Find the Domain 16x^2-9y^2=1

$16 x^{2} - 9 y^{2} = 1$

Step 1

Subtract $16 x^{2}$ from both sides of the equation.

$- 9 y^{2} = 1 - 16 x^{2}$

Step 2

Divide each term in $- 9 y^{2} = 1 - 16 x^{2}$ by $- 9$ and simplify.

Divide each term in $- 9 y^{2} = 1 - 16 x^{2}$ by $- 9$ .

$\frac{- 9 y^{2}}{- 9} = \frac{1}{- 9} + \frac{- 16 x^{2}}{- 9}$

Simplify the left side.

Cancel the common factor of $- 9$ .

Cancel the common factor.

$\frac{- 9 y^{2}}{- 9} = \frac{1}{- 9} + \frac{- 16 x^{2}}{- 9}$

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

$y^{2} = \frac{1}{- 9} + \frac{- 16 x^{2}}{- 9}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y^{2} = - \frac{1}{9} + \frac{- 16 x^{2}}{- 9}$

Dividing two negative values results in a positive value.

$y^{2} = - \frac{1}{9} + \frac{16 x^{2}}{9}$

Step 3

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

$y = \pm \sqrt{- \frac{1}{9} + \frac{16 x^{2}}{9}}$

Step 4

Simplify $\pm \sqrt{- \frac{1}{9} + \frac{16 x^{2}}{9}}$ .

Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

$y = \pm \sqrt{- \frac{1}{9} + \frac{{(4 x)}^{2}}{9}}$

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

$y = \pm \sqrt{- \frac{1}{9} + \frac{{(4 x)}^{2}}{3^{2}}}$

Rewrite $\frac{{(4 x)}^{2}}{3^{2}}$ as ${(\frac{4 x}{3})}^{2}$ .

$y = \pm \sqrt{- \frac{1}{9} + {(\frac{4 x}{3})}^{2}}$

Simplify the expression.

Rewrite $\frac{1}{9}$ as ${(\frac{1}{3})}^{2}$ .

$y = \pm \sqrt{- {(\frac{1}{3})}^{2} + {(\frac{4 x}{3})}^{2}}$

Reorder $- {(\frac{1}{3})}^{2}$ and ${(\frac{4 x}{3})}^{2}$ .

$y = \pm \sqrt{{(\frac{4 x}{3})}^{2} - {(\frac{1}{3})}^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \frac{4 x}{3}$ and $b = \frac{1}{3}$ .

$y = \pm \sqrt{(\frac{4 x}{3} + \frac{1}{3}) (\frac{4 x}{3} - \frac{1}{3})}$

Simplify terms.

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{4 x + 1}{3} (\frac{4 x}{3} - \frac{1}{3})}$

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{4 x + 1}{3} \cdot \frac{4 x - 1}{3}}$

Multiply $\frac{4 x + 1}{3}$ by $\frac{4 x - 1}{3}$ .

$y = \pm \sqrt{\frac{(4 x + 1) (4 x - 1)}{3 \cdot 3}}$

Multiply $3$ by $3$ .

$y = \pm \sqrt{\frac{(4 x + 1) (4 x - 1)}{9}}$

Rewrite $\frac{(4 x + 1) (4 x - 1)}{9}$ as ${(\frac{1}{3})}^{2} ((4 x + 1) (4 x - 1))$ .

Factor the perfect power $1^{2}$ out of $(4 x + 1) (4 x - 1)$ .

$y = \pm \sqrt{\frac{1^{2} ((4 x + 1) (4 x - 1))}{9}}$

Factor the perfect power $3^{2}$ out of $9$ .

$y = \pm \sqrt{\frac{1^{2} ((4 x + 1) (4 x - 1))}{3^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} ((4 x + 1) (4 x - 1))}{3^{2} \cdot 1}$ .

$y = \pm \sqrt{{(\frac{1}{3})}^{2} ((4 x + 1) (4 x - 1))}$

Pull terms out from under the radical.

$y = \pm \frac{1}{3} \sqrt{(4 x + 1) (4 x - 1)}$

Combine $\frac{1}{3}$ and $\sqrt{(4 x + 1) (4 x - 1)}$ .

$y = \pm \frac{\sqrt{(4 x + 1) (4 x - 1)}}{3}$

Step 5

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.

$y = \frac{\sqrt{(4 x + 1) (4 x - 1)}}{3}$

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

$y = - \frac{\sqrt{(4 x + 1) (4 x - 1)}}{3}$

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

$y = \frac{\sqrt{(4 x + 1) (4 x - 1)}}{3}$

$y = - \frac{\sqrt{(4 x + 1) (4 x - 1)}}{3}$

$y = \frac{\sqrt{(4 x + 1) (4 x - 1)}}{3}$

$y = - \frac{\sqrt{(4 x + 1) (4 x - 1)}}{3}$

Step 6

Set the radicand in $\sqrt{(4 x + 1) (4 x - 1)}$ greater than or equal to $0$ to find where the expression is defined.

$(4 x + 1) (4 x - 1) \geq 0$

Step 7

Solve for $x$ .

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

$4 x + 1 = 0$

$4 x - 1 = 0$

Set $4 x + 1$ equal to $0$ and solve for $x$ .

Set $4 x + 1$ equal to $0$ .

$4 x + 1 = 0$

Solve $4 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$4 x = - 1$

Divide each term in $4 x = - 1$ by $4$ and simplify.

Divide each term in $4 x = - 1$ by $4$ .

$\frac{4 x}{4} = \frac{- 1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 1}{4}$

Divide $x$ by $1$ .

$x = \frac{- 1}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{4}$

Set $4 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Solve $4 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

The final solution is all the values that make $(4 x + 1) (4 x - 1) \geq 0$ true.

$x = - \frac{1}{4}, \frac{1}{4}$

Use each root to create test intervals.

$x < - \frac{1}{4}$

$- \frac{1}{4} < x < \frac{1}{4}$

$x > \frac{1}{4}$

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 $x < - \frac{1}{4}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{1}{4}$ and see if this value makes the original inequality true.

$x = - 3$

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

$(4 (- 3) + 1) (4 (- 3) - 1) \geq 0$

The left side $143$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{1}{4} < x < \frac{1}{4}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{1}{4} < x < \frac{1}{4}$ and see if this value makes the original inequality true.

$x = 0$

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

$(4 (0) + 1) (4 (0) - 1) \geq 0$

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > \frac{1}{4}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{1}{4}$ and see if this value makes the original inequality true.

$x = 3$

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

$(4 (3) + 1) (4 (3) - 1) \geq 0$

The left side $143$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - \frac{1}{4}$ True

$- \frac{1}{4} < x < \frac{1}{4}$ False

$x > \frac{1}{4}$ True

$x < - \frac{1}{4}$ True

$- \frac{1}{4} < x < \frac{1}{4}$ False

$x > \frac{1}{4}$ True

The solution consists of all of the true intervals.

$x \leq - \frac{1}{4}$ or $x \geq \frac{1}{4}$

Step 8

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

Interval Notation:

$(- \infty, - \frac{1}{4}] \cup [\frac{1}{4}, \infty)$

Set-Builder Notation:

${x | x \leq - \frac{1}{4}, x \geq \frac{1}{4}}$

Step 9

Find the Domain 16x^2-9y^2=1

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