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

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.