Algebraic Manipulations to solve for q If $p-q=\sqrt3$ and $\sqrt{\sqrt{3}-\sqrt{p}}=q$, find $q$.
Some work
$$\bigg(\sqrt{\sqrt{3}+\sqrt{p}}
\bigg)\bigg(\sqrt{\sqrt{3}-\sqrt{p}}\bigg)=\sqrt{3-p}$$
$$\sqrt{\sqrt{3}+\sqrt{p}}=\frac{\sqrt{3-p}}{q}$$
$$q^2\sqrt3+q^2\sqrt{p}=3-p$$
$$(\sqrt3+\sqrt{p})q^2+p-3=0$$
I can't find any way to use the first fact. 

Square the second equation and rearrange
\begin{eqnarray*}
\sqrt{3}-q^2 = \sqrt{p}.
\end{eqnarray*}
Square this and subtract $q$ and we can use the second equation to get 
\begin{eqnarray*}
q^4 -2 q^2 \sqrt{3}-q +3 -\sqrt{3}=0.
\end{eqnarray*}
Now this can be factorised !
\begin{eqnarray*}
(q^2-q - \sqrt{3})(q^2+q +1 -\sqrt{3})=0.
\end{eqnarray*}
Should be a doddle from here ?
