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Math Help - Finding variable for fraction = 0

  1. #1
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    Finding variable for fraction = 0

    What approach should i take to finding the values of x for this equation


    \frac{x^3-3x^2+1}{x^3}=0
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  2. #2
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    Find all x for which the the numerator equals 0 and the denominator doesn't.
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  3. #3
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    Quote Originally Posted by derfleurer View Post
    Find all x for which the the numerator equals 0 and the denominator doesn't.
    Could you give some hints on how i could do that
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  4. #4
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     <br />
\frac{x^3-3x^2+1}{x^3}=0<br />

    multiply both side by x^3

     <br />
x^3-3x^2+1=0<br />

    now use the factor theorem

    if f(a)=0 then (x-a) is a factor
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  5. #5
    No one in Particular VonNemo19's Avatar
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    First multiply both sides by x^3 getting

    x^3-3x^2+1=0

    Then recall that the zeros of any polynomial equation of the form a_nx^n+a_{n-1}x^{n-1}......a_1x+a_0=0 can be expressed as a ratio of \frac{p}{q} where p are all possible factors of a_0 and q are the possible factors of the leading coefficient a_n.

    Obviously some of the zeros are of a repeating nature, so you can narrow the field down a little bit. In more complicated expressions, you would normally use synthetic division to test the roots, but here it should be evident which ones are the right choice.

    This could help.




    Box 5.3a: The Cubic Formula

    1997 by Karl Hahn

    --------------------------------------------------------------------------------

    Solving a Cubic
    Four centuries before European mathematicians solved the problems of the cubic and quartic equations, Indian mathematician and astronomer, Bhaskara, had solved these problems. You can see a biography of Bhaskara by clicking here.

    The information on this page is now available in pdf format. Click here to get document (pdf viewer required.
    There are also pdf converters available online if you need to have any pdf document in another format.)

    Note that this material is not at all likely to be on the exam. The formula is not even all that useful. It's given here merely to satisfy your curiosity. So you may skip this box if you like.

    If you are given a cubic equation in the form of


    x3 + px2 + qx + r = 0 eq. 5.3a-1

    and need to solve for x, then the first thing you do is substitute variables. Everywhere you see an x in the cubic, replace it with
    p
    x = u - eq. 5.3a-2
    3

    When you get done squaring and cubing this expression, then substituting stuff back in and gathering like terms, you will get
    u3 + au + b = 0 eq. 5.3a-3a

    where
    p2
    a = q - eq. 5.3a-3b
    3

    and
    pq 2p3
    b = r - + eq. 5.3a-3c
    3 27

    Now compute A and B by

    A =





    eq. 5.3a-4a





    and


    B =





    eq. 5.3a-4b





    Then you have the following solutions for u

    u = A + B eq. 5.3a-5a
    ____
    u = -(1/2)(A + B) + √-3/4 (A - B) eq. 5.3a-5b
    ____
    u = -(1/2)(A + B) - √-3/4 (A - B) eq. 5.3a-5c

    Of course, you know how convert from the u solutions to the x solutions (hint: look at 5.3a-2).

    You may be troubled by the expression, √-3/4, which is not a real number, but is a complex number. Note that A and B will also nonreal complex numbers whenever

    b2 a3
    + < 0 eq. 5.3a-6
    4 27

    So the cubic formula does require that you understand the arithmetic of complex numbers.

    And if you are not up on complex numbers, don't worry. We will be reviewing them in a later section.
    Last edited by VonNemo19; May 17th 2009 at 04:01 PM.
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  6. #6
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    It confuses me that on the graph of this function, where y = 0, x does not = \frac{1}{3}
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