Solution
Perfect square of a quadratic equation is of form $$\left(pa\ +\ q\right)^2$$
Let 'k' be the no. added to make the required equation as a perfect square
$$\left(pa\ +\ q\right)^2=p^2a^2\ +\ 2apq\ +\ q^2=16a^2-12a\ +\ k$$
Comparing the coefficients, we get $$p^2=16,\ p=\pm\ 4.$$
$$2apq=-12a.\ So,\ pq=-6,\ \ $$ which gives q=$$\ \pm\frac{\ 3}{2}$$
k=$$\ q^2$$=$$\ \frac{\ 9}{4}$$
Option A is the answer.
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