$(x^2-4)^2\cdot(x^2-6x-7)=0$
$S.S.=\{x_1,x_2,...,x_n\}$
$\Rightarrow n=?$
Answer is given as $4$. I think it should be $6$ because of multiplicity of the roots. I debated this problem with my classmates but we can't reach a verdict.
What is the number of elements in the Solution Set of this problem?
This seems to be a matter of semantics. When counted with multiplicity, the equation has $6$ roots. But the set of roots has only four elements, because there are only four distinct roots. That is to say, $$\{-2,-2,1,2,2,7\}=\{-2,1,2,7\},$$ because by definition a set does not contain duplicate elements.