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Math Help - Biquadratic equation, arithmetic progression

  1. #1
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    Biquadratic equation, arithmetic progression

    Greeings,

    The equation below is given and the problem is to find all values of m, for which the roots of the equation form an arithmetic progression.

    Though, I am pretty sure Vieta's formulae are to be used, I am puzzled by the fact they are usable in quadratic equations only..
    I've considered a substitution of the x^2=y like, but then I loose the link between the substituted equation's roots and the biquadratic equation's roots..

    Thank you.
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  2. #2
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    Quote Originally Posted by Logic View Post
    Greeings,

    The equation below is given and the problem is to find all values of m, for which the roots of the equation form an arithmetic progression.

    Though, I am pretty sure Vieta's formulae are to be used, I am puzzled by the fact they are usable in quadratic equations only..
    I've considered a substitution of the x^2=y like, but then I loose the link between the substituted equation's roots and the biquadratic equation's roots..

    Thank you.
    Your equation is:

    x^4-(3m+5)x^2+(m+1)^2=0

    Assume this has four distinct real roots (as it must if they are to be in arithmetic progression since it can only have 0, 2 or 4 real roots and so you need 4 to have an arithmetic progression), and as this is a quadratic in y=x^2, they are of the form: -b,\ -a,\ a,\ b;\ \ b > a > 0.

    So for these to be in arithmetic progression means that:

     <br />
a=\frac{1}{2}(b-a)<br />

    where a and b are the distinct positive roots of y^2-(3m+5)y+(m+1)^2=0 and a is the smaller of the two.
    Last edited by Constatine11; December 26th 2008 at 09:19 AM.
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