Determine if Continuous x^2-5y^2=6

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Trigonometry

$x^{2} - 5 y^{2} = 6$

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

$- 5 y^{2} = 6 - x^{2}$

Divide each term by $- 5$ and simplify.

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Divide each term in $- 5 y^{2} = 6 - x^{2}$ by $- 5$ .

$\frac{- 5 y^{2}}{- 5} = \frac{6}{- 5} + \frac{- x^{2}}{- 5}$

Cancel the common factor of $- 5$ .

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Cancel the common factor.

$\frac{- 5 y^{2}}{- 5} = \frac{6}{- 5} + \frac{- x^{2}}{- 5}$

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

$y^{2} = \frac{6}{- 5} + \frac{- x^{2}}{- 5}$

Simplify each term.

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Move the negative in front of the fraction.

$y^{2} = - \frac{6}{5} + \frac{- x^{2}}{- 5}$

Dividing two negative values results in a positive value.

$y^{2} = - \frac{6}{5} + \frac{x^{2}}{5}$

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

$y = \pm \sqrt{- \frac{6}{5} + \frac{x^{2}}{5}}$

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

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Simplify the right side of the equation.

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Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{- 6 + x^{2}}{5}}$

Rewrite $\sqrt{\frac{- 6 + x^{2}}{5}}$ as $\frac{\sqrt{- 6 + x^{2}}}{\sqrt{5}}$ .

$y = \pm \frac{\sqrt{- 6 + x^{2}}}{\sqrt{5}}$

Multiply $\frac{\sqrt{- 6 + x^{2}}}{\sqrt{5}}$ by $\frac{\sqrt{5}}{\sqrt{5}}$ .

$y = \pm \frac{\sqrt{- 6 + x^{2}}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$

Combine and simplify the denominator.

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Multiply $\frac{\sqrt{- 6 + x^{2}}}{\sqrt{5}}$ and $\frac{\sqrt{5}}{\sqrt{5}}$ .

$y = \pm \frac{\sqrt{- 6 + x^{2}} \sqrt{5}}{\sqrt{5} \sqrt{5}}$

Raise $\sqrt{5}$ to the power of $1$ .

$y = \pm \frac{\sqrt{- 6 + x^{2}} \sqrt{5}}{\sqrt{5} \sqrt{5}}$

Raise $\sqrt{5}$ to the power of $1$ .

$y = \pm \frac{\sqrt{- 6 + x^{2}} \sqrt{5}}{\sqrt{5} \sqrt{5}}$

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

$y = \pm \frac{\sqrt{- 6 + x^{2}} \sqrt{5}}{{\sqrt{5}}^{1 + 1}}$

Add $1$ and $1$ .

$y = \pm \frac{\sqrt{- 6 + x^{2}} \sqrt{5}}{{\sqrt{5}}^{2}}$

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$y = \pm \frac{\sqrt{- 6 + x^{2}} \sqrt{5}}{{(5^{\frac{1}{2}})}^{2}}$

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

$y = \pm \frac{\sqrt{- 6 + x^{2}} \sqrt{5}}{5^{\frac{1}{2} \cdot 2}}$

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

$y = \pm \frac{\sqrt{- 6 + x^{2}} \sqrt{5}}{5^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$y = \pm \frac{\sqrt{- 6 + x^{2}} \sqrt{5}}{5^{\frac{2}{2}}}$

Divide $1$ by $1$ .

$y = \pm \frac{\sqrt{- 6 + x^{2}} \sqrt{5}}{5}$

Evaluate the exponent.

$y = \pm \frac{\sqrt{- 6 + x^{2}} \sqrt{5}}{5}$

Combine using the product rule for radicals.

$y = \pm \frac{\sqrt{(- 6 + x^{2}) \cdot 5}}{5}$

Reorder factors in $\pm \frac{\sqrt{(- 6 + x^{2}) \cdot 5}}{5}$ .

$y = \pm \frac{\sqrt{5 (- 6 + x^{2})}}{5}$

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

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First, use the positive value of the $\pm$ to find the first solution.

$y = \frac{\sqrt{5 (- 6 + x^{2})}}{5}$

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

$y = - \frac{\sqrt{5 (- 6 + x^{2})}}{5}$

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

$y = \frac{\sqrt{5 (- 6 + x^{2})}}{5}$

$y = - \frac{\sqrt{5 (- 6 + x^{2})}}{5}$

$y = \frac{\sqrt{5 (- 6 + x^{2})}}{5}$

$y = - \frac{\sqrt{5 (- 6 + x^{2})}}{5}$

$y = \frac{\sqrt{5 (- 6 + x^{2})}}{5}$

$y = - \frac{\sqrt{5 (- 6 + x^{2})}}{5}$

Set the radicand in $\sqrt{5 (- 6 + x^{2})}$ greater than or equal to $0$ to find where the expression is defined.

$5 (- 6 + x^{2}) \geq 0$

Solve for $x$ .

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Divide each term by $5$ and simplify.

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Divide each term in $5 (- 6 + x^{2}) \geq 0$ by $5$ .

$\frac{5 (- 6 + x^{2})}{5} \geq \frac{0}{5}$

Cancel the common factor of $5$ .

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Cancel the common factor.

$\frac{5 (- 6 + x^{2})}{5} \geq \frac{0}{5}$

Divide $- 6 + x^{2}$ by $1$ .

$- 6 + x^{2} \geq \frac{0}{5}$

Divide $0$ by $5$ .

$- 6 + x^{2} \geq 0$

Add $6$ to both sides of the inequality.

$x^{2} \geq 6$

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

$x \geq \pm \sqrt{6}$

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

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First, use the positive value of the $\pm$ to find the first solution.

$x \geq \sqrt{6}$

Next, use the negative value of the $\pm$ to find the second solution. Since this is an inequality, flip the direction of the inequality sign on the $-$ portion of the solution.

$x \leq - \sqrt{6}$

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

$x \geq \sqrt{6}$ or $x \leq - \sqrt{6}$

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

Interval Notation:

$(- \infty, - \sqrt{6}] \cup [\sqrt{6}, \infty)$

Set-Builder Notation:

${x | x \leq - \sqrt{6}, x \geq \sqrt{6}}$

Since the domain is not all real numbers, $x^{2} - 5 y^{2} = 6$ is not continuous over all real numbers.

Not continuous

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