1. [SOLVED] Simplifying an expression with radicals

The problem is the following:

$\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}$

I have to simplify it, but I have no idea how to start. Any help is appreciated. Thanks!

Pharod

2. Hello Pharod

Welcome to Math Help Forum!
Originally Posted by Pharod
The problem is the following:

$\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}$

I have to simplify it, but I have no idea how to start. Any help is appreciated. Thanks!

Pharod
Let $x = \sqrt{a+b}+\sqrt{a-b}$, where $a = 8$ and $b = 2\sqrt{10+2\sqrt5}$

Then $x^2 = a+b+2\sqrt{a^2-b^2}+a-b$
$=2a+2\sqrt{a^2-b^2}$

$=16+2\sqrt{24-8\sqrt5}$

$=4(4+\sqrt{6-2\sqrt5})$
Also $6-2\sqrt5 = (\sqrt5 - 1)^2$

$\Rightarrow \sqrt{6-2\sqrt5} = \sqrt5 - 1$

$\Rightarrow x^2 = 4(4+\sqrt5-1)$

$\Rightarrow x = 2\sqrt{3+\sqrt5}$

I don't think it will simplify any more.

$\vdots$
$\Rightarrow x = 2\sqrt{3+\sqrt5}$

I don't think it will simplify any more.
It will go a bit further: $(1+\sqrt5)^2 = 6+2\sqrt5$, and so $2\sqrt{3+\sqrt5} = \sqrt2\sqrt{6+2\sqrt5} = \sqrt2(1+\sqrt5) = \sqrt2+\sqrt{10}$

4. Hey, thank you both for leading me. I knew it had to be something with squaring, I just couldn't get to it for some reason.

Pharod

$x \:=\: 2\sqrt{3+\sqrt5}$

I don't think it will simplify any more.

$3 + \sqrt{5} \:=\:\frac{(1 + \sqrt{5})^2}{2} \;=\;2\cdot\frac{(1+\sqrt{5})^2}{4}$

. . . . . . $\;=\;2\left(\underbrace{\frac{1+\sqrt{5}}{2}}_{\te xt{Golden Mean}}\right)^2 \;=\;2\,\phi^2$

Therefore: . $x \;=\;2\sqrt{2}\,\phi$