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Thread: Very simple concept is bugging me.

  1. #1
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    Very simple concept is bugging me.

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    Lets say we have Sqrt[(-3)^2]

    if i solve it the way i know i'll get -3. But recently I was watching a math video online and it said that the solution would be +3. Why is that? I must add that the way I solve this expression is as follows.


    Is this the wrong way of solving the square root?

    Please help


    Regards
    Attached Thumbnails Attached Thumbnails Very simple concept is bugging me.-untitled.jpg  
    Last edited by shadoWizard; 11-07-2012 at 07:23 PM.

  2. #2
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    I think You can not bring out -3 from radical unless you square it.
    your way is correct and because of that when we have (9)1/2----> we have two root -3 and 3 because both make your equation correct.
    (3)^2=9
    (-3)^2=9
    so (9)^(0.5)= 3 or -3
    But you are not eligible for bringing out -3 below the radical. so (-3)^0.5 is not possible.you should square -3 below the radical and then bringing it out. this time you are sure the root was -3 not 3 because you have seen it before squaring it.
    Best

  3. #3
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    lil2011 thanks for the answer. But this is what i needed to ask that why am i not eligible to bring out -3 as you said?

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    Its not defined - as you know the radical of negative number is imaginary
    are you familiar with this kind of numbers: 3+3i
    actually this is 3+3(-1)^(1/2)
    we don't have this concept at GRE exam. its related to advanced math. so that you should only think about real number.
    Best

  5. #5
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    But when I multiply the 2 and 1/2 (in exponents) both the twos are cancelled out so we never touch -3. So before we reach the -ve number, sq. root is cancelled... isn't it?

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    When you omit radical you obtain (1/2) they are not there simultaneously.So that you can not do that with negative number.negative number can not come out unless you say : first I square (-3) so it would be 9 below the radical then I square root from 9 which is +3 and come out.

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