Editing Quintic function (section)

===Roots of a solvable quintic===
A polynomial equation is solvable by radicals if its [[Galois group]] is a [[solvable group]]. In the case of irreducible quintics, the Galois group is a subgroup of the [[symmetric group]] {{math|''S''<sub>5</sub>}} of all permutations of a five element set, which is solvable if and only if it is a subgroup of the group {{math|''F''<sub>5</sub>}}, of order {{math|20}}, generated by the cyclic permutations {{math|(1 2 3 4 5)}} and {{math|(1 2 4 3)}}.

If the quintic is solvable, one of the solutions may be represented by an [[algebraic expression]] involving a fifth root and at most two square roots, generally [[nested radical|nested]]. The other solutions may then be obtained either by changing the fifth root or by multiplying all the occurrences of the fifth root by the same power of a [[root of unity|primitive 5th root of unity]], such as
:<math>\frac{\sqrt{-10-2\sqrt{5}}+\sqrt{5}-1}{4}.</math>

In fact, all four primitive fifth roots of unity may be obtained by changing the signs of the square roots appropriately; namely, the expression
:<math>\frac{\alpha\sqrt{-10-2\beta\sqrt{5}}+\beta\sqrt{5}-1}{4},</math>
where <math> \alpha, \beta \in \{-1,1\}</math>, yields the four distinct primitive fifth roots of unity.

It follows that one may need four different square roots for writing all the roots of a solvable quintic. Even for the first root that involves at most two square roots, the expression of the solutions in terms of radicals is usually highly complicated. However, when no square root is needed, the form of the first solution may be rather simple, as for the equation {{math|''x''<sup>5</sup> − 5''x''<sup>4</sup> + 30''x''<sup>3</sup> − 50''x''<sup>2</sup> + 55''x'' − 21 {{=}} 0}}, for which the only real solution is

: <math>x=1+\sqrt[5]{2}-\left(\sqrt[5]{2}\right)^2+\left(\sqrt[5]{2}\right)^3-\left(\sqrt[5]{2}\right)^4.</math>

An example of a more complicated (although small enough to be written here) solution is the unique real root of {{math|''x''<sup>5</sup> − 5''x'' + 12 {{=}} 0}}. Let {{math|''a'' {{=}} {{sqrt|2''φ''<sup>−1</sup>}}}}, {{math|''b'' {{=}} {{sqrt|2''φ''}}}}, and {{math|''c'' {{=}} {{radic|5|4}}}}, where {{math|''φ'' {{=}} {{sfrac|1+{{sqrt|5}}|2}}}} is the [[golden ratio]]. Then the only real solution {{math|''x'' {{=}} −1.84208...}} is given by

: <math>-cx = \sqrt[5]{(a+c)^2(b-c)} + \sqrt[5]{(-a+c)(b-c)^2} + \sqrt[5]{(a+c)(b+c)^2} - \sqrt[5]{(-a+c)^2(b+c)} \,,</math>

or, equivalently, by

:<math>x = \sqrt[5]{y_1}+\sqrt[5]{y_2}+\sqrt[5]{y_3}+\sqrt[5]{y_4}\,,</math>

where the {{math|''y<sub>i</sub>''}} are the four roots of the [[quartic equation]]

:<math>y^4+4y^3+\frac{4}{5}y^2-\frac{8}{5^3}y-\frac{1}{5^5}=0\,.</math>

More generally, if an equation {{math|1=''P''(''x'') = 0}} of prime degree {{math|''p''}} with rational coefficients is solvable in radicals, then one can define an auxiliary equation {{math|1=''Q''(''y'') = 0}} of degree {{math|''p'' − 1}}, also with rational coefficients, such that each root of {{math|''P''}} is the sum of {{math|''p''}}-th roots of the roots of {{math|''Q''}}. These {{math|''p''}}-th roots were introduced by [[Joseph-Louis Lagrange]], and their products by {{math|''p''}} are commonly called [[Lagrange resolvent]]s. The computation of {{math|''Q''}} and its roots can be used to solve {{math|1=''P''(''x'') = 0}}. However these {{math|''p''}}-th roots may not be computed independently (this would provide {{math|''p''<sup>''p''−1</sup>}} roots instead of {{math|''p''}}). Thus a correct solution needs to express all these {{math|''p''}}-roots in term of one of them. Galois theory shows that this is always theoretically possible, even if the resulting formula may be too large to be of any use.

It is possible that some of the roots of {{math|''Q''}} are rational (as in the first example of this section) or some are zero. In these cases, the formula for the roots is much simpler, as for the solvable [[de Moivre]] quintic{{anchor|de Moivre quintic}}

:<math>x^5+5ax^3+5a^2x+b = 0\,,</math>

where the auxiliary equation has two zero roots and reduces, by factoring them out, to the [[quadratic equation]]

:<math>y^2+by-a^5 = 0\,,</math>

such that the five roots of the de Moivre quintic are given by

:<math>x_k = \omega^k\sqrt[5]{y_i} -\frac{a}{\omega^k\sqrt[5]{y_i}},</math>

where ''y<sub>i</sub>'' is any root of the auxiliary quadratic equation and ''ω'' is any of the four [[primitive root of unity|primitive 5th roots of unity]].  This can be easily generalized to construct a solvable [[septic equation|septic]] and other odd degrees, not necessarily prime.