maths

if alpha and bita are the zeroes of ax^2+bx+c then evaluate alpha^4+bita^4

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asked by keshav
  1. I will use m and n for alpha and beta

    from the equation we know
    m+n = -b/a
    mn = c/a

    (m+n)^2 = m^2 + n^2 + 2mn
    so m^2 + n^2 = (m+n)^2 - 2mn

    also
    (m^2 + n^2)^2 = m^4 + n^4 + 2m^2n^2

    so m^4 + n^4 = (m^2 + n^2)^2 - 2(mn)^2
    =[(m+n)^2 - 2mn]^2 - 2(mn)^2
    = [(-b/a)^2 - 2c/a]^2 - 2c^2/a^2
    = [b^2/a^2 - 2c/a]^2 - 2c^2/a^2

    let's test it with x^2 - 7x + 12 = 0
    roots m and n , where m = 4, n=3
    m+n= 7
    mn = 12
    according to my answer
    m^4 + n^4 = [(49/1-2(12)]^2 - 2(144))/1 = 25^2 - 288 = 337

    actual m^4 +n^4 = 4^4 + 3^4 = 256 + 81 = 337

    YEAH!

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    posted by Reiny
  2. For ax^2+bx+c=0
    the zeroes are:
    α,β = (-b±sqrt(b²-4ac))/2a
    =(p±q)
    where
    p=-b/2a
    q=sqrt(b²-4ac)/2a

    So to evaluate
    α^4+β^4
    =(p+q)^4+(p-q)^4
    =2(p^4+6p²q²+q^4)
    =2[(-b/2a)^4+6(-b/2a)²(sqrt(b²-4ac)/2a)²+(sqrt(b²-4ac)/2a)^4]

    Expand and simplify to get:
    2c²/a²-4b²c/a³+(b/a)^4

    Check me.

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