Solve the Inequality for x (x^2-9x)/(x-5)<0

$\frac{x^{2} - 9 x}{x - 5} < 0$

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

$x^{2} - 9 x = 0$

$x - 5 = 0$

Factor $x$ out of $x^{2} - 9 x$ .

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

$x \cdot x - 9 x = 0$

Factor $x$ out of $- 9 x$ .

$x \cdot x + x \cdot - 9 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 9$ .

$x (x - 9) = 0$

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

$x = 0$

$x - 9 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

The final solution is all the values that make $x (x - 9) = 0$ true.

$x = 0, 9$

Add $5$ to both sides of the equation.

$x = 5$

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

$x = 0, 9$

$x = 5$

Consolidate the solutions.

$x = 0, 9, 5$

Find the domain of $\frac{x^{2} - 9 x}{x - 5}$ .

Set the denominator in $\frac{x^{2} - 9 x}{x - 5}$ equal to $0$ to find where the expression is undefined.

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

$(- \infty, 5) \cup (5, \infty)$

Use each root to create test intervals.

$x < 0$

$0 < x < 5$

$5 < x < 9$

$x > 9$

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 $x < 0$ to see if it makes the inequality true.

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

$x = - 2$

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

$\frac{{(- 2)}^{2} - 9 \cdot - 2}{(- 2) - 5} < 0$

The left side $- 3.\overline{142857}$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 2$

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

$\frac{{(2)}^{2} - 9 \cdot 2}{(2) - 5} < 0$

The left side $4.\overline{6}$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $5 < x < 9$ to see if it makes the inequality true.

Choose a value on the interval $5 < x < 9$ and see if this value makes the original inequality true.

$x = 7$

Replace $x$ with $7$ in the original inequality.

$\frac{{(7)}^{2} - 9 \cdot 7}{(7) - 5} < 0$

The left side $- 7$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 9$ to see if it makes the inequality true.

Choose a value on the interval $x > 9$ and see if this value makes the original inequality true.

$x = 12$

Replace $x$ with $12$ in the original inequality.

$\frac{{(12)}^{2} - 9 \cdot 12}{(12) - 5} < 0$

The left side $5.\overline{142857}$ is not less than the right side $0$ , which means that the given statement is false.

False

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

$x < 0$ True

$0 < x < 5$ False

$5 < x < 9$ True

$x > 9$ False

$x < 0$ True

$0 < x < 5$ False

$5 < x < 9$ True

$x > 9$ False

The solution consists of all of the true intervals.

$x < 0$ or $5 < x < 9$

The result can be shown in multiple forms.

Inequality Form:

$x < 0 or 5 < x < 9$

Interval Notation:

$(- \infty, 0) \cup (5, 9)$

Solve the Inequality for x (x^2-9x)/(x-5)<0

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