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 for Real Coefficients, we know from

$D = Q^3 + R^2 = -121 < 0$

that all roots are real and unequal.


From Roots of Complex Number/Examples/Cube Roots of 2+11i, we have:

$\sqrt [3] {\paren {2 + 11 i } } = \set {2 + i, -1 + \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 - \sqrt 3 }, -1 - \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 + \sqrt 3}}$

From Roots of Complex Number/Examples/Cube Roots of 2-11i, we have:

$\sqrt [3] {\paren {2 - 11 i } } = \set {2 - i, -1 + \dfrac {\sqrt 3} 2 + i \paren {\dfrac 1 2 + \sqrt 3 }, -1 - \dfrac {\sqrt 3} 2 + i \paren {\dfrac 1 2 - \sqrt 3}}$


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

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

Our $3$ solutions are:

\(\ds \paren{2 + i } + \paren{2 - i }\) \(=\) \(\ds 4\)
\(\ds \paren{-1 + \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 - \sqrt 3 } } + \paren{-1 + \dfrac {\sqrt 3} 2 + i \paren {\dfrac 1 2 + \sqrt 3 } }\) \(=\) \(\ds -2 + \sqrt 3\)
\(\ds \paren{-1 - \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 + \sqrt 3} } + \paren{-1 - \dfrac {\sqrt 3} 2 + i \paren {\dfrac 1 2 - \sqrt 3 } }\) \(=\) \(\ds -2 - \sqrt 3\)


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


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