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Roots and factor theorem

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data
    knowing a,b and c are roots 3x^3-x^2-10x+8=0
    show that:
    1) 1/a+1/b+1/c=5/4
    2)a^2+b^2+c^2=61/9

    2. Relevant equations
    factor theorem --> (x-a)(x-b)(x-c)


    3. The attempt at a solution
    can only use factor theorem:
    therefore (x-a)(x-b)(x-c)--> up to: x^3-x^2(a+b+c)+x(ac+bc+ab)-abc
    no idea where to go now!
     
    Last edited: Oct 1, 2012
  2. jcsd
  3. Oct 1, 2012 #2

    tiny-tim

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    welcome to pf!

    hi safat! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)

    (1) = what / what ?

    (2) = (a+b+c)2 minus … ? :wink:
     
  4. Oct 2, 2012 #3
    thanks tiny-tim.

    that's what I got by confrontating the original equation (for the record:3(x3-1/3x2-10/3x+8/3)=0 )
    and x3-x2(a+b+c)+x(ac+bc+ab)-abc=0

    1/a+1/b+1/c=(ac+bc+ab)/abc=(-10/3)/(-8/3)=5/4

    About part two:
    should I use the quadratic formula or can I go by guessing?
    I got (a+b+c)2-2(ac+bc+ab)=(-1/3)2-2(-10/3)=61/9
     
  5. Oct 2, 2012 #4

    tiny-tim

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    hi safat! :smile:

    (btw, no such word as "confrontate" … i think you mean "compare"! :wink:)
    yup! :biggrin:
    (i'm not sure what you mean by the quadratic formula, but anyway …)

    yes, guessing is fine, so long as you check that your guess works! :smile:
     
  6. Oct 2, 2012 #5
    thanks again tiny-tim!
    you also provide a really efficient proof reading service! :cool:

    About the second part..
    tbh, I have done it by guessing. But it looks like the great majority of my peers used the quadratic formula (x=-b+-(√b2-4ac)/2a)
    I was wondering how it could be done that way.
     
  7. Oct 2, 2012 #6

    tiny-tim

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    beats me! :rolleyes:

    your way is better :smile:
     
  8. Oct 2, 2012 #7

    uart

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    It's not really guessing. You have made use of some simple relationships between the roots and the coefficients (which you derived and which can be fairly easily memorized). The object of this type of question is generally to solve it without explicitly finding the roots.

    Of course if it happens that you can explicitly find all the roots then that is a viable option. In this particular example the roots turned out to be easy to find, but what if the question was slightly different, say [itex]3 x^3 - x^2 -10x +7 = 0[/itex]. Your method would still work just as easily, but your class mates would presumably find the going much tougher.
     
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