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.
 A: There is only one answer which is $x=1$ as you said. It might be possible however that the book erroneously proceeded with the following steps thus leading them to believe that there were 2 answers.
$${1\over x^2}-1={1\over x}-1$$
$${1\over x^2}={1\over x}$$
$$x^2=x$$
$$x^2-x=0$$
$$x(x-1)=0$$
$$x=1, x=0$$
However by plugging in $0$ into the original equation we get ${1\over 0}-1={1\over 0}-1$. We can therefore discard 0 which leaves us only with 1.
A: The only other solution I can think of is infinity (working in the '(projective) extended real number line')
A: Another way to see it is
$$\frac{1}{x^2} - 1 = \frac{1}{x} -1 \iff \frac{1}{x^2} = \frac{1}{x} \iff x^2 = x \longrightarrow x = 1$$
Where the case $x=0$ has be excluded manually. Your solution is correct.
A: 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...
A: $$\frac{1}{x^2} - 1 = \frac{1}{x} -1$$
$$\frac{1}{x^2}  = \frac{1}{x} $$
$$x^2-x=0$$
$$x(x-1)=0$$
The solutions are $x=1$ or $x=0$.
A: $$\frac{1}{x^2} - 1 = \frac{1}{x} -1$$
$$\frac{1}{x^2}  = \frac{1}{x} $$
multiply by x suppose that $x \ne0$
$$x^2-x=0$$
$$x(x-1)=0$$
The unique solution is  $x=1$ solution $x=0$ is excluded.
