Solve the Inequality for z 64z^2-9<0

$64 z^{2} - 9 < 0$

Step 1

Add $9$ to both sides of the inequality.

$64 z^{2} < 9$

Step 2

Divide each term in $64 z^{2} < 9$ by $64$ and simplify.

Divide each term in $64 z^{2} < 9$ by $64$ .

$\frac{64 z^{2}}{64} < \frac{9}{64}$

Simplify the left side.

Cancel the common factor of $64$ .

Cancel the common factor.

$\frac{64 z^{2}}{64} < \frac{9}{64}$

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

$z^{2} < \frac{9}{64}$

Step 3

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

$\sqrt{z^{2}} < \sqrt{\frac{9}{64}}$

Step 4

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$| z | < \sqrt{\frac{9}{64}}$

Simplify the right side.

Simplify $\sqrt{\frac{9}{64}}$ .

Rewrite $\sqrt{\frac{9}{64}}$ as $\frac{\sqrt{9}}{\sqrt{64}}$ .

$| z | < \frac{\sqrt{9}}{\sqrt{64}}$

Simplify the numerator.

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

$| z | < \frac{\sqrt{3^{2}}}{\sqrt{64}}$

Pull terms out from under the radical.

$| z | < \frac{| 3 |}{\sqrt{64}}$

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

$| z | < \frac{3}{\sqrt{64}}$

Simplify the denominator.

Rewrite $64$ as $8^{2}$ .

$| z | < \frac{3}{\sqrt{8^{2}}}$

Pull terms out from under the radical.

$| z | < \frac{3}{| 8 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $8$ is $8$ .

$| z | < \frac{3}{8}$

Step 5

Write $| z | < \frac{3}{8}$ as a piecewise.

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

$z \geq 0$

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

$z < \frac{3}{8}$

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

$z < 0$

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

$- z < \frac{3}{8}$

Write as a piecewise.

${\begin{cases} z < \frac{3}{8} & z \geq 0 \\ - z < \frac{3}{8} & z < 0 \end{cases}$

Step 6

Find the intersection of $z < \frac{3}{8}$ and $z \geq 0$ .

$0 \leq z < \frac{3}{8}$

Step 7

Solve $- z < \frac{3}{8}$ when $z < 0$ .

Divide each term in $- z < \frac{3}{8}$ by $- 1$ and simplify.

Divide each term in $- z < \frac{3}{8}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- z}{- 1} > \frac{\frac{3}{8}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{z}{1} > \frac{\frac{3}{8}}{- 1}$

Divide $z$ by $1$ .

$z > \frac{\frac{3}{8}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\frac{3}{8}}{- 1}$ .

$z > - 1 \cdot \frac{3}{8}$

Rewrite $- 1 \cdot \frac{3}{8}$ as $- \frac{3}{8}$ .

$z > - \frac{3}{8}$

Find the intersection of $z > - \frac{3}{8}$ and $z < 0$ .

$- \frac{3}{8} < z < 0$

Step 8

Find the union of the solutions.

$- \frac{3}{8} < z < \frac{3}{8}$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$- \frac{3}{8} < z < \frac{3}{8}$

Interval Notation:

$(- \frac{3}{8}, \frac{3}{8})$

Step 10

Solve the Inequality for z 64z^2-9<0

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