Find the sum of radicals without squaring, Is that impossible? 
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$

My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
So I was wondering about a way to find this sum without squaring?  It seems impossible, but I still want to ask.

I want to solve the problem with a geometric approach. Consider a right triangle $ABC$ as below:

If we suppose that $BH = \sqrt{3 - \sqrt{5}}$ and $HC = \sqrt{3 + \sqrt{5}}$, then we have $AH^2 = BH.HC = \sqrt{9 - 5} = 2$. So $AH = \sqrt{2}$. Now use Pythagorean theorem:
\begin{align*}
\overset{\triangle}{ABH}&: AB^2 = AH^2 + BH^2 = 2 + 3 - \sqrt{5} = 5 - \sqrt{5} \implies AB = \sqrt{5 - \sqrt{5}}\\
\overset{\triangle}{ACH}&: AC^2 = AH^2 + HC^2 = 2 + 3 + \sqrt{5} = 5 + \sqrt{5} \implies AC = \sqrt{5 + \sqrt{5}}\\
\overset{\triangle}{ABC}&: BC^2 = AB^2 + AC^2 = 5 - \sqrt{5} + 5 + \sqrt{5} = 10 \implies BC = \sqrt{10}
\end{align*}
Therefore:
$$\bbox[5px, border: 2px solid magenta]{\sqrt{10} = BC = BH + HC = \sqrt{3 - \sqrt{5}} + \sqrt{3 + \sqrt{5}}}$$
