Solve the Inequality for x x^3-27x<0

$x^{3} - 27 x < 0$

Step 1

Convert the inequality to an equation.

$x^{3} - 27 x = 0$

Step 2

Factor $x$ out of $x^{3} - 27 x$ .

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} - 27 x = 0$

Factor $x$ out of $- 27 x$ .

$x \cdot x^{2} + x \cdot - 27 = 0$

Factor $x$ out of $x \cdot x^{2} + x \cdot - 27$ .

$x (x^{2} - 27) = 0$

Step 3

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

$x = 0$

$x^{2} - 27 = 0$

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

Set $x^{2} - 27$ equal to $0$ and solve for $x$ .

Set $x^{2} - 27$ equal to $0$ .

$x^{2} - 27 = 0$

Solve $x^{2} - 27 = 0$ for $x$ .

Add $27$ to both sides of the equation.

$x^{2} = 27$

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

$x = \pm \sqrt{27}$

Simplify $\pm \sqrt{27}$ .

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$x = \pm \sqrt{9 (3)}$

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2} \cdot 3}$

Pull terms out from under the radical.

$x = \pm 3 \sqrt{3}$

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.

$x = 3 \sqrt{3}$

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

$x = - 3 \sqrt{3}$

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

$x = 3 \sqrt{3}, - 3 \sqrt{3}$

Step 6

The final solution is all the values that make $x (x^{2} - 27) = 0$ true.

$x = 0, 3 \sqrt{3}, - 3 \sqrt{3}$

Step 7

Use each root to create test intervals.

$x < - 3 \sqrt{3}$

$- 3 \sqrt{3} < x < 0$

$0 < x < 3 \sqrt{3}$

$x > 3 \sqrt{3}$

Step 8

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 $x < - 3 \sqrt{3}$ to see if it makes the inequality true.

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

$x = - 8$

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

${(- 8)}^{3} - 27 \cdot - 8 < 0$

The left side $- 296$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 \sqrt{3} < x < 0$ to see if it makes the inequality true.

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

$x = - 3$

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

${(- 3)}^{3} - 27 \cdot - 3 < 0$

The left side $54$ is not less than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 3$

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

${(3)}^{3} - 27 \cdot 3 < 0$

The left side $- 54$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 8$

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

${(8)}^{3} - 27 \cdot 8 < 0$

The left side $296$ 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.

$x < - 3 \sqrt{3}$ True

$- 3 \sqrt{3} < x < 0$ False

$0 < x < 3 \sqrt{3}$ True

$x > 3 \sqrt{3}$ False

$x < - 3 \sqrt{3}$ True

$- 3 \sqrt{3} < x < 0$ False

$0 < x < 3 \sqrt{3}$ True

$x > 3 \sqrt{3}$ False

Step 9

The solution consists of all of the true intervals.

$x < - 3 \sqrt{3}$ or $0 < x < 3 \sqrt{3}$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 \sqrt{3} or 0 < x < 3 \sqrt{3}$

Interval Notation:

$(- \infty, - 3 \sqrt{3}) \cup (0, 3 \sqrt{3})$

Step 11

Solve the Inequality for x x^3-27x<0

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