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Math Help - how to solve cubic function.

  1. #1
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    how to solve cubic function.

    The question says comupte the area by integrating, BUT in order to integrate I have to find the find the intersaction points. How do I find the intersection point of this CUBIC FUNCTION-- X^3+X^2-12. Can anyone help me please?

    Additional Info--

    Well it's area between two curves problem. So In order to do that I have to figure out where the two curves intersect over y- axis.

    I set functions y= x^3-6 and y=x^2+6 equal to each other to find the intersection points.

    then I got x^3-x^2-12=0

    Now I don't know how to solve. I was thinking about quadratic formula, but not sure if that can be used in this case.

    Even if it can, Is there any other ways to factor this x^3-x^2-12=0 function to find the x values?
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    Last edited by skboss; February 25th 2010 at 03:43 PM.
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  2. #2
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    Quote Originally Posted by skboss View Post
    The question says comupte the area by integrating, BUT in order to integrate I have to find the find the intersaction points. How do I find the intersection point of this CUBIC FUNCTION-- X^3+X^2-12. Can anyone help me please?
    intersection with what?
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  3. #3
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    Well it's area between two curves problem. So In order to do that I have to figure out where the two curves intersect over y- axis.

    I set functions y= x^3-6 and y=x^2+6 equal to each other to find the intersection points.

    then I got x^3-x^2-12=0

    Now I don't know how to solve. I was thinking about quadratic formula, but not sure if that can be used in this case.

    Even if it can, Is there any other ways to factor this x^3-x^2-12=0 function to find the x values?
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  4. #4
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    Quote Originally Posted by skboss View Post
    Well it's area between two curves problem. So In order to do that I have to figure out where the two curves intersect over y- axis.

    I set functions y= x^3-6 and y=x^2+6 equal to each other to find the intersection points.

    then I got x^3-x^2-12=0

    Now I don't know how to solve. I was thinking about quadratic formula, but not sure if that can be used in this case.

    Even if it can, Is there any other ways to factor this x^3-x^2-12=0 function to find the x values?
    the curves y = x^3 - 6 and y = x^2 + 6 intersect at only a single point , x \approx 2.676

    solving that cubic equation by hand is not easy.
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  5. #5
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    Thanks. Yeah, I figured that by looking at the calculator, but wondering if someone knew how to do it by hand. But thanks anyone.
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