Algebra Examples

| 2 - 2 \sqrt{3} i |

$| 2 - 2 \sqrt{3} i |$

Step 1

Use the formula

| a + b i | = \sqrt{a^{2} + b^{2}}

$| a + b i | = \sqrt{a^{2} + b^{2}}$ to find the magnitude.

\sqrt{2^{2} + {(- 2 \sqrt{3})}^{2}}

$\sqrt{2^{2} + {(- 2 \sqrt{3})}^{2}}$

Step 2

Raise

2

$2$ to the power of

2

$2$ .

\sqrt{4 + {(- 2 \sqrt{3})}^{2}}

$\sqrt{4 + {(- 2 \sqrt{3})}^{2}}$

Step 3

Apply the product rule to

- 2 \sqrt{3}

$- 2 \sqrt{3}$ .

\sqrt{4 + {(- 2)}^{2} {\sqrt{3}}^{2}}

$\sqrt{4 + {(- 2)}^{2} {\sqrt{3}}^{2}}$

Step 4

Raise

- 2

$- 2$ to the power of

2

$2$ .

\sqrt{4 + 4 {\sqrt{3}}^{2}}

$\sqrt{4 + 4 {\sqrt{3}}^{2}}$

Step 5

Rewrite

{\sqrt{3}}^{2}

${\sqrt{3}}^{2}$ as

3

$3$ .

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Step 5.1

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt{3}

$\sqrt{3}$ as

3^{\frac{1}{2}}

$3^{\frac{1}{2}}$ .

\sqrt{4 + 4 {(3^{\frac{1}{2}})}^{2}}

$\sqrt{4 + 4 {(3^{\frac{1}{2}})}^{2}}$

Step 5.2

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

\sqrt{4 + 4 \cdot 3^{\frac{1}{2} \cdot 2}}

$\sqrt{4 + 4 \cdot 3^{\frac{1}{2} \cdot 2}}$

Step 5.3

Combine

\frac{1}{2}

$\frac{1}{2}$ and

2

$2$ .

\sqrt{4 + 4 \cdot 3^{\frac{2}{2}}}

$\sqrt{4 + 4 \cdot 3^{\frac{2}{2}}}$

Step 5.4

Cancel the common factor of

2

$2$ .

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Step 5.4.1

Cancel the common factor.

\sqrt{4 + 4 \cdot 3^{\frac{2}{2}}}

$\sqrt{4 + 4 \cdot 3^{\frac{2}{2}}}$

Step 5.4.2

Rewrite the expression.

\sqrt{4 + 4 \cdot 3^{1}}

$\sqrt{4 + 4 \cdot 3^{1}}$

\sqrt{4 + 4 \cdot 3^{1}}

$\sqrt{4 + 4 \cdot 3^{1}}$

Step 5.5

Evaluate the exponent.

\sqrt{4 + 4 \cdot 3}

$\sqrt{4 + 4 \cdot 3}$

\sqrt{4 + 4 \cdot 3}

$\sqrt{4 + 4 \cdot 3}$

Step 6

Multiply

4

$4$ by

3

$3$ .

\sqrt{4 + 12}

$\sqrt{4 + 12}$

Step 7

Add

4

$4$ and

12

$12$ .

\sqrt{16}

$\sqrt{16}$

Step 8

Rewrite

16

$16$ as

4^{2}

$4^{2}$ .

\sqrt{4^{2}}

$\sqrt{4^{2}}$

Step 9

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

4

$4$

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