If 2 is a root of the equation x2 + bx + 12 = 0 and the

2 is the common roots given quadric equation are $x^{2}+b x+12=0$, and $x^{2}+b x+q=0$

Then find the value of q.

Here, $x^{2}+b x+12=0$.......(1)

$x^{2}+b x+q=0 \ldots \ldots(2)$

Putting the value of $x=2$ in equation (1) we get

$2^{2}+b \times 2+12=0$

$4+2 b+12=0$

$2 b=-16$

$b=-8$

Now, putting the value of $b=-8$ in equation (2) we get

$x^{2}-8 x+q=0$

Then,

$a_{2}=1, b_{2}=-8$ and,$c_{2}=q$

As we know that $D_{1}=b^{2}-4 a c$

Putting the value of $a_{2}=1, b_{2}=-8$ and, $c_{2}=q$

$=(-8)^{2}-4 \times 1 \times q$

$=64-4 q$

The given equation will have equal roots, if $D=0$

$64-4 q=0$

$4 q=64$

$q=\frac{64}{4}$

$q=16$

Thus, the correct answer is $(c)$