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    (Original post by oShahpo)
    The answer is correct, but the method is wrong.
    Square root of x^2 is not x but + or - x. The way to deal with this is to understand that sqrt of -1 means absolutely nothing if you don't treat as i, because it is not defined in terms of anything but i.
    Therefore, sqrt(x) * sqrt(x) = sqrt(x^2) indeed, but then if you plug x = -1 you get sqrt(1) which is 1 or -1. You see the confusion here?
     \sqrt{x^2}  = |x| .
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    (Original post by JLegion)
    I'm not sure I understand where you are coming from.
    You can't use the word "the" because you're implying uniqueness, there is no unique square root of -1. Hence you cannot say "the square root of -1". Let me know if I can make it clearer. :-)
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    (Original post by timebent)
    so the bit where it says a,b is greater than or equal to 0 surely you applied up there in your first paragraph where you said "a" and "b" are both -1, your second step of \sqrt -1x-1 should thus equal 1 in the end.
    Huh? I said "I know you're thinking..." - I'm not saying that the equation I wrote in my first paragraph is correct, indeed it is as false as they come.

    However it seems that if you use negative numbers you still use the negative regardless
    Not sure what you mean by this...

    so in this case then \sqrt -2 \sqrt 2 should then equal [/latex]\sqrt 4[/latex] right?

    so if i'm right

    then what is the outcome of \sqrt 2 \sqrt -2??
    That would usually be \pm 4i (remember that there are two square roots).
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    (Original post by Zacken)
    You can't use the word "the" because you're implying uniqueness, there is no unique square root of -1. Hence you cannot say "the square root of -1". Let me know if I can make it clearer. :-)
    Makes perfect sense, just a poor choice of words I suppose
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    (Original post by Zacken)
    No, this is wrong. \sqrt{x\cdot x} \neq \sqrt{x}\sqrt{x} when x < 0. Indeed, the definition of i is i^2 = -1, any approach you try to use to derive this will be flawed because it's a definition.
    I perfectly understand that. The reason I subbed in x = -1 is to show that if this was a defined operation it would lead to results which contradict other results obtained using i. In other words I was just trying to show that the operation of rootx * rootx = rootx^2 only works for x >=0.
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    (Original post by Zacken)
    Huh? I said "I know you're thinking..." - I'm not saying that the equation I wrote in my first paragraph is correct, indeed it is as false as they come.



    Not sure what you mean by this...



    That would usually be \pm 4i (remember that there are two square roots).
    Ah right i see, gotta change my brain lol.

    Where 2 positive real numbers which are roots, are multiplied they give a positive answer. so positivexpositive=positive
    but where 2 imaginary number which are roots are multiplied they give a negative answer not a positive(which i thought was right) so negativexneative=negative

    right so it can be either or thanks
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    (Original post by oShahpo)
    The answer is correct, but the method is wrong.
    Square root of x^2 is not x but + or - x. The way to deal with this is to understand that sqrt of -1 means absolutely nothing if you don't treat as i, because it is not defined in terms of anything but i.
    Therefore, sqrt(x) * sqrt(x) = sqrt(x^2) indeed, but then if you plug x = -1 you get sqrt(1) which is 1 or -1. You see the confusion here?
    The reason for this confusion is that sqrt(x) is not defined for negative numbers, if you are using negative numbers you must break it down in terms of i. So in this case it becomes sqrt(x^2) = sqrt(x*x)= sqrt(x) * sqrt(x) = sqrt(-1) ^2 = i^2 = -1 only.
    sqrt(x^2) is |x| not + or - x
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    (Original post by usainlightning)
    sqrt(x^2) is |x| not + or - x
    Not it isn't? (x)^2 = x^2, (-x)^2 = x^2.
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    (Original post by oShahpo)
    Not it isn't? (x)^2 = x^2, (-x)^2 = x^2.
    No, he's right. \sqrt{x^2} = |x|.
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    (Original post by Zacken)
    No, he's right. \sqrt{x^2} = |x|.
    so does that mean that x=|x|??
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    (Original post by timebent)
    so does that mean that x=|x|??
     x=|x| if  x\geq 0 .  x=-|x| if  x<0 .
    Contrary to what you may believe or have been told, you cannot just say that  \sqrt{ (f(x))^2} = f(x) . In actual fact  \sqrt{(f(x))^2}=|f(x)| .
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    (Original post by Zacken)
    No, he's right. \sqrt{x^2} = |x|.
    But isn't -5 a root of 25? How can root 5^2 equal modulus(5) when -5 is also a root?
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    (Original post by B_9710)
     x=|x| if  x\geq 0 .  x=-|x| if  x<0 .
    Contrary to what you may believe or have been told, you cannot just say that  \sqrt{ (f(x))^2} = f(x) . In actual fact  \sqrt{(f(x))^2}=|f(x)| .
    Ah i see the first 2 statements you made were true but the third isn't :/

    So if i square x then whatever comes out is always positive? and modulus always returns a positive number so it works?(my version doesn't but yours does?)
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    (Original post by timebent)
    Ah i see the first 2 statements you made were true but the third isn't :/

    So if i square x then whatever comes out is always positive? and modulus always returns a positive number so it works?(my version doesn't but yours does?)
    All the statements I made are true. You cannot square root a real number and get a negative result.
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    (Original post by B_9710)
    All the statements I made are true. You cannot square root a real number and get a negative result.
    ok thanks
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    (Original post by oShahpo)
    But isn't -5 a root of 25? How can root 5^2 equal modulus(5) when -5 is also a root?
    -5 is a square root of 25, yes. But the \sqrt{} \, \colon\, \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0} function is the principal square root of 25.
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    (Original post by Zacken)
    -5 is a square root of 25, yes. But the \sqrt{} \, \colon\, \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0} function is the principal square root of 25.
    Oh I see, thanks.
 
 
 
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