Trigonometry Examples

$y^{2} - 4 x^{2} = 16$

Step 1

Solve for $y$ .

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Step 1.1

Add $4 x^{2}$ to both sides of the equation.

$y^{2} = 16 + 4 x^{2}$

Step 1.2

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

$y = \pm \sqrt{16 + 4 x^{2}}$

Step 1.3

Simplify $\pm \sqrt{16 + 4 x^{2}}$ .

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Step 1.3.1

Factor $4$ out of $16 + 4 x^{2}$ .

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Step 1.3.1.1

Factor $4$ out of $16$ .

$y = \pm \sqrt{4 \cdot 4 + 4 x^{2}}$

Step 1.3.1.2

Factor $4$ out of $4 \cdot 4 + 4 x^{2}$ .

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

Step 1.3.2

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

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Step 1.3.2.1

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

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

Step 1.3.2.2

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

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

Step 1.3.3

Pull terms out from under the radical.

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

Step 1.3.4

Raise $2$ to the power of $2$ .

$y = \pm 2 \sqrt{4 + x^{2}}$

Step 1.4

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

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Step 1.4.1

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

$y = 2 \sqrt{4 + x^{2}}$

Step 1.4.2

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

$y = - 2 \sqrt{4 + x^{2}}$

Step 1.4.3

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

$y = 2 \sqrt{4 + x^{2}}$

$y = - 2 \sqrt{4 + x^{2}}$

$y = 2 \sqrt{4 + x^{2}}$

$y = - 2 \sqrt{4 + x^{2}}$

$y = 2 \sqrt{4 + x^{2}}$

$y = - 2 \sqrt{4 + x^{2}}$

Step 2

Find the domain to determine if the expression is continuous.

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Step 2.1

Set the radicand in $\sqrt{4 + x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$4 + x^{2} \geq 0$

Step 2.2

Solve for $x$ .

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Step 2.2.1

Subtract $4$ from both sides of the inequality.

$x^{2} \geq - 4$

Step 2.2.2

Since the left side has an even power, it is always positive for all real numbers.

All real numbers

Step 2.3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

Since the domain is all real numbers, $2 \sqrt{4 + x^{2}}, - 2 \sqrt{4 + x^{2}}$ is continuous over all real numbers.

Continuous

Step 4