Convert to Set Notation 9w^2-4>=0

$9 w^{2} - 4 \geq 0$

Step 1

Solve $w \leq - \frac{2}{3} or w \geq \frac{2}{3}$ .

Add $4$ to both sides of the inequality.

$9 w^{2} \geq 4$

Divide each term in $9 w^{2} \geq 4$ by $9$ and simplify.

Divide each term in $9 w^{2} \geq 4$ by $9$ .

$\frac{9 w^{2}}{9} \geq \frac{4}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 w^{2}}{9} \geq \frac{4}{9}$

Divide $w^{2}$ by $1$ .

$w^{2} \geq \frac{4}{9}$

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{w^{2}} \geq \sqrt{\frac{4}{9}}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| w | \geq \sqrt{\frac{4}{9}}$

Simplify the right side.

Simplify $\sqrt{\frac{4}{9}}$ .

Rewrite $\sqrt{\frac{4}{9}}$ as $\frac{\sqrt{4}}{\sqrt{9}}$ .

$| w | \geq \frac{\sqrt{4}}{\sqrt{9}}$

Simplify the numerator.

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

$| w | \geq \frac{\sqrt{2^{2}}}{\sqrt{9}}$

Pull terms out from under the radical.

$| w | \geq \frac{| 2 |}{\sqrt{9}}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| w | \geq \frac{2}{\sqrt{9}}$

Simplify the denominator.

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

$| w | \geq \frac{2}{\sqrt{3^{2}}}$

Pull terms out from under the radical.

$| w | \geq \frac{2}{| 3 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$| w | \geq \frac{2}{3}$

Write $| w | \geq \frac{2}{3}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$w \geq 0$

In the piece where $w$ is non-negative, remove the absolute value.

$w \geq \frac{2}{3}$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$w < 0$

In the piece where $w$ is negative, remove the absolute value and multiply by $- 1$ .

$- w \geq \frac{2}{3}$

Write as a piecewise.

${\begin{cases} w \geq \frac{2}{3} & w \geq 0 \\ - w \geq \frac{2}{3} & w < 0 \end{cases}$

Find the intersection of $w \geq \frac{2}{3}$ and $w \geq 0$ .

$w \geq \frac{2}{3}$

Divide each term in $- w \geq \frac{2}{3}$ by $- 1$ and simplify.

Divide each term in $- w \geq \frac{2}{3}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- w}{- 1} \leq \frac{\frac{2}{3}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{w}{1} \leq \frac{\frac{2}{3}}{- 1}$

Divide $w$ by $1$ .

$w \leq \frac{\frac{2}{3}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\frac{2}{3}}{- 1}$ .

$w \leq - 1 \cdot \frac{2}{3}$

Rewrite $- 1 \cdot \frac{2}{3}$ as $- \frac{2}{3}$ .

$w \leq - \frac{2}{3}$

Find the union of the solutions.

$w \leq - \frac{2}{3}$ or $w \geq \frac{2}{3}$

Step 2

Use the inequality $w \leq - \frac{2}{3} or w \geq \frac{2}{3}$ to build the set notation.

${w | w \leq - \frac{2}{3} or w \geq \frac{2}{3}}$

Step 3

Convert to Set Notation 9w^2-4>=0

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