# Eikonal Blog

## Cubic equation

The roots of 3rd order algebraic equation $x^3 + bx^2+cx+d=0$ are given by the Tartaglia-Cardano formulas:

• $x_0 = X_0 + X_{+} + X_{-}$,
• $x_1 = X_0 + \omega X_{+} + \omega^2 X_{-}$,
• $x_2 = X_0 + \omega^2 X_{+} + \omega X_{-}$,

where:

• $\omega:\equiv e^{i\frac{2\pi}{3}} = \frac{-1+i\sqrt{3}}2$,
• $X_0 :\equiv -\frac{b}3$,
• $X_{\pm} :\equiv \left\{ - \left(\frac{27 d -9bc+2b^3}{54}\right) \pm \sqrt{\left(\frac{27 d -9bc+2b^3}{54}\right)^2 - \left( \frac{b^2-3c}{9}\right)^3} \right\}^{\frac13}$,
• $X_{+} X_{-} = \frac{b^2-3c}{9}$,
• $X_{+}^2 + X_{-}^2 = - \frac{27 d -9bc+2b^3}{54}$.

More:

### Special case: reduced cubic equation (Cardano’s formula)

Change of variables $x = y -\frac{b}3$ reduces the general third-order algebraic equation $x^3 + bx^2+cx+d=0$ to the reduced form $y^3+p y + q = 0$. This one can be solved using the Cardan’s ansatz

• $y = \sqrt{A+B} + \sqrt{A-B}$ for some not yet determined $A$ and $B$.
• This gives: $y^3 = 2A+ 3y \sqrt{A^2-B^2}$
• Comparing coefficients with the $y^3 = -p y - q$ gives $2A=q$ and $3 \sqrt{A^2-B^2} = p$
• Their solution is: $A=\frac{q}2$, $B = \sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}3\right)^3}$.

So, finally, one root of the reduced equation $y^3+p y + q = 0$ is (the Cardano’s formula): $y_1 = \sqrt{\frac{q}2+\sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}3\right)^3}} + \sqrt{\frac{q}2-\sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}3\right)^3}}$

The remaining two roots are solutions of the quadratic equation $y^2 + y_1 y + y_1^2+p=0$: $y_{2,3} = \frac12(-y_1 \pm \sqrt{-3 y^2_1 -4 p})$.

## Quartic equation

### Special case: $y^4+y=x$

The solution is $y = \frac12 \epsilon \sqrt{A} + \frac12 \epsilon' \sqrt{B}$ for $\epsilon, \epsilon' = \pm 1$, where $B = -A - \frac{2\epsilon}{\sqrt{A}}$, and $A$ is any of three roots of the cubic equation $A^3+4xA-1=0$. For example, we may take the Cardan’s solution: $A_1 = \sqrt{a+b} + \sqrt{a-b}$ where $a=\frac12$ and $b = \frac1{2\sqrt{3}}\sqrt{27+256 x^3}$. Putting all elements together, $y = \frac12 \epsilon \sqrt{\sqrt{\frac12+\frac1{2\sqrt{3}}\sqrt{27+256 x^3}} + \sqrt{\frac12-\frac1{2\sqrt{3}}\sqrt{27+256 x^3}}} + \frac12 \epsilon' \sqrt{- \sqrt{\frac12+\frac1{2\sqrt{3}}\sqrt{27+256 x^3}} - \sqrt{\frac12-\frac1{2\sqrt{3}}\sqrt{27+256 x^3}} -\frac{2\epsilon}{\sqrt{\sqrt{\frac12+\frac1{2\sqrt{3}}\sqrt{27+256 x^3}} + \sqrt{\frac12-\frac1{2\sqrt{3}}\sqrt{27+256 x^3}}}}}$

Derivation: Start from the ansatz $y = \frac12 \epsilon \sqrt{A} + \frac12 \epsilon' \sqrt{B}$ for yet unknown A & B (Q: How? Why? ;-). Then

• $y^2 = \frac14 (A+B) + \frac{\epsilon\epsilon'}2 \sqrt{AB}$,
• $y^4 = \frac1{16} (A+B)^2 + \frac{\epsilon\epsilon'}4 (A+B) \sqrt{AB} + \frac14 AB$,
• Require that $A+B = \frac{\alpha}{\sqrt{A}}$ (Again: How? Why?)
• Then $y^4 = \frac{\alpha^2}{16 A} -\frac{A^2}4 + \frac{\epsilon\epsilon'\alpha}4 \sqrt{B} + \frac{\alpha}4 \sqrt{A}$,
• and $y^4 + y = \frac{\alpha^2}{16 A} -\frac{A^2}4 + \left(\frac{\epsilon\epsilon'\alpha}4+\frac{\epsilon'}2\right) \sqrt{B} + \left(\frac{\alpha}4 +\frac{\epsilon}2\right)\sqrt{A}$.
• Setting $\alpha=-2\epsilon$ simplifies that equation by eliminating roots: $y^4 + y = \frac{\alpha^2}{16 A} -\frac{A^2}4 = x$
• which is equivalent to the cubic equation $A^3+4xA-1=0$ solvable via Cardano’s procedure.

Example: A simple (and trivial) example is for $x=0$, when $y^4+y=0$ has roots $\{y_1=0\} \cup \{y_{n}=e^{i\frac{\pi}3(n-1)} | n=2,3,4\}$. Here $b=a=\frac12$, $A_1=1$, $B_1=-1-2\epsilon$ and $y_{\epsilon,\epsilon'}(A_1) = \frac{\epsilon+\epsilon' \sqrt{-1-2\epsilon}}2$, i.e.

• $y_{+,+} = \frac{1+i\sqrt{3}}2 \equiv y_2$
• $y_{+,-} = \frac{1-i\sqrt{3}}2 \equiv y_3$
• $y_{-,+} = 0 \equiv y_1$
• $y_{-,-} = -1 \equiv y_4$

More:

More: