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

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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

\(\ds x^3 - 15x - 4\) \(=\) \(\ds 0\)

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:

\(\ds Q\) \(=\) \(\ds \dfrac {3 a c - b^2} {9 a^2}\)
\(\ds \) \(=\) \(\ds \dfrac {3 \times 1 \times \paren {-15} - 0^2} {9 \times 1^2}\)
\(\ds \) \(=\) \(\ds -5\)

and:

\(\ds R\) \(=\) \(\ds \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}\)
\(\ds \) \(=\) \(\ds \dfrac {9 \times 1 \times 0 \times \paren {-15} - 27 \times 1^2 \times \paren{-4 } - 2 \times 0^3} {54 \times 1^3}\)
\(\ds \) \(=\) \(\ds 2\)

Thus:

\(\ds x\) \(=\) \(\ds S + T\) putting $x = S + T$
\(\ds \) \(=\) \(\ds \sqrt [3] {R + \sqrt {Q^3 + R^2} } + \sqrt[3] {R - \sqrt {Q^3 + R^2} }\) substituting for $S$ and $T$
\(\ds \) \(=\) \(\ds \sqrt [3] {R + \sqrt {\paren {-5}^3 + R^2} } + \sqrt [3] {R - \sqrt {\paren {-5}^3 + R^2} }\) substituting for $Q$
\(\ds \) \(=\) \(\ds \sqrt [3] {2 + \sqrt {\paren { {-5}^3 + 2^2 } } } + \sqrt [3] {2 - \sqrt {\paren { {-5}^3 + 2^2 } } }\) substituting for $R$
\(\ds \) \(=\) \(\ds \sqrt [3] {2 + \sqrt {-121 } } + \sqrt [3] {2 - \sqrt {-121 } }\)
\(\ds \) \(=\) \(\ds \sqrt [3] {\paren {2 + 11 i } } + \sqrt [3] {\paren {2 - 11 i } }\)


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:

\(\ds \) \(=\) \(\ds \sqrt {2^2 + 11^2 }\) Definition of Complex Modulus
\(\ds \) \(=\) \(\ds \sqrt {125 }\)


The argument under the first cube root is:

\(\ds \map \cos {\arg z}\) \(=\) \(\ds \dfrac 2 { \sqrt {125 } }\) Definition of Polar Form of Complex Number
\(\ds \map \sin {\arg z}\) \(=\) \(\ds \dfrac {11 } { \sqrt {125 } }\) Definition of Polar Form of Complex Number
\(\ds z\) \(=\) \(\ds \map \arctan {\dfrac {11} 2}\) Quadrant 1 - Roughly $79.7 \degrees$
\(\ds \) \(=\) \(\ds \map \arctan {5.5}\)


The argument under the second cube root is:

\(\ds \map \cos {\arg z}\) \(=\) \(\ds \dfrac 2 { \sqrt {125 } }\) Definition of Polar Form of Complex Number
\(\ds \map \sin {\arg z}\) \(=\) \(\ds \dfrac {-11 } { \sqrt {125 } }\) Definition of Polar Form of Complex Number
\(\ds z\) \(=\) \(\ds \map \arctan {\dfrac {-11} 2}\) Quadrant 4 - Roughly $280.3 \degrees$
\(\ds \) \(=\) \(\ds \map \arctan {-5.5}\)


We can now rewrite our solution as follows:

\(\ds \sqrt [3] {\paren {2 + 11 i } } + \sqrt [3] {\paren {2 - 11 i } }\) \(=\) \(\ds \sqrt [3] {\sqrt {125 } } \paren {\sqrt [3] {\map \cis {\map \arctan {5.5} } } + \sqrt [3] {\map \cis {\map \arctan {-5.5} } } }\)
\(\ds \) \(=\) \(\ds \sqrt 5 \paren {\sqrt [3] {\map \cis {\map \arctan {5.5} } } + \sqrt [3] {\map \cis {\map \arctan {-5.5} } } }\) Power of Power


$\sqrt [3] {\map \cis {\map \arctan {5.5} } }$ 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} } } } $


\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \sqrt 5 \paren { \paren {\map \cos {\dfrac {\map \arctan {5.5} } 3} + i \map \sin {\dfrac {\map \arctan {5.5} } 3} } + \paren {\map \cos {\dfrac {\map \arctan {-5.5} } 3} + i \map \sin {\dfrac {\map \arctan {-5.5} } 3} } }\) No. $\paren {\map \sin {\dfrac {\map \arctan {5.5} } 3} + \map \sin {\dfrac {\map \arctan {-5.5} } 3} } > 0$
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \sqrt 5 \paren { \paren {\map \cos {\dfrac {\map \arctan {5.5} } 3} + i \map \sin {\dfrac {\map \arctan {5.5} } 3} } + \paren {\map \cos {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {2 \pi} 3} + i \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {2 \pi} 3} } }\) No. $\paren {\map \sin {\dfrac {\map \arctan {5.5} } 3} + \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {2 \pi} 3} } < 0$
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds \sqrt 5 \paren { \paren {\map \cos {\dfrac {\map \arctan {5.5} } 3} + i \map \sin {\dfrac {\map \arctan {5.5} } 3} } + \paren {\map \cos {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {4 \pi} 3} + i \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {4 \pi} 3} } }\) Yes. $\paren {\map \sin {\dfrac {\map \arctan {5.5} } 3} + \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {4 \pi} 3} } = 0$
\(\text {(4)}: \quad\) \(\ds \) \(=\) \(\ds \sqrt 5 \paren { \paren {\map \cos {\dfrac {\map \arctan {5.5} } 3 + \dfrac {2 \pi} 3} + i \map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {2 \pi} 3} } + \paren {\map \cos {\dfrac {\map \arctan {-5.5} } 3} + i \map \sin {\dfrac {\map \arctan {-5.5} } 3} } }\) No. $\paren {\map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {2 \pi} 3} + \map \sin {\dfrac {\map \arctan {-5.5} } 3} } > 0$
\(\text {(5)}: \quad\) \(\ds \) \(=\) \(\ds \sqrt 5 \paren { \paren {\map \cos {\dfrac {\map \arctan {5.5} } 3 + \dfrac {2 \pi} 3} + i \map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {2 \pi} 3} } + \paren {\map \cos {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {2 \pi} 3} + i \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {2 \pi} 3} } }\) Yes. $\paren {\map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {2 \pi} 3} + \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {2 \pi} 3} } = 0$
\(\text {(6)}: \quad\) \(\ds \) \(=\) \(\ds \sqrt 5 \paren { \paren {\map \cos {\dfrac {\map \arctan {5.5} } 3 + \dfrac {2 \pi} 3} + i \map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {2 \pi} 3} } + \paren {\map \cos {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {4 \pi} 3} + i \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {4 \pi} 3} } }\) No. $\paren {\map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {2 \pi} 3} + \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {4 \pi} 3} } > 0$
\(\text {(7)}: \quad\) \(\ds \) \(=\) \(\ds \sqrt 5 \paren { \paren {\map \cos {\dfrac {\map \arctan {5.5} } 3 + \dfrac {4 \pi} 3} + i \map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {4 \pi} 3} } + \paren {\map \cos {\dfrac {\map \arctan {-5.5} } 3} + i \map \sin {\dfrac {\map \arctan {-5.5} } 3} } }\) Yes. $\paren {\map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {4 \pi} 3} + \map \sin {\dfrac {\map \arctan {-5.5} } 3} } = 0$
\(\text {(8)}: \quad\) \(\ds \) \(=\) \(\ds \sqrt 5 \paren { \paren {\map \cos {\dfrac {\map \arctan {5.5} } 3 + \dfrac {4 \pi} 3} + i \map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {4 \pi} 3} } + \paren {\map \cos {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {2 \pi} 3} + i \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {2 \pi} 3} } }\) No. $\paren {\map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {4 \pi} 3} + \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {2 \pi} 3} } < 0$
\(\text {(9)}: \quad\) \(\ds \) \(=\) \(\ds \sqrt 5 \paren { \paren {\map \cos {\dfrac {\map \arctan {5.5} } 3 + \dfrac {4 \pi} 3} + i \map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {4 \pi} 3} } + \paren {\map \cos {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {4 \pi} 3} + i \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {4 \pi} 3} } }\) No. $\paren {\map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {4 \pi} 3} + \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {4 \pi} 3} } < 0$

Our $3$ solutions are:

\(\ds \sqrt 5 \paren { \paren {\map \cos {\dfrac {\map \arctan {5.5} } 3} + i \map \sin {\dfrac {\map \arctan {5.5} } 3} } + \paren {\map \cos {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {4 \pi} 3} + i \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {4 \pi} 3} } }\) \(=\) \(\ds \sqrt 5 \paren { \paren {\dfrac 2 {\sqrt 5} + i \dfrac 1 {\sqrt 5} } + \paren {\dfrac 2 {\sqrt 5} - i \dfrac 1 {\sqrt 5} } }\)
\(\ds \) \(=\) \(\ds 4\)
\(\ds \sqrt 5 \paren { \paren {\map \cos {\dfrac {\map \arctan {5.5} } 3 + \dfrac {2 \pi} 3} + i \map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {2 \pi} 3} } + \paren {\map \cos {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {2 \pi} 3} + i \map \sin {\dfrac {\map \arctan {-5.5} } 3 + \dfrac {2 \pi} 3} } }\) \(=\) \(\ds \sqrt 5 \paren { \paren {\dfrac {-2 - \sqrt 3} {2 \sqrt 5} + i \dfrac {\sqrt {13 - 4 \sqrt 3} } {2 \sqrt 5} } + \paren {\dfrac {-2 - \sqrt 3} {2 \sqrt 5} - i \dfrac {\sqrt {13 - 4 \sqrt 3} } {2 \sqrt 5} } }\)
\(\ds \) \(=\) \(\ds -2 - \sqrt 3\)
\(\ds \sqrt 5 \paren { \paren {\map \cos {\dfrac {\map \arctan {5.5} } 3 + \dfrac {4 \pi} 3} + i \map \sin {\dfrac {\map \arctan {5.5} } 3 + \dfrac {4 \pi} 3} } + \paren {\map \cos {\dfrac {\map \arctan {-5.5} } 3} + i \map \sin {\dfrac {\map \arctan {-5.5} } 3} } }\) \(=\) \(\ds \sqrt 5 \paren { \paren {\dfrac {-2 + \sqrt 3} {2 \sqrt 5} + i \dfrac {\sqrt {13 + 4 \sqrt 3} } {2 \sqrt 5} } + \paren {\dfrac {-2 + \sqrt 3} {2 \sqrt 5} - i \dfrac {\sqrt {13 + 4 \sqrt 3} } {2 \sqrt 5} } }\)
\(\ds \) \(=\) \(\ds -2 + \sqrt 3\)


Hence:

\(\ds x^3 - 15 x - 4\) \(=\) \(\ds \paren {x - 4} \paren {x - \paren {-2 - \sqrt 3} } \paren {x - \paren {-2 + \sqrt 3} }\)
\(\ds \) \(=\) \(\ds \paren {x - 4} \paren {x^2 + 4x + 1}\)

The result follows.

$\blacksquare$


Sources