Cardano's Formula/Real Coefficients

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Theorem

Let $P$ be the cubic equation:

$a x^3 + b x^2 + c x + d = 0$

with $a \ne 0$.

Let $a, b, c, d \in \R$.


Let $D$ be the discriminant of $P$:

$D := Q^3 + R^2$

where:

$Q = \dfrac {3 a c - b^2} {9 a^2}$
$R = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}$


Then:

$(1): \quad$ If $D > 0$, then one root is real and two are complex conjugates.
$(2): \quad$ If $D = 0$, then all roots are real, and at least two are equal.
$(3): \quad$ If $D < 0$, then all roots are real and unequal.


Proof

From Cardano's Formula, the roots of $P$ are:

\(\text {(1)}: \quad\) \(\ds x_1\) \(=\) \(\ds S + T - \dfrac b {3 a}\)
\(\text {(2)}: \quad\) \(\ds x_2\) \(=\) \(\ds -\dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}\)
\(\text {(3)}: \quad\) \(\ds x_3\) \(=\) \(\ds -\dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}\)


where:

\(\ds S\) \(=\) \(\ds \sqrt [3] {R + \sqrt {Q^3 + R^2} }\)
\(\ds T\) \(=\) \(\ds \sqrt [3] {R - \sqrt {Q^3 + R^2} }\)


Zero Discriminant

First the easy case: $D = 0$.

Hence $S = T = \sqrt [3] R$, and so $S + T = 2 \sqrt [3] R, S - T = 0$.

From the above, this gives us:

\(\text {(1)}: \quad\) \(\ds x_1\) \(=\) \(\ds 2 \sqrt [3] R - \dfrac b {3 a}\)
\(\text {(2)}: \quad\) \(\ds x_2\) \(=\) \(\ds -\sqrt [3] R - \dfrac b {3 a}\)
\(\text {(3)}: \quad\) \(\ds x_3\) \(=\) \(\ds -\sqrt [3] R - \dfrac b {3 a}\)

Thus the roots $x_2$ and $x_3$ are equal, and all three roots are real.

They are all equal when $R = 0$.

$\Box$


Positive Discriminant

Let $D = Q^3 + R^2 > 0$.

Then $S = R + \sqrt{Q^3 + R^2}$ and $T = R - \sqrt{Q^3 + R^2}$ are wholly real and distinct.

Therefore, so are $S + T$ and $S - T$.

Hence:

$\dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}$

and

$\dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}$

are complex conjugates.

$\Box$


Negative Discriminant

Let $D = Q^3 + R^2 < 0$.

Then $\sqrt D = \pm i \left|{Q^3 + R^2}\right| = \pm i E$, say, where $E > 0$.

Thus $S^3 = R + i E, T^3 = R - i E$.

Let $\sqrt [3] {R + i E} = p + i q$, and so $\sqrt [3] {R - i E} = p - i q$.

Hence $S + T = 2 p, S - T = 2 i q$.

So:

\(\ds y_1\) \(=\) \(\ds S + T\)
\(\ds \) \(=\) \(\ds 2 p\)
\(\ds y_2\) \(=\) \(\ds -\dfrac {S + T} 2 + \dfrac {i \sqrt 3} 2 \paren {S - T}\)
\(\ds \) \(=\) \(\ds -p - \sqrt 3 q\) after algebra
\(\ds y_3\) \(=\) \(\ds -\dfrac {S + T} 2 - \dfrac {i \sqrt 3} 2 \paren {S - T}\)
\(\ds \) \(=\) \(\ds -p + \sqrt 3 q\) after algebra


Subtracting $\dfrac b {3 a}$ from the above, we obtain the three distinct real solutions:

\(\text {(1)}: \quad\) \(\ds x_1\) \(=\) \(\ds 2 p - \dfrac b {3 a}\)
\(\text {(2)}: \quad\) \(\ds x_2\) \(=\) \(\ds -p - \sqrt 3 q - \dfrac b {3 a}\)
\(\text {(3)}: \quad\) \(\ds x_3\) \(=\) \(\ds -p + \sqrt 3 q - \dfrac b {3 a}\)

$\blacksquare$


Sources

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