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Math Help - Show that the equation have one real root

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    Show that the equation have one real root

    Show that the cubic equation x^3 + 3x -15 = 0 has only one real root.
    Step-by-step guide is appreciated. Thanks
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    Re: Show that the equation has only one real root

    \Delta=-\left[4(3)^3+27(-15)^2\right]<0.
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    Re: Show that the equation has only one real root

    Quote Originally Posted by Nehushtan View Post
    \Delta=-\left[4(3)^3+27(-15)^2\right]<0.
    Sorry, I don't understand at all, can you explain, please?
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    Junior Member Nehushtan's Avatar
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    Re: Show that the equation has one real root

    \Delta=-(4p^3+27q^2) is the discriminant of the cubic equation x^3+px+q=0.

    • If \Delta > 0 the equation has three distinct real roots.

    • If \Delta=0 the equation has three real roots, not all distinct.

    • If \Delta<0 the equation has one real root and two complex roots.
    Last edited by Nehushtan; March 28th 2013 at 01:08 PM.
    Thanks from alexander9408
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    Re: Show that the equation have one real root

    Alexander
    get the function f(x)=x^3+3x-15
    the first derivative is f'(x) = 3x^2+3 which is > 0 for all real numbers x .this means that the function f(x) =x^3+3x -15 is a monotone increasing function and it will meet once and only once the x-axis ..Thus the equation x^3+3x-15 = 0 has one and only one real root.
    the other two roots wil be conjugate complex numbers .
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