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sqrt(x^2) = |x|, but doesn't it also = x from power law?

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by shstylo » Mon Mar 28, 2011 8:57 pm
so I got stuck at a question which uses sqrt(x^2)=|x| which I understand. However, if I'm correct in my power laws isn't it also x?

sqrt(x^2)=(x^2)^1/2, then multiplying in the exponents = x^1 which is x? but clearly this isn't the case?

maybe I'm doing something wrong and the answer could be very obvious.
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by 6983manish » Mon Mar 28, 2011 9:31 pm
shstylo wrote:so I got stuck at a question which uses sqrt(x^2)=|x| which I understand. However, if I'm correct in my power laws isn't it also x?

sqrt(x^2)=(x^2)^1/2, then multiplying in the exponents = x^1 which is x? but clearly this isn't the case?

maybe I'm doing something wrong and the answer could be very obvious.
As per the laws of exponents , value of even power of any number can have the base as +ve or -ve.

As in case of X^2 , value of X can be +ve or -ve.

Please post the exact problem to get the precise help out of the forum.
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by confused mind » Wed Apr 13, 2011 11:45 am
There is a difference in the two although it's bit confusing..
let's say-
x^2=4 => x= 2,-2
but sqrt(4)=2
It would be more clear by this e.g.
sqrt(sqrt(16))=sqrt(4)=2 { bcoz sqrt for negative numbers is not defined)
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