A very nice olympiad maths question | x^6=(x-1)^6 | You need to know this trick | Algebra
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- เผยแพร่เมื่อ 8 พ.ค. 2024
- See the way I breakdown the solution of this question. There is a lot you can learn from this video.
How to solve x^6=(x-1)^6
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You need to know these tricks: Substitute x=y+1/2 into the given equation and rearrange to (y+1/2)^6-(y-1/2)^6=0. Pascal's triangle 1 6 15 20 15 6 1
2[6(1/2)y^5+20(1/2)^3*y^3+6(1/2)^5*y]=0. Expand: 6y^5+5y^3+(3/8)y=0. Divide by y and multiply by 8: 48y^4+40y^2+3=0.
Apply quadratic formula: y^2=(-40±32)/96=-1/12 or -3/4. Take square roots: y=±i√3/6 or ±i√3/2. Also note that y=0.
Thus, x=y+1/2=(3±i√3)/6 or (1±i√3)/2 or 1/2.
extremely neat! thanks!
Also thought about Pascals triangle
very good
Thank you. Glad you enjoyed it.
oho ur great
I would lije ti apperciat ur methods
Thanks. I am grateful.
I always use b/2 = arithmetic mean of roots, A, and c = geometric mean squared of roots, G, with (x)2 + b*x + c for x = A +- sqrt((A)2 - G). It is a much quicker route to quadratic roots.
By inspection x=½ --> x-1=-½ is one of the root, as raising both of them to an even number yield the same positive number.
x/(x -1) = e ^i(k π/3) for k = 1 2, 3, 4,5..There are 5 roots as this is a polynomial of degree 5.
Excellent
Almost exactly my thought when I saw the problem.
By visual inspection, 0.5 is clearly a solution. Also note that if we expand (x-1)^6 it will lead off with x^6, which will cancel. Therefore we only expect five roots. The four roots other than 0.5 will be complex. To find them we proceed as follows.
Divide both sides by x^6 to get
(x-1)^6 / x^6 = 1
[(x-1)/x]^6 = 1
(1 - 1/x)^6 = 1
To satisfy this, 1 - 1/x is one of the sixth roots of unity. These of course are at magnitude 1, angle a multiple of 60 degrees: Also, given the symmetry of the problem negating the right hand side doesn't change the solution set, and makes further work slightly cleaner. We have
-1 + 1/x = 1
-1 + 1/x = -1
-1 + 1/x = cos(60) + j*sin(60)
-1 + 1/x = cos(60) - j*sin(60)
-1 + 1/x = -cos(60) + j*sin(60)
-1 + 1/x = -cos(60) - j*sin(60)
Now add 1 to both sides of every equation:
1/x = 2
1/x = 0
Excellent
Given the symmetry, the only real root has to be 0.5. obvious. Using your transformation the other four roots are the four corners on the unit circle. Excellent.
L'équation est d'autant plus intéressante, qu'au premier coup d'œil, elle paraît impossible dans R, comme l'équation x = x - 1.
(Dans un esprit purement constructif) quand la correction est trop longue, l'exercice paraît artificiellement compliqué.
Il me semble que votre correction peut être allégée en considérant que si votre interlocuteur s'intéresse à une équation du 6e degré, alors on peut supposer qu'il connaît les identités remarquables, qu'il sait résoudre une équation du second degré, et qu'il est à l'aise avec les calculs algébriques.
Un exemple extrait de votre corrigé, l'expression:
( x -1 )^2 + x ( x - 1 ) se factorise aisément en
( x - 1 ) ( x + 1 )
= x^2 -1
sans explication nécessaire et sans développement.
x=0,5 0,5 - 1= - 0,5
@@habeebalbarghothy6320what
Ok, going for the representation of the roots, we get that there are 5 roots, you showed 4 wich are on the complex plane, how would we find the one on the real plane? I can guess it's +1/2, but how do we prove it?
In the video, I have one real root and 4 complex roots, making it five roots altogether. You can go ahead and watch how I got 1/2 as the real root.
Easy, the solution is x = ω. Any limit ordinal could work here really. Because any limit ordinal minus 1 is just that limit ordinal itself since it doesn't have any predecessor. As a result, there's more than 1 solution because we can't put a value of the number of limit ordinals there are. Keep in mind that limit ordinals are multiples of ω.
That's excellent, brother.
x^6 그래프를 상상해보면, y축을 기준으로 좌우 대칭이니 그 그래프를 오른쪽으로 1만큼 이동시킨 그래프와의 교점의 x좌표가 0과 1의 1:1 내분점일꺼라 생각할래요. 그편이 훨씬 간단하고 저에겐 더 아름답게 느껴지거든요
That's fantastic, brother.
Very good. But speed up speed up. You are toooooo slow.
Okay, thanks for the tip.
@@SpencersAcademy An Olympiad aspirant is very much aware of the basic identities. So, no need to reiterate those. You can directly jump redundant steps.
Thank you.
@@tirtharajbanerjee Point noted. Thanks for the tips
Lấy căn bậc 6 hai vế,ta có x=x-1 hay x-x=-1 hãy x-x=-1 hay0x=-1h ấy x=-1/0.Phương trình vô nghiệm
Why not solve 1=(1-y)^6 where y=1/x
x^6=(x-1)^6 then x=+/-(x-1) x=x-1 no solution other x=-(x-1) hence x=1/2
Esse trem em português já é complicado imagina em grego. Fala português pra facilitar.
the easier way is to apply square root on both sides of the equation.
x=1/2....
x^6=(x-1)^6 positive for all x . Then ((x-1)/x)^6 = 1. (x-1)/x is 6th root of unity. Then, (x-1)/x = e^(i2kpi/6)=> x=1/(1- e^(i2kpi/6)) for k=0,1..,5. For k=0, there's no solution. For k=3, x=1/2. Substitute k=1, 2, 4,5 and obtain two pairs of complex conjugates for x.
Fantastic. You nailed it.💯
@@SpencersAcademy Okay I nail your Olympiad math problems. Let's see if you can nail our 9th grade regular maths problems. Solve sin(9x)=sin(5x)+sin(3x). You may record a video if you wish.
You must have 6 solutions. I contend in the X yi plane, you have two coincidental roots at -.5.
In a question like this, there exist only five roots.
Let me explain it this way:
Let's say you expand the RHS using binomial expansion. You'll have something like this:
X^6 = x^6 - 6x^5 +15x^4 - 20x^3 + 15x^2 -6x +1.
Subtract x^6 from both sides makes the equation 5th-degree polynomial. And that's why we've got five roots.
7:22 "your trick" is more confusing than just applying the formula.
Otherwise an excellent video, the most challanging problem I saw for today 😉
Glad it helped!
Shud be a quintic equation
No real root. X cannot equal x minus 2
X=1/2
If u take the 6th root, u get x=plus/minus( x-1). The plus sign gives u no solution but the minus sign gives x=-x+1 so x=1/2
That's fantastic. There are other solutions for x. You just got one. Which is the only real value.
1/2
why you don't take the 6th root for both side at first?
When you do that, you're gonna miss other vital solutions for x
It will become x = x+1 which has no solution
@@dante_loves_plzzaIf power is even |x|=x+1
@@cosmolbfu67 that's |x^2n| = (x+1)^2n, I'm talking about x = x+1
If you do this, some of the results will be not available
When would this equation to solve appear in a "real-world" situation? 🤔
Probably the stuff used to make the concepts of the equasions used to make the components for the machine you are typing that question on.
Quite a lot . Work on QM , and EM.
What are the acronyms QM and EM? Please clarify.
Should have worked on 6 answers.
Why? The polynomial has degree 5
X is to the power of 6!
@@mk-jl3zd yep, but there is a subtraction
Valore assoluto di x = valore assoluto di ( x -1) e finisce in un minuto
You would o ly be having just one value of x. There are still other values