Roots of this sextic

Question

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.

Answer 1

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).

Answer 2

$\newcommand{\R}{\mathbb R}$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*}

Answer 3

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.