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Math Help - cubic equation proof needed

  1. #1
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    Question cubic equation proof needed

    Say the equation is

    ax3+bx2+cx+d=0

    suppose that he roots of this equation are α,β, and γ.
    How to prove that,
    1)
    α+β+γ = -b/a
    2)
    αβγ=d/a
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  2. #2
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    Quote Originally Posted by callkalpa View Post
    Say the equation is

    ax3+bx2+cx+d=0

    suppose that he roots of this equation are α,β, and γ.
    How to prove that,
    1) α+β+γ = -b/a
    2) αβγ=d/a
    Start by reading this: Vieta's Formulas -- from Wolfram MathWorld

    Then use Google.
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  3. #3
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    Hello, callkalpa!

    Say the equation is: . ax^3+bx^2+cx+d\:=\:0

    Suppose that the roots of this equation are: . \alpha,\:\beta,\:\gamma

    Prove that:

    . . 1)\;\;\alpha + \beta + \gamma \:=\:-\dfrac{b}{a}
    . . 2)\;\;\alpha\beta\gamma \:=\:-\dfrac{d}{a}

    Divide the equation by a\!:\;\;x^3 + \dfrac{b}{a}x^2 + \dfrac{c}{a}x + \dfrac{d}{a} \:=\:0 .[1]


    Since \alpha,\,\beta,\,\gamma are roots, the equation has the form:
    . . (x-\alpha)(x-\beta)(x-\gamma) \:=\:0

    which simplifies to:
    . . x^2 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \alpha\gamma)x - \alpha\beta\gamma \:=\:0 .[2]


    Equate coefficients of [1] and [2]:

    . . \begin{array}{ccc}-(\alpha + \beta + \gamma) &=& \dfrac{b}{a} \\ \\[-3mm] \alpha\beta + \beta\gamma + \alpha\gamma &=& \dfrac{c}{a} \\ \\[-3mm] -\alpha\beta\gamma &=& \dfrac{d}{a} \end{array}


    Therefore: . \begin{Bmatrix}\alpha+\beta+\gamma &=& \text{-}\,\dfrac{b}{a} \\ \\[-3mm] \alpha\beta + \beta\gamma + \alpha\gamma &=& \;\dfrac{c}{a} \\ \\[-3mm]  \alpha\beta\gamma &=& \text{-}\,\dfrac{d}{a} \end{Bmatrix}

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