The number of values of k for which x2−(k−2)x+k2×x2+kx+(2k−1) is a perfect square is

For given situation, x2−(k−2)x+k2=0 and x2+kx+2k−1=0should have both roots common or each should have equalroots. If both roots are common, then11=−(k−2)k=k22k−1⇒ k=−k+2 and 2k−1=k2⇒k=1If both the equations have equal roots, then      (k−2)2−4k2=0  and  k2−4(2k−1)=0⇒(3k−2)(−k−2)=0 and k2−8k+4=0There is no common value of k.Therefore, k = 1 is the only possible value.