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Math Help - Help Needed

  1. #1
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    Help Needed

    First on a fuction f(x)=2x^2-3
    How do I find x such that f(x)=x

    How do I factorise 3x^2y-6xy+12xy^2

    And solve the equation
    1-3y=2(y-8)
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by t=0m View Post
    First on a fuction f(x)=2x^2-3
    How do I find x such that f(x)=x
    You are looking for a value of x such that
    2x^2 - 3 = x

    or
    2x^2 - x - 3 = 0

    See what you can come up with. (Hint: the quadratic factors.)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by t=0m View Post
    How do I factorise 3x^2y-6xy+12xy^2
    All three terms contain a common 3xy. So:
    3x^2y-6xy+12xy^2 = 3xy(x - 2 + 4y)
    is the best we can do.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by t=0m View Post
    And solve the equation
    1-3y=2(y-8)
    Come now. If you are supposed to be able to do the other two you should be able to do this one.

    I'll start you off:
    1-3y=2(y-8)

    1 - 3y = 2y - 16

    Can you take it from here?

    -Dan
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  5. #5
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    The funtion is the top one and you need to make The number you put in (x) = x
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    You are looking for a value of x such that
    2x^2 - 3 = x

    or
    2x^2 - x - 3 = 0

    See what you can come up with. (Hint: the quadratic factors.)

    -Dan
    Quote Originally Posted by t=0m View Post
    The funtion is the top one and you need to make The number you put in (x) = x
    Isn't that what I did?

    -Dan
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