Given that $a > 0$, if $f(g(a)) = 8$, where $f(x) = x^2 + 8$ and $g(x) = x^2 - 4$, what is the value of $a$?
I can start by plugging in $g(a)$ into $f(x)$ and simplifying: $f(g(a)) = f(a^2 - 4) = (a^2 - 4)^2 + 8$. Now I can set this equal to 8 and solve for $a$: $(a^2 - 4)^2 + 8 = 8$. Subtracting 8 from both sides gives me $(a^2 - 4)^2 = 0$. Taking the square root of both sides gives me $a^2 - 4 = 0$. Adding 4 to both sides gives me $a^2 = 4$. Taking the square root of both sides again gives me $a = \pm 2$. However, I know that $a > 0$, so I can eliminate the negative possibility and conclude that $a = 2$. 