$a^2\phi=2$ , what is the value of $2a(\phi+1)$? 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 ?
 A: I think the way you dealt with the problem is already rather direct and probably as "quick" as one could expect to solve the problem "cold" on a timed exam.
Here's another way to approach the calculation that forestalls dealing with a square-root until it is needed to make the answer choice.  As you note, the essential golden-ratio relation is $ \ \phi^2 \ = \ \phi + 1 \ \ . $  We wish to "evaluate" $ \ 2a · (\phi + 1) \ = \ 2a · \phi^2 \ \ ; $ let's square this for now and work with $ \ 4a^2 · \phi^4 \ \ . $  From the given "definition" for $ \ a \ \ , $ this is equal to $ \ 4 · \left(\frac{2}{\phi} \right) · \phi^4  \ = \ 8 · \phi^3 \ \ ; $  this is what we will need to obtain a square-root for (as you had found).
Since the $ \ (m + n\sqrt5) \ $ sum occurs in all of the answer choices in some fashion, we would like to express our result in those terms. We may write $ \ \psi \ = \  1 + \sqrt5 \ \ , $ making $  \ \phi \ = \ \frac{\psi}{2} \ $ and $ \ 8 · \phi^3 \ = \ 8 · \left(\frac{\psi}{2} \right)^3 \ = \ \psi^3 \ \ . $  It is reasonably simple to calculate $$ (1 \ + \ \sqrt5)^3 \ \ = \ \ 1^3 \ + \ 3·1^2 · \sqrt5 \ + \ 3 · 1 · (\sqrt5)^2 \ + \ (\sqrt5)^3 \ \ = \ \ 1 \ + \ 3\sqrt5 \ + \ 15 \ + \ 5\sqrt5  $$ $$ = \ \ 16 \ + \ 8\sqrt5 \ \ . $$
Hence,  $ \ 2a · (\phi + 1) \ = \ \sqrt{\psi^3} \ = \ \sqrt{16 \ + \ 8\sqrt5} \ = \  2 · \sqrt{4 \ + \ 2\sqrt5} \ \ , $ which is choice $ \ \mathbf{(1)} \ \ . $
This makes the irrational arithmetic a little quicker, but I couldn't say that this argument would occur to many people under "test conditions".
A: Just solve for $a$ and plug it in.
Assuming $a\ge 0$ we have
$a^2 \phi = 2$
$a^2 = \frac 2 {\phi}$
$a= \sqrt{\frac 2\phi } = \frac {\sqrt 2}{\sqrt \phi}$.
So
$2a(\phi +1)= 2\frac {\sqrt 2}{\sqrt \phi} (\phi+1)=2\sqrt 2(\sqrt \phi + \frac 1{\sqrt \phi})=....$
Okay here we can use $\phi^2 = \phi + 1$.
$2\sqrt 2\frac {\phi+1}{\sqrt \phi}= 2\sqrt 2\frac {\phi^2}{\phi^{\frac 12}}= 2\sqrt 2\phi^{\frac 32}=2\sqrt 2\phi \sqrt \phi=$
and given the choices
$2\sqrt{2\phi^2 \phi} = 2\sqrt{2(\phi+1)\phi}=2\sqrt {2\phi^2 + 2\phi}=$
$2\sqrt{2(\phi + 1) + 2\phi}=2\sqrt {4\phi + 2}=2\sqrt{2(1+\sqrt 5) + 2}=$
$2\sqrt{2\sqrt 5 + 4}$.
.....
I guess that wasn't that much quick.  But it was more directed.
I wonder if we can put $\phi^2 = \phi + 1$ if we can find a similar expresion for $\sqrt \phi$....
$\sqrt \phi = \sqrt{\phi^2 - 1} = \sqrt{(\phi -1)(\phi + 1)}=$
$\sqrt{\frac {(\sqrt 5-1)(\sqrt 5+3)}4}$...
Yeah... that could do it
$2a(\phi + 1) = 2\frac {\sqrt 2}{\sqrt{\frac {(\sqrt 5-1)(\sqrt 5+3)}4}}= 2\frac {2\sqrt 2}{\sqrt{(\sqrt 5-1)(\sqrt 5+3)} }=...$
That'll get you there but I would say it was easier....
