If $$(a^2+\frac{1}{a^2} )=18$$  then find ($$a-\frac{1}{a}$$​)  ?

Correct option is D

Given:

​$$(a^2+\frac{1}{a^2} )=18$$ 

Formula Used: 

​$$a - \frac{1}{a} =k$$​​

Then , $$a^2 + \frac{1}{a^2} = k^2 + 2$$​​

Solution: 

$$k^2 + 2 = 18$$ 

$$k^2 = 16 \\k = 4$$

Then the value of  $$a - \frac{1}{a} =k = 4$$