Evaluate square root of -49- square root of -9

$\sqrt{- 49} - \sqrt{- 9}$

Step 1

Simplify each term.

Rewrite $- 49$ as $- 1 (49)$ .

$\sqrt{- 1 (49)} - \sqrt{- 9}$

Rewrite $\sqrt{- 1 (49)}$ as $\sqrt{- 1} \cdot \sqrt{49}$ .

$\sqrt{- 1} \cdot \sqrt{49} - \sqrt{- 9}$

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

$i \cdot \sqrt{49} - \sqrt{- 9}$

Rewrite $49$ as $7^{2}$ .

$i \cdot \sqrt{7^{2}} - \sqrt{- 9}$

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

$i \cdot 7 - \sqrt{- 9}$

Move $7$ to the left of $i$ .

$7 \cdot i - \sqrt{- 9}$

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

$7 i - \sqrt{- 1 (9)}$

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

$7 i - (\sqrt{- 1} \cdot \sqrt{9})$

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

$7 i - (i \cdot \sqrt{9})$

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

$7 i - (i \cdot \sqrt{3^{2}})$

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

$7 i - (i \cdot 3)$

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

$7 i - (3 \cdot i)$

Multiply $3$ by $- 1$ .

$7 i - 3 i$

Step 2

Subtract $3 i$ from $7 i$ .

$4 i$