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    I worked out this question just fine all the way through until the last part, when subbing in the value x = 4, why do they disregard the negative solution of root 9? Would I be wrong in giving the answer +/- 13 instead of just positive 13 for this question? I can't think of any way you would just 'know' to not use -3 as the root of 9.

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    If you had to work it out at the point (4, 36) then there is no need to sub in the other value, but I can't give a proper answer as I don't know what the question is asking you to do. What is the question?

    This was a stupid post, I didn't read the question properly.
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    9^{-1/2}=\frac{1}{3} not \pm \frac{1}{3}
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    (Original post by BabyMaths)
    9^{-1/2}=\frac{1}{3} not \pm \frac{1}{3}
    Since when? I'm really confused now.
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    (Original post by claret_n_blue)
    If you had to work it out at the point (4, 36) then there is no need to sub in the other value, but I can't give a proper answer as I don't know what the question is asking you to do. What is the question?
    The question and a large part of the solution are attached to the original post.
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    (Original post by Ronove)
    Since when? I'm really confused now.
    Can't blame you there. Plenty of teachers seem to think that \sqrt{4}=\pm 2. It doesn't.
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    (Original post by BabyMaths)
    9^{-1/2}=\frac{1}{3} not \pm \frac{1}{3}
    Since you said this (which shook my understanding to the core by the way ) I found this page http://gmatclub.com/forum/square-roo...ve-114114.html which has blown my mind. Thanks for the response, haha!
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    (Original post by BabyMaths)
    Can't blame you there. Plenty of teachers seem to think that \sqrt{4}=\pm 2. It doesn't.
    Well I am my own teacher, haven't had a maths teacher since doing my GCSEs in 2004! I have no idea how I would ever have found this out without you saying it.
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    (Original post by Ronove)
    Well I am my own teacher, haven't had a maths teacher since doing my GCSEs in 2004! I have no idea how I would ever have found this out without you saying it.
    To clarify if you are asked to solve x^2=4 you write down x = \pm 2.

    If you are asked to evaluate \sqrt 4 then the answer is 2.
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    (Original post by BabyMaths)
    Can't blame you there. Plenty of teachers seem to think that \sqrt{4}=\pm 2. It doesn't.
    Woah! What?
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    (Original post by Mr M)
    To clarify if you are asked to solve x^2=4 you write down x = \pm 2.

    If you are asked to evaluate \sqrt 4 then the answer is 2.
    Thank you!
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    (Original post by claret_n_blue)
    Woah! What?
    What?
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    (Original post by TenOfThem)
    What?
    Why is it just 2?

    Or is this just for exams, in real life it still plus or minus 2 isn't it?
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    Couple of long threads on the maths teachers forum at TES explain why students aren't clear on this.


    http://community.tes.co.uk/forums/p/...4.aspx#7840374

    http://community.tes.co.uk/forums/p/...3.aspx#7132713
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    (Original post by claret_n_blue)
    Why is it just 2?

    Or is this just for exams, in real life it still plus or minus 2 isn't it?
    No

    It never has been

    -2 squared does still =4

    But the square root of 4 is 2
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    (Original post by claret_n_blue)
    Why is it just 2?

    Or is this just for exams, in real life it still plus or minus 2 isn't it?

    \pm \sqrt{4}=\pm 2

     \sqrt{4}= 2
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    (Original post by claret_n_blue)
    Why is it just 2?

    Or is this just for exams, in real life it still plus or minus 2 isn't it?
    No. The square root of x is defined to be the POSITIVE number y which satisfies y^2=x
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    (Original post by TenOfThem)
    -2 squared does still =4

    But the square root of 4 is 2
    Why? How can you not go back the other way?

    (Original post by james22)
    No. The square root of x is defined to be the POSITIVE number y which satisfies y^2=x
    No, I thought the definition was the number which when multiplied by itself gives your original number. This doesn't matter if its positive or negative as (-y)*(-y) = y * y
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    (Original post by claret_n_blue)
    No, I thought the definition was the number which when multiplied by itself gives your original number. This doesn't matter if its positive or negative as (-y)*(-y) = y * y
    You thought wrong then.
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    (Original post by Mr M)
    You thought wrong then.
    http://whatis.techtarget.com/definit...re-root-symbol
 
 
 
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