Write in Standard Form (-2+ square root of -8)+(7- square root of -52)

$(- 2 + \sqrt{- 8}) + (7 - \sqrt{- 52})$

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 $- 8$ as $- 1 (8)$ .

$- 2 + \sqrt{- 1 (8)} + 7 - \sqrt{- 52}$

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$- 2 + \sqrt{- 1} \cdot \sqrt{8} + 7 - \sqrt{- 52}$

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

$- 2 + i \cdot \sqrt{8} + 7 - \sqrt{- 52}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$- 2 + i \cdot \sqrt{4 (2)} + 7 - \sqrt{- 52}$

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

$- 2 + i \cdot \sqrt{2^{2} \cdot 2} + 7 - \sqrt{- 52}$

Pull terms out from under the radical.

$- 2 + i \cdot (2 \sqrt{2}) + 7 - \sqrt{- 52}$

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

$- 2 + 2 \cdot i \sqrt{2} + 7 - \sqrt{- 52}$

Rewrite $- 52$ as $- 1 (52)$ .

$- 2 + 2 i \sqrt{2} + 7 - \sqrt{- 1 (52)}$

Rewrite $\sqrt{- 1 (52)}$ as $\sqrt{- 1} \cdot \sqrt{52}$ .

$- 2 + 2 i \sqrt{2} + 7 - (\sqrt{- 1} \cdot \sqrt{52})$

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

$- 2 + 2 i \sqrt{2} + 7 - (i \cdot \sqrt{52})$

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$- 2 + 2 i \sqrt{2} + 7 - (i \cdot \sqrt{4 (13)})$

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

$- 2 + 2 i \sqrt{2} + 7 - (i \cdot \sqrt{2^{2} \cdot 13})$

Pull terms out from under the radical.

$- 2 + 2 i \sqrt{2} + 7 - (i \cdot (2 \sqrt{13}))$

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

$- 2 + 2 i \sqrt{2} + 7 - (2 \cdot i \sqrt{13})$

Multiply $2$ by $- 1$ .

$- 2 + 2 i \sqrt{2} + 7 - 2 i \sqrt{13}$

Step 3

Simplify the expression.

Add $- 2$ and $7$ .

$2 i \sqrt{2} + 5 - 2 i \sqrt{13}$

Reorder $2 i \sqrt{2}$ and $5$ .

$5 + 2 i \sqrt{2} - 2 i \sqrt{13}$