### Instruction: 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 $a$?### Response: I need to find $a$ such that $f(g(a)) = 8$.
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 $a = 2$.