Cardano's Formula/Examples/x^3 - 15x - 4

Example of Use of Cardano's Formula

 * $\ds x^3 - 15x - 4 = 0$

has solutions $x = 4$, $x = -2 + \sqrt 3$ and $x = -2 - \sqrt 3$.

Proof
This is in the form:
 * $a x^3 + b x^2 + c x + d = 0$

where:
 * $a = 1$
 * $b = 0$
 * $c = -15$
 * $d = -4$

From Cardano's Formula:
 * $x = S + T$

where:
 * $S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }$
 * $T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }$

where:

and:

Thus:

Using Cardano's Formula/Real Coefficients, we know from
 * $D = Q^3 + R^2 = -121 < 0$

that all roots are real and unequal.

The complex modulus under the cube root is:

The argument under the first cube root is:

The argument under the second cube root is:

We can now rewrite our solution as follows:


 * $\sqrt [3] {\map \cis {\map \arctan {5.5} } }$ has $3$ unique cube roots.

$\tuple { \map \cis {\dfrac {\map \arctan {5.5} } 3}, \map \cis {\dfrac {\map \arctan {5.5} } 3 + \dfrac {2 \pi } 3}, \map \cis {\dfrac {\map \arctan {5.5} } 3 + \dfrac {4 \pi } 3 } }$


 * $\sqrt [3] {\map \cis {\map \arctan {-5.5} } }$ also has $3$ unique cube roots.

$\tuple { \map \cis {\dfrac {\map \arctan {-5.5} } 3}, \map \cis {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {2 \pi } 3}, \map \cis {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {4 \pi } 3 } }$

We need to investigate $9$ sums and find solutions where the complex part vanishes.

$ \sqrt 5 \paren {\sqrt [3] {\map \cis {\map \arctan {5.5} } } + \sqrt [3] {\map \cis {\map \arctan {-5.5} } } } $

Our $3$ solutions are:

Hence:

The result follows.