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When solving for roots to a cubic equation, the sign of the $\Delta$ tells us when there will be 3 distinct real roots (as long as the first terms coefficient, $a$, is non-zero.) Namely when $\Delta$ is positive.

The equations to find the 3 roots are:

  • $x_1 = -\frac{1}{3a}(b + C + \frac{\Delta_0}{C})$
  • $x_2 = -\frac{1}{3a}(b + \frac{C(-1 + i\sqrt 3)}{2} + \frac{2\Delta_0}{C(-1 + i\sqrt 3)})$
  • $x_3 = -\frac{1}{3a}(b + \frac{C(-1 - i\sqrt 3)}{2} + \frac{2\Delta_0}{C(-1 - i\sqrt 3)})$

Where: $$C = \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}}{2}}$$

By the given equation, $\Delta_1^2 - 4\Delta_0^3 = -27a^2\Delta$, I know that when $\Delta$ is positive the square root in $C$ will produce an $i$. So when $\Delta$ is positive $C$ is effectively: $$C = \sqrt[3]{\frac{\Delta_1 + i\sqrt{4\Delta_0^3 - \Delta_1^2}}{2}}$$

So obviously the $i$ in $C$ cancels with the $i$s in the $x_2$ and $x_3$ roots, and we get 3 real roots. But for the life of me I cannot work out how. Can someone help me break those steps down?

Jonathan Mee
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3 Answers3

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You are dealing with the https://en.wikipedia.org/wiki/Casus_irreducibilis

In the case of three real irrational roots, one may express them using cube roots of complex numbers that have both real and imaginary real part. There is no way to separate the real and imaginary parts of these cube roots in any sort of closed form. All you know is that the cube root of $\alpha$ plus the cube root of $\bar{\alpha}$ is real.

This topic is discussed in many Galois theory books; I think it predates Galois, not sure. Maybe not: in the book by Rotman, he says that Cardano and others of the time grew frustrated when they could not separate out the real and imaginary parts of the roots of such a cubic, and they tried all sorts of involved trickery. Then he says they were doomed to failure, and proves it.

Will Jagy
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  • So according to the article my only hope in this case is to resort to solving by trigonometry? – Jonathan Mee Nov 18 '15 at 19:17
  • @JonathanMee, yes, trig or numerical solution, unless one of the three real roots turns out to be rational after all. – Will Jagy Nov 18 '15 at 19:19
  • Which cannot happen in the case where $\Delta$ is negative, unless I missed something. – Jonathan Mee Nov 18 '15 at 19:34
  • Tartaglia had the method in 1530. He told it to Cardano ,who published it. See Cubic Equation, Wikipedia. – DanielWainfleet Nov 18 '15 at 20:01
  • @JonathanMee I'm curious to know where you saw that. I went through it myself, if we begin with rational $B,C$ and cubic $$ (x-B)(x^2 + Bx + C) $$ we get $$ \Delta = - (C + 2 B^2)^2(4 C - B^2). $$ To have all real roots we need $4 C \leq B^2,$ so $4C - B^2 \leq 0.$ With that condition, $\Delta> 0$ unless, with rational $\gamma,$ we have $$ (x + 2 \gamma)(x - \gamma)^2 $$ with a repeated root and $\Delta = 0.$ News to me. – Will Jagy Nov 18 '15 at 21:54
  • Well I was depending upon the statement from http://www.wikipedia.org: "If Δ > 0, then the equation has three distinct real roots." (I thought I'd proved that false, but, they're just Casus Irreducibilis again. – Jonathan Mee Nov 19 '15 at 11:57
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For example, let's try the equation $x^3 - 2 x^2 - x + 2 = 0$: $a= 1$, $b = -2$, $c = -1$, $d=2$, $\Delta_0 = 7$, $\Delta_1 = 20$.

$$C = \sqrt[3]{10 + 9 i \sqrt{3}} = \sqrt{7} e^{i\theta} = \sqrt{7}(\cos\theta + i \sin\theta)$$ where $$ \theta = \dfrac{1}{3} \arctan(9 \sqrt{3}/10)$$ Then $$ \Delta_0/C = \sqrt{7} e^{-i\theta} = \sqrt{7} (\cos \theta - i \sin \theta)$$ so $$ \eqalign{x_1 &= -\dfrac{1}{3} \left( -2 + \sqrt{7} (\cos\theta + i \sin \theta) + \sqrt{7} (\cos\theta - i \sin\theta)\right)\cr &= \dfrac{2}{3} - \dfrac{2\sqrt{7}}{3} \cos \theta}$$ $$ \eqalign{x_2 &= -\dfrac{1}{3} \left( - 2 + \sqrt{7} (\cos(\theta + 2 \pi/3) + i \sin(\theta + 2 \pi/3)) + \sqrt{7} (\cos(\theta - 2 \pi/3) + i \sin(\theta + 2 \pi/3))\right)\cr &= \dfrac{2}{3} - \dfrac{2\sqrt{7}}{3} \cos(\theta + 2\pi/3)}$$ and similarly $$ x_3 = \dfrac{2}{3} - \dfrac{2\sqrt{7}}{3} \cos(\theta - 2\pi/3)$$ It turns out that $x_1 = -1$, $x_2 = 2$, $x_3 = 1$ (which is no accident: I chose the polynomial to have those roots).

Robert Israel
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  • So wait a sec, you're saying this will fail if and only if, we get a formula which does not have 3 roots, but if there are 3 roots this is guaranteed to work? – Jonathan Mee Nov 18 '15 at 20:31
  • Where did I say that? When $\Delta < 0$ there are always $3$ distinct real roots. – Robert Israel Nov 19 '15 at 00:21
  • There are 3 conditions on this working according to https://en.wikipedia.org/wiki/Cubic_function#Trigonometric_method_for_three_real_roots: 1. $p < 0$ 2. $\Delta > 0$ 3. $-1 < q\sqrt{\frac{-27}{8p^3}} < 1$ The problem is 3. I don't believe we can make any guarantees about that and consequentially, I'm not convinced that this method is always capable of returning all the roots in the event that 2 is met :( – Jonathan Mee Nov 19 '15 at 12:08
  • For the depressed cubic $x^3 + p x + q$ with real coefficients, the discriminant is $\Delta = -4 p^3 - 27 q^2$. In order to have $\Delta > 0$ you must have $p < 0$ and $-1 < q \sqrt{\dfrac{-27}{4p^3}} < 1$. I don't know where you're getting that $8$ instead of $4$. – Robert Israel Nov 19 '15 at 18:18
  • Right you are. This is the right solution for my problem. Thanks for bringing it to my attention. – Jonathan Mee Nov 19 '15 at 19:47
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Given $x^3+a x^2+b x+c x+d=0$, substitute $x=y-a/3. $ This gives $y^3+p y+q=0$ where $p,q$ are determined by $a,b,c,d$ and the substitution. Now let $y=u+v$. We have $$0=(u+v)^3+p(u+v)+q=(u^3+v^3+q)+(u+v)(3u v+p).$$ Suppose there are non-zero $u,v$ such that $$u^3+v^3+q=0=3u v+p.$$ Substituting $v=-p/3u$ into $0=u^3+v^3+q$ and letting $w=u^3$, we have $$w^2+q w+(-p^3/27)=u^3(u^3+q+v^3)=0.$$ Solve the quadratic equation in $w . $ Let $u$ be any complex cube root of (either value of) $w$. Then $x=y-a/3=u+v-a/3=u-p/3u-a/3$ solves the original cubic....When there are 3 real solutions for $x$ and they are not an arithmetic sequence, the values of $w$ will not be real numbers and you need the cube root of one of them. One way is to put $w=r. cis (t)=r(\cos t+i\sin t)$ with $r>0$ and $t\in (0,2\pi). $ The cube roots of $w$ are $r^{1/3}cis ((t+2\pi n)/3)$ for $n\in \{0,1,2\}.$.....The main idea in this method of solving the cubic equationis to change the variable to a sum $u+v$ and see if the equation can be split into 2 simpler equations.

DanielWainfleet
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  • $q$ is not defined by $a$, $b$, and $c$. It also needs $d$. – Jonathan Mee Nov 19 '15 at 12:12
  • It looks like you're defining Vieta's Substitution. Now, I can get both roots of $w$, and I can prove they will be real, since $\Delta > 0$ requires $27q^2 + 4p^3 > 0$. But Vieta's solution talks about finding the three cube roots of $u$, is that what you're trying to say in your closing paragraph? Cause I can't understand that part of your answer. – Jonathan Mee Nov 19 '15 at 12:57
  • yes i left out d.... typo....the last paragraph is about the cube roots of u. Agreed that w may be real.but not in the cae I mentioned..Made changes...Trying to get a cube root of a non-real by 16th century methods leads to another cubic, and needs another non-real cube root, which leads to....etc – DanielWainfleet Nov 19 '15 at 18:31
  • Yeah, looks like the best way to handle this is the solution offered by Robert Israel – Jonathan Mee Nov 19 '15 at 19:49