TITLE: struggle simplifying $\sqrt{9+\sqrt{5}}$
QUESTION [6 upvotes]: I need to simplify $\sqrt{9+\sqrt{5}}$
I already do this (proven it) $\sqrt{9-4\sqrt{5}}=2-
\sqrt{5}$
But I couldn't when apply to $\sqrt{9+\sqrt{5}}=\sqrt{9-4\sqrt{5}+5\sqrt{5}}=\sqrt{(2+\sqrt{5})^2+5\sqrt{5}}$
PLEASE help me out

REPLY [4 votes]: Suppose
$\sqrt{9+\sqrt{5}}
= a+\sqrt{b}
$.
Squaring both sides,
$9+\sqrt{5}
= a^2+b+2a\sqrt{b}
= a^2+b+\sqrt{4a^2b}
$.
Equating the parts,
$9 = a^2+b$
and
$5 = 4a^2b$.
From the second,
$a^2 = 5/(4b)$,
so, from the first,
$9 = 5/(4b)+b$,
or $4b^2-36b+5 = 0$.
The discriminant is
$d = 36^2-4\cdot 4\cdot 5
=16(9^2-5)
=16\cdot 76
=64\cdot 39
$.
This is not a square of an integer,
so there is no integer (or rational)
expression in a simplified form.
You could, of course,
write
$\sqrt{9+\sqrt{5}}
= 3\sqrt{1+\sqrt{5}/9}
$,
but this doesn't seem 
to be worth much.