macjas wrote:
How many of the integers that satisfy the inequality \(\frac{(x+2)(x+3)}{(x-2)} \geq 0\) are less than 5?
A. 1
B. 2
C. 3
D. 4
E. 5
OG 2019 PS14203
We can see that, when x = -2 and x = -3, the fraction evaluates to be ZERO.
So, we already have two values that satisfy the given inequality:
x = -2 and
x = -3Also, when x = 2, the fraction is UNDEFINED
So, we have
three critical points: x = -2, x = -3 and x = 2
These 3 points divide the number line into 4 regions:
Region i: x < -3
Region ii: -3 < x < -2
Region iii: -2 < x < 2
Region iv: x > 2
Let's see what happens with the values of x in each region...
Region i: x < -3If x < -3, then (x+2), (x+3), and (x-2) are all NEGATIVE, which means the fraction evaluates to be a NEGATIVE value
Since we want the fraction to be greater than or equal to 0, none of the values in this region satisfy the inequality.
Region ii: -3 < x < -2If -3 < x < -2, then (x+2) is NEGATIVE, (x+3) is POSITIVE, and (x-2) is NEGATIVE, which means the fraction evaluates to be a POSITIVE value
However, since there are no integers within this range, there's nothing to count here.
Region iii: -2 < x < 2If -2 < x < 2, then (x+2) is POSITIVE, (x+3) is POSITIVE, and (x-2) is NEGATIVE, which means the fraction evaluates to be a NEGATIVE value
Since we want the fraction to be greater than or equal to 0, none of the values in this region satisfy the inequality.
Region iv: x > 2If x > 2, then (x+2) is POSITIVE, (x+3) is POSITIVE, and (x-2) is POSITIVE, which means the fraction evaluates to be a POSITIVE value
In other words, ALL x-values greater than 2 satisfy the given inequality.
However, since we're looking for integer values less than 5, the only x values that satisfy the given inequality are
x = 3 and
x = 4In total, there are
FOUR solutions Answer: D
Cheers,
Brent