anishasjkaul wrote:

Hi, Bunuel

The trick for inequalities written in the first post of this link, is it always true??... Will it work for all inequalities? Even third degree equation of X ( for example) with an inequality sign?

Yes, it is always true. But a more easy approach is as follows:

Lets have an inequality of degree n of the form

(x-1)(x-2)(-3-x)(x-4)<=0,

Now here, we see that coefficient of highest power of x is negative. My approach is as follows:

Step 1: Make coefficient of highest power positive by changing the sign of inequality

(x-1)(x-2)(x+3)(x-4)>=0

Step 2: Identify critical point.

Four critical points are there -3,1,2,4

Since you have already made coefficient of highest power of x positive in first step, the inequality if always be positive beyond the right most critical point i.e. beyond 4 it is always positive

Step3: Plot the regions.

Now the +ve and -ve regions will alternate

>4 it is positive

2<x<4 it is negative

1<x<2 it is positive

-3<x<1 it is negative

x<-3 it is positive.

In this way you can solve the inequality.

In case we have an even power of a term as shown below:

(x-1)(x-2)^2(x-3)(x-4)>=0

The regions at the critical point 2 will not alternate. We will again start from right most critical point:

x>4 is positive

3<x<4 is negative

2<x<3 is positive

1<x<2 is positivex<1 is negative

In case of an odd power the signs alternate normally.

Hope it makes sense!!!

Kudos if you like!!!