I reduced a problem to this:
We have $a^2\phi=2$ where $a>0$ what is the value of $2a(\phi+1)$ ?
$1)2\sqrt{2\sqrt5+4}\qquad\qquad2)2\sqrt{\sqrt5+4}\qquad\qquad3)2\sqrt{2\sqrt5+1}\qquad\qquad4)2\sqrt{\sqrt5+1}$
Where $\phi$ is golden ratio ($\frac{1+\sqrt5}2)$.
This is a problem from a timed exam, so I should solve it quickly. Here I used $\phi^2=\phi+1$ several times to get the answer:
$$2a(\phi+1)=2a\phi^2=\sqrt{4a^2\phi^4}=\sqrt{8\phi^3}=\sqrt{8\phi(\phi+1)}=\sqrt{8(2\phi+1)}=2\sqrt{2(\sqrt5+2)}$$ Hence the answer is first choice. although the method I used is quick, but are there other approaches to get the answer (preferably) faster ?