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Roots of an equation (help)

  1. Feb 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Heres the question I'm tackling:
    <b> the height of a prism is x-1, width x-2, length x+3 and the volume is 42cm cubed. Find the dimensions</b>


    2. Relevant equations
    Quadratic formula, common factor, family of functions



    3. The attempt at a solution
    I know that 7, 2 and 3 are multiples of 42 but i do not know how to show my work... I have this so far...
    42=(x-1)(x-2)(x+3)
    42=(x-1)(x^2-x-6)
    42=(x^3-x^2-6x-x^2+x+6)
    42=(x^3-2x^2-5x+6)

    and I do not know if im doing things right...
     
    Last edited: Feb 20, 2007
  2. jcsd
  3. Feb 20, 2007 #2
    Is that a rectangular, triangular, or what kind of prism..
     
  4. Feb 20, 2007 #3
    Rectangular prism... sorry for the misleading question
     
  5. Feb 20, 2007 #4
    42=(x-1)(x-2)(x+3)
    42=(x-1)(x^2-x-6)

    ! That should be +x. Makes rest of your calculations wrong.

    I got it to be
    (x^3)-7x-36=0
    or
    (x^3)-7x+6=42

    And that's all I got so far..
    ---
    PS, 42 cubed? That adds another dyanmic..
    Maybe (x^3)-7x+6=(42^3)
    ?
     
  6. Feb 20, 2007 #5
    HEY, using simple random "pluggin in random numbers", I think the "x" is 4..
    Therefore, the sides would be
    (4-1), (4-2), (4+3)
    or
    3, 2, 7

    Now, someone else explain an algebraic way of doing this:D
     
  7. Feb 20, 2007 #6
    Ya well im not really sure how to do all this... but thats a good theory.. It is a rectangular prism and 42 is 42cm^2. Although, I'm not even sure if I'm doing all this right
     
  8. Feb 20, 2007 #7
    Yo ill tinks That works!!! THANKS
     
  9. Feb 20, 2007 #8

    @/@

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  10. Feb 20, 2007 #9
    Although through the pluggin thoery i need to know how to use the right methods to get to 4
     
  11. Feb 20, 2007 #10
    Say you graphed it and that was the only solution
     
  12. Feb 20, 2007 #11
    4 is the correct answer I just do not know how to get to it... My teacher vaguely told me anything about questions like this.
     
  13. Feb 20, 2007 #12
    Graph..
    y=(x^3)-7x-36

    See where the 0 is..at 4.. What does that mean logically I don't know.:D
     
  14. Feb 20, 2007 #13

    cristo

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    To solve the equation in this case, you could use the graphical method; i.e. find the roots of the equation x^3-7x-36=0. However, since this is a real problem, looking for a real, positive length, there are certain conditions that the solution must satisfy. Firstly, we are not interested in complex solutions. We note from above that x cannot be negative, or zero, (since then at least one of the sides has negative length, which has no meaning), and x cannot be 1 or 2 (since in each case this gives us a side with zero length)

    So, simply start plugging in values starting with 3; you won't have to put many values in to obtain the result!
     
    Last edited: Feb 20, 2007
  15. Feb 20, 2007 #14
    I dont have a graphing calculator...
     
  16. Feb 20, 2007 #15

    @/@

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  17. Feb 20, 2007 #16

    cristo

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    Staff Emeritus
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    IMHO, it is not necessary to learn these methods of finding roots to a cubic equation for these simple cases. At least one of the roots of problems of this level will be a small number, so "guessing" a solution is enough. Of course, if one wanted to find all the roots, and has found one a, say, using the "trial and error" method, then the equation can be reduced to a quadratic by dividing (x-a) into it. The remaining polynomial can be solved using the usual quadratic formula.
     
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