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  1. #1
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    roots

    Find the equation that roots inverted roots of the equation
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    Quote Originally Posted by dapore View Post
    Find the equation that roots inverted roots of the equation
    Say the roots of 2x^4+9x^3+14x^2-9x+2 = 0 are \alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}.
    Is what you want the equation whose roots are \frac{1}{\alpha_{1}}, \frac{1}{\alpha_{2}}, \frac{1}{\alpha_{3}}, \frac{1}{\alpha_{4}}? See here...
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  3. #3
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    Quote Originally Posted by dapore View Post
    Find the equation that roots inverted roots of the equation
    Just replace \normalsize x by 1/x in the equation: 2x^{-4}+9x^{-3}+14x^{-2}-9x^{-1}+2=0. Then multiply through by x^4. (The new equation is very similar to the original one.)
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    Quote Originally Posted by Opalg View Post
    Just replace \normalsize x by 1/x in the equation: 2x^{-4}+9x^{-3}+14x^{-2}-9x^{-1}+2=0. Then multiply through by x^4.
    Why does this work, may I ask?
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    Quote Originally Posted by TheCoffeeMachine View Post
    Why does this work, may I ask?
    If alpha is a solution, then (x-\alpha) is a factor.

    \displaystyle\frac{1}{x}-\alpha=0\Rightarrow x=\frac{1}{\alpha}

    Do you see how the relation is the same as starting with \frac{1}{\alpha}\mbox{?}
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    I believe this works simply because (\frac{1}{\alpha})^{-1}=\alpha
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