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Solution for equation of 2,3,4, degree

  1. Mar 26, 2014 #1
    I noticed that is always possible to find the roots of polynomials of kind:
    ##A\lambda^2 + B##
    ##A\lambda^3 + B##
    ##A\lambda^4 + B##
    ##A\lambda^5 + B##

    So I thought to transform the quadratic, the cubic, the quartic and the quintic function in the forms above, look:

    ##ax^3+bx^2+cx+d=0##

    ##(x=\lambda+u+v)##

    ##a(\lambda+u+v)^3+b(\lambda+u+v)^2+c(\lambda+u+v)+d=0##

    ##\\a \left ( \begin{matrix}
    \lambda^3 & +3\lambda^2u & +3\lambda u^2 & +u^3\\
    +3\lambda^2v & +6\lambda u v & +3u^2 v & \\
    +3\lambda v^2 & +3uv^2 & & \\
    +v^3 & & & \\
    \end{matrix} \right )

    + b \left ( \begin{matrix}
    \lambda^2 & +2\lambda u & +u^2\\
    +2\lambda v & +2 u v & \\
    +v^2 & &
    \end{matrix} \right )

    + c (\lambda + u + v) + d = 0##

    ##a\lambda^3+(3a(u+v)+b)\lambda^2+(3a(u+v)^2 + 2b(u+v) + c)\lambda + (a(u+v)^3 + b(u+v)^2 + c(u+v) + d) = 0##

    So, comparing the equations:

    ##A\lambda^3 + B = a\lambda^3+(3a(u+v)+b)\lambda^2+(3a(u+v)^2 + 2b(u+v) + c)\lambda + (a(u+v)^3 + b(u+v)^2 + c(u+v) + d) = 0##

    implies that:
    ##\begin{cases}
    A=a\\
    0=(3a(u+v)+b)\\
    0=(3a(u+v)^2 + 2b(u+v) + c)\\
    B=(a(u+v)^3 + b(u+v)^2 + c(u+v) + d)\\
    \end{cases}##

    So,
    ##A\lambda^3 + B = 0\;\;\;\Rightarrow\;\;\;\;\lambda = \sqrt[3]{\frac{-B}{A}} \;\;\;\Rightarrow\;\;\;\; x-u-v = \sqrt[3]{\frac{-B}{A}} \;\;\;\Rightarrow\;\;\;\; x = \sqrt[3]{\frac{-B}{A}} + (u+v)\;\;\;\Rightarrow\;\;\;\; x = \sqrt[3]{\frac{-(a(u+v)^3 + b(u+v)^2 + c(u+v) + d)}{a}} + (u+v)##

    Happens that
    ##0=3a(u+v)+b\;\;\;\Rightarrow\;\;\;\;u+v = \frac{-b}{3a}##

    Conclusion:
    [tex]x=\sqrt[3]{\frac{d}{a}-\frac{2b^3}{3^3a^3}-\frac{bc}{3a^2}}-\frac{b}{3a}[/tex]

    This ideia is correct?
     
  2. jcsd
  3. Mar 26, 2014 #2

    micromass

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