Q:

Why the solution set of the equation $\sqrt{x-1}(x^2-1)$ doesn’t include $-1$?

Why the solution set of the equation $\sqrt{x-1}(x^2-1)$ doesn’t include $-1$. If I take $x=-1$ I get $2i\times 0$ which equals $0$ and hence $-1$ should also be considered the solution (along with $+1$)

A:

The function $\sqrt{x-1}(x^2-1)$ is defined so as to take only real values : in other words, its domain does not include $-1$, for which $\sqrt{x-1}$ does not take a real value (and therefore, it cannot be multiplied by $x^2-1$ even if $x^2-1 = 0$, because we don't remain in the domain of reals, so a priori do not know that multiplication is defined there.) 
Therefore, $-1$ cannot be considered to be a root of the equation.

