We search then to write this as (Descartes method) :
(x^{2}+px-u)(x^{2}-px-v) = 0
Expanding and equating the two expressions results into :
-a = (u+v)+p^{2} and (u+v)^{2}=(a+p^{2})^{2}
b = p(u-v) and b^{2} = p^{2}(u-v)^{2} = p^{2}((u+v)^{2}-4uv)
c = uv
Let Z = p^{2} and eliminate (u+v)^{2} and uv from the previous system, this gives :
b^{2} = Z((a+Z)^{2}-4c)
And the resolvant third degree equation :
Z^{3}+2aZ^{2}+(a^{2}-4c)Z-b^{2} = 0 |
Each of the roots of this equation gives a set of values for p, u and v.
But globally, this is the same as exchanging the roots of the 4^{th} degree equation,
and we can choose any root, namely Z_{1}.
Then (x_{1},x_{2}) are roots of x^{2}+px-u = 0 and
(x_{3},x_{4}) roots of x^{2}-px-v = 0
Specifically x_{1}+x_{2} = -p = -√Z_{1}
and x_{3}+x_{4} = p = √Z_{1}
Exchanging the roots gives similarily :
x_{1}+x_{3} = -√Z_{2},
x_{2}+x_{4} = √Z_{2}
and
x_{1}+x_{4} = -√Z_{3},
x_{2}+x_{3} = √Z_{3}
By linear combination of these expressions we get the roots of the 4^{th} degree equation :
x_{1} = 1/2 (-√Z_{1}-√Z_{2}-√Z_{3})
x_{2} = 1/2 (-√Z_{1}+√Z_{2}+√Z_{3}) x_{3} = 1/2 (√Z_{1}-√Z_{2}+√Z_{3}) x_{4} = 1/2 (√Z_{1}+√Z_{2}-√Z_{3}) with Z_{i} solutions of Z^{3}+2aZ^{2}+(a^{2}-4c)Z-b^{2} = 0 |
As for the 3^{rd} degree equation, these formulaes are useless in practice.
Practically, solving a 4^{th} degree equation is through numeric approximations,
except specific cases or trivial solutions.
A script for solving through Newton method.
Note also that although formulaes with square and cubic roots exist for the generic equations of degree 2, 3 and 4, there can't be any such formula for generic equations of degree 5 and more (see Abel and Galois theory).