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Math Help - Polynomial Question

  1. #1
    Newbie
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    Polynomial Question

    Hey there,

    I was reading an article, where the author claims that the following polynomial:
    x^{2}-(1+\frac{1}{\theta }+\frac{1}{\theta\alpha  })x+\frac{1}{\theta }=0<br />

    has two roots, first one root is \lambda , which is between 0 and 1, and the other can be expressed as \frac{1}{\theta\lambda  }

    note that this polynomial is obtained after factoring a second order difference equation with forward and lag operators. Now I have no idea how he arrives at that identity, could anyone explain me the trick behind it?

    And while you are at it, could you tell me a similar identity between the roots of the following polynomial:

    x^{2}-(1+\frac{1}{\beta }+\frac{\gamma\kappa  }{\beta })x+\frac{1}{\beta }=0

    So for example let \lambda , be the root between 0 and 1 of this polynomial, what is the second root in terms of \lambda ?

    Thanks a lot

    Burak
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  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
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    someplace
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    Quote Originally Posted by ichoosetonotchoosetochoos View Post
    Hey there,

    I was reading an article, where the author claims that the following polynomial:
    x^{2}-(1+\frac{1}{\theta }+\frac{1}{\theta\alpha  })x+\frac{1}{\theta }=0<br />

    has two roots, first one root is \lambda , which is between 0 and 1, and the other can be expressed as \frac{1}{\theta\lambda  }

    note that this polynomial is obtained after factoring a second order difference equation with forward and lag operators. Now I have no idea how he arrives at that identity, could anyone explain me the trick behind it?

    And while you are at it, could you tell me a similar identity between the roots of the following polynomial:

    x^{2}-(1+\frac{1}{\beta }+\frac{\gamma\kappa  }{\beta })x+\frac{1}{\beta }=0

    So for example let \lambda , be the root between 0 and 1 of this polynomial, what is the second root in terms of \lambda ?

    Thanks a lot

    Burak
    The constant term in a monic quadratic (one that has coefficien of  $$ x^2 equal to $$ 1) is the product of the two roots. So for a monic quadratic if the constant term is $$ c and one root is $$ \lambda then the other root is c/\lambda

    CB
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