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Math Help - Palindromic problem, please help!

  1. #1
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    Palindromic problem, please help!

    Question: Find the minimum possible value of a^{2}+b^{2}such that the quartic polynomial x^{4}+ax^{3}+bx^{2}+ax+1 has at least 1 real root.

    Attempt: Since the equation is palindromic, the standard technique is:
    x^{4}+ax^{3}+bx^{2}+ax+1\Rightarrow x^{2}+ax+b+\frac{a}{x}+\frac{1}{x^{2}}
    Let X=x+\frac{1}{x}\therefore X^{2}+pX+q=0

    Since \frac{f(x))}{x^{4}}=f(\frac{1}{x})=0the real roots must be 1 or -1 correct? But when I solve simultaneous equations using 1 and -1 as roots I get a^{2}+b^{2}=4which is too large. The glitch I ran into was excluding the case that at the minimal value of a^2+b^2there might be two real roots of multiplicity 2. That made things a little ugly.

    I also tried this approach: Suppose r is a simple multiplicity 1 root and a^2+b^2 is at it's minimal value. The graph of the polynomial near the root the just looks like a line crossing the x-axis, right? That means if I move the graph just a little by varying a and b, there will still be a real root because the graph will still cross the x-axis, just in a slightly different spot. That in turn means I can make a or b just a little smaller so I decrease a^2+b^2and still have a root. So that would contradict a^2+b^2being the smallest value where you still have a real root.
    That means at the minimal value, all of the real roots must have multiplicity 2 or greater but I think I am stuck here...
    Last edited by smokesalot; March 18th 2013 at 05:53 AM.
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  2. #2
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    Re: Palindromic problem, please help!

    Bump!
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  3. #3
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    Re: Palindromic problem, please help!

    CONTRADICTION!!!
    This is the maths "help" forum and no body helps.
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