Editing Quartic function (section)

====Special cases of the formula====
*If <math>\Delta > 0,</math> the value of <math>Q</math> is a non-real complex number. In this case, either all roots are non-real or they are all real. In the latter case, the value of <math>S</math> is also real, despite being expressed in terms of <math>Q;</math> this is [[casus irreducibilis]] of the cubic function extended to the present context of the quartic. One may prefer to express it in a purely real way, by using [[trigonometric functions]], as follows:
::<math>S = \frac{1}{2} \sqrt{-\frac23\ p+\frac{2}{3a}\sqrt{\Delta_0}\cos\frac{\varphi}{3}}</math>
:where
::<math>\varphi = \arccos\left(\frac{\Delta_1}{2\sqrt{\Delta_0^3}}\right).</math>

*If <math>\Delta \neq 0</math> and <math>\Delta_0 = 0,</math> the sign of <math>\sqrt{\Delta_1^2 - 4 \Delta_0^3}=\sqrt{\Delta_1^2} </math>  has to be chosen to have <math>Q \neq 0,</math> that is one should define <math>\sqrt{\Delta_1^2}</math> as <math>\Delta_1,</math> maintaining the sign of <math>\Delta_1.</math>
*If <math>S = 0,</math> then one must change the choice of the cube root in <math>Q</math> in order to have <math>S \neq 0.</math> This is always possible except if the quartic may be factored into <math>\left(x+\tfrac{b}{4a}\right)^4.</math> The result is then correct, but misleading because it hides the fact that no cube root is needed in this case. In fact this case{{Clarify|reason=Both kinds or specific?|date=March 2024}} may occur only if the [[numerator]] of <math>q</math> is zero, in which case the associated [[#Converting to a depressed quartic|depressed quartic]] is biquadratic; it may thus be solved by the method described [[#Biquadratic equation|below]].
*If <math>\Delta = 0</math> and <math>\Delta_0 = 0,</math> and thus also <math>\Delta_1 = 0,</math> at least three roots are equal to each other, and the roots are [[rational function]]s of the coefficients. The triple root <math>x_0</math> is a common root of the quartic and its second derivative <math>2(6ax^2+3bx+c);</math> it is thus also the unique root of the remainder of the [[Euclidean division]] of the quartic by its second derivative, which is a linear polynomial. The simple root <math>x_1</math> can be deduced from <math>x_1+3x_0=-b/a.</math>
*If <math>\Delta=0</math> and <math> \Delta_0 \neq 0,</math> the above expression for the roots is correct but misleading, hiding the fact that the polynomial is [[irreducible polynomial|reducible]] and no cube root is needed to represent the roots.