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Cubic equations - determining the nature of roots

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 ax3+bx2+cx+d=0ax^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
Reply 1
Avatar for BabyMaths
BabyMaths
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 x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−xz−yz)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 x3+3x2+5x−1=0x^3+3x^2+5x-1=0 can be transformed into X3+2X−4=0X^3+2X-4=0 using the substitution x=X−1x=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.
Reply 2
Avatar for plusC
plusC
OP
Original post by BabyMaths
In this case you have a depressed cubic. You can solve it using the identity x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−xz−yz)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 x3+3x2+5x−1=0x^3+3x^2+5x-1=0 can be transformed into X3+2X−4=0X^3+2X-4=0 using the substitution x=X−1x=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
Reply 3
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BabyMaths
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.
Reply 4
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plusC
OP
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
Reply 5
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BabyMaths
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?
Reply 6
Avatar for plusC
plusC
OP
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?
Reply 7
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BabyMaths
Original post by plusC
Is this right?


Yes.
Reply 8
Avatar for RichE
RichE
15
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 ax3+bx2+cx+d=0ax^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?
Reply 9
Avatar for plusC
plusC
OP
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/prospective-students/undergraduate/background quite interesting and helpful

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