struggle simplifying $\sqrt{9+\sqrt{5}}$

Question

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

Answer 1

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.

Answer 2

Let $\sqrt{9+\sqrt{5}}=A+B \sqrt{5}$. Square each side: $9+\sqrt{5} = A^2 + 2AB \sqrt{5} + 5 B^2$. Now we get two equations and two unknowns and solve for $A$ and $B$...