Trigonometry Examples

$z^{4} + 4 \sqrt{2} z^{2} + 16 = 0$

Step 1

Substitute $u = z^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} + 4 \sqrt{2} u + 16 = 0$

$u = z^{2}$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = 4 \sqrt{2}$ , and $c = 16$ into the quadratic formula and solve for $u$ .

$\frac{- 4 \sqrt{2} \pm \sqrt{{(4 \sqrt{2})}^{2} - 4 \cdot (1 \cdot 16)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Apply the product rule to $4 \sqrt{2}$ .

$u = \frac{- 4 \sqrt{2} \pm \sqrt{4^{2} {\sqrt{2}}^{2} - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Step 4.1.2

Raise $4$ to the power of $2$ .

$u = \frac{- 4 \sqrt{2} \pm \sqrt{16 {\sqrt{2}}^{2} - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Step 4.1.3

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

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Step 4.1.3.1

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

$u = \frac{- 4 \sqrt{2} \pm \sqrt{16 {(2^{\frac{1}{2}})}^{2} - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Step 4.1.3.2

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

$u = \frac{- 4 \sqrt{2} \pm \sqrt{16 \cdot 2^{\frac{1}{2} \cdot 2} - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Step 4.1.3.3

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

$u = \frac{- 4 \sqrt{2} \pm \sqrt{16 \cdot 2^{\frac{2}{2}} - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Step 4.1.3.4

Cancel the common factor of $2$ .

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Step 4.1.3.4.1

Cancel the common factor.

$u = \frac{- 4 \sqrt{2} \pm \sqrt{16 \cdot 2^{\frac{2}{2}} - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Step 4.1.3.4.2

Rewrite the expression.

$u = \frac{- 4 \sqrt{2} \pm \sqrt{16 \cdot 2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Step 4.1.3.5

Evaluate the exponent.

$u = \frac{- 4 \sqrt{2} \pm \sqrt{16 \cdot 2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Step 4.1.4

Multiply $16$ by $2$ .

$u = \frac{- 4 \sqrt{2} \pm \sqrt{32 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Step 4.1.5

Multiply $- 4 \cdot 1 \cdot 16$ .

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Step 4.1.5.1

Multiply $- 4$ by $1$ .

$u = \frac{- 4 \sqrt{2} \pm \sqrt{32 - 4 \cdot 16}}{2 \cdot 1}$

Step 4.1.5.2

Multiply $- 4$ by $16$ .

$u = \frac{- 4 \sqrt{2} \pm \sqrt{32 - 64}}{2 \cdot 1}$

Step 4.1.6

Subtract $64$ from $32$ .

$u = \frac{- 4 \sqrt{2} \pm \sqrt{- 32}}{2 \cdot 1}$

Step 4.1.7

Rewrite $- 32$ as $- 1 (32)$ .

$u = \frac{- 4 \sqrt{2} \pm \sqrt{- 1 \cdot 32}}{2 \cdot 1}$

Step 4.1.8

Rewrite $\sqrt{- 1 (32)}$ as $\sqrt{- 1} \cdot \sqrt{32}$ .

$u = \frac{- 4 \sqrt{2} \pm \sqrt{- 1} \cdot \sqrt{32}}{2 \cdot 1}$

Step 4.1.9

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

$u = \frac{- 4 \sqrt{2} \pm i \cdot \sqrt{32}}{2 \cdot 1}$

Step 4.1.10

Rewrite $32$ as $4^{2} \cdot 2$ .

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Step 4.1.10.1

Factor $16$ out of $32$ .

$u = \frac{- 4 \sqrt{2} \pm i \cdot \sqrt{16 (2)}}{2 \cdot 1}$

Step 4.1.10.2

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

$u = \frac{- 4 \sqrt{2} \pm i \cdot \sqrt{4^{2} \cdot 2}}{2 \cdot 1}$

Step 4.1.11

Pull terms out from under the radical.

$u = \frac{- 4 \sqrt{2} \pm i \cdot (4 \sqrt{2})}{2 \cdot 1}$

Step 4.1.12

Move $4$ to the left of $i$ .

$u = \frac{- 4 \sqrt{2} \pm 4 i \sqrt{2}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$u = \frac{- 4 \sqrt{2} \pm 4 i \sqrt{2}}{2}$

Step 4.3

Simplify $\frac{- 4 \sqrt{2} \pm 4 i \sqrt{2}}{2}$ .

$u = - 2 \sqrt{2} \pm 2 i \sqrt{2}$

Step 5

The final answer is the combination of both solutions.

$u = - 2 \sqrt{2} + 2 i \sqrt{2}, - 2 \sqrt{2} - 2 i \sqrt{2}$

Step 6

Substitute the real value of $u = z^{2}$ back into the solved equation.

$z^{2} = - 2 \sqrt{2} + 2 i \sqrt{2}$

${(z^{2})}^{1} = - 2 \sqrt{2} - 2 i \sqrt{2}$

Step 7

Solve the first equation for $z$ .

$z^{2} = - 2 \sqrt{2} + 2 i \sqrt{2}$

Step 8

Solve the equation for $z$ .

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Step 8.1

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

$z = \pm \sqrt{- 2 \sqrt{2} + 2 i \sqrt{2}}$

Step 8.2

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

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Step 8.2.1

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

$z = \sqrt{- 2 \sqrt{2} + 2 i \sqrt{2}}$

Step 8.2.2

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

$z = - \sqrt{- 2 \sqrt{2} + 2 i \sqrt{2}}$

Step 8.2.3

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

$z = \sqrt{- 2 \sqrt{2} + 2 i \sqrt{2}}, - \sqrt{- 2 \sqrt{2} + 2 i \sqrt{2}}$

Step 9

Solve the second equation for $z$ .

${(z^{2})}^{1} = - 2 \sqrt{2} - 2 i \sqrt{2}$

Step 10

Solve the equation for $z$ .

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Step 10.1

Remove parentheses.

$z^{2} = - 2 \sqrt{2} - 2 i \sqrt{2}$

Step 10.2

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

$z = \pm \sqrt{- 2 \sqrt{2} - 2 i \sqrt{2}}$

Step 10.3

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

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Step 10.3.1

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

$z = \sqrt{- 2 \sqrt{2} - 2 i \sqrt{2}}$

Step 10.3.2

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

$z = - \sqrt{- 2 \sqrt{2} - 2 i \sqrt{2}}$

Step 10.3.3

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

$z = \sqrt{- 2 \sqrt{2} - 2 i \sqrt{2}}, - \sqrt{- 2 \sqrt{2} - 2 i \sqrt{2}}$

Step 11

The solution to $z^{4} + 4 \sqrt{2} z^{2} + 16 = 0$ is $z = \sqrt{- 2 \sqrt{2} + 2 i \sqrt{2}}, - \sqrt{- 2 \sqrt{2} + 2 i \sqrt{2}}, \sqrt{- 2 \sqrt{2} - 2 i \sqrt{2}}, - \sqrt{- 2 \sqrt{2} - 2 i \sqrt{2}}$ .

$z = \sqrt{- 2 \sqrt{2} + 2 i \sqrt{2}}, - \sqrt{- 2 \sqrt{2} + 2 i \sqrt{2}}, \sqrt{- 2 \sqrt{2} - 2 i \sqrt{2}}, - \sqrt{- 2 \sqrt{2} - 2 i \sqrt{2}}$