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Math Help - Roots of a Sixth-Degree Polynomial

  1. #1
    Senior Member Sambit's Avatar
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    Roots of a Sixth-Degree Polynomial

    Consider the equation x^6 - 5x^4 + 16x^2 - 72x +9 = 0. I calculate its 2nd derivative 30x^4 - 60x^2 + 32 which is always >0. So f '(x) is monotonic increasing for all x. So f '(x) has only 1 real root. This suggests f(x) has either 0 or 2 real roots; and this is where I am stuck. Since Descarte's rule applied on f(x) says there is 0 or 2 or 4 +ve roots and 0 or 2 -ve roots, it does not help me anyway. How can I go further?
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by Sambit View Post
    Consider the equation x^6 - 5x^4 + 16x^2 - 72x +9 = 0. I calculate its 2nd derivative 30x^4 - 60x^2 + 32 which is always >0. So f '(x) is monotonic increasing for all x. So f '(x) has only 1 real root. This suggests f(x) has either 0 or 2 real roots; and this is where I am stuck. Since Descarte's rule applied on f(x) says there is 0 or 2 or 4 +ve roots and 0 or 2 -ve roots, it does not help me anyway. How can I go further?
    Think Intermediate Value Theorem. (What happens when x=1?)
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  3. #3
    Senior Member Sambit's Avatar
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    Intermediate value theorem states that if a function is continuous in an interval [a,b] and if f(a)=/=f(b) then it assumes every value between a and b. Also f(1)<0. Then what? Should I search for values of x such that f(x) becomes +ve?
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  4. #4
    MHF Contributor
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    Quote Originally Posted by Sambit View Post
    Intermediate value theorem states that if a function is continuous in an interval [a,b] and if f(a)=/=f(b) then it assumes every value between a and b. Also f(1)<0. Then what? Should I search for values of x such that f(x) becomes +ve?
    Yes. (And they shouldn't be hard to find. In fact, you only need one such value, to ensure that the polynomial has at least one root. You already know that the number of roots is either 0 or 2, so that will settle it.)
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  5. #5
    Senior Member Sambit's Avatar
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    Yes I got it. Besides, i realised just now that since f ''(x)>0 for all x, the function is convex. So there is either no root or 2 roots. Thank you anyway.
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