# Roots of this sextic

I'm looking for the roots of the sextic equation in $$x$$ $$x^6 - (3 m) x^5 + 5 m^2 x^4 - (5 m^3) x^3 + 3 m^4 x^2 - m^5 x + L = 0.$$ I know that at most two of the roots of this are real when $$m$$ and $$L$$ are positive integers. Also mathematica finds a closed form for all the roots (surprisingly). They are

x = 1/2 (m - sqrt(2 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3) + (4 2^(1/3) m^4)/(3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3) - 5 m^2)/sqrt(3))

x = 1/2 (sqrt(2 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3) + (4 2^(1/3) m^4)/(3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3) - 5 m^2)/sqrt(3) + m)

x = 1/2 (m - sqrt(m^2 - 4 (((1 - i sqrt(3)) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3))/(6 2^(1/3)) + ((1 + i sqrt(3)) m^4)/(3 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3)) + (2 m^2)/3)))

x = 1/2 (m + sqrt(m^2 - 4 (((1 - i sqrt(3)) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3))/(6 2^(1/3)) + ((1 + i sqrt(3)) m^4)/(3 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3)) + (2 m^2)/3)))

x = 1/2 (m - sqrt(m^2 - 4 (((1 + i sqrt(3)) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3))/(6 2^(1/3)) + ((1 - i sqrt(3)) m^4)/(3 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3)) + (2 m^2)/3)))


and

x = 1/2 (m + sqrt(m^2 - 4 (((1 + i sqrt(3)) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3))/(6 2^(1/3)) + ((1 - i sqrt(3)) m^4)/(3 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3)) + (2 m^2)/3)))\$.


What I need to know is, which two of these roots are the ones that are candidates to be real, when $$m$$ and $$L$$ are positive integers? Also, how does one determine this (without substituting values)?

Pardon the messy values, it's more worth it to just feed it into mathematica.

• They are excellent. 2 days ago

If $$m=0$$, then your sextic is $$x^6+L$$. For this to have a real root, you need $$L\leq 0$$, in which case those real roots are the real sixth roots of $$-L$$.

We may now assume $$m\neq 0$$. After dividing your sextic by $$m^6$$, and taking $$y:=x/m$$ and $$K:=L/m^6$$, you have a new (simpler) sextic $$f(y):=y^6-3y^5+5y^4-5y^3+3y^2-y+K$$ with only one parameter. This new sextic has real roots exactly when the old one did (just scaled by $$m$$).

The derivative of $$f$$ has only a single real root, at $$y=1/2$$. Thus, the minimum of $$f$$ occurs at $$f(1/2)=-\frac{9}{64}+K$$. In order for there to be a real root, you need $$K\leq \frac{9}{64}$$. If you run

Manipulate[Plot[y^6 - 3 y^5 + 5 y^4 - 5 y^3 + 3 y^2 - y + K == 0, {y, 0, 1}], {K, 0, 9/64}]


in Mathematica, you can visually see how the two roots behave, as $$K$$ varies.

Because the minimum is at $$y=1/2$$, we make the linear shift $$z:=y-1/2$$. Then, the sextic transforms to $$z^6+\frac{5}{4}z^4+\frac{3}{16}z^2+\left(K-\frac{9}{64}\right).$$

Notice that this is really a cubic in $$z^2$$, with a negative constant term. Multiplying through by $$64$$, to remove the denominator, replacing $$4z^2$$ with $$w$$, and $$64K-9$$ with $$J$$, we are really just looking for the (unique) nonnegative real root of $$w^3+5w^2+3w+J$$ when $$J\leq 0$$.

Thus, it is not surprising after all that Mathematica found solutions to the original sextic, since cubics are always solvable using radicals.

Running the code

Solve[w^3 + 5 w^2 + 3 w + J == 0, w]


you get three expressions involving radicals. When $$J=0$$, only one of these expressions is nonnegative (and is equal to $$0$$). That expression is the one that will ultimately give you your nonnegative real root, because that expression is continuous as a function in $$J\leq 0$$.

Now that you have your solution for $$w$$ in terms of $$J$$, you can transform everything back to your original system (if you like).

$$\newcommand{\R}{\mathbb R}$$The roots are \begin{equation*} \begin{aligned} x_1&:=\frac{1}{2} \left(m-\frac1{\sqrt3}\sqrt{\frac{4 \sqrt[3]{2} m^4}{t}-5 m^2+2^{5/3} t}\right), \\ x_2&:=\frac{1}{2} \left(m+\frac1{\sqrt3}\sqrt{\frac{4 \sqrt[3]{2} m^4}{t}-5 m^2+2^{5/3} t}\right), \\ x_3&:=\frac{1}{2} \left(m-\sqrt{m^2-4 \left(\frac{\left(1+i \sqrt{3}\right) m^4}{3\ 2^{2/3} t}+\frac{2 m^2}{3}+\frac{\left(1-i \sqrt{3}\right) t}{6 \sqrt[3]{2}}\right)}\right), \\ x_4&:=\frac{1}{2} \left(m+\sqrt{m^2-4 \left(\frac{\left(1+i \sqrt{3}\right) m^4}{3\ 2^{2/3} t}+\frac{2 m^2}{3}+\frac{\left(1-i \sqrt{3}\right) t}{6 \sqrt[3]{2}}\right)}\right), \\ x_5&:=\frac{1}{2} \left(m-\sqrt{m^2-4 \left(\frac{\left(1-i \sqrt{3}\right) m^4}{3\ 2^{2/3} t}+\frac{2 m^2}{3}+\frac{\left(1+i \sqrt{3}\right) t}{6 \sqrt[3]{2}}\right)}\right), \\ x_6&:=\frac{1}{2} \left(m+\sqrt{m^2-4 \left(\frac{\left(1-i \sqrt{3}\right) m^4}{3\ 2^{2/3} t}+\frac{2 m^2}{3}+\frac{\left(1+i \sqrt{3}\right) t}{6 \sqrt[3]{2}}\right)}\right), \end{aligned} \end{equation*} where $$\begin{equation*} t:=\sqrt[3]{3 \sqrt{3} \sqrt{L \left(27 L-4 m^6\right)}-27 L+2 m^6}. \end{equation*}$$

Clearly, $$x_1\in\R\iff x_2\in\R$$, $$x_3\in\R\iff x_4\in\R$$, and $$x_5\in\R\iff x_6\in\R$$, since $$m$$ is real.

Using the Mathematica Reduce command, we get that

• $$x_3$$ and $$x_5$$ are never real and hence $$x_4$$ and $$x_6$$ are never real (if $$m$$ and $$L$$ are positive integers)

• $$x_1\in\R\iff x_2\in\R\iff m\ge z_2$$, where $$z_2$$ is the positive root $$z$$ of the equation $$9z^6=64L$$.

Thus, only $$x_1$$ and $$x_2$$ can ever be real roots of your sextic equation, and they are actually real (for given positive integers $$m$$ and $$L$$) iff $$\begin{equation*} L\le\frac9{64}\,m^6. \end{equation*}$$

Starting from @Pace Nielsen answer $$w^3+5w^2+3w+J=0$$ The discriminant is $$\Delta =-(J+9) (27 J-13)$$

If $$J\lt -9$$, the real solution write $$w=\frac{1}{3} \left(-5+8 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(-\frac{27J+115}{128} \right)\right)\right)$$

For $$-9 < J <0$$, for $$k=0,1,2$$ $$w_k=\frac{1}{3} \left(-5+8 \sin\left( \frac{(4 k+1)\pi}{6}+\frac 13 \cos ^{-1}\left(\frac{27J+115}{128} \right)\right)\right)$$ whcih are much nicer than the radicals.