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Math Help - algebra

  1. #1
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    algebra

    i have this question

    its just for fun, i DONT want the answer just want someone to tell me what approach is good for such questions

    so

    a+b+c=6

    0<a<b<c

    ab+bc+ca=9

    prove 0<a<1<b<3<c<4

    ------

    i just started on it and i got these ideas

    ab should be < b

    and ab< both ac and bc
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  2. #2
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    Hi

    One possible approach is to consider a, b and c as solutions of a 3rd degree polynomial equation

    We know that if a, b and c are solutions of
    x^3 + \alpha x^2 + \beta x + \gamma = 0
    then
    \alpha = - (a + b + c)
    \beta = ab + bc + ca
    \gamma = - abc

    a, b and c are therefore solutions of
    x^3 -6 x^2 + 9x + \gamma = 0 with \gamma < 0

    Study the variations of the function f(x) = x^3 -6 x^2 + 9x + \gamma and especially the conditions to have 3 roots and you will get your answer
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  3. #3
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    see this is like an 'interview question' which someone told me about

    and it should be solved in a minute with pen and paper

    so im trying to find a simple answer

    thanks tho
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  4. #4
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    also forgive my ignorance, but if you do assume a cubic polynomial, how do you know the solutions for alpha,beta and gamma are those you showed. how did you get those
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  5. #5
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    Because if a, b and c are solutions of
    x^3 + \alpha x^2 + \beta x + \gamma = 0
    then
    x^3 + \alpha x^2 + \beta x + \gamma = (x-a) (x-b) (x-c)

    Develop the second member and you will find
    \alpha = - (a + b + c)
    \beta = ab + bc + ca
    \gamma = - abc
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  6. #6
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    Quote Originally Posted by chogo View Post
    see this is like an 'interview question' which someone told me about

    and it should be solved in a minute with pen and paper

    so im trying to find a simple answer

    thanks tho
    My method allows to find the solution in a minute with pen and paper
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  7. #7
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    ok i did some reading, and i now understand the number of solutions cant exceed the degree

    alpha beta and gamma are the solutions and they are such because of the degrees on the member

    so you get the equation you gave me. Now how should i solve that, i could put it through matlab, but that cheating
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  8. #8
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    Maybe my advice is not relevant because I do not know which is your level in mathematics

    a, b and c being solutions of
    x^3 -6 x^2 + 9x + \gamma = 0 with \gamma < 0

    if you study the variations of the function f(x) = x^3 -6 x^2 + 9x + \gamma by finding the derivative, you will see that f :
    is increasing from 0 to 1 with f(0) = -abc < 0 and f(1) = 4 - abc which must be > 0
    is decreasing between 1 and 3 with f(1) = 4 - abc > 0 and f(3) = -abc < 0
    in increasing between 3 and 4 with f(3) = -abc < 0 and f(4) = 4-abc > 0

    Therefore one of the solutions is between 0 and 1, one is between 1 and 3 and one is between 3 and 4
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  9. #9
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    curious

    and what values do u consider for gamma? isn't there a way to find it?
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  10. #10
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    Here is the graphs of the function f(x) = x^3 -6 x^2 + 9x + \gamma for several values of \gamma between -5 and 1

    The value of \gamma has no impact on the form of the curve because no impact on the derivative, it is only a translation on y axis

    You can see that \gamma must be between -4 and 0 in order to have 3 roots and in this case the roots are between 0 and 1, between 1 and 3 and between 3 and 4

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  11. #11
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    thanks for the help

    one last question

    any recommended topics i should read to further understand this?
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  12. #12
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    and when i open (x-a)(x-b)(x-c)

    i get the coeffs as

    alpha=-c
    beta=ab+bc
    and gamma = -abc

    is that now how i get the members?

    sorry for taking up so much of your time. dont bother to reply if you dont have time, ill try and figure it out for myself
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  13. #13
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    Don' t worry !
    (x-a)(x-b)(x-c) = (x-ax-bx+ab)(x-c)
    (x-a)(x-b)(x-c) = (x-(a+b)x+ab)(x-c)
    (x-a)(x-b)(x-c) = x^3-cx-(a+b)x+(a+b)c+abx-abc
    (x-a)(x-b)(x-c) = x^3-(a+b+c)x+(ac+bc+ab)x-abc
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