Solve the Inequality for x 2x^3-18x<x^2-9

$2 x^{3} - 18 x < x^{2} - 9$

Subtract $x^{2}$ from both sides of the inequality.

$2 x^{3} - 18 x - x^{2} < - 9$

Convert the inequality to an equation.

$2 x^{3} - 18 x - x^{2} = - 9$

Add $9$ to both sides of the equation.

$2 x^{3} - 18 x - x^{2} + 9 = 0$

Factor the left side of the equation.

Reorder terms.

$2 x^{3} - x^{2} - 18 x + 9 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{3} - x^{2}) - 18 x + 9 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{2} (2 x - 1) - 9 (2 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x^{2} - 9) = 0$

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

$(2 x - 1) (x^{2} - 3^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$(2 x - 1) ((x + 3) (x - 3)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (x + 3) (x - 3) = 0$

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

$2 x - 1 = 0$

$x + 3 = 0$

$x - 3 = 0$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $(2 x - 1) (x + 3) (x - 3) = 0$ true.

$x = \frac{1}{2}, - 3, 3$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < \frac{1}{2}$

$\frac{1}{2} < x < 3$

$x > 3$

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 $x < - 3$ to see if it makes the inequality true.

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

$x = - 6$

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

$2 {(- 6)}^{3} - 18 \cdot - 6 < {(- 6)}^{2} - 9$

The left side $- 324$ is less than the right side $27$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < \frac{1}{2}$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 0$

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

$2 {(0)}^{3} - 18 \cdot 0 < {(0)}^{2} - 9$

The left side $0$ is not less than the right side $- 9$ , which means that the given statement is false.

False

Test a value on the interval $\frac{1}{2} < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $\frac{1}{2} < x < 3$ and see if this value makes the original inequality true.

$x = 2$

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

$2 {(2)}^{3} - 18 \cdot 2 < {(2)}^{2} - 9$

The left side $- 20$ is less than the right side $- 5$ , which means that the given statement is always true.

True

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

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

$x = 6$

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

$2 {(6)}^{3} - 18 \cdot 6 < {(6)}^{2} - 9$

The left side $324$ is not less than the right side $27$ , which means that the given statement is false.

False

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

$x < - 3$ True

$- 3 < x < \frac{1}{2}$ False

$\frac{1}{2} < x < 3$ True

$x > 3$ False

$x < - 3$ True

$- 3 < x < \frac{1}{2}$ False

$\frac{1}{2} < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$x < - 3$ or $\frac{1}{2} < x < 3$

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or \frac{1}{2} < x < 3$

Interval Notation:

$(- \infty, - 3) \cup (\frac{1}{2}, 3)$

Solve the Inequality for x 2x^3-18x<x^2-9

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