Math Olympiad - Find all roots of a cubic equation | Be Careful

[@kpdywo848]

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

[@marianondrejkovic2084]

Imaginary parts multiplied by 3 forgotten

[@piotrstrzelczyk5013]

@@marianondrejkovic2084 :D

[@motivationer7184]

Simply saying 3,3omega, 3 omega square.

[@Grim_Reaper_from_Hell]

Once you get 3 the other 2 solutions are simply 3*e^(±2πi/3). (e^(2π/3))³=e^(2π)

[@ShivamSharma20480]

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

[@ClaudioBrogliato]

If I remember well putting those solutions on a complex plane should result in the vertex of an equilateral triangle.

[@ClarkPotter]

That's pleasing.

[@donaldasayers]

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.

[@Grim_Reaper_from_Hell]

The word Olympiad is a clickbait

[@SPDLand]

Huh?

[@joeldurugbo3147]

So simply multiply the real root by the primitve cube roots of unity to get the two complex roots

[@tassiedevil2200]

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.

[@mohamad-hu9vv]

Great!

[@sagarrajak3826]

Yes one root is 3 and the other two are complex i.e complex conjugate

[@parthasarathidas2360]

Using cube roots of unity concept . its in iit jee syllabus

[@RealQinnMalloryu4]

(x ➖ 3ix+3i )

[@MadScientyst]

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....🤔

[@LloydCash-he1qv]

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

[@LloydCash-he1qv]

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!

[@abneryokum]

@@LloydCash-he1qv This guy is a Bozo. All his solutions are long and tedious.

[@Misha-g3b]

3, -1,5+_1,5iV3.

[@Yahya2-mz3jm]

Why he doesn't use demoaver?

[@PravinKumar-v6x]

Is there anyone from INDIA❤

[@rastikigrov8454]

but what about the root x = -3i^2/3?

[@gregoryian108]

wow!

[@poovizhisaravanan6421]

3^3=27

[@minewithsayo4220]

Quite interesting!

[@Mister_Mouse7]

:/ x³=27 x=3√27=>3

[@vaichu44u]

I just did the x^3=1 version of this today 😂😂

[@RafaAssyifa]

X=3

[@1234larry1]

Once you get the difference of squares, difference of cubes and sum of cubes down, its not that hard.

[@bjornfeuerbacher5514]

Not necessary for solving this equation.

[@williamgraham2468]

Abraham de Moivre would like a word...

[@АндрейЛюбавин-э4щ]

3

[@DedMatveev]

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.

[@joso5554]

Thanks for pointing this. These YT videos almost never bother to mention this essential input.

[@stvp68]

“Find all roots” implies complex roots, doesn’t it?

[@bendunselman]

​​@@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....

[@stvp68]

@@bendunselman ah, okay, thanks!

[@kitkann7865]

@@joso5554លៀ😢៩ផ៥ហឱអ:យ🎉លេីៀ+​ឱ🎉ន🎉ៀល៤ឬៀនរ៥ល😊្

[@Parsi-Aria]

۳×۳×۳=۲۷ I feel like Nikolas tesla 😊

[@mariapiazza-od8ib]

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 😊😊

[@dhruvtakiar]

Who cares when he was truly a genius

[@azhanahmedali9851]

You forgot that there exist somethings like log .

[@bjornfeuerbacher5514]

Why should one use log for this equations???

[@azhanahmedali9851]

@@bjornfeuerbacher5514 it makes it much easier than ever .

[@bjornfeuerbacher5514]

@@azhanahmedali9851 ??? Sorry, I don't see it. Please explain to me how one solves this equation using log.

[@azhanahmedali9851]

@@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).

[@bjornfeuerbacher5514]

@@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?

[@rewwilh6435]

-3i

[@tombufford136]

The complex roots seem unmeaningful.

[@bjornfeuerbacher5514]

Why?

[@tombufford136]

@@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.

[@bjornfeuerbacher5514]

@@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"?

[@Muslem_x5i]

Anyone from syria 🤎✨

[@antonygattini3073]

😄😄🤣🤣🤣🤣😂😂

[@AnhNguyenNhat-rk3mm]

unvaliable bommmmmmmmmmmmmmm

[@markphc99]

Be careful of what? This is a straightforward problem , not really higher maths.

[@Peterthepainter66]

3x3x3=27 that's it.

[@nsy4652]

No way

[@gopagon2114]

X=9

[@szvetozargligorovics9431]

Hope you not teaching mathematic because your x it look like Coco Chanel sign

[@halcon2134]

Try a limit but that has no solution

[@bmrw]

9(3)=27 so its Simple 9+9+9=27

[@wMexso]

you forgot about -3

[@DedMatveev]

(-3)^3=-27

[@wMexso]

i forgor ​@@DedMatveev

[@RealQinnMalloryu4]

(x ➖ 3ix+3i)

[@johnsmith1953x]

Math is harrrdd.....

[@zaphodbeeblebrox-fz5fh]

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

[@СергійКузьменко-ы8м]

Just 3

[@richardrowe1538]

3