Question

8 If x2 + kor+1=0 has a root x=1 then k= ___​

Answer 1

Answer:

perfect square = equal roots =Δ=0

∴(2(k+2))2−4(4−k)(8k+1)=0

4(k2+4+4k)−4(32k+4−8k2−k)=0 

⇒k2+4+4k−32k−4+8k2+k=0

⇒9k2−27k=0

⇒k2−3k=0

⇒k(k−3)=0

⇒k=0 or k=3