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Math Help - Completing the square

  1. #1
    ADY
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    Completing the square

    Hi Guys,

    If i've got the completed square : f(x) = \frac{1}{9}(x - 3)^2 - 4

    How do i get this back into the form f(x) = ax^2 + bx + c

    having a slow afternoon,

    Thanks

    Last edited by ADY; October 25th 2009 at 08:08 AM. Reason: Better coding
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  2. #2
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    You would:

    <br />
\frac{1}{9}(x^2 - 6x + 9) - 4

    \frac{1}{9}x^2 - \frac{2}{3}x + 1 - 4

    \frac{1}{9}x^2 - \frac{2}{3}x - 3

    Now complete the square.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by ADY View Post
    Hi Guys,

    If i've got the completed square : f(x) = \frac{1}{9}(x - 3)^2 - 4

    How do i get this back into the form f(x) = ax^2 + bx + c

    having a slow afternoon,

    Thanks

    HI

    Expand the whole things again .

    f(x)=\frac{1}{9}(x^2-6x+9)-4

    =\frac{1}{9}x^2-\frac{2}{3}x+1-4

    then simplify from here .
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  4. #4
    ADY
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    I'm trying to find the x coordinates, would it be easier to factorise rather than use completing the squares? by graphing i know the answers to be -3 and 9, its just showing that algebraically
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by ADY View Post
    I'm trying to find the x coordinates, would it be easier to factorise rather than use completing the squares? by graphing i know the answers to be -3 and 9, its just showing that algebraically
    I find the quadratic formula easiest - especially when a \neq 1. To find the x intercepts set f(x) to 0

    x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

    However if you've got it in completed square form it would be simpler to use that form.

    <br /> <br />
f(x) = \frac{1}{9}(x - 3)^2 - 4 = 0<br />

    (x-3)^2 = 36

    x-3 = \pm 6

    x = 9 or x=-3 as you said
    Last edited by e^(i*pi); October 25th 2009 at 09:22 AM. Reason: tidying up latex
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