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Math Help - need help understanding this....

  1. #1
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    need help understanding this....

    This is similar to my last post....

    g over g-1 equals 8 over g+2

    Ok now what I did was the following;

    g(g+2) = 8(g-1)

    as you can tell I cross multiplied....

    g^2+2g = 8g-8

    now my dilemma is about the g^2. Since I have to solve for g the only thing I see I can do is move the the 2g to the left and the 8 to the right so that it would look like this;

    g^2+8=8g-2g

    I can then combine the 8g and the 2g to get 6g but what about the g^2+8....

    Thanks in advance for all the help!!
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  2. #2
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    Quote Originally Posted by Morzilla View Post
    This is similar to my last post....

    g over g-1 equals 8 over g+2

    Ok now what I did was the following;

    g(g+2) = 8(g-1)

    as you can tell I cross multiplied....

    g^2+2g = 8g-8

    ...
    Hello,

    all your considerations and calculations are OK - only the final step is missing:

    g^2+2g = 8g-8~\iff~g^2-6g=-8 . Now calculate a complete square at the LHS of the equation:

    g^2-6g {\color{red} + 9}=-8{\color{red} + 9}~\iff~(g-3)^2=1 . Calcualte the square-root on both sides:

    |\ g-3\ | = 1~\iff~g=3 \pm 1 ~\iff~g=4~\vee~g=2
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  3. #3
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    Quote Originally Posted by earboth View Post
    Hello,

    all your considerations and calculations are OK - only the final step is missing:

    g^2+2g = 8g-8~\iff~g^2-6g=-8 . Now calculate a complete square at the LHS of the equation:

    g^2-6g {\color{red} + 9}=-8{\color{red} + 9}~\iff~(g-3)^2=1 . Calcualte the square-root on both sides:

    |\ g-3\ | = 1~\iff~g=3 \pm 1 ~\iff~g=4~\vee~g=2
    OK two questions....

    1) what is the LHS

    2) where did the nine come from!?

    Thanks once again!!

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  4. #4
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    Quote Originally Posted by Morzilla View Post
    OK two questions....

    1) what is the LHS

    2) where did the nine come from!?

    Thanks once again!!

    Hi,

    1) LHS = left hand side

    2) I assumed that you are familiar with the method of completing the square:
    As you may know

    (a \pm b)^2 = a^2 \pm 2ab+b^2

    If you have

    g^2 - 6g this is the beginning of a binomial formula. You need a third summand so that you get a complete square. This 3rd summand is here \left(\frac{-6}{2}\right)^2 = (-3)^2 = 9 .
    Thus

    g^2-6g + 9 = (g-3)^2
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  5. #5
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    Quote Originally Posted by earboth View Post
    Hi,

    1) LHS = left hand side

    2) I assumed that you are familiar with the method of completing the square:
    As you may know

    (a \pm b)^2 = a^2 \pm 2ab+b^2

    If you have

    g^2 - 6g this is the beginning of a binomial formula. You need a third summand so that you get a complete square. This 3rd summand is here \left(\frac{-6}{2}\right)^2 = (-3)^2 = 9 .
    Thus

    g^2-6g + 9 = (g-3)^2

    Yes I do remember somewhat about completing the square! ahhhhh ok ok I see......Thanks!!!

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