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# How do you solve $\dfrac{{{x}^{2}}+5x}{x-3}\ge 0$ using a sign chart?

Last updated date: 18th Jun 2024
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Hint: Factorize the numerator of the given expression and substitute both the numerator and denominator equal to 0. Find the values of x and represent them on the number line. Now, find the interval of values of x for which the given inequality statement holds true. Take the union sets of values of x thus obtained to get the answer.

Complete step by step solution:
Here, we have been provided with the inequality $\dfrac{{{x}^{2}}+5x}{x-3}\ge 0$ and we are asked to solve it using a sign chart. That means we have to find the solution set of x.
Now, factoring the numerator of the given inequality we have, taking x common,
$\Rightarrow \dfrac{x\left( x+5 \right)}{x-3}\ge 0$
Substituting the numerator of the above expression equal to 0, we get,
$\Rightarrow x\left( x+5 \right)=0$
$\Rightarrow x=0$ or $x+5=0$
$\Rightarrow x=0$ or $x=-5$
Substituting the denominator of the above expression equal to 0, we get,
$\Rightarrow x-3=0$
$\Rightarrow x=3$
In the next step we have to represent the above obtained values of x on a number line, so we have,

Now, we have to check the sign of the expression $\dfrac{x\left( x+5 \right)}{x-3}$ for different intervals of the values of x. So, let us check them one – by – one.
1. When $-\infty Let us consider the value of x = -6 and check the sign of the function \[\dfrac{x\left( x+5 \right)}{x-3}$. So, at x = -6, we have,
\begin{align} & \Rightarrow \dfrac{x\left( x+5 \right)}{x-3}=\dfrac{-6\left( -6+5 \right)}{-6-3} \\ & \Rightarrow \dfrac{x\left( x+5 \right)}{x-3}=\dfrac{6}{-9} \\ & \Rightarrow \dfrac{x\left( x+5 \right)}{x-3}=\dfrac{-2}{3} \\ \end{align}
Clearly, the function has a negative value in this interval of values of x.
2. When $-5Let us consider the value of x = -1 and check the sign of the function \[\dfrac{x\left( x+5 \right)}{x-3}$. So, at x = -1, we have,
\begin{align} & \Rightarrow \dfrac{x\left( x+5 \right)}{x-3}=\dfrac{-1\left( -1+5 \right)}{-1-3} \\ & \Rightarrow \dfrac{x\left( x+5 \right)}{x-3}=\dfrac{-4}{-4} \\ & \Rightarrow \dfrac{x\left( x+5 \right)}{x-3}=1 \\ \end{align}
Clearly, the function has a positive value in this interval of values of x.
3. When $0Let us consider the value of x = 1 and check the sign of the function \[\dfrac{x\left( x+5 \right)}{x-3}$. So, at x = 1, we have,
\begin{align} & \Rightarrow \dfrac{x\left( x+5 \right)}{x-3}=\dfrac{1\left( 1+5 \right)}{1-3} \\ & \Rightarrow \dfrac{x\left( x+5 \right)}{x-3}=\dfrac{6}{-2} \\ & \Rightarrow \dfrac{x\left( x+5 \right)}{x-3}=-3 \\ \end{align}
Clearly, the function has a negative value in this interval of values of x.
4. When x > 3.
Let us consider the value of x = 4 and check the sign of the function $\dfrac{x\left( x+5 \right)}{x-3}$. So, at x = 4, we have,
\begin{align} & \Rightarrow \dfrac{x\left( x+5 \right)}{x-3}=\dfrac{4\left( 4+5 \right)}{4-3} \\ & \Rightarrow \dfrac{x\left( x+5 \right)}{x-3}=36 \\ \end{align}
Clearly, the function has a positive value in this interval of values of x.
Now, representing the sign of the function on the above drawn number line for the given intervals, we get,

So, the value of the expression $\dfrac{{{x}^{2}}+5x}{x-3}$ is greater than 0 in the interval $-5 3. As we can see that the given inequality in the question also contains an ‘equal to’ sign in addition to the ‘greater than’ sign. So, we need to include the values for which the function \[\dfrac{{{x}^{2}}+5x}{x-3}$ is zero. So, this function we be 0 at the values of x for which the numerator will be 0, i.e., at x = 0 and x = -5. So, the solution set of the given inequality can be given as: -
$\Rightarrow x\in \left[ -5,0 \right]\cup \left( 3,\infty \right)$

Note:
One may note that we cannot include the value x = 3 in our solution set because at this value of x the function $\dfrac{{{x}^{2}}+5x}{x-3}$ will become undefined as the denominator will become 0. Note that there is no formula to solve the above question, so we need to follow the above approach and process to get the solution. Remember that if the ‘equal to’ sign would not have been present in the given inequality then we would not have considered x = 0 and x = -5 in our solution set.