How would you show that $\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$? I've recently seen a Highschool problem and I was wondering, how would you show that
$$\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$$
Thank you for your time,

Since $14+4\sqrt{10}\gt14-4\sqrt{10}\gt0$, the difference of the square roots is equal to something positive, i.e.,
$$\sqrt{14+4\sqrt{10}}-\sqrt{14-4\sqrt{10}}=u\gt0$$
If we square both sides, we have
$$\begin{align}
u^2&=(14+4\sqrt{10})-2\sqrt{14+4\sqrt{10}}\sqrt{14-4\sqrt{10}}+(14-4\sqrt{10})\\
&=28-2\sqrt{14^2-4^2\cdot10}\\
&=28-2\sqrt{36}\\
&=16
\end{align}$$
Since $u\gt0$ was already established, we can conclude $u=4$.
