Evaluate square root of -9- square root of -36

$\sqrt{- 9} - \sqrt{- 36}$

Step 1

Simplify each term.

Rewrite $- 9$ as $- 1 (9)$ .

$\sqrt{- 1 (9)} - \sqrt{- 36}$

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$\sqrt{- 1} \cdot \sqrt{9} - \sqrt{- 36}$

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

$i \cdot \sqrt{9} - \sqrt{- 36}$

Rewrite $9$ as $3^{2}$ .

$i \cdot \sqrt{3^{2}} - \sqrt{- 36}$

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

$i \cdot 3 - \sqrt{- 36}$

Move $3$ to the left of $i$ .

$3 \cdot i - \sqrt{- 36}$

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

$3 i - \sqrt{- 1 (36)}$

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

$3 i - (\sqrt{- 1} \cdot \sqrt{36})$

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

$3 i - (i \cdot \sqrt{36})$

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

$3 i - (i \cdot \sqrt{6^{2}})$

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

$3 i - (i \cdot 6)$

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

$3 i - (6 \cdot i)$

Multiply $6$ by $- 1$ .

$3 i - 6 i$

Step 2

Subtract $6 i$ from $3 i$ .

$- 3 i$