## Quartic equations

Composite degrees n are better solved by polynomial decomposition.  However, for $n = 2^m$, there is an alternative method.  Thus, we’ll give two methods to solve the quartic.  Given the depressed quartic, $x^4+px^2+qx+r = 0$

Its resolvent cubic is simply, $y^3+2py^2+(p^2-4r)y-q^2=0$

Method 1:

Given any root y of the cubic, then, \begin{aligned} x_{1,2} &=\tfrac{1}{2} \left(\sqrt{y} \pm \sqrt{-(2p+y)-\tfrac{2q}{\sqrt{y}}}\,\right)\\ x_{3,4} &= \tfrac{1}{2} \left(-\sqrt{y} \pm \sqrt{-(2p+y)+\tfrac{2q}{\sqrt{y}}}\, \right)\end{aligned}

For convenience, one can choose the real root y.  Notice that { $x_3, x_4$} are the negative case of $\pm \sqrt{y}$.  The formula can easily be proven by squaring the variable y, forming the quadratic in { $x_1,\,x_2$}, and eliminating y between them using resultants, $\text{Factor}[\text{Resultant}[y^6+2py^4+(p^2-4r)y^2-q^2,\,x^2+xy+\frac{y^3+py-q}{2y},\,y]]\\ = (x^4+px^2+qx+r)^3$

proving any root of the sextic (a cubic in $y^2$) will suffice.

Method 2:

Given the three roots { $y_1, y_2, y_3$} of the resolvent cubic, then, \begin{aligned} x_{1,2} &= \frac{\sqrt{y_1} \pm (\sqrt{y_2}+\sqrt{y_3}\,)}{2}\\ x_{3,4} &= \frac{-\sqrt{y_1} \pm (\sqrt{y_2}-\sqrt{y_3}\,)}{2}\end{aligned}

where signs are chosen such that, $\sqrt{y_1}\sqrt{y_2}\sqrt{y_3} = -q$

Proof:  Expanding, $(x-x_1)(x-x_2)(x-x_3)(x-x_4) = 0$

using the formulas, we get, $x^4-\frac{1}{2}(y_1+y_2+y_3)x^2-\sqrt{y_1}\sqrt{y_2}\sqrt{y_3}\,x\\+\frac{1}{16}(y_1^2+y_2^2+y_3^2-2(y_1y_2+y_1y_3+y_2y_3)) = 0$

From  and the properties of elementary symmetric polynomials, we find that it is in fact equal to, $x^4+px^2+qx+r = 0$

A similar method was used to solve the octic $x^8-x^7+29x^2+29 = 0$ in this post, though now one had to take square roots of its resolvent septic.