Example:
\(\sqrt x(x+1)=0\)
This suggests two solutions, \(x=0\) and \(x=-1\).
However, upon plugging these solutions back into the equation, you get
\(\sqrt0(0+1)=0(1)=0\)
which checks out, but
\(\sqrt{-1}(1-1)=0\)
does not because \(\sqrt x\) is defined only for \(x\ge0\) (assuming you're looking for real solutions only). So, we call \(x=-1\) an extraneous solution, and the complete solution set (over the real numbers) is \(x=0\).
