Solve the Inequality for x (x-3)/(x^2-16)<0

$\frac{x - 3}{x^{2} - 16} < 0$

Step 1

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

$x - 3 = 0$

$x^{2} - 16 = 0$

Step 2

Add $3$ to both sides of the equation.

$x = 3$

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.

$x = 3$

$x = 4, - 4$

Step 8

Consolidate the solutions.

$x = 3, 4, - 4$

Step 9

Find the domain of $\frac{x - 3}{x^{2} - 16}$ .

Set the denominator in $\frac{x - 3}{x^{2} - 16}$ 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.

$x < - 4$

$- 4 < x < 3$

$3 < x < 4$

$x > 4$

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.

$\frac{(- 6) - 3}{{(- 6)}^{2} - 16} < 0$

The left side $- 0.45$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 4 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$\frac{(0) - 3}{{(0)}^{2} - 16} < 0$

The left side $0.1875$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $3 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 4$ and see if this value makes the original inequality true.

$x = 3.5$

Replace $x$ with $3.5$ in the original inequality.

$\frac{(3.5) - 3}{{(3.5)}^{2} - 16} < 0$

The left side $- 0.1\overline{3}$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

$\frac{(6) - 3}{{(6)}^{2} - 16} < 0$

The left side $0.15$ 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.

$x < - 4$ True

$- 4 < x < 3$ False

$3 < x < 4$ True

$x > 4$ False

$x < - 4$ True

$- 4 < x < 3$ False

$3 < x < 4$ True

$x > 4$ False

Step 12

The solution consists of all of the true intervals.

$x < - 4$ or $3 < x < 4$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x < - 4 or 3 < x < 4$

Interval Notation:

$(- \infty, - 4) \cup (3, 4)$

Step 14

Solve the Inequality for x (x-3)/(x^2-16)<0

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