Find the Domain x^2-2y^2=18

$x^{2} - 2 y^{2} = 18$

Step 1

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

$- 2 y^{2} = 18 - x^{2}$

Step 2

Divide each term in $- 2 y^{2} = 18 - x^{2}$ by $- 2$ and simplify.

Divide each term in $- 2 y^{2} = 18 - x^{2}$ by $- 2$ .

$\frac{- 2 y^{2}}{- 2} = \frac{18}{- 2} + \frac{- x^{2}}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 y^{2}}{- 2} = \frac{18}{- 2} + \frac{- x^{2}}{- 2}$

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

$y^{2} = \frac{18}{- 2} + \frac{- x^{2}}{- 2}$

Simplify the right side.

Simplify each term.

Divide $18$ by $- 2$ .

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

Dividing two negative values results in a positive value.

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

Step 3

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

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

Step 4

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

To write $- 9$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Combine $- 9$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

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

Multiply $- 9$ by $2$ .

$y = \pm \sqrt{\frac{- 18 + x^{2}}{2}}$

Rewrite $\sqrt{\frac{- 18 + x^{2}}{2}}$ as $\frac{\sqrt{- 18 + x^{2}}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{- 18 + x^{2}}}{\sqrt{2}}$

Multiply $\frac{\sqrt{- 18 + x^{2}}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{- 18 + x^{2}}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{- 18 + x^{2}}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{- 18 + x^{2}} \sqrt{2}}{\sqrt{2} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

$y = \pm \frac{\sqrt{- 18 + x^{2}} \sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

$y = \pm \frac{\sqrt{- 18 + x^{2}} \sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = \pm \frac{\sqrt{- 18 + x^{2}} \sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Add $1$ and $1$ .

$y = \pm \frac{\sqrt{- 18 + x^{2}} \sqrt{2}}{{\sqrt{2}}^{2}}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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

$y = \pm \frac{\sqrt{- 18 + x^{2}} \sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

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

$y = \pm \frac{\sqrt{- 18 + x^{2}} \sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$y = \pm \frac{\sqrt{- 18 + x^{2}} \sqrt{2}}{2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y = \pm \frac{\sqrt{- 18 + x^{2}} \sqrt{2}}{2^{\frac{2}{2}}}$

Rewrite the expression.

$y = \pm \frac{\sqrt{- 18 + x^{2}} \sqrt{2}}{2^{1}}$

Evaluate the exponent.

$y = \pm \frac{\sqrt{- 18 + x^{2}} \sqrt{2}}{2}$

Combine using the product rule for radicals.

$y = \pm \frac{\sqrt{(- 18 + x^{2}) \cdot 2}}{2}$

Reorder factors in $\pm \frac{\sqrt{(- 18 + x^{2}) \cdot 2}}{2}$ .

$y = \pm \frac{\sqrt{2 (- 18 + x^{2})}}{2}$

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.

$y = \frac{\sqrt{2 (- 18 + x^{2})}}{2}$

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

$y = - \frac{\sqrt{2 (- 18 + x^{2})}}{2}$

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

$y = \frac{\sqrt{2 (- 18 + x^{2})}}{2}$

$y = - \frac{\sqrt{2 (- 18 + x^{2})}}{2}$

$y = \frac{\sqrt{2 (- 18 + x^{2})}}{2}$

$y = - \frac{\sqrt{2 (- 18 + x^{2})}}{2}$

Step 6

Set the radicand in $\sqrt{2 (- 18 + x^{2})}$ greater than or equal to $0$ to find where the expression is defined.

$2 (- 18 + x^{2}) \geq 0$

Step 7

Solve for $x$ .

Simplify $2 (- 18 + x^{2})$ .

Apply the distributive property.

$2 \cdot - 18 + 2 x^{2} \geq 0$

Multiply $2$ by $- 18$ .

$- 36 + 2 x^{2} \geq 0$

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x = - 3 \sqrt{2}, 3 \sqrt{2}$

Use each root to create test intervals.

$x < - 3 \sqrt{2}$

$- 3 \sqrt{2} < x < 3 \sqrt{2}$

$x > 3 \sqrt{2}$

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 $x < - 3 \sqrt{2}$ to see if it makes the inequality true.

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

$x = - 7$

Replace $x$ with $- 7$ in the original inequality.

$2 (- 18 + {(- 7)}^{2}) \geq 0$

The left side $62$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$2 (- 18 + {(0)}^{2}) \geq 0$

The left side $- 36$ is less than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 7$

Replace $x$ with $7$ in the original inequality.

$2 (- 18 + {(7)}^{2}) \geq 0$

The left side $62$ 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.

$x < - 3 \sqrt{2}$ True

$- 3 \sqrt{2} < x < 3 \sqrt{2}$ False

$x > 3 \sqrt{2}$ True

$x < - 3 \sqrt{2}$ True

$- 3 \sqrt{2} < x < 3 \sqrt{2}$ False

$x > 3 \sqrt{2}$ True

The solution consists of all of the true intervals.

$x \leq - 3 \sqrt{2}$ or $x \geq 3 \sqrt{2}$

Step 8

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

Interval Notation:

$(- \infty, - 3 \sqrt{2}] \cup [3 \sqrt{2}, \infty)$

Set-Builder Notation:

${x | x \leq - 3 \sqrt{2}, x \geq 3 \sqrt{2}}$

Step 9

Find the Domain x^2-2y^2=18

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