<s>[INST] 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.
 [/INST] Write,
$$\sqrt {3\pm\sqrt 5}=\sqrt a\pm\sqrt b,\, a\ge b$$
and we obtain
$$A=\sqrt {3+\sqrt 5}+\sqrt {3-\sqrt 5}=2\sqrt a$$
Then, we want to find ratinal $a,b$ such that:
$$\sqrt {3\pm\sqrt 5}=\sqrt a\pm\sqrt b,\, a>b$$ holds.
We have
$$3\pm \sqrt 5=a+b+2\sqrt {ab}\\
\implies \begin{cases}a+b=3\\ ab=\frac 54\end{cases}$$
Using the  Vieta's formulas, we have
$$t^2-3t+\frac 54=0\\
4t^2-12t+5=0\\
t_{1,2}=\frac {6\pm4}{4}\\
\implies a=\frac 52\\
\implies b=\frac 12.$$
This gives,
$$\sqrt {3\pm\sqrt 5}=\sqrt \frac 52\pm \sqrt \frac 12$$
and
$$A=2\sqrt a=\sqrt {10}.$$
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