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Math Help - How many real roots does this equations have?

  1. #1
    Senior Member Sambit's Avatar
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    How many real roots does this equations have?

    Suppose I have an equation f(x) = x^6-x^3+2x^2-3x-1 = 0. How to determine the EXACT number of real roots in this equation? I can say from Descarte's rule of sign that f(x) has 1 negative real root and 1 or 3 positive real roots. But how can I know the EXACT number of real roots?
    Not only this one, I am having problem with many other equations as well. So I want a general method which will work for all algebraic equations.

    Thanks in advance.
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  2. #2
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    Generally the way you solve it is find all rational roots. You should have learned how to find all candidates for rational roots already, by looking at the leading coefficient and the constant term. Test each. When you've found one, call it a, synthetically divide the polynomial by (x-a). Repeat the process until you've found all rational roots. At this point you should be able to take the remaining polynomial and either factor it or use the quadratic to find any remaining roots.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Sambit View Post
    Suppose I have an equation f(x) = x^6-x^3+2x^2-3x-1 = 0. How to determine the EXACT number of real roots in this equation? I can say from Descarte's rule of sign that f(x) has 1 negative real root and 1 or 3 positive real roots. But how can I know the EXACT number of real roots?
    Not only this one, I am having problem with many other equations as well. So I want a general method which will work for all algebraic equations.

    Thanks in advance.
    A real root of a polynomial lies between a adjacent pair of extrema or above the largest extrema or below the smallest extrema.

    So we differentiate to get:



    This by Descartes' rule of signs has one or three positive roots and no negative roots.

    Differentiating again gives:



    which has no negative roots and zero or two positive roots. Now we can show this last has no real roots since it has a single extrema at and it is positive there.

    Hence is increasing and so has exactly one real root, and so the original function has either zero or two real roots, and as it has at least one such it must have exactly two real roots.

    CB
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  4. #4
    Senior Member Sambit's Avatar
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    Thanks. I just have to see whether it applies to all equations in general.
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