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?

  • $\begingroup$ Did you try plugging some $x<-1$ into the original inequality? If you do that, I think you will agree that it doesn't work. $\endgroup$ Apr 12, 2019 at 10:51
  • $\begingroup$ The answer you accepted doesn't address your question ! $\endgroup$
    – user65203
    Apr 12, 2019 at 12:15

5 Answers 5


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 $(3,\infty)$.


Ok I get your point.

Let's first assume



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


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,


This is true when $A<0$ and $B>0$ because $-ve\times +ve=-ve$ or when $A>0$ and $B<0$ because $+ve\times -ve=-ve$.



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 $x\in (-1,3)$

  • $\begingroup$ @UbaidHassan its not sensible to divide by $x$ until and unless you know whether $x$ is positive, negative or zero. And even if you somehow know what x is, i don't know how you will get -3<0? $\endgroup$
    – user585765
    Apr 13, 2019 at 7:16


Completing the square:


$(x-1)^2 <4;$

$|x-1| <2;$

$-1 < x <3.$


$$ab<0\iff a<0\land b<0$$ 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.


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.


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