<s>[INST] $a^2\phi=2$ , 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 ?
 [/INST] 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....
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