Why is $x^2 - 2x > 0$ the same as $x<0\lor x>2$ Why is "$x^2 - 2x > 0$" the same as $x<0\lor x>2$ and "$x^2 - 2x < 0$" same as $0<x<2$? 
 A: In light of @M. Strochyk's, we finally can see what happens for $p$:

A: Notice, the product of any two (non-zero) real numbers will be positive if both the numbers are either positive or both negative $$x^2-2x>0$$ $$x(x-2)>0$$
consider the cases, 


*

*if $x<0$ $\implies (x-2)<0$
hence, $$x(x-2)>0$$

*if $x>2$ $\implies (x-2)>0$
hence, $$x(x-2)>0$$
Similarly, the product of any two (non-zero) real numbers will be negative if both the numbers have opposite signs 

*if $0<x<2$ $\implies (x-2)<0$
hence, $$x(x-2)<0$$
A: My interpretation of your question is that you are wondering why the asymmetry between the "or" of the first and the "and" of the second expression.
As @Ian points out, the first step is to decompose the equation into $x(x-2)\lessgtr 0$ which breaks into four cases:
$$x<0,\ x<2\implies x(x-2)>0$$
$$x<0,\ x>2\implies x(x-2)<0$$
$$x>0,\ x<2\implies x(x-2)<0$$
$$x>0,\ x>2\implies x(x-2)>0$$
Thus, before accounting for the relative magnitude of $0$ and $2$ here we get the half-simplified solutions:
$$x^2-2x>0\iff 0<x>2\ \lor\ 0>x<2$$
$$x^2-2x<0\iff 0<x<2\ \lor\ 0>x>2$$
But now we should consider the effect of our extra knowledge that $0<2$. In the first part we have $x>2$ implies $0<2<x$ so $0<x>2$ is equivalent to $x>2$, and similarly $0>x<2$ is equivalent to $x<0$, while in the second part we have $0<x<2$ which can't be simplified, and $0>x>2>0$ is a contradiction. Thus we get the "official" solutions:
$$x^2-2x>0\iff x<0\lor x>2$$
$$x^2-2x<0\iff 0<x<2$$
Note that the only reason we can perform this simplification is because we know $0<2$. If we were solving $x^2-ax>0$ we would have to stick with the more complicated expression $0<x>a\lor 0>x<a$ (which can also be written $x<\min(0,a)\lor x>\max(0,a)$).
