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Math Help - prove the Square root

  1. #1
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    prove the Square root

    i need to use with Rolle's theorem to prove that this Polynomial has no then 2 Square root, Regardless of the values m and n.


    i not find any Square root is can be?
    someone can show me the correct way to Solve it ?


    thanks.
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  2. #2
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    This problem, as stated, does not require you to find any roots, just determine that roots exist. (And not "square roots". Square roots are specifically roots to the equation x^2= a^2.) It's not clear to me what you mean by "no then 2 roots". What does the "then" mean? Do you mean that, for all m and n, it has either 0 or 2 roots, but never 1, 3, or 4?

    Rolle's theorem says that if f(a)= f(b)= 0 then there exist some number, c, between a and b, such that f'(c)= 0. You want to show that x^4- 2x^3+ 6x^2+ mx+ n= 0. If you take that to be f', then f(x)= \frac{1}{5}x^5- \frac{1}{2}x^4+ 2x^3+ \frac{m}{2}x^2+ mx+ C for some number C. You can always choose C to make f= 0 for a given x.
    Last edited by HallsofIvy; December 10th 2010 at 02:17 AM.
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  3. #3
    Senior Member BAdhi's Avatar
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    functions are not printed well sir,

    x^4- 2x^3+ 6x^2+ mx+ n= 0
    f(x)= \frac{1}{5}x^5- \frac{1}{2}x^4+ 2x^3+ \frac{m}{2}x^2+ mx+ C
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  4. #4
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    well yes i mean than 2 root.

    but i not Understand what you Write: "If you take that to be f', then [tex]f(x)= \frac{1}{5}x^5- \frac{1}{2}x^4+ 2x^3+ \frac{m}{2}x^2+ mx+ C[tex]: what is mean?
    did you mean f'=4x^3+6x^2+12x+m?

    thanks
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  5. #5
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    how did you get it ?
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