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Math Help - Finding solution of x

  1. #1
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    Finding solution of x

    Here is the problem: x^(2) ^(1/2) (That's the square root of x squared) = -x consists of ?

    A. zero only
    B. nonpositive real numbers only
    C. positive real numbers only
    d. all real numbers
    e. no real numbers

    Explanation: Choice B is correct. Every positive number n hast two square roots, one positive and the other negative, but the square root of n denotes the positive number whose square is n.

    Why does it only denote the positive number? I am used to puttingi + or - whenever I take the square root of something. I just want to understand this clearly. If I wanted to simplify it more, could I also do that to solve? Thanks.
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  2. #2
    Super Member Quacky's Avatar
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    Re: Finding solution of x

    I hate this with a livid passion myself, but in function theory, in order to define something as a function, it has to 'take' one starting value and 'generate' one ending value. Taking the square root is "one-to-many" - it takes one value, and generates two, and therefore needs to be restricted in order to be classed as a function.

    Other examples include \sqrt[3]{a+ib} and arcsin(x)

    Many say that \sqrt{x^2}=|x| which is the easiest way to "play the mediator" as it were.
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  3. #3
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    Re: Finding solution of x

    Quote Originally Posted by benny92000 View Post
    Here is the problem: x^(2) ^(1/2) (That's the square root of x squared) = -x consists of ?

    A. zero only
    B. nonpositive real numbers only
    C. positive real numbers only
    d. all real numbers
    e. no real numbers

    Explanation: Choice B is correct. Every positive number n hast two square roots, one positive and the other negative, but the square root of n denotes the positive number whose square is n.

    Why does it only denote the positive number? I am used to puttingi + or - whenever I take the square root of something. I just want to understand this clearly. If I wanted to simplify it more, could I also do that to solve? Thanks.
    You say "I am used to putting + or - whenever I take the square root of something." For example, if the problem were to solve " x^2= a" you would write " x= \pm\sqrt{a}". And you would write that precisely because " \sqrt{a}" does NOT include both.
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  4. #4
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    Re: Finding solution of x

    That makes more sense now. Thanks!
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