Solving $x^2-2x-3<0$ If i have to solve $x^2-2x-3<0$ I would do
$$x+1 < 0, \quad  x-3<0$$
and end up getting  $x<-1$ and $x<3$. Why is it wrong to use the same inequality sign? Shouldn’t both of the linear equations formed from factorising the quadratic use the same inequality sign as the quadratic? 
 A: Hint:  Sketch the graph of $x²-2x -3$. Notice that it is zero when $x=-1$ and $x=3$. Then consider the following areas: $(-\infty,-1)$, $(-1,3)$ and $(3,\infty)$.
A: Ok I get your point.
Let's first assume 
$$x^2-2x-3=0$$
$$(x+1)(x-3)=0$$
This is true when $x=-1$ or when $x=3$. That is, by equating each of the linear term with $0$ one by one.
Now let's come to 
$$(x+1)(x-3)<0$$
So you are asking, to solve this why don't we do the same thing again, i.e. one by one making both the linear terms less than $0$.
But that's not the correct way to do it, because multiplying two negative numbers yields a positive number.
Suppose there are two variables $A$ and $B$ and it is given that, 
$AB<0$
This is true when $A<0$ and $B>0$ because $-ve\times +ve=-ve$ or when $A>0$ and $B<0$ because $+ve\times -ve=-ve$.
Similarly 
$$(x+1)(x-3)<0$$
Either when $x+1<0$ and $x-3>0$ or when $x+1>0$ and $x-3<0$. The former statement is absurd as there exist no real number which is less than $-1$ as well as greater than $3$. Hence the answer is $x\in (-1,3)$
A: Option:
Completing the square:
$(x-1)^2-4<0;$
$(x-1)^2 <4;$
$|x-1| <2;$
$-1 < x <3.$
A: $$ab<0\iff a<0\land b<0$$ is wrong. You can easily find counterexamples.
The truth is
$$ab<0\iff (a<0\land b>0)\lor(a>0\land b<0).$$
A product is negative when the factors have opposite signs.
A: No, because of the rule of signs. 
Anyway, there's a theorem on the sign of quadratic polynomials:

Let $ax^2+bx+c\enspace (a\ne0$) a quadrattic polynomial with real coefficients. Then $ax^2+bx+c$ has the sign of $a$, except between its (real) roots, if any. 

