Math Olympiad - Find all roots of a cubic equation | Be Careful
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- Опубліковано 8 вер 2024
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Easy: x^3=27 in complex form : 27 exp[2k*i Pi[. So, 3 roots are: 3 exp[2k*i Pi/3]: for k=0 Root = 3, for k=1 root = -3/2+i Sqrt(3)/2, for k=2 root =-3/2-i Sqrt(3)/2
Imaginary parts multiplied by 3 forgotten
@@marianondrejkovic2084 :D
Olympiad problem? I think not, O-level further maths for me. Roots of a number are always distributed equally around a circle in the Argand plane.
The word Olympiad is a clickbait
Huh?
If I remember well putting those solutions on a complex plane should result in the vertex of an equilateral triangle.
That's pleasing.
Once you get 3 the other 2 solutions are simply 3*e^(±2πi/3). (e^(2π/3))³=e^(2π)
So simply multiply the real root by the primitve cube roots of unity to get the two complex roots
This is one case where the aficionados of the polar approach could quickly get to the same answer since most would know Cos[2 pi/3] = -1/2, Sin[2 pi/3] Sqrt[3]/2 etc.
The author, as usual, "forgot" to say when posing the problem in which SET he wants to search for x: among real or complex values. Both the solution method and the answer depend on this.
Thanks for pointing this. These YT videos almost never bother to mention this essential input.
“Find all roots” implies complex roots, doesn’t it?
@@stvp68No, not necessarily. There may well be solutions in other algebras over other rings/fields/domains. For instance in a Clifford algebra over the quaternions, octonions, split complex numbers, extended complex numbers, dual numbers or a Z[27] field. (But I did not think of solving any of those) The maths world has a lot more space to navigate....
@@bendunselman ah, okay, thanks!
@@joso5554លៀ😢៩ផ៥ហឱអ:យ🎉លេីៀ+ឱ🎉ន🎉ៀល៤ឬៀនរ៥ល😊្
To the uninitiated, once the word ALL roots are mentioned in a Math problem, the alarm bell sould sound off that the soultion set MUST involve the Complex Domain, eg. 1 can also be represented as 1+0i & even 0 as 0+0i, so a typical algebraic equation can have both Real & Complex solutions....🤔
Yes one root is 3 and the other two are complex i.e complex conjugate
wow!
Great!
It's simple look X^3=27 find the value of x=? 1step x^3=27, 2step x=rootunder27, 3step x=3 because 3 is the cube number of 27
Once you get the difference of squares, difference of cubes and sum of cubes down, its not that hard.
Not necessary for solving this equation.
Is there anyone from INDIA❤
Using cube roots of unity concept . its in iit jee syllabus
EINSTEIN in the thumbnail ? 😢😢 Apmk he's NOT a mathematician, nor was he a CHAMP in maths 😮😮 PLEASE check it out.. and let us know 😊😊
Quite interesting!
x^3 = 27
x^3 = 3^3.e^(i.2nπ)
x = 3.e^(i.2nπ/3)
x = 3 or x = -3/2 + i.3√3/2 or x = -3/2 - i.3√3/2
The complex solution is best arrived at using Euler's formula. The author's algebraic method is long and tedious as well as essentially unnecessary!
@@LloydCash-he1qv This guy is a Bozo. All his solutions are long and tedious.
(x ➖ 3ix+3i )
3, -1,5+_1,5iV3.
:/ x³=27
x=3√27=>3
I just did the x^3=1 version of this today 😂😂
X=3
۳×۳×۳=۲۷
I feel like Nikolas tesla 😊
Try a limit but that has no solution
3x3x3=27 that's it.
-3i
3
You forgot that there exist somethings like log .
Why should one use log for this equations???
@@bjornfeuerbacher5514 it makes it much easier than ever .
@@azhanahmedali9851 ??? Sorry, I don't see it. Please explain to me how one solves this equation using log.
@@bjornfeuerbacher5514 To convert the equation x³ = 27 into logarithmic form, we can use the property of logarithms that states:
logₐ(b) = c ⇔ a^c = b
In this case, we can rewrite the equation as:
x³ = 27
logₓ(27) = 3
Or, alternatively:
3 logₓ(x) = log₇(27)
Since 27 is 3³, we can also write:
3 logₓ(x) = 3 log₇(3)
logₓ(x) = log₇(3)
Note that the base of the logarithm can be any positive real number, but in this case, I've used base 7 (since 27 = 3³) and base x (to maintain the original variable).
@@azhanahmedali9851 "logₓ(27) = 3"
This does not help for solving the equation.
"3 logₓ(x) = log₇(27)"
??? Where did you get that from? That's simply not true!
"Since 27 is 3³
Err, as soon as you recognize that, you have found the (real) solution of the equation x³ = 27. So you just demonstrated yourself that log does not help in any way for solving this equation - you only have to see that 3³ = 27, and you are done!
"but in this case, I've used base 7 (since 27 = 3³)"
????? There is no reason at all for using base 7 here! Did you perhaps mean base 27?
The complex roots seem unmeaningful.
Why?
@@bjornfeuerbacher5514 X^3 is not a circular or repeating function, it tends to infinity. There is one intersection of x^3 and 27 when x = 3. the cube of a negative number is negative and 27 is positive.
@@tombufford136 Yes so all what you have written. But so what??? What does all of that have to do with your assertion that the complex roots are "unmeaningful"?
😄😄🤣🤣🤣🤣😂😂
9(3)=27 so its Simple 9+9+9=27
Be careful of what? This is a straightforward problem , not really higher maths.
you forgot about -3
(-3)^3=-27
i forgor @@DedMatveev
(x ➖ 3ix+3i)
Math is harrrdd.....
Not an olympiad question at all. And an horrible error in line 2 on the board, because (inside the complex numbers, which are needed anyway), x = 3 is _not_ the only solution of the equation. Do not present it like this
Just 3
3