The quadratic equations $x^2-6x+a=0$ and $x^2-cx+6=0$ have one root in common. The other roots of the first and second equations are integers in the ratio $4:3$. Then the common root is:

Let the common root be $\alpha$. Let the other roots be $4k$ and $3k$ ($k$ is an integer).For $x^2-6x+a=0$, roots are $\alpha ,4k$:$\alpha +4k=6$ ... (1)For $x^2-cx+6=0$, roots are $\alpha ,3k$:$\alpha \cdot 3k=6⟹\alpha k=2$ ... (2)From (2), $k=2/\alpha$. Substituting in (1):$\alpha +4(2/\alpha )=6⟹\alpha +8/\alpha =6$${\alpha }^2-6\alpha +8=0$$(\alpha -2)(\alpha -4)=0⟹\alpha =2,4$.If $\alpha =2$, then $k=2/2=1$. The other roots are $4(1)=4$ and $3(1)=3$. Both are integers.If $\alpha =4$, then $k=2/4=0.5$. The other roots are $4(0.5)=2$ and $3(0.5)=1.5$. $1.5$ is not an integer.So, the common root is $\alpha =2$.