If α, β, γ, σ are the roots of the equation x4+4x3−6x2+7x-9=0 then the value of 1+α21+β21+γ21+σ2 is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
9
b
11
c
13
d
5
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Since, α,β, γ, σ are the roots of the given equation, we havex4+4x3−6x2+7x−9=(x−α)(x−β)(x−γ)(x−σ)Putting x = i and then x = -i , we get 1−4i+6+7i−9=(i−α)(i−β)(i−γ)(i−σ)and 1+4i+6−7i−9=(−i−α)(−i−β)(−i−γ)(−i−σ)Multiplying these two equations, we get (−2+3i)(−2−3i)=1+α21+β21+γ21+σ2or 13=1+α21+β21+γ21+σ2