Write in Standard Form 5 square root of -16+3 square root of -81

$5 \sqrt{- 16} + 3 \sqrt{- 81}$

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

$5 \sqrt{- 1 (16)} + 3 \sqrt{- 81}$

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

$5 (\sqrt{- 1} \cdot \sqrt{16}) + 3 \sqrt{- 81}$

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

$5 (i \cdot \sqrt{16}) + 3 \sqrt{- 81}$

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

$5 (i \cdot \sqrt{4^{2}}) + 3 \sqrt{- 81}$

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

$5 (i \cdot 4) + 3 \sqrt{- 81}$

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

$5 (4 \cdot i) + 3 \sqrt{- 81}$

Multiply $4$ by $5$ .

$20 i + 3 \sqrt{- 81}$

Rewrite $- 81$ as $- 1 (81)$ .

$20 i + 3 \sqrt{- 1 (81)}$

Rewrite $\sqrt{- 1 (81)}$ as $\sqrt{- 1} \cdot \sqrt{81}$ .

$20 i + 3 (\sqrt{- 1} \cdot \sqrt{81})$

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

$20 i + 3 (i \cdot \sqrt{81})$

Rewrite $81$ as $9^{2}$ .

$20 i + 3 (i \cdot \sqrt{9^{2}})$

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

$20 i + 3 (i \cdot 9)$

Move $9$ to the left of $i$ .

$20 i + 3 (9 \cdot i)$

Multiply $9$ by $3$ .

$20 i + 27 i$

Step 3

Add $20 i$ and $27 i$ .

$47 i$