Let the common root be $r$.

Then write the roots as:

• For $x^2-6x+a=0$: roots are $r$ and $m$
• For $x^2-cx+6=0$: roots are $r$ and $n$

We are told that the other roots are integers in the ratio $4:3$. So $m:n = 4:3$.

Use Vieta's formulas for the first equation: $r+m=6$ Hence, $m=6-r$

Use Vieta's formulas for the second equation: $rn=6$ Hence, $n=\frac{6}{r}$

Now use the ratio condition: $\frac{m}{n}=\frac{4}{3}$ Substituting $m=6-r$ and $n=\frac{6}{r}$, $\frac{6-r}{6/r}=\frac{4}{3}$

Simplify: $\frac{r(6-r)}{6}=\frac{4}{3}$ $r(6-r)=8$ $6r-r^2=8$ $r^2-6r+8=0$ $(r-2)(r-4)=0$

So, $r=2 \text{ or } r=4$

Check the integer ratio condition carefully.

• If $r=2$: $m=6-2=4, \quad n=\frac{6}{2}=3$ So the other roots are $4$ and $3$, which are in the ratio $4:3$. This works.

• If $r=4$: $m=6-4=2, \quad n=\frac{6}{4}=\frac{3}{2}$ Here $n$ is not an integer, so this does not satisfy the condition.

Therefore, the common root is $\boxed{2}$

Comparing with the stored correct answer:

• Derived answer: $2$ (Option D)
• Stored correct answer: D

They agree.