solve the following problem?

if α and β are zeroes of the polynomial 6x^2 + x + 1 , then find the value of :-

(i) α^3 β + α β^3

(ii) α^2 + β^2

(iii) α^2/β + β^2/α

2 Answers

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  • 7 years ago

    (α+ß)=-b/a=-1/6, and αß=c/a=1/6

    (i) α^3 β + α β^3

    =αß(α²+ß²)

    =αß[(α+ß)²-2αß]

    =(1/6)[1/36 - 1/3]

    =-11/216

    (ii)α²+ß²

    =(α+ß)²-2αß

    =1/36-1/3

    =-11/36

    (iii) α²/ß+ß²/α

    =(α³+ß³)/(αß)

    =(α+ß)((α+ß)²-3αß)/(αß)

    =17/36

  • moe
    Lv 7
    7 years ago

    6x^2 + x + 1 = 0

    Factorize to get:

    (6x+1)(x+1) = 0

    so,

    if(6x+1) = 0 then x = -1/6 or

    else (x+1) = 0 then x = -1

    so, roots are: -1 (=α) and -1/6(=β)

    in which case:

    (i) α^3 β + α β^3 =(-1)^3(-1/6) + (-1)(-1/6)^3 = 0.1723

    (ii) α^2 + β^2 = (-1)^2 + (1/6)^2 = 1.0278

    (iii) α^2/β + β^2/α = (-1)^2/(-1/6) + (-1/6)^2/(-1) = -6.0278

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