# 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?

• 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. Apr 12, 2019 at 10:51
– user65203
Apr 12, 2019 at 12:15

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 $$(3,\infty)$$.

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 $$x\in (-1,3)$$

• @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?
– user585765
Apr 13, 2019 at 7:16

Option:

Completing the square:

$$(x-1)^2-4<0;$$

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

$$|x-1| <2;$$

$$-1 < x <3.$$

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