In this video, I solved an interesting olympiad cubic equation.
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@mathboy81885 ай бұрын
Your presentation is damn-near perfect: slow clear enunciation, straight legible & well-spaced writing, good eye contact, not standing too long in front of your writing (barely at all in fact), explaining your thought process and then the math steps to work them out, and most of all, enthusiasm for the material! People who've never done it probably don't appreciate just how hard this is to do well. When I was teaching math, my mind was going a mile a minute, but almost none of it was math (which was the trivial part). It was about monitoring the issues above and more. Am I making eye contact with everyone? How's my time? Is my voice loud enough for those in the back, but not too loud? Is this explanation too high or too low? Have I offered enough high and low insights for the outliers who find the topic too easy/hard? Is an interesting tangential observation worth the time and deviation? And so on. If you aren't a teacher, then you _must_ become one in some capacity, as it is absolutely your calling. And if you are a teacher as I assume, then your students are very lucky.
@lucmacot54965 ай бұрын
Treat a known value as an unknown to be able to use algebraic identities: Bravo! Beautiful!
@luisclementeortegasegovia86035 ай бұрын
Professor, you have the great ability to manage the algebra beautifully, and those substitutions are a master piece!
@adammohamed52565 ай бұрын
I dunno why your videos don't pop up here for more than 3 months !! Gr8 work! Keep it up bro.
@dalesmart98814 ай бұрын
Hay I like your videos, but I must say that the when I first saw the problem, it posed no difficulty because I was able to figure out what to do within one minute (using a different approach). I used factor theorem to determine f(root x) =0 and conclude that (x - root x) is a factor. This meant that the other factor will be quadratic, so I used coefficient comparison to determine the quadratic factor then solve using the quadratic equation. Hence all the solutions were determined.
@apone28205 ай бұрын
Your channel is such a hidden jewel man, I love your videos.
@user-ow7vd4ju1sАй бұрын
As a follower of physics, I am/was good at math and enjoyed it a lot, but the skills/tricks I learned were directed at solving problems found in the physical world. The most interesting problems you present are not from that realm and are all fresh to me. So many new tricks to learn!
@Blade.57865 ай бұрын
By observation, x = √3 It's easy to find the other solutions from there.
@dandeleanu36485 ай бұрын
At olympiad there is't solution by observation!
@koenth23595 ай бұрын
@@dandeleanu3648Why not? Observation is legitimate, as long as you prove it's correct.
@ppbuttocks20155 ай бұрын
so just say by observation as the proof then as the top comment says@@koenth2359
@isabelshurmanfeitoza68985 ай бұрын
@@dandeleanu3648 se você faz os cálculos não está errado não
@mathboy81885 ай бұрын
That's the ideal way to solve it. The question is how to proceed if you don't see that.
@vietdungle12375 ай бұрын
2:29 from there it's clear to see that x-sqrt(3) is a common factor. By the way, your solution is very interesting but complicated for this particular problem. That method usually is used for higher degree of x (like x^5) because cubic equations literally have a formula or usually in exam have an easy to find common factor
@yessinegebssi162Ай бұрын
yes he could have used the Horner's method, since sqrt of 3 is a clear solution. however, his methode is way too good , i like it .
@andrewjames66765 ай бұрын
You remind me of the best teacher I ever had (Physics, England, 1957).
@creature_from_Nukualofa5 ай бұрын
this can be solved as a quadratic where the "variable" is sqrt(3) - i .s. (1-x) (sqrt(3)^2) + x sqrt(3) + x^3 - then a= (1-x), b= x and c= x^3 - plugging this to the quadratic formula one gets the first solution fast without guessing - then this can be reduced to a normal quadratic eq. given one solution is known. I really like you channel and your way of explaining !!
@Hrishi020055 ай бұрын
Simply when we put X=√3 Then the equation x³-(3+√3)x+3=0 satisfied. So x-√3 is a factor of the above eqn We get x²+√3x-√3 from fx=qx.gx So X=√3, [{-√3±(√3+4√3)}/2].
@rainerzufall425 ай бұрын
One should always check claims that were made... So I put the equation into Wolfram Alpha and got: >> x = sqrt(3) >> x = 1/2 (-sqrt(3) - sqrt(3 + 4 sqrt(3))) >> x = 1/2 (sqrt(3 + 4 sqrt(3)) - sqrt(3))
@atanubiswas.50984 ай бұрын
The question was so easy to me, but it was your presentation style which made me a fan of yours❤ Thank you sir 🥰
@thekingtheking143119 күн бұрын
Ty for all these videos !!! Love them 😍
@Rai_Te2 ай бұрын
Very elegant solution. ... I also tried it, and saw that sqrt(3) is a solution from the beginning. So I just did a polynominal division (orignal formula / (x - sqrt(3)) which gave me the quadratic remainder. However, a solution where you find one answer 'by inspection' (which is just a nice way to say 'i guessed until I found something') is always inferior to a solution by formula, so kudos to you.
@user-jh7pn9bo3z4 ай бұрын
I plugged that in Wolfram Alpha and I get the correct answers. Make sure you've written the equation properly and everything should be fine.
@nothingbutmathproofs71505 ай бұрын
I think that it would have been better to factor out sqrt(3) from -sqrt(3)+3. Real nice job as usual.
@arsessahra-yb9zb5 ай бұрын
با درود،یک خلاقیت در حل این مسیله به خرج دادی،بسیار سپاسگزارم
@francaisdeuxbaguetteiii73165 ай бұрын
Depressed cubic formula 😂
@abraham52765 ай бұрын
🤓
@hvok993 ай бұрын
My first thought as well 😂
@ben_adel34374 ай бұрын
Thats so cool idk how but when i tried to solve it i just say that x=sqrt(3) and then just divided by x-sqrt(3) and found the other ones but this is really helpful because in most cases i can't just see it
@voice4voicelessKrzysiek5 ай бұрын
Nice! After distributing (3+sqrt(3))x into 3x + sqrt(3)x I got the same answer without substitution since I started seeing a way out of the problem, not without earlier reassurance from you that there is a way out, though 😁
@jamesharmon49945 ай бұрын
What an elegant solution!
@allegrobas5 ай бұрын
Wow !!! Good work !!!!
@lukaskamin7554 ай бұрын
brilliant solution, I just wanted to mention that when you make that inference when the product equals zero, then one of the multipliers equals zero, while THE OTHER EXISTS (or defined). That doesn't make issues in this particular problem, but it might in other cases like irrational equation of type A(x)*sqrt(B(x))=0 (there might be more than one irrational factor)
@PrimeNewtons4 ай бұрын
I'm thinking about this
@88kgs5 ай бұрын
Very nice video The real name of quadratic formula is.....Shreedha Ahcharya formula . Regards 🙏
@rutamupadhye18285 ай бұрын
teacher, your channel is great
@JeffreyLWhitledge5 ай бұрын
The HP Prime gets the same answer you did. The TI-nspire CX II CAS decided to give the decimal approximations. The Casio CG-500 gives a crazy answer that looks like the Wolfram Alpha version. Prime Newtons and HP Prime win this round.
@PrimeNewtons5 ай бұрын
That's an interesting convergence
@JeffreyLWhitledge5 ай бұрын
@@PrimeNewtons ha ha, true!
@reamartin64585 ай бұрын
Top notch 👍
@kingtown95805 ай бұрын
Can you make video on the theory and make a playlist so people can learn new concepts of math which they don't know
@moonwatcher20015 ай бұрын
Awesome, mate!
@juliovasquezdiaz24324 ай бұрын
Gracias. Me gustó el video Saludos
@kangsungho17525 ай бұрын
Awesome method!
@avalagum79574 ай бұрын
The most difficult part is to know that x = sqrt(3) is a solution. Then the rest is easy: x^3 - (3+sqrt(3))x + 3 = (x - sqrt(3)) * a quadratic equation
@BartBuzz4 ай бұрын
Very clever solution. The first solution you showed was very non-Olympiad!
@jacobgoldman57805 ай бұрын
Unfortunate that sqrt(3+4sqrt(3)) doesnt seem to simplify nicely unless I miss something obvious.
@emremokoko5 ай бұрын
no it doesnt, unfortunately.
@roger7341Ай бұрын
We know a cubic has at least one real root, and there is more than one way to skin a cat. A real value of x is around 2 so try fixed-point iteration, starting with x=2. x←∛[(3+√3)x-3] returns x=1.73205..., which is close to x=√3, the exact root. Divide the given equation by (x-√3): x^2+√3x-√3=0 yields x=[-√3±√(3+4√3)]/2
@butterflyeatsgrapesАй бұрын
🦋I SMILED THROUGHOUT THE WHOLE VIDEOOO THANK U SO MUCHHHHHHH🦋
@Ether.215 ай бұрын
amazing solution
@basqye95 ай бұрын
excellent!
@sonaraghavan94544 ай бұрын
Awesome presentation.
@sonaraghavan94544 ай бұрын
When I first saw your problem, I applied factor theorem and figured out that cubic function becomes zero at f(√3). So for sure one root is √3. Then apply synthetic division and find the remaining two factors.
@user-xw6ky8ob4l5 ай бұрын
Excellent, you live on the edges of the undiscovered Mathematics . This is an enviable attribute, checked at x^1/3.This monstrosity produced by Fulkram must be rejected as a bad joke!
@VittorioBalbi19625 ай бұрын
Brilliant thinking Watch out y squared is 3 not radical 3 so the solution might be even simpler
@ayan.rodrigo5 ай бұрын
Fucking FANTASTIC, my friend
@cliffordabrahamonyedikachi81753 ай бұрын
Simply the quadratic formular were left the same way as the solution.
@kangsungho17525 ай бұрын
What an Idea!
@user_math20235 ай бұрын
Super ❤❤❤❤❤❤
@lbwmessenger-solascriptura56984 ай бұрын
i think the question was easy great fan
@alexanderkonieczka25925 ай бұрын
when i put it in wolfram i got your same answer.... maybe a typo on entry or they fixed it?
@janimed92665 ай бұрын
Very good
@TomCruz1425 ай бұрын
(x+√3) is a factor...very easy
@weo94735 ай бұрын
I like your smile
@Mohamed2023LaayouneАй бұрын
,Plz what is the name of the method that you use to find the solution 2 and 3, who know it , he can answer
@v8torque9325 ай бұрын
I don’t watch for the math I watch to see a black guy stare at me with no audio
@shaswatadutta44515 ай бұрын
This is a really easy problem.
@lukaskamin7552 ай бұрын
Why you say the first factoring you mentioned doesn't work? I tried and it worked perfectly: x(x²-3)-√3(x-√3)=0, than ( x-√3)(x²+x√3-√3)=0. So x1=√3, x2,3=½(-√3±√(3+4√3)) IMHO it's too easy for an Olympiad 😊
@cluedohere5 ай бұрын
oh, i am surprised that the wolfram alpha got that answer. it's not because the answer is complicate but wrong(see the approximate value they gave).
@NadiehFan4 ай бұрын
Don't know what's wrong with your WolframAlpha. If I enter your equation like this: x^3 - (3 + sqrt(3))x + 3 = 0 I get exactly the answers you get, not those intractable expressions you show in the video. Also, I already saw this equation earlier on other channels like this one: kzbin.info/www/bejne/j5-9q6OPiK6ci7c Of course, once you hit upon the idea to write 3 as (√3)² and rewrite the equation as x³ − (√3)²x − √3·x + (√3)² = 0 it is easy to see that we can do factoring by grouping to get x(x² − (√3)²) − √3·(x − √3) = 0 and then we see that we can take out a factor (x − √3) since x² − (√3)² = (x − √3)(x + √3). To make this more transparent you can of course first replace √3 with a variable y to get x³ − y²x − yx + y² = 0 and then do factoring by grouping to get x(x² − y²) − y(x − y) = 0 where we can take out a factor (x − y) which is what you do in the video. A slightly different approach consists in rewriting x³ − y²x − yx + y² = 0 as (1 − x)y² − xy + x³ = 0 which we can consider as a quadratic equation ay² + by + c = 0 with a = 1 − x, b = −x, c = x³. The discriminant of this quadratic in y is D = b² − 4ac = (−x)² − 4(1 − x)x³ = x² − 4x³ + 4x⁴ = (2x² − x)² and using the quadratic formula y = (−b ± √D)/2a we therefore get y = (x − (2x² − x))/(2(1 − x)) ⋁ y = (x + (2x² − x))/(2(1 − x)) which gives y = x ⋁ y = x²/(1 − x) and replacing y with √3 again this gives x = √3 ⋁ x²/(1 − x) = √3 Of course, x²/(1 − x) = √3 gives x² + √3·x − √3 = 0 which is exactly the quadratic we get by factoring.
@PrimeNewtons4 ай бұрын
Of course!
@evbdevy3525 ай бұрын
You could be a great actor.
@PrimeNewtons5 ай бұрын
I am! The board is my stage.
@evbdevy3525 ай бұрын
@@PrimeNewtons Congratulations.I wish you success.Thanks a lot.
@loggerkey69055 ай бұрын
6:22 😂😂
@user-nd7th3hy4l5 ай бұрын
Une solution unique x comprise entre -1 et 0.
@subhashchandra-yo4rb19 күн бұрын
You could do it without substituting √ 3 as y😊
@BozskaCastica5 ай бұрын
Yeah did it before I watched the video. But I didn't do the substitution.
@bhchoi83575 ай бұрын
Love you
@egondanemmanueltchicaya10895 ай бұрын
😂😂😂😂 impressive
@dante224real15 ай бұрын
D=eSnu+5 /(0-nuR)^CHin find intergers that fit D=N+CHi=69
You have some strange Wolfram Alpha! Normal Wolfram Alpha gives a perfectly good short solution! Why are you misleading your subscribers?
@rainerzufall425 ай бұрын
At least two people out there thinking, that this shouldn't be too complicated for Wolfram Alpha...
@rainerzufall425 ай бұрын
BTW: What is this video other than guessing the root x = sqrt(3) and finding the other two roots? For example: x³-(3+sqrt(3))x+4=0 has two complex roots, that are ugly, although I just wrote 4 instead of 3. The real root is: x = -(3 + sqrt(3) + 3^(1/3) (6 - sqrt(2 (9 - 5 sqrt(3))))^(2/3))/(3^(2/3) (6 - sqrt(2 (9 - 5 sqrt(3))))^(1/3))