Is there a way to solve this equation for x?
$\sqrt{x^2+8}-x-4\arctan\left(\frac{x+3}{x^2+x+6}\right) = 0$
Numerically, I get $x \approx 1.39$, but is there a way I can get an exact solution?
Is there a way to solve this equation for x?
$\sqrt{x^2+8}-x-4\arctan\left(\frac{x+3}{x^2+x+6}\right) = 0$
Numerically, I get $x \approx 1.39$, but is there a way I can get an exact solution?
The solutions can't be algebraic numbers, because if $x$ was algebraic the same would be true of $y = (\sqrt{x^2+8} - x)/4 $ and $z = (x+3)/(x^2 + x + 6)$. Now your equation says $\tan y = z$, and thus $\cos^2 y = \frac{1}{z^2+1}$, so $\cos y$ would be algebraic. But by Lindemann's theorem $\cos y$ is transcendental for any nonzero algebraic $y$.
Thanks to Wims Function Calculator, the solution is $ x \approx 1.39370747885618057089248282982, $ and is not recognized by the Inverse Symbolic Calculator.
EDIT: Wims Function Calculator further gives $ x \approx 1.3937074788561805708924828298190336450650412142437367971287022786222374715488762515547338431838298094439414865124475411133191132591019104894926923243497788034792713333649381694869738035510413209296893. $ However, the result from the Inverse Symbolic Calculator indicates that a "closed-form" solution is not likely to exist.