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Cubic equations - determining the nature of roots Watch

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    Hi so I've been working on the Oxford uni maths bridging material and was wondering if there is a simpler way to determine the nature of roots of a standard cubic ax^3+bx^2+cx+d=0

    I know that a discriminant exists for cubics,

    but this seems tricky to remember and I believe it would be better to ''understand'' why this formula is true, or use graphs, turning points etc to determine the nature of roots (if that is possible)

    tl;dr can someone tell me how I'd go about proving this discriminant or how I'd use other methods to determine the natural of roots of a polynomial like x^3+px+q=0
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    (Original post by plusC)
    tl;dr can someone tell me how I'd go about proving this discriminant or how I'd use other methods to determine the natural of roots of a polynomial like x^3+px+q=0
    In this case you have a depressed cubic. You can solve it using the identity x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz).

    You can use a substitution of the form x=X-A to transform any cubic equation into a depressed cubic.

    For example x^3+3x^2+5x-1=0 can be transformed into X^3+2X-4=0 using the substitution x=X-1.

    On the other hand, I don't think anyone expects you to memorise the discriminant. If you ever need it you can look it up.
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    (Original post by BabyMaths)
    In this case you have a depressed cubic. You can solve it using the identity x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz).

    You can use a substitution of the form x=X-A to transform any cubic equation into a depressed cubic.

    For example x^3+3x^2+5x-1=0 can be transformed into X^3+2X-4=0 using the substitution x=X-1.

    On the other hand, I don't think anyone expects you to memorise the discriminant. If you ever need it you can look it up.
    Oh I see and from here -3yz=p and q=y^3+z^3 yes? and then presumably I can use these to find the values of y and z in terms of p and q (or do I need more?)

    Apart from that useful tip, how would I determine the nature of roots of a depressed cubic, take the discriminant of the quadratic formed and compare it to the other factor? or use the formula obviously
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    (Original post by plusC)
    Oh I see and from here -3yz=p and q=y^3+z^3 yes? and then presumably I can use these to find the values of y and z in terms of p and q (or do I need more?)

    Apart from that useful tip, how would I determine the nature of roots of a depressed cubic, take the discriminant of the quadratic formed and compare it to the other factor? or use the formula obviously
    You could also think about stationary points.
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    (Original post by BabyMaths)
    You could also think about stationary points.
    And lastly say if you calculated some stationary points but the solution to dy/dx=0 was actually imaginary e.g. x^2=-p/3 for p>0 which yields imaginary solutions, how would you then determine the number of roots the cubic has if its turning points aren't real? thanks
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    (Original post by plusC)
    And lastly say if you calculated some stationary points but the solution to dy/dx=0 was actually imaginary e.g. x^2=-p/3 for p>0 which yields imaginary solutions, how would you then determine the number of roots the cubic has if its turning points aren't real? thanks
    I've been assuming that your a,b,c and d are real.

    If there's no real x such that dy/dx=0 what does the graph of y=ax^3+bx^2+cx+d look like?

    How many real roots are there then? How many complex roots?
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    (Original post by BabyMaths)
    I've been assuming that your a,b,c and d are real.

    If there's no real x such that dy/dx=0 what does the graph of y=ax^3+bx^2+cx+d look like?

    How many real roots are there then? How many complex roots?
    Yeah they are real, I'm not quite sure but if it doesn't turn would it be like a curvy straight line, or are there points of inflexion to consider

    I'm going to take a guess and say a complex conjugate and one real root (there must be at least one real solution so crosses an axis once but never turns)

    Is this right?
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    (Original post by plusC)
    Is this right?
    Yes.
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    (Original post by plusC)
    Hi so I've been working on the Oxford uni maths bridging material and was wondering if there is a simpler way to determine the nature of roots of a standard cubic ax^3+bx^2+cx+d=0

    I know that a discriminant exists for cubics,

    but this seems tricky to remember and I believe it would be better to ''understand'' why this formula is true, or use graphs, turning points etc to determine the nature of roots (if that is possible)

    tl;dr can someone tell me how I'd go about proving this discriminant or how I'd use other methods to determine the natural of roots of a polynomial like x^3+px+q=0
    If you've eliminated the x^2 term by a translation, then the discriminant is a lot easier to remember and for such a cubic is easy to derive by looking at where the cubic's stationary points are (in relation to the x-axis)..

    Another method of solution though - for such a depressed cubic - is to substitute in x = k cos(t) where k is chosen so that the resulting expression in a multiple of cos(3t). This method works fine for complex roots as well provided you know a little about the complex trig functions.

    What bridging material are you using by the way?
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    (Original post by RichE)
    If you've eliminated the x^2 term by a translation, then the discriminant is a lot easier to remember and for such a cubic is easy to derive by looking at where the cubic's stationary points are (in relation to the x-axis)..

    Another method of solution though - for such a depressed cubic - is to substitute in x = k cos(t) where k is chosen so that the resulting expression in a multiple of cos(3t). This method works fine for complex roots as well provided you know a little about the complex trig functions.

    What bridging material are you using by the way?
    Oh interesting alternate method - will be sure to keep it in mind and try it out later, I'm using this:

    https://www.maths.ox.ac.uk/prospecti...ate/background quite interesting and helpful
 
 
 
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