Spain | A Nice Algebra Problem | Math Olympiad

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  • Опубліковано 12 жов 2024
  • math olympiad
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    Find the value of x?
    How to solve √1 + √1+x = ∛x
    In this video, we'll show you How to Solve Math Olympiad Question A Nice Radical Equation √1 + √1+x = ∛x in a clear , fast and easy way. Whether you are a student learning basics or a professtional looking to improve your skills, this video is for you. By the end of this video, you'll have a solid understanding of how to solve math olympiad exponential equations and be able to apply these skills to a variety of problems.
    #matholympiad #maths #math #algebra

КОМЕНТАРІ • 17

  • @逸園-無毒果園
    @逸園-無毒果園 3 дні тому +7

    x=0 and x=-1 are not solutions
    x=8 is the only solution

    • @SALogics
      @SALogics  2 дні тому +3

      You are right! ❤

    • @melissajenner8068
      @melissajenner8068 2 дні тому +1

      Yes, verification needed.

    • @ChrisManton
      @ChrisManton День тому

      True, but the math is correct. How to explain ?

    • @CheesedogIsWhat
      @CheesedogIsWhat 22 години тому +2

      @@ChrisManton It's called an extraneous solution. It is like a fake root resulted from a transformed equation which is not directly the same as the original equation. Take ✓x = -1 for example, squaring both sides gives us x = 1, but we know ✓1 is 1, not -1, so 1 is an extraneous solution. So there it is, that's the answer.

    • @sohamb4944
      @sohamb4944 10 годин тому

      @@CheesedogIsWhat I always wished sqrt(x) was a direct inverse of x^2

  • @Magn0sC4rLs0n
    @Magn0sC4rLs0n День тому +4

    Spoiler blocker 🛡️ 🛡️ 🛡️

  • @sy8146
    @sy8146 День тому +3

    Thank you for explaining. The solution is only x=8.

    • @SALogics
      @SALogics  20 годин тому +1

      you are welcome! ❤

  • @CharlesChen-el4ot
    @CharlesChen-el4ot День тому +1

    (1+x) =( x^(2/3) - 1)^2
    1+x = x^(4/3) - 2x^(2/3) + 1
    x = x * x ^(1*3) - 2x^(2/3)
    1= x ^(1/3)- 2*x^(-1/3)
    Let y = x^(1/3)
    1 = y - 2 /y
    y^2 - y - 2 =0
    (y-2)(y+1)=0
    Solution 1
    y = 2; x^(1/3)= 2; x=8
    Solution 2
    y = -1 ; x^(1/3)=-1;
    x= -1 帶入原式不合
    非解 !

    • @SALogics
      @SALogics  20 годин тому +1

      Very nice trick! ❤

  • @key_board_x
    @key_board_x 2 дні тому +1

    First solution: x = - 1
    √[1 + √(1 + x)] = ³√x → where: x = - 1
    √[1 + √(1 - 1)] ? ³√(- 1)
    √[1 + 0] ? (- 1)^(1/3)
    √1 ? - 1
    1 ≠ - 1 → x = - 1 must be rejected
    Second solution: x = 0
    √[1 + √(1 + x)] = ³√x → where: x = 0
    √[1 + √(1 - 0)] ? ³√(0)
    √[1 + √(1)] ? 0
    √[1 + 1] ? 0
    √2 ≠ 0 → x = 0 must be rejected

    • @SALogics
      @SALogics  День тому +1

      Very nice! I really appreciate that ❤

  • @АндрейПергаев-з4н
    @АндрейПергаев-з4н 2 дні тому +2

    Не верно
    Если подставьте х=0
    0=sqrt 2,
    Если подставить х=-1
    1=-1, что тоже не верно
    Поэтому корень один х=8, о чём не сказано, значит корня получается три. Ответ полностью не верно

    • @SALogics
      @SALogics  День тому +1

      большое спасибо!
      x = 8 включено в решения, но я забыл проверить значения x. Извините за неудобства! ❤

  • @taniacsibi6879
    @taniacsibi6879 День тому +1

    Din cele 3. Sol.doar x=8 verifica ec. Data

    • @SALogics
      @SALogics  20 годин тому +1

      Da, ai dreptate 8 este singura solutie ❤