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Roots of a cubic polynomial

  1. Aug 29, 2007 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    In the equation [tex]x^3+ax^2+bx+c=0[/tex]
    the coefficients a,b and c are all real. It is given that all the roots are real and greater than 1.
    (i) Prove that [tex]a<-3[/tex]
    (ii)By considering the sum of the squares of the roots,prove that [tex]a^2>2b+3[/tex]
    (iii)By considering the sum of the cubes of the roots,prove that [tex]a^3<-9b-3c-3[/tex]


    2. Relevant equations

    If the roots are A,B and C then A+B+C = a/1=a
    ABC= -c/a
    AB+AC+BC= b/a

    3. The attempt at a solution

    I do not know if there are any other formula for the squares/cubes of roots other than the ones i stated above; If there are any simpler ones please tell me.
    I got out parts (ii) by taking (A+B+C)=a and appropriately squaring it, but I was unable to get out parts (i) and (iii), could someone please help me prove it..thanks
     
    Last edited: Aug 29, 2007
  2. jcsd
  3. Aug 29, 2007 #2
    Since the roots are real we know the polynomial factors into [itex](x-r_1)(x-r_2)(x-r_3)[/itex]. Look at how [itex](x-r_1)(x-r_2)(x-r_3)[/itex] multiplies out and look at a b and c in terms of the roots. For example, we know that c must be negative as [itex]-r_1r_2r_3=c<0[/itex]. We actually know [itex]c<-1[/itex] as each of these roots are greater than 1.
     
    Last edited: Aug 29, 2007
  4. Aug 29, 2007 #3

    rock.freak667

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    thanks, I will re-try it and see now
     
  5. Aug 29, 2007 #4

    dextercioby

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  6. Aug 29, 2007 #5

    dextercioby

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    The point iii) is really tricky.

    [tex] (A+B+C)^3 = A^3 +B^3 +C^3 -3ABC +3(A+B+C)(AB+AC+BC) [/tex]

    which means

    [tex] -a^3 =A^3 +B^3 +C^3 +3c -3ab > 3+3c+9b [/tex] ,

    where i used the fact that the sum of the cubes is larger than 3 and the fact that a is smaller than -3.

    Multiply by -1 and you're done.
     
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