If $${\left[ {a + \frac{1}{a}} \right]^2} - 2\left[ {a - \frac{1}{a}} \right] = 12,$$      then which of the following is a value of 'a'?

$$\eqalign{ & {\left[ {a + \frac{1}{a}} \right]^2} - 2\left[ {a - \frac{1}{a}} \right] = 12 \cr & {\left[ {a - \frac{1}{a}} \right]^2} + 4 - 2\left[ {a - \frac{1}{a}} \right] = 12 \cr & {\left[ {a - \frac{1}{a}} \right]^2} - 2\left[ {a - \frac{1}{a}} \right] - 8 = 0 \cr & {\text{Let }}a - \frac{1}{a} = x \cr & \therefore \,{x^2} - 2x - 8 = 0 \cr & {x^2} - 4x + 2x - 8 = 0 \cr & \left( {x - 4} \right)\left( {x + 2} \right) = 0 \cr & x = 4,\,x = - 2 \cr & a - \frac{1}{a} = - 2,\,a - \frac{1}{a} = 4 \cr & \therefore \,a + \frac{1}{a} = \sqrt {{4^2} + 4} \cr & a + \frac{1}{a} = 2\sqrt 5 \cr & a - \frac{1}{a} = 4 \cr & \therefore \,2a = 2\sqrt 5 + 4 \cr & a = 2 + \sqrt 5 \cr} $$