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Math Help - Need help solving.....squares..

  1. #1
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    Exclamation Need help solving.....squares..

    This shouldn't be hard, but I can't ever seem to remember what to do with a problem like so:

    c^2 + 12 = 7c and it asks me to solve it. Do I square root both sides? Or what?

    same goes for ones like:

    (y+4)^2 - 81 = 0

    Thank you for the help.
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  2. #2
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    Alright, I managed to use the quadratic on those, though I am having trouble with one similar to this:

    x^2 + 1 = 4x


    Thanks for the help again.
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  3. #3
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    The second is easier as there are less steps;

    (y+4)^2-81=0
    \implies (y+4)^2=81
    \implies y+4=\pm 9
    \implies y=-4 \pm 9

    Now for the first;

    c^2+12=7c
    \implies c^2-7c+12=0

    We want this in the same form as above i.e. A(c+B)^2+C
    \implies A(c^2+2Bc+B^2)+C
    \implies Ac^2+2ABc+AB^2+C

    Now compare co-effecients;

    c^2=Ac^2 \implies A=1
    7c=2ABc \implies B=\frac {-7}{2}

    12=AB^2+C \implies C=12-(\frac{-7}{2})^2 \implies C=12-\frac {49}{4}
    \implies C=-\frac{1}{4}

    Hence c^2-7c+12=(c-\frac{7}{2})^2-\frac{1}{4}=0

    Now solve like number 2.
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  4. #4
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    Quote Originally Posted by tmanderson View Post
    Alright, I managed to use the quadratic on those, though I am having trouble with one similar to this:

    x^2 + 1 = 4x


    Thanks for the help again.
    So you are allowed to use the quadratic formula? Then you should just make one side equal 0 then use it.

    In my previous post i presumed you wanted to know how to solve the square.
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