Algebra Examples

$(\sqrt{20} - \sqrt{- 3}) + (\sqrt{5} - \sqrt{- 12})$

Step 1

Remove parentheses.

$\sqrt{20} - \sqrt{- 3} + \sqrt{5} - \sqrt{- 12}$

Step 2

Simplify each term.

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Step 2.1

Rewrite $20$ as $2^{2} \cdot 5$ .

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Step 2.1.1

Factor $4$ out of $20$ .

$\sqrt{4 (5)} - \sqrt{- 3} + \sqrt{5} - \sqrt{- 12}$

Step 2.1.2

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

$\sqrt{2^{2} \cdot 5} - \sqrt{- 3} + \sqrt{5} - \sqrt{- 12}$

Step 2.2

Pull terms out from under the radical.

$2 \sqrt{5} - \sqrt{- 3} + \sqrt{5} - \sqrt{- 12}$

Step 2.3

Rewrite $- 3$ as $- 1 (3)$ .

$2 \sqrt{5} - \sqrt{- 1 (3)} + \sqrt{5} - \sqrt{- 12}$

Step 2.4

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$2 \sqrt{5} - (\sqrt{- 1} \cdot \sqrt{3}) + \sqrt{5} - \sqrt{- 12}$

Step 2.5

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

$2 \sqrt{5} - (i \cdot \sqrt{3}) + \sqrt{5} - \sqrt{- 12}$

Step 2.6

Rewrite $- 12$ as $- 1 (12)$ .

$2 \sqrt{5} - i \sqrt{3} + \sqrt{5} - \sqrt{- 1 (12)}$

Step 2.7

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$2 \sqrt{5} - i \sqrt{3} + \sqrt{5} - (\sqrt{- 1} \cdot \sqrt{12})$

Step 2.8

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

$2 \sqrt{5} - i \sqrt{3} + \sqrt{5} - (i \cdot \sqrt{12})$

Step 2.9

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 2.9.1

Factor $4$ out of $12$ .

$2 \sqrt{5} - i \sqrt{3} + \sqrt{5} - (i \cdot \sqrt{4 (3)})$

Step 2.9.2

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

$2 \sqrt{5} - i \sqrt{3} + \sqrt{5} - (i \cdot \sqrt{2^{2} \cdot 3})$

Step 2.10

Pull terms out from under the radical.

$2 \sqrt{5} - i \sqrt{3} + \sqrt{5} - (i \cdot (2 \sqrt{3}))$

Step 2.11

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

$2 \sqrt{5} - i \sqrt{3} + \sqrt{5} - (2 \cdot i \sqrt{3})$

Step 2.12

Multiply $2$ by $- 1$ .

$2 \sqrt{5} - i \sqrt{3} + \sqrt{5} - 2 i \sqrt{3}$

Step 3

Simplify by adding terms.

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Step 3.1

Add $2 \sqrt{5}$ and $\sqrt{5}$ .

$- i \sqrt{3} + 3 \sqrt{5} - 2 i \sqrt{3}$

Step 3.2

Subtract $2 i \sqrt{3}$ from $- i \sqrt{3}$ .

$- 3 i \sqrt{3} + 3 \sqrt{5}$

Step 3.3

Reorder $- 3 i \sqrt{3}$ and $3 \sqrt{5}$ .

$3 \sqrt{5} - 3 i \sqrt{3}$