gmat800live
VeritasKarishma apologies for bothering you. I read your inequality series and it's been enlightening.
I was wondering, for a problem such as this one where you have (x+1)(x-2) on numerator and (x-5)(x+3) on denominator... how would you approach it? Clearly the "wavy" method works... but I wonder how would you tackle this with the method you explain in your series "Inequalities with complications".
Am I allowed to multiply by 1 since it's positive? For instance I multiply (x-5)/(x-5) and (x+3)/(x+3) on left side to get (x+1)(x-2)(x-5)(x+3) on numerator and then (x-5)^2 and (x+3)^2 on denominator! This means that I can ignore the denominator since it will be always positive, and hence I end with a structure similar to the ones you tackled on your post (x+1)(x-2)(x-5)(x+3) > 0.
Would that work? Thank you.
You don't need to make it so complicated though what you suggest works too. The reason why factors in the denominator work is simply this:
When is abcd positive? When a, b, c and d all are positive. Or when a and b are positive and c and d are negative. etc etc
When is ab/cd positive? When a, b, c and d all are positive. Or when a and b are positive and c and d are negative. etc etc
Aren't the two cases the same? (Except that c and d cannot be 0 in the second case) Anyway, since the expression needs to be positive, a, b, c and d cannot be 0 in first case either.
Does it matter whether c and d are in numerator or denominator? No.
The signs of a, b, c and d decide the sign of the expression irrespective of whether they are in numerator or denominator. The case is the same here.