Graph -9 square root of 2x+1>-18

$- 9 \sqrt{2 x + 1} > - 18$

To remove the radical on the left side of the inequality, square both sides of the inequality.

${(- 9 \sqrt{2 x + 1})}^{2} > {(- 18)}^{2}$

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x + 1}$ as ${(2 x + 1)}^{\frac{1}{2}}$ .

${(- 9 {(2 x + 1)}^{\frac{1}{2}})}^{2} > {(- 18)}^{2}$

Simplify the left side.

Simplify ${(- 9 {(2 x + 1)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- 9 {(2 x + 1)}^{\frac{1}{2}}$ .

${(- 9)}^{2} {({(2 x + 1)}^{\frac{1}{2}})}^{2} > {(- 18)}^{2}$

Raise $- 9$ to the power of $2$ .

$81 {({(2 x + 1)}^{\frac{1}{2}})}^{2} > {(- 18)}^{2}$

Multiply the exponents in ${({(2 x + 1)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$81 {(2 x + 1)}^{\frac{1}{2} \cdot 2} > {(- 18)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$81 {(2 x + 1)}^{\frac{1}{2} \cdot 2} > {(- 18)}^{2}$

Rewrite the expression.

$81 {(2 x + 1)}^{1} > {(- 18)}^{2}$

Simplify.

$81 (2 x + 1) > {(- 18)}^{2}$

Apply the distributive property.

$81 (2 x) + 81 \cdot 1 > {(- 18)}^{2}$

Multiply.

Multiply $2$ by $81$ .

$162 x + 81 \cdot 1 > {(- 18)}^{2}$

Multiply $81$ by $1$ .

$162 x + 81 > {(- 18)}^{2}$

Simplify the right side.

Raise $- 18$ to the power of $2$ .

$162 x + 81 > 324$

Solve for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Subtract $81$ from both sides of the inequality.

$162 x > 324 - 81$

Subtract $81$ from $324$ .

$162 x > 243$

Divide each term in $162 x > 243$ by $162$ and simplify.

Divide each term in $162 x > 243$ by $162$ .

$\frac{162 x}{162} > \frac{243}{162}$

Simplify the left side.

Cancel the common factor of $162$ .

Cancel the common factor.

$\frac{162 x}{162} > \frac{243}{162}$

Divide $x$ by $1$ .

$x > \frac{243}{162}$

Simplify the right side.

Cancel the common factor of $243$ and $162$ .

Factor $81$ out of $243$ .

$x > \frac{81 (3)}{162}$

Cancel the common factors.

Factor $81$ out of $162$ .

$x > \frac{81 \cdot 3}{81 \cdot 2}$

Cancel the common factor.

$x > \frac{81 \cdot 3}{81 \cdot 2}$

Rewrite the expression.

$x > \frac{3}{2}$

Find the domain of $- 9 \sqrt{2 x + 1} + 18$ .

Set the radicand in $\sqrt{2 x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$2 x + 1 \geq 0$

Solve for $x$ .

Subtract $1$ from both sides of the inequality.

$2 x \geq - 1$

Divide each term in $2 x \geq - 1$ by $2$ and simplify.

Divide each term in $2 x \geq - 1$ by $2$ .

$\frac{2 x}{2} \geq \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{- 1}{2}$

Divide $x$ by $1$ .

$x \geq \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{1}{2}$

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

$[- \frac{1}{2}, \infty)$

Use each root to create test intervals.

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

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

$x > \frac{3}{2}$

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 $x < - \frac{1}{2}$ to see if it makes the inequality true.

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

$x = - 3$

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

$- 9 \sqrt{2 (- 3) + 1} > - 18$

The left side is not equal to the right side, which means that the given statement is false.

False

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

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

$x = 0$

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

$- 9 \sqrt{2 (0) + 1} > - 18$

The left side $- 9$ is greater than the right side $- 18$ , which means that the given statement is always true.

True

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

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

$x = 4$

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

$- 9 \sqrt{2 (4) + 1} > - 18$

The left side $- 27$ is not greater than the right side $- 18$ , which means that the given statement is false.

False

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

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

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

$x > \frac{3}{2}$ False

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

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

$x > \frac{3}{2}$ False

The solution consists of all of the true intervals.

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

Graph -9 square root of 2x+1>-18

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