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

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  • Опубликовано: 5 фев 2025
  • In this video, I'll be showing you step by step on how to solve this Olympiad Maths Algebra problem using a simple trick.
    Please feel free to share your ideas in the comment section.
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Комментарии • 17

  • @thebeast4216
    @thebeast4216 Месяц назад +9

    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
      @ronseymour4976 Месяц назад +1

      Yes, this method is unintuitive and very long winded.

  • @YongHowChin
    @YongHowChin Месяц назад +4

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

  • @sairosulaiman
    @sairosulaiman Месяц назад

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

  • @hassnaabraim6318
    @hassnaabraim6318 Месяц назад +3

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

    • @grassytramtracks
      @grassytramtracks Месяц назад

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

  • @pot_kivach160
    @pot_kivach160 Месяц назад

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

  • @chrismcgowan3938
    @chrismcgowan3938 Месяц назад

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

    • @vorpal22
      @vorpal22 Месяц назад +1

      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.

  • @hangthuy458
    @hangthuy458 Месяц назад

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

    • @vorpal22
      @vorpal22 Месяц назад +1

      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
    @philippedelaveau528 Месяц назад

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

  • @christiaanvanhyfte1873
    @christiaanvanhyfte1873 Месяц назад

    Sorry. I will not comment.

  • @themieljadida4459
    @themieljadida4459 Месяц назад

    Nul .... nul.

  • @sairosulaiman
    @sairosulaiman Месяц назад

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

  • @sairosulaiman
    @sairosulaiman Месяц назад

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

  • @sairosulaiman
    @sairosulaiman Месяц назад

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