Math Olympiad | A Nice Algebra Problem | How to solve for X in this problem ?
Vložit
- čas přidán 23. 01. 2024
- Hello My Dear Family😍😍😍
I hope you all are well 🤗🤗🤗
If you like this video about
How to solve this math problem
please Like & Subscribe my channel as it helps me alot ，🙏🙏🙏🙏
Here, we have
x⁶ = (x-2)⁶ so we have following situations:
(i) x=x-2 i.e. 0= -2 that is not possible
(ii) both sides the power is even so x and (x-2) are additive
inverse of each other.
-x = x-2
-2x = -2
So x=1
nice video but I have to point it it can become confusing if you alter between x and • for multiplying. if you have algebra in the question, you should never use x for multiplying
x^6=(x-2)^6 | ^1/6
|x|=|x-2|
x=x-2 or -x=x-2
0=-2 x=1
x=1
No one needs complex numbers
X=1 is the only real root. The other four are complex.
Great solution 👌👌👌
Brill 👍🏿
This problem isn't stated properly. It should specify *which* solutions to find, or, equivalently, *how many* solutions to find. The way it is stated, it makes it seem like the problem is asking for a single value for x, in which case x = 1 is the most obvious and natural solution. A more appropriate way to state this problem would be "find all complex numbers x satisfying the equation x^6 = (x-2)^6".
Still, nice video! :)
А если разложить как разность кубов сначала
(х^2-(х-2)^2)*(х^4+(х(х-2))^4+(х-2)^4)=0
Легко доказать что второй множитель всегда больше чем 0, т.е. нет таких действительных значений х тогда остаётся только
Х^2-(х-2)^2=0, что решается легко
X=-(X-2) X=1
Veel te ingewikkeld
Can be solved much easier in less than a minut.
X^6=(X-2)^6
X^(6/3)=(X-2)^(6/3)
X^2=(x-2)^2
X^2=X^2-4X+4
0=-4X+4
4X=4
X=1
Orig.eq.
1^6=(1-2)^6 ==> 1=(-1)^6
1=1
Решается в уме за 10 секунд !
вы нашли все комплексные корни за 10 секунд? Браво.
Зд. быстро можно найти только действительные корни.
Х=1 - только я за пару сек решил это уравнерие устно🤔?ико ту ван Карл, после стакана красного - икату ван
Just look at it. Try 1.
Clearly x ≠0, so dividing by x we get: (1-2/x)^6 = 1; so:
1-2/x = exp(2nπ i /6) whr n= 0,1,...5 and i = sqrt (-1).
Another way, in the last step is to write:
(1 - 2/x) = ± ( 1, ω, ω²)
X=1
Х = 1
Done in few seconds.
Where is the sixth solution!
Notice that x to the power of 6 cancel out upon binomial expansion so this is essentially a polynomial of degree 5.
1
Thanks! But...
why is the square root of (x-2)⁶ equal to (x-2)³ and not (2-x)³?
So, √((x-2)⁶) = |(x-2)³l, and not (x-2)³ .
Is his answer correct?
Surely there should be 6 solutions?
The x to the 6th terms cancel out after multiplying out (x-2) to the 6th, so the polynomial ends up being of degree 5, not 6.
@@Jon60987 Point. Mea culpa,
1 (here real).
X est négatif c'est - 2
If you make the substitution x = y + 1, the solution simplifies very quickly
А где шестой корень потерял?
По изв. теореме число корней р этого уравн. не >6 => (р=6 V p
x=1 lol
1 is the only real root
If you are going to find complex roots, you can do it a little differently.
(1) x/(x-2)=t ; (2) x=2*t/(t-1) t=NO=1; (3) x^6/(x-2)^6=t^6=1 ; (4) t^6=e^(2*pi*i*n ; n - integer . So : (5) t=e^(i*pi*n/3) . (6) n=0 : t1=e^(i*0)=e^0=1 NO !! ; n=1 :
t2=e^(i*pi/3)=cos(pi/3)+i*sin(pi/3)=0,5*(1+i*sqrt(3) ) ; n=2 : t3=e^(2*i*pi/3)=…=-0,5*(1-i*sqrt(3) ) ; n=3 : t4=e^(3*i*pi/3)=-1 ; ………. n=5 : t6=e^(5*i*pi/3)=….=0,5*(1-sqrt(3) ) ; n=6 : t7=t1 …..
We substitute (6) in (2) - we get Your answer.
With respect , Lidiy
Congratulations. You just refuted the fundamental theorem of algebra🤣🤣🤣. " Every, non zero, single variable, degree n polynomial with complex coefficients has, counted with mutiplicity, exactly n complex roots". You are bigger than Gauss 😂😂😂😂
In fact, this equation has order 5.
@@gthrjkzwbz I know, I only said it as a joke, ironically.😉
Very boring! Such a simple maths, again and again talk too much and write the basic formulas!
Shit math question