Question

Solve the following equations:
$x^{-2}-2x^{-1}=8$.

Answer 1

$x^{-2}-2x^{-1}=8\frac{1}{x^2}-\frac{2}{x}=8$

Put $t=\frac{1}{x}$

$t^2-2t-8=0t^2-4t+2t-8=0t(t-4)+2(t-4)=0(t+2)(t-4)=0$

$t=-2,4$

But $x=\frac{1}{t}$

$\therefore x=-\frac{1}{2},\frac{1}{4}$