Math Olympiad | A Nice Algebra Problem | How to solve for X in this problem ?
2024 ж. 23 Қаң.
54 003 Рет қаралды
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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
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=1 is the only real root. The other four are complex.
Brill 👍🏿
Great solution 👌👌👌
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! :)
X=-(X-2) X=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
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
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)
If you make the substitution x = y + 1, the solution simplifies very quickly
Х = 1 Done in few seconds.
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?
1
X=1
X est négatif c'est - 2
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.
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).
А где шестой корень потерял?
По изв. теореме число корней р этого уравн. не >6 => (р=6 V p
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.😉
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
x=1 lol
Shit math question