Global North | Math Olympiad Algebra Problem

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  • Опубліковано 21 жов 2024

КОМЕНТАРІ • 19

  • @benyseus6325
    @benyseus6325 2 місяці тому +1

    One of the few times I would sanction the use of the cubic equation. Just put 80x^3 -x +1 = 0 and solve. The answer you get is x= -0.25 meanwhile the other two roots are complex. Don’t bother with binomial expansion stuff which takes up a lot of time.

  • @__Mike137
    @__Mike137 2 місяці тому +3

    Hello, I am sorry but where did you pull that formula for the quadratic equation from???? First of all it is not 2C it is 2a. Second you are writing the equation for 1/x not X. The correct formula is X= 2a/[-b + - Sqrt(b*2 - 4ac)]. Replace 1/X by Y to minimize confusion from the beginning. Y*3 - Y*2 +80=0

    • @superacademy247
      @superacademy247  2 місяці тому +1

      Check out on this rare formula ua-cam.com/video/TbCH_TMZEcM/v-deo.htmlsi=o_ybAOSLZFya3fM2

  • @SidneiMV
    @SidneiMV 2 місяці тому +3

    (1/x)² - (1/x)³ = 80
    (1/x)³ - (1/x²) + 80 = 0
    1/x = u => x = 1/u
    u³ - u² + 80 = 0
    u³ + 64 - (u² - 16) = 0
    (u + 4)(u² - 4u + 16) - (u + 4)(u - 4) = 0
    (u + 4)(u² - 5u + 20) = 0
    u = - 4 => *x = -1/4*
    u² - 5u + 20 = 0 => complex roots

    • @BruceLee-io9by
      @BruceLee-io9by 2 місяці тому +1

      I solved it the same way.

    • @stefans.7681
      @stefans.7681 2 місяці тому +1

      Nice and elegant solution.
      I did also the Substitution like you.
      I factored to u^2 (1-u) = 80 = 16 * 5, so u = -4 (x = …) is a solution.
      Then I use polynomial Division by (u + 4) and Proof, that the polynomial reminder has no more roots.

  • @cemdeve78
    @cemdeve78 2 місяці тому +1

    One look and I can see -1/4 is one of the roots.

  • @key_board_x
    @key_board_x 2 місяці тому +1

    (1/x)² - (1/x)³ = 80 → let: a = 1/x → where: x ≠ 0
    a² - a³ = 80
    a³ - a² + 80 = 0
    a³ - (5a² - 4a²) + 80 = 0
    a³ - 5a² + 4a² + 80 = 0
    a³ - 5a² + 4a² + 80 + (20a - 20a) = 0
    a³ - 5a² + 4a² + 80 + 20a - 20a = 0
    a³ - 5a² + 20a + 4a² - 20a + 80 = 0
    (a³ - 5a² + 20a) + (4a² - 20a + 80) = 0
    a.(a² - 5a + 20) + 4.(a² - 5a + 20) = 0
    (a + 4).(a² - 5a + 20) = 0
    First case: (a + 4) = 0
    a + 4 = 0
    a = - 4 → recall: a = 1/x
    → x = - 1/4
    Second case: (a² - 5a + 20) = 0
    a² - 5a + 20 = 0
    Δ = (- 5)² - (4 * 20) = 25 - 80 = - 55 = 55i²
    a = (5 ± i√55)/2 → recall: a = 1/x
    x = 2/(5 ± i√55)
    First possibility: x = 2/(5 + i√55)
    x = 2.(5 - i√55)/[(5 + i√55).(5 - i√55)]
    x = 2.(5 - i√55)/[25 + 55]
    x = 2.(5 - i√55)/80
    → x = (5 - i√55)/40
    Second possibility: x = 2/(5 - i√55)
    x = 2.(5 + i√55)/[(5 - i√55).(5 + i√55)]
    x = 2.(5 + i√55)/[25 + 55]
    x = 2.(5 + i√55)/80
    → x = (5 + i√55)/40

  • @CTJ2619
    @CTJ2619 2 місяці тому +1

    you made an error on the quadratic equation - it isn’t divided by 2c , it should be 2a

    • @ponnappakurup5601
      @ponnappakurup5601 2 місяці тому +1

      It is correct. Multipy the quadratic eqn. by x^2.Eqn. will be cx^2+bx+a=0.

  • @anasanasa645
    @anasanasa645 2 місяці тому +1

    Discriminant over 2a not 2c

    • @superacademy247
      @superacademy247  2 місяці тому

      Please check out this video. ua-cam.com/video/TbCH_TMZEcM/v-deo.htmlsi=FUtW057kevM07O_b

  • @CTJ2619
    @CTJ2619 2 місяці тому +1

    please redo this video - your videos are generally quite good however this one you failed to correct the error with the quadratic equation

    • @superacademy247
      @superacademy247  2 місяці тому

      Check out on this rare formula ua-cam.com/video/TbCH_TMZEcM/v-deo.htmlsi=o_ybAOSLZFya3fM2

  • @wow-lp3138bg
    @wow-lp3138bg 2 місяці тому

    2c? what?

    • @superacademy247
      @superacademy247  2 місяці тому

      please check out this video ua-cam.com/video/TbCH_TMZEcM/v-deo.htmlsi=o_ybAOSLZFya3fM2

  • @kennethlombardi3003
    @kennethlombardi3003 2 місяці тому

    You are pronouncing the letter B as the letter P!