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Math Help - Imaginary roots factoring help

  1. #1
    Member diddledabble's Avatar
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    r^{2}-6r+6=0
    This might result in an i root. I am not sure just stuck.

    This is the determinant of my matrix and I need to find the eigenvalues. No square roots or fractions allowed. Must be a whole number or something like 2i
    Last edited by mr fantastic; July 29th 2009 at 06:15 PM. Reason: Merged posts
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  2. #2
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    Quote Originally Posted by diddledabble View Post
    r^{2}-6r+6=0

    This might result in an i root. I am not sure just stuck.
     r^{2}-6r+6=0

    I would complete the square here

     (r^{2}-6r+9)+6-9=0

     (r^{2}-6r+9)-3=0

     (r-3)^{2}-3=0

     (r-3)^{2}=3

    Now solving for r will determine the nature of the roots.

     (r-3)^{2}=3

     r-3=\sqrt{3}

     r=3\pm\sqrt{3}

    These look real to me.
    Last edited by mr fantastic; July 29th 2009 at 06:13 PM. Reason: Added quote
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by diddledabble View Post
    r^{2}-6r+6=0
    This might result in an i root. I am not sure just stuck.
    [I take this to be the characteristic equation from a DE]

    To find r, use the quadratic formula. You should get r=3\pm\sqrt{3}.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by diddledabble View Post
    This is the determinant of my matrix and I need to find the eigenvalues. No square roots or fractions allowed. Must be a whole number or something like 2i
    What is the matrix you're working with?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by diddledabble
    \mathbf{x}^{\prime}(t)=\begin{bmatrix}6&-3\\2&1\end{bmatrix}{\color{red}\mathbf{x}(t)} with initial value of \mathbf{x}(0)=\begin{bmatrix}-10\\-6\end{bmatrix}
    I think you left out an important part (in red)

    Let \mathbf{A}=\begin{bmatrix}6&-3\\2&1\end{bmatrix}.

    Then to find eigenvalues, we set \det\left(\mathbf{A}-\lambda \mathbf{I}\right)=0.

    In this case, it turns out that \mathbf{A}-\lambda \mathbf{I}=\begin{bmatrix}6-\lambda & -3\\2&1-\lambda\end{bmatrix}.

    It now follows that \det\left(\mathbf{A}-\lambda \mathbf{I}\right)=\left(6-\lambda\right)\left(1-\lambda\right)-\left(2\right)\left(-3\right)=6-7\lambda+\lambda^2+6=\lambda^2-7\lambda+12

    Thus, \det\left(\mathbf{A}-\lambda \mathbf{I}\right)=0\implies\lambda^2-7\lambda+12=0\implies\left(\lambda-4\right)\left(\lambda-3\right)=0\implies \lambda=4 or \lambda=3.
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