Simplify square root of -9+3 square root of -16

$\sqrt{- 9} + 3 \sqrt{- 16}$

Step 1

Simplify each term.

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

$\sqrt{- 1 (9)} + 3 \sqrt{- 16}$

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

$\sqrt{- 1} \cdot \sqrt{9} + 3 \sqrt{- 16}$

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

$i \cdot \sqrt{9} + 3 \sqrt{- 16}$

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

$i \cdot \sqrt{3^{2}} + 3 \sqrt{- 16}$

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

$i \cdot 3 + 3 \sqrt{- 16}$

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

$3 \cdot i + 3 \sqrt{- 16}$

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

$3 i + 3 \sqrt{- 1 (16)}$

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

$3 i + 3 (\sqrt{- 1} \cdot \sqrt{16})$

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

$3 i + 3 (i \cdot \sqrt{16})$

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

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

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

$3 i + 3 (i \cdot 4)$

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

$3 i + 3 (4 \cdot i)$

Multiply $4$ by $3$ .

$3 i + 12 i$

Step 2

Add $3 i$ and $12 i$ .

$15 i$