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
One approach is to determine the CRITICAL POINTS: the values where the lefthand side is EQUAL TO 0 or is UNDEFINED.
The lefthand side is equal to 0 when x=-2 and x=-3.
The lefthand side is undefined when x=2.
We already know that x=-2 and x=-3 are valid solutions because they are where (x+2)(x+3) / x-2 = 0.
To determine the range where (x+2)(x+3) / x-2 > 0, try one integer value to the left and right of each critical point.
x < -3:
Plugging x=-4 into (x+2)(x+3) / x-2 > 0, we get:
(-4+2)(-4+3)/(-4-2) > 0
2/-6 > 0.
Doesn't work.
This means that no value less than -3 will work.
-3<x<-2:
No integer values in this range.
-2<x<2:
Plugging x=0 into (x+2)(x+3) / x-2 > 0, we get:
(0+2)(0+3)/(0-2) > 0
-3 > 0.
Doesn't work.
This means that no value between -2 and 2 will work.
x>2:
Plugging x=3 into (x+2)(x+3) / x-2 > 0, we get:
(3+2)(3+3)/(3-2) > 0
30 > 0.
This works.
This means that ANY VALUE greater than 2 will work.
There are only two integer values between 2 and 5:
3 and 4.
Thus, there are four integer values less than 5 that satisfy the inequality: -3, -2, 3 and 4.
The correct answer is
D.
Other problems that I've solved with the critical point approach:
https://www.beatthegmat.com/inequality-c ... 89518.html
https://www.beatthegmat.com/knewton-q-t89317.html
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