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Principal Quintic Form

A general Quintic Equation

\begin{displaymath}
a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2+ a_1 x + a_0 = 0
\end{displaymath} (1)

can be reduced to one of the form
\begin{displaymath}
y^5 + b_2 y^2 + b_1 y + b_0 = 0,
\end{displaymath} (2)

called the principal quintic form.


Newton's Relations for the Roots $y_j$ in terms of the $b_j$s is a linear system in the $b_j$, and solving for the $b_j$s expresses them in terms of the Power sums $s_n(y_j)$. These Power sums can be expressed in terms of the $a_j$s, so the $b_j$s can be expressed in terms of the $a_j$s. For a quintic to have no quartic or cubic term, the sums of the Roots and the sums of the Squares of the Roots vanish, so

$\displaystyle s_1(y_j)$ $\textstyle =$ $\displaystyle 0$ (3)
$\displaystyle s_2(y_j)$ $\textstyle =$ $\displaystyle 0.$ (4)

Assume that the Roots $y_j$ of the new quintic are related to the Roots $x_j$ of the original quintic by
\begin{displaymath}
y_j = {x_j}^2 +\alpha x_j + \beta.
\end{displaymath} (5)

Substituting this into (1) then yields two equations for $\alpha$ and $\beta$ which can be multiplied out, simplified by using Newton's Relations for the Power sums in the $x_j$, and finally solved. Therefore, $\alpha$ and $\beta$ can be expressed using Radicals in terms of the Coefficients $a_j$. Again by substitution into (4), we can calculate $s_3(y_j)$, $s_4(y_j)$ and $s_5(y_j)$ in terms of $\alpha$ and $\beta$ and the $x_j$. By the previous solution for $\alpha$ and $\beta$ and again by using Newton's Relations for the Power sums in the $x_j$, we can ultimately express these Power sums in terms of the $a_j$.

See also Bring Quintic Form, Newton's Relations, Quintic Equation



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© 1996-9 Eric W. Weisstein
1999-05-26