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!
 A: Hint:
Try to write $\;\sqrt{14+8\sqrt 3}=a+b\sqrt 3$. This means
$$(a+b\sqrt3)^2=14+8\sqrt 3.$$
This relation will be satisfied if $\;\begin{cases}a^2+3b^2=14,\\ab=4.\end{cases}$
If $a^2$ and $b^2$ are integers, there are not so many possibilities to have $a^2+3b^2=14$. Try to check if any of these possibilities also satisfies $ab=4$.
A: This can be computed by a very simple $ $ Square Root Denesting Rule:
Simple Denesting Rule $\rm\ \ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \  \color{brown}{divide\ out}\ \sqrt{trace} $
$\ 14+8\sqrt 3\ $ has norm $= 4.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = 2\,\ $ yields $\,\  12+8\sqrt 3\:$
with $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{24}\, =\, 2\sqrt 6.\ \ \ \rm \color{brown}{Dividing\ it\ out}\ $ of  above  yields $\,\ \sqrt 8 + \sqrt 6$ 
Remark $\ $ Many more worked examples are in prior posts on this denesting rule.
A: 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 $a^2-b=4$.
A: By Bernard's answer: $$\;\begin{cases}a^2+3b^2=14,\\ab=4.\end{cases}$$ So $$a^4+3(ab)^2=14a^2 \Rightarrow a^4-14a^2+3\cdot4^2=0\Rightarrow u^2-14u+6\cdot 8=0 \Rightarrow (u-6)(u-8)=0$$ where $u=a^2$. Now we can find all possibilities.
A: 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:
$(2 \sqrt{2} + \sqrt{6})^2 = 8 + 2 \cdot 2 \sqrt{2} \cdot \sqrt{6} + 6 = 14 + 8 \sqrt{3}.$
A: Note
$$\sqrt {14 + 8\sqrt {3}}=\sqrt{2(7+2\sqrt{12})}
=\sqrt{2(\sqrt4+\sqrt3)^2}=\sqrt2(2+\sqrt3)
$$
