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Coefficients of polinom

  1. Mar 10, 2008 #1
    1. The problem statement, all variables and given/known data

    What should the coefficients a,b,c of the polinom [tex]P(x)=x^3+ax^2+bx+c[/tex] be, so his roots [tex]x_1+x_2=x_3[/tex] ?

    2. Relevant equations

    [tex]P(x)=a_n(x-c_1)(x-c_2)...(x-c_n_-_1)(x-c_n)[/tex]

    [tex]c_1+c_2+...+c_n= -\frac{a_n_-_1}{a_n}[/tex]

    [tex]c_1c_2+c_2c_3+...+c_n_-_1c_n= \frac{a_n_-_2}{a_n}[/tex]

    [tex]c_1c_2c_3+c_1c_2c_4+...+c_n_-_2c_n_-_1c_n=-\frac{a_n_-_3}{a_n}[/tex]

    ...........................................................................................

    [tex]c_1c_2...c_n_-_1 + c_1c_2...c_n_-_2c_n+......+c_2c_3...c_n=(-1)^n^-^1 \frac{a_1}{a_n}[/tex]

    [tex]c_1c_2c_3...c_n= (-1)^n \frac{a_0}{a_n}[/tex]

    3. The attempt at a solution

    I don't know where to start from. Anybody have any idea? Thnx for the help.
     
    Last edited: Mar 10, 2008
  2. jcsd
  3. Mar 10, 2008 #2
    I think you forgot parts 2 and 3. ;)
     
  4. Mar 10, 2008 #3

    rock.freak667

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    Normally you should show some working or at least some idea of what you think you must do. But if [itex]P(x)=x^3+ax^2+bx+c[/itex], I can think of many cubic polynomials in which the roots will be [itex]\alpha,\beta,\alpha+\beta[/itex]. I am assuming that you are to find a cubic polynomial with leading coefficient 1 and has roots [itex]\alpha,\beta,\alpha+\beta[/itex].

    Start with the relations of the roots to the coefficients.

    For a cubic polynomial of the form [itex]ax^3+bx^2+cx+d=0[/itex]
    [tex]\sum \alpha = \frac{-b}{a}[/tex]

    [tex]\sum \alpha\beta = \frac{c}{a}[/tex]

    [tex]\sum \alpha\beta\gamma =\frac{d}{a} [/tex]

    (Note: [itex]\sum \alpha[/itex] denotes the sum of the roots taking one at a time)
     
  5. Mar 10, 2008 #4
    Ok... Sorry... Is it ok, now?
     
  6. Mar 10, 2008 #5

    rock.freak667

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    Yes that is how to start but as I was saying before, there are many polynomials whose roots can be [itex]\alpha,\beta,\alpha+\beta[/itex]. Try solving for the roots to be in terms of the coefficients of the polynomial.
     
  7. Mar 10, 2008 #6
    Can you please start just a little bit, with solving, so I can go on?
     
  8. Mar 10, 2008 #7

    rock.freak667

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    [tex]\sum \alpha=\frac{-b}{a}[/tex]

    For P(x),a=1 and b=a

    so that

    [tex]\sum \alpha=\frac{-a}{1}=-a[/tex]

    [tex]\sum \alpha =\alpha+\beta+\gamma=-a[/tex]

    and you know that [itex]\gamma=\alpha+\beta[/tex]

    so for [itex]\sum \alpha[/itex] you really get

    [tex]2\gamma =-a[/tex]

    [tex]\gamma=\frac{-a}{2}[/tex]

    Can you go on from here?
     
  9. Mar 10, 2008 #8
    Ok.
    [tex]x_1+x_2+x_3=-a[/tex]

    [tex]x_1+x_1x_3+x_2x_3=b[/tex]

    [tex]x_1x_2x_3=-c[/tex]

    ------------------------------------

    [tex]x_3+x_3=a[/tex]

    [tex]x_3(x_1+x_2)+(x_3-x_2)x_2=b[/tex]

    [tex](x_3-x_2)x_3=-c[/tex]

    ----------------------------------------

    [tex]2x_3=a[/tex]

    [tex]x^2_3+x_3x_2-x^2_2=b[/tex]

    [tex]x^2_3-x_2x_3=-c[/tex]

    I am stuck here. Can you help me please?
     
  10. Mar 10, 2008 #9

    tiny-tim

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    Theofilius, you haven't used x1 + x2 + x3 …

    I believe in trying to find an easy way.

    We know P(x) only has three roots.

    So P(x) = (x+p)(x+q)(x+r), with r = p + q.

    So P(x) = (x+p)(x+q)(x+p+q).

    So a = 2p + 2q, b = … , c = … ? :smile:
     
  11. Mar 10, 2008 #10
    Don't know how to continue... Can you please give some hint more :D?
     
  12. Mar 10, 2008 #11

    tiny-tim

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    P(x) = (x+p)(x+q)(x+p+q)

    So P(x) =
    x^3
    + x^2(p + q + (p + q))
    + x(pq + p(p+q) + q(p+q))
    + pq(p+q)​

    But we are given that P(x) = x^3 + ax^2 + bx + c.

    So the two expressions must be equal.

    So a = 2p + 2q, b = … , c = … ? :smile:
     
  13. Mar 10, 2008 #12
    [tex]a=2p+2q, b=p^2+q^2+3pq , c=p+q [/tex]

    What is next?
     
  14. Mar 10, 2008 #13

    tiny-tim

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    Careful!

    It's actually [tex]a=2p+2q, b=p^2+q^2+3pq , c=pq(p+q)\,. [/tex]

    And it's simpler if you leave the middle one as [tex]b=(p+q)^2+pq\,.[/tex]

    Anyway, next you divide c by a, giving you pq = 2c/a;

    and then you fiddle around with b … :smile:
     
  15. Mar 10, 2008 #14
    Can you give me just the solution, so I will understand what you do. Like this part by part, I can't "pack" the parts.
     
  16. Mar 10, 2008 #15

    tiny-tim

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    Theofilius, I haven't seen any sign that you do understand what I do.

    You're really bad at algebra.

    It's no good understanding what other people do unless you can do it yourself.

    This is a very simple problem, but you're stiil having difficulty with it.

    You need the practice!
     
  17. Mar 11, 2008 #16
    Ok, how will I practice, when I don't know how to solve this problem. Can you please tell me the principle of solving, so I can do it. What's next?
     
  18. Mar 11, 2008 #17
  19. Mar 12, 2008 #18
    I came up with this system.

    [tex]2x_3 = -a[/tex]
    [tex]x_3^2 - \frac{c}{x_3}=b[/tex]
    [tex]x_1x_2 = - \frac{2c}{a}[/tex]

    I sow in my text book results, and the result is:

    [tex]a^3 - 4ab + 8c = 0[/tex]

    Please help! Thanks.
     
  20. Mar 12, 2008 #19

    tiny-tim

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    Yes, that looks fine. :smile:

    Your only problem is you've lost track of what your target is.

    Your target is to find an equation involving a b and c, but not x1 x2 or x3.

    Can you see how to use the three equations above to get you an equation involving only a b and c? :smile:

    (Try … if you can't do it, I'll help you … but it's actually very easy, so take some time before giving up …)
     
  21. Mar 12, 2008 #20
    Substitute for [itex]x_3[/tex] from the 1st into the 2nd equation, like this:

    [tex]x_3 = - \frac{a}{2}[/tex]

    [tex]\frac{a^2}{4} + \frac{2c}{a} - b =0[/tex]

    Now multiply by 4a, and problem solved.:smile:
     
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