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

If α and β are the roots of the equation x27x+1=0 then find the value 1(α7)2+1(β7)2

Open in App
Solution

α and β are the roots of the equation x27x+1=0
Here, a=1,b=7,c=1
αβ=ca=11=1 ----- ( 1 )
α2β2=1 ----- ( 2 )
α+β=ba=71=7 ----- ( 3 )
(α+β)2=α2+β2+2αβ
(7)2=α2+β2+2(1) [ Using ( 1 ) and ( 3 ) ]
49=α2+β2+2
α2+β2=47 ------- ( 4 )
Now,
1(α7)2+1(β7)2=(β7)2+(α7)2(α7)2(β7)2

=β214β+49+α214α+49(β214β+49)(α214α+49)

=α2+β214(α+β)+98α2β214αβ2+49β214α2β+196αβ686β+49α2686α+2401

=4714(7)+98α2β214αβ(α+β)+49(α2+β2)+196αβ686(α+β)+2401 [ Using ( 3 ) and ( 4 ) ]
=47114(1)(7)+49(47)+196(1)686(7)+2401

=47198+2303+1964802+2401

=471
1(α7)2+1(β7)2=47

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Solving QE by Factorisation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon