Eikonal Blog


Roots of algebraic equations

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_{-},


  • \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}.


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[3]{A+B} + \sqrt[3]{A-B} for some not yet determined A and B.
  • This gives: y^3 = 2A+ 3y \sqrt[3]{A^2-B^2}
  • Comparing coefficients with the y^3 = -p y - q gives 2A=q and 3 \sqrt[3]{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[3]{\frac{q}2+\sqrt{\left(\frac{q}{2}\right)^2-\left(\frac{p}3\right)^3}} + \sqrt[3]{\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[3]{a+b} +  \sqrt[3]{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[3]{\frac12+\frac1{2\sqrt{3}}\sqrt{27+256 x^3}} +  \sqrt[3]{\frac12-\frac1{2\sqrt{3}}\sqrt{27+256 x^3}}} +  \frac12 \epsilon' \sqrt{- \sqrt[3]{\frac12+\frac1{2\sqrt{3}}\sqrt{27+256 x^3}} -  \sqrt[3]{\frac12-\frac1{2\sqrt{3}}\sqrt{27+256 x^3}} -\frac{2\epsilon}{\sqrt{\sqrt[3]{\frac12+\frac1{2\sqrt{3}}\sqrt{27+256 x^3}} +  \sqrt[3]{\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


Quintic equation


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