Solve the Inequality for x (7x)/(x^2-16)>=0

$\frac{7 x}{x^{2} - 16} \geq 0$

Step 1

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

$7 x = 0$

$x^{2} - 16 = 0$

Step 2

Divide each term in $7 x = 0$ by $7$ and simplify.

Divide each term in $7 x = 0$ by $7$ .

$\frac{7 x}{7} = \frac{0}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{0}{7}$

Divide $x$ by $1$ .

$x = \frac{0}{7}$

Simplify the right side.

Divide $0$ by $7$ .

$x = 0$

Step 3

Add $16$ to both sides of the equation.

$x^{2} = 16$

Step 4

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{16}$

Step 5

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 4$

Step 6

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 4$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 4$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 4, - 4$

Step 7

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

$x = 0$

$x = 4, - 4$

Step 8

Consolidate the solutions.

$x = 0, 4, - 4$

Step 9

Find the domain of $\frac{7 x}{x^{2} - 16}$ .

Set the denominator in $\frac{7 x}{x^{2} - 16}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 16 = 0$

Solve for $x$ .

Add $16$ to both sides of the equation.

$x^{2} = 16$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 4$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 4$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 4$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 4, - 4$

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

$(- \infty, - 4) \cup (- 4, 4) \cup (4, \infty)$

Step 10

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 0$

$0 < x < 4$

$x > 4$

Step 11

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 $x < - 4$ to see if it makes the inequality true.

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

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

$\frac{7 (- 6)}{{(- 6)}^{2} - 16} \geq 0$

The left side $- 2.1$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 4 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

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

$\frac{7 (- 2)}{{(- 2)}^{2} - 16} \geq 0$

The left side $1.1\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

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

$\frac{7 (2)}{{(2)}^{2} - 16} \geq 0$

The left side $- 1.1\overline{6}$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

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

$\frac{7 (6)}{{(6)}^{2} - 16} \geq 0$

The left side $2.1$ 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.

$x < - 4$ False

$- 4 < x < 0$ True

$0 < x < 4$ False

$x > 4$ True

$x < - 4$ False

$- 4 < x < 0$ True

$0 < x < 4$ False

$x > 4$ True

Step 12

The solution consists of all of the true intervals.

$- 4 < x \leq 0$ or $x > 4$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$- 4 < x \leq 0 or x > 4$

Interval Notation:

$(- 4, 0] \cup (4, \infty)$

Step 14

Solve the Inequality for x (7x)/(x^2-16)>=0

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