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Math Help - quick explanation of a step in completing the square

  1. #1
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    quick explanation of a step in completing the square

    Let's say we have 3n^2+12n=k ---> 3(n^2+4n)=k
    I've noticed that I cannot take that three from the left side and move it to the right (my dividing) before completing the square and using that three to multiply the negative number I get 3(n^2+4n+4-4)=k ----> 3*-4 .....
    But I'm not understanding why I cannot move that 3 over before hand. Can someone explain this in the simple terms I would understand. I'm decent with my math, but .... Well, use your own explanatory discretion.
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  2. #2
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    Quote Originally Posted by mike_302 View Post
    Let's say we have 3n^2+12n=k ---> 3(n^2+4n)=k
    I've noticed that I cannot take that three from the left side and move it to the right (my dividing) before completing the square and using that three to multiply the negative number I get 3(n^2+4n+4-4)=k ----> 3*-4 .....
    But I'm not understanding why I cannot move that 3 over before hand. Can someone explain this in the simple terms I would understand. I'm decent with my math, but .... Well, use your own explanatory discretion.
    Are you solving for n? If so, there's no reason why you can't first divide both sides by 3:

    n^2 + 4n = k/3

    => n^2 + 4n + 4 - 4 = k/3

    => (n + 2)^2 - 4 = k/3

    => (n + 2)^2 = (k/3) + 4

    etc.
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    hmm.. Now I really wish I had the question before where I had received an incorrect answer for, what I had assumed, dividing first. I then went back to that step and reworked the question and got the correct answer. I'll try to find it again, but no promises. Maybe I just messed up a calculation somewhere. Oh well. If you say it can be done then I'll do it that way and see how it works out for me. I trust you and all, but I've just been leaving it in the equation for so long now, in my routine practice, but it would be much easier in many cases to move that coefficient of the square over to the right by division. thanks!
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    Quote Originally Posted by mr fantastic View Post
    Are you solving for n? If so, there's no reason why you can't first divide both sides by 3:

    n^2 + 4n = k/3

    => n^2 + 4n + 4 - 4 = k/3

    => (n + 2)^2 - 4 = k/3

    => (n + 2)^2 = (k/3) + 4

    etc.
    \Rightarrow n + 2 = \pm \sqrt{\frac{k}{3} + 4} = \pm \sqrt{\frac{k + 12}{3}}


    \Rightarrow n = -2 \pm \sqrt{\frac{k + 12}{3}}.
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