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.

Use Vieta's formulas:
Let, $$x_1=\sqrt {3-\sqrt 5},\,\,\,x_2=\sqrt {3+\sqrt 5}$$
$$x^2-px+2=0\\p=\frac{x^2+2}{x}$$
and we have,
$$p=\frac{5+\sqrt 5}{\sqrt {3+\sqrt 5}}=\frac{5-\sqrt 5}{\sqrt {3-\sqrt 5}}$$
Then, using the rule
$$p=\frac ab=\frac cd\implies p=\frac{a+c}{b+d}$$
We get
\begin{aligned}p&= \frac{5+\sqrt 5+5-\sqrt 5}{\sqrt {3+\sqrt 5}+\sqrt {3-\sqrt 5}}\\
&=\frac {10}{p}=p\\
&\implies p=\sqrt {10}. \end{aligned}
