The integral values of m for which the roots of the equation mx2+(2m−1)x+(m−2)=0 are rational are given by the expression [where n is integer]

Discriminant D=(2m−1)2−4(m−2)m=4m+1must be perfect square. Hence,4m+1=k2, for some k∈I⇒ m=(k−1)(k+1)4Clearly, k must be odd. Let k=2n+1∴ m=2n(2n+2)4=n(n+1),n∈I