Solving Cubic when There are Known to be 3 Real Roots

Question

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?

Answer 1

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).

Answer 2

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.

Answer 3

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.