jmc

The roots of $a{x}^2+bx+c$ are rational if and only if the discriminant $$b^2-4ac$$is a perfect square.

Similarly, the roots of $4a{x}^2+12bx+kc=0$ are rational if and only if its discriminant $$(12b{)}^2-4(4a)(kc)=144b^2-16kac$$is a perfect square.

To narrow down the possible values of $k,$ we look at specific examples. Take $a=1,$ $b=6,$ and $c=9.$ Then $b^2-4ac=0$ is a perfect square, and we want $$144b^2-16kac=5184-144k=144(36-k)$$to be a perfect square, which means $36-k$ is a perfect square. We can check that this occurs only for $k=11,$ 20, 27, 32, 35, and 36.

Now, take $a=3,$ $b=10,$ and $c=3.$ Then $b^2-4ac=64$ is a perfect square, and we want $$144b^2-16kac=14400-144k=144(100-k)$$to be a perfect square, which means $100-k$ is a perfect square. We can check that this occurs only for $k=19,$ 36, 51, 64, 75, 84, 91, 96, 99, 100. The only number in both lists is $k=36.$

And if $b^2-4ac$ is a perfect square, then $$144b^2-16kac=144b^2-576ac=144(b^2-4ac)$$is a perfect square. Hence, the only such value of $k$ is $\boxed{36}.$