is to just skip to the end and solve it directly, each one is more shocked than the last at how very easy this all is. And they are truly shocked when I show them how easy it is when a doesn't equal 1. I mean... look at what happened with x^2 + 9x + 20, fractions were introduced to what should have been HORRIBLY SIMPLE! Stop babying these kids with all these "factoring methods", they can handle the guessing and checking (not to mention, it is actually the far easier method). *frustrated rant over, I had a day...*
I generalized it with the roots adding up to -b/a and their product being c/a, I did the same thing where one root is essentially the x value for the vertex plus h, the other is the x value for the vertex minus h m = -b/2a + h n = -b/2a - h Their product should be c/a m . n = (-b/2a + h)(-b/2a - h) m . n = b²/4a² - h² b²/4a² - h² = c/a At this point you may have realized that it's building up the main formula to solve for quadratics h² = b²/4a² - c/a We multiply c/a times 4a/4a h² = b²/4a² - 4ac/4a² h² = (b² - 4ac)/4a² h = √(b² - 4ac)/2a Yes, the discriminant shows up here, being b² - 4ac Let's denote the discriminant by using ∆ h = √∆/2a The main formula is x = (-b ± √∆)/2a So if we just replace √∆/2a for h x = -b/2a ± h I should point out that -b/2a is the x value for the vertex in every quadratic, where f(-b/2a) is the y value for this vertex of this same quadratic f(x) This in a way proves that each of the roots is the same distance away from the x value of the vertex, I believe this applies for every polynomial that has an even degree
What i have done for similar problems is right m=b-n then n(b-n)=c so n^2-bn+c=o then use x=(-b±squareroot(b^2-4ac))/(2a) for n^2-bn+c as it would allow me to right ax^2+bx+c=0 in form of (x+n)(x+m)=0
Nice visualization of the solutions of quadratic formula for a=1. And you showed it: all quadratic equations can be rewrite to the pq formula. Your solutions m and n are the negative solutions of this formula x = -b/2 ± sqrt[(-b/2)² - c]. so m/n = b/2 ± sqrt[(b/2)² - c]. I removed the minus under the square, as it is not important, if you square it anyways... but it is your solution, you showing for different b and c.
Why people hear in the comment section are confused between quadratic formula and an factorising method In quadratic formula we find the exact value of 'x' Whereas this is a factorising method where we find the "factors" of a quadratic equation
This method is what I figured out in deriving the quadratic formula which I thought was that's how you actually derive it instead of completing the square. Although this ensures that you get its factors, I don't think it's the method I'm going to use mainly in my factoring, because to factor quickly, a method that allows you to being able to do it mentally or at least minimal writing would be my bet. But this method would require me a lot of thinking, writing, and/or storing numbers in my head. This method actually gave me an idea. instead of just using the quadratic formula, just use the formula derived from this method directly because comparing it to the quadratic formula, it is easier to compute the square root. when a = 1 m = b/2 + √( (b/2)² - c ) n = b/2 - √( (b/2)² - c ) it is essentially the same as the method, but for me using the formula is faster than doing all the work
I have a different method in solving this. I just take the factors of 192, then find the factors that adds to 32, which is the 8 and 24. It might not be applicable for all the problems involving this, but I believe it will solve most of the problem.
Don't factor out the a. Multiply a and c and do this process as if a=1 and c=ac. When you have m and n, fill in (ax+m)(ax+n) and toss out anything common for a and m or a and n, the product of which is a. (It is necessary to factor out a common from the quadratic first.) This is fantastic. I've always hated the standard instructions to just find the numbers that work. I devised a way that uses the factor pairs of AC. It worked well for my students, but I had to admit that it could be time consuming. This is better. Thanks for the excellent video.
Who invented this method? It's great. I'm 76. I got a math degree many years ago. I never worked in math but did use simple math and algebra on occasion. But never saw this method.
So the general case for this is: (b/2 + n), (b/2 -n) And the general case for n is: √((b/2)^2 - C) Putting all that together we get a simple and elegant single equation: b/2 +- √((b/2)^2 - C) Then we just take the negative to get the value of X, and we get: - b/2 +- √((b/2)^2 - C) ... Also known as, the (reduced) quadratic formula
If you look very very very closely at the procedure, Its just the quadratic formula but in step by step method of solving the determinat to get C and adding and subtracting that to b/2a which will give the two roots
x^2 + 32x + 192 = 0 :: average of roots is r. roots are (r+p) and (r-p); sum of roots = 32, product of roots = 192. Sum:: (r+p)+(r-p) = 2r = 32 so r = 16. Product:: (r+p)*(r-p) = 192 r^2 - p^2 = 192 256 - p^2 = 192 p^2 = 256-192 p^2 = 64 so p = √64 = 8 the roots are r ± p or 16 ± 8 or 24 and 8
Haha i remember one day one student I was tutoring bitching about where the quadratic formula came from. So I derived it on the blackboard for him. I would suggest as an exercise (or a future video in this series) that you can do it too. Thanks for this . I like it just as much as the quadratic formula because it's cute, it's easy and it's novel. One thing this could have embellished a bit is the form of the graph and it's roots.
Isn't it just the same as Quadratic Formula? And probably more faster using Quadratic Formula For equation: ax^2 + bx + c = 0; the values of x that satisfy: x = ( -b +- \sqrt{ b^2 - 4ac } )/( 2a ) Tweak a little bit inside the root: \sqrt{ b^2 - 4ac } = \sqrt{ ( ( ( b^2 )/4 ) - ( ( 4ac )/4) ) * 4 } = \sqrt{ ( b/2 )^2 - ac } * \sqrt{ 4 } = 2 * \sqrt{ ( b/2 )^2 - ac } Formula rn: x = ( -b +- 2* \sqrt{ ( b/2 )^2 - ac } )/( 2a ); tweak more become x = ( -b )/( 2a ) +- ( \sqrt{ (b/2)^2 - ac } )/( a ) And if you make it become factor, just multiply by -1 Because: ( x+m ) * ( x+n ) = 0; x+m = 0 or x+n = 0; x =--m or x = -n x^2 + bx + c = 0; ( x + ( ( b/ ( 2a ) ) + ( \sqrt{ (b/2)^2 - ac } )/( a ) ) ) * ( x + ( ( b/ ( 2a ) ) - ( \sqrt{ (b/2)^2 - ac } )/( a ) ) ) *Note: +- mean plus or minus \sqrt{ } mean square root of ...
The presentation in this video is clear and interesting, but the method described is essentially the same as what you do when you solve a quadratic equation by Completing The Square.
It is Indian old method. If m+n=k then both m,n are equal,k/2. If m,n are unequal then one is more than k/2,other is less than k/2 m=k/2+u,n=k/2-u so that sum is k. mn=k^2/4-u^2 u^2=k^2/4 -mn This is reduced form of quadratic formula.
Or this c is all about expressing n and m witj only 1 variable. You could also just say that if m + n = v then n = v - m and boom you got it. Also another thing you could do is if m + n = i, m*n = j, then (m+n )squared is m2 + 2mn a n2 and you know 2mn so you get the value of m2 and n2, then calculate m-n squared, get the root, and boom you got it
Factoring with this rule, which ill write as M, n = b/2 ± sqrt(b²/4 - c) Is smart, but if you want to know x, instead of just factoring to simplify fractions or smth, the quadratic formula is probably still faster to calculate
Thanks for the effort of presenting it this way, but that's literally the general rule of the quadratic equation. Instead of presenting it in a one-step formula: x = (-b+-sqrt(b^2-4*a*c))/(2a), you had to break down into simpler, but more, steps.
Nice video! A good way especially when the 3rd term has many factors But I would suggest another sneaky way: 1- know that you can also find the factors from the roots (i.e. use the quadratic formula & move in the opposite direction) ex.: If the roots are 2 & 1 then the factors are: (x-1)(x-2) 2- multiply the above factors & compare them to the original to make sure you have all the constants Ex.: you found x=1 & x=2 Then: (x-1)(x-2) So you have: x^2-3x+6 But the original expression is: 2(x^2)-6x+12 Comparing to: x^2-3x+6 you need to multiply by 2: 2(x-1)(x-2)
I just looked at the quadratic equation on the thumbnail and (x+24)(x+8). Took me all of about 15 seconds. The last time I factorised a quadratic equation: 1993.
Yes, of course, it is. It's ridiculous to call this a new method, or a method with a certain name attached. This is common knowledge probably since Bronce Age. 😂 Edit: Just using the memorized quadratic formula works even faster for me when doing the calculation in the head.
The problem suggested on top looks kinda ugly so I personally would not even bother checking values, but use one of the following: 1. completing the square. For this it is obvious 2b=32 so b=16 b^2=256 => (x+16)^2+192-256 then a^2-b^2 ftw 2. quadratic formula
Okay, wow. This method should've been taught in schools. Incredible!
Edit: I am actually lost for words. Sounds like glazing but it isn't
It was. Do I use it? No. I'm faster with the quadratic formula. This method provides some insight to the attentive student, but no gain in speed over the quadratic formula for real-world numbers
this proves how we try to overcomplicate things in attempts to solve them, while all we needed was a simple thinking process.
@@wernerviehhauser94I agree the quadratic formula technically works for all these problems and provides even more.
The need to guess and check just gets eliminated in all it’s entirety.
@@wernerviehhauser94 I have taught math. I recommend using the quadratic formula as a last resort. They are too many ways you can make a mistake.
@@saftheartist6137 I have taught math. I recommend using the quadratic formula as a last resort. They are too many ways you can make a mistake.
My only suggestion is that "c" is used to define two different things: The third "element" of the quadradic equation and the "constant value" added to or subtracted from the second element of the quadratic, b. To help prevent confusion, I would suggest calling the second constant value "k", so we don't have two c's with different meanings.
Right, mathematical rigor must be provided
k= sqrt{(b/2)-(c)}; m = (b/2)+(k); n =(b/2)-(k). Final factors: (x +/- m)(x +/- n). k must be within bracket to take care of sign e.g. x**2 + 5 x - 14 provides k = 4.5, m=2.5+4.5=7, n=2.5-4.5= (-2). Factors are (x+7)(x-2)
Simpler to remember than the quadratic formula, arguably easier to calculate. But as others pointed out, it’s essentially the same method.
This is pretty much just how the quadratic formula works. This is how the quadratic formula should be taught, in my opinion; by starting with this as motivation
The quadratic formula is derived from the completing the square method. Rookie mistake.
@@gregorymorse8423 You trolling? This method is the same steps as completing the square.
@@gregorymorse8423formulas can be derived from a multiple of methods. Rookie mistake.
It is called the Poh Shen Loh method. A math professor named Poh Shen Loh discovered this method. The math professor himself said he found this method buried in a very very old Math book.
Here is the link
ua-cam.com/video/XKBX0r3J-9Y/v-deo.htmlsi=LiskkjE6oaHBAPZY
i've heard it called Hindi Method or Sridhara method. We used to learn it as an application of Vieta. Nonetheless, all I use is the quadratic formula - it's faster than everything else for real-world numbers in physics problems.
@wernerviehhauser94 this is why I don't give a shit about how a formula is named because it's only about politics
Well it is poh Shen loh method but it is explained better
It’s the ancient Babylonians’ method. They were very advanced in math and astronomy. If NJ Wildberger ever gets his channel restored, you’ll be able to see his demonstration of a real, Babylonian worked problem using this method.
@@glennschexnayder3720 Professor Loh basically condensed the method and made it easier to understand. This is especially true when explained with a graph.
I came here to refresh my memory on factoring so I can begin teaching my girls (8th & 6th grades).
I love this method because I hate ever having to say “guess and test”. I always look for methods/process or creat one.
This is excellent. I can’t wait to teach it.
I bet they aren’t teaching this in their school.
For small numbers this method is sheer time consuming. But for bigger numbers it is helpful.
Dude, teach them the generic way to solve quadratic equations, not this magic shit
@@angelviloria4966I would assume the same. However, just as others advised, please don't skip the formula taught in school. Remember the roots and (-b +- sqrt (b²-4ac)) / 2a
You can also do it in an another way, which in the end is essentially equivalent to the one in the video: first complete the square, x² + 32x + 192 = x² + 32x + 16² - 16² + 192 = (x + 16)² - 64. Then write this as a difference of squares: (x + 16)² - 64 = (x + 16)² - 8² = ( (x + 16) + 8 ) ( (x + 16) - 8 ) = (x + 24) (x + 8).
It always comes back to completing the square
You lost me at completing the square lmao I hate doing that. Guess and check with calculator is by far the fastest and easiest way to solve these
Do it with with x² + 4x + 10
@@lookiii1It works well. You complete the square, then add and subtract the square root of the completing adjustment (X+2+√6)(X+2-√6)
Your equation furnishes complex roots. Here they are: -2 + 2.4494897427832i and -2 - 2.4494897427832i Please check.
This is just a manual execution of the abc or pq formula (there are many names for it). Still impressive.
For simplification I use a = 1 => pq formula for x² + px + q = 0
=> x_{1/2} = - p/2 +/- sqrt( (p/2)² - q )
(x - x_1)(x - x_2) = 0 => m = -x_1 and n = -x_2. That's p/2 -/+ sqrt( (p/2)² - q ).
So completing the square is also called PQ method.
Extra bonus is giving a result in vertex form.
In the thumbnail it's clearly written a new amazing "factoring method". That is to find the factors.....
Not a new method of finding the value of x in the quadratic equation
Which we all know the best is the quadratic formula 😊 hope it explains the confusion
Maths teaches- if something's similar doesn't mean it's congruent.
If something approaches a place doesn't mean it's at the place- Calculus
If this seems like the quadratic formula it doesn't mean it actually is it or completing the square method
Clearly visualise: It is a good method of finding the factors while the quadratic formula is for finding the value of x.
❤❤❤love your finding efforts
i've had thoughts of this but never found a way to make them useful to my math classes, well until i found this video i could actually do something with it. THanks!
Great method, and VERY well-presented! Just wordy enough without feeling like you're dragging out minor details. Concise, but not sparse. Thank you!
You can also reverse the quadratic formula so that you don't need to work with numbers of such significant magnitude:
x = (-b ± √(b² - 4ac))/2a
x = -b/2a ± √(b² - 4ac)/2a
x = -b/2a ± √((b² - 4ac)/4a²)
x = -b/2a ± √(b²/4a² - c/a)
(-b)² = b²
x = -b/2a ± √((-b)²/(2a)² - c/a)
x = -b/2a ± √((-b/2a)² - c/a)
q = -b/2a
x = q ± √(q² - c/a)
Another nice thing about this variant is the fact that you only need to calculate _q_ once.
As an example of these conveniences, consider
8x² + 14x - 15:
q = -14/2(8) = -7/8
x = -7/8 ± √((-7/8)² - (-15)/8)
x = -7/8 ± √(49/64 + 120/64)
x = -7/8 ± √(169/64)
x = -7/8 ± 13/8
x = 6/8 , -20/8
x = 3/4 , -5/2
8x² + 14x - 15
= (x - 3/4)(x + 5/2)
= (4x-3)(2x+5)
= 4x(2x) + 4x(5) + (-3)(2x) + (-3)(5)
= 8x² + 20x - 6x - 15
= 8x² + 14x - 15
compared to:
x = [-14 ± √((-14)² - 4(8)(-15))]/2(8)
x = [-14 ± √(196 + 480)]/16
x = (-14 ± √676)/16
x = [-14 ± √(4•169)]/16
x = [-14 ± 2(13)]/16
x = (-7 ± 13)/8
x = 6/8 , -20/8
x = 3/4 , -5/2
The reason you can multiply things is because you're using the roots of the expression, meaning when you set the expression equal to 0, you are effectively multiplying 0 by whatever factors you choose, which doesn't affect the truth of the equation.
While the use of _m_ and _n_ are important to understand the underlying concepts shown in the video, it's not truly necessary once you can use an algorithm like this one, and i definitely prefer this algorithm over the standard quadratic formula.
Okay I'm a little disappointed because the thumbnail said this was new, but for everyone saying to use the quadratic formula instead... that's like adding five three times when asked to do five times three. It might be faster when you first learn, but oh my god is it slow. when I looked at the first problem, my first instinct was that it was eight, and that's not some weird boast like "oh I can do factors slightly faster than you," that's the bare minimum. It's honestly shocking to me that so many people are plugging these numbers into the quadratic formula thinking it's faster. Now that you've watched this video, do this until you have it down perfectly easily. memorizing the quadratic formula won't help you. Learn to factor.
Wonderful video. Since this amounts to using the quadratic formula for writing the factors of the quadratic, you could have just gone for "How to write the factors of a quadratic by using the quadratic formula". And *that* would have been a wonderful video. And nobody would have guessed or checked anything.
FANTASTIC! I've suspected there was a formula or algorithm for finding the factors, but never found one. Thank you for the explanation!
Love this. So much of boring guessing work of stone age in school.
m = sum/2 + sqrt((sum/2)^2 - product)
n = sum/2 - sqrt((sum/2)^2 - product)
I don't know about the other parts of the world, but here in the Czech republic, some 35 years ago, we were not taught to guess, but we learned the standard formula for the roots, to which this "new" approach is equivalent.
Sending this to every math teacher I’ve ever had 🔥
This is a fun video, I found the general form for this method by not plugging anything in where x^2+bx+c=(x+((b/2)+sqrt(((b/2)^2)-c))*(x+((b/2)-sqrt(((b/2)^2)-c))
If there is a coefficient on x^2 divide it by every term and put it on the outside of the formula
a(x^2+(b/a)x+(c/a))
We can treat b/a as “b” and c/a as “c” for the formula
Very clever! x^2+bx+c=0 x=-b/2(+/-)(b^2-4c)^.5/2 Someone noticed that the roots where b/2 +/- the same number. Then it becomes obvious how you developed your approach. Telling the students your thought processes is the way math was supposed to be taught. It was not intuitively obvious to me until I looked at the quadratic formula.
I just wanted to let you know that I never learned that before! Many thanks. It's another pain in the back removed with a quadratic polynomial!
And this is exactly what is called the p-q formula (at least here in Germany) for solving quadratic equations: x^2 + px + q = 0 x = -p/2 +/- sqrt((p/2)^2 -q). It’s just not displayed as a single formula but broken down into two steps.
Mathematics is the basis of all sciences and you explained it well, thank you
Yall this is literally the quadratic formula broken down into steps. Fire video tho!
This is scarily accurate. Even with irrational radicals
Hmmm... I love this method, since back in the day I love Factoring than Quadratic formula. This method gives me a new ways. Thank you.
I havent seen this before. Absolutely brilliant! I wish i knew this in the 90´s !
This method should go viral, we have to make children's lives easier. My school years would've been so not depressing if this method was around my time.
Most methods were there in books outside of the ones we were told to follow. However, the teachers who could think of explaining how to approach a given problem from different angles weren't around in some respectable number. They followed a particular book and stopped at that. There has been a very asked cha ge in how children learn these days. So many follow good channels on UA-cam as well. No wonder then, the depressing days had to be endured by us.
x = [-b±√(b²-4ac)]/2a
This is introduced in many math classes, but alas too many parents will claim this is more difficult or "I never learned it that way" and blame this on Common Core.
Isn't the method shown no less effort than factorising ax^2 + bx + c by calculating D = b^2 - 4ac and then writing the factorisation as a.( x + (b + √D)/2a) ).( x + (b - √D)/2a )?
That has the advantage of using a formula that's either already familiar, or at least will have further applications later on.
If you are worried by big numbers, just calculate d = (b/2)^2 - ac and write the factorisation as a.( x + (b/2 + √d)/a ).( x + (b/2 - √d)/a) which scales thing down by a factor of 4 (like the video does).
Yes. It's just an alternate form for the quadratic formula. This is the famous Po Shen Loh method, and 3B1B has a good video covering it in a little more depth
Brilliant. I am sure my 8th grade math teacher in my school will find this interesting.
This actually really cool and helpful, I might use this next time instead of the quadratic equation
Watching this ticks me off because not only does the more efficient methid make sense, but the OG method makes more sense than what I was taught.
My math education sucked noodles. I took calculous in HS, and was fumbling in the dark the whole time.
we are literally solving a quadratic , to solve another quadratic . Although the approach is nice.
Well it's always a single term quadratic which is much easier than multi term quadratic equation
@@airking2883yes, but completing the square is not only already familiar, but is equivalent.
In fact I invented this method (or very similar) some decades ago (now retired from high school). But often with much less work. Take x^2 - 6x - 91 = 0. Only integer factoring of 91 is 7 x 13, so I write (x - 7) (x - 13), yes - both with a minus. Now it is easy to see if any - must be made into a + (which is "easier" than the opposite). Since 7 - 13 = -6, we must have + on the 7. Thus: (x + 7) (x - 13).
I've seen OutlierOrg but different approach but same principle.
Thank you so much.
You earn a subscribe
Nice explanation of Po Shen Loh's method, publish about 5 years ago. Po does say that the steps to doing this method have been know for a very long time, but the combination used here wasn't well recorded elsewhere, so he is standing on the shoulders of others.
I am very grateful for your method. It helps me a lot, during my exams and worksheets. I think this should be known further, you are an inventor.
Another way to think of this is that the solutions to x^2 + b x + c = 0 are x = - b/2 +/- sqrt( (b/2)^2 - c ) which is just a rewriting of the quadratic formula.
Played this on x2.6 speed and quadratic formula got the answer first with plenty to spare. While this certainly gives a cool perspective that seems to speak to students, we should probably spend more time pondering what to do about the fact that the formula (by definition the easiest way to do something) is so intimidating to students.
Mind blowing method yet so intuitive and simple!
I wish this was taught in my classes!! Thank you 😊
Amazing, this method should include in our curriculum .
My daughter is a high school math teacher. She likes this method, but can't change the approved curriculum. This is why things don't get better in US schools.
IMO, factoring is a waste of time.
Surely there's nothing against teaching it after teaching the 'approved' method? It could be done easily, in a 40 min class time!
I taught HS math, and always found time for additional niceties like this. Doesn't matter if it is approved, although she may want to avoid quizzing students on it if it really is a backward administration. But for the test that wraps up all methods, i.e., solve using any method, students can use this method if they wish.
@@jamesharmon4994 you've clearly never had to integrate anything requiring partial fraction deconposition
@@jamesharmon4994respectfully, your opinion is wrong unless you are implying there is a faster way
You can also factor by grouping 2x^2-7x-4 = 2x^2 -8x+1x -4 = (2x^2-8x) + (1x-4) = 2x*(x-4) + (x-4) -> factor (x-4) from each group making -> (2x+1)(x-4)
I double checked this method and it certainly works. It also leads to the standard equation taught in schools sqrt(b^2 - 4ac) etc. I am puzzled for why it works though. Who would have thought in the first place? It is quite magical in its own way. Well it was magical until You explained it so well :-) You are a good teacher. There is always the How and Why. Feynman says shut up and calculate. So faced with these equations just apply the standard formula and sattisfy the 'How.' You have explained The 'Why' however. That is quite meritorious :-) Physics is in the doldrums, brain dead from not thinking and just calculating. Maybe expalin Math to them :--)
To get to the quadratic equation (which you have most of there), is done by completing the square of the general quadratic equation "ax^2+bx+c" and simplifying.
@@alemswazzu Oh I get that. I am not good at grammar. My puzzle was with this new method working in the first place :-)
Wish I knew this wayy earlier.
Thanks to this video I just realized that you can actually also use the Quadratic Formula for factoring! (by replacing -b to positive)
Oh and you can derive the process/formula used in the video from the Quadratic Formula(where -b is positive) and vice versa, which makes sense since both ultimately gives the same answer
Compelling method. Reminds me of "completing the square" since you're taking half the b value and factoring out an a>1. A "benefit" of using the c value and listing factors is that students can generate a table of y = c/x in their graphing calculators and look for integer coordinate pairs. I love the link to multiplying conjugate pairs in this method though! I'll stick with the AC method for a>1 for now.
Very nice video. Small suggestion: use c' in the m/n formulas or another name to avoid any confusion with c in the quadratic equation.
Excellent Explanation Sir Ji Thankyou Sir
In India, this method is taught to 12-13 year student (8th level).
My grandfather taught me this in my childhood who had learned it in 1920s.
Since I was not taught this, I was using the quadratic formula if I can't guess the factors. Thanks.
Excellent and simple method, thanks for bringing it up
Thank you! I have been writing a maths "book" for my teenage son and my father who are both doing their secondary school maths (my father as a hobby project) and neither would say they have a natural feel for this kind of thing. My book is an attempt to provide the clearest explanations for everything they encounter in the curriculum. And, as is the invariable way with maths, when you think you have something down pat, someone like you comes along with a splendid video that makes me think, "why the heck didn't I think of that?". I hope you don't mind if I steal your explanation for my book.
This is fundamentally Poh Shen Loh factoring, derived differently; but you've missed an opportunity to combine with the Australian (i.e. "down under", referring to the placement of a in the denominator) technique for factoring quadratics with a 1.
When b is odd, we encounter an unusual circumstance where the decimal fraction 4.5 may be more convenient than the common fraction 9/2, using this shortcut for squaring a half integer:
1.5² = 0.25 + 100 * (1 * 2) = 2.25
2.5² = 0.25 + 100 * (2 * 3) = 6.25
3.5² = 0.25 + 100 * (3 * 4) = 12.25
4.5² = 0.25 + 100 * (4 * 5) = 20.25
5.5² = 0.25 + 100 * (5 * 6) = 30.25
:
9.5² = 0.25 + 100 * (9 *10) = 90.25
:
The Australian method for handling a 1 proceeds by initially ignoring that a 1 as, using your example:
3.5² - c² = (2)(-4) = - 8
becomes
c² = 12.25 + 8 = 20.25 = 4.5²
and then factoring as
[ (2x - 3.5 - 4.5) (2x - 3.5 + 4.5) ] / 2
= [ (2x - 8) (2x + 1) ] / 2
= [ (2x - 8) / 2 ] (2x + 1)
= (x - 4) (2x + 1).
Notice how the a value of 2 was initially placed in both factors, as well as "down under" in the denominator. When a is composite and must be factored across both final factors, this trick delays having to determine that factoring and distribution until a final step, not interfering with the rest. The factoring of a reduces to identifying the common factor in each numerator factor - much simpler than previous methods.
Po Shen Loh, not Poh Shen Loh.
In this question 15:37
2x²-7x-4=0 it will written as
*X²-7x-8* =0 after finding Root
(x-8)(x+1)=0 then divide 2 both side
(x-4)(x+1/2)=0
This also easy by their method
*Maine isse aur aasan bna diya*
Hi there 🎉good good job
If you in put as a(x-m)(x-n)=0
and your factor form
as ax^2+bx+c=0 and find the ROOTS as m&n then plug into first equation a(x-m)(x-n)=0 it's much more easier 😊
Good method, and well presented.
(The rest of this is not for you, but more for the existence of this method in the first place) Ridiculous though, factoring is not a difficult process, and this method is not only completely unnecessary but much more difficult than just factoring it. I speak to you as a math tutor, and when the kids come to me being confused by yet another factoring method to make things "easier", and I show them how very easy it is to just skip to the end and solve it directly, each one is more shocked than the last at how very easy this all is. And they are truly shocked when I show them how easy it is when a doesn't equal 1. I mean... look at what happened with x^2 + 9x + 20, fractions were introduced to what should have been HORRIBLY SIMPLE! Stop babying these kids with all these "factoring methods", they can handle the guessing and checking (not to mention, it is actually the far easier method). *frustrated rant over, I had a day...*
I generalized it with the roots adding up to -b/a and their product being c/a, I did the same thing where one root is essentially the x value for the vertex plus h, the other is the x value for the vertex minus h
m = -b/2a + h
n = -b/2a - h
Their product should be c/a
m . n = (-b/2a + h)(-b/2a - h)
m . n = b²/4a² - h²
b²/4a² - h² = c/a
At this point you may have realized that it's building up the main formula to solve for quadratics
h² = b²/4a² - c/a
We multiply c/a times 4a/4a
h² = b²/4a² - 4ac/4a²
h² = (b² - 4ac)/4a²
h = √(b² - 4ac)/2a
Yes, the discriminant shows up here, being b² - 4ac
Let's denote the discriminant by using ∆
h = √∆/2a
The main formula is
x = (-b ± √∆)/2a
So if we just replace √∆/2a for h
x = -b/2a ± h
I should point out that -b/2a is the x value for the vertex in every quadratic, where f(-b/2a) is the y value for this vertex of this same quadratic f(x)
This in a way proves that each of the roots is the same distance away from the x value of the vertex, I believe this applies for every polynomial that has an even degree
What i have done for similar problems is right m=b-n then n(b-n)=c so n^2-bn+c=o then use x=(-b±squareroot(b^2-4ac))/(2a) for n^2-bn+c as it would allow me to right ax^2+bx+c=0 in form of (x+n)(x+m)=0
Very thanks sir i got 5/5 in section of quadratic equations in my bank manager exam. I'm from India and a aspirant of govt bank exams
I have to do 35 questions in 20 min which also include 5 questions from quadratic equations
This is amazing and when I become a teacher I am so teaching my students this
Nice visualization of the solutions of quadratic formula for a=1. And you showed it: all quadratic equations can be rewrite to the pq formula. Your solutions m and n are the negative solutions of this formula x = -b/2 ± sqrt[(-b/2)² - c]. so m/n = b/2 ± sqrt[(b/2)² - c]. I removed the minus under the square, as it is not important, if you square it anyways... but it is your solution, you showing for different b and c.
Why people hear in the comment section are confused between quadratic formula and an factorising method
In quadratic formula we find the exact value of 'x'
Whereas this is a factorising method where we find the "factors" of a quadratic equation
This method is what I figured out in deriving the quadratic formula which I thought was that's how you actually derive it instead of completing the square.
Although this ensures that you get its factors, I don't think it's the method I'm going to use mainly in my factoring, because to factor quickly, a method that allows you to being able to do it mentally or at least minimal writing would be my bet. But this method would require me a lot of thinking, writing, and/or storing numbers in my head.
This method actually gave me an idea.
instead of just using the quadratic formula, just use the formula derived from this method directly because comparing it to the quadratic formula, it is easier to compute the square root.
when a = 1
m = b/2 + √( (b/2)² - c )
n = b/2 - √( (b/2)² - c )
it is essentially the same as the method, but for me using the formula is faster than doing all the work
Or perhaps solving one equation for a variable and plugging it into the other equation
I have a different method in solving this. I just take the factors of 192, then find the factors that adds to 32, which is the 8 and 24. It might not be applicable for all the problems involving this, but I believe it will solve most of the problem.
This method already taught to me 🔥🔥🔥
1:50 he did not considered 8 and 24 because that's the answer
Don't factor out the a. Multiply a and c and do this process as if a=1 and c=ac. When you have m and n, fill in (ax+m)(ax+n) and toss out anything common for a and m or a and n, the product of which is a. (It is necessary to factor out a common from the quadratic first.)
This is fantastic. I've always hated the standard instructions to just find the numbers that work. I devised a way that uses the factor pairs of AC. It worked well for my students, but I had to admit that it could be time consuming. This is better. Thanks for the excellent video.
6:26 m= 32 - n
Replace m in the product
Check if n and m is negative or not
Who invented this method? It's great. I'm 76. I got a math degree many years ago. I never worked in math but did use simple math and algebra on occasion. But never saw this method.
A professor named Po Shen Loh.
@@trien30to give him credit for the invention is generous. Popularization, sure.
After seeing this method presented a few ways, I realize it's about parametrizing using the sum equation instead of the product equation.
So the general case for this is: (b/2 + n), (b/2 -n)
And the general case for n is:
√((b/2)^2 - C)
Putting all that together we get a simple and elegant single equation:
b/2 +- √((b/2)^2 - C)
Then we just take the negative to get the value of X, and we get:
- b/2 +- √((b/2)^2 - C) ...
Also known as, the (reduced) quadratic formula
Best math video ever!
I will be teaching quadratic today. Will add this concept as well.
If you look very very very closely at the procedure, Its just the quadratic formula but in step by step method of solving the determinat to get C and adding and subtracting that to b/2a which will give the two roots
x^2 + 32x + 192 = 0 :: average of roots is r. roots are (r+p) and (r-p); sum of roots = 32,
product of roots = 192. Sum:: (r+p)+(r-p) = 2r = 32 so r = 16. Product:: (r+p)*(r-p) = 192
r^2 - p^2 = 192 256 - p^2 = 192 p^2 = 256-192 p^2 = 64 so
p = √64 = 8 the roots are r ± p or 16 ± 8 or 24 and 8
this method should get more popular and get in school.
I don’t think so, the quadratic formula completing the square and ‘just guessing’ are imo all faster. Honestly just guessing is pretty easy to do
Mind blown. Thank you so much!!!!
Haha i remember one day one student I was tutoring bitching about where the quadratic formula came from. So I derived it on the blackboard for him.
I would suggest as an exercise (or a future video in this series) that you can do it too. Thanks for this . I like it just as much as the quadratic formula because it's cute, it's easy and it's novel. One thing this could have embellished a bit is the form of the graph and it's roots.
This method is brilliant. I love it.
I mean, there's a reason why I prefer completing the square.
Realizing this is completing the square
I have in fact seen that method and I am happy I did
Mid term splitting, i was indeed taught this in school
Isn't it just the same as Quadratic Formula? And probably more faster using Quadratic Formula
For equation: ax^2 + bx + c = 0; the values of x that satisfy: x = ( -b +- \sqrt{ b^2 - 4ac } )/( 2a )
Tweak a little bit inside the root: \sqrt{ b^2 - 4ac } = \sqrt{ ( ( ( b^2 )/4 ) - ( ( 4ac )/4) ) * 4 } = \sqrt{ ( b/2 )^2 - ac } * \sqrt{ 4 } = 2 * \sqrt{ ( b/2 )^2 - ac }
Formula rn: x = ( -b +- 2* \sqrt{ ( b/2 )^2 - ac } )/( 2a ); tweak more become x = ( -b )/( 2a ) +- ( \sqrt{ (b/2)^2 - ac } )/( a )
And if you make it become factor, just multiply by -1
Because: ( x+m ) * ( x+n ) = 0; x+m = 0 or x+n = 0; x =--m or x = -n
x^2 + bx + c = 0; ( x + ( ( b/ ( 2a ) ) + ( \sqrt{ (b/2)^2 - ac } )/( a ) ) ) * ( x + ( ( b/ ( 2a ) ) - ( \sqrt{ (b/2)^2 - ac } )/( a ) ) )
*Note:
+- mean plus or minus
\sqrt{ } mean square root of ...
This algorithm is just great! How did I not figure it on my own?
The presentation in this video is clear and interesting, but the method described is essentially the same as what you do when you solve a quadratic equation by Completing The Square.
Too nicely explained thanks a lot
It is Indian old method.
If m+n=k then both m,n are equal,k/2.
If m,n are unequal then one is more than k/2,other is less than k/2
m=k/2+u,n=k/2-u so that sum is k.
mn=k^2/4-u^2
u^2=k^2/4 -mn
This is reduced form of quadratic formula.
Didnt quite understand it
It's the same thing taught in the video just written in very compact from @@dendaGulliLapoch
THE WAY I WAS ABLE TO ALGEBRAICALLY DO THE 2ND ONE IN MY HEAS IN THE SAME TIME IT WOULDVE TAKEN ME TO GUESS AND CHECK AFTER TAKING CALC BC IS CRAZYY
Great method! You can also use shri dharacharya formula.
Or this c is all about expressing n and m witj only 1 variable. You could also just say that if m + n = v then n = v - m and boom you got it. Also another thing you could do is if m + n = i, m*n = j, then (m+n )squared is m2 + 2mn a n2 and you know 2mn so you get the value of m2 and n2, then calculate m-n squared, get the root, and boom you got it
Wow, amazing method
Thanks to Poh shen Loh for making our work easier.
His name is Po Shen Loh, not Poh Shen Loh.
Factoring with this rule, which ill write as
M, n = b/2 ± sqrt(b²/4 - c)
Is smart, but if you want to know x, instead of just factoring to simplify fractions or smth, the quadratic formula is probably still faster to calculate
Thanks for the effort of presenting it this way, but that's literally the general rule of the quadratic equation.
Instead of presenting it in a one-step formula: x = (-b+-sqrt(b^2-4*a*c))/(2a), you had to break down into simpler, but more, steps.
Nice video! A good way especially when the 3rd term has many factors
But I would suggest another sneaky way:
1- know that you can also find the factors from the roots (i.e. use the quadratic formula & move in the opposite direction)
ex.: If the roots are 2 & 1 then the factors are:
(x-1)(x-2)
2- multiply the above factors & compare them to the original to make sure you have all the constants
Ex.: you found x=1 & x=2
Then: (x-1)(x-2)
So you have: x^2-3x+6
But the original expression is:
2(x^2)-6x+12
Comparing to: x^2-3x+6 you need to multiply by 2:
2(x-1)(x-2)
Jesus Christ this is the greatest thing I've ever seen
I just looked at the quadratic equation on the thumbnail and (x+24)(x+8). Took me all of about 15 seconds. The last time I factorised a quadratic equation: 1993.
tysm this is a lifesaver
Isn't that just the quadratic formula...
Same ! I was also thinkin about it. 🧐
Yes, of course, it is. It's ridiculous to call this a new method, or a method with a certain name attached. This is common knowledge probably since Bronce Age. 😂
Edit: Just using the memorized quadratic formula works even faster for me when doing the calculation in the head.
Absolutely, you are right! But in a simpler way. In fact i can mentally do it, the simpler ones.
Yes.
No, this is a method which works _without_ using the quadratic formula explicitly. Or which can be used to _derive_ the quadratic formula.
The problem suggested on top looks kinda ugly so I personally would not even bother checking values, but use one of the following:
1. completing the square. For this it is obvious 2b=32 so b=16 b^2=256 => (x+16)^2+192-256 then a^2-b^2 ftw
2. quadratic formula