Solve the Inequality for x (x^2+6x)/(x-2)>0

$\frac{x^{2} + 6 x}{x - 2} > 0$

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

$x^{2} + 6 x = 0$

$x - 2 = 0$

Factor $x$ out of $x^{2} + 6 x$ .

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

$x \cdot x + 6 x = 0$

Factor $x$ out of $6 x$ .

$x \cdot x + x \cdot 6 = 0$

Factor $x$ out of $x \cdot x + x \cdot 6$ .

$x (x + 6) = 0$

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

$x = 0$

$x + 6 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $x (x + 6) = 0$ true.

$x = 0, - 6$

Add $2$ to both sides of the equation.

$x = 2$

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

$x = 0, - 6$

$x = 2$

Consolidate the solutions.

$x = 0, - 6, 2$

Find the domain of $\frac{x^{2} + 6 x}{x - 2}$ .

Set the denominator in $\frac{x^{2} + 6 x}{x - 2}$ equal to $0$ to find where the expression is undefined.

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$(- \infty, 2) \cup (2, \infty)$

Use each root to create test intervals.

$x < - 6$

$- 6 < x < 0$

$0 < x < 2$

$x > 2$

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 $x < - 6$ to see if it makes the inequality true.

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

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

$\frac{{(- 8)}^{2} + 6 (- 8)}{(- 8) - 2} > 0$

The left side $- 1.6$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 6 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 6 < x < 0$ and see if this value makes the original inequality true.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

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

The left side $1.8$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 2$ and see if this value makes the original inequality true.

$x = 1$

Replace $x$ with $1$ in the original inequality.

$\frac{{(1)}^{2} + 6 (1)}{(1) - 2} > 0$

The left side $- 7$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$\frac{{(4)}^{2} + 6 (4)}{(4) - 2} > 0$

The left side $20$ 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.

$x < - 6$ False

$- 6 < x < 0$ True

$0 < x < 2$ False

$x > 2$ True

$x < - 6$ False

$- 6 < x < 0$ True

$0 < x < 2$ False

$x > 2$ True

The solution consists of all of the true intervals.

$- 6 < x < 0$ or $x > 2$

The result can be shown in multiple forms.

Inequality Form:

$- 6 < x < 0 or x > 2$

Interval Notation:

$(- 6, 0) \cup (2, \infty)$

Solve the Inequality for x (x^2+6x)/(x-2)>0

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