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    How do I show that a cubic equation only has on real root?
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    (Original post by hpblcparaboloid)
    How do I show that a cubic equation only has on real root?
    Factorise it into (x-b)f(x) where x=b is a real root and f(x) is a quadratic which has discriminant less than 0.
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    The question came up in fp1 numerical methods, am I supposed to use a numerical method??
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    If the curve has no stationary points then f has only one real root.
    You could find stationary the points (if any exist at all) and show using the geometry of the curve that the configuration of the stationary points mean that the curve can only intersect the x axis once.
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    (Original post by hpblcparaboloid)
    The question came up in fp1 numerical methods, am I supposed to use a numerical method??
    Most likely then. Depends on the question. Post it.
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    (Original post by RDKGames)
    Most likely then. Depends on the question. Post it.
    f(x) = x^3 + 8x - 19
    Show that the equation f(x) = 0 has only one real root

    Would differentiating it help?
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    (Original post by hpblcparaboloid)
    f(x) = x^3 + 8x - 19
    Show that the equation f(x) = 0 has only one real root

    Would differentiating it help?
    No- differentiating helps find the gradient, whilst you're trying to find where the line intercepts the x-axis.
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    (Original post by hpblcparaboloid)
    f(x) = x^3 + 8x - 19
    Show that the equation f(x) = 0 has only one real root

    Would differentiating it help?
    It's easy to see that the equation f'(x)=0 has no real solutions and so the curve y=f(x) has no stationary points. This is sufficient to conclude that f has only one real root - can you see why?
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    (Original post by B_9710)
    It's easy to see that the equation f'(x)=0 has no real solutions and so the curve y=f(x) has no stationary points. This is sufficient to conclude that f has only one real root - can you see why?
    why does the fact that f'(x) has no real solutions mean that there are no stationary points?
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    (Original post by hpblcparaboloid)
    why does the fact that f'(x) has no real solutions mean that there are no stationary points?
    Well, what do you normally do when you are trying to find stationary points?
 
 
 
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