Write in Standard Form square root of -9+ square root of -4- square root of 16

$\sqrt{- 9} + \sqrt{- 4} - \sqrt{16}$

Step 1

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

$a x^{2} + b x + c$

Step 2

Simplify each term.

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

$\sqrt{- 1 (9)} + \sqrt{- 4} - \sqrt{16}$

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

$\sqrt{- 1} \cdot \sqrt{9} + \sqrt{- 4} - \sqrt{16}$

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

$i \cdot \sqrt{9} + \sqrt{- 4} - \sqrt{16}$

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

$i \cdot \sqrt{3^{2}} + \sqrt{- 4} - \sqrt{16}$

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

$i \cdot 3 + \sqrt{- 4} - \sqrt{16}$

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

$3 \cdot i + \sqrt{- 4} - \sqrt{16}$

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

$3 i + \sqrt{- 1 (4)} - \sqrt{16}$

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

$3 i + \sqrt{- 1} \cdot \sqrt{4} - \sqrt{16}$

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

$3 i + i \cdot \sqrt{4} - \sqrt{16}$

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

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

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

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

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

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

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

$3 i + 2 i - \sqrt{4^{2}}$

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

$3 i + 2 i - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$3 i + 2 i - 4$

Step 3

Simplify by adding terms.

Add $3 i$ and $2 i$ .

$5 i - 4$

Reorder $5 i$ and $- 4$ .

$- 4 + 5 i$