help

$x-1/x=3\\ x^2-1=3x\\ x^2-3x-1=0\\ x=\frac{3\pm\sqrt{13}}{2}\\ $

So

x+1/x

=$\frac{3+\sqrt{13}}{2}+\frac{1}{3+\frac{\sqrt{13}}{{2}}}\\ =\sqrt{13}$

x+1\x

$\frac{3-\sqrt{13}}{2}+\frac{1}{3-\frac{\sqrt{13}}{{2}}}\\ =-\sqrt{13}$

So I believe the possible values of x+1/x are $\boxed{\sqrt{13}\ \text{and}\ -\sqrt{13}.}$