Can anyone tell me where I messed up (not getting the correct answer):

What is the 4th order Maclaurin of: $\displaystyle f(x) = \sqrt{9 + x^2}$

$\displaystyle 1 + u = 9 + x^2 \implies$

$\displaystyle \sqrt{1 + u} = 1 + \frac{u}{2} - \frac{u^2}{2} + O(x^3) \implies$

$\displaystyle \sqrt{9 + x^2} = 1 + \frac{8 + x^2}{2} - \frac{(8 + x^2)^2}{8} + O(x^6)$

EDIT: The substitution $\displaystyle u = \frac{x^2}{9}$ works better. Then I can just use $\displaystyle 3\sqrt{1 + \frac{x^2}{9}}$