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.
 A: A polynomial approach.
Note that $a_1,a_2=\sqrt{3\pm\sqrt 5}$ are two of the roots of $x^4-6x^2+4=0.$ The other roots are $-a_1,-a_2.$
So $$x^4-6x^2+4=(x-a_1)(x-a_2)(x+a_1)(x+a_2).$$
Now, $a_1a_2=\sqrt{4}=2.$ If $S=a_1+a_2,$ then this factorization becomes:
$$x^4-6x^2+4=(x^2-Sx+2)(x^2+Sx+2)=x^4+(4-S^2)x^2+4.$$
So $4-S^2=-6,$ or $S^2=10.$
I guess that's sort of squaring, but we never actually numerically square $S.$
More generally, given the roots of $x^4-bx^2+c^2,$ there are two roots $a_1,a_2$ with $a_1a_2=c,$ and then you get a similar result:
$$x^4-bx^2+c^2=(x^2-Sx+c)(x^2+Sx+c)=x^4+(2c-S^2)x^2+c^2,$$ so $2c+b=S^2.$
So this means, at least if $c\geq 0,$ that $$\sqrt{\frac{b+\sqrt{b^2-4c^2}}2}+\sqrt{\frac{b-\sqrt{b^2-4c^2}}2}=\pm\sqrt{2c+b}$$
Multiplying by $\sqrt{2},$ this gives:
$$\sqrt{b+\sqrt{b^2-4c^2}}+\sqrt{b-\sqrt{b^2-4c^2}}=\pm\sqrt{4c+2b}\tag1$$
You also get:
$$\sqrt{b+\sqrt{b^2-4c^2}}-\sqrt{b-\sqrt{b^2-4c^2}}=\pm\sqrt{2b-4c},\tag 2$$ since in this case $a_1a_2=-c.$
The case $b=3,c=1$ in (1) gives your result, since $3\pm\sqrt 5$ are both real and positive, so we know the result has to be positive.
In both $(1)$ and $(2)$ you get the positive sign if $b,c$ and all the square roots are real. If the square roots are complex, you are stuck figuring out the sign.
But if $b,c$ are real, and $0\leq 2c\leq b,$ we can solve $(1)$ and $(2)$ to get:
$$\sqrt{b+\sqrt{b^2-4c^2}}=\frac{\sqrt{2b+4c}+\sqrt{2b-4c}}{2}=\frac{\sqrt{b+2c}+\sqrt{b-2c}}{\sqrt 2}$$
and similarly:
$$\sqrt{b-\sqrt{b^2-4c^2}}=\frac{\sqrt{b+2c}-\sqrt{b-2c}}{\sqrt 2}$$
If $d=b^2-4c^2,$ this gives us:
$$\sqrt{b+\sqrt{d}}=\frac{\sqrt{b+\sqrt{b^2-d}}+\sqrt{b-\sqrt{b^2-d}}}{\sqrt2}$$ when $0\leq d\leq b^2.$
A: Trick:
$$
\sqrt {3-\sqrt{5}}=\frac{\sqrt {6-2\sqrt{5}}}{\sqrt 2}=\frac{\sqrt {(\sqrt 5)^2-2\sqrt{5}+1}}{\sqrt 2}=\frac{\sqrt {(\sqrt 5-1)^2}}{\sqrt 2}
=\frac{\sqrt 5-1}{\sqrt 2}.
$$
Similarly we see $\sqrt {3+\sqrt{5}}
=\frac{\sqrt 5+1}{\sqrt 2}$. Thus the sum is $2\cdot \frac{\sqrt 5}{\sqrt 2}=\sqrt {10}$.
A: 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}
A: In this answer I tried to generalize the answer given above.
Generalization:
Let,
$$\sqrt {a\pm\sqrt b}=\sqrt m\pm\sqrt n,\,\,\,m\ge n$$
and
$$A=\sqrt {a+\sqrt b}+\sqrt {a-\sqrt b}=2\sqrt m$$
Then we have
$$\sqrt {a\pm\sqrt b}=\sqrt m\pm\sqrt n\\
\begin{cases} m+n=a\\mn =\frac {b}{4}\end{cases}\\
t^2-at+\frac b4=0\\
4t^2-4at+b=0\\
m=\frac {a+\sqrt{a^2-b}}{2}\\
n=\frac {a-\sqrt{a^2-b}}{2}$$
This gives,
$$\sqrt {a\pm\sqrt b}=\sqrt{\frac {a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac {a-\sqrt{a^2-b}}{2}}$$
and
\begin{aligned}A&=2\sqrt m=\sqrt {4m}\\
&=\boxed{\sqrt {2\left(a+\sqrt {a^2-b}\right)}}.\end{aligned}
$\text {QED}.$
A: 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
$$A=2\sqrt a=\sqrt {10}.$$
A: 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:
$$\bbox[5px, border: 2px solid magenta]{\sqrt{10} = BC = BH + HC = \sqrt{3 - \sqrt{5}} + \sqrt{3 + \sqrt{5}}}$$
