Evaluate square root of -2^2+ cube root of -5^3

$\sqrt{- 2^{2}} + \sqrt[3]{- 5^{3}}$

Step 1

Simplify each term.

Raise $2$ to the power of $2$ .

$\sqrt{- 1 \cdot 4} + \sqrt[3]{- 5^{3}}$

Multiply $- 1$ by $4$ .

$\sqrt{- 4} + \sqrt[3]{- 5^{3}}$

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

$\sqrt{- 1 (4)} + \sqrt[3]{- 5^{3}}$

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

$\sqrt{- 1} \cdot \sqrt{4} + \sqrt[3]{- 5^{3}}$

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

$i \cdot \sqrt{4} + \sqrt[3]{- 5^{3}}$

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

$i \cdot \sqrt{2^{2}} + \sqrt[3]{- 5^{3}}$

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

$i \cdot 2 + \sqrt[3]{- 5^{3}}$

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

$2 \cdot i + \sqrt[3]{- 5^{3}}$

Raise $5$ to the power of $3$ .

$2 i + \sqrt[3]{- 1 \cdot 125}$

Multiply $- 1$ by $125$ .

$2 i + \sqrt[3]{- 125}$

Rewrite $- 125$ as ${(- 5)}^{3}$ .

$2 i + \sqrt[3]{{(- 5)}^{3}}$

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

$2 i - 5$

Step 2

Reorder $2 i$ and $- 5$ .

$- 5 + 2 i$