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Math Help - Roots and calcul

  1. #1
    Super Member dhiab's Avatar
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    Roots and calcul

    \alpha ,\beta ,\gamma roots of equation :

    x^3 - x - 1 = 0
    Calculate :
    \frac{{1 + \alpha }}{{1 - \alpha }} + \frac{{1 + \beta }}{{1 - \beta }} + \frac{{1 + \gamma }}{{1 - \gamma }}
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let y_1=\frac{1+\alpha}{1-\alpha}, \ y_2=\frac{1+\beta}{1-\beta}, \ y_3=\frac{1+\gamma}{1-\gamma}.

    We'll find the equation whose roots are y_1, y_2, y_3

    For this let y=\frac{1+x}{1-x}\Rightarrow x=\frac{y-1}{y+1}

    Replace x in the equation:

    \left(\frac{y-1}{y+1}\right)^3-\frac{y-1}{y+1}-1=0\Rightarrow y^3+7y^2-y+1=0

    Then y_1+y_2+y_3=-7
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  3. #3
    Senior Member pacman's Avatar
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    red_dog bashed the olympiad caliber problem again, whew!
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  4. #4
    Newbie I4talent's Avatar
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    x^3-x-1 = 0

    (x-\alpha)(x-\beta)(x-\gamma) = 0

    \left(\frac{1+x}{1-x}-\frac{1-\alpha}{1+\alpha}\right) \left(\frac{1+x}{1-x}-\frac{1-\beta}{1+\beta}\right) \left(\frac{1+x}{1-x}-\frac{1-\gamma}{1+\gamma}\right) = 0<br />

    \left(y-\frac{1-\alpha}{1+\alpha}\right) \left(y-\frac{1-\beta}{1+\beta}\right) \left(y-\frac{1-\gamma}{1+\gamma}\right) = 0<br />

    <br />
\left (\frac{y-1}{1+y}-\alpha\right) \left (\frac{y-1}{1+y}-\beta\right) \left (\frac{y-1}{1+y}-\gamma\right) = 0<br />

    \therefore \left (\frac{y-1}{1+y}\right)^3-\left (\frac{y-1}{1+y}\right)-1 = 0

    <br />
\frac{(y-1)(y-1)(y-1)}{(y+1)(y+1)(y+1)}-\left(\frac{y-1}{1+y}\right)-1 = 0

    (1+y)(1+y)(1+y) \frac{(y-1)(y-1)(y-1)}{(y+1)(y+1)(y+1)}- (1+y)(1+y)(1+y)\left(\frac{y-1}{1+y}\right)-(1+y)(1+y)(1+y) = 0

    (y-1)(y-1)(y-1)-(1+y)(1+y)(y-1)-(1+y)(1+y)(1+y) = 0

    -1+y^3-3y^2+3y-(-1-y+y^2+y^3)-(1+3y+3y^2+y^3) = 0

    -1+y^3-3y^2+3y+1+y-y^2-y^3-1-3y-3y^2-y^3 = 0

    y^3-y^3-y^3-3y^2-y^2-3y^2+3y+y-3y+1-1-1 = 0<br />

    -y^3-7y^2+y-1 = 0

    y^3+7y^2-y+1 = 0

    \therefore\frac{{1 + \alpha }}{{1 - \alpha }} + \frac{{1 + \beta }}{{1 - \beta }} + \frac{{1 + \gamma }}{{1 - \gamma }} = \frac{-7}{1} = \boxed{-7}
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  5. #5
    Super Member dhiab's Avatar
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    Quote Originally Posted by red_dog View Post
    Let y_1=\frac{1+\alpha}{1-\alpha}, \ y_2=\frac{1+\beta}{1-\beta}, \ y_3=\frac{1+\gamma}{1-\gamma}.

    We'll find the equation whose roots are y_1, y_2, y_3

    For this let y=\frac{1+x}{1-x}\Rightarrow x=\frac{y-1}{y+1}

    Replace x in the equation:

    \left(\frac{y-1}{y+1}\right)^3-\frac{y-1}{y+1}-1=0\Rightarrow y^3+7y^2-y+1=0

    Then y_1+y_2+y_3=-7
    Hello: Are you the details for y1 + y 2+ y 3 =-7 Thanks
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  6. #6
    Senior Member pacman's Avatar
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    i told you dhiab, red_dog's method is great. It was even beautiful than the official solution, or that of Andreescu and Feng.

    Neat , and concise
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  7. #7
    Senior Member pacman's Avatar
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    thanks
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