jmc

Let $k=x+\frac{1}{x}$. Notice that $k^2=x^2+2+\frac{1}{x^2}$, so $x^2+\frac{1}{x^2}=k^2-2$. Substituting this into the equation gives $(k^2-2)+3\cdot (k)=26$, or $k^2+3k-28=(k+7)(k-4)=0$. Since $x$ is positive, then $k>0$, so $k=4$. Substituting back, $x+\frac{1}{x}=4⟹x^2-4x+1=0⟹x=\frac{4\pm \sqrt{16-4}}{2}=2\pm \sqrt{3}$. To match the desired form, we take the solution $x=2+\sqrt{3}$, and the answer is $\boxed{5}$.