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Math Help - Disguised quadratic equation

  1. #1
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    Question Disguised quadratic equation

    I've had a go at this but I can't seem to get anywhere with it.


    2x^4 + x^3 - 6x^2 + x + 2 = 0

    Use the substitution p = x + 1/x to get a quadratic in p.


    I've tried just substituting in x = p - 1/x but the x's don't cancel. How should I approach this problem?

    Thanks for your help
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  2. #2
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    Re: Disguised quadratic equation

    Well, x+2 sure looks good as a factor....so try long division...
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  3. #3
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    Re: Disguised quadratic equation

    Quote Originally Posted by kinhew93 View Post
    I've had a go at this but I can't seem to get anywhere with it.


    2x^4 + x^3 - 6x^2 + x + 2 = 0

    Use the substitution p = x + 1/x to get a quadratic in p.


    I've tried just substituting in x = p - 1/x but the x's don't cancel. How should I approach this problem?

    Thanks for your help
    1. If the coefficients of the equation are symmetric to the central summand then divide through by x^2. You'll get:

    2x^2 + x - 6 + \frac1x + \frac2{x^2}=0

    2. If p = x + \frac1x then p^2 = x^2 + 2 + \frac1{x^2} ~\implies~ p^2-2 = x^2 + \frac1{x^2}

    3. Group the equation of #1 to:

    2 \left(x^2+\frac1{x^2}\right) + \left(x+\frac1x\right)-6=0

    Now replace the terms in brackets by the terms of #2:

    2(p^2-2)+p-6=0~\implies~2p^2+p-10=0

    which yields: p = 2~\vee~p=-\frac52

    4. Now replace p in the equation of #2 and solve for x:

    2 = x+\frac1x~\vee~-\frac52 = x+\frac1x

    x^2-2x+1 = 0~\vee~x^2+\frac52 x +1 = 0

    x=1~\vee~x=-\frac12\vee x=-2
    Thanks from kinhew93
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  4. #4
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    Re: Disguised quadratic equation

    Thanks Wilmer I'm sure that would work but I'm looking to solve it using the substitution given.
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