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Roots of equations

Discussion in 'General Discussion' started by TheCenturion, 22 Apr 2011.

  1. TheCenturion

    Capodecina

    Joined: 18 Mar 2008

    Posts: 12,755

    Harro everyone

    The textbook doesn't help. We haven't covered it in class. No one is online on facebook that can help.

    OcUK is my last resort :o

    The equation z^3 + 4z^2 - 3z + 1 = 0 has roots a, b and c

    Find the values of:

    a + b + c
    ab + bc + ac
    abc


    Somehow I need to factorise the equation, cos then I'll get values for -a, -b and -c. But how? This question comes up in every exam paper (further pure 1) and I've got no idea how to do it

    Thanks friends :)
     
  2. muon

    Capodecina

    Joined: 8 Nov 2006

    Posts: 20,419

    Location: London

  3. marc_howarth

    Gangster

    Joined: 29 Jul 2005

    Posts: 445

    Location: Matlock

    That's a horrible question.

    Usually there's an obvious root say -1 or 1 which you can use.

    When there's an obvious root to be found, say a. You can divide the cubic by (x - a) and then factorise the remaining quadratic.
     
  4. TheCenturion

    Capodecina

    Joined: 18 Mar 2008

    Posts: 12,755

    I've gotten a + b + c

    That's - 4/1 = -4

    Not sure how to proceed from there
     
  5. TheCenturion

    Capodecina

    Joined: 18 Mar 2008

    Posts: 12,755

    It's a standard FP1 question I'm afraid. There's specific rules for identities that you have to use. But in the text book, these rules are only explained for quadratics, and not cubics :rolleyes:

    Still using the rule, I got the answer to a + b + c, now need to try to get the others.
     
  6. marc_howarth

    Gangster

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    Location: Matlock

    It's a horrible question in terms of having to factorise it. If there are identities that you can use to avoid factorising I guess it's not too bad. Been a while since I did FP1 though
     
  7. Six7th

    Gangster

    Joined: 25 Jan 2011

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    Location: Kent

    Does seem pretty nasty. Usually you'd just put a number in for z until it equals zero, then you know it is a root and the factor is (x minus whatever the number is). Using that you can then create a quadratic factor, and factorise that to get the other two.

    Edit: beaten..and realised it's FP1 not C1 :p
     
  8. TheCenturion

    Capodecina

    Joined: 18 Mar 2008

    Posts: 12,755

    I think I've found the rules. No thanks to the textbook or my teacher :rolleyes:

    a+b+c = -b/a
    ab+bc+ac = c/a
    abc = -d/a

    For the equation ax^3 + bx^2 + cx + d

    so a + b + c = -4/1 = -4
    ab + bc + ac = -3/1 = -3
    abc = -1/1 = -1

    Them's the answers apparently. Still, I'm thinking wtf :o
     
  9. marc_howarth

    Gangster

    Joined: 29 Jul 2005

    Posts: 445

    Location: Matlock

    It's confusing in the way that you are using a, b, c as the roots of the equations as well as the coeffecients of each term.

    think of:
    ax^3 + bx^2 + cx + d = 0 having roots f, g and h:
    then
    f + g + h = -b/a
    fg + gh + fh = c/a
    fgh = -d/a
     
  10. Six7th

    Gangster

    Joined: 25 Jan 2011

    Posts: 287

    Location: Kent

  11. TheCenturion

    Capodecina

    Joined: 18 Mar 2008

    Posts: 12,755

    Yeah that's true. It's only because it was alpha beta gamma initially, but that's very long winded to write.


    Wow thanks mate, they've listed all the identities there! :)
     
  12. muon

    Capodecina

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    Location: London

    When did they start doing rule based mathematics at A-level? How is that pure maths?

    I hope they teach you to prove those rules.
     
    Last edited: 22 Apr 2011
  13. Six7th

    Gangster

    Joined: 25 Jan 2011

    Posts: 287

    Location: Kent

    Unfortunately on me exam board (OCR), there's seems to be very little of deriving things from first principles. They teach you in the book but it's never needed in the exam.
     
  14. TheCenturion

    Capodecina

    Joined: 18 Mar 2008

    Posts: 12,755

    I hate fp1. It's a pile of ****

    I need to get an A but...
     
  15. bhavv

    PermaBanned

    Joined: 14 Nov 2009

    Posts: 13,645

    Maths threads need to be BANNED from these forums for making me feel like a stupid :(
     
  16. Aod

    Sgarrista

    Joined: 7 Oct 2004

    Posts: 8,675

    Location: London

    agreed!
     
  17. Philtor

    Soldato

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    You've put me back on the drink again.
     
    Last edited: 22 Apr 2011
  18. BetaNumeric

    Gangster

    Joined: 6 Mar 2011

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    Location: Scotland, UK

    It's about algebraic identities, how is it not pure mathematics, in the A Level sense?

    It's trivial to prove. In fact you can prove the general result in precisely the same manner but due to lack of LaTeX support I'll do the cubic case.

    f(x) = ax^3 + bx^2 + cx + d. Let the three roots be A, B, C, ie f(A) = f(B) = f(C) = 0. By factorisation theorems if f(A) = 0 then f(x) is exactly divisible by x-A and likewise for B and C. Thus f(x) is exactly divisible by (x-A)(x-B)(x-C). The coefficient of x^3 in that is 1 so we overall scale by a to get f(x) = a(x-A)(x-B)(x-C). This has all been done without doing actual expansions and factorisations. We expand this form of f(x) and equate the two formulae, f(x) = a(x-A)(x-B)(x-C) = ax^3 + a(A+B+C)x^2 + a(AB+BC+CA)x + a(ABC) = ax^3 + bx^2 + cx + d. Therefore b = a(A+B+C), c = a(AB+BC+CA), d = a(ABC). QED.

    The generalise result for order N polynomials is done in a likewise manner.

    Nice signature, though it isn't gauge invariant :p
     
  19. muon

    Capodecina

    Joined: 8 Nov 2006

    Posts: 20,419

    Location: London

    But simply using the identities given in the formula book shows no understanding? It's not even applying the factor theorem. At A-level you begin to understand everything from first principles.

    Thanks. That is a question which requires understanding.
     
    Last edited: 23 Apr 2011
  20. Philtor

    Soldato

    Joined: 20 Nov 2009

    Posts: 5,986

    No s*** :D Surely the bloody computer can work it out? lol !