You are Here: Home >< Maths

# Cubic equations - determining the nature of roots Watch

1. 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

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
2. (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 .

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

For example can be transformed into using the substitution .

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.
3. (Original post by BabyMaths)
In this case you have a depressed cubic. You can solve it using the identity .

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

For example can be transformed into using the substitution .

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
4. (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.
5. (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
6. (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?
7. (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?
8. (Original post by plusC)
Is this right?
Yes.
9. (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

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?
10. (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:

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: July 26, 2014
Today on TSR

### Degrees to get rich!

... and the ones that won't

### Women equal with Men?

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.