Analytical solution of a polynomial with non integer order Can anyone think of a possible analytical solution of the following equation?
$x\left(1-0.2x^2\right)^{5/2}=constant$
I am not a mathematician, but, it seems to me that only numerical methods can help.
Thanks in advance.
 A: $x\left(1 - 0.2x^{2}\right)^{5/2} = c\,\sqrt{5\,} = \mbox{constant}$. I
${\bf guess}$ $\left(1 - 0.2x^{2}\right) \geq 0$. Then,
$\left\vert x\right\vert \leq \sqrt{5\,}\,$ and
${\rm sgn}\left(x\right) = {\rm sgn}\left(c\right)$. Define
$\xi\ \ni\ 0 \leq \xi \leq 1$ and
$x \equiv {\rm sgn}\left(c\right)\,\sqrt{5\xi\,}\,$:
$$
\sqrt{\xi\,}\,\left(1 - \xi\right)^{5/2} = \left\vert c\right\vert\,,
\qquad
\xi\left(1 - \xi\right)^{5} = c^{2}
$$
The function ${\rm f}\left(\xi\right) \equiv \xi\left(1 - \xi\right)^{5}$ has its maximum value at $\xi_{M} = 1/6$.
${\rm f}\left(\xi_{M}\right) = 5^{5}/6^{6} \approx 0.0670$. When
$\left\vert c\right\vert > 5^{5/2}/6^{3} \approx 0.2588$, there isn't any real solution for $\xi$. When $\left\vert c\right\vert < 5^{5/2}/6^{3} \approx 0.2588$ there are two solutions for $\xi$: 1) $< 1/6$ and 2) $> 1/6$.
There are four solutions which can be found approximately:
$$
\begin{array}{ll}
\left\vert c\right\vert \gtrsim 0\,, 
& \qquad &
\xi \approx c^{2}\quad\mbox{and}\quad \xi \approx 1 - \left\vert c\right\vert^{2/5}
\\[3mm]
\left\vert c\right\vert \lesssim{5^{5/2} \over 6^{3}}\,,
& \qquad &
\xi \approx \xi_{M}\,\,\,
\mp\,\,\,
\sqrt{2\left[{\rm f}\left(\xi_{M}\right) - c^{2}\right] \over -{\rm f}''\left(\xi_{M}\right)}
\end{array}
$$
With this information, the numerical approach should be the $Bisection\ Method$.
