Evaluate square root of -36- square root of -16

$\sqrt{- 36} - \sqrt{- 16}$

Step 1

Simplify each term.

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

$\sqrt{- 1 (36)} - \sqrt{- 16}$

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

$\sqrt{- 1} \cdot \sqrt{36} - \sqrt{- 16}$

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

$i \cdot \sqrt{36} - \sqrt{- 16}$

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

$i \cdot \sqrt{6^{2}} - \sqrt{- 16}$

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

$i \cdot 6 - \sqrt{- 16}$

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

$6 \cdot i - \sqrt{- 16}$

Rewrite $- 16$ as $- 1 (16)$ .

$6 i - \sqrt{- 1 (16)}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$6 i - (\sqrt{- 1} \cdot \sqrt{16})$

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

$6 i - (i \cdot \sqrt{16})$

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

$6 i - (i \cdot \sqrt{4^{2}})$

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

$6 i - (i \cdot 4)$

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

$6 i - (4 \cdot i)$

Multiply $4$ by $- 1$ .

$6 i - 4 i$

Step 2

Subtract $4 i$ from $6 i$ .

$2 i$