Editing Quartic function (section)

====Euler's solution====
A variant of the previous method is due to [[Leonhard Euler|Euler]].<ref>{{Citation|last = van der Waerden|first=Bartel Leendert|author-link = Bartel Leendert van der Waerden|title = [[Moderne Algebra|Algebra]]|volume = 1|publisher=[[Springer Science+Business Media|Springer-Verlag]]|edition = 7th|isbn = 0-387-97424-5|year = 1991|section = The Galois theory: Equations of the second, third, and fourth degrees|zbl = 0724.12001}}</ref><ref>{{Citation|last = Euler|first = Leonhard|author-link = Leonhard Euler|title = [[Elements of Algebra]]|chapter= Of a new method of resolving equations of the fourth degree|publisher=[[Springer Science+Business Media|Springer-Verlag]]|orig-year = 1765|year = 1984|zbl = 0557.01014|isbn = 978-1-4613-8511-0}}</ref> Unlike the previous methods, both of which use ''some'' root of the resolvent cubic, Euler's method uses all of them. Consider a depressed quartic {{math|''x''<sup>4</sup> + ''px''<sup>2</sup> + ''qx'' + ''r''}}. Observe that, if
* {{math|''x''<sup>4</sup> + ''px''<sup>2</sup> + ''qx'' + ''r'' {{=}} (''x''<sup>2</sup> + ''sx'' + ''t'')(''x''<sup>2</sup> − ''sx'' + ''v'')}},
* {{math|''r''<sub>1</sub>}} and {{math|''r''<sub>2</sub>}} are the roots of {{math|''x''<sup>2</sup> + ''sx'' + ''t''}},
* {{math|''r''<sub>3</sub>}} and {{math|''r''<sub>4</sub>}} are the roots of {{math|''x''<sup>2</sup> − ''sx'' + ''v''}},
then
* the roots of {{math|''x''<sup>4</sup> + ''px''<sup>2</sup> + ''qx'' + ''r''}} are {{math|''r''<sub>1</sub>}}, {{math|''r''<sub>2</sub>}}, {{math|''r''<sub>3</sub>}}, and {{math|''r''<sub>4</sub>}},
* {{math|''r''<sub>1</sub> + ''r''<sub>2</sub> {{=}} −''s''}},
* {{math|''r''<sub>3</sub> + ''r''<sub>4</sub> {{=}} ''s''}}.
Therefore, {{math|(''r''<sub>1</sub> + ''r''<sub>2</sub>)(''r''<sub>3</sub> + ''r''<sub>4</sub>) {{=}} −''s''<sup>2</sup>}}. In other words, {{math|−(''r''<sub>1</sub> + ''r''<sub>2</sub>)(''r''<sub>3</sub> + ''r''<sub>4</sub>)}} is one of the roots of the resolvent cubic (''{{EquationNote|2}}'') and this suggests that the roots of that cubic are equal to {{math|−(''r''<sub>1</sub> + ''r''<sub>2</sub>)(''r''<sub>3</sub> + ''r''<sub>4</sub>)}}, {{math|−(''r''<sub>1</sub> + ''r''<sub>3</sub>)(''r''<sub>2</sub> + ''r''<sub>4</sub>)}}, and {{math|−(''r''<sub>1</sub> + ''r''<sub>4</sub>)(''r''<sub>2</sub> + ''r''<sub>3</sub>)}}. This is indeed true and it follows from [[Vieta's formulas]]. It also follows from Vieta's formulas, together with the fact that we are working with a depressed quartic, that {{math|''r''<sub>1</sub> + ''r''<sub>2</sub> + ''r''<sub>3</sub> + ''r''<sub>4</sub> {{=}} 0}}. (Of course, this also follows from the fact that {{math|''r''<sub>1</sub> + ''r''<sub>2</sub> + ''r''<sub>3</sub> + ''r''<sub>4</sub> {{=}} −''s'' + ''s''}}.) Therefore, if {{math|''α''}}, {{math|''β''}}, and {{math|''γ''}} are the roots of the resolvent cubic, then the numbers {{math|''r''<sub>1</sub>}}, {{math|''r''<sub>2</sub>}}, {{math|''r''<sub>3</sub>}}, and {{math|''r''<sub>4</sub>}} are such that
:<math>\left\{\begin{array}{l}r_1+r_2+r_3+r_4=0\\(r_1+r_2)(r_3+r_4)=-\alpha\\(r_1+r_3)(r_2+r_4)=-\beta\\(r_1+r_4)(r_2+r_3)=-\gamma\text{.}\end{array}\right.</math>
It is a consequence of the first two equations that {{math|''r''<sub>1</sub> + ''r''<sub>2</sub>}} is a square root of {{math|''α''}} and that {{math|''r''<sub>3</sub> + ''r''<sub>4</sub>}} is the other square root of {{math|''α''}}. For the same reason,
* {{math|''r''<sub>1</sub> + ''r''<sub>3</sub>}} is a square root of {{math|''β''}},
* {{math|''r''<sub>2</sub> + ''r''<sub>4</sub>}} is the other square root of {{math|''β''}},
* {{math|''r''<sub>1</sub> + ''r''<sub>4</sub>}} is a square root of {{math|''γ''}},
* {{math|''r''<sub>2</sub> + ''r''<sub>3</sub>}} is the other square root of {{math|''γ''}}.
Therefore, the numbers {{math|''r''<sub>1</sub>}}, {{math|''r''<sub>2</sub>}}, {{math|''r''<sub>3</sub>}}, and {{math|''r''<sub>4</sub>}} are such that
:<math>\left\{\begin{array}{l}r_1+r_2+r_3+r_4=0\\r_1+r_2=\sqrt{\alpha}\\r_1+r_3=\sqrt{\beta}\\r_1+r_4=\sqrt{\gamma}\text{;}\end{array}\right.</math>
the sign of the square roots will be dealt with below. The only solution of this system is:
:<math>\left\{\begin{array}{l}r_1=\frac{\sqrt{\alpha}+\sqrt{\beta}+\sqrt{\gamma}}2\\[2mm]r_2=\frac{\sqrt{\alpha}-\sqrt{\beta}-\sqrt{\gamma}}2\\[2mm]r_3=\frac{-\sqrt{\alpha}+\sqrt{\beta}-\sqrt{\gamma}}2\\[2mm]r_4=\frac{-\sqrt{\alpha}-\sqrt{\beta}+\sqrt{\gamma}}2\text{.}\end{array}\right.</math>
Since, in general, there are two choices for each square root, it might look as if this provides {{math|8 ({{=}} 2<sup>3</sup>)}} choices for the set {{math|{''r''<sub>1</sub>, ''r''<sub>2</sub>, ''r''<sub>3</sub>, ''r''<sub>4</sub>}}}, but, in fact, it provides no more than {{math|2}}&nbsp;such choices, because the consequence of replacing one of the square roots by the symmetric one is that the set {{math|{''r''<sub>1</sub>, ''r''<sub>2</sub>, ''r''<sub>3</sub>, ''r''<sub>4</sub>}}} becomes the set {{math|{−''r''<sub>1</sub>, −''r''<sub>2</sub>, −''r''<sub>3</sub>, −''r''<sub>4</sub>}}}.

In order to determine the right sign of the square roots, one simply chooses some square root for each of the numbers {{math|''α''}}, {{math|''β''}}, and {{math|''γ''}} and uses them to compute the numbers {{math|''r''<sub>1</sub>}}, {{math|''r''<sub>2</sub>}}, {{math|''r''<sub>3</sub>}}, and {{math|''r''<sub>4</sub>}} from the previous equalities. Then, one computes the number {{math|{{sqrt|''α''}}{{sqrt|''β''}}{{sqrt|''γ''}}}}. Since {{math|''α''}}, {{math|''β''}}, and {{math|''γ''}} are the roots of (''{{EquationNote|2}}''), it is a consequence of Vieta's formulas that their product is equal to {{math|''q''<sup>2</sup>}} and therefore that {{math|{{sqrt|''α''}}{{sqrt|''β''}}{{sqrt|''γ''}} {{=}} ±''q''}}. But a straightforward computation shows that
:{{math|{{sqrt|''α''}}{{sqrt|''β''}}{{sqrt|''γ''}} {{=}} ''r''<sub>1</sub>''r''<sub>2</sub>''r''<sub>3</sub> + ''r''<sub>1</sub>''r''<sub>2</sub>''r''<sub>4</sub> + ''r''<sub>1</sub>''r''<sub>3</sub>''r''<sub>4</sub> + ''r''<sub>2</sub>''r''<sub>3</sub>''r''<sub>4</sub>.}}
If this number is {{math|−''q''}}, then the choice of the square roots was a good one (again, by Vieta's formulas); otherwise, the roots of the polynomial will be {{math|−''r''<sub>1</sub>}}, {{math|−''r''<sub>2</sub>}}, {{math|−''r''<sub>3</sub>}}, and {{math|−''r''<sub>4</sub>}}, which are the numbers obtained if one of the square roots is replaced by the symmetric one (or, what amounts to the same thing, if each of the three square roots is replaced by the symmetric one).

This argument suggests another way of choosing the square roots:
* pick ''any'' square root {{math|{{sqrt|''α''}}}} of {{math|''α''}} and ''any'' square root {{math|{{sqrt|''β''}}}} of {{math|''β''}};
* ''define'' {{math|{{sqrt|''γ''}}}} as <math>-\frac q{\sqrt{\alpha}\sqrt{\beta}}</math>.
Of course, this will make no sense if {{math|''α''}} or {{math|''β''}} is equal to {{math|0}}, but {{math|0}} is  a root of (''{{EquationNote|2}}'') only when {{math|''q'' {{=}} 0}}, that is, only when we are dealing with a [[Quartic function#Biquadratic equation|biquadratic equation]], in which case there is a much simpler approach.