Precalculus Examples

$\frac{a^{2} + 4 a + 4}{16 - b^{4}} \div \frac{4 - a^{2}}{4 + b^{2}}$

Step 1

Set the denominator in $\frac{a^{2} + 4 a + 4}{16 - b^{4}}$ equal to $0$ to find where the expression is undefined.

$16 - b^{4} = 0$

Step 2

Solve for $a$ .

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Step 2.1

Subtract $16$ from both sides of the equation.

$- b^{4} = - 16$

Step 2.2

Divide each term in $- b^{4} = - 16$ by $- 1$ and simplify.

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Step 2.2.1

Divide each term in $- b^{4} = - 16$ by $- 1$ .

$\frac{- b^{4}}{- 1} = \frac{- 16}{- 1}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Dividing two negative values results in a positive value.

$\frac{b^{4}}{1} = \frac{- 16}{- 1}$

Step 2.2.2.2

Divide $b^{4}$ by $1$ .

$b^{4} = \frac{- 16}{- 1}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $- 16$ by $- 1$ .

$b^{4} = 16$

Step 2.3

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

$b = \pm \sqrt[4]{16}$

Step 2.4

Simplify $\pm \sqrt[4]{16}$ .

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Step 2.4.1

Rewrite $16$ as $2^{4}$ .

$b = \pm \sqrt[4]{2^{4}}$

Step 2.4.2

Pull terms out from under the radical, assuming positive real numbers.

$b = \pm 2$

Step 2.5

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

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Step 2.5.1

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

$b = 2$

Step 2.5.2

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

$b = - 2$

Step 2.5.3

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

$b = 2, - 2$

Step 3

Set the denominator in $\frac{4 - a^{2}}{4 + b^{2}}$ equal to $0$ to find where the expression is undefined.

$4 + b^{2} = 0$

Step 4

Solve for $a$ .

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Step 4.1

Subtract $4$ from both sides of the equation.

$b^{2} = - 4$

Step 4.2

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

$b = \pm \sqrt{- 4}$

Step 4.3

Simplify $\pm \sqrt{- 4}$ .

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Step 4.3.1

Rewrite $- 4$ as $- 1 (4)$ .

$b = \pm \sqrt{- 1 (4)}$

Step 4.3.2

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$b = \pm \sqrt{- 1} \cdot \sqrt{4}$

Step 4.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$b = \pm i \cdot \sqrt{4}$

Step 4.3.4

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

$b = \pm i \cdot \sqrt{2^{2}}$

Step 4.3.5

Pull terms out from under the radical, assuming positive real numbers.

$b = \pm i \cdot 2$

Step 4.3.6

Move $2$ to the left of $i$ .

$b = \pm 2 i$

Step 4.4

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

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Step 4.4.1

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

$b = 2 i$

Step 4.4.2

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

$b = - 2 i$

Step 4.4.3

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

$b = 2 i, - 2 i$

Step 5

Set the denominator in $\frac{a^{2} + 4 a + 4}{16 - b^{4}} \div \frac{4 - a^{2}}{4 + b^{2}}$ equal to $0$ to find where the expression is undefined.

$\frac{4 - a^{2}}{4 + b^{2}} = 0$

Step 6

Solve for $a$ .

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Step 6.1

Set the numerator equal to zero.

$4 - a^{2} = 0$

Step 6.2

Solve the equation for $a$ .

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Step 6.2.1

Subtract $4$ from both sides of the equation.

$- a^{2} = - 4$

Step 6.2.2

Divide each term in $- a^{2} = - 4$ by $- 1$ and simplify.

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Step 6.2.2.1

Divide each term in $- a^{2} = - 4$ by $- 1$ .

$\frac{- a^{2}}{- 1} = \frac{- 4}{- 1}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{a^{2}}{1} = \frac{- 4}{- 1}$

Step 6.2.2.2.2

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

$a^{2} = \frac{- 4}{- 1}$

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Divide $- 4$ by $- 1$ .

$a^{2} = 4$

Step 6.2.3

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

$a = \pm \sqrt{4}$

Step 6.2.4

Simplify $\pm \sqrt{4}$ .

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Step 6.2.4.1

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

$a = \pm \sqrt{2^{2}}$

Step 6.2.4.2

Pull terms out from under the radical, assuming positive real numbers.

$a = \pm 2$

Step 6.2.5

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

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Step 6.2.5.1

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

$a = 2$

Step 6.2.5.2

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

$a = - 2$

Step 6.2.5.3

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

$a = 2, - 2$

Step 7

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

Interval Notation:

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

Set-Builder Notation:

${a | a \neq 2, - 2}$