Problem Solving

kamalj 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

OAD

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 three 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 x = -2 or when x = -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 the 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 any integer less than -4, the numerator again will be 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.

Alternative solution:

If we don't have any clue of how to solve the inequality algebraically, we can solve it by trial and error.

Since we are given that x has to be an integer less than 5, we can first test x = 4:

(4+2)(4+3)/(4-2) = (6)(7)/(2) = 21 ≥ 0 → True!

Since we only need to determine whether the expression (x+2)(x+3)/(x-2) is positive (or negative or 0) when compared to 0, we only need to know the sign of the expression. For example, when we plug in 4 for x, we can say that the resulting factors 6, 7, and 2 are positive, and let's denote the expression as (+)(+)/(+) = (+). (Notice here we are using the notation (+) for positive. We will use (-) if the factor is negative and 0 if the factor is 0. Let's now test x = 3:

(+)(+)/(+) = (+) ≥ 0 → True!

x = 2:

(+)(+)/(0) = undefined → Since we can't have 0 as the denominator, x can't be 2.

x = 1:

(+)(+)/(-) = (-) ≥ 0 → False!

x = 0:

(+)(+)/(-) = (-) ≥ 0 → False!

x = -1:

(+)(+)/(-) = (-) ≥ 0 → False!

x = -2:

(0)(+)/(-) = 0 ≥ 0 → True!

x = -3:

(-)(0)/(-) = 0 ≥ 0 → True!

x = -4:

(-)(-)/(-) = (-) ≥ 0 → False!

From this point, we can see that if x is -5 or less, the top two factors will always be (-) and the bottom factor will also be (-), and therefore, the quotient will always be (-) and will never be greater than or equal to 0. So we have only 4 integer values less than 5 for x that satisfy the inequality, namely, 4, 3, -2 and -3.

Answer: D