CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If α,β are the roots of the equation x2p(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 x2px(p+c)=0
α+β=p and αβ=(p+c)
Now, (α+1)(β+1)=αβ+α+β+1=1c
Consider, α2+2α+1α2+2α+c+β2+2β+1β2+2β+c
=(α+1)2(α+1)2(1c)+(β+1)2(β+1)2(1c)
=(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)(αβ)+(β+1)2(β+1)(βα)
=α+1αββ+1αβ
=αβαβ=1.
α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Algebra of Limits
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon