If α,β are the roots of the equation x2−p(x+1)−c=0, then the value of α2+2α+1α2+2α+c+β2+2β+1β2+2β+c is
A
1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
-1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A 1 Since, α,β are the roots of x2−px−(p+c)=0 α+β=p and αβ=−(p+c) Now, (α+1)(β+1)=αβ+α+β+1=1−c Consider, α2+2α+1α2+2α+c+β2+2β+1β2+2β+c =(α+1)2(α+1)2−(1−c)+(β+1)2(β+1)2−(1−c) =(α+1)2(α+1)2−(α+1)(β+1)+(β+1)2(β+1)2−(α+1)(β+1). =(α+1)2(α+1)(α−β)+(β+1)2(β+1)(β−α) =α+1α−β−β+1α−β =α−βα−β=1. ⇒α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=1