Question

How many of the integers that satisfy the inequality \[\dfrac{{\left( {x + 2} \right)\left( {x + 3} \right)}}{{x - 2}} \geqslant \;0\] are less than \[5\] ?
A. \[1\]
B. \[2\]
C. \[3\]
D. \[4\]
E. \[5\]

Answer 1

Hint:In the above given question, we are given an inequality written as \[\dfrac{{\left( {x + 2} \right)\left( {x + 3} \right)}}{{x - 2}} \geqslant \;0\] . We have to find how many integers which are less than \[5\] can satisfy the given inequality. In order to approach the solution, first we have to find the values of \[x\] for which the inequality gives us the value equal to zero. After that we can find the remaining values of \[x\] , which are less than \[5\] , for which the inequality is greater than zero.

Complete step by step answer:
Given that, the inequality which is written as,
\[ \Rightarrow \dfrac{{\left( {x + 2} \right)\left( {x + 3} \right)}}{{x - 2}} \geqslant \;0\]
We have to find \[x \in \mathbb{Z}\] such that \[x < 5\] .
Now, let us consider the inequality when it is equal to zero.
Therefore, we have the equation
\[ \Rightarrow \dfrac{{\left( {x + 2} \right)\left( {x + 3} \right)}}{{x - 2}} = \;0\]
Multiplying both sides by \[\left( {x - 2} \right)\] , we get
\[ \Rightarrow \left( {x + 2} \right)\left( {x + 3} \right) = \;0\]
This is only possible when \[x = - 2\] or \[x = - 3\] .

Now, let us consider the inequality only considering it greater than zero.That gives us the inequality as,
\[ \Rightarrow \dfrac{{\left( {x + 2} \right)\left( {x + 3} \right)}}{{x - 2}} > 0\]
The inequality is greater than zero only if both the numerator and denominator are either positive or negative.Now, the numerator is only positive when \[x > - 1\]. And the denominator is only positive when \[x > 2\]. But it is given that \[x < 5\].
Hence, we have \[ - 1 < 2 < x < 5\]
This gives the possible values for \[x \in \mathbb{Z}\] as \[x = 3\] and \[x = 4\] .
Therefore, we have the values of \[x \in \mathbb{Z}\] for \[x < 5\] as \[x = - 2, - 3,3,4\] .
Hence, there are four integers less than \[5\] which satisfy the given inequality.

So the correct option is D.

Note:The given inequality \[\dfrac{{\left( {x + 2} \right)\left( {x + 3} \right)}}{{x - 2}} \geqslant \;0\] is not defined when we take the value of \[x\] as \[x = 2\]. This is because, if we take the value \[x = 2\] , then the denominator becomes zero and the numerator is non zero. And we know that if a non zero number is divided by zero then it becomes undefined.