Indeed, but I was not talking about the official answer and hence my disclaimer and the apology

If x=2, then \(\frac{(x+2)(x+3)}{x-2}\) would be undefined and not \(\geq{0}\) because the denominator would become zero. We cannot divide by zero.

Does this make sense?
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Algebraic approach is too long here. I've just picked numbers <5 and stopped by -3 (considering some restrictions as X>2 and X<5 and also testing +/- cases.
But I'm nevertheless interested in the algebraic approach - Bunuel can you check it please.

Expression = 0
Solutions x= -2; -3 and x=2 can not be a solution here - we can not divide by 0

Expression > 0
Case 1 Denominator/Numerator are both positive +/+
x>-2; -3; 2 ---> x>2 which gives us two solutions 3 and 4 (because x must be < 5)

Case 2 Denominator/Numerator are both negativ -/-
Case A (x+2)>0; (x+3)<0; (x-2)<0
x>-2, x<-3, x<2

Case b (x+2)<0; (x+3)>0; (x-2)<0
x<-2, x>-3, x<2

There is no solution for Case A & B - no intersections, so we can't make this expression negativ
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

For these inequality questions, the best way is to represent numbers on a number line. By representing on a number line, we will be easily able to compare the sign of the equations and will result in faster solving of inequalities.

Representing (x+2) on number line

x+2 is +ve (when x>-2)
x+2 is -ve (when x<-2)



Representing (x+3) on number line

x+3 is +ve (when x>-3)
x+3 is -ve (when x<-3)



Representing (x-2) on number line

x-2 is +ve (when x>2)
x-2 is -ve (when x<2)



Combining all these equations.
The representation of these equations on the number line is shown in the figure attached.



From the figure, we can easily deduce that the given equation \((x+2)(x+3)/(x-2) >= 0\) will be satisfied when
1) All three equations are positive
2) Two equations are negative and one equation is positive.
3) Equations give the result as zero.

Such a condition is satisfied for the following integers.

At integers 3 and 4 all the equations are positive. Hence satisfy the inequality
At integers -2 and -3 the value of the equation is zero, hence satisfies the inequality.

So overall 4 integers satisfy the inequality. So option (D) is correct.

MATH REVOLUTION VIDEO SOLUTION:


_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Before diving into this problem we want to first analyze the given inequality:

(x+2)(x+3)/(x-2) ≥ 0

This means that (x+2)(x+3) divided by x-2 is equal to or greater than zero. This provides us with 3 options for the signs of the numerator and denominator to yield a final answer that is either positive or 0:

1) (+)/(+) = positive

2) (-)/(-) = positive

3) 0/any nonzero number = 0

The above options will allow the inequality to hold true. We must be strategic in the numbers that we test. The easiest course of action is to start with option 3. We know that when n = -2 or when n = -3, the numerator of our inequality, (x+2)(x+3), will be zero. Thus, we have found two integer values for x less than 5 that fulfill the inequality. Next let’s focus our attention on option 1.

Option 1 tells us that both the numerator and denominator of (x+2)(x+3)/(x-2), must be positive. We see that when x = 4 or x = 3, we have a positive numerator and a positive denominator. At this point we should notice that we have already found 4 values of x that fulfill the inequality and, at most (according to the answer choices), there could be 5. So let’s consider option 2 to see whether we can find any values for x that make both the numerator and denominator negative. To do this we concentrate on the denominator of the fraction, x-2. We can see right away that the only integers that will make “x – 2” negative are 1, 0, -1, -2, -3, -4, and so on. That is, if x is an integer less than 2, then the denominator will be negative. However, when plugging 1, 0, or -1, into the numerator, we see that the numerator will remain positive and thus the entire fraction will be a negative value. When we plug in -2 or -3 for x, the numerator is 0 and entire fraction is 0. (This is actually option 3 above and we have already included -2 and -3 as part of the solutions.) Lastly, when we plug in -4 or integers less than -4, the numerator will be again positive and thus the entire fraction will result in a negative value. So with this information we can sufficiently determine that there are only 4 integer values less than 5 that fulfill the inequality.

The answer is D.
_________________

if I take x=2 then >=0 !
why not its satisfied ?

Hello,

Here is a short video to explain a smart way of solving this question. If you get the concept explained in the video, the question can be solved in 60 seconds or less.

https://youtu.be/ZIgKyR7oN88

I have covered this concept in detail on my conceptual post today-
https://gmatclub.com/forum/gmat-quant-o ... l#p1836874

All the best!
Maxximus
_________________

GMAT Preparation Online- World's Most Exhaustive GMAT Prep Program | Gain 7-days demo access (includes a full length test)

Here I find it easier to use a method of picking numbers. We have the following range
-4, -3, -2, -1, 0, 1, 2, 3, 4.

1. x= - 4, => (x+2)*(x+3)/(x-2) => (-4+2)(-4+3)/-6 = - 1/3 < 0, doesn’t work

2. x=- 3, => x= - 4, => (x+2)*(x+3)/(x-2) => 0/… = 0, works.

3. x= -2 => 0/-4, => works

4. x = -1 => - 2/3 < 0, doesn’t work

5. x = 0, = > -3, < 0, doesn’t work

6. x = 1, => (-1 +2)*(-1 + 3)/-3 = - 2/3 < 0, doesn’t work

7. x = 2, => (4)*(5)/0 => doesn’t work

8. x = 3, => 5*6/1 = 30 > 0, works,

9. x = 4, => 6*7/2 = 21 > 0, works

So here we have 4 integers (-3, -2, 3 and 4) that are less than 5 and satisfying the inequality.

THE BEST Approach

How many of the integers that satisfy the inequality are less than 5?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Solution:
CASE 1: If x-2> 0 or x > 2
(x+2) (x+3) >= 0
(x+3) (x+2) >= 0
(x-(-3)) (x-(-2)) >= 0
It means either x is less than -2 or greater than 2 but we cannot take negative in this case.
Possible values of x will be 3, 4 again we remember we must take integers less than 2.

CASE 2: If x-2 < 0 or x < - 2
−3≤x≤−2
It is given x<5
Thus, integral solutions would be x = -3, -2
Final Solution are… -3, -2, 3, 4, It is four integers.
The Correct Answer D: