Simplification of $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$ I was trying to simplify $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$. Numerical evaluation suggested that the answer is $\sqrt{2}$ and it checked out when I substituted $\sqrt{2}$ in the equation $x= \sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$.
But I still cannot simplify the initial expression by any obvious means (squaring, multiplying by conjugate etc). Any idea how to do it? Can it be done?
 A: The nested radical $\sqrt{16 - 4 \sqrt{7}}$ is equal to $\sqrt{14}-\sqrt{2}$. Simply square both sides to prove that. Using $(a-b)^2=a^2+b^2-2ab$, we get$$(\sqrt{14}-\sqrt{2})^2=14+2-2\sqrt{2}\sqrt{14}=16-2\sqrt{2}\sqrt{2}\sqrt{7}=16-4 \sqrt{7}$$
You can read more about denesting on Wikipedia.
EDIT: Lucian's answer is explicitly solving the denesting equation
$$\sqrt{a-b \sqrt{c}\ } = \sqrt{y}-\sqrt{x}$$
for this particular case in which half the answer ($y=14$) was given to you, that's why he gets an equation in a single unknown. If you can simply guess/know both $x$ and $y$, then you can just verify the solution by squaring both sides as I did.
The general solution formula is also obtained by squaring both sides and doing an identification, yielding the system $a = x+y$ and $b^2c= 4xy$. This is actually exactly the same system that you would get when solving $\sqrt{a+b \sqrt{c}\ } = \sqrt{y}+\sqrt{x}$, the full proof for which is given in the Wikipedia article I linked. So the general solution for $\sqrt{a-b \sqrt{c}\ } = \sqrt{y}-\sqrt{x}$ is also $x= \frac{a-\Delta}{2},$ $y=\frac{a+\Delta}{2}$ where $\Delta=\sqrt{a^2 - b^2c}$ must be a rational number, otherwise there's no solution in rationals for $x$ and $y$. Applying this general method to $a= 16$, $b= 4$ and $c=7$ yields $\Delta=\sqrt{16^2-4^2\cdot7} = \sqrt{256-112}=12$, so $x=\frac{16-12}{2}=2$ and $y=\frac{16+12}{2}=14$.
Also, if you already know $y$ (in your problem $y=14$), then from the equation $a=x+y$, i.e. in your problem $16=x+14$, it follows immediately that the only possible solution is with $x=2$, so you can just check that the second equation $b^2c=4xy$ is verified, i.e. $4^2\cdot7=4\cdot2\cdot14$.
A: $x=\sqrt{14}-\sqrt{16-4\sqrt7}\iff x-\sqrt{14}=-\sqrt{16-4\sqrt7}~=>~\big(x-\sqrt{14}\big)^2=16-4\sqrt7$
$\iff x^2+14-2x~\sqrt{2\cdot7}~=~\big(2+14\big)-4\sqrt7.~$ Can you take it from here ? :-)
A: Suppose that
$$
\left(\sqrt{a}\pm\sqrt{b}\right)^2
=\overbrace{a+b\vphantom{\sqrt4}}^c\pm\overbrace{\sqrt{4ab}}^{\sqrt{d}}
$$
Then
$$
\begin{align}
c^2-d
&=a^2+2ab+b^2-4ab\\
&=(a-b)^2
\end{align}
$$
Thus,
$$
a=\frac{c+\sqrt{c^2-d}}2
\qquad\text{and}\qquad
b=\frac{c-\sqrt{c^2-d}}2
$$

If $c=4$ and $d=7$, we get
$$
4-\sqrt{7}=\left(\sqrt{\frac72}-\sqrt{\frac12}\right)^2
$$
Therefore,
$$
\begin{align}
\sqrt{14}-\sqrt{16-4\sqrt7}
&=\sqrt{14}-2\sqrt{4-\sqrt7}\\
&=\sqrt{14}-\left(\sqrt{14}-\sqrt2\right)\\
&=\sqrt2
\end{align}
$$
A: \begin{align}
\Big(\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}\Big)^2&=14+16-4\sqrt{7}-2\sqrt{14}\sqrt{16 - 4 \sqrt{7}}\\
&=30-4\sqrt{7}-4\sqrt{14}\sqrt{4 -  \sqrt{7}}\\
&=30-4\sqrt{7}-4\sqrt{56 -  14\sqrt{7}}\\
&=30-4\sqrt{7}-4\sqrt{49 -  2\times 7 \sqrt{7}+(\sqrt{7})^2}\\
&=30-4\sqrt{7}-4\sqrt{(7- \sqrt{7})^2}\\
&=30-4\sqrt{7}-4(7- \sqrt{7})\\
&=2
\end{align}
