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Algebra problem using graphing calc

  1. Feb 26, 2013 #1
    Hey guys I was given an assignment as part of my intro to calculus course, but im not sure what the question is actually asking me to find or solve

    1. The problem statement, all variables and given/known data
    use a graphics calculator to solve the following equation
    2x^3 - 3x^2 - 11x + 6 = 0

    am I looking for the zeros of the function or am I trying to solve for x? and how do I use a graphics calculator to solve this sort of equation? I can solve for x algebraically and I can use my calculator to find the x-intercepts but I don't know how to solve this problem with my calculator.

    3. The attempt at a solution
    I know the zero's of the functions are y= 0 when x= -2, 0.5 and 3 but I dont think this is what the question is asking for??
     
  2. jcsd
  3. Feb 26, 2013 #2
    Well when you have a question like this you will have to factor it out and see where it equals zero ( which values of x make it equal to zero).
     
  4. Feb 26, 2013 #3
    ok so factorising its zeros I would get
    (x-3)(2x-1)(x+2)
     
    Last edited: Feb 26, 2013
  5. Feb 26, 2013 #4
    or is it...

    (x-3)(2x-1)(x+2)
    ---> (2x2 - x - 6x + 3)(x + 2)
    ---> (2x3 - 7x - 3)(x + 2)
    ---> 2x3 + 4x2 - 7x2 - 14x + 3x + 6 = 0

    THEN by collecting like terms I get the original equation...

    ---> 2x3 - 3x2 - 11x + 6 = 0

    so this equation has roots of x= -2, 0.5 and 3
     
    Last edited: Feb 26, 2013
  6. Feb 26, 2013 #5

    eumyang

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    They mean the same thing.

    Assuming you have a TI-84, graph the function, change the window so that all x-intercepts are shown, and hit [2ND]->[TRACE]->Zero. Provide the left and right bounds, and you'll get the zero. Repeat the process as needed until you get all zeros. Check the calculator's manual.
     
  7. Feb 26, 2013 #6

    Mark44

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    You started with an equation, so each subsequent step should involve an equation. The two sides of an equation are separated by =, not --->.
     
  8. Feb 26, 2013 #7

    SammyS

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    This is essentially what eumyang said, but I'll say it again:

    Solving the equation, [itex]\displaystyle \ \ 2x^3 - 3x^2 - 11x + 6 = 0 \,, \ [/itex] for x,

    is equivalent to finding the zeros of the function defined by [itex]\displaystyle \ \ f(x) = 2x^3 - 3x^2 - 11x + 6 \ .[/itex]
     
  9. Feb 26, 2013 #8

    Ray Vickson

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    Which is the same as finding the *roots* of ##2x^3 - 3x^2 - 11x + 6.##
     
  10. Feb 26, 2013 #9

    epenguin

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    And all this, I like to reflect, all polynomial equation solving, a lot of algebra, what fraction would you say Ray Vickson or anyone, of algebra, of math, of useful math? - is down to the peculiar property of 0, the only number that remains itself when multiplied by every other number.
     
  11. Feb 26, 2013 #10
    thanks for the replies everyone, as expected it was as simple as using my TI to find the zeros as eumyang mentioned which I know how to do. It was the wording of the problem which confused me as I was not sure what I was "solving"

    Usually problems of this kind that I have encountered previously are worded something along the lines of "find the zeros of the function y = ........................... or f(x) = .............................."

    and mark44 I know that each side of the equation is seperated by =, the ---> was simply an quick easy way of saying "goes to", it's bad mathematical notation I know so in future Ill be sure to use the correct symbols when posting on the forum :)

    thanks again everyone
    -5ym
     
  12. Feb 27, 2013 #11

    HallsofIvy

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    No, technically, it is not. Functions, including polynomials, have "zeros" but only equations have "roots".
     
  13. Feb 27, 2013 #12

    Ray Vickson

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