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Math Help - Third degree polynomial equations

  1. #1
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    Third degree polynomial equations

    Hi

    I am stuck on a problem :

    [ x^3 + 7 ] / [x^2 + 1 ] = 5

    Question is : How many real roots does this equation have?

    IN this I cant use a graphing calculator. Nor do i have access to even a simple calculator.

    I think this can be done using the Intermediate value theorm (if the value of the function is positive and negative for some values of x then it must pass through X axis because it is continous.) But this does not seem a very scientific method to me. Moreover, this is hit and try.

    Any "proper" method to solve such questions?
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  2. #2
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    Quote Originally Posted by champrock View Post
    Hi

    I am stuck on a problem :

    [ x^3 + 7 ] / [x^2 + 1 ] = 5

    Question is : How many real roots does this equation have?

    IN this I cant use a graphing calculator. Nor do i have access to even a simple calculator.

    I think this can be done using the Intermediate value theorm (if the value of the function is positive and negative for some values of x then it must pass through X axis because it is continous.) But this does not seem a very scientific method to me. Moreover, this is hit and try.

    Any "proper" method to solve such questions?
    Not hard to see (and without calculator) from the IVT that x^3 -5x^2 + 2 has roots in the intervals (-1, 0), (0, 1) and (4, 5).
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  3. #3
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    how did u get those intervals?

    What i had done was that I had differentiated the function  x^3 -5x^2 + 2 and got the points where slope is zero as x=10/3 and x=0.

    At x=0 the function takes the value 2
    and at x=10/3 the function takes the value -16.518.


    So, there must exist atleast one real root where the functions value is 0.

    Any input will be appreciated.
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  4. #4
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    Quote Originally Posted by champrock View Post
    how did u get those intervals?

    What i had done was that I had differentiated the function  x^3 -5x^2 + 2 and got the points where slope is zero as x=10/3 and x=0.

    At x=0 the function takes the value 2
    and at x=10/3 the function takes the value -16.518.


    So, there must exist atleast one real root where the functions value is 0.

    Any input will be appreciated.
    I don't think you need to differentiate:

    x^3 + 7  = 5x^2 +5

    and so x^3 -5x^2 + 2 = 0

    and then follow on from mr fantastic
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  5. #5
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    If you cannot see it easily, use the [HTML]http://en.wikipedia.org/wiki/Synthetic_division[/HTML]
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  6. #6
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    Quote Originally Posted by champrock View Post
    how did u get those intervals?

    What i had done was that I had differentiated the function  x^3 -5x^2 + 2 and got the points where slope is zero as x=10/3 and x=0.

    At x=0 the function takes the value 2
    and at x=10/3 the function takes the value -16.518.


    So, there must exist atleast one real root where the functions value is 0.

    Any input will be appreciated.
    Trying the interval (4, 5) is suggested by the coefficient of x, since 5^3 = 5 (5^2) and so the cubic will equal 2.

    The other intervals involve simple values of x and are easily seen by inspection.
    Last edited by mr fantastic; March 1st 2009 at 03:19 AM. Reason: Corrected typo: Quadratic --> Cubic
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