The greatest value of $|z|$ if $\Big|z+\frac{1}{z}\Big|=3$ where $z\in\mathbb{C}$ 
$\bigg|z+\dfrac{1}{z}\bigg|=3$ then the greatest value of $|z|$ is ___________

My Attempt
$$
\bigg|z+\frac{1}{z}\bigg|=\bigg|\dfrac{z^2+1}{z}\bigg|=\frac{|z^2+1|}{|z|}=3\\
\bigg|z+\frac{1}{z}\bigg|=3\leq|z|+\frac{1}{|z|}\implies |z|^2-3|z|+1\ge0\\
|z|=\frac{3\pm\sqrt{5}}{2}\implies\color{red}{|z|\in(-\infty,\frac{3-\sqrt{5}}{2}]\cup[\frac{3+\sqrt{5}}{2},+\infty)}
$$
$$
\bigg|z+\frac{1}{z}\bigg|=3\geq\bigg||z|-\frac{1}{|z|}\bigg|\\
|z|^2-3|z|-1\leq0\text{ or }|z|^2+3|z|-1\geq0\\
|z|=\frac{3\pm\sqrt{13}}{2}\text{ or }|z|=\frac{-3\pm\sqrt{13}}{2}\\
\color{red}{|z|\in[\frac{3-\sqrt{13}}{2},\frac{3+\sqrt{13}}{2}]}\text{ or }\color{red}{|z|\in(-\infty,\frac{-3-\sqrt{13}}{2}]\cup[\frac{-3+\sqrt{13}}{2},+\infty)}
$$
The solution given in my reference is $\dfrac{3+\sqrt{13}}{2}$, why am I not able to find it in my attempt ?
Note: A similar question has been asked before, If $∣z+\frac{1}{z}∣=a$ then find the minimum and maximum value of $|z|$, but it does not have any answer which address why we choose a specific version of the triangle inequality.

The Joukowsky transform $z\to z+1/z$ transforms circles of radius $r>1$ into ellipses of semi-axes $r-1/r$ and $r+1/r$. The max value of $|z|$ corresponds to the radius of the circle such that the smallest semi-axis $r-1/r=3$. Solving this equation $r^2-3r-1=0$ for $r>0$ you get the result $r=(3+\sqrt{13})/2$ which agrees with yours.
The minor semi axes correspond to the purely imaginary values that jmerry found.
