Find the greatest integer not exceeding $A$ if $A = \sqrt{x^2+\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}}$ and $x=20062007$. 
Find the greatest integer not exceeding $A$ if $$A = \sqrt{x^2+\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}}$$ and $x=20062007$.

The nested square roots are very tricky and I am not able to find any pattern or something except that every $x^2$ term is a perfect square. As far as I think, it is not plausible to denest the radical by calculating the square roots.  
Is there any simpler approach ? I am interested in the general case when there is no particular given value of $x$. I have tried my best, but there is nothing I can come up with.
Thanks in Advance ! :-)

A).
Show that $A>20062008$:
$$E:=100x^2+39x+\sqrt{3}>64x^2+16x+1 = (8x+1)^2;$$
$$D:=16x^2 + \sqrt{E} > 16x^2 + (8x+1) = (4x+1)^2;$$
$$C:=4x^2+ \sqrt{D} > 4x^2+(4x+1) = (2x+1)^2;$$
$$B:=x^2+\sqrt{C} > x^2+(2x+1) = (x+1)^2;$$
$$A=\sqrt{B}>x+1.$$
B). Show that $A<20062009$. More rough estimation:
$$E<(10x+2)^2;$$
$$D<16x^2+(10x+2)<(4x+2)^2;$$
$$C<4x^2+(4x+2)<(2x+2)^2;$$
$$B<x^2+(2x+2)<(x+2)^2;$$
$$A<x+2.$$
