Roots of $2x^3-4x+1$

Question

I'm having difficulty getting the solution to the cubic equation $2x^3-4x+1=0$ and from http://www2.trinity.unimelb.edu.au/~rbroekst/MathX/Cubic%20Formula.pdf it claims that the general solution to $Ax^3+Bx^2+Cx+D=0$ is $$p+q+r=-B/A$$ $$pq+qr+rp=C/A$$ $$pqr=-D/A$$

where $p,q,r$ are the roots.

I also tried http://www2.trinity.unimelb.edu.au/~rbroekst/MathX/Cubic%20Formula.pdf and very carefully followed their technique (which looks at first glance different but must obviously be equivalent) however it didn't line up with what I got from Wolframalpha. So I'm just looking to see where I messed up or how others would solve this.

Here's my attempt:

$$p+q+r=0$$ $$pq+qr+rp=-2$$ $$pqr=-1/2$$

From the first we get that $$p=-q-r$$ then from the second $$p(q+r) = -2-qr$$ then from the two of these we get $$-p^2 = -2-qr$$ equivalently $$0=p^3 -2p+1/2$$

Wait! What?! A cubic to solve a cubic..... cubics all the way down!? Obviously not.... so if it's possible to solve this apart from numerical techniques, I would be really interested in such an answer. Thank you

Answer 1

If you use what is described here (since $\Delta=404$ implies three real roots), you should end with $$x_k=2 \sqrt{\frac{2}{3}} \cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos ^{-1}(-\frac{3\sqrt 3}{8 \sqrt 2})\right)\qquad \text{with} \qquad k=0,1,2$$

Answer 2

Let $x=\frac{2\sqrt2}{\sqrt3}t.$

Thus, $$\frac{32\sqrt2}{3\sqrt3}t^3-\frac{8\sqrt2}{\sqrt3}t+1=0$$ or $$4t^3-3t+\frac{3\sqrt3}{8\sqrt2}=0.$$ Now, let $t=\cos\theta$.

Hence, $$\cos3\theta=-\frac{3\sqrt3}{8\sqrt2}$$ or $$\theta=\pm\left(60^{\circ}-\frac{1}{3}\arccos\frac{3\sqrt3}{8\sqrt2}\right)+120^{\circ}k,$$ where $k$ is an integer number.

Id est, we got three following roots: $$x_1=\frac{2\sqrt2}{\sqrt3}\cos\left(60^{\circ}-\frac{1}{3}\arccos\frac{3\sqrt3}{8\sqrt2}\right),$$ $$x_2=\frac{2\sqrt2}{\sqrt3}\cos\left(60^{\circ}+\frac{1}{3}\arccos\frac{3\sqrt3}{8\sqrt2}\right)$$ and $$x_3=-\frac{2\sqrt2}{\sqrt3}\cos\left(\frac{1}{3}\arccos\frac{3\sqrt3}{8\sqrt2}\right).$$

Answer 3

We have a cubic equation $2x^3 - 4x + 1 = 0$, what $x$ is its root ?

May you can solve this by formula $\cos 3t = 4\cos^3 t - 3\cos t$. But I hope that you understood the means and enjoys from apply them on many cubic equations. Indeed, you will solve this cubic equation.

The cubic equation is $x^3 - 2x + \frac{1}{2} = 0$ which have form $t^3 + pt + q = 0$. A cubic equation of this class can be got from $x^3 + ax^2 + bx + c = 0$ if replace the $x$ by $t - \frac{a}{3}$ and then the coefficients $$p = \frac{3b - a^2}{3}\quad \mathrm{and}\quad q = \frac{2a^2 - 9ab + 27c}{27}.$$

Just as $(x + a)^3 = x^3 + 3ax^2 + 3a^2x + a^3$, if $z = e^{it} = x + iy$, then $$z^3 = e^{i3t}= x^3 + i3yx^2 - 3y^2x - iy^3 = (4\cos^3 t - 3\cos t) + i(-4\sin^3 t + 3\sin t).$$

If $z = e^{it} = x + iy$ and $\zeta = x^3 - iy^3$, then it is $4\zeta - 3\overline{z} - z^3 = 0$. Just as this, if $z = re^{it} = rx + iry$ and $\zeta = r^3x^3 - ir^3y^3$, then $4\zeta^3 - 3r^2\overline{z} - z^3 = 0$.

For your cubic equation $4x^3 - 8x + 2 = 0$, you know $$r = \sqrt{\frac{8}{3}}\quad \mathrm{and} \quad t = \frac{1}{3}\arg(-2\sqrt{\frac{3}{8}}^3) + \frac{2n\pi}{3} = \frac{1}{3}\arg(-\frac{3}{8}\sqrt{\frac{3}{2}})+ \frac{2n\pi}{3}$$ where $n \in \mathbb{Z}$.

Therefore $$x = \sqrt{\frac{8}{3}}\exp(i(\frac{1}{3}\arg(-\frac{3}{8}\sqrt{\frac{3}{2}})+\frac{2n\pi}{3})),\quad n \in \mathbb{Z}.$$ But $x \in \mathbb{R}$ and the term under $\arg$ function is real, so the root $$x = \sqrt{\frac{8}{3}}\cos(\frac{1}{3}\cos^{-1}(-\frac{3}{8}\sqrt{\frac{3}{2}})+\frac{2n\pi}{3})),\quad n \in \mathbb{Z}.$$