The correct option is B 5
(x−1)(x−k)=−4
Converting into standard from
x2−kx−x+k=−4
x2−(k+1)x+(k+4)=0
a=1,b=−(k+1),c=k+4
Since the quadratic equation has real and equal roots,
D=b2−4ac=0
(−(k+1))2−4×1×(k+4)=0
k2+2k+1−4K−16=0
k2−2k−15=0
k2−5k+3k−15=0
(k−5)(k+3)=0
k=5,−3
Since, we have to find the positive value of k.
Hence, k =5
(b) is correct