You use the quadratic formula all the time, but where did it come from?

[@TaimTeravel]

Now prove the cubic formula 😂

[@bprpmathbasics]

You got it! kzbin.info/www/bejne/i33WgZuNotabr8Usi=v-8aw9bjPSCqA2Fc

[@TaimTeravel]

@@bprpmathbasicsYou don't know how much this is helping me. Thank you so much 💟 😭

[@nssinger4276]

This is class 10 maths​@@TaimTeravel

[@RikiFaridoke]

To proof root of cubic formula is so easy guys.

[@Adenfall]

This guy makes learning math easy and fun. Why can't all teachers be like him

[@ErikOosterwal]

Other teachers can't be like him because they don't have black and red markers. 🤣

[@hookahsupplier.5155]

Not all teachers have the merit to teach math, not only does it stipulate a formidable prowess, but also a level of self-composure and today we have a dearth of that type of teachers. Should only be thankful to have such people.

[@kavinesh_the_legend]

Those students are lucky

[@MikehMike01]

Most teachers are incompetent and overpaid (unions)

[@blockyboxhead]

cuz hes him

[@ronaldjensen2948]

0:12 Better to say "a cannot be zero because if it is, you don't have a quadratic equation, you have a line"

[@anglaismoyen]

Even better: a line is just a degenerate quadratic:D

[@Anik_Sine]

Taking the limit of the quadratic formula as a approaches 0 gives x = -c/b or -infinity, one of which is indeed the solution for bx +c = 0.

[@yoylecake313]

@@anglaismoyen does that mean a quadratic is a degenerate cubic

[@anglaismoyen]

@@yoylecake313 Why not?

[@Wutheheooooo]

@@yoylecake313 a cube in 2 dimension is a square

[@stephenhill4492]

I was aware of where the formula came from, but it’s great to see it explained in such an effective way.

[@ruspg3d701]

same haha

[@limwanjin361]

Same

[@Limited_Light]

kzbin.info/www/bejne/o4esZXuJrL-Hm7M

[@PatClevenger0709]

49 years old, and i always accepted the quadratic equation without question. Thank you for showing the proof.

[@matteovillella1482]

It made me so happy as the familiar numbers started becoming apparent. As my math teacher would say: it gave me a “warm fuzzy feeling”

[@ErimlRGG]

I had a professor do this about 10 years ago but he used another method. Is one of my favorite moments ever, not even school wise. I don't know why, probably because in school you frequently are supposed to take rules and formulas and accept them and rarely do they get explanations. He did a few mistakes and took him almost two whole whiteboards with his method but we got there and the whole class was invested in it and double checking his math

[@Nikioko]

You can also use the pq formula: x₁,₂ = −p/2 ± √[(p/2)² − q] Advantage: Way easier to memorize. Disadvantage: You might have to divide the whole equation by a factor to get x² + px + q = 0.

[@LqwAbidingCitizen]

the quadratic formula isn't even that hard to memorize. my teacher always labelled the √b²-4ac as ∆ and then used ∆ instead, made it way easier.

[@illadiel6049]

A short video explaining the logic behind the magic number would be handy

[@AusterEngineer]

Idk if you already saw it but he did a video on that recently Edit: it was called solving an unfactorable quadratic equation from 7 days ago

[@silver6054]

I agree. The idea is to derive the formula and for those that care, showing the detail would be good. And it's all pretty trivial. You have x^2 + (b/a)x and you want to make that look like (x+K)^2 = x^2 + 2Kx +K^2. So matching the x term you must have 2K = (b/a) which quickly yields the magic thing that needs to be aded

[@Michael-pp8lz]

I love these kinds of videos. I think understand the "theory" about these mathematical equations makes learning math much less stressful. Many of my instructors, at both the HS and college level, jumped straight into the application and left theory out.

[@bprpmathbasics]

A faster way to prove the quadratic formula: 👉 kzbin.info/www/bejne/aXW2f3WjgJyKbpY

[@glorymanheretosleep]

We all know where it came from. It came from YOU bprpmathbasics!

[@leonardobarrera2816]

Did you remember the partial derrivative for the quadratic formula on blackpenredpen’s channel!!! That was awesome

[@bprpmathbasics]

@@leonardobarrera2816 thanks!

[@reminderIknows]

how was this comment 1 month ago?? was it unlisted?

[@leonardobarrera2816]

@@bprpmathbasics you are well come

[@Qermaq]

This is the best DIY channel. Learned to make my own quadratic equation on KZbin today ;)

[@BulletWulf]

Very Under-rated channel. Keep up the good work! I always wondered why the quadratic equation is true and never considered deriving it from the coefficients. Half way through I was wondering why the magic number was that value and you explained that too. Thanks for the cool lesson!

[@bprpmathbasics]

Thank you!

[@highfall8145]

@@bprpmathbasics how did you upload the video and then theres comments over 4 months olkd

[@amr0733]

​@@highfall8145it used to be unlisted

[@SamusSelf-Destruct]

I love watching these videos because watching all of the numbers fall into place is just very soothing for my autistic brain. It’s like giving my overactive brain a massage. It’s why I love math and science so much. This is the first time I’ve actually learned something, though, and it is so, so satisfying. Thank you so much.

[@muffinandme1]

I went to high school from 1973 to 78, and excelled in maths. Recently I tried reviewing hs maths with the aid of my son's old text books and was going well until I hit logarithms ans trig. You see we learned both of these using log tables. Scientific calculators were not allowed to be used until the final part of 6th Form (or Year 12 as it had changed to). I found I had absolutely no clue how to proceed using a scientific calculator, and the text books unhelpfully assumed you already had that knowledge. I would give alot to have my old textbooks back from then, and a set of log tables of course!

[@maxhagenauer24]

I always remembered it a different way that is faster. ax^2 + bx + c = 0 ax^2 + bx = -c 4a^2x^2 +4abx = -4ac 4a^2x^2 + 4abx + b^2 = -4ac + b^2 (2ax + b)^2 = b^2 - 4ac 2ax + b = +- sqrt(b^2 - 4ac) 2ax = -b +- sqrt(b^2 - 4ac) x = [-b +- sqrt(b^2 - 4ac)] / (2a)

[@modemramachandraiah8038]

Very good

[@MathPhysicsEngineering]

On my chanel you can find a playlist with the derivation of formulas for all polynomial equations from order 2 up to order 4. I don't derive the full formula for the 4th-order equation, but show how can it be derived in principle. Anyone who watches my video for the 4-th order equation will be able to derive the formula by himslef given that he will be willing to put in the required patience and effort. I show 2 methods for the 4-th order equation.

[@maxhagenauer24]

@@brandonsteele2826 So I could factor the left side.

[@hyanparaiso7986]

This is by far the most amazing video explaining the quadratic formula that i have ever seen, thank you for making math a fun subject!!

[@HenryUesaka]

We all know how to solve X^2=A The problem is how to transform X^2+B/A*X+C/A=0 now Since (A+B)(A-B)=A^2-B^2 And X^2+B/A*X+C/A=X(X+B/A)+C/A just let Y=X+B/2A the equation be= (Y-B/2A)(Y+B/2A)+C/A=Y^2-B^2/4A^2+C/A----->>>> Y^2=B^2/4A^2-C/A Then we get the formula

[@russelllomando8460]

astounding - thanks 72 & keep learning

[@عثمان-محمد-ج7د]

I was using it for many years but I didn't know why is it true I thought that it comes from square root completing but I wasn't sure Thanks for your work Your Chunnel is my favourite channel ( or channels ) in the math

[@hbplayz4428]

when i started this chapter i viewed its derivation first, it makes a lot more sense after watching its derivation!

[@hv1785]

Graduate degree in engineering… but never knew the derivation of this. Then again, I think most of us engineers are happy just knowing something works. But it’s really cool after all these years to find out where it comes from. Thanks!

[@amoghsod2212]

Well I proved the quadratic equation when it was taught to us a little differently, we have ax²+ bx +c = 0 Firstly I would assume a ,b ,c are all real numbers So then we can tranform the equation by splitting the middle term , where let p + q = b , so ax² + px + qx + c = 0 , so if we assume that by splitting the middle term we can get all the answer of the quadratic (it will get in handy later) , we would get ax(x +p/a)+q(x+c/q) = 0 , which would get us (ax+q)(x+p/a) = 0 or( ax + q)(x + c/q) since we can split the terms to find the solutions the expression x+p/a and x+c/q must be equal , thus pq = ac , this would explain why when finding terms that would split we expect the coefficients of the split terms to = ac , but it is beneficial to us to use the former form , so we convert the form into (x+p/a)(x+q/a) so x = -p/a,-q/a But we know that p + q = b So by squaring, p²+q² + 2pq = b² , now lets subtract 4pq from both sides to get p² + q² -2pq = b² - 4pq , by applying some algebraic identities, (p-q)² = b² -4ac , since pq = ac So p - q = +or-√(b²-4ac) This is how we find if an equation has a real solution or not , since p + q = b a real number we also require p-q to be a real number since during addition the imaginary part of p and q may get cancelled so for the answers to be real p-q must be real and thus , +or-√(b²-4ac) must be a real number so b²-4ac>= 0 , for all real solutions . Now p + q = b p-q = I am not going to write it again so p = b +or- √(b² -4ac) So q would be identical as +or- decides which is p and which is q but we require x to be -p/a and -q/a So we do that and get x = {-b +or-√(b²-4ac)}/2a You can try and do that with equations of cubic and quartic it is going to get exponentially difficult or tetrationaly difficult, and if you are brave you may try for quintic , but you will not get it since galois theory , remember our assumption that we assume we can get all solutions by splitting the middle term , you would find if you want a complete solution I mean every solution, you won't get by basic arithmetic operations , since all the solutions of quintic equation aren't in the radical extension if the field of rationals 😀 Edit :- Here you go for better reference kzbin.info/www/bejne/sHS4aqeQZ7uejcUsi=It38BrUP6lEqrg_y

[@ROCKY-pu1zt]

Sir, Thank you. I will need to go over a few times, but thank you!

[@giantswingallday7001]

Never have I appreciated the tic-tac-toe method in high school algebra, neither with the completing the square because it was so hard. That’s why I preferred the quadratic formula when it comes to solving these equations. Thanks so much sir!

[@andreonofre2625]

Very nice! I used the formula so many times but never had seen how to derive it.

[@darthrevan8376]

When I was in 8th grade I worked ahead in the math book because I was bored and came across a question which would require the quadratic formula. Since I did not know that it existed I ended up creating pretty much the formula but did not reach the last step of dividing both both under the 2a.

[@M1551NGN0]

Fun fact: it's also known as the Sridharacharya formula. Sridharacharya Formula is a mathematical formula that is used to solve quadratic equations. The Sridharacharya formula is commonly also known as the quadratic formula. Sridharacharya gave a method to solve the quadratic equations and hence it is named after the great mathematician and is called the Sridharacharya Formula. Source: cuemath

[@drashokkumar9209]

Shridharacharya multiplied the LHS by 4 . We get square of 2x . Completion of square looks easier then .

[@M1551NGN0]

@@drashokkumar9209 yeah that's true though

[@GOUTAM-wi3wi]

Certainly this formula was the creation of SREEDHAR ACHARYA

[@BuleriaChk]

There is no solutoin for x=0 because the three terms are affine vectors (thr polynimials of different orders span a vector space with real positive coefficients. Also, there are no negative numbers -c = a-b, b>a iff b-c=a, a-a =0, a=a If there are no negative numbers, there are no square roots of negative numbers.

[@TheTallRaver]

Fantastic explanation! I was just wandering about that this week and could not find a good video on KZbin that would describe how this equation came about. Thank you and keep up the good work!👍👍

[@davidwoodhead1037]

Great explanation 😊

[@Rao_Sahab.700]

It is shridharacharya formula

[@deanambrose1889]

श्री धराचार्य सूत्र 🇮🇳

[@someonespadre]

I’m just a Grandpa making a hobby of relearning basic algebra, it’s fascinating.

[@bprpmathbasics]

That’s wonderful! Cheers!

[@StephenMarkTurner]

I do this derivation every year or to, just to keep it fresh. I think I first learned it in 1968 as a 12 year old.

[@stevenranck5478]

There’s a better form of the quadratic formula that does allow ‘a’ to be 0, and it should be taught in schools instead: X = -2c / (b + - sqrt( b^2 - 4ac ))

[@ninjacodertech]

i literally thought about this yesterday!

[@gleysonsantos3960]

You are amazing, thanks for making math enjoyable for us.

[@wes9627]

I had been using "the quadratic formula" for decades without ever thinking about how it came into existence. So a while back I set out to derive it from an entirely different perspective. Assume that r and s are the two roots of ax^2+bx+c=0 or x^2+(b/a)x+(c/a)=0=(x-r)(x-s)=x^2-(r+s)x+rs. Thus, (r+s)=-(b/a) and rs=(c/a). How can we solve these two equations for r and s in terms of a, b, and c? Let u=(r+s)/2 and v=(r-s)/2. Then r=u+v and s=u-v. What does that get us? (r+s)=2u=-(b/a) so u=-b/(2a). rs=(u+v)(u-v)=u^2-v^2=c/a, which can be solved for v^2=u^2-c/a=[-b/(2a)]^2-c/a=(b^2-4ac)/(4a^2). From this equation we have v=√(b^2-4ac)/(2a). Finally, we have r=(u+v)=-b/(2a)+√(b^2-4ac)/(2a)=[-b+√(b^2-4ac)]/(2a) and s=(u-v)=-b/(2a)-√(b^2-4ac)/(2a)=[-b-√(b^2-4ac)]/(2a).

[@the.wealthversity]

One of the best math teachers!

[@Aussiesnrg]

I wish someone explained it that well when I was in high school

[@Iomhar]

How about inserting this value of x into the original formula to see if it works?

[@bprpmathbasics]

Here’s an example kzbin.info/www/bejne/oHSsl6KBlq11jpYsi=SBBLgabSe0J8znHM

[@Yadavharsh.]

It was given by Indian mathematician named shridhar acharya we learned it in India in class 7 as quadratic formula or so called shridhar acharya rule !!

[@drashokkumar9209]

Another , and probably better , approach was suggested by Shridhara Acharya ( 8th century , India ) . Multiply the LHS by 4 . Then divide by a . You get square of 2x . Then, complete the square .

[@jigglyCroissant]

Literally the same thing . And an even better approach is to multiply both sides by 4a . Not boasting , just sayin'

[@prywatne4733]

Here in Poland we are taught the quadratic formula a bit differently and I think it makes it easier for students to remember. so for ax² + bx + c = 0, we calculate the "Δ", Δ = b² - 4ac, if Δ > 0 then calculate the square root of this Δ, and then put it in x₁ = (-b-√Δ)/2a x₂ = (-b+√Δ)/2a if Δ = 0, just calculate x = -b/2a and if Δ < 0 then there is no answer

[@Wutheheooooo]

every equation have the same number of solution as their degree. I would rather say they have "no real solution" as there won't be confusion when we go further.

[@Soltris_yt]

This channel deserves billions of subscribers and views. 💜

[@navi2710]

If every student had you as a math teacher humanity would be experiencing a golden age.

[@funprog]

There is no magic number 2(b/2a)x is clearer to explain, it is the the product 2ab of the (a+b)^2 so for the b=b/2a => (x+ b/2a)^2. From that it is easy to note which term to add to complete the square i.e (b/2a)^2 The formula is derived easily and it should be practised more to derive it. Saying magic numbers etc help you to forget it after a while if you dont know why you are doing something. 1. Divide by a 2. Make 2ab => 2/2 (b/a) 3. Make clear the b factor 2x(b/2a) 4. Add (b/2a)^2 to both sides 5. Move the constant factors to the right side 6. On the left you have a perfect square 7. X^2+ (b/2a)^2 + 2x(b/2a) = -c/a + (b/2a)^2=> 8. (X + b/2a)^2 = (b^2)4a^2 - ca 9. Take the root etc...

[@JaywantWani]

This Formula is given by Indian Mathematician Sridharacharya in 8th century

[@raghvendrasingh1289]

❤ suppose that roots are p & q then p+q = - b/a (p+b/2a)+(q+b/2a) = 0 p & q are values of x if we put y = x+b/2a then we will find an equation in which sum of roots = 0 hence there will be no linear term x = y - b/2a ax^2+bx+c = 0 a(y^2 - yb/a+b^2/4a^2)+b(y - b/2a)+c = 0 ay^2 = (b^2/4a)- c y^2 = (b^2 - 4ac)/4a^2 y = √D/2a , - √D/2a where D = b^2 - 4ac is known as discriminant now x = y - b/2a hence x = (- b + √D)/2a , ( - b - √D)/2a

[@danielnuncio4711]

Just let out the biggest “BRRRRUUUUHHHHH” on the quiet floor of the library when you drew the last rectangle. 🤯

[@robertloveless4938]

I learned this solution back in 1964. Never bothered to remember it. Jusf know it works EVERY..... TIME.

[@Literallyeveryonealive]

I always teach my students how to do completing the square then at the end go “oops we found the quadratic formula”

[@dennissvensson7933]

ax² + bx + c = 0 ax² + bx = -c 4a²x² + 4abx = -4ac (2ax)² + 4abx + b² = b² + 4ac (2ax + b)² = b² + 4ac Is a more Nice way

[@rgsboys3423]

It was given by mathematician named Shridharacharya in 8th century

[@michaeledwards2251]

An alternative method would be 1. Assume the roots are equal, giving B/2A. 2. Investigate the case of different roots, B/2A +/- delta 3. Determine delta. I use this method, as it can be done by KISS, Keep It Simple. There are certainly many other methods : completing the square is only one example.

[@noo6423]

i thought it was a bit more complicated than just moving everything around to make it x=, thank you for showing me i was wondering how about a week ago, thank you for showing me!

[@TheJaguar1983]

I remember this from Year 11 Maths Methods. I basically forgot the derivation of the formula and just used it.

[@bushra3820]

anyone else notice the huge marker collection

[@m.h.6470]

I always dislike it, when people say that in √x² the √ and ² cancel each other. This is incorrect and leads to the common mistake, that people believe, that √ leads to ±, which it doesn't! The reality is, that √x² = |x| So if you have something like: x² = 25 and you do √ on both sides, you don't end up with: x = ±5 you actually end up with: |x| = 5 This equation explains far better, what is going on and it has the same two solutions (x = 5 and x = -5), but it doesn't confuse anyone, that √25 = ±5, which simply isn't true!

[@ZacharyBlue]

I think you meant to write 5 instead of 25 in some of the lines

[@m.h.6470]

@@ZacharyBlue yes, you are correct, I fixed it.

[@ImCalebRosengard]

But where has 1/2 (magic number) come from? What am I supposed to do in order to get to that number for starters?

[@Budgeman83030]

An instructor I had said many formulas we have came from a bunch of guys drinking beer and smoking ‘things’ while gathered around a bon fire

[@mimzim7141]

A good exercise in also to plug back the solution for x in the quadratic and check that you do get zero.

[@david_porthouse]

The other version of the formula is x = 2c / (-b +- sqrt(b*b - 4ac)). Use whatever version avoids an arithmetic subtraction and does not lead to nonsense as a -> 0. In the real world, quadratic equations are often only slightly quadratic and we ought to know about this other version.

[@sayedkashifkashifhussain2616]

Great sir

[@rcarioca]

How did they figure out to add (1/2 B) squared to both sides ?

[@habl844]

See the Wikipedia page for "Completing the square", there's a nice geometric version as well. Back in the day it was much more common to write math with geometry.

[@mahmoudsaleh1292]

Excellent bro 💪👏👏👏

[@mateuszpaczynski6195]

thank you so much :)

[@toferg.8264]

Awesome!

[@Anotherperson-m5b]

Nostalgic ❤

[@katathoombs]

I _might_ have taken my chances in the national abiturient test on maths during my upper secondary schooling, if we had had maths-tubers like this back then...

[@shimizu111]

Thanks bro ❤

[@tathagatusc4244]

Lovely👍

[@asheraley779]

8 minute video and it seems simple now that I'm learning series in calculus, but why wasn't I taught this?

[@jbreezy101]

0:12-0:17 then it would be linear, and you wouldn’t have to use the quadratic formula. You could just solve for x in terms of b & c.

[@juancarlossanchezveana1812]

Excellent

[@RikiFaridoke]

Please, proof it by complex analysis method, can you do it sir?

[@SuryaBudimansyah]

Sir this channel is called "bprp math basics"

[@88kgs]

This is Shri DharaCharya formula, from Vedic maths

[@Ultranger]

I’ve been trying to find a similar formula but for cubic functions which is hard

[@remus_lupin]

Check Mathologer. It's a very complex equation, which he specializes in, often with visuals. Might need to scroll for a while, so do a search, instead.

[@carultch]

Mathologer does an excellent job showing where the cubic formula comes from, and how to use it: kzbin.info/www/bejne/hF6uiYaqqtWqqcU

[@bushra3820]

how does he use 3 colors after only holding 2

[@rykehuss3435]

When will you do something difficult like complex analysis or algebraic geometry?

[@ludoviclemaignen9432]

It is a long time since I left school, so I don't remember how to derive it, but I do remember that my math teacher showed us how to derive it. I am surprised that you feel the need to do a video about it. Does that mean that modern teachers just drop it on students without explaining it?

[@teatimea]

i have one question. u said that u want 1x^2+bx so u have to put the c/a in the other side. Why?

[@teatimea]

oops i have one more other question mb. in timeline 8:13, you mentioned how we should just leave the square root of b^2-4ac. why? thank you so much and I'm so sorry TT

[@khalidelgazzar]

Awesome 😎👍

[@LearnerSupriya07]

This formula was given by Indian Mathematician name "Shridharacharya" Proud to be Indian.❤

[@samrudhb7855]

Awesome 👌

[@katathoombs]

Okay, so that's where it comes from. But...why do we want to to complete the square? • Why do we always want X²+BX on one side O__o why do we do that? Who came up with the need? How? The proof was shown even way back when in school, but I had, and have, no idea of the _why_ we want it. Tbh I don't even have any idea of how or why we come across with any quadratic functions. And I'm not even talking of the "real life use" sense, but the "how'd we end up pondering this from addition and multiplication" :D

[@sparshsharma5270]

I remember learning Completing the Square method 6 years back in 9th grade.

[@JayTemple]

I was self-taught, and I memorized the formula before I could complete the square!

[@RikiFaridoke]

Hi guys, please try quarternion method to proof it, can you do it guys?

[@bellykumar1984]

sqrt (2x+3)=x^2-6, how to solve it

[@fulltimeslackerii8229]

Basically, you get the solution by completing the square on the generic formula. It’s so cool.

[@notar2123]

There is a much longer, but far more intuitive approach to prove the quadratic formula. In fact, you prove something more general that the quadratic formula, namely the fact that any polynomial of the form ax²+bx+c can be written as a(x-x₁)(x-x₂). Just put in the quadratic formula for x₁ and x₂, using the negative value of √(b² -4ac) for x₁ and the positive value for x₂. Now expand everything, and after a lot of calculation and cancelation, you will end up with ax²+bx+c, which is exactly what we wanted to prove. Thus, once you set ax²+bx+c = 0, you can rewrite the equation as a(x-x₁)(x-x₂) = 0, which is trivial and the only solutions are x=x₁ and x=x₂. Would I recommend this approach? Of course not, but it is interesting none the less.

[@NadiehFan]

There is a logical flaw in your reasoning, because if you start from the expressions for x₁ and x₂ in terms of a, b, c and show that a(x − x₁)(x − x₂) = 0 implies ax² + bx + c = 0 this does not imply that the converse is true, unless you also show that all steps are reversible (which they are). But this still would be insatisfactory, because then you only have demonstrated that the formula is true, not how it can be found if you don't already know it. But the idea to derive the quadratic formula from (properties of) the roots x₁ and x₂ is perfectly feasible and instructive. According to the factor theorem a polynomial P(x) has a factor (x − x₁), that is, P(x) = (x − x₁)Q(x) for some polynomial Q(x), _if and only if_ x = x₁ is a root of P(x), that is, P(x₁) = 0. With a repeated application of this theorem we can show that if a quadratic equation ax² + bx + c = 0 (a, b, c real, a ≠ 0) has the roots x₁ and x₂, then we must have ax² + bx + c = a(x − x₁)(x − x₂) Expanding the right hand side of this identity we have ax² + bx + c = ax² − a(x₁ + x₂)x + ax₁x₂ Two polynomials are identical if and only if their corresponding coefficients are identical (this is also a theorem) so we have b = − a(x₁ + x₂) and c = ax₁x₂ and since a ≠ 0 this implies x₁ + x₂ = −b/a x₁x₂ = c/a These are of course Vieta's formulas for the roots of the quadratic equation ax² + bx + c = 0. Now, the question is if we can derive expressions for x₁ and x₂ in terms of a, b, c from these two relations between the roots and the coefficients of the quadratic equation, and the answer is yes. The idea is to first find an expression for the _difference_ x₁ − x₂ in terms of a, b, c, because if we have expressions for both x₁ + x₂ and x₁ − x₂ then x₁ and x₂ will be easy to find by adding and subtracting, because we have (x₁ + x₂) + (x₁ − x₂) = 2x₁ and (x₁ + x₂) − (x₁ − x₂) = 2x₂ so one root will be half the sum of x₁ + x₂ and x₁ − x₂ and the other root will be half the difference between x₁ + x₂ and x₁ − x₂. To get an expression for x₁ − x₂ in terms of a, b, c we can use the _identity_ (p − q)² = (p + q)² − 4pq with p = x₁ and q = x₂ to find (x₁ − x₂)² = (x₁ + x₂)² − 4x₁x₂ = (−b/a)² − 4c/a = b²/a² − 4ac/a² = (b² − 4ac)/a² This is interesting, because a² is positive since a ≠ 0 and if x₁ and x₂ and therefore also x₁ − x₂ must be real, then its square (x₁ − x₂)² cannot be negative so b² − 4ac cannot be negative if the equation ax² + bx + c = 0 (a, b, c real, a ≠ 0) is to have real roots. So we have already found what is known as the _discriminant_ b² − 4ac of the quadratic before we have even found a formula for its roots. Now, supposing that b² − 4ac is nonnegative, it follows that we can either have x₁ − x₂ = √(b² − 4ac)/a or x₁ − x₂ = −√(b² − 4ac)/a But we only need a single value for x₁ − x₂, because inverting the sign of x₁ − x₂ simply amounts to swapping the values of x₁ and x₂. So, let's take the first expression for x₁ − x₂, then we have 2x₁ = (x₁ + x₂) + (x₁ − x₂) = −b/a + √(b² − 4ac)/a = (−b + √(b² − 4ac))/a 2x₂ = (x₁ + x₂) − (x₁ − x₂) = −b/a + √(b² − 4ac)/a = (−b − √(b² − 4ac))/a and so we find that x₁ = (−b + √(b² − 4ac))/2a x₂ = (−b − √(b² − 4ac))/2a and we have derived the quadratic formula for the solutions of ax² + bx + c = 0 from the known sum and product of its roots.

[@notar2123]

@@NadiehFan Yes I am aware of what you wrote and that math teachers don't like it when a students use a regressive proof. In such a case, however, I find it pointless to insist on that, since all a student would have to do is rewrite the same steps in the reverse order and bam - the assertion follows from the presumption (I apologize if I don't use the correct terms, I did not study math in the English language). Also, your approach requires the use of the factor theorem, so either one would have to prove it, or its use would have to be explicitly allowed. The point of my original comment was, that is the easiest way an actual student could prove the formula - 0 brainpower needed, 0 creativity needed, it is a very naive approach that requires only the knowledge of the actual quadratic formula beforehand (you could call it a proof for dummies). Easy in terms of actual thinking, ridiculously hard in terms of calculations. That's also why I said I would never recommend someone to use it, I just found it to be interesting in a way. I am no mathematician, however.

[@NadiehFan]

@@notar2123 Of course proving Vieta's formulas without using the quadratic formula requires the use of the factor theorem, but there are elementary proofs for that which are accessible to high school students. In fact in my country this was taught in high school in my time, but that was a long time ago. I wouldn't recommend any high school math teacher to prove or derive the quadratic formula by starting from Vieta's formulas, but in my opinion that would still be preferable to working back from the quadratic formula to a quadratic polynomial. But all this is really a moot issue, because there is a far easier method to derive the quadratic formula which is known as Sridhara's method and which is perfectly suitable for high school usage: ax² + bx + c = 0 ax² + bx = −c 4a²x² + 4abx = −4ac 4a²x² + 4abx + b² = b² − 4ac (2ax + b)² = b² − 4ac 2ax + b = √(b² − 4ac) ⋁ 2ax + b = −√(b² − 4ac) 2ax = −b + √(b² − 4ac) ⋁ 2ax = −b − √(b² − 4ac) x = (−b + √(b² − 4ac))/2a ⋁ x = (−b − √(b² − 4ac))/2a This is accesible to high school students who have had some exposure to solving quadratics by completing the square (which is also used for finding the coordinates of the vertex of a parabola which is the graph of a quadratic function) and indeed all steps are reversible. The advantage of Sridhara's approach over the conventional derivation which starts by converting the quadratic into a monic quadratic equation (by dividing both sides by a) is that this derivation avoids the use of fractions until the very last step by first multiplying both sides by 4a.

[@Gozieaaa]

I simply reverse enginnered the quadratic equation from the formula and then reversed the steps to get my proof since I'm stupid and had no idea how to prove the equation 😅

[@youssefchihab1613]

Shouldn't it be an equivalence instead of an implication ?

[@marsisgr8]

This man is a genius wtf

[@adriminecraft7000]

"How exciting"

[@champu823]

Its the perfect square method !!!!

[@AFSMG]

Brillante

[@Wandering_Horse]

Bravo! I am going to go through that on my white board until it's burned into memory!

[@nithinsirimanne2924]

Simply the target is to find x