Editing Quartic function (section)

====Converting to a depressed quartic====
For solving purposes, it is generally better to convert the quartic into a '''depressed quartic''' by the following simple change of variable. All formulas are simpler and some methods work only in this case. The roots of the original quartic are easily recovered from that of the depressed quartic by the reverse change of variable.

Let
:<math> a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0 </math>
be the general quartic equation we want to solve.

Dividing by {{math|''a''<sub>4</sub>}}, provides the equivalent equation {{math|''x''<sup>4</sup> + ''bx''<sup>3</sup> + ''cx''<sup>2</sup> + ''dx'' + ''e'' {{=}} 0}}, with {{math|''b'' {{=}} {{sfrac|''a''<sub>3</sub>|''a''<sub>4</sub>}}}}, {{math|''c'' {{=}} {{sfrac|''a''<sub>2</sub>|''a''<sub>4</sub>}}}}, {{math|''d'' {{=}} {{sfrac|''a''<sub>1</sub>|''a''<sub>4</sub>}}}}, and {{math|''e'' {{=}} {{sfrac|''a''<sub>0</sub>|''a''<sub>4</sub>}}}}.
Substituting {{math|''y'' − {{sfrac|''b''|4}}}} for {{mvar|x}} gives, after regrouping the terms, the equation {{math|''y''<sup>4</sup> + ''py''<sup>2</sup> + ''qy'' + ''r'' {{=}} 0}},
where
:<math>\begin{align}
p&=\frac{8c-3b^2}{8} =\frac{8a_2a_4-3{a_3}^2}{8{a_4}^2}\\
q&=\frac{b^3-4bc+8d}{8} =\frac{{a_3}^3-4a_2a_3a_4+8a_1{a_4}^2}{8{a_4}^3}\\
r&=\frac{-3b^4+256e-64bd+16b^2c}{256}=\frac{-3{a_3}^4+256a_0{a_4}^3-64a_1a_3{a_4}^2+16a_2{a_3}^2a_4}{256{a_4}^4}.
\end{align}
</math>

If {{math|''y''<sub>0</sub>}} is a root of this depressed quartic, then {{math|''y''<sub>0</sub> − {{sfrac|''b''|4}}}} (that is {{math|''y''<sub>0</sub> − {{sfrac|''a''<sub>3</sub>|4''a''<sub>4</sub>}})}} is a root of the original quartic and every root of the original quartic can be obtained by this process.

====Ferrari's solution====
As explained in the preceding section, we may start with the ''depressed quartic equation''
:<math> y^4 + p y^2 + q y + r = 0. </math>
This depressed quartic can be solved by means of a method discovered by [[Lodovico Ferrari]]. The depressed equation may be rewritten (this is easily verified by expanding the square and regrouping all terms in the left-hand side) as
:<math> \left(y^2 + \frac p2\right)^2 = -q y - r + \frac{p^2}4. </math>
Then, we introduce a variable {{mvar|m}} into the factor on the left-hand side by adding {{math|2''y''<sup>2</sup>''m'' + ''pm'' + ''m''<sup>2</sup>}} to both sides. After regrouping the coefficients of the power of {{mvar|y}} on the right-hand side, this gives the equation 
{{NumBlk|:|<math> \left(y^2 + \frac p2 + m\right)^2 = 2 m y^2 - q y + m^2 + m p + \frac{p^2}4 - r, </math>|{{EquationRef|1}}}}
which is equivalent to the original equation, whichever value is given to {{mvar|m}}.

As the value of {{mvar|m}} may be arbitrarily chosen, we will choose it in order to [[complete the square]] on the right-hand side. This implies that the [[discriminant]] in {{mvar|y}} of this [[quadratic equation]] is zero, that is {{mvar|m}} is a root of the equation
:<math> (-q)^2 - 4 (2m)\left(m^2 + p m + \frac{p^2}4 - r\right) = 0,\,</math>
which may be rewritten as

{{NumBlk|:|<math>8m^3+ 8pm^2 + (2p^2 -8r)m- q^2 =0.</math>|{{EquationRef|1a}}}}

This is the [[resolvent cubic]] of the quartic equation. The value of {{mvar|m}} may thus be obtained from [[Cubic equation#Cardano's method|Cardano's formula]]. When {{mvar|m}} is a root of this equation, the right-hand side of equation (''{{EquationNote|1}}'') is the square
:<math>\left(\sqrt{2m}y-\frac q{2\sqrt{2m}}\right)^2.</math>
However, this induces a division by zero if {{math|''m'' {{=}} 0}}. This implies {{math|''q'' {{=}} 0}}, and thus that the depressed equation is bi-quadratic, and may be solved by an easier method (see above). This was not a problem at the time of Ferrari, when one solved only explicitly given equations with numeric coefficients. For a general formula that is always true, one thus needs to choose a root of the cubic equation such that {{math|''m'' ≠ 0}}. This is always possible except for the depressed equation {{math|''y''<sup>4</sup> {{=}} 0}}.

Now, if {{mvar|m}} is a root of the cubic equation such that {{math|''m'' ≠ 0}}, equation (''{{EquationNote|1}}'') becomes
:<math> \left(y^2 + \frac p2 + m\right)^2 = \left(y\sqrt{2 m}-\frac{q}{2\sqrt{2 m}}\right)^2. </math>

This equation is of the form {{math|''M''<sup>2</sup> {{=}} ''N''<sup>2</sup>}}, which can be rearranged as {{math|''M''<sup>2</sup> − ''N''<sup>2</sup> {{=}} 0}} or {{math|(''M'' + ''N'')(''M'' − ''N'') {{=}} 0}}. Therefore, equation (''{{EquationNote|1}}'')  may be rewritten as
:<math> \left(y^2 + \frac p2 + m + \sqrt{2 m}y-\frac q{2\sqrt{2 m}}\right) \left(y^2 + \frac p2 + m - \sqrt{2 m}y+\frac q{2\sqrt{2 m}}\right)=0.</math>

This equation is easily solved by applying to each factor the [[quadratic formula]]. Solving them we may write the four roots as
:<math>y={\pm_1\sqrt{2 m} \pm_2 \sqrt{-\left(2p + 2m \pm_1 {\sqrt 2q \over \sqrt{m}} \right)} \over 2},</math>
where {{math|±<sub>1</sub>}} and {{math|±<sub>2</sub>}} denote either {{math|+}} or {{math|−}}. As the two occurrences of {{math|±<sub>1</sub>}} must denote the same sign, this leaves four possibilities, one for each root.

Therefore, the solutions of the original quartic equation are
:<math>x=-{a_3 \over 4a_4} + {\pm_1\sqrt{2 m} \pm_2 \sqrt{-\left(2p + 2m \pm_1 {\sqrt2q \over \sqrt{m}} \right)} \over 2}.</math> 
A comparison with the [[#General_formula_for_roots|general formula]] above shows that {{math|{{sqrt|2''m''}} {{=}} 2''S''}}.