Solving quadratic equation $$\frac{1}{x^2} - 1 = \frac{1}{x} -1$$
Rearranging it I get: $1-x^2=x-x^2$, and so $x=1$. But the question Im doing says to find 2 solutions. How would I find the 2nd solution?
Thanks.

I think it should be emphasised what the salient point is here:
Given the equation
$$
\Phi =\Psi
$$
you may multiply both sides by the same non-zero number $a$ to obtain the equivalent equation
$$
a\Phi =a\Psi.
$$
Multiplying both sides of an equation by 0 may give an equation that's not equivalent to the original equation. 
With your equation, eventually you'll get to the point where you have
$$
\tag{1}{1\over x^2}= {1\over x}.
$$
At this point, if you want to "cancel the $x$'s", you could multiply both sides by
$x^2$ as long as $x^2\ne0$. You need to consider what happens when $x=0$ separately.
$x=0$ is not a solution of (1) in this case, so the solutions of (1) are the non-zero solutions of
$$
1=x.
$$
If you multiplied both sides of (1) by $x^3$, the solutions would be the non-zero solutions of
$$
x=x^2.
$$ 
Your text made an error, most probably, at this stage...
