Hi all,

Today I have been troubled by Q12 from Dr.Siklos's book.

OK, I don't have the Siklos booklet, but I think I can help you on this one anyway.

If you've got (24/b³ )z³ - (6/b)z² - 1 = 0

then if you divide the equation by two and multiply it by 3^0.5 , then letting b = 3^0.5 gives you the equation in the form that you want, with k = 3^0.5 /2

If you didn't know what value of b to choose, then you could work this out by looking at the simultaneous equations given by comparing the coefficients of the polynomial in z, that is

24/b³ = 4c

and

6/b = 3c

(You need to introduce a constant c because you can multiply through the whole new equation by any constant you like.)

So what you've shown is that if you solve the equation 4z³ - 3z = 3^0.5 /2, then the solutions to equation (*) are given by x = z/b - a for the three values of z.

I don't know what it means about the (4/3)s I'm afraid without seeing the question.

Now notice that you can use the first part of the question to help you solve the second. (In these two part questions you almost always need the answer to the first part to let you solve the second.)

Hope that helps (as always, let me know if it doesn't),

James.

James,

$x^3+\alpha x^2+\beta x+\gamma$

can be solved by this method.''
If you could provide a little explanation of what the question wants from us, it would also be helpful towards my understanding.

Once again, thank for your help.

Hal.

Hal,

I'm afraid you're answers aren't quite correct: you need to add 1 to each of the values for x (what you've given are the solutions for y).

For the final part of the question, I'm not really sure what it means. I expect that the situation when it won't work is when the cubic has a pair of non-real solutions, in which case you will run into problems when trying to take the inverse cosine of a complex number.

Try following though this method with the (reasonably arbitrarily chosen) example equation

x³ + x² + x + 1 = 0

and see what happens. (Can you see what the roots of this equation are?)

James.

James,

Yes, (x + 1) is a factor: if you then divide by (x + 1) then you'll get a quadratic and you can find the roots of that. (I take it you *do* know about complex numbers?)

But I suggest you try out this question using that cubic equation - then find out exactly what happens and why the method doesn't work, and then you'll probably understand what the question wants you to say.

James.

Hi James,

Thank you, I see how to factorise the cubic. Is this correct?
If (x+1) is a factor of f(x), then
f(x) = (x + 1)(x² + 1)
when f(x) = 0,
x = -1, and x² = -1, (which gives us a non-real roots, complex root, x = i, -i, right?)

In answer to your query, yep I know about complex numbers, but have still yet to do it thoughly. Is there any thing specific I need to know about complex numbers in regard to this question?

if x has 1 real solution, and 2 complex, it means in terms of our question?:

when $a=-1$ and $b=3^{1/2}$

hence $z=3^{1/2}(\pm i=1)=\cos \theta$,

Ummm, this shows us that $\cos \theta =$ complex, and can't have real roots? Not sure of what to do next?
Thank you again.
Hal.

x = (z/3^0.5 ) + 1,
we have also f(x) = (x + 1)(x² + 1)
= after tidy up = (z³ /3.3^0.5 ) + z² + (z/3^0.5 ) + 1 = 0
hence it is not in the form of 4z³ - 3z = k, since the one we got has a quadratic in it.
Does this answer the question?? Is it ok?

Hal.

That's right, the roots are -1, i and -i.

You don't need to know anything specifically about complex numbers, just to be aware of their existence.

You will find that there's a problem when you get cos(3 theta) = complex - that's why "not all cubic equations of the form ... can be solved by this method"!

Hovever, there's no reason why you should take a=-1 and b=3^½ in general. What I was trying to say, is attempt to do the following question, and see where it goes wrong:

"Show how the cubic equation x³ + x² + x + 1 = 0 (*)
can be arranged to form : 4x³ - 3z = k
by means of substitution y = x + a and z = by, for suitable values of the constants a and b.
Hence find the three roots of the equation (*)"

There's nothing special about the values -1 and 3^½ - they're just what you happen to need in the particular example given in the quesiton.

The point is, that you *can* use this method to solve lots of cubics - in fact I think you can probably use it for *any* cubic with three real roots. Try it, say, for the equation 16x³ - 120x² + 297x - 243 = 0. Provided I've just done my arithmetic correctly, this should work in the same way as before, but with differrent (but nice) values for a, b and theta.

But the method runs into difficulty when you have complex roots. Try these both and see.

James.

James.

Thank you James,
I'll give the one you set a go.
I'll be back with what I get soon.

Hal.

Hi James,

I am doing the x³ + x² + x + 1 = 0 at the moment and have hit a small problem.
using the subs x = y - a, the equation became:
y³ + (2/3)y = (20/27) ???? - is this correct??? (I got a = 1/3)??
so when you sub in y = (z/b), you get:
(z³ /b³ ) + (2/3)*(z/b) = (20/27), compare this to the form that we want it: 4z³ - 3z = k,
we have two equations (like the way you showed me):
(1/b³ ) = 4c, and
(2/3b) = -3c ????? (is this correct???)

then when you solve it you get:
b² = -9/8 - which tells me that something has gone wrong?

Hal.

OK, first I get y³ + (2/3)y = -(20/27), but that is just a minor error.

Then you do get b² = -9/8. This is not really a problem - you can just let b = i3/(2^3/2 ).

The biger problem is when you substitute this value of b in to get 4z³ - 3z = k. You'll get a complex value of k, and you don't know how to take inverse cosines of a complex number.

The point here is that it *shouldn't* work - that's what the last part of the question is asking about. It's not that you've done something wrong.

On the other hand, the other example *will* work.

James.

Thank you James, I'll look through my workings to see where it went Dodo.

I'll now continue to do the other one you set.

You help is much appreciated James.

Hal.

Hi James,

Here is what I got for the cubic, 16x³ - 120x² + 297x - 243 = 0
After the substitition x = y - a, it became:
16y³ - 3y = 0.5?? (got a = -2.5)??
Then sub in y = (z/b)
Hence, (16z³ /b³ ) - (3z/b) = 0.5, and you also have 4z² - 3z = k,
Solving, (16/b³ ) = 4c and (3/b) = ,
gave b = ±2, but lets choose b=+2,
We get (after tidy up) 4z³ - 3z = 1 ???
Hence k = 1.

So, now try to get solutions.

$3\theta =0$, $2\pi$, $4\pi$,

$\theta =0$, $(2\pi /3)$, $(4\pi /3)$

After rearranging and subbing, you get roots as:

$x=(5/2)+(\cos \theta )/2$, where $\theta =0$, $(2\pi /3)$, $(4\pi /3)$.
Is this ok?

Hal.

That's absolutely correct.

So, do you understand what the question's getting on about now?

James.

Thank You James.

I think so. If I'm right, then what you've just written is spot on. However, I'm not entirely convinced that I'm not missing the point myself. But let's not worry about that - there certainly is a problem in the case of non-real roots. It's also possible that there could be a problem in other types of cubics with all real roots, but I don't have time to look into that at the moment.

As far as choosing the values goes, what I *should* have done was to pick the roots -1, i and -i and then multiply out (x + 1)(x - i)(x + i). What I did do wasn't quite as direct at first.

James.

Hi James,

Thanks for you all your time and help for this question. You have increased my understanding of this type of question and methods.

One minor point, I am not sure why we multiplied by constant c, eg: 24³ /b³ = 4c and 6/b = 3c. Could you expand a little on why we do this? Do we need the c, as it cancels out anyway? What would it mean if we left the c out?

Hal.

Hal,

Think about what we were trying to do: we wanted to pick a value of b such that the equation 'was in the form of 4z³ - 3z = k'.

So we had (24/b³ )z³ - (6/b)z = 1. It is clear that no value of b we choose is going to make this equation in exactly the one above. But we don't need to do that, because we know that we can multiply through the equation by a constant c, so our equation is actually equivalent to

c(24/b³ )z³ - c(6/b)z = c

which is where the 'c' comes from.

James.

James,

After some thinking about what you said. I think I understand why the 'c' was needed. But since the c cancels out, would you be wrong to leave it out?

Thanks.
Hal.

If you can get by without it, then ignore it.

My point it, that if you try to convert

(24/b³ )z³ - (6/b)z = 1

into

4z³ - 3z = k

just by choosing a value of b, you *cannot* do this. This should be obvious. Introducing another parameter c was just my way of taking this into account.

Sorry if I am confusing things.

James.

Thanks for all your help James!