Denesting a square root: $\sqrt{7 + \sqrt{14}}$ Write:
$$\sqrt{7 + \sqrt{14}} = a + b\sqrt{c}$$
Form. $$7 + \sqrt{14} = a^2 + 2ab\sqrt{c} + b^2c$$
$a^2 + b^2c = 7$ and $2ab = 1$, and $c = 14$
But that doesn't seem right as $a, b,$ wont be integers?
 A: A simple way to see if a double radical $\sqrt{a\pm \sqrt{b}}$  can be denested is to check if $a^2-b$ is a perfect square. In this case we have:
$$
\sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}}
$$
(you can easely verify this identity).
In this case $a^2-b=35$ is not a perfect square.

Note that if $\sqrt{a+\sqrt{b}}$ can be denested than $a^2-b$ must be positive since, by:
$$
\sqrt{a+\sqrt{b}})= \sqrt{p}+\sqrt{q}
$$
we have (squaring)
$$
a+\sqrt{b}=p+q+2\sqrt{pq}
$$
and for $a,b,q,p \in \mathbb{Q}$ this implies:
$$
p+q=a \qquad \land \qquad \sqrt{b}=2\sqrt{pq} \iff pq=b/4
$$
this means that $p$ and $q$ are solutions of the equation $ x^2-ax+b/4=0$ that has rational solutions only if $\Delta=a^2-b>0$
A: Set $r=\sqrt{7+\sqrt{14}}$; then $r^2=7+\sqrt{14}$ and so
$$
14=r^4-14r^2+49
$$
or
$$
r^4-14r^2+35=0
$$
The polynomial $X^4-14X^2+35=0$ is irreducible over the rational numbers by Eisenstein's criterion (with $7$), so the degree of $r$ over the rationals is $4$. A number of the form $a+b\sqrt{c}$ with rational $a,b,c$ has degree $2$ over the rationals.
Therefore you can't find rational $a,b,c$ that satisfy your request.
