Harvard University Admission Interview Trick | How To Solve!!! |You Should Know This Trick!

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  • Опубліковано 27 січ 2025

КОМЕНТАРІ • 5

  • @stpat7614
    @stpat7614 Місяць тому

    Including complex roots:
    a^6 - 4 = 0
    a^(3*2) - 4 = 0
    (a^3)^2 - 4 = 0
    (a^3)^2 - 2^2 = 0
    (a^3 - 2)(a^3 + 2) = 0
    (a^3 - [2^(1/3)]^3)(a^3 + [2^(1/3)]^3) = 0
    (a - 2^[1/3])(a^2 + a*2^[1/3] + [2^(1/3)]^2)(a + 2^[1/3])(a^2 - a*2^[1/3] + [2^(1/3)]^2) = 0
    (a - 2^[1/3])(a^2 + a*2^[1/3] + 2^[1/3*2])(a + 2^[1/3])(a^2 - a*2^[1/3] + 2^[1/3*2]) = 0
    (a - 2^[1/3])(a^2 + 2^[1/3]*a + 2^[2/3])(a + 2^[1/3])(a^2 - 2^[1/3]*a + 2^[2/3]) = 0
    (a - 2^[1/3])(a + 2^[1/3])(a^2 + 2^[1/3]*a + 2^[2/3])(a^2 - 2^[1/3]*a + 2^[2/3]) = 0
    a - 2^(1/3) = 0, or a + 2^(1/3) = 0, or 1*a^2 + 2^(1/3)*a + 2^(2/3) = 0, or 1*a^2 - 2^(1/3)*a + 2^(2/3) = 0
    a - 2^(1/3) + 2^(1/3) = 0 + 2^(1/3), or a + 2^(1/3) - 2^(1/3) = 0 - 2^(1/3), a = (-2^[1/3] +/- sqrt[(2^[1/3])^2 - 4*1*2^(2/3)]) / (2*1), or a = (-(-2^[1/3]) +/- sqrt[(-2^[1/3])^2 - 4*1*2^(2/3)]) / (2*1)
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- sqrt[(2^[1/3])^2 - 4*2^(2/3)]) / (2), or a = (2^[1/3] +/- sqrt[(-2^[1/3])^2 - 4*2^(2/3)]) / (2)
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- sqrt[(2^2)^(1/3) - 4*(2^2)^(1/3)]) / 2, or a = (2^[1/3] +/- sqrt[(-1)^2*(2^2)^(1/3) - 4*(2^2)^(1/3)]) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- sqrt[4^(1/3) - 4*4^(1/3)]) / 2, or a = (2^[1/3] +/- sqrt[1*4^(1/3) - 4*4^(1/3)]) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- sqrt[4^(1/3) - 4*4^(1/3)]) / 2, or a = (2^[1/3] +/- sqrt[4^(1/3) - 4*4^(1/3)]) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- sqrt[4^(1/3)*(1 - 4)]) / 2, or a = (2^[1/3] +/- sqrt[4^(1/3)*(1 - 4)]) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- sqrt[4^(1/3)*(-3)]) / 2, or a = (2^[1/3] +/- sqrt[4^(1/3)*(-3)]) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- sqrt[(2^2)^(1/3)*3*(-1)]) / 2, or a = (2^[1/3] +/- sqrt[(2^2)^(1/3)*3*(-1)]) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- sqrt[2^(2*1/3)*3*(-1)]) / 2, or a = (2^[1/3] +/- sqrt[2^(2*1/3)*3*(-1)]) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- sqrt[2^(2/3)*3*(-1)]) / 2, or a = (2^[1/3] +/- sqrt[2^(2/3)*3*(-1)]) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- [2^(2/3)*3*(-1)]^[1/2]) / 2, or a = (2^[1/3] +/- [2^(2/3)*3*(-1)]^[1/2]) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- [2^(2/3)]^[1/2]*3^[1/2]*[-1]^[1/2]) / 2, or a = (2^[1/3] +/- [2^(2/3)]^[1/2]*3^[1/2]*[-1]^[1/2]) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- 2^[(2/3)*(1/2)]*3^[1/2]*i) / 2, or a = (2^[1/3] +/- 2^[(2/3)*(1/2)]*3^[1/2]*i) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = (-2^[1/3] +/- 2^[1/3]*3^[1/2]*i) / 2, or a = (2^[1/3] +/- 2^[1/3]*3^[1/2]*i) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = 2^(1/3)*(-1 +/- 3^[1/2]*i) / 2, or a = 2^(1/3)*(1 +/- 3^[1/2]*i) / 2
    a = 2^(1/3), or a = - 2^(1/3), or a = 2^(1/3)*(-1 +/- 3^[1/2]*i) / 2^(3/3), or a = 2^(1/3)*(1 +/- 3^[1/2]*i) / 2^(3/3)
    a = 2^(1/3), or a = - 2^(1/3), or a = 2^([1/3]-[3/3])*(-1 +/- 3^[1/2]*i), or a = 2^([1/3]-[3/3])*(1 +/- 3^[1/2]*i)
    a = 2^(1/3), or a = - 2^(1/3), or a = 2^(-2/3)*(-1 +/- 3^[1/2]*i), or a = 2^(-2/3)*(1 +/- 3^[1/2]*i)
    a = 2^(1/3), or a = -2^(1/3), or a = 2^(-2/3)*(-1 + 3^[1/2]*i), or a = 2^(-2/3)*(-1 - 3^[1/2]*i), or a = 2^(-2/3)*(1 + 3^[1/2]*i), or a = 2^(-2/3)*(1 - 3^[1/2]*i)
    a = 2^(1/3), or a = -2^(1/3), or a = -2^(-2/3)*(1 - 3^[1/2]*i), or a = -2^(-2/3)*(1 + 3^[1/2]*i), or a = 2^(-2/3)*(1 + 3^[1/2]*i), or a = 2^(-2/3)*(1 - 3^[1/2]*i)
    a1 = 2^(1/3)
    a2 = -2^(1/3)
    a3 = -2^(-2/3)*(1 - 3^[1/2]*i)
    a4 = -2^(-2/3)*(1 + 3^[1/2]*i)
    a5 = 2^(-2/3)*(1 + 3^[1/2]*i)
    a6 = 2^(-2/3)*(1 - 3^[1/2]*i)

  • @larbibenmrad1968
    @larbibenmrad1968 Місяць тому

    a^6 - 4 = 0
    pour a : positif
    a^6 = 4
    6 log a = log 4 ....log a = (log4) / 6
    alors ...10^log a =a = 10^ (log 4)/6
    a = 1.26 (1.259
    puis la puissance de 6 : nombre pair
    - 1.26 est une autre solution
    1.26 = sq(3 ; 2)
    - 1.26 = -sq(3 ; 2)
    vérif : 1.26^6 = 4.0001
    (-1.26)^6 = 4.0001

    • @cesarjules3590
      @cesarjules3590 4 дні тому

      Solution que j'ai choisie sauf que (-1.26)^6=-4.0015 donc une seule solution a=1.26 pour 1.26^6=4.0015. C'était une méthode plus rapide et moins usine à gaz, bravo quand même.

    • @cesarjules3590
      @cesarjules3590 4 дні тому

      erreur de ma calculette y'a bien a=-1.26...