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!

First, we guess that $\sqrt{14 + 8 \sqrt{3}}$ can be written in the form $a + b$, where $a$ and $b$ are simple numbers that may involve square roots. (For example, if we square $1 + \sqrt{3}$, then we get $(1 + \sqrt{3})^2 = 1 + 2 \sqrt{3} + 3 = 4 + 2 \sqrt{3}$, so $\sqrt{4 + 2 \sqrt{3}} = 1 + \sqrt{3}$.)
So we have that
$a + b = \sqrt{14 + 8 \sqrt{3}}.$
Squaring both sides, we get
$a^2 + 2ab + b^2 = 14 + 8 \sqrt{3}.$
To solve for $a$ and $b$, we must obtain at least two equations, so we try setting $a^2 + b^2 = 14$ and $2ab = 8 \sqrt{3}$. Then $ab = 4 \sqrt{3}$. Squaring this equation, we get $a^2 b^2 = (4 \sqrt{3})^2 = 48$.
So we have the sum and product of $a^2$ and $b^2$. Then the quadratic whose roots are $a^2$ and $b^2$ is
[(x - a^2)(x - b^2) = x^2 - (a^2 + b^2) x + a^2 b^2 = x^2 - 14x + 48 = 0.]
This quadratic factors as $(x - 6)(x - 8) = 0$, so the roots are 6 and 8.
This tells us that $a^2$ and $b^2$ are equal to 6 and 8 in some order. Therefore, we suppose that
\begin{align*}
\sqrt{14+8\sqrt{3}} &= \pm \sqrt{6} \pm \sqrt{8} \\
&= \pm \sqrt{6} \pm 2\sqrt{2}.
\end{align*}
We can easily see that $\sqrt{14+8\sqrt 3}$ is large enough that both signs above must be $+$. Thus
$\sqrt{14 + 8 \sqrt{3}} = \boxed{2 \sqrt{2} + \sqrt{6}}.$
We can check this answer by squaring it:
$(2 \sqrt{2} + \sqrt{6})^2 = 8 + 2 \cdot 2 \sqrt{2} \cdot \sqrt{6} + 6 = 14 + 8 \sqrt{3}.$
