nao173

$x(x+6) > 16+4x$

$x^2 + 6x >16 + 4x$

$x^2 + 2x > 16$

$x^2 + 2x - 16 > 0$

$(x+1)^2 > 17$ (understand why)

Solve for x, should be pretty straightforward.

$x - \frac{1}{x} = 3$

$x^2 - 1 = 3x$

$x^2 - 3x -1 = 0$

$x = \frac{3\pm\sqrt{13}}{2}$

Because Triangle ADB is a right triangle, DB = $\sqrt{81-16} = \sqrt{65}$. 

$sin^{-1}(\frac{1}{2}) = \angle{DCA} = 30$, $DC = 4\sqrt{3}$.

$\angle{B} \cong \angle{B}, \angle{D} = \angle{E},$ Triangle BDA is congruent to Triangle BEC. 

$\frac{x}{CB} = \frac{4}{9} = sin(\angle{B})$,

 $CB = \frac{16\sqrt{3}+4\sqrt{65}}{9}$. Use the Pythagorean Theorem to solve for AE. 

Tried my best. Please correct me if I did anything wrong.