Hi,
May be this approach can help to eliminate options:
Its given that x and k are Integers. Also x/6= Integer. Thus x= 6I
So expression becomes squareroot of 288*k*6I
i.e. 12*2* squareroot of 3*k*I
Thus k can be 3 and I can be k or 1.
This will eliminate all but not B.
If k and x are positive integers and x is divisible by 6
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saurabhdhakad
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