1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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

6
$\begingroup$

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 $(3,\infty)$.

$\endgroup$
4
$\begingroup$

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 $x\in (-1,3)$

$\endgroup$
1
  • $\begingroup$ @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
3
$\begingroup$

Option:

Completing the square:

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

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

$|x-1| <2;$

$-1 < x <3.$

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .