Japanese | Math Olympiad | A Nice Algebra Problem

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  • Опубликовано: 18 ноя 2024

Комментарии • 6

  • @doichecabano
    @doichecabano Месяц назад +2

    2.5; (5+5i)/2; (5-5i)/2

  • @wes9627
    @wes9627 Месяц назад

    Always look for substitutions to make the problem solution easier. Substitute x=y+5/2 into the given equation and rearrange to:
    (y+5/2)^4-(y-5/2)^4=2*4[y^3(5/2)+y(5/2)^3]=0
    Divide by 8(5/2)y to obtain y^2+(5/2)^2=0, which has roots y=±i5/2. Also y=0 is a root.
    Thus, x=y+5/2=(5±5i)/2 or x=5/2

  • @عبدالواسع-س8م
    @عبدالواسع-س8م Месяц назад

    Well done.
    Thanks so much!

  • @shaozheang5528
    @shaozheang5528 Месяц назад

    Xpand

  • @hakanerci4372
    @hakanerci4372 Месяц назад

    Why bother with complex numbers? Are real numbers not enough?

    • @user-eb7lp7wj5k
      @user-eb7lp7wj5k Месяц назад

      Bro, 2x² - 10x +25 = 0 or 25-10x = 0, because a * 0 = 0🤓🤓🤓