sajibfin06 wrote:How many of the integers that satisfy the inequality (x+2)(x+3)/(x-2)≥0 are less than 5 ?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Can you also tell me whether (x+2)(x+3)/(x-2)≥0 and (x+2)(x+3)≥0 mean the same thing ?

If question were asked as what the possible range of value of (x+2)(x+3)/(x-2)≥0 , how do we solve them , please explain in detail .
Thnaks in advance .

x + 2 >= 0 OR x + 3 >= 0

x >=-2 or x = -2 or x > -3 or x =-3

x = -1,1,2 doesn't fit

By plugin integer values below 5

possible values of x are 4,3,-2 ,-3


Answer is D = 4

Hi sajibfin06,

To answer your question about "the possible range of values":

(1) Note that x can't be 2; otherwise we have zero in the denominator.
(2) For the fraction to be equal to zero, x could be -2 or -3.
(3) For the fraction to be positive (> 0), the numerator and denominator must have the same sign. Let's use a number line:

x <--------(-3)---(-2)---(-1)---(0)---(1)---(2)--->
x+2 ---------------0+++++++++++++++++++
x+3 --------0++++++++++++++++++++++++
(x+2)(x+3) ++0-----0+++++++++++++++++++
x-2 -------------------------------------0+++

So we see two ranges where (x+2)(x+3) and (x-2) have the same sign: -2 > x > -3, and x > 2.

Putting this together, the possible ranges for x are -2 ≥ x ≥ -3, and x > 2.

And the integer values less than 5 where the inequality is true are x = -3, -2, 3, and 4 (as nikhilgmat31 has said correctly).

Also, as you can see from the number line exercise, (x+2)(x+3)/(x-2) ≥ 0 and (x+2)(x+3) ≥ 0 are not the same thing. For (x+2)(x+3) ≥ 0, either -3 ≥ x or x ≥ -2. After all, since we multiply both sides of the inequality by (x-2), we'd have to flip the inequality sign if x-2 < 0.

Let me know if that answers your questions or not!
Spencer

Thank you -Spencer@Prep4GMAT- very much . I get my problem cleared.

To determine the valid solutions for x, we can use the CRITICAL POINTS approach:
https://www.beatthegmat.com/inequalities ... 76123.html