Here's a video I posted 4 years ago, when the channel was just starting: The Quadratic Formula - Why Do We Complete The Square? ruclips.net/video/EBbtoFMJvFc/видео.html I was still new to making videos as the channel was just starting. It took so long to do the animations, and I recorded the entire video in a single take so the video is not as polished. So I was overwhelmed by the positive comments, like: "This video should be presented in every high school and middle school algebra class. "
Great video and animations. Helpful for my cousin. ♥️ It's easier to remember than Brahmagupta's formula, but one thing, in engineering level, this method will take a hell of a time to solve.... Brahmagupta's formula will help at that moment. Brilliant idea anyways.
If you divide out the equation so "a" is 1, then the quadratic solution is: -½b ± sqrt(¼b²-c). The innovation is the normalization of a=1, means you can solve quadratics faster. This is presumably important for timed math competitions.
@@bramkivenko9912 The thing is, in *both cases* you're dividing two numbers by a. The only difference is if you're dividing by a at the start or at the beginning. This isn't any faster.
The whole comment section: This is "new"? Here in Germany/Russia/India/Greece/... we learn it in 8th grade. It's known as the "pq-Formel" / "Viète's theorem" / "Middle Term Split" / "S(um) and P(roduct) method"...
yup middle term split here in India....we learn it in 8th Standard and then we learn quadratic formula in 10th Standard....as we cannot solve every quadratic equation by middle term split (easily).
Precisely how I feel. I would rather memorize a formula than a bunch of steps. I feel most students I taught this to wouldn't remember the motivation for most of these steps and instead just memorize. It's better to just teach how to derive the quadratic equation, with variables and with numbers.
@@jamieg2427 but with the method we can clearly see the logic of solving it and it more intuitive. If we use formula that we memorize, we only can solve it but never had the deep understanding of why that formula works in the firsr place, and that is just like a robot,like here are the number now crunch that into this formula. I prefer the concept so that when we found a harder or unique problem, we still can solve it and not run out of ideas because we dont have a formula for it
Well of course it is. All methods must eventually lead to the correct answer, but some methods/formulae might just be easier/quicker/more intuitive to use for different people.
@@ravindrawiguna8681 I mean that's why you learn where the quadratic formula comes from how it's derived. But after you see that there's no need to pretty much derive it every time you want to use it. Just apply it directly.
My list of steps for both methods, separated into trivial parts: Formula Video 1. b^2 Divide a 2. 2a -B/2 3. Twice the result * c Square the result then - C 4. Square root the difference with step 1 Square root the result (roots are obtained here for both) 5. Simplify the fractions if applicable Notes: Steps 2 and 3 of the formula can be very quick but are necessary for each result. Whenever I've had to deal with the formula in (grade) school, I was always told to simplify the result as much as possible to receive full credit, otherwise why do any of the work to begin with. My college math courses didn't care about simplifying unless it was explicitly stated. Dividing the a can be done across steps 2 and 3 of the video method. When your work needs to be shown, there's fewer places for error because you've already performed the simplification along the way to the answer. You'll notice the steps are the same for both methods, but the division is done first instead of last.
@@NirateGoel 4ac is the same as doing 2a then that result * 2c. Again, I broke things down into trivial parts that mattered, hence why the subtractions are combined with the roots step. Calculating 2a is a necessary step in calculating 4ac, whose value is also required elsewhere in the formula. I even stated in the notes that this would be a reasonably quick computation compared to all the other steps.
Yeah actually we learned this before the other formula, but we only used it when there was no coefficient in front of x^2. They didn't tell us we could just facture out the coefficient :(
This is just 'completing the square'. Completing the square for a general quadratic equation is precisely how you derive the quadratic equation. Soooo... nothing new here.
Check again. It is the individual steps that is used in the process. 'Z' is a new variable in itself. completing square uses number functions and the discriminant method to get the roots
They both work, but this is definitely not completing the square. 2 ways to get the same answer, but I like this one more since it builds on what we know about factoring. To complete the square you try to get a polynomial from standard form to (x + n)^2 + k Point being that (x + n)^2 is a completed square, so you can move k to the other side and take the square root. With this method instead you want a polynomial to go from standard form to something like a factored polynomial. (x - r1)(x - r2) Multiplying that out to standard form gives Vieta's formula which gives a straightforward 2 equations and 2 variables -B = r1 + r2 C = r1 r2 That's made simpler by showing the roots are in the form r1 = -B/2 + z r2 = -B/2 - z So then it's just 1 variable and 1 equation.
The quadratic formula is easier to memorize. This method(in the video) is intuitive(easier for beginners) but still requires memorization of steps. The beauty of quadratic equation formula is the √(b^2 -4ac) part of it: This part tells you whether the roots are real, imaginary root or single root.(Discriminant) This formula also gives us the shape of the parabola that the equation forms. For beginners memorising the formula is difficult but as you dive deeper into mathematics, this formula looks elegant.
It's called the PQ-formula and it's way easier to remember than the Quadratic formula. You re-arrange the second degree polynomial into the form of: x^2+px+q hence to get the roots, simply plug in the coefficient "p" into the main pq-formula which follows below: x=-p/2±√((p/2)^2-q) If p < 0 then you'll have p/2 before the square root, it doesn't matter what what the whether p is negative or positive in the square root since it will always become a positive number after being squared in side the square root. If q < 0 then it will be written as +q in the square root and if q > 0 then it will be written as -q. For more detailed overview just check out the link of an image that I have attached below: images.app.goo.gl/fEm2U8BJXsa48Z7j9
"You don't have to memorize anything" Except a multi-step process with much more opportunity to forget some little, but vital, step. Way, way harder than just memorizing Brahmagupta's formula!!
I find it way easier to remember as - p/2 +- sqrt( (p/2)^2 - q) xD It's also way faster to tell if there's a solution in R for the equation without the need of a calculator since u can just approximate it x^2 + px + q
frankjohnson123 That makes absolute no sense. If two methods are identical, then they have the same room for error. That is implied by the definition of identical.
When I saw this, I thought for a moment: "I wonder if it's just going to be a complicated way to do the PQ-formula". A few minutes later, lo and behold.
nowonmetube ah I love the PQ-FORMEL (speak it out loud in a very german accent). Ich mag die Mitternachtsformel mehr als die PQ-Formel (auch wenn's irgendwie das gleiche ist...)
Yeah I remember I used to do it this way in 8th standard. My tuition teacher had taught me this and I clearly remember it was 8th grade coz that was the first time I joined a tuition. Those days were great!
You’re memorizing a whole method that explains exactly what the quadratic equation is DOING. In fact it sounds even harder to memorize as it’s wordier. It’s much easier to know the quadratic equation
@@theraytech54 It is not similar. It is exactly the same. Just all the individual steps combined into 1 formula. The derivation of the pq formula however is usually thought in a more geometrical way. But if you think of it, what is done here is not that different from "completing the square" in an algebraic way. I am not to say, that this derivation is bad or anything like that. It is a nice derviation. But to call it "new" is way beyond what it should be called.
@@wockhardt69 When you google remember Sundar Pichai an Indian is the CEO of it. Whenever you count money remember 0 is given by Aryabhatta an Indian. Trigonometry was given by Indians, Baudhayan theorem was taken by Pythagoras and took the patent. So, stop lecturing us, India is the best. AUM 🙏🏻 MAY BUDDHA, NARAYANA, SHIVA give you some buddhi (intellect) to think 😁
In India you never learn the quadratic formula till 10th class (high school) Before that the children are taught the product and sum method, but not motivated to generalise it
Inteseting mthod . But if the forget the quadratic formula, I will do completing the square which is not harder than this method x^2-8x+15=0 x^2-8x=-15 x^2-8x+16=-15+16 (x-4)^2=1 x-4 = 1 or x-4 = -1 x=5 or x=3
Zack West I agree. The ‘common’ teached way is easier to memorise. This formula is indeed easier for beginners, but you’ll need to meorise more steps, and that’s not what mathematicians want to do.
Mwexim yeah. There are already so many easy methods to find x for quadratics. If they’re going to find an easier way of memorizing types of functions they should do it for higher degree polynomials
I'm sorry to be butthurt over here but this is just like that time I thought I had figured out a general equation to calculate the sum of natural numbers only to realize some Gauss dude did that a few centuries back. All the more power to anyone helping people that don't yet know this. Lord knows math needs to be made fun for all.
When I was 12 we started doing geometry and learned pythagoras. I realized about connection between angles and opposite sides. I had full notebook of my findings, mostly ratios between longer and shorter side of right triangle. I also had some breakthroughs when I started drawing circles around right angle triangles. My teacher told me that I was on a good path and too keep working. It lasted for a month or so. 2 years later I started the high school, we started doing trig and it all flashed back to me. Basically I (kind of) reinvented tan function as a 12 year old.
Thanks sir. Your videos are great inspiration for me and my students to explore new horizons of maths as I have been also involved for preparing students for class8 maths Olympiad at my school level and we have won last year gold for our campus amongst 18 campuses participants.
I still find the method I've been tought over 30 years ago the easiest to remember and provide. What you need is to always remember two equations: (x+a)^2= x^2+2ax+a^2 and (x+a)*(x-a)=x^2-a^2. Simple example is x^2+6x+8=0 x^2+6x+9-1=0 (x+3)^2-1=0 (x+3-1)(x+3+1)=0 (x+2)(x+4)=0 So we have the solution.You may notice, that if you try to solve a general equation ax^2+bx+c=0 using this method, you will get the quadtratic formula, which was always hard to remember to me...
People should stop calling this a new method. Anyone who has graphed a parabola can see that the roots are going to be of the form m +- n where m is the x-coordinate of the vertex of the parabola, given by m = - B/2 in your notation. Yes, in the centuries people have been solving quadratic equations, people have tried plugging in - B/2 +- n and solving for n. If they know a little more they'd set the product (-B/2 + n)(-B/2 -n) = C and solve for n. Seriously, this needs to stop being treated as something new. All these fancy degrees being shown off just make the situation more infuriating.
In Italy we are taught a similar thing as a shorter formula for quadratic equations when b is even. If a=1 (which is in this case done by dividing everything by a) you get exactly that formula. Also, the new formula is not so easier to remember than the normal one
I’m italian too, and i can tell that this is off topic. I mean, it is true that you can simplify the quadratic equation but this has nothing to do with the method explained in the video
It is not just similar to what's taught in Germany. It is exactly what I was taught in my German high school. But our teachers explain it easier😂 You can use the stuff in the square root for finding out if the function is a passant, tangent, or secant line as well. In the complex method the factor a is just integrated, but that confuses students and/or lead to small mistakes you do even if you know how it's done.
This is also known as the PQ-Formula, we were told to memorize it (in my case). However I've also seen students remember this formula with some rules of thumb. (0. make sure the equation is x²+px+q=0, and not let's say 2x² or so) 1. divide the middle term (p) by -2 2. add the +- symbol and draw the square root, then square the term from step 1 and put it inside the square root 3. subtract q (also inside the square root) formula will look something like this x = (-p/2) +- sqrt((p/2)²-q)
I'am from Germany and we get taught the first part of this method under the name "PQ"-Formla. x = - (p/2) +- sqrt( (p/2)^2 - q ) I always were mad that we not got taught about the abc-formula because for some equations the division with the a factor wasn't that easy. This method really helps me, because I like the quadratic approach is rasiert than the root approach.
From my experience as a math teacher, even though students know both methods, most will prefer Brahmaguptas Formula because they can just plug everything in and the solution falls out. They will even use it if the solution basically jumps at them. Mostly students for which math classes aren't a constant state of hardship will take a second to consider what might be a quicker or easier way. Cheers to all the people in the comments who are vocal about finding this trivial. Thanks for letting everyone know.
Any method that solves a quadratic equation can be used to prove Sridhacharya's formula, since the resulting solutions will be equivalent. The way we should judge methods like these is by how easy they are to remember and execute. This method, while more intuitive and easier to remember than Sridhacharya's formula, takes a few more steps to do. Therefore, neither is inherently better. They are both useful, for different circumstances and mathematicians.
Cripes, any article I read on this mess was incomprehensible. This video makes it make /a lot/ more sense. Thanks Presh!. I genuinely don't know how much of this process and or the cheat steps I may have been taught in school. Been too long and I figured out how to get my ti-89 to solve the tricker ones for me quite early.
I saw this article a few weeks ago, nice to see a video on this :) Though, my biggest qualm with this method is the fact that if the coefficient A is not equal to 1 you have to factor the function into that form which can result in B and C becoming cumbersome fractions. Other than that it's a really nice way to be able to think about the roots and the mathematical intuition behind them.
This method also allows you to find the middle (aka top value) of the function. I knew there was possibility to find x-values through addition/subtraction of the x value of the top
I feel like the Quadratic formula is much more useful. Having studied maths until the end of high school, and now having calculus and algebra in university, I feel like having the discriminant is quite useful. As you delve deeper into maths, the formula is more useful than this alternative method which might seem more intuitive for beginners.
I wish i knew this method for some of my pure maths classes in diff eqs when the type of roots didnt really matter than much and we were forced to solve them no matter what. (And we werent allowed calculators so i think this method woulda been easier to do by hand than the quadratic eqn)
R2D2 from Star Trek I prefer the quadratic formula. It’s a formula and it’s good to solve any quadratic equation. I’ve used it so many times it’s almost impossible for me to forget.
In that video at 1:10, Achrotone noticed the typo "back-to-bank" which is now corrected: ruclips.net/video/lDbQA4euAbY/видео.html&lc=UgwYAupZwmfkh8Nfred4AaABAg
@Shailesh Kumar true, and the formal description is easy for everyone to do it theirselves. I'm from Germany and we learn this formula from 5th grade (10yo) and I never knew this wasn't taught in the entire world 🤷♂️
May I complement Po-Shen-Loh on a brilliant exposition of a bit of basic, fundamental mathematics. A good example of plain honesty, simple truth and easy understanding. I first watched the video a day or so ago and it was only a day later that the penny dropped. As per Leonhard Euler's 'Elements of Algebra' (x-a)(x-b) = x^2 - (a-b)x + ab where as we know a & b are the roots of the quadratic. Taking (a+b)^2 and (a-b)^2 [ i.e. props. 4 & 7 from book 2 of Euclid's 'Elements' ] then expanding and subtracting we get the answer 4ab hence we have (a+b)^2 - (a-b)^2 = 4ab. This is a theorem, prop.8 of book 2 of the 'Elements' and for some unknown reason demoted to a RULE alias 'The Quarter Squares Rule'. After a bit of simplification we end with [(a+b)/2]^2 - [(a-b)/2]^2 = ab. The algorithm given in the video then amounts to [(a+b)/2]^2 - ab = [(a-b)/2]^2 which taking the square root leaves (a-b)/2. So (a/2+b/2+a/2-b/2)=a & (a/2+b/2 -a/2 -[-b/2])=b. It is still highly commendable that the 'QSR' has been derived by another route and has been admirably utilised for the factoring of quadratic equations. What I find a bit astounding and some what sad is that together with the hits on the 3blue1brown and MindYourDecisions videos on the same topic a combined total of around 1,453,000 views no one else seems to have spotted the connection. Finally if we change a & b to x^m & x^n then the answer (ab) becomes x^(m+n) hence all integers raised to a power above the second are the difference of two squares . Further more the bigger the power the more DoS solutions there are for any one integer raised to that power! What does this mean for Fermat's Last Theorem.
I am from Germany. I'm 36 years of age and even my mom learned this "new" method at school. In Germany every student older than 16 knows this method as the "PQ Formula". And this is since...I don't know...Kaiser Wilhelm, I think 😂😘
4:06 I'm German and here the formula isn't ax*2+bx+c but ax*2+px+q. Then we can build the next formula: X1,2= -p:2 +- the root of ((p:2)*2 -q). If you solve this you get the two solutions. Then you can check them with the "Satz des Vieta" like you showed in the video. X1+X2=-p and X1*X2=q. It's very simple if youve done it often enough.
actually, when you're dealing with parameters and very small numbers which you mostly will it just becomes more complicated, thats why you teach the formula, this method is only good for convenient numbers in easy problems.
this is exactly the quadratic equation and how it came to be. when first taught the teacher shows how we came about this formula so we can use it confidently knowing why it works
Simple visual depictions of completing the square can readily be found online and are far easier to understand for factoring novices. However, this is still interesting. Thank you for posting it.
Here's a much more interesting approach in my opinion. A neat result is that for a quadratic Q with turning point T we have: Q(T+-t)=Q(T)+at^2. If the solutions are T+t and T-t, we obtain at^2=-Q(T) which gives t=sqrt(-Q(T)/a). Thus the solutions are x=T+-sqrt(-Q(T) /a)
I believe this would make more intuitive sense starting with the symmetric property of a parabola, specifically that 2 solutions equidistant from some value m would show f(m+d) = f(m-d) = 0. In other words, f(x) = a(x-(m+d))(x-(m-d)). Then expanding would result in f(x) = a(x^2 - 2mx + m^2 - d^2). So as long B = neg (2m) or m = neg(B/2) and C = m^2 - d^2 or d = sqrt(m^2 - C), then we have our solutions of neg(B/2) (+/-) sqrt(m^2-C). The scaling affect of 'a' has no impact here, as sliding values of 'a' does not affect zeroes. Of course, the assumption of solutions here does require FTA, but Gauss took care of that for us :)
This is just quadratic formula with extra variables and steps, @4:09 it's literally exactly the same thing only it has been chosen to write C in place of c/a and B instead of b/a. I feel like I would be less upset if this wasn't called "a new method" (i.e a way to completely avoid the quadratic formula) but instead called something like "a new way to derive the quadratic formula" or something similar.
At 3:27 in the video the righthand side lost the negative signs on B. I realize that this doesn't matter, essentially you left out a step where you multiplied both equations by negative 1, but some people might miss this. Note that the "new way" works with equations where a=1. If you start with this assumption then the standard formula for solving quadratic equations becomes somewhat simpler. If you look at the steps in the "new way" and start writing them down in a way to solve for x then the result looks suspiciously like the standard formula. I did this when I first saw one of these videos pop up on the internet recently. It appears that you took the standard formula, dropped out the a terms since a=1, and moved the 2 from the denominator up into the radical. So rather than consider this a "new way" to solve a quadratic equation, it should be viewed as the derivation of the standard formula. At 3:6 in the video the equations on the righthand side of the screen are exactly the standard formula with the 2 in the denominator moved around. Granted if a person was alone on a desert island and couldn't remember the standard formula, they could use the "new way" to reason through to a solution. I'm an engineer and I have to solve quadratic equations frequently. It is second nature for me to plug the terms into the standard formula to get the solution.
This is just a derivation of Brahmagupta's quadratic formula. At the 4:10 mark you can substitute back for B=b/a and C=c/a and recover the standard formula with some straightforward factorisation. The one thing I have learnt from the video is how the Brahmagupta's formula is derived. Something that should be taught alongside memorizing a formula (one that is still stuck in my head after more than 35 years).
In Germany, we learned in school first something we called "quadratische Ergänzung", which could be translated to quadratic completion. So, one adds and subtracts a number, so that the formula becomes (x-d)^2+e=0. In the next lesson, we derived the quadratic formula out of this.
Fun Fact: We call the quadratic formula "midnight formula" because it is so important, that you must be able to recall it, even when you are woken up at midnight.
@wise ol' man yeah that's what I meant to say, but I can't see how it's easier than just knowing the quadratic formula, it feels like u added some extra steps to it
In India, even 8th class students know about this method as factorization method. We have even made various tricks to make this way ever faster. Hence, nothing new for me. 😂
x^+11x+28=0 has two roots (-4 and -7), because 4 and 7 are factors of 28 and add up to 11. Now try using Po Shen Loh's method. Our mid-point has to be -11/2. This has to be squared, giving 121/4. Now we must subtract 112/4 (that's 28) to get u^2 (that's 9/4). Now we take the square root to get u (that's 3/2). Now we must subtract 3/2 from -11/2 (that's -8/2=-4) to get our first root. Then we must add 3/2 to -11/2 (that's -14/2=-7) to get our second root. That's an awful lot of work involving positive and negative fractions for such a simple quadratic. Try it!
But wait......... Looking the -B/2 +/-z, wouldn't it actually also be that z= sqrt(B^2 -4C)/2? Considering that the a of the quadratic formula in this situation is always =1 then this is still the quadratic equation.(I mean, duh) If B^2/4 - z^2 = C, then z^2 = B^2/4 -C or (B^2-4C)/4. Then z = +/-sqrt(B^2-4C)/2. That's an interesting way to think about it. So in the end, this method is simply a quadratic formula through a new perspective that are more intuitive, or simply also a way to make sense of what the quadratic formula actually represents. In other words, quadratic formula can be thought of as the average of B +/- an identity that is related to B and can give a product of C.
Recently ran into a YT video showing something very similar to this method (starting with the middle x-term, then dealing with the constant term), with the 1st example problem: x² - 60x + 899 = 0. Heck, who wants to deal with the large numbers associated with factors of 899, or have to calculate the discrimination & its square root from the quadratic formula??! A bit of inspection, and we easily notice that 30² = 900. If we write 899 = 900 - 1 , then our quadratic becomes: x² - 60x + 900 - 1 = 0, we can write the 1st 3 terms as a perfect square, and the constant 1 can also be written as the perfect square 1² (if the resulting constant q isn't a perfect square, then just write it as one: √(q²) ) (x - 30)² - 1² = 0 , which is a difference of two squares, and can be written as the product of the sum & difference of the square roots: [(x - 30) + 1][(x - 30) - 1] = 0 , and, on simplification becomes (x - 29)(x + 31) = 0 , then, the zero a product property yields solutions x = 29 and x = -31.
Another way of expressing this idea is that we are doing a substitution x = z - B/2. This is cool because substitution is commonly used in much more advanced maths too.
@@mvpistakenbyme818 its the same tho. Try using CTS in the problem and you'll see the difference. Even QF is CTS with a memorizable formula. There is generally only 2 ways to solve Quadratic Equations. The factoring method and Completing the Square Method 😂🤣
That was my thought - it’s the method of “completing the square” which then works out to be the quadratic formula anyway. In exams when asked to derive the quadratic equation, this is the method I used
@@keyboardcorrector2340 This is nothing.. We start perp for IIT JEE from junior kg and till 4th we're on top of the world.... Sarvashaktishaali Gaitonde🤣
I would set one separate variable m to be the midpoint between the two roots, m=-B/2, and d to be the distance between each root and m. Then the roots are m+-d, and their product C is m^2-d^2, so d^2=m^2-C. This yields the roots via a formula that looks much simpler, m+-sqrt(m^2-c).
To all those saying that this is not new, and has been taught to schoolchildren I've seen most comments point to India. As a Indian student, YES we were taught that if you find 2 numbers that multiply to C and add to B, you have your factors. (called the split the middle term method) This method is about HOW TO find those 2 numbers. After that its normal. But what has NOT been taught to us(simply because in exams they don't give HUGE numbers) is HOW TO find those numbers. Like yes, for the equation: x^2 + 11x + 28 you can easily find out that the roots are 4 and 7. How about for this: x^2 -60x + 899? Can you still do it? Except the method Po-Shen Lo told here CAN actually easily solve the 2nd one You know that 1/2 of 60 is 30. You can just simply do 30^2 = 900 in your head, and see that the diff is 1(well, you see that the square of the diff is 1, but its the same thing). Which means the roots are 29 and 31. And this is also IGNORING all those quadratic equations that DO NOT have integer roots. Like look at this eqn 2x^2 -16x + 26. most people will divide by 2, that's a integral step in every method. So this comes out to x^2 -8x + 13. Can you split the middle term easily here? Nope. Except if you used the method shown above, you can easily solve this It simply comes out to 16 - (diff)^2 = 13, which you can easily see means that the diff^2 is 3, and the diff is root 3. Which means your roots are 4 +- root 3. Like yes, for easier smaller equations this is not that time saving. But for those bigger ones with numbers you arent even sure are prime, this helps a ton. And if you ACTUALLY understand WHY everything works(there's a good vid on it by Outlier.org) You can literally do it in your mind.
I've been to school in Germany and I can't recall ever having learned a method to solve a quadratic equation that involves guessing. I could never remember the ABC formula when I was in school so I used the completion method and it is still my preferred method today.
But... why would I do all of that process when is easier and faster to just do the formula? It's not really that difficult to learn and it's way quicker to remember 1 simple equation that all of that method.
Because when I was in school, I had to show my work. "That process" is the same steps for both methods, but the division is done first instead of last.
I would say it would be important to know how something work. Take daily life as example, yes we know we turn on the stove fire come out, but it would be better to know more about why is there fire or things related like combustion. Another example would be we know stepping on the paddle the car would move forward, but would be better to know about how an engine work.
At 9th grade we were taught three methods to solve quadratic equations. Out of three ,two are described in the video. If you know the third one then like the comment😁😁😁😁😁😎😎 Hint: alpha squared /b1c2-c1b2
I prefer to factorise the constant and factorise the sum of the coefficients knowing that the factorisation of the coefficients is 1 greater for both factors. X^2 + 5X + 6 = 0 Constant: 6 Coff Sum: 12 6=2x3 12=3x4 Done. (X+2)(X+3)
Love the content Presh, but could you please do videos in dark mode? It hurts my eyes when I see your videos at night. Inverting the colours should suffice
No, because completing the square doesn't utilize the concept of finding two numbers that are symmetrically positioned on each side of a given "average" m . Completing the square: re-writes the quadratic equation into the form (x+p)² - q = 0 This method: re-writes the quadratic equation into the form (x-u)(x-v) = 0 where u = m+d and v = m-d , for some fixed values of m and d. However, they are related by the insight that (x-p)² - q = 0 can be directly re-written as (x+p-√q)(x+p+√q) = 0 which means p coincides with -m , and q coincides with d² . (And of course they must be somehow related, because they are solving the same type of equation.)
I was honestly watching the video to see a new way to solve quadratic equations, but while explaining this "new" way, on time 3:50 you actually gave the exact same quadratic equation I have been using all time long!!!!!. The only thing this "new" way does is to use 2 formulas instead of 1. The only value I see is that you are able to get some results without calculator, which is cool, but after that, not very new.
This is basically the completing the square method with a little more clear teaching. At the end, you do reach the quadratic formula. This is just a derivation for the formula itself.
Here's a video I posted 4 years ago, when the channel was just starting: The Quadratic Formula - Why Do We Complete The Square? ruclips.net/video/EBbtoFMJvFc/видео.html
I was still new to making videos as the channel was just starting. It took so long to do the animations, and I recorded the entire video in a single take so the video is not as polished. So I was overwhelmed by the positive comments, like: "This video should be presented in every high school and middle school algebra class.
"
Great video and animations. Helpful for my cousin. ♥️ It's easier to remember than Brahmagupta's formula, but one thing, in engineering level, this method will take a hell of a time to solve.... Brahmagupta's formula will help at that moment. Brilliant idea anyways.
Keep the initially assumed values of B and C in the final result and you will come back to the same old quadratic formula
@@trailokyatripathy4341 yep.
MindYourDecisions These come from the vietta formulas . They’ve been known for years 🤷🏼♂️
MindYourDecisions
I wish I knew this 2 year ago
In India (and many other countries) there are three ways to do it:
1. Split the middle term
2. Completeting Square Method
3. Quadratic Formula
They( American ) just changed our well known Quadratic formula 😅 . For just getting some credit !
i think that's world wide
Correct
I hate completing square method
@@yasirarafat7654 it's ez just watch some yt vids
This is literally a derivation of the quadratic formula
If you divide out the equation so "a" is 1, then the quadratic solution is:
-½b ± sqrt(¼b²-c).
The innovation is the normalization of a=1, means you can solve quadratics faster. This is presumably important for timed math competitions.
I thought common derivation of the quadratic formula is by completing the square. Of course the two methods are identical but at least I’m new to this
@@bramkivenko9912 The thing is, in *both cases* you're dividing two numbers by a. The only difference is if you're dividing by a at the start or at the beginning. This isn't any faster.
@@Felixr2 I understand, but I think a person can perform this faster while less likely to introduce errors. To each his own.
It is the other way around
The whole comment section:
This is "new"? Here in Germany/Russia/India/Greece/... we learn it in 8th grade.
It's known as the "pq-Formel" / "Viète's theorem" / "Middle Term Split" / "S(um) and P(roduct) method"...
yup middle term split here in India....we learn it in 8th Standard and then we learn quadratic formula in 10th Standard....as we cannot solve every quadratic equation by middle term split (easily).
@@shyamparihar4071 how is this the same as "middle term split" method? He literally used that in the starting of the vid followed by the "new" method
Also here in Nepal
And China
I’m in Canada and we learnt this in grade 7😂😂
Old: chicken produce eggs
America's latest: eggs are produced by chicken .
🤣🤣🤣🤣🤣
In Europe and US our education is not that hard
You know it's a Chinese guy that came up with this, right?
😂 🤣 Lmfao... Underrated
What does that mean?
It ends up being the same thing, except that instead of memorizing a formula, you're memorizing a method that ends up in the same formula
Precisely how I feel. I would rather memorize a formula than a bunch of steps. I feel most students I taught this to wouldn't remember the motivation for most of these steps and instead just memorize.
It's better to just teach how to derive the quadratic equation, with variables and with numbers.
@@jamieg2427 but with the method we can clearly see the logic of solving it and it more intuitive. If we use formula that we memorize, we only can solve it but never had the deep understanding of why that formula works in the firsr place, and that is just like a robot,like here are the number now crunch that into this formula.
I prefer the concept so that when we found a harder or unique problem, we still can solve it and not run out of ideas because we dont have a formula for it
Well of course it is. All methods must eventually lead to the correct answer, but some methods/formulae might just be easier/quicker/more intuitive to use for different people.
@@ravindrawiguna8681 Most students won't need to have a deep understanding and they don't want it either.
@@ravindrawiguna8681 I mean that's why you learn where the quadratic formula comes from how it's derived. But after you see that there's no need to pretty much derive it every time you want to use it. Just apply it directly.
In the end, we end up with the same process but in a different route.. but it does look more lengthy than the usual way tho..
Very true.
My list of steps for both methods, separated into trivial parts:
Formula Video
1. b^2 Divide a
2. 2a -B/2
3. Twice the result * c Square the result then - C
4. Square root the difference with step 1 Square root the result (roots are obtained here for both)
5. Simplify the fractions if applicable
Notes:
Steps 2 and 3 of the formula can be very quick but are necessary for each result. Whenever I've had to deal with the formula in (grade) school, I was always told to simplify the result as much as possible to receive full credit, otherwise why do any of the work to begin with. My college math courses didn't care about simplifying unless it was explicitly stated.
Dividing the a can be done across steps 2 and 3 of the video method. When your work needs to be shown, there's fewer places for error because you've already performed the simplification along the way to the answer.
You'll notice the steps are the same for both methods, but the division is done first instead of last.
Expect for the -b you were unfair with steps 2&3 4ac is one step not 2.
@@NirateGoel 4ac is the same as doing 2a then that result * 2c. Again, I broke things down into trivial parts that mattered, hence why the subtractions are combined with the roots step. Calculating 2a is a necessary step in calculating 4ac, whose value is also required elsewhere in the formula. I even stated in the notes that this would be a reasonably quick computation compared to all the other steps.
A: The quadratic formula is not difficult.
B: This method isnt easier.
It is easier if u get good numbers
its been there for years its easier and in india its more famous
@@anmolsekhon3545 which one is easy for quadratic formula is easy
This one has less to remember and it's easier to use without writing anything down
It is harder. It’s the old method with the added thing of dividing by a.
I have known this formula as the p q formula
And yes, I am from Germany
Exactly bro
Here in Germany we learn it as a quite simple formula in 9th grade. I learned it in 1999 and still teach it that way.
Yeah actually we learned this before the other formula, but we only used it when there was no coefficient in front of x^2.
They didn't tell us we could just facture out the coefficient :(
Same in Czech republic:) p,q as well. And imho it works well only on nice behaving a, so I prefer the standard formula anyway.
Sweden too bro, we calle it PQ-formeln just like on Germany.
Presh : Mathematicians found a NEW way for solving quadratic equations
The Comments : Nope
This is just 'completing the square'. Completing the square for a general quadratic equation is precisely how you derive the quadratic equation. Soooo... nothing new here.
Check again.
It is the individual steps that is used in the process. 'Z' is a new variable in itself. completing square uses number functions and the discriminant method to get the roots
@@Abhinav-ss3th no...unless we have a different definition of "completing the square"
Tom Sharpe it's a little different than completing the square
The original is like finding a product and a sum
And the new one is finding a power 2
They both work, but this is definitely not completing the square. 2 ways to get the same answer, but I like this one more since it builds on what we know about factoring.
To complete the square you try to get a polynomial from standard form to
(x + n)^2 + k
Point being that (x + n)^2 is a completed square, so you can move k to the other side and take the square root.
With this method instead you want a polynomial to go from standard form to something like a factored polynomial.
(x - r1)(x - r2)
Multiplying that out to standard form gives Vieta's formula which gives a straightforward 2 equations and 2 variables
-B = r1 + r2
C = r1 r2
That's made simpler by showing the roots are in the form
r1 = -B/2 + z
r2 = -B/2 - z
So then it's just 1 variable and 1 equation.
This is literally Diophante's method, step by step. "Ancient mathematicians didn't do it like this" well yes they did
No
Lol tartaglia, Ferrari, cardano, lagrange, they all knew this and did it better honestly hahaha
Shri dharacharya invented this... Not some diophante guy
@@daywill8849 They are talking about the new method, not the old one
@@daywill8849 hey@Daywill,He do not invented it he discovered it.....AND he is not the only one...😇
The quadratic formula is easier to memorize. This method(in the video) is intuitive(easier for beginners) but still requires memorization of steps.
The beauty of quadratic equation formula is the √(b^2 -4ac) part of it:
This part tells you whether the roots are real, imaginary root or single root.(Discriminant)
This formula also gives us the shape of the parabola that the equation forms.
For beginners memorising the formula is difficult but as you dive deeper into mathematics, this formula looks elegant.
Nerd
Err, every parabola has the same shape. The only difference is that they are scaled with different constants.
b^2-4ac is called as discriminant not determinant!!
It's called the PQ-formula and it's way easier to remember than the Quadratic formula.
You re-arrange the second degree polynomial into the form of: x^2+px+q
hence to get the roots, simply plug in the coefficient "p" into the main pq-formula which follows below:
x=-p/2±√((p/2)^2-q)
If p < 0 then you'll have p/2 before the square root, it doesn't matter what what the whether p is negative or positive in the square root since it will always become a positive number after being squared in side the square root.
If q < 0 then it will be written as +q in the square root and if q > 0 then it will be written as -q.
For more detailed overview just check out the link of an image that I have attached below:
images.app.goo.gl/fEm2U8BJXsa48Z7j9
@Adam Romanov My point was it shows where the graph cuts on the x-axis. I thought it was implied in my answer. My bad.
3:45 Nice!
You just derived the pq formula!
This is the pq formula.
Bruh
"You don't have to memorize anything" Except a multi-step process with much more opportunity to forget some little, but vital, step. Way, way harder than just memorizing Brahmagupta's formula!!
At this point you might aswell make an algorithm for finding the root for any type of function.
I find it way easier to remember as - p/2 +- sqrt( (p/2)^2 - q) xD It's also way faster to tell if there's a solution in R for the equation without the need of a calculator since u can just approximate it
x^2 + px + q
I was always told it was called *The* Quadratic Formula.
@@piman9280 Hmm, it was discovered by that man, so its sometimes also called brahmagupta's formula
Y’all like “only Americans didn’t know this” but i got taught this before the quadratic formula
True
Same lol
Me2
Same lol A and B were alpha and beta.
Lol same they taught us this method in 8th grade and quadratic one in 9th grade ig
This method really only is new for Americans 😂
@@C4pt41nN3m0
Yeah cause this method is way easier to remember and can be used in more situations 👌
It’s not taught because it’s just the same method with more room for error
frankjohnson123 That makes absolute no sense. If two methods are identical, then they have the same room for error. That is implied by the definition of identical.
Videos like these really imply that americans aren't very intelligent 🤔
@@C4pt41nN3m0 esatto, si fa quando B è pari
"Mathematician finds a new and easier way to calculate quadratic equations"
Yeah no it definitely wasnt him that found it
And it's definitely not easier lol.
@@Ignasimp y tho
....It was tho
@@genieinthepot2455 however important they were, they did not invent factoring
When I saw this, I thought for a moment: "I wonder if it's just going to be a complicated way to do the PQ-formula".
A few minutes later, lo and behold.
To me actually the beginning formula looks exactly like the pq-Formel, say what? O_o
God damn you and your pun
nowonmetube ah I love the PQ-FORMEL (speak it out loud in a very german accent). Ich mag die Mitternachtsformel mehr als die PQ-Formel (auch wenn's irgendwie das gleiche ist...)
Americans- Whoa that’s a new way to solve Quadratic Equations !
Meanwhile in India- Grade 8 students solve via this method.
Bhai konse school m ho ?
grade 10*
@@SAsquirtle grade 8
@@kavithajagdeesha8999 10 for cbse
Bruh I'm in 8th and I do the middle term split
The "New Way" in this video is what they teach us in school in 9th grade.
Ha ha ha sooooo true..👍
Exactly
It was taught in 7th to us
Yeah I remember I used to do it this way in 8th standard. My tuition teacher had taught me this and I clearly remember it was 8th grade coz that was the first time I joined a tuition. Those days were great!
@@siddharthdoshi4858 agree🙋🏻♀️
You’re memorizing a whole method that explains exactly what the quadratic equation is DOING. In fact it sounds even harder to memorize as it’s wordier. It’s much easier to know the quadratic equation
3:46 I recognized the classic quadratic formula when he got to ±√ thing.
No, it's not. It's easier to know a formula but it's better to learn a method. Always. That's because embedded within the method is the formula.
Should be great to teach kids about this instead since mathematics is all about following logic and thinking not purely memorizing.
Here in Germany this method is very well known as the "pq-Formel" ("pq-Formula") and definitely nothing new.
It's only similar and he talks about it
Or "Vieta's theorem"
@@theraytech54
It is not similar. It is exactly the same.
Just all the individual steps combined into 1 formula.
The derivation of the pq formula however is usually thought in a more geometrical way. But if you think of it, what is done here is not that different from "completing the square" in an algebraic way.
I am not to say, that this derivation is bad or anything like that. It is a nice derviation. But to call it "new" is way beyond what it should be called.
When you say "everyone" is talking about this new formula, presumably you mean the tiny minority of us who go to RUclips to muse about maths.
Mathematicians: This is new way of solving quadratic equations
Indian teachers (teaching this to students for decades): Hold my chalk
Indian teachers op 😂😂
Yah -b+_(b²-4ac)^½
--------------------------
2a
@@roastedpeanuts694 this is a certified good classic
@@wockhardt69 And still 45% of your NASA employees are Indian, PM of UK is Indian. 😂😂😎😎
@@wockhardt69 When you google remember Sundar Pichai an Indian is the CEO of it.
Whenever you count money remember 0 is given by Aryabhatta an Indian. Trigonometry was given by Indians, Baudhayan theorem was taken by Pythagoras and took the patent.
So, stop lecturing us, India is the best. AUM 🙏🏻
MAY BUDDHA, NARAYANA, SHIVA give you some buddhi (intellect) to think 😁
In India you never learn the quadratic formula till 10th class (high school)
Before that the children are taught the product and sum method, but not motivated to generalise it
Here in Vietnam, students learn the quadratic formula in the 2nd semester of 9th grade. It's pretty close.
Yeah and this method is called splitting the middle term
I am from india and i learnt quadratic formula in 9th grade
@@tumhregfkahusband8725 he is talking about cbse and not icse or state board
Same here in Sri Lanka
Inteseting mthod . But if the forget the quadratic formula, I will do completing the square which is not harder than this method
x^2-8x+15=0
x^2-8x=-15
x^2-8x+16=-15+16
(x-4)^2=1
x-4 = 1 or x-4 = -1
x=5 or x=3
This is the method that should be in the video ! 😂
"Complete squares" is like the ancient proof for life, time and everything! :)
yeah, our algebra professor showed us that way
Sami can someone explain where the 16 came from?
Sami is it a method to make -15 become 1 or 8x2
"an easier way to find the roots." :/
Zack West I agree. The ‘common’ teached way is easier to memorise. This formula is indeed easier for beginners, but you’ll need to meorise more steps, and that’s not what mathematicians want to do.
Mwexim yeah. There are already so many easy methods to find x for quadratics. If they’re going to find an easier way of memorizing types of functions they should do it for higher degree polynomials
@@mwexim7132 This way at least you understand what's going on. Memorizing a formula isn't learning much...
@@Ibakecookiess you don't really understand much lol. It's literally the way you derive the formula, nothing else
@@zwest808 The problem is, it is proven that for 5th degree polynomials, you can't find a method for discriminants, so that's indeed interesting.
I'm sorry to be butthurt over here but this is just like that time I thought I had figured out a general equation to calculate the sum of natural numbers only to realize some Gauss dude did that a few centuries back. All the more power to anyone helping people that don't yet know this. Lord knows math needs to be made fun for all.
When I was 12 we started doing geometry and learned pythagoras. I realized about connection between angles and opposite sides.
I had full notebook of my findings, mostly ratios between longer and shorter side of right triangle. I also had some breakthroughs when I started drawing circles around right angle triangles.
My teacher told me that I was on a good path and too keep working. It lasted for a month or so.
2 years later I started the high school, we started doing trig and it all flashed back to me. Basically I (kind of) reinvented tan function as a 12 year old.
As a child what i did
Sum of infinite nos is -1/12 and also general formula is n(n+1)/2 and equal them.
@@aarushrathore1276
lim(n -> infinity) {n(n+1)/2} = -1/12
I remember that some years ago I had discovered that a²=(a-1)²+a+a-1
Eg 6²=5²+6+5.
Then I realized it was just an "application" of (a+b)²
Me too as 7 years old figured out to calculate sum of continuous natural numbers by adding first and last term and for others..
Nothing new, already using it for last 3 years
So you took your content from such places and then teach that to your students on your name
It's a shameful act
Hii sir
Thanks sir. Your videos are great inspiration for me and my students to explore new horizons of maths as I have been also involved for preparing students for class8 maths Olympiad at my school level and we have won last year gold for our campus amongst 18 campuses participants.
This approach looks more time demanding and it's not a new thing. I learnt in back in Secondary school in late 90s. And in Nigeria🇳🇬
I gave a LIKE just because the last line: and in Nigeria...
@@tamirerez2547 mentioning Nigeria makes any comment funny!
Can you make a video on solving cubic equation without guessing any solution. I want you to do that because your explanation is good.
You don't guess in cubic there are formulas and other methods to solve them.
x^2+bx+c = 0
x^2+bx+(b/2)^2-(b/2)^2+c = 0
(x+b/2)^2+c-(b/2)^2 =0
That's even easier than remembering the formula, cause this is just common sense
@@vasileiospapazoglou2362 yeah but that's so complicated, and guessing the roots seems like a better method to go, although not really effective
There is the cardanic formula for this.
@@0000-z4z suppose I forgot the formula. Now there should be a straight forward method to find the solution.
I still find the method I've been tought over 30 years ago the easiest to remember and provide. What you need is to always remember two equations: (x+a)^2= x^2+2ax+a^2 and (x+a)*(x-a)=x^2-a^2.
Simple example is x^2+6x+8=0
x^2+6x+9-1=0
(x+3)^2-1=0
(x+3-1)(x+3+1)=0
(x+2)(x+4)=0
So we have the solution.You may notice, that if you try to solve a general equation ax^2+bx+c=0 using this method, you will get the quadtratic formula, which was always hard to remember to me...
3:56 Substitute the value of B and C as -b/a and c/a and voila, you have the quadratic formula :)
Bruh, I was so utterly disappointed watching this video, thought he was gonna talk about some revolutionary way of finding roots
I had never been taught this before, but now that I have seen it, it seems so obvious. Well done explanation of this process.
People should stop calling this a new method. Anyone who has graphed a parabola can see that the roots are going to be of the form m +- n where m is the x-coordinate of the vertex of the parabola, given by m = - B/2 in your notation. Yes, in the centuries people have been solving quadratic equations, people have tried plugging in - B/2 +- n and solving for n. If they know a little more they'd set the product (-B/2 + n)(-B/2 -n) = C and solve for n. Seriously, this needs to stop being treated as something new. All these fancy degrees being shown off just make the situation more infuriating.
In Italy we are taught a similar thing as a shorter formula for quadratic equations when b is even. If a=1 (which is in this case done by dividing everything by a) you get exactly that formula.
Also, the new formula is not so easier to remember than the normal one
I’m italian too, and i can tell that this is off topic. I mean, it is true that you can simplify the quadratic equation but this has nothing to do with the method explained in the video
YOLO Zemp the topic is indeed different and we use different demonstrations, but the formula is still the same
In poche parole hanno scoperto l'acqua calda
lorenzo timpone esattamente 😂
Si, la formula del “B Mezzi” con il B pari
It is not just similar to what's taught in Germany. It is exactly what I was taught in my German high school. But our teachers explain it easier😂 You can use the stuff in the square root for finding out if the function is a passant, tangent, or secant line as well. In the complex method the factor a is just integrated, but that confuses students and/or lead to small mistakes you do even if you know how it's done.
You have changed my life by telling is method.
This is also known as the PQ-Formula, we were told to memorize it (in my case). However I've also seen students remember this formula with some rules of thumb.
(0. make sure the equation is x²+px+q=0, and not let's say 2x² or so)
1. divide the middle term (p) by -2
2. add the +- symbol and draw the square root, then square the term from step 1 and put it inside the square root
3. subtract q (also inside the square root)
formula will look something like this
x = (-p/2) +- sqrt((p/2)²-q)
I have a question.
Where exactly is this regarded as a new method? Because I learnt this exact technique 5 years in high school...
Where did you go to high school? I'm pretty sure they don't teach it in most parts of the US but they do teach it in some countries.
@@MrHatoi I actually went to high school in Jamaica and that's how they teach it across the island.
@@kobe11111 That's interesting to know. I went to school in the US and I never learned this.
@@MrHatoi live and you learn I guess 🤷🏾♂️
I'am from Germany and we get taught the first part of this method under the name "PQ"-Formla.
x = - (p/2) +- sqrt( (p/2)^2 - q )
I always were mad that we not got taught about the abc-formula because for some equations the division with the a factor wasn't that easy.
This method really helps me, because I like the quadratic approach is rasiert than the root approach.
From my experience as a math teacher, even though students know both methods, most will prefer Brahmaguptas Formula because they can just plug everything in and the solution falls out. They will even use it if the solution basically jumps at them. Mostly students for which math classes aren't a constant state of hardship will take a second to consider what might be a quicker or easier way.
Cheers to all the people in the comments who are vocal about finding this trivial. Thanks for letting everyone know.
Well said.
Here is nothing new.....you have just solve it by using shreedharacharya's formula.....just a another proof of the shreedharacharya's formula
Could you plz share any information about it( Shreedharacharya's formula), I'd be grateful to you.
Any method that solves a quadratic equation can be used to prove Sridhacharya's formula, since the resulting solutions will be equivalent. The way we should judge methods like these is by how easy they are to remember and execute. This method, while more intuitive and easier to remember than Sridhacharya's formula, takes a few more steps to do. Therefore, neither is inherently better. They are both useful, for different circumstances and mathematicians.
Are u from Agartala?
Cripes, any article I read on this mess was incomprehensible. This video makes it make /a lot/ more sense. Thanks Presh!.
I genuinely don't know how much of this process and or the cheat steps I may have been taught in school. Been too long and I figured out how to get my ti-89 to solve the tricker ones for me quite early.
I saw this article a few weeks ago, nice to see a video on this :)
Though, my biggest qualm with this method is the fact that if the coefficient A is not equal to 1 you have to factor the function into that form which can result in B and C becoming cumbersome fractions. Other than that it's a really nice way to be able to think about the roots and the mathematical intuition behind them.
Or, as Indians are taught, for the quadratic equation ax²+bx+c, find p,q such that p+q=b and pq=ac. Thus, -p and -q are the roots of the equation.
fractions aren't that hard to deal with. Math can throw a lot worse at you than a couple ratios
This method also allows you to find the middle (aka top value) of the function. I knew there was possibility to find x-values through addition/subtraction of the x value of the top
I feel like the Quadratic formula is much more useful. Having studied maths until the end of high school, and now having calculus and algebra in university, I feel like having the discriminant is quite useful.
As you delve deeper into maths, the formula is more useful than this alternative method which might seem more intuitive for beginners.
You don’t have to throw the discriminant out the window though. This method is simpler.
I wish i knew this method for some of my pure maths classes in diff eqs when the type of roots didnt really matter than much and we were forced to solve them no matter what. (And we werent allowed calculators so i think this method woulda been easier to do by hand than the quadratic eqn)
R2D2 from Star Trek I prefer the quadratic formula. It’s a formula and it’s good to solve any quadratic equation. I’ve used it so many times it’s almost impossible for me to forget.
I already have seen this 5 min ago
In that video at 1:10, Achrotone
noticed the typo "back-to-bank" which is now corrected: ruclips.net/video/lDbQA4euAbY/видео.html&lc=UgwYAupZwmfkh8Nfred4AaABAg
@Shailesh Kumar true, and the formal description is easy for everyone to do it theirselves. I'm from Germany and we learn this formula from 5th grade (10yo) and I never knew this wasn't taught in the entire world 🤷♂️
@@fix5072 yes...& That formula is invented in India....& now used in all over the world 😃
@@AshutoshIIT ok boomer
@Shailesh Kumar hey Bro, already our quadratic formula was derived by dividing 'a'😄
May I complement Po-Shen-Loh on a brilliant exposition of a bit of basic, fundamental mathematics. A good example of plain honesty, simple truth and easy understanding.
I first watched the video a day or so ago and it was only a day later that the penny dropped.
As per Leonhard Euler's 'Elements of Algebra' (x-a)(x-b) = x^2 - (a-b)x + ab where as we know a & b are the roots of the quadratic.
Taking (a+b)^2 and (a-b)^2 [ i.e. props. 4 & 7 from book 2 of Euclid's 'Elements' ] then expanding and subtracting we get the answer 4ab hence we have (a+b)^2 - (a-b)^2 = 4ab. This is a theorem, prop.8 of book 2 of the 'Elements' and for some unknown reason demoted to a RULE alias 'The Quarter Squares Rule'. After a bit of simplification we end with [(a+b)/2]^2 - [(a-b)/2]^2 = ab.
The algorithm given in the video then amounts to [(a+b)/2]^2 - ab = [(a-b)/2]^2 which taking the square root leaves (a-b)/2.
So (a/2+b/2+a/2-b/2)=a & (a/2+b/2 -a/2 -[-b/2])=b.
It is still highly commendable that the 'QSR' has been derived by another route and has been admirably utilised for the factoring of quadratic equations. What I find a bit astounding and some what sad is that together with the hits on the 3blue1brown and MindYourDecisions videos on the same topic a combined total of around 1,453,000 views no one else seems to have spotted the connection.
Finally if we change a & b to x^m & x^n then the answer (ab) becomes x^(m+n) hence all integers raised to a power above the second are the difference of two squares . Further more the bigger the power the more DoS solutions there are for any one integer raised to that power! What does this mean for Fermat's Last Theorem.
If memorizing a formula is the hardest part of the quadratic formula, maybe math isn't for you.
Like many other people pointed out:
IT'S JUST _LEAN_ COMPETING THE SQUARE METHOD
I am from Germany. I'm 36 years of age and even my mom learned this "new" method at school. In Germany every student older than 16 knows this method as the "PQ Formula". And this is since...I don't know...Kaiser Wilhelm, I think 😂😘
Haha same..I'm from Germany too..
Dear author! Where were you with your video 30 years ago when I was at school! That's great method, damn!
4:06 I'm German and here the formula isn't ax*2+bx+c but ax*2+px+q. Then we can build the next formula: X1,2= -p:2 +- the root of ((p:2)*2 -q). If you solve this you get the two solutions. Then you can check them with the "Satz des Vieta" like you showed in the video. X1+X2=-p and X1*X2=q. It's very simple if youve done it often enough.
actually, when you're dealing with parameters and very small numbers which you mostly will it just becomes more complicated, thats why you teach the formula, this method is only good for convenient numbers in easy problems.
Exactly, the examples given here were pretty convenient numbers
I've seen the method, but didn't fully understand it until now. I do find the solution to finding the equation very elegant.
“But the key insight in this method is that you dont have to memorize anything”
Yeah sure...
You people are using calculator
So,
No need to memorize tables
No need to memorize square root.
No need to memorize algebraic identity (a-b) (a+b)
this is exactly the quadratic equation and how it came to be. when first taught the teacher shows how we came about this formula so we can use it confidently knowing why it works
Simple visual depictions of completing the square can readily be found online and are far easier to understand for factoring novices. However, this is still interesting. Thank you for posting it.
It's the commonest method of solving a quadratic equation in India.
I derived the quadratic formula myself when I wasn't able to memorize the formula ... And then I didn't needed the formula to get roots.
This is brahmagupta formula this is sridharacharya formula
@Victor Yago you too brutus
Here's a much more interesting approach in my opinion.
A neat result is that for a quadratic Q with turning point T we have:
Q(T+-t)=Q(T)+at^2.
If the solutions are T+t and T-t, we obtain at^2=-Q(T) which gives t=sqrt(-Q(T)/a).
Thus the solutions are x=T+-sqrt(-Q(T) /a)
I believe this would make more intuitive sense starting with the symmetric property of a parabola, specifically that 2 solutions equidistant from some value m would show f(m+d) = f(m-d) = 0. In other words, f(x) = a(x-(m+d))(x-(m-d)). Then expanding would result in f(x) = a(x^2 - 2mx + m^2 - d^2). So as long B = neg (2m) or m = neg(B/2) and C = m^2 - d^2 or d = sqrt(m^2 - C), then we have our solutions of neg(B/2) (+/-) sqrt(m^2-C). The scaling affect of 'a' has no impact here, as sliding values of 'a' does not affect zeroes.
Of course, the assumption of solutions here does require FTA, but Gauss took care of that for us :)
when it's Christmas break, but Math is still haunting me from my recommendations :3
Mathematicians : how to solve equations
Engeneers : *DØ ÅPRŌXĮMĀTÎØNS*
This is just quadratic formula with extra variables and steps, @4:09 it's literally exactly the same thing only it has been chosen to write C in place of c/a and B instead of b/a. I feel like I would be less upset if this wasn't called "a new method" (i.e a way to completely avoid the quadratic formula) but instead called something like "a new way to derive the quadratic formula" or something similar.
At 3:27 in the video the righthand side lost the negative signs on B. I realize that this doesn't matter, essentially you left out a step where you multiplied both equations by negative 1, but some people might miss this.
Note that the "new way" works with equations where a=1. If you start with this assumption then the standard formula for solving quadratic equations becomes somewhat simpler.
If you look at the steps in the "new way" and start writing them down in a way to solve for x then the result looks suspiciously like the standard formula. I did this when I first saw one of these videos pop up on the internet recently. It appears that you took the standard formula, dropped out the a terms since a=1, and moved the 2 from the denominator up into the radical. So rather than consider this a "new way" to solve a quadratic equation, it should be viewed as the derivation of the standard formula. At 3:6 in the video the equations on the righthand side of the screen are exactly the standard formula with the 2 in the denominator moved around.
Granted if a person was alone on a desert island and couldn't remember the standard formula, they could use the "new way" to reason through to a solution.
I'm an engineer and I have to solve quadratic equations frequently. It is second nature for me to plug the terms into the standard formula to get the solution.
This is just a derivation of Brahmagupta's quadratic formula.
At the 4:10 mark you can substitute back for B=b/a and C=c/a and recover the standard formula with some straightforward factorisation.
The one thing I have learnt from the video is how the Brahmagupta's formula is derived. Something that should be taught alongside memorizing a formula (one that is still stuck in my head after more than 35 years).
To be honest I can't see how this makes it easier, u just took the quadratic formula and simplified it
In Germany, we learned in school first something we called "quadratische Ergänzung", which could be translated to quadratic completion. So, one adds and subtracts a number, so that the formula becomes (x-d)^2+e=0. In the next lesson, we derived the quadratic formula out of this.
Fun Fact: We call the quadratic formula "midnight formula" because it is so important, that you must be able to recall it, even when you are woken up at midnight.
@wise ol' man yeah that's what I meant to say, but I can't see how it's easier than just knowing the quadratic formula, it feels like u added some extra steps to it
@@0000-z4z we never learned something like that here, we just learned the quadratic formula straight away
0000000 0000000 in the USA (and maybe other english speaking countries we call it “completing the square”
.
I know it, cz im in 10th
Even 7th class student know this
I am doing engineering and i didn't know it 🤣🤣
@@dhruvbhargav1547 shame on me 🤣
@@dhruvbhargav1547 This formula is pretty good but doesn't gives us the discriminant which is important.
You arts with maths 😧 you are pretty brave
This is not any "new" way.It just resulted from the property of quadratic coefficients.Everyone who knows Brahmgupta's formula knows this.
In India, even 8th class students know about this method as factorization method. We have even made various tricks to make this way ever faster.
Hence, nothing new for me. 😂
lol right, let me know when India starts beating the US in the math olympiad
@@petrosprastakos 🤣🤣
@@petrosprastakos good morning
@@petrosprastakos lol let me know when American firms stop hiring Indian CEO.
@@aamitanandd Nice burn! :)
First of all it's not bhram Gupta's quadratic formula it's Shridharacharya's quadratic formula
Both were from India , so kindly shut up ....
@@Sailed_away well its like saying George Washington and Donald Trump are both america's presidents hence they are the same guys
Exactly, this is sridhar acharya
@@pragul1999 ,😂😂😂😂😂
You made my day
Better still, *THE* quadratic formula.
I find it easier to complete the square. Maybe just because I'm used to it 😁
Me: does this method
Teacher:Well yes but actually no.
I've seen this method a lot
Thats cool, and how about videos that I have for several topics ?
In this lockdown period this video was one of the best that i have seen for revision 😃😄video was awesome 😃😄
x^+11x+28=0 has two roots (-4 and -7), because 4 and 7 are factors of 28 and add up to 11. Now try using Po Shen Loh's method. Our mid-point has to be -11/2. This has to be squared, giving 121/4. Now we must subtract 112/4 (that's 28) to get u^2 (that's 9/4). Now we take the square root to get u (that's 3/2). Now we must subtract 3/2 from -11/2 (that's -8/2=-4) to get our first root. Then we must add 3/2 to -11/2 (that's -14/2=-7) to get our second root. That's an awful lot of work involving positive and negative fractions for such a simple quadratic. Try it!
In the UK everyone that does Further Maths learn this in a topic called roots of polynomials.
Very true, Hi, I am making Videos for O\A levels as well. Please feel free to check them and do share your feedback.
When you had known this method before you watched the video
*A Genius*
But then you looked at the comments.
Nope,this is just a simple mathematics at 9th grade
@@humanityisnumberone6008 Not in my country ;), It's just a joke man.
@@humanityisnumberone6008 my teachers have not thought me this method
Absolutely it I knew this when I was 13
@@pallabgoswami2451 Same as I. But in Poland we aren't taught this method, It is not in our schools.
But wait.........
Looking the -B/2 +/-z, wouldn't it actually also be that z= sqrt(B^2 -4C)/2? Considering that the a of the quadratic formula in this situation is always =1 then this is still the quadratic equation.(I mean, duh)
If B^2/4 - z^2 = C,
then z^2 = B^2/4 -C or (B^2-4C)/4. Then z = +/-sqrt(B^2-4C)/2.
That's an interesting way to think about it. So in the end, this method is simply a quadratic formula through a new perspective that are more intuitive, or simply also a way to make sense of what the quadratic formula actually represents. In other words, quadratic formula can be thought of as the average of B +/- an identity that is related to B and can give a product of C.
Yeah this is pretty similar to the quadratic formula 😅
Recently ran into a YT video showing something very similar to this method (starting with the middle x-term, then dealing with the constant term), with the 1st example problem:
x² - 60x + 899 = 0.
Heck, who wants to deal with the large numbers associated with factors of 899, or have to calculate the discrimination & its square root from the quadratic formula??!
A bit of inspection, and we easily notice that 30² = 900.
If we write 899 = 900 - 1 , then our quadratic becomes:
x² - 60x + 900 - 1 = 0, we can write the 1st 3 terms as a perfect square, and the constant 1 can also be written as the perfect square 1² (if the resulting constant q isn't a perfect square, then just write it as one: √(q²) )
(x - 30)² - 1² = 0 , which is a difference of two squares, and can be written as the product of the sum & difference of the square roots:
[(x - 30) + 1][(x - 30) - 1] = 0 , and, on simplification becomes
(x - 29)(x + 31) = 0 , then, the zero a product property yields solutions x = 29 and x = -31.
Another way of expressing this idea is that we are doing a substitution x = z - B/2. This is cool because substitution is commonly used in much more advanced maths too.
This is basically "Completing the Square" in a nutshell. 😄
..... or not using a sledgehammer to crack a nut.
Bruh no both are entirely different things 😂😂
@@mvpistakenbyme818 its the same tho. Try using CTS in the problem and you'll see the difference. Even QF is CTS with a memorizable formula. There is generally only 2 ways to solve Quadratic Equations. The factoring method and Completing the Square Method 😂🤣
Sridhracharaya formula 🙏🙏🙏🙏🙏
I just tried to say that name and now the world is upside down
@@antipro4483 it is simple, but not easy!!!
Like in icse, we know this in class 7. Thats like when we were 12 year olds.
I was reading Wittgenstein and solving advanced integral calculus problems at age four my guy.
@@keyboardcorrector2340 lmao
Ikr
That was my thought - it’s the method of “completing the square” which then works out to be the quadratic formula anyway. In exams when asked to derive the quadratic equation, this is the method I used
@@keyboardcorrector2340 This is nothing.. We start perp for IIT JEE from junior kg and till 4th we're on top of the world.... Sarvashaktishaali Gaitonde🤣
I would set one separate variable m to be the midpoint between the two roots, m=-B/2, and d to be the distance between each root and m. Then the roots are m+-d, and their product C is m^2-d^2, so d^2=m^2-C. This yields the roots via a formula that looks much simpler, m+-sqrt(m^2-c).
To all those saying that this is not new, and has been taught to schoolchildren
I've seen most comments point to India.
As a Indian student, YES we were taught that if you find 2 numbers that multiply to C and add to B, you have your factors. (called the split the middle term method)
This method is about HOW TO find those 2 numbers. After that its normal. But what has NOT been taught to us(simply because in exams they don't give HUGE numbers) is HOW TO find those numbers.
Like yes, for the equation: x^2 + 11x + 28 you can easily find out that the roots are 4 and 7.
How about for this: x^2 -60x + 899? Can you still do it?
Except the method Po-Shen Lo told here CAN actually easily solve the 2nd one
You know that 1/2 of 60 is 30. You can just simply do 30^2 = 900 in your head, and see that the diff is 1(well, you see that the square of the diff is 1, but its the same thing). Which means the roots are 29 and 31.
And this is also IGNORING all those quadratic equations that DO NOT have integer roots. Like look at this eqn
2x^2 -16x + 26.
most people will divide by 2, that's a integral step in every method.
So this comes out to x^2 -8x + 13. Can you split the middle term easily here? Nope. Except if you used the method shown above, you can easily solve this
It simply comes out to 16 - (diff)^2 = 13, which you can easily see means that the diff^2 is 3, and the diff is root 3.
Which means your roots are 4 +- root 3.
Like yes, for easier smaller equations this is not that time saving. But for those bigger ones with numbers you arent even sure are prime, this helps a ton. And if you ACTUALLY understand WHY everything works(there's a good vid on it by Outlier.org) You can literally do it in your mind.
I've been to school in Germany and I can't recall ever having learned a method to solve a quadratic equation that involves guessing. I could never remember the ABC formula when I was in school so I used the completion method and it is still my preferred method today.
But... why would I do all of that process when is easier and faster to just do the formula? It's not really that difficult to learn and it's way quicker to remember 1 simple equation that all of that method.
Because when I was in school, I had to show my work. "That process" is the same steps for both methods, but the division is done first instead of last.
I would say it would be important to know how something work. Take daily life as example, yes we know we turn on the stove fire come out, but it would be better to know more about why is there fire or things related like combustion. Another example would be we know stepping on the paddle the car would move forward, but would be better to know about how an engine work.
Mr. Money it depends sometimes you don’t always have to do the long formula method just use your head
Quadratic formula is computationally faster than ph shen lol this for computers
At 9th grade we were taught three methods to solve quadratic equations.
Out of three ,two are described in the video.
If you know the third one then like the comment😁😁😁😁😁😎😎
Hint: alpha squared /b1c2-c1b2
Factorization method (Splitting middle term), Completing square method, and Quadratic formula
In Eq :- x^2 + 8x + 15 =0.
Take lcm of last no.i.e 15
Select no.s in such a way that it gets matched to 8
3×5
I prefer to factorise the constant and factorise the sum of the coefficients knowing that the factorisation of the coefficients is 1 greater for both factors.
X^2 + 5X + 6 = 0
Constant: 6
Coff Sum: 12
6=2x3
12=3x4
Done.
(X+2)(X+3)
Love the content Presh, but could you please do videos in dark mode? It hurts my eyes when I see your videos at night. Inverting the colours should suffice
Daniel, you can control this using your "Accessibility" settings on your phone/computer. You can invert the colors to achieve this effect. :-)
This is a massively convoluted method. No wonder I've never heard of it
Isn’t this literally explaining how completing the square works lol?
No, because completing the square doesn't utilize the concept of finding two numbers that are symmetrically positioned on each side of a given "average" m .
Completing the square: re-writes the quadratic equation into the form
(x+p)² - q = 0
This method: re-writes the quadratic equation into the form
(x-u)(x-v) = 0
where u = m+d and v = m-d , for some fixed values of m and d.
However, they are related by the insight that (x-p)² - q = 0 can be directly re-written as
(x+p-√q)(x+p+√q) = 0
which means p coincides with -m , and q coincides with d² .
(And of course they must be somehow related, because they are solving the same type of equation.)
Yeah its basically completing square method. Also love the profile picture btw!😂 I love Brendon urie😂
I was honestly watching the video to see a new way to solve quadratic equations, but while explaining this "new" way, on time 3:50 you actually gave the exact same quadratic equation I have been using all time long!!!!!.
The only thing this "new" way does is to use 2 formulas instead of 1.
The only value I see is that you are able to get some results without calculator, which is cool, but after that, not very new.
This is basically the completing the square method with a little more clear teaching. At the end, you do reach the quadratic formula. This is just a derivation for the formula itself.