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Math Help - Square Root Method

  1. #1
    Member ~berserk's Avatar
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    Square Root Method

    I have the problem: (x+1)^2=3

    the answers are: -1-\sqrt{3}, -1+\sqrt{3}

    I need to know how to arrive at the answer
    Last edited by ~berserk; October 25th 2009 at 02:08 PM.
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    Quote Originally Posted by ~berserk View Post
    I have the problem: (x+1)^2=3

    the answers are: -1-\sqrt{3}, 1-\sqrt{3}

    I need to know how to arrive at the answer
    one of your "answers" is incorrect ...

    (x+1)^2 = 3

    x+1 = \pm \sqrt{3}

    x = -1 \pm \sqrt{3}
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  3. #3
    Member ~berserk's Avatar
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    Corrected, thank you
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by ~berserk View Post
    I have the problem: (x+1)^2=3

    the answers are: -1-\sqrt{3}, -1+\sqrt{3}

    I need to know how to arrive at the answer
    I shan't go through the answer as skeeter has done that (and it's the best method for solving the equation when it's set up as in the question)

    As (a)^2 = (-a)^2 \: , \: a \in \mathbb{R} we cannot be sure if we are getting the negative solution or the positive one. In fact both are equally valid.
    If it is easier you can use the difference of two squares to show this

    a^2 = b^2 \: \rightarrow \: a = b - this is not entirely true:

    If a^2=b^2 then it follows that a^2-b^2=0. By the difference of two squares:

    a^2-b^2 = (a-b)(a+b)=0 \: \therefore a = \pm b
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