If $$x = 3 + \sqrt 8 ,$$   then $${x^2} + \frac{1}{{{x^2}}}$$  is equal to = ?

Solution (By Examveda Team)

$$\eqalign{ & \because x = 3 + \sqrt 8 \cr & \Rightarrow {x^2} = {\left( {3 + \sqrt 8 } \right)^2} \cr & \Rightarrow {x^2} = {3^2} + {\left( {\sqrt 8 } \right)^2} + 2 \times 3 \times \sqrt 8 \cr & \Rightarrow {x^2} = 9 + 8 + 6\sqrt 8 \cr & \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr} $$
$$\therefore {x^2} + \frac{1}{{{x^2}}}$$
$$ = \left( {17 + 12\sqrt 2 } \right)$$   $$ + \frac{1}{{\left( {17 + 12\sqrt 2 } \right)}}$$   $$ \times \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{\left( {17 - 12\sqrt 2 } \right)}}$$
$$\eqalign{ & = \left( {17 + 12\sqrt 2 } \right) + \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{289 - 288}} \cr & = \left( {17 + 12\sqrt 2 } \right) + \left( {17 - 12\sqrt 2 } \right) \cr & = 17 + 17 \cr & = 34 \cr} $$