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,
 A: $$\begin{align}\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} &= \sqrt{\left(\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}}\right)^2} \\
&= \sqrt{14 + 4\sqrt{10} + 14 - 4\sqrt{10} - 2\sqrt{(14 + 4\sqrt{10})(14-4\sqrt{10})}} \\
&= \sqrt{28 - 2\sqrt{14^2 - 4^2\cdot 10}} \\
&= \sqrt{28 - 2\cdot \sqrt{196 - 160}} \\
&= \sqrt{28 - 2\sqrt{36}} \\
&= \sqrt{28 - 12} \\
&= \sqrt{16} \\
&= 4\end{align}$$ 
In the above, you're applying $(a-b)^2 = a^2 + b^2 - 2ab$ and $(a+b)(a-b) = a^2 - b^2$.
A: As an alternative way note that $$\eqalign{\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} &= \sqrt{\sqrt{10}^2+2\cdot2\sqrt{10}+2^2} - \sqrt{\sqrt{10}^2-2\cdot2\sqrt{10}+2^2}},$$ then use the identity $a^2\pm2ab+b^2=(a\pm b)^2$, and simplify.
A: $$\sqrt{ 14 \pm 4 \sqrt{10}} = \sqrt{10} \pm 2 $$
since $\sqrt{10} \pm 2>0 $ and $(\sqrt{10} \pm 2)^2 =  14 \pm 4 \sqrt{10} $.
A: This was quite easy sum , here I used the general solution for (a+2b^1/2)^1/2=x^1/2 +y^1/2
 
A: Both numbers are positive, so the equality is true if, and only if its squares are equal. By squaring both sides we get
$$14+4\sqrt{10}-2\sqrt{(14+4\sqrt10)(14-4\sqrt{10})}+14-4\sqrt{10} = 4^2$$
$$14-2\sqrt{(14^2-16 \cdot 10)}+14 = 16$$
$$28-12 = 16$$
$$16 = 16$$
A: Denote $\sqrt{14+4\sqrt{10}}=a, \sqrt{14-4\sqrt{10}}=b$, then
$$a^2+b^2=28, ab=6$$
So $(a-b)^2=16$ and $a-b=4$.
A: 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 $u=4$.
