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Thread: square root problem

  1. #1
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    square root problem

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    This is a very basic question.

    when we have sq. root of x2, then we (by law) first square the x and then apply the root. But up till now, I have been dealing with this situation differently. When I see sqrt(x2), I turn the sq. root sign to 1/2 and multiply it with the square on the x. Both the 2s in the numerator and denominators cancel out. But we all know that this is not the right way.
    My question is WHY? Is it PEMDAS? or something else....

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    The identity (x^a)^b = x^(ab) for real numbers x, a, b is only defined for positive x, i.e. when you write sqrt(x^2) = x, you necessarily assume (by definition!) that x > 0, but in the original function sqrt(x^2) x does not have to be positive, therefore functions y=sqrt(x^2) and {y=x AND x >0} are not equivalent (they differ in their domains). An equivalent function to sqrt(x^2) is abs(x).

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    Thanks MikeS.

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