Pritishd wrote:

Can someone help me with the process to find the solution to:

\(\frac{(x+2)(x+3)}{(x-2)} >=0\)

The result I have is \(x>2, x<-3, -3<=x<=-2\) and \(-2<x<2\)

I am able to understand the first two solutions but not the last two.

Any inputs would be appreciated. Thanks.

This is really a positive/negative problem, not an algebra problem. You know that because it tells you something is greater than (or equal to) 0. Whenever you see that on the GMAT, you want to translate it in your head as 'is positive'.

Okay, so (x+2)(x+3)/(x-2) is positive (or zero). What does that mean?

When a fraction is positive, then either the top and bottom are both positive, or the top and bottom are both negative.

So, top and bottom are both positive. (x+2)(x+3) is positive, and (x-2) is also positive. When does that happen? Only when x>2. Otherwise, x-2 would come out negative.

Other case: top and bottom are both negative. (x+2)(x+3) is negative, and (x-2) is also negative. That's tricky. If x-2 is negative, then x<2. But when is (x+2)(x+3) negative? Only when one of those terms is negative, and the other one is positive. That

only happens when x is between -2 and -3. So our second scenario happens when x<2, and also -3<=x<=-2. However, we don't have to say 'x<2', because if x is between -3 and -2, you already know that it's less than 2. So that's redundant info.

So we've got two solutions: in the case where top and bottom are both positive, x>2.

In the case where top and bottom are both negative, -3 <= x <= -2.

Try plugging some numbers in there to check!

_________________