Solve the Inequality for x (x^2+5x)/(x-3)>=0

$\frac{x^{2} + 5 x}{x - 3} \geq 0$

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x^{2} + 5 x = 0$

$x - 3 = 0$

Factor $x$ out of $x^{2} + 5 x$ .

Factor $x$ out of $x^{2}$ .

$x \cdot x + 5 x = 0$

Factor $x$ out of $5 x$ .

$x \cdot x + x \cdot 5 = 0$

Factor $x$ out of $x \cdot x + x \cdot 5$ .

$x (x + 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 5 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $x (x + 5) = 0$ true.

$x = 0, - 5$

Add $3$ to both sides of the equation.

$x = 3$

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 0, - 5$

$x = 3$

Consolidate the solutions.

$x = 0, - 5, 3$

Find the domain of $\frac{x^{2} + 5 x}{x - 3}$ .

Set the denominator in $\frac{x^{2} + 5 x}{x - 3}$ equal to $0$ to find where the expression is undefined.

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The domain is all values of $x$ that make the expression defined.

$(- \infty, 3) \cup (3, \infty)$

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 0$

$0 < x < 3$

$x > 3$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

$\frac{{(- 8)}^{2} + 5 (- 8)}{(- 8) - 3} \geq 0$

The left side $- 2.\overline{18}$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 5 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

$\frac{{(- 2)}^{2} + 5 (- 2)}{(- 2) - 3} \geq 0$

The left side $1.2$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$\frac{{(2)}^{2} + 5 (2)}{(2) - 3} \geq 0$

The left side $- 14$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\frac{{(6)}^{2} + 5 (6)}{(6) - 3} \geq 0$

The left side $22$ is greater than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 5$ False

$- 5 < x < 0$ True

$0 < x < 3$ False

$x > 3$ True

$x < - 5$ False

$- 5 < x < 0$ True

$0 < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

$- 5 \leq x \leq 0$ or $x > 3$

The result can be shown in multiple forms.

Inequality Form:

$- 5 \leq x \leq 0 or x > 3$

Interval Notation:

$[- 5, 0] \cup (3, \infty)$

Solve the Inequality for x (x^2+5x)/(x-3)>=0

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