<s>[INST] Fully simplify $\sqrt {14 + 8\sqrt {3}}$. Please help me with this problem: Fully simplify $\sqrt{14 + 8\sqrt {3}}$.
I've tried to assume that the radical is in the form of $a+b\sqrt{3}$, so $a+b\sqrt{3}=\sqrt {14 + 8\sqrt {3}}$, then I squared both sides and got $$a^2+3b^2+2ab\sqrt{3}=14+8\sqrt{3}.$$ 
So I assumed that $a^2+3b^2$ must be equalto $14$, and $2ab$ must be equal to $-8$. So $ab$ must equal to $-4$. So now I know that $a$ and $b$ must be factors of $-4$. But now I am stuck because the cases $a=-1$, $b=4$, $a=1$, $b=-4$, $a=4$, $b=-1$, and $a=-4$, $b=1$ do not work because of the fact that $a^2+3b^2$ must equal to $14$. (We have that $(-1)^2+3\cdot4^2=1+3\cdot16$ which is clearly not $14$, same goes with $a=1$, $b=-4$; $4^2+3(-1)^2=16+3=19$ which is also not $14$.) 
So now I am out of ideas to simplify this expression, or maybe this is already fully simplified? Please help!
 [/INST] Write
$$
\sqrt{14+8\sqrt{3}}=\sqrt{x\vphantom{b}}+\sqrt{\smash{y}\vphantom{b}}
$$
Then
$$
14+8\sqrt{3}=x+y+2\sqrt{\smash{xy}\vphantom{b}}
$$
so we can look for
$$
xy=48,\qquad x+y=14
$$
that reduces to finding the roots of the equation
$$
z^2-14z+48=0
$$
that are $8$ and $6$. So we can take $x=8$ and $y=6$:
$$
\sqrt{14+8\sqrt{3}}=\sqrt{6}+\sqrt{8}
$$
Alternatively, recall the identity
$$
\sqrt{a+\sqrt{b}}=
\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}
$$
In this case $a^2-b=4$.
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