Let, $\alpha$ be the common root and the other roots of the equations be $4\beta$ and $3\beta$ respectively. Then,
$\alpha +4\beta =6,\alpha \left(4\beta \right)=a$
$\alpha +3\beta =c,\alpha \left(3\beta \right)=6\Rightarrow \frac{4}{3}=\frac{a}{6}\Rightarrow \alpha =8$
The first equation is $x^2-6x+8=0$
Whose roots are $2$ and $4$
If $\alpha =2\Rightarrow \beta =1$
So, roots of first equation is $2,4$ and that of second equation is $2,3$
If $\alpha =4\Rightarrow \beta =\frac{1}{2}\Rightarrow 3\beta =\frac{3}{2},4\beta =2$
Here the roots are not integers $\Rightarrow \alpha =2$