$$\sqrt{7 + \sqrt{14}} = a + b\sqrt{c}$$

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?


2 Answers 2


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

  • $\begingroup$ 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? $\endgroup$ Commented Sep 28, 2015 at 13:21
  • $\begingroup$ @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 :( $\endgroup$ Commented Sep 28, 2015 at 18:51
  • 1
    $\begingroup$ @NajibIdrissi: I've added a proof to my answer. $\endgroup$ Commented Oct 3, 2015 at 19:29

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