Concept: 1 Mark
Steps: 2 Marks
Answer: 1 Mark
We have,
(k−12)x2+2(k−12)x+2=0
Here, a=k−12,b=2(k−12) and c=2
∴D=b2−4ac=4(k−12)2−4(k−12)×2
⇒D=4(k−12){(k−12)−2}
⇒D=4(k−12)(k−14)
The given equation will have equal roots, if
D=0⇒4(k−12)(k−14)=0
⇒k−12=0 or, k−14=0
⇒k=12 or, k=14