Let f(x)=(x−α)(x−β) It is given that f(0)=p⇒αβ=p and f(1)=‌
1
3
⇒(1−α)(1−β)=‌
1
3
Now, let us assume that, α is the common root of f(x)=0 and fofofof (x)=0 fofofof (x)=0 ⇒‌ fofof ‌(0)=0 ⇒fof(p)=0 So, f(p) is either α or β. (p−α)(p−β)=α (αβ−α)(αβ−β)=α⇒(β−1)(α−1)β=1(∵α≠0) So, β=3 (1−α)(1−3)=‌