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Completing the square - Please Help watch

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    Fellow Mathematicians

    I've recently stumbled upon a question which I'm not entirely sure of:

    Complete the square on 6x - x^2

    Someone said that you put a minus sign in front of the brackets, but don't know whatsoever as to why this is so. I have never seen such a quadratic which is ordered in this fashion. Can someone please help me solve this and provide help for me to solve something in similar form if i see one again?

    All comments are so much appreciated! Many thanks
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    You can factorize out a "-1", like so: 6x - x^2 = -[x^2 - 6x]. Then complete the square as usual on the parenthesized terms, expanding at the end.
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    You didn't have to come up with a new thread about the same topic.

    If you put a minus in front of the expression, and then change the pluses or minuses within the bracket to the opposite, you get this: -1(x^2 - 6x).

    As you can see, the value of the expressions is the same (you can check by expanding) but now the bit in the brackets is more familiar and you can now complete the square for it.
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    Hey Rupetj.

    So then it becomes -1 (x - 3)^2 and then -1 (x - 3) - 9?
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    (Original post by nick198929)
    Fellow Mathematicians

    I've recently stumbled upon a question which I'm not entirely sure of:

    Complete the square on 6x - x^2

    Someone said that you put a minus sign in front of the brackets, but don't know whatsoever as to why this is so. I have never seen such a quadratic which is ordered in this fashion. Can someone please help me solve this and provide help for me to solve something in similar form if i see one again?

    All comments are so much appreciated! Many thanks
    Someone was me, and I don't know why you've started an identical thread.

    If you had x^2-6x

    I presume you can convert it to (x-3)^2-9

    If it is within a bracket, then it's still the same thing.

    In this case:

    -[x^2-6x]

    goes to

    -[\;(x-3)^2-9\;]

    then just run the minus back through to get:

    -(x-3)^2+9

    and that is completed.

    Hope that helps.

    Edit: Changed bracket type to make it clearer.
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    (Original post by nick198929)
    Hey Rupetj.

    So then it becomes -1 (x - 3)^2 and then -1 [(x - 3)^2 - 9]?
    If you look at the quote, the bit in bold is the bit you are missing.

    This is because everything within the brackets is being multiplied by -1.
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    Thanks Rupertj. GhostWalker, why does it go from - (x - 3)^2 - 9 to - (x - 3) + 9 ? The 9 changes from negative to positive.
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    Basic algebra:

    -(a-b) = -a+b
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    (Original post by nick198929)
    Thanks Rupertj. GhostWalker, why does it go from - (x - 3)^2 - 9 to - (x - 3) + 9 ? The 9 changes from negative to positive.
    Because as I told you in my last post, everything is multiplied by -1, so when you expand, the -9 becomes a +9 as you are multiplying two negatives.
 
 
 
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