If α β and γ are the roots of the equation x3−7x+7=0, then 1α4+1β4+1γ4 is

If α β and γ are the roots of the equation x37x+7=0, then 1α4+1β4+1γ4 is

  1. A

    7/3

  2. B

    3/7

  3. C

    4/7

  4. D

    7/4

    Fill Out the Form for Expert Academic Guidance!l



    +91



    Live ClassesBooksTest SeriesSelf Learning



    Verify OTP Code (required)

    I agree to the terms and conditions and privacy policy.

    Solution:

    Here, Σα=0,Σαβ=7,αβγ=7
     1α4+1β4+1γ4=α4β4+β4γ4+γ4α4α4β4γ4
    =Σα4β4α4β4γ4…………………………..(i)
    Now ,ΣαβΣαβΣαβΣαβ=(Σαβ)2(Σαβ)2
    (7)4=α2β2+β2γ2+γ2α2+2αβγ(α+β+γ)α2β2+β2γ2+γ2α2+2αβγ(α+β+γ)=α2β2+β2γ2+γ2α2α2β2+β2γ2+γ2α2[Σα=α+β+γ=0]
    =α4β4+β4γ4+γ4α4+2α4β2γ2+2α2β4γ2+2α2β2γ4=Σα4β4+2α2β2γ2α2+β2+γ2=Σα4β4+2α2β2γ2(Σα)22Σαβ=Σα4β4+2α2β2γ2[02×(7)]=Σα4β4+2(7)2(2×7) Σα4β4=(7)4+4(7)3 Σα4β4=(7)3(7+4)=3(7)3
    On putting this value in Eq. (i), we get
    1α4+1β4+1γ4=3(7)3(7)4=37=37

    Chat on WhatsApp Call Infinity Learn

      Talk to our academic expert!



      +91


      Live ClassesBooksTest SeriesSelf Learning




      Verify OTP Code (required)

      I agree to the terms and conditions and privacy policy.