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    GRRRR i dont liek quartic roots. i mean i am find with quadratic and cubic but quartic!!!! GRRRRRRRRRR

    anyway pls help

    the equation x^4 + 4x²+3x-5= 0 has four roots alpha, beta, gamma, detlta

    find a quatric equation with interger coefficeints which has roots

    alpha²-1, beta²-1, gamma²-1 and delta²-1

    pls help using method of sum of roots, product in 1 pair, etc... thanks

    another similiar q

    the equation x^4-x³ + 2x²+5x-1= 0 has four roots alpha, beta, gamma, detlta

    find a quatric equation with interger coefficeints which has roots

    alpha², beta², gamma² and delta².


    Try breaking it into two quadratic equations, eg (ax^2 + bx + c)(dx^2 + ex + f) = 0, and go from there. That's what I'd do. Do it by trial and error. the last term (cf), can only be a few options, and so can the adx^4, as you have ad = 1, and cf = -5. therefore you already have

    (x^2 + bx + 1)(x^2 + ex - 5)
    (x^2 + bx - 1)(x^2 + ex + 5)

    then all you gotta do is work out the b and e.

    x4 + 4x²+3x-5= 0 lets say it has roots µ, π, λ and ∂

    So the equation with roots µ²-1,....etc

    ω = x²-1
    x= √(ω-1)

    sub ω into now,

    (√(ω-1))4 +4(√(ω-1))² + 3(√(ω-1)) - 5 = 0
    (ω-1)² + 4(ω-1) +3(√(ω-1)) - 5 = 0
    ω² + 2ω - 10 = -3(√(ω-1))

    square both sides

    ω4+4ω³-20ω²-40ω+4ω² + 100 = 9ω-9
    ω4 + 4ω³ -16ω² -49ω + 9 =0

    x4 -x³ + 2x²+5x-1= 0

    roots are sqrd

    so if √ω = x

    (√ ω )4 - (√ ω )³ + 2(√ ω )² +5( √ ω ) + 9 = 0
    ω² - ω3/2 + 2ω +5ω1/2 + 9 =0
    ω² + 2ω + 9 = (ω3/2 - 5ω1/2 )
    sq both sides
    ω4 + 4ω³ + 18ω² + 4ω² + 36ω + 81 = ω³ -10ω² + 25ω
    ω4 + 3ω³ + 32ω² + 9ω + 81 = 0
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