A Very Nice Math Olympiad Problem | Solve for the value of x? | Algebra Equation

[@thebeast4216]

To avoid such difficult calculation we can use concept of nth root of unity which says that if x^n=1 then x=e^(2πki/n) where k=0,1,2,3......(n-1) If you put k gretaer than that then roots will starts repeating and also we can simplify them in much nicer form by euler identity which says e^(ix)=cosx+isinx

[@ronseymour4976]

Yes, this method is unintuitive and very long winded.

[@YongHowChin]

At a glance, x needs to be 1 to satisfy the equation,

[@sairosulaiman]

√9=±3 and √8=±2√2....so there are complex roots too

[@hassnaabraim6318]

this is not math. it's something else 😂😂

[@grassytramtracks]

Luckily for you this is a needlessly complicated method. Just use the nth roots of unity

[@chrismcgowan3938]

x = 1 and very likley solutions with i .....

[@vorpal22]

The only integer solution is 1, and the others are going to occur in pairs as complex numbers. Whenever you have x^n + .... and you're solving, since polynomials split completely over C, you will get n solutions. An even number of them will be complex, and the rest of them will be real.

[@pot_kivach160]

When i find something out of bound, i hit Thumb down button. And everything makes sense then.

[@hangthuy458]

X^5=1=1^5X=1

[@vorpal22]

x = 1 is one solution. Polynomials split completely over C, so there are guaranteed to be five solutions, and since 5 is prime, they will all be unique, and four of them will be complex, occurring in two pairs.

[@philippedelaveau528]

Racine niéme de l’unité. Réponse immédiate sans calcul.

[@christiaanvanhyfte1873]

Sorry. I will not comment.

[@themieljadida4459]

Nul .... nul.

[@sairosulaiman]

√9=±3 and √8=±2√2....so there are complex roots too

[@sairosulaiman]

√9=±3 and √8=±2√2....so there are complex roots too

[@sairosulaiman]

√9=±3 and √8=±2√2....so there are complex roots too