Simplifying radicals inside radicals: $\sqrt{24+8\sqrt{5}}$ Simplify: $\sqrt{24+8\sqrt{5}}$
I removed the common factor 4 out of the square root to obtain $2\sqrt{6+2\sqrt{5}}$, but the answer key says it is $2+2\sqrt{5}$.
Am I missing out on some general rule here?

To see whether $\sqrt {24+8\sqrt 5 }=A+B\sqrt 5$ with rational $A, B,$ suppose $$24+8\sqrt 5=(A+B\sqrt 5)^2=(A^2+5 B^2)+2 A B\sqrt 5.$$ Since $\sqrt 5$ is not rational, this requires $$24=A^2+5B^2 \;\text {and }\; 2 A B=8.$$ From $2 A B=8$ we have $B=4/A.$ (Note that we cannot have $A=0$ else $24+8\sqrt 5=5 B^2$ is rational, implying $\sqrt 5$ is rational.)
Substituting $4/A$ for $B$ in $24=A^2+5 B^2,$ and multiplying through by $A^2,$ and re-arranging, we get $$(A^2)^2-24 A^2+80=0.$$ This is quadratic in $A^2,$ with solutions $$A^2\in \{4, 20\}.$$ But $A^2=20$ implies $A$ is not rational. This leaves $A^2=4,$ so $A=\pm 2.$ Since we have $B=4/A,$ this implies $(A,B)=(2,2)$ or $(A,B)=(-2,-2).$ Since $A+B\sqrt 5>0,$ we have $A=B=2.$
