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.
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
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].
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
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: th-cam.com/video/YnZzpYSIiUU/w-d-xo.html 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.
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.
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
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))
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
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))
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!
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.
As always, I love your presentation-clarity, enthusiasm, and great blackboard technique. The cubic formula leads solutions. It can also lead to nested radicals which can be tricky to simplify. The form of this equation suggests trying x=sqrt(3) as a solution.
I havent watched the video yet just clicked on it and here is my solution: By hit and trial, x=√3 is a solution. So x³-(√3+3)x+3= x³-√3x -3x +3= 0 => (Adding and subtracting √3x² and reareanging the equation), x³-√3x²+√3x²-3x-√3x+3=0 =>( x²+√3x-√3)(x-√3)=0 From here, x=√3 is a solution. If we see quadratic, D= b²-4ac= 3+4√3 Hence x= (-√3±√(3+4√3))/2 Hence we get three solutions of the equation: x=√3, (-√3±√(3+4√3))/2
Replace x = √3 in the polynom and calculate ,P(√3)=0 , x = √3 is a root of the polynom divide (Ecludian division) the polynom per (x-√3) and obtain a second degree polynom resolve the second degree polynom =0.
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 😊
O did the same, but i did not use y. Just presume It was a Square, and made The math with Square 3 anyway. Your solution is Just more beautiful to watch. 😂
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!
It was really useful. But my take would be assuming the three possible roots and going forth with their sum, product and sum of products to find the value of x.
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 !!
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 😁
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)
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.
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.
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.
Depressed cubic formula 😂
🤓
My first thought as well 😂
Treat a known value as an unknown to be able to use algebraic identities: Bravo! Beautiful!
have you checked the answers given by the computer ? You got 3 real roots, while the computer solutions seem to be 3 complex roots.
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
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 .
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].
Unfortunate that sqrt(3+4sqrt(3)) doesnt seem to simplify nicely unless I miss something obvious.
no it doesnt, unfortunately.
I always see to make it perfect square inside the root so that it comes out, but unfortunately here it doesn't work
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.
hello
I wonder how it is possible with wolfram alfa it is a machine
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
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:
th-cam.com/video/YnZzpYSIiUU/w-d-xo.html
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.
Of course!
I like your smile
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.
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
You remind me of the best teacher I ever had (Physics, England, 1957).
Professor, you have the great ability to manage the algebra beautifully, and those substitutions are a master piece!
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))
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).
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
Your channel is such a hidden jewel man, I love your videos.
I think that it would have been better to factor out sqrt(3) from -sqrt(3)+3. Real nice job as usual.
Very clever solution. The first solution you showed was very non-Olympiad!
You have some strange Wolfram Alpha! Normal Wolfram Alpha gives a perfectly good short solution! Why are you misleading your subscribers?
At least two people out there thinking, that this shouldn't be too complicated for Wolfram Alpha...
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))
when i put it in wolfram i got your same answer.... maybe a typo on entry or they fixed it?
Yeah did it before I watched the video. But I didn't do the substitution.
By observation,
x = √3
It's easy to find the other solutions from there.
At olympiad there is't solution by observation!
@@dandeleanu3648Why not? Observation is legitimate, as long as you prove it's correct.
so just say by observation as the proof then as the top comment says@@koenth2359
@@dandeleanu3648 se você faz os cálculos não está errado não
That's the ideal way to solve it. The question is how to proceed if you don't see that.
Brilliant thinking
Watch out y squared is 3 not radical 3 so the solution might be even simpler
🦋I SMILED THROUGHOUT THE WHOLE VIDEOOO THANK U SO MUCHHHHHHH🦋
I don’t watch for the math I watch to see a black guy stare at me with no audio
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!
D=eSnu+5 /(0-nuR)^CHin
find intergers that fit D=N+CHi=69
Very nice video
The real name of quadratic formula is.....Shreedha Ahcharya formula .
Regards 🙏
Fucking FANTASTIC, my friend
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.
That's an interesting convergence
@@PrimeNewtons ha ha, true!
As always, I love your presentation-clarity, enthusiasm, and great blackboard technique. The cubic formula leads solutions. It can also lead to nested radicals which can be tricky to simplify. The form of this equation suggests trying x=sqrt(3) as a solution.
با درود،یک خلاقیت در حل این مسیله به خرج دادی،بسیار سپاسگزارم
Gracias. Me gustó el video
Saludos
X=3^0,5 et on divise pour les deux restes.
The question was so easy to me, but it was your presentation style which made me a fan of yours❤ Thank you sir 🥰
I havent watched the video yet just clicked on it and here is my solution:
By hit and trial, x=√3 is a solution.
So x³-(√3+3)x+3= x³-√3x -3x +3= 0
=> (Adding and subtracting √3x² and reareanging the equation), x³-√3x²+√3x²-3x-√3x+3=0
=>( x²+√3x-√3)(x-√3)=0
From here, x=√3 is a solution.
If we see quadratic, D= b²-4ac= 3+4√3
Hence x= (-√3±√(3+4√3))/2
Hence we get three solutions of the equation:
x=√3, (-√3±√(3+4√3))/2
i think the question was easy great fan
For me it was easier to "see" that sqrt(3) was a root, then getting the 2nd degree polynomial, than the substitution+difference_of_squares
Replace x = √3 in the polynom and calculate ,P(√3)=0 , x = √3 is a root of the polynom divide (Ecludian division) the polynom per (x-√3) and obtain a second degree polynom
resolve the second degree polynom =0.
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 😊
Actually, by observation x = √3 looks like a solution and indeed it is 😆.
So, divide the cubic by x - √3 to find the quadratic formular.
I dunno why your videos don't pop up here for more than 3 months !!
Gr8 work! Keep it up bro.
Une solution unique x comprise entre -1 et 0.
Une solution unique x comprise entre -1 et 0.
O did the same, but i did not use y. Just presume It was a Square, and made The math with Square 3 anyway. Your solution is Just more beautiful to watch. 😂
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!
6:22 😂😂
It was really useful. But my take would be assuming the three possible roots and going forth with their sum, product and sum of products to find the value of x.
😂😂😂😂 impressive
I treated x as constant 3^1/2 as variable (p say) and solved for p. Things fell in place easily.
The method is so smart that I couldn't sleep thinking of how smart was that method
I got it right.
nice night
teacher, your channel is great
Love you
Can you make video on the theory and make a playlist so people can learn new concepts of math which they don't know
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 !!
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 😁
Simply the quadratic formular were left the same way as the solution.
This is a really easy problem.
Very good
Super ❤❤❤❤❤❤
What an Idea!
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)
I'm thinking about this
,Plz what is the name of the method that you use to find the solution 2 and 3, who know it , he can answer
Awesome, mate!
What an elegant solution!
( x - √3) ( x^2 + x √3 + √3) = 0
x = √ 3 , ( - √3 + √ ( 3/4 - √3))
- ( √3 + √ ( 3/4 - √3))
x = -√3 , other solution can be fine easily
My a level maths teacher exects me to do this in 30 seconds
Wow !!! Good work !!!!
x₁ + x₂ + x₃ = 0, interesting solutions.
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.
Ty for all these videos !!! Love them 😍
excellent!
Awesome method!
Top notch 👍
amazing solution
You could do it without substituting √ 3 as y😊
Awesome presentation.
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.
You could be a great actor.
I am! The board is my stage.
@@PrimeNewtons Congratulations.I wish you success.Thanks a lot.
(x+√3) is a factor...very easy