Evaluate square root of -4+ square root of -36

$\sqrt{- 4} + \sqrt{- 36}$

Step 1

Simplify each term.

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

$\sqrt{- 1 (4)} + \sqrt{- 36}$

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

$\sqrt{- 1} \cdot \sqrt{4} + \sqrt{- 36}$

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

$i \cdot \sqrt{4} + \sqrt{- 36}$

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

$i \cdot \sqrt{2^{2}} + \sqrt{- 36}$

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

$i \cdot 2 + \sqrt{- 36}$

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

$2 \cdot i + \sqrt{- 36}$

Rewrite $- 36$ as $- 1 (36)$ .

$2 i + \sqrt{- 1 (36)}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$2 i + \sqrt{- 1} \cdot \sqrt{36}$

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

$2 i + i \cdot \sqrt{36}$

Rewrite $36$ as $6^{2}$ .

$2 i + i \cdot \sqrt{6^{2}}$

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

$2 i + i \cdot 6$

Move $6$ to the left of $i$ .

$2 i + 6 i$

Step 2

Add $2 i$ and $6 i$ .

$8 i$