Find the Domain 3x^2-2y^2=12

$3 x^{2} - 2 y^{2} = 12$

Step 1

Subtract $3 x^{2}$ from both sides of the equation.

$- 2 y^{2} = 12 - 3 x^{2}$

Step 2

Divide each term in $- 2 y^{2} = 12 - 3 x^{2}$ by $- 2$ and simplify.

Divide each term in $- 2 y^{2} = 12 - 3 x^{2}$ by $- 2$ .

$\frac{- 2 y^{2}}{- 2} = \frac{12}{- 2} + \frac{- 3 x^{2}}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 y^{2}}{- 2} = \frac{12}{- 2} + \frac{- 3 x^{2}}{- 2}$

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

$y^{2} = \frac{12}{- 2} + \frac{- 3 x^{2}}{- 2}$

Simplify the right side.

Simplify each term.

Divide $12$ by $- 2$ .

$y^{2} = - 6 + \frac{- 3 x^{2}}{- 2}$

Dividing two negative values results in a positive value.

$y^{2} = - 6 + \frac{3 x^{2}}{2}$

Step 3

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

$y = \pm \sqrt{- 6 + \frac{3 x^{2}}{2}}$

Step 4

Simplify $\pm \sqrt{- 6 + \frac{3 x^{2}}{2}}$ .

To write $- 6$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = \pm \sqrt{- 6 \cdot \frac{2}{2} + \frac{3 x^{2}}{2}}$

Simplify terms.

Combine $- 6$ and $\frac{2}{2}$ .

$y = \pm \sqrt{\frac{- 6 \cdot 2}{2} + \frac{3 x^{2}}{2}}$

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{- 6 \cdot 2 + 3 x^{2}}{2}}$

Simplify the numerator.

Factor $3$ out of $- 6 \cdot 2 + 3 x^{2}$ .

Factor $3$ out of $- 6 \cdot 2$ .

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

Factor $3$ out of $3 x^{2}$ .

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

Factor $3$ out of $3 (- 2 \cdot 2) + 3 (x^{2})$ .

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

Multiply $- 2$ by $2$ .

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

Rewrite $- 4 + x^{2}$ in a factored form.

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

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

Reorder $- 2^{2}$ and $x^{2}$ .

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

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

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

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

Rewrite $\sqrt{\frac{3 (x + 2) (x - 2)}{2}}$ as $\frac{\sqrt{3 (x + 2) (x - 2)}}{\sqrt{2}}$ .

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

Multiply $\frac{\sqrt{3 (x + 2) (x - 2)}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Combine and simplify the denominator.

Multiply $\frac{\sqrt{3 (x + 2) (x - 2)}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Simplify the numerator.

Combine using the product rule for radicals.

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

Multiply $2$ by $3$ .

$y = \pm \frac{\sqrt{6 (x + 2) (x - 2)}}{2}$

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.

$y = \frac{\sqrt{6 (x + 2) (x - 2)}}{2}$

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

$y = - \frac{\sqrt{6 (x + 2) (x - 2)}}{2}$

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

$y = \frac{\sqrt{6 (x + 2) (x - 2)}}{2}$

$y = - \frac{\sqrt{6 (x + 2) (x - 2)}}{2}$

$y = \frac{\sqrt{6 (x + 2) (x - 2)}}{2}$

$y = - \frac{\sqrt{6 (x + 2) (x - 2)}}{2}$

Step 6

Set the radicand in $\sqrt{6 (x + 2) (x - 2)}$ greater than or equal to $0$ to find where the expression is defined.

$6 (x + 2) (x - 2) \geq 0$

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 $0$ .

$x + 2 = 0$

$x - 2 = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $6 (x + 2) (x - 2) \geq 0$ true.

$x = - 2, 2$

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 2$

$x > 2$

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 $x < - 2$ to see if it makes the inequality true.

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

$x = - 4$

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

$6 ((- 4) + 2) ((- 4) - 2) \geq 0$

The left side $72$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < 2$ to see if it makes the inequality true.

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

$x = 0$

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

$6 ((0) + 2) ((0) - 2) \geq 0$

The left side $- 24$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

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

$6 ((4) + 2) ((4) - 2) \geq 0$

The left side $72$ 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.

$x < - 2$ True

$- 2 < x < 2$ False

$x > 2$ True

$x < - 2$ True

$- 2 < x < 2$ False

$x > 2$ True

The solution consists of all of the true intervals.

$x \leq - 2$ or $x \geq 2$

Step 8

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

Interval Notation:

$(- \infty, - 2] \cup [2, \infty)$

Set-Builder Notation:

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

Step 9

Find the Domain 3x^2-2y^2=12

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