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I have to explain the problem I am thinking first:

The solutions of $x^3+6x-20=0$ are $x = 2$, $-1+3i$ and $-1-3i$.

The cubic formula for these solutions is:

$$x = \sqrt[3]{10 + \sqrt{108}} + \sqrt[3]{10 + \sqrt{108}}$$

First $\sqrt{108} \approx -10.392304845413$, $10.392304845413$ has two values.

And $\sqrt[3]{10 + -10.392304845413}$ has three values $v_1\approx 0.3660 + 0.6339745i$, $v_2\approx -0.7320508$, $v_3\approx 0.3660254 - 0.6339745i$.

And $\sqrt[3]{10 + 10.392304845413}$ has three values $u_1\approx 2.7320508$, $u_2\approx -1.366 + 2.366$, $u_3\approx -1.366 - 2.366i$.

So there are nine possible values of $x$:

  • $v_1 + u_1 = 3.098 + 0.63397i$
  • $v_1 + u_2 = -1 + 3i$
  • $v_1 + u_3 = -1 - 1.732i$
  • $v_2 + u_1 = 2$
  • $v_2 + u_2 = -2.098 + 2.366i$
  • $v_2 + u_3 = -2.098 - 2.366i$
  • $v_3 + u_1 = 3.098 - 0.63397i$
  • $v_3 + u_2 = -1 + 1.732$
  • $v_3 + u_3 = -1 - 3i$

but the correct ones are $v_1 + u_2$, $v_2 + u_1$, $v_3 + u_3$.

Edit Here is a table of 3vu for Arturo Magidin:

  • $3 v_1 u_1 = 3 + 5.196152i$
  • $3 v_1 u_2 = -6$
  • $3 v_1 u_3 = 3 - 5.196152i$
  • $3 v_2 u_1 = -6$
  • $3 v_2 u_2 = 3 - 5.196152i$
  • $3 v_2 u_3 = 3 + 5.196152i$
  • $3 v_3 u_1 = 3 - 5.196152i$
  • $3 v_3 u_2 = 3 + 5.196152i$
  • $3 v_3 u_3 = -6$

My questions are:

  • What are the others values of $x$ (those which are not solutions)?
  • Is there a better notation for taking roots that points out (for example) that the cube roots need to be synchronized?
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A method that avoids the ambiguity is to express roots $x_{2},x_{3}$ as a function of $x_{1}=\sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}$ as follows $x_{2}=-\frac{x_{1}}{2}+\sqrt{\frac{x_{1}^{2}}{4}+\frac{q}{x_{1}}}$ and $x_{3}=-\frac{x_{1}}{2}-\sqrt{\frac{x_{1}^{2}}{4}+\frac{q}{x_{1}}}$. Since $\sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}=2$ and $\frac{x_{1}^{2}}{4}+\frac{q}{x_{1}}=\frac{x_{1}^{2}}{4}+\frac{-20}{x_{1}}=-9 $ we have $x_{2}=-1+3i$ $x_{3}=-1-3i$ – Américo Tavares Dec 28 '10 at 13:46
@quanta: There is a typo in the solution $x=\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-6\sqrt{108}}$. – Américo Tavares Dec 28 '10 at 13:52
Americo, no that was intentional - I explained my usage of radicals. – quanta Dec 28 '10 at 21:12
quanta, a root of the equation $x^{3}+px+q=0$ is obtained writing $x=u+v$. We get $S=u^{3}+v^{3}=-q$ and $P=u^{3}v^{3}=-\frac{p^{3}}{27}$. These numbers ($u^{3}$ and $v^{3}$) are the roots of $Y^{2}-SY+P=0$: $Y_{+}=\frac{S+\sqrt{S^{2}-4P}}{2}=-\frac{q}{2}+\frac{1}{2}\sqrt{q^{2}+\frac{4p^‌​{3}}{27}}$ and $Y_{-}=\frac{S-\sqrt{S^{2}-4P}}{2}=-\frac{q}{2}-\frac{1}{2}\sqrt{q^{2}+\frac{4p^‌​{3}}{27}}$ See my post… – Américo Tavares Dec 28 '10 at 22:33
There are simpler, less error-prone and more intuitive ways to find roots than $p,q$-formulas. I suggest that you learn those instead. – You Dec 29 '10 at 15:52
up vote 9 down vote accepted

When applying Cardano's formulae, you must take specific branches of the square and cubic root functions. Also, note that the square root of 108 is justs a single value; while $x^2 - 108=0$ has two (real) solutions, the expression $\sqrt{108}$ is a single value: the nonnegative solution to $x^2-108=0$.

In any case, remember that if you have a depressed cubic, $x^3 + px + q = 0$, we introduce two variables $u$ and $v$ subject to the condition that $u+v = x$ and $3uv + p = 0$. This requires the product of $u$ and $v$ to be a real number. Eventually, you arrive at $u$ and $v$ being the cubic roots in question.

But remember that you are assuming that $uv$ is a real number. This is why you cannot just take all possible combinations with all possible roots. In other words, all those "other values" of $u+v$ are values that do not satisfy the system of equations you have: \begin{align*} (u+v)^3 + p(u+v) + q &=0\\ 3uv + p &=0 \end{align*} There is no general notation that tells you which branches of the complex-valued cubic and square root functions you must take, you just have to remember that the product of the two cubic roots has to be a real number. This is what cuts down from nine to three values.

There is a standard way of listing all solutions: once you find one value of $u$ and $v$ such that $uv$ is a real number, then the three solutions are given by $u+v$, $\omega u + \omega^2 v$, and $\omega^2u + \omega v$, where $\omega = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$ is a primitive cubic root of unity. (Notice that if $u$ is one value of the cubic root, then the other two values are $\omega u$ and $\omega^2u$, and likewise with $v$).

It seems that you are not quite aware of all the assumptions you are making about $u$ and $v$ when you use Cardano's formula. For one thing, if you review the derivation of the formula, say, in Wikipedia, you will see that the condition that $3uv+p=0$ is worked into the derivation. Here's the method: remember that $x=u+v$ is supposed to be a solution of $$x^3 + px + q = 0.$$ Plugging in and doing a bit of work, we get: \begin{align*} 0 &= (u+v)^3 + p(u+v) + q\\ &= u^3 + 3u^2v + 3uv^2 + v^3 + p(u + v) + q\\ &= u^3 + v^3 + 3uv(u+v) + p(u+v) + q\\ &= u^3 + v^3 + (3uv+p)(u+v) + q. \end{align*} It is at this point that the extra condition $3uv+p=0$ is introduced, so that the equation reduces to $u^3+v^3 + q = 0$, Then $u^3+v^3 = -q$, and $(uv)^3 = -\frac{p^3}{27}$. Now remember that if $a$ and $b$ are two numbers, then $(z-a)(z-b) = z^2 - (a+b)z + ab$, so $a$ and $b$ are always the roots of the quadratic that has linear coefficient minus their sum, and constant coefficient their product. Applying this to $u^3$ and $v^3$, since we know their sum and their product, we deduce that $u^3$ and $v^3$ are the two roots of the quadratic: $$z^2 +qz - \frac{p^3}{27}=0$$ and from this, using the quadratic formula, we obtain that $$u^3 = -\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}, \qquad v^3 = -\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}.$$ Therefore, $$u = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}},\qquad v = \sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}},$$ so that the solution $x$ is given by the formula you have: $$x = u+v = \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$$ But the assumption that $3uv+p=0$ was instrumental in obtaining this formula. The very fact that you write that $x$ is equal to the sum of the two cubic roots already assumes that the product of those two (complex-valued) cubic roots will be a real number, specifically, $-\frac{p}{3}$.

While you may perhaps not have made that assumption consciously, Cardano certainly did, and by using his formula you are also making the assumption, whether you were aware of it or not.

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He's lucky his starting cubic is already depressed... :) – J. M. Dec 28 '10 at 1:20
@J.M.: Actually, I'm lucky; otherwise, I would have had to write the $3uv+p=0$ condition in terms of the original coefficients... – Arturo Magidin Dec 28 '10 at 1:30
@quanta: Yes you are. If the condition $3uv=-p$ is to be true, since $p$ is the coefficient of a polynomial with real coefficients, the product $uv$ is necessarily real. – J. M. Dec 29 '10 at 8:35
Thank you. I understand now: 3uv = -p is a condition which tells us how to synchronize the cube roots (you proved this in your update). At first I was using f(u+v) = 0 as my condition for synchronizing the roots. Your condition erases the problem because we can use v = -p/3u instead of having two cube roots! – quanta Dec 30 '10 at 0:07
@quanta: It's not "my" condition. It's Cardano's. – Arturo Magidin Dec 30 '10 at 0:38

Here's a pretty clean version of Cardano I use (adapted from Numerical Recipes). I assume here that the principal values for both the square and cube root (the result is in the interval $\left(-\frac{\pi}{n},\frac{\pi}{n}\right]$, which as Isaac says applies for most computing systems) are taken.

For $x^3+bx+c=0$, letting

$q=\frac{b}{3}$ and $r=\frac{c}{2}$



$$B=\begin{cases}\frac{q}{A}&\text{ if }A\neq0\\0&\text{ if }A=0\end{cases}$$

the three roots can be expressed as

$$-\frac{A-B}{2}\pm i\frac{\sqrt{3}}{2}(A+B)$$

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More of an extended comment than an answer:

Usually, the way I've seen the radical notation used, it is mean to imply only the "principal" root, for some definition of "principal"—therein lies the trouble. I have seen principal $n$th root defined as the $n$th root with the least non-negative argument or as the $n$th root with argument in $(-\frac{\pi}{n},\frac{\pi}{n}]$ (and, in fact, the latter is the only definition that I've seen implemented in calculators and computer algebra systems). Most of the instances where I've seen the cubic formula (or quartic formula), it is intended for the first definition, though it is possible to write the formula (both) in such a way that takes the $n$th root only once and thus avoids the need for synchronizing the choice of principal root across multiple expressions (such as my answer here for quartics).

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Rewrite the equation as $$x = \omega^a \sqrt[3]{10 + \sqrt{108}} - \frac{6}{3 \omega^a \sqrt[3]{10 + \sqrt{108}}}$$

All three values of a = 0,1,2 produce a solution to the equation with no ambiguity.

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You were given complete answers above. What is it you found unsatisfying about them? – Arturo Magidin Dec 29 '10 at 6:38
Now I have a complete solution! I accept your answer instead of this since it explains how to find it. Thank you. – quanta Dec 30 '10 at 0:12

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