Question

Solve $(x^3-4x)\sqrt{x^2-1}=0$.

Answer 1

Given equation is solvable for $x^2-1\ge 0$
or $x\in (-\infty ,-1][1,\infty )$
$(x^3-4x)\sqrt{x^2-1}=0$
or $x(x-2)(x+2)\sqrt{x^2-1}=0$
or $x=0,-2,2,-1,1$
But $x\in (-\infty ,-1]\cup [1,\infty )$
$\to x=\pm 1,\pm 2$.