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
In actuality, this method is exactly equivalent to a variation of completing the square. • Divide by the leading coefficient if it isn't 1, to get x² + bx + c = 0 • Square half the linear coeff., (½b)² = B, and add & subtract that to the LHS, to get x² + bx + B + c - B = (x + ½b)² - [B - c] • Now treat the LHS as a diff of squares, which then factors into the product of (x + ½b ± √[B - c]) If you apply this to the examples in the video, I think you'll see that it amounts to the same process. In fact, all quad. soln. methods are essentially equivalent. That said, the method in the video is quite good. Fred
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).
Very nice. Someone was smart enough to think if this and formalize, removing guess work. I would still factorize mentally in terms of sum and product, and in the case of large numbers that are difficult for me to break down that way, I would use this procedure. Appreciate this new and simple way.
I'm an Algebra tracher and Factoring is the next unit we're going to start. I have always had a bit of a dislike for using guess-and-check for factoring, so I think I'll introduce my class to this method. I also appreciate that it can be used to remove "factor by grouping" altogether. That's another one that students have struggled with in the past. The only concern might be with students who still struggle with fractions, but I feel like there is always going to be a habdful of those that come down the pipeline. Thank you for this very informative and easy to follow video!
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.
I would say that this video is and explanation in disguise of how the quadratic formula was devised: k = sqrt(b^2/4 - c), m = b/2+k and, n = b/2-k iff a = 1. Work that out for x and you will end up with the quadratic formula.
I saw this method once, but after time went by, I completely forgot it existed, and when I wanted to search it, I couldn’t find it anywhere, so thanks for bringing it up again
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 ).
You have stumbled on to Vieta's formulas. To solve a cubic, there is a sum, a sum of products, and a product: (l+m+n),(lm+ln+nm), and lastly (lmn). Solve a quartic,....on and on. Enumeration algebra is an overlooked gem.
1:09 product of C and A not only C, And also another method is that you can factorize 192 like we do in HCF and LCM and then multiply them to make 2 numbers that do the job
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
This is is a nice alternative method to know, but I don't think it's necessarily superior to "guessing" methods for factoring. For practical applications in physics, chemistry, etc, the quadratic formula is better. Actually I don't see the product/sum method as guessing. If there are rational factors, I teach my students to find them using number sense, even when 'a' is not 1. If the factors are irrational, you might as well use the quadratic formula, in my opinion.
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
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.
This method seems very effective for calculating large numbers. However, I have a question about the step at 15:37 where you multiplied 2 with (x + 1/2). At the beginning, you factored out 2 from the entire polynomial (2x² - 7x - 4). But when simplifying the quadratic equation, you only multiplied 2 with (x + 1/2). Shouldn't you multiply 2 with both brackets, (x + 1/2)(x - 4)? Please correct me if I'm misunderstanding!
If you multiply both terms by 2, then check by calculating the result of those two terms - you will get the wrong answer. (2x+1)(x-4) = 2x^2 - 7x -4 which is correct (2x+1)(2x-8)=4x^2 -14x -8 which is incorrect Example: if you have say 2 x 3 x 4 - you only use each number once to get the result of 24
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
True, but in general students that barely understand functions have zero insight in the quadratic formula. I think this method is much closer to their skills.
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!
This is a cool method to notice and to understand how it works, but it's really just a rearrangement of completing the square, which is itself embedded in the quadratic formula. I don't think there'd be a lot of utility in teaching it as its own method, because as soon as you find solutions by any method you can just put those back in the factored form. I suppose it could be useful to a student who can remember a sequence of simple steps better than they can remember a single more complex formula.
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.
I agree with making factoring easier, which is why an even better shortcut is to "Factor a quadratic by completing the square and then rewriting the result as a difference of squares." Just Google what I put in quotes. You're welcome. 😎
Other commenters have mentioned that this technique is basically (just) how the quadratic equation works. While, strictly speaking, this may be true I still think that this video is extremely valuable in the way that it views the QE from rather different direction. In short, you can never too many tools in your toolbox. ;-) (More to the point I think that in some cases this technique could be quicker/handier/less cumbersome than using the QE. For example, if your goal were to factor the equation rather than simply obtaining its roots -say as a sub-step in dealing with a larger problem.)
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.
I mean instead of complicating things just use the standard quadratic formula. But sure maybe you guys have your own tricks. I always like to keep things simple.
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.
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.
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.
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.
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.
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.
Thanks so much for this method! I teach in highschool and even though I've explored alternatives, I never thought of something so elegant. I especially like that it takes away the arbitrariness of using the quadratic roots equation, or the frustrating guesswork of assigning numbers and seeing if they work.
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.
Like others have said, this is really just a version of the quadratic formula. The two factors could also be found by this formula: [b+/- sqrt(b² - 4ac)] / 2a It's just the quadratic formula but without the negative on the first "b." Great video!
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.
This is not entirely correct. Memorizing the quadratic formula WILL help you. Factoring won’t help you when the factors don’t turn out to be nice whole numbers. In those cases (some ugly fraction, an irrational number or a complex one) the quadratic formula is very nice to have memorized. However I do agree that mindlessly memorizing the quadratic formula without learning how to factor is a bad idea. People should learn both. Specifically, factoring first and the formula second.
quadratic formula helps when you’re not doing easy factoring like basically everything in this video. imagine having irrational numbers for a b and c? you deal with that a lot in harder classes
Bro's like those annoying high school teachers. Don't tell somebody what to do. You have to realize there's more than 1 way of solving problems. Quadratic formula may be slow for you but it is always a guarantee
When simplifying all of the steps, you end up just getting the quadratic formula. However, this is a fantastic way of building a more intuitive understanding of the quadratic formula especially since it initially seems like such a mess and a whole bunch of mumbo jumbo!
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.
It is called the poh shen loh method. A math professor named Poh shen loh is the man who discovered this method. If you google "poh shen loh method" you will find it.
How about applying quadratic formula! It should be first and the best method for factoring quadratic equations. AC method works only on certain equations with a simple numbers. In real life situations it can be any numbers.
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.
Absolutely beautiful. 🥰 No need to memorize any formula, not even the (a+b)(a+c) or the (a+b)+(a-b) or (a+b)(a-b), because you can easily complete them on the next line. What are the odds that you remember the quadratic formula some 30-40 years from now? I'm 76 years old, and even though I graduated from university with maths as one of my majors, I don't (because I almost never use it).
Great job! The method is powerful, try its generalization. For instance, - -if you have x^3+bx^2+cx+d, and x1, x2, x3 are solutions, then x1+x2+x3=b, x1.x2 + x1.x3 + x2.x3 =c and x1.x2.x3=c. -if you have x^4+bx^3+cx^2+dx+e and x1, x2, x3, x4 are solutions, then x1+x2+x3+x4=b, x1.x2 + x1.x3 + x1.x4 + x2.x3 + x2.x4 + x3.x4=c and x1.x2.x3 + x1.x2.x4 + x2.x3.x4=c and x1.x2.x3.x4 = e💙💜💙💜💙💜💙
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.
Please stop with the "wow", "they should teach it in school". I mean, it's ok, it works, but is it really usefull? This is useless if you really understand the expressions of the roots: (-b+/-sqrt(b^2-4ac))/(2a). It's easy with the CANONICAL FORM. That's it, no need to push it with new expressions like m, n, c and so on. PS: in France, we teach our students how to find the roots with a formula, how to find that formula by hand (the canonical form) and that's it. And if they know how to find roots by guessing, good for them.
In my experience, most students who struggle with multiplication struggle with fractions even more. It’s probably more efficient to stick to finding factors of ac first, and for more challenging problems, make an organized list of the factor pairs. What does help from this method is to recognize that larger b values (compared to ac) mean that the numbers are further apart. This understanding helps with guess and check.
I agree, but then we are teachers. Teaching this method to the masses (especially Foundation Tier) would baffle them. Higher Tier GCSE students with strong algebra skills who enjoy generalising results could take to this way. They also appreciate being shown multiple methods and are able to select the most appropriate or efficient method when necessary.
try applying it to 0.7*t^2-27.78*t+800=0 (that's an actual equations my students got when trying to find the time and coordinate where two accellerating objects meet). Try both methods. The quadratic formula is way superior, especially if the second question is what kind of a head start one of the objects needed so that the objects don't meet.
Analytically, the extreme value (minimum or maximum) of the function f(x)= ax^2+bx+c is at 2ax+b=0, i.e. at x=-b/2a. The solutions of the equation are to the left and right of this, at an equal distance from this. In fact, the solving formula also expresses this.
I don't see the 'amazingness' in this, since you can do the same thing by solving the quadratic equation. ax²+bx+c=a(x-p)(x-q), where p and q are the roots of the quadratic.
I agree with you absolutely. Completing the square has been used to derive the quadratic formula. That is already systematic enough. If the ultimate objective is to get the solution, guessing and quadratic formula would do the job faster than this method. In addition, using the quadratic formula helps you to know immediately if the equation have rational roots or not (just by checking if b2-4ac>0 or
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, this process is easier to follow than using the formula. Really good as long as the formula is also presented and understood for use in purely algebraic manipulations
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
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*
Earlier in my days, we used to do prime factorisation of the constant which simplifies the guessing game with massive amount. But yes, this is amazing.
Why can't you just use the quadratic roots formula? It doesn't need any guesswork. It will give 2 roots values (m,n) which you can write as (x-m)(x-n). This approach is literally that quadratic formula manipulated. (-b +- D)/2a. As you can see, the same number D (discriminant) is added and subtracted from -b. D is the playing the same role as "c" in this approach.
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.
The same suggestion of gpt4 who doesn't know this universal method to my amazement!!!
Brasil
@@thomasdemilio6164clarity.
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.
@@gregorymorse8423 This is literally completing the square. He's doing so by finding c in 16² - c² = 192.
That is nothing like completing the square
In actuality, this method is exactly equivalent to a variation of completing the square.
• Divide by the leading coefficient if it isn't 1, to get x² + bx + c = 0
• Square half the linear coeff., (½b)² = B, and add & subtract that to the LHS, to get x² + bx + B + c - B = (x + ½b)² - [B - c]
• Now treat the LHS as a diff of squares, which then factors into the product of (x + ½b ± √[B - c])
If you apply this to the examples in the video, I think you'll see that it amounts to the same process.
In fact, all quad. soln. methods are essentially equivalent.
That said, the method in the video is quite good.
Fred
yes indeed
thanks for the insight fred!!
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
@@ranjittyagi9354 How did you square the “b”?? …Thank you in advance!!
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.
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)
that's basically quadratic formula
Very nice. Someone was smart enough to think if this and formalize, removing guess work.
I would still factorize mentally in terms of sum and product, and in the case of large numbers that are difficult for me to break down that way, I would use this procedure.
Appreciate this new and simple way.
I'm an Algebra tracher and Factoring is the next unit we're going to start.
I have always had a bit of a dislike for using guess-and-check for factoring, so I think I'll introduce my class to this method. I also appreciate that it can be used to remove "factor by grouping" altogether. That's another one that students have struggled with in the past.
The only concern might be with students who still struggle with fractions, but I feel like there is always going to be a habdful of those that come down the pipeline.
Thank you for this very informative and easy to follow video!
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.
I would say that this video is and explanation in disguise of how the quadratic formula was devised: k = sqrt(b^2/4 - c), m = b/2+k and, n = b/2-k iff a = 1. Work that out for x and you will end up with the quadratic formula.
I saw this method once, but after time went by, I completely forgot it existed, and when I wanted to search it, I couldn’t find it anywhere, so thanks for bringing it up again
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.
Yall this is literally the quadratic formula broken down into steps. Fire video tho!
The method we use at our schools are very lengthy and consumes a lot of time. Good that I got to learn this method! Thanks a lot!
You have stumbled on to Vieta's formulas. To solve a cubic, there is a sum, a sum of products, and a product: (l+m+n),(lm+ln+nm), and lastly (lmn). Solve a quartic,....on and on. Enumeration algebra is an overlooked gem.
1:09 product of C and A not only C, And also another method is that you can factorize 192 like we do in HCF and LCM and then multiply them to make 2 numbers that do the job
Great method, and VERY well-presented! Just wordy enough without feeling like you're dragging out minor details. Concise, but not sparse. Thank you!
Excellent. When 'c' is a large number then it really helps solving this way. Thanks for jogging my memory, after a hiatus of 53 years.
This is the best trick i have ever seen.thankss
Babe wake up new factoring method just dropped
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.
Well the difference is.. x²=k type quadratic isn't exactly solving something.
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
Sending this to every math teacher I’ve ever had 🔥
This is is a nice alternative method to know, but I don't think it's necessarily superior to "guessing" methods for factoring. For practical applications in physics, chemistry, etc, the quadratic formula is better. Actually I don't see the product/sum method as guessing. If there are rational factors, I teach my students to find them using number sense, even when 'a' is not 1. If the factors are irrational, you might as well use the quadratic formula, in my opinion.
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
You could go a step further and you will end up with k = sqrt(b^2/4 - c), m = b/2+k and, n = b/2-k iff a = 1.
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 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!
You explained so good , I am astonished to see this ,we never learnt this in school .it is really amazing . thanks .
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.
This method seems very effective for calculating large numbers. However, I have a question about the step at 15:37 where you multiplied 2 with (x + 1/2). At the beginning, you factored out 2 from the entire polynomial (2x² - 7x - 4). But when simplifying the quadratic equation, you only multiplied 2 with (x + 1/2). Shouldn't you multiply 2 with both brackets, (x + 1/2)(x - 4)? Please correct me if I'm misunderstanding!
If you multiply both terms by 2, then check by calculating the result of those two terms - you will get the wrong answer.
(2x+1)(x-4) = 2x^2 - 7x -4 which is correct
(2x+1)(2x-8)=4x^2 -14x -8 which is incorrect
Example: if you have say 2 x 3 x 4 - you only use each number once to get the result of 24
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
True, but in general students that barely understand functions have zero insight in the quadratic formula. I think this method is much closer to their skills.
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!
1:50 he did not considered 8 and 24 because that's the answer
probably because he wants to show this method as “easier”
This is a cool method to notice and to understand how it works, but it's really just a rearrangement of completing the square, which is itself embedded in the quadratic formula.
I don't think there'd be a lot of utility in teaching it as its own method, because as soon as you find solutions by any method you can just put those back in the factored form.
I suppose it could be useful to a student who can remember a sequence of simple steps better than they can remember a single more complex formula.
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.
I agree with making factoring easier, which is why an even better shortcut is to "Factor a quadratic by completing the square and then rewriting the result as a difference of squares." Just Google what I put in quotes. You're welcome. 😎
FANTASTIC! I've suspected there was a formula or algorithm for finding the factors, but never found one. Thank you for the explanation!
Other commenters have mentioned that this technique is basically (just) how the quadratic equation works.
While, strictly speaking, this may be true I still think that this video is extremely valuable in the way that it views the QE from rather different direction.
In short, you can never too many tools in your toolbox. ;-)
(More to the point I think that in some cases this technique could be quicker/handier/less cumbersome than using the QE. For example, if your goal were to factor the equation rather than simply obtaining its roots -say as a sub-step in dealing with a larger problem.)
I haven’t even been taught factorisation of quadratic formulas yet, but found this video flabbergastingly interesting and really easy to understand.
Bro is 6 years old
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.
I mean instead of complicating things just use the standard quadratic formula.
But sure maybe you guys have your own tricks. I always like to keep things simple.
@@uncannyroaches5933 yeah fr the quadratic formulae is so simple why no one uses it
This is nothing but another form of completing the square. Though I liked the way you explained.
I've seen OutlierOrg but different approach but same principle.
Thank you so much.
You earn a subscribe
Same here
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.
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.
This is awesome method. Thanks for showing all possible situations for generaliy
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
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.
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.
Just awesome
I love how much easy you have done the factorization
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
Thanks so much for this method! I teach in highschool and even though I've explored alternatives, I never thought of something so elegant.
I especially like that it takes away the arbitrariness of using the quadratic roots equation, or the frustrating guesswork of assigning numbers and seeing if they work.
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 :-)
Like others have said, this is really just a version of the quadratic formula. The two factors could also be found by this formula:
[b+/- sqrt(b² - 4ac)] / 2a
It's just the quadratic formula but without the negative on the first "b."
Great video!
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.
This is not entirely correct. Memorizing the quadratic formula WILL help you. Factoring won’t help you when the factors don’t turn out to be nice whole numbers. In those cases (some ugly fraction, an irrational number or a complex one) the quadratic formula is very nice to have memorized.
However I do agree that mindlessly memorizing the quadratic formula without learning how to factor is a bad idea. People should learn both. Specifically, factoring first and the formula second.
quadratic formula helps when you’re not doing easy factoring like basically everything in this video. imagine having irrational numbers for a b and c? you deal with that a lot in harder classes
Using quadratic formula to find the factors lol.
Bro's like those annoying high school teachers. Don't tell somebody what to do. You have to realize there's more than 1 way of solving problems. Quadratic formula may be slow for you but it is always a guarantee
Good enough for 6th graders
When simplifying all of the steps, you end up just getting the quadratic formula. However, this is a fantastic way of building a more intuitive understanding of the quadratic formula especially since it initially seems like such a mess and a whole bunch of mumbo jumbo!
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.
You can also use sridharachaya method( quadratic formula) that is x= -b/2a +-( b² - 4ac)/2a
What is the name of this method?
It is called the poh shen loh method. A math professor named Poh shen loh is the man who discovered this method. If you google "poh shen loh method" you will find it.
Factoring by difference of squares
Poh Shen Loh method
Quadratic formula with extra steps.
How about applying quadratic formula! It should be first and the best method for factoring quadratic equations. AC method works only on certain equations with a simple numbers. In real life situations it can be any numbers.
Indian students learned it in 9th class it's completing square method 🗿
Yes
n = root((b/2)^2/ac) where n is the number added/subtracted from b/2.
Credit to Po Shen Lo.
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.
God bless you so much for this. I have been looking for something like this.
Absolutely beautiful. 🥰
No need to memorize any formula, not even the (a+b)(a+c) or the (a+b)+(a-b) or (a+b)(a-b), because you can easily complete them on the next line.
What are the odds that you remember the quadratic formula some 30-40 years from now? I'm 76 years old, and even though I graduated from university with maths as one of my majors, I don't (because I almost never use it).
by ur logic what are the odds that you have to SOLVE a quadratic formulae 30 years from now ?? this is litrally basic maths
for euations having large c value this is good
Great job! The method is powerful, try its generalization. For instance, -
-if you have x^3+bx^2+cx+d, and x1, x2, x3 are solutions, then x1+x2+x3=b, x1.x2 + x1.x3 + x2.x3 =c and x1.x2.x3=c.
-if you have x^4+bx^3+cx^2+dx+e and x1, x2, x3, x4 are solutions, then x1+x2+x3+x4=b, x1.x2 + x1.x3 + x1.x4 + x2.x3 + x2.x4 + x3.x4=c and x1.x2.x3 + x1.x2.x4 + x2.x3.x4=c and x1.x2.x3.x4 = e💙💜💙💜💙💜💙
I know this one. It is the same as Poshen lo’s.
Wow, just wow. This needs to be taught ASAP in schools.
I mean, there's a reason why I prefer completing the square.
Realizing this is completing the square
Brilliant. I am sure my 8th grade math teacher in my school will find this interesting.
It feels like this is just deriving the quadratic formula in a way.
All maths are similar in a way
I havent seen this before. Absolutely brilliant! I wish i knew this in the 90´s !
I have exams in 2 weeks, hopefully this is gonna help
Edit: finished the video, this is actually pretty sick😂
Same here, just subtract one week😅
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.
Since the ultimate goal is to find the value of x, I go straight to the quadratic formula, not caring what m and n are.
All too often, it's not factorable into integers, which means a lot of wasted time.
For example... change the 192 to 191.
It gets more complicated if "a" is not 1.
Try to factor this:
5x^2 + 73x + 191 = 0
@@jamesharmon4994 Check the discriminant first if there is a doubt.
@@jamesharmon4994 -- 191 is prime, so it does not factor. The trinomial is prime.
Haven't seen this method before, thanks for letting us know such an easy way to tackle any quadratic equation...😊
Thanks
I am 64 years old
but this the first time I know it
Just completing the square and using the solutions of the function equaling zero as the factors which is what they are.
I dont know if this is new for the west, but Indians have been always learning it like this :)
Thank god for this video, I've always struggled with something so basic
Please stop with the "wow", "they should teach it in school". I mean, it's ok, it works, but is it really usefull?
This is useless if you really understand the expressions of the roots: (-b+/-sqrt(b^2-4ac))/(2a). It's easy with the CANONICAL FORM. That's it, no need to push it with new expressions like m, n, c and so on.
PS: in France, we teach our students how to find the roots with a formula, how to find that formula by hand (the canonical form) and that's it. And if they know how to find roots by guessing, good for them.
In my experience, most students who struggle with multiplication struggle with fractions even more. It’s probably more efficient to stick to finding factors of ac first, and for more challenging problems, make an organized list of the factor pairs.
What does help from this method is to recognize that larger b values (compared to ac) mean that the numbers are further apart. This understanding helps with guess and check.
I agree, but then we are teachers. Teaching this method to the masses (especially Foundation Tier) would baffle them. Higher Tier GCSE students with strong algebra skills who enjoy generalising results could take to this way. They also appreciate being shown multiple methods and are able to select the most appropriate or efficient method when necessary.
This makes me furious about how much time I wasted memorizing the quadratic formula to handle "non-factorable" cases.
try applying it to 0.7*t^2-27.78*t+800=0 (that's an actual equations my students got when trying to find the time and coordinate where two accellerating objects meet). Try both methods. The quadratic formula is way superior, especially if the second question is what kind of a head start one of the objects needed so that the objects don't meet.
So, you can determine irrational factors without the quadratic formula?
This is literally the quadratic formula, though.
In truth, you just got a good teacher for the first time
You did *not* waste time. There is also completing the square.
Analytically, the extreme value (minimum or maximum) of the function f(x)= ax^2+bx+c is at 2ax+b=0, i.e. at x=-b/2a. The solutions of the equation are to the left and right of this, at an equal distance from this. In fact, the solving formula also expresses this.
I don't see the 'amazingness' in this, since you can do the same thing by solving the quadratic equation. ax²+bx+c=a(x-p)(x-q), where p and q are the roots of the quadratic.
Yes, but you have to see that before discovering the method! Lots of people won't - as proven by the fact that most don't!
I agree with you absolutely. Completing the square has been used to derive the quadratic formula. That is already systematic enough. If the ultimate objective is to get the solution, guessing and quadratic formula would do the job faster than this method. In addition, using the quadratic formula helps you to know immediately if the equation have rational roots or not (just by checking if b2-4ac>0 or
this is absolutely freaking amazing! thank you! i have always found the guessing method to be flawed. this is an awesome exact method.
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).
This actually really cool and helpful, I might use this next time instead of the quadratic equation
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.
Nice, this process is easier to follow than using the formula.
Really good as long as the formula is also presented and understood for use in purely algebraic manipulations
Ive been looking for a way to solve factoring quadratics without guessing for months now so thank you !
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
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*
Earlier in my days, we used to do prime factorisation of the constant which simplifies the guessing game with massive amount. But yes, this is amazing.
Great method! You can also use shri dharacharya formula.
Excellent and simple method, thanks for bringing it up
Ifk if this was taught in school, but I used to do this while calculating because I somehow discovered this theory myself 😂
Bro u are an absolute life saver the amount of time i can save instead of guessing is astronomical
Why can't you just use the quadratic roots formula? It doesn't need any guesswork. It will give 2 roots values (m,n) which you can write as (x-m)(x-n). This approach is literally that quadratic formula manipulated. (-b +- D)/2a. As you can see, the same number D (discriminant) is added and subtracted from -b. D is the playing the same role as "c" in this approach.
😂😂😂BRO U JUST LITERALLY APPLIED QUADRATIC FORMULA IT'S SAME
I APPRECIATE YOUR EFFERTS 👍👍