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Question: Why isn't -b+sqrt(b^2)=0?

Question:Why isn't -b+sqrt(b^2)=0?

Jeff 75 Maple 14
algebra
August 19 2013
1

I tried the following, but did not get 0. Why not?

>restart;

>b:=3*beta*Pi*(3-4*cos(z)+cos(2*z))/(16*z);

>-b+sqrt(b^2)

 

I know I can do this by hand, but this is part of a larger equation which I narrowed down to this issue.

 
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