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.