acceptable maths solution

eoins23456

(x-t)^2 is a factor of x^3+3px+c
show that
p=-t^2
c=2t^3

for the question above(2000 q1c) Could you use the fact that if you have a double root like (x-t)^2 then (x-t) is a root of a derivative of the cubic or would this not be an acceptable solution?As in are you allowed use differentiation in an algebra question in an instance like this?Probably a stupid question

jumpguy

Is this HL or OL?

What you do is square out (x-t)^2 to get x^2 - 2tx + t^2

Divide this into x^3 + 3px + c This is long division in algebra...so it's abit hard to do in one's head on the internet, but that's what you do.

The next bit is hard to explain, but basically, your long division has to turn out with zero (no remainder) as it is a factor of x^3 + 3px + c

I shall attempt to explain it though, when you divide a few times and things don't nicely cancel... (THIS ISN'T YOUR QUESTION, I'M MAKING THIS ONE UP)

x^3 + 5tpx + 6tp
x^3 + 4tx + 4p

Change the signs of the bottom line (in this case they'll all be -), so x^3 cancel. The 5tpx - 4tx must equal zero, as it is a factor. The x is common so there's no need to include it.
5tp - 4t = 0
Find t on it's own
5tp = 4t
p = 4t/5t

The 6tp - 4p must also turn out to be zero
6tp - 4p = 0
6tp = 4p
6t = 4
t = 6/4 = 3/2

Look at this for basic algebra long division:
http://www.mathsrevision.net/alevel/pages.php?page=1


This is better explained in T&T4, Algebra 2 chapter, with some good examples there done from start to finish.


I apologise for that poor explanation...it's late.

Podge_irl

eoins23456 wrote: »

Could you use the fact that if you have a double root like (x-t)^2 then (x-t) is a root of a derivative of the cubic or would this not be an acceptable solution?As in are you allowed use differentiation in an algebra question in an instance like this?Probably a stupid question

You are perfectly entitled to use differentiation in an algebra question, however what you have stated isn't true. Having a double root in no way implies that it is also a root of the derivative.

**Timbuk2**

You can't use differentiation in this case, as it isn't Mathematically correct (although I see where you are coming from)

The way jumpguy posted is correct. You will get a remainder of something like (a + pt)x + (2q + pt) [these are made up]

The remainder should be zero, or in this case 0x + 0, therefore equate the co-effecients and let each one equal zero

MathsManiac

The technique is certainly correct. It is indeed true that if (x-t) is a double root of a cubic, then it is also a root of the derivative. (This can easily be shown if you note that the cubic can then be expressed as (ax+b)(x-t)^2. Differentiate this as a product and its clear that (x-t) is a factor.)

I used to correct this paper, although I don't any more, and it's my view that the approach you describe would have been regarded as perfectly satisfactory.

MathsManiac

..and by the way, it is also the technique suggested by John Brennan in his set of solutions to this paper:
http://www.leavingcertsolutions.com/mall/leavingcertsolutions/pdf/leaving_cert_Higher_Paper1_Maths_Solutions.pdf

eoins23456

ah right thats grand so cos the double root has come up in 97,2000 n 2005 and its a handy method.

Ruski

jumpguy wrote: »

The 6tp - 4p must also turn out to be zero
6tp - 4p = 0
6tp = 4p
6t = 4
t = 6/4 = 3/2

It's actually 2/3, since the 6 moves below the line to make 4 divide by 6.

eoins23456

http://www.youtube.com/watch?v=ELoeyQFDi5E

solution to the 2008 algebra question.

Do those viete formulas apply to every cubic equation?

MathsManiac

eoins23456 wrote: »

Do those viete formulas apply to every cubic equation?

Yes. It's the cubic equivalent of the quadratic "alpha and beta stuff".

By the way, that video makes a right meal of part (b). If you're solving a cubic equation that has an integer root (which are the only ones on the LC course) the easiest way to find the integer root is to note that it has to be a factor of the constant term. So in that case, (i.e. the case in the video,) it has to be plus or minus 1, 3 or 9. Once you divide the corresponding factor, you can just solve the remaining quadratic.