1. ## Absolute Inequalities

Hello

I got a question here about absolute values in inequalities that's been bugging me for a while.

The questions is: Solve, for x in terms of a, the inequality

|x^2-3ax+2a^2|<|x^2+3ax-a^2|

where x is a real number, a is a real number and a is not equal to 0.

I tried the graphical method, where i drew two parabolas representing the two quadratic equations, and then compare their values in different segments of the graph when the absolute function has been removed for both graphs, as well as considering two cases when a>0 or a<0.

Eventually I arrived at the solution, which happens to be x>a/2>0 OR x<a/2<0.

However that took me two full pages of working to get the answer. For a simple school assignment question like this, I can't help but wonder if there is any simpler, or non-graphical, method to derive the answer?

Any thoughts?

Thanks.

2. Originally Posted by KocaKola
|x^2-3ax+2a^2|<|x^2+3ax-a^2|
Square both sides,
$(x^2-3ax+2a^2)^2 < (x^2+3ax-a^2)^2$
Rewrite,
$(x^2-3ax+2a^2)^2 - (x^2+3ax-a^2)^2 <0$
Difference of two squares,
$(x^2-3ax+2a^2 - x^2 - 3ax+a^2)(x^2-3ax+2a^2+x^2+3ax-a^2) <0$
Hence,
$(3a^2 - 6ax)(2x^2+a^2) <0$
Thus,
$3a^2 - 6ax >0 \mbox{ and }2x^2+a^2>0$
OR,
$3a^2 - 6ax <0 \mbox{ and }2x^2+a^2 <0$

The rest of the details is up to thee.

3. Thanks a lot!!! That cleared things up. Cheers!