Solving a Cubic equation Using an Algebraic Trick
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- Опубліковано 30 жов 2020
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I love how you explain how to think in these challenging problems in stead of just doing everything correct right away, because many solutions can involve steps that most people wouldn't come up with themselves.
Thank you!
That's depressing why do you say most ppl woukdnt come up with themselves..are most ppl not very smart then?
Coming up with something like this in your own is something only extremely smart people are likely to figure out.
To get all the solution, when you have 2x^3 - (x-2)^3=0, you can factor out by using a^3 - b^3 = (a-b)(a^2 + ab + b^2) and we know that a = crt2*x and b = x-2 so we can substitute : (crt2*x-x+2)((crt2*x)^2 -crt2*x(x-2) + (x-2)^2) = 0 => [(crt2 - 1)x + 2][(crt2)^2*x^2 - crt2*x^2 + 2crt2*x + x^2 - 4x + 4] = 0 -> [(crt2 - 1)x + 2][(crt2^2 - crt2 + 1)*x^2 + (2crt2 - 4)*x + 4] = 0. And then, we get (crt2 - 1)x + 2 = 0 or (crt2^2 - crt2 +1)x^2 + (2crt2 - 4)x + 4 = 0
So for the first equation, we get x = - 2/(crt2-1) and for second, delta = 4*crt2^2 - 16*crt2 + 16 - 16crt2^2 + 16crt2 - 16 = - 12*crt2^2 so x1= - [crt2^2 - crt2 + 1 + i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2) and x2 = - [crt2^2 - crt2 + 1 - i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2)
So the answer are x1 = - 2/(crt2-1), x2= - [crt2^2 - crt2 + 1 + i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2) and x3 = - [crt2^2 - crt2 + 1 - i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2)
Crt = cuberoot and sqrt = squareroot
At 8:40 you take the cube root of both sides. Let w = (-1+sqrt(-3))/2. We can conclude cbrt(2) = x-2, but we could also conclude w*cbrt(2)=x-2, as well as w^2*cbrt(2) = x-2, and this gives the other two roots.
Nice!
@Arthur So the idea is to only put positive numbers in a square root sign. Seems good to me. Since the square root symbol means the positive root, that should help.
Teaching people the thought process to search for a solution. Very good!
This is smart!
Thanks :)
Nice trick! It was the third one I tried. Finding the complex roots isn't too hard, but writing them on the screen would make everyone's eyes glaze over.
That’s is so amazing trick and clearly how to do it ,great explaining!
8:18 There are two more solutions. 2x^3-(x-2)3=[2^(1/3)x-(x-2)][2^(2/3)x^2+2^(1/3)x(x-2)+(x-2)^2]=0 First term is real solution (SyberMath solution). Second term giver two complex conjugate solutions.
I usually ignore the complex solutions but you're right!
Beautiful way of solving the problem,enriching my knowledge.🙏🙏
It's my pleasure!
Solving this equation using lagrange’s resolvent and cardano’s formula in another video would be great sir!
Sir, you have showed only the real solution and rationalised the denominator but there are also two other complex roots(also because by definition, a cubic equation has three roots) consisting of omega(cube root of 1), thank you for this solution....
No problem. Yes there are complex solutions!
Just to understand: what’s omega?
Following my standard cubic method was less grungy than usual with this equation. I wound up with that answer, but I tried in vain to simplify it. I think that it took about as long as your method.
...nice job explaining the real number answer. Very smooth!
Thanks!
Hands-down the best mathematician on UA-cam he and his problems are unmatched and every single problem applies so many different concepts. His explanation is always on point.
He's actually not a mathematician - he just enjoys solving math problems. However, he would be very flattered!
Magnificent mathematical innovation!
A great one!
Great solution. Thank you. I learned one more.
Great to hear!
@@SyberMath So far I got that this expression equals (x+2)^3 -12x but NOW WHAT..surely this is what most very intelligent people wkuld do..i havent watched your solution yet but what then do you have (x +2)^3= 24x and take the cube root of both sides?? What else i don't see why wohld multiply the equation by something because there's no pattern tnere or to dividie by anything..hope you can resoo d..
You are amazing !
Can you do more cubic equation with all type of solution please?
x³+6x²-12x+8=0
The coefficients hint at a perfect cube, dividing by the binomial coefficients ( 1 3 3 1)
we get 1 2 -4 8 neatly powers of two (signs are a mess) then, as per video for 3 steps
(x+2)³=x³+6x²+12x+8
(x-2)³=x³-6x²+12x-8
so 2x³-(x-2)³=x³+6x²-12x+8=0
2x³=(x-2)³
or ( (∛2.x)/(x-2) )³=1 (NB ∛ is the cube root - it doesn't look very readable on screen)
Euler's formula: e^iθ = cosθ + i sinθ
1=e^(2inπ) has (3rd) roots at n=0,1,2 i.e. e^0, e^(2iπ/3), e^(4iπ/3) so 1, -(1/2)±(i√3/2)
for a solution s (one of the above)
(∛2.x)/(x-2)=s
x(∛2-s)=-2s
x=2s/(s-∛2)
s=1 => 2/(1-∛2) × (1+∛2+(∛2)²)/(1+∛2+(∛2)²)
x=2(1+∛2+(∛2)²)/(1-2)
x=-2-2^(4/3)-2^(5/3)
s=-(1/2)±(i√3/2)
x=2*(-(1/2)±(i√3/2))/(-(1/2)±(i√3/2)-∛2)
=2*(-(1/2)±(i√3/2))/(-(1/2+∛2)±(i√3/2))
after a whole mess of rearranging, the other solutions are
x = -2+∛2(1+i√3)+(∛2)²(1-i√3)
x = -2+∛2(1-i√3)+(∛2)²(1+i√3)
Awesome!
Nice trick!
One can divide the equation by 8 and let y = -x/2. So the equation becomes: -y^3 + 3y^2 + 3y + 1 = 0. 2y^3 = y^3 + 3y^2 + 3y + 1 = (y + 1)^3.
2 = (1 + 1/y)^3. Taking cubic root, 1 + 1/y = r, where r = cubicRoot(2). There are 3 cubic roots of 2. r represents any of them.
y = 1 / [r - 1] = [r^2 + r + 1 ]. As y = -x/2, x = -2 * [r^2 + r + 1 ].
...
Beatifull amazing maths is beaty
Great!
Very nice. If you don’t have a calculator handy you can approximate the root using recursion and the system y=24x, y=(x+2)^3 with x[0] less than zero.
More succinctly, just use the recursive relationship x[i+1]=-(1/24)(-x[i]+2)^3 with x[0]
Nice!
Moço, você é maravilhoso. Conhece demais.(Brasil). (Young man, you are wonderful. You know too much.)
How about this method of using (x+y)^3 pattern ?
(x+2)^3 = x^3 + 3(x)(2)(x+2) + 2^3 = 24x = 24(x+2) - 48
Let y = x+2 and m = -8 and n = -24. Then, y^3 + 3my = 2n.
So, cubic-discriminant D = n^2 + m^3 = 576 - 512 = 64 = 8^2. Therefore x+2 = y = ∛(n + √D) + ∛(n - √D) = ∛(-24 + √64) + ∛(-24 - √64) = - ∛32 - ∛16
Hence, *x = -2 - ∛32 - ∛16*
All these different seemingly random equations with all these different tricks.I think this is crazy in one sense at least.Yes,algebraic skill is developed and that is laudable,but it makes me feel that something is amiss with all this stuff.Structural relations between different objects to find equation solving universality I crave.I pray to the mathematical gods to let me get to the bottom of all this.
5:45 This line seems like my native language dialogue 😂
Especial thanks to thinking loudly for solving this. I like this method becsuse of being normal .thanks again.
Most welcome 😊
I lobe this sooooo much :D
Thanks!
Thats great
why the crossing out. the factor of (x-2) is clear. so (x-2)(x^2-4) +6x(x-2)=0
So beautiful! 😇👏👏👏
Could you solve this equation using cardano’s formula sir? Thank you a lot sir
thx again dear teacher ..
You're welcome 😊
Alt.
(2-x)^3+2x^3=0
B/c you can flip the 2 to the front part of the equation. It's a bit more cleaner to.
How would you know to arrange it that way?
David Flanders Let 2=y, and notice how in a binomial expansion the signs alternate. Variables are just #'s.
Can you suggest a book which has problems in algebra of this level
Hard to find one book but I'm compiling a list of good books and will share them in the community tab soon. Feel free to remind me if I forget!
ua-cam.com/channels/W4czokv40JYR-w7u6aXZ3g.htmlcommunity
So neat.
Thanks!
Unbelievable
I am bangladeshi .but your class is very nice so I see and flow your lecture
Many many thanks. Greetings from the United States! 💖
smarter than smartest...
i got solution
thanks ❤️
Thank you for the kind words! 💖
That's ONE or the solutions. What about the other 2?
Simply multiply whole equation with (-1). Further the given equation must change in to (a-b) ^3 form
Tjat doesn't do enough..that just gives you something that equals-(x -2)^3 but then what? I don't see what you can do
Sub y = x + 2 to get y³ - 24y + 48 = 0 and Cardano it. I see the trick you were trying to show, but the messing around with the irrational denominator undid all the good work.
Nice!
Very good
Thank you
-2 is also a real solution how to find that
awesome
Thank you, Antoine! 🤩
You hsce x minus a multiply x minus b multiply x minus c is equal to that cubic equstiion too
You are best
Thank you! 😍
Why didn’t you just find one factor then use long division?? Would have been quicker than all this mucking about.
-7.69
👍
Abrakadabra....🤣
😂 Ehehe! This is like magic, isn't it?
👍👍👍
Tricks work if somebody gives you a cubic where they work. Chang one coefficient or a sign and you may have to throw in the towel. Give me something that works ALL the time, without tricks.
Please don’t write at the bottom of the screen. We can’t see under the subtitles.
I keep forgetting 😁
it was lengthy problem and not tricky
3X+12X-12X+8=0
In exam you do not have so much time to try all that
Right but you have time if you're not taking a test
apenii😂xpahamla
Mathemagic👍
Hehe! Thanks! 😁
This is wrong, x should have 3 answers. Not one answer. x=2, x=2(sqrt(3)-2) , x=-2(sqrt(3)+2)
All the cubic equations are required to have three answers?
I got x=-2.
How?
-(2*2*2)+6*-1(2*2)-12(-2)+8=0. That's how I got x=-2. Check it out.
It's -8-24+24+8=0
@@carloshuertas4734 doesn't work
@@carloshuertas4734 you multiplied wrong on the x² term
For God’s sake remove the subtitles especially that you are talking English!
You can turn off the subtitles