Discriminant help plz

$x= {-b \pm \sqrt{b^2-4ac} \over 2a}={-10\pm \sqrt{10^2-4k^2} \over 2k}$ If  rational solutions means x real so must $10^2-4k^2> 0$

So $k=1,2,3,4$ 

Total 4 positive integer values of k

Hope this helps! 

If it is 5, then the disciminant is 0, but the problem says solutions, which implies that there is more than one. Am I just being dumb or do we actually have to consider this?

Wait.. if k=1, the discriminat is 100-4*1^2=96, which is not a perfect square... so the solutions wouldn't be rational...right?