# 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?

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.

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 $\Delta=a^2-b>0$

• You have proven that if $a^2 - b$ is a perfect square, then the square root can be "denested". What about the converse? If $a^2 - b$ isn't a perfect square, why cannot the square root be denested? – Najib Idrissi Sep 28 '15 at 13:21
• @NajibIdrissi: Wikipedia(en.wikipedia.org/wiki/Nested_radical#Denesting_nested_radicals) indicates an ''if and only if'' condition... but really I' ve not a simple proof :( – Emilio Novati Sep 28 '15 at 18:51
• @NajibIdrissi: I've added a proof to my answer. – Emilio Novati Oct 3 '15 at 19:29