CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If α,β,γ,δR satisfy (α+1)2+(β+1)2+(γ+1)2+(δ+1)2α+β+γ+δ=4.
If the equation a0x4+a1x3+a2x2+a3x+a4=0 has the roots (α+1β1),(β+1γ1),(γ+1δ1),(δ+1α1),   then the value of a2a0 is


Solution

(α1)2+(β1)2+(γ1)2+(δ1)2=0
α=β=γ=δ=1
So, all the roots are (1+1-1)=1
So, the equation is (x1)4=0
x44x3+6x24x+1=0
a2a0=61=6

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...



footer-image