Evaluate square root of -18+ square root of -72

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Trigonometry

$\sqrt{- 18} + \sqrt{- 72}$

Simplify each term.

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Rewrite $- 18$ as $- 1 (18)$ .

$\sqrt{- 1 (18)} + \sqrt{- 72}$

Rewrite $\sqrt{- 1 (18)}$ as $\sqrt{- 1} \cdot \sqrt{18}$ .

$\sqrt{- 1} \cdot \sqrt{18} + \sqrt{- 72}$

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

$i \cdot \sqrt{18} + \sqrt{- 72}$

Rewrite $18$ as $3^{2} \cdot 2$ .

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Factor $9$ out of $18$ .

$i \cdot \sqrt{9 (2)} + \sqrt{- 72}$

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

$i \cdot \sqrt{3^{2} \cdot 2} + \sqrt{- 72}$

Pull terms out from under the radical.

$i \cdot (3 \sqrt{2}) + \sqrt{- 72}$

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

$3 \cdot i \sqrt{2} + \sqrt{- 72}$

Rewrite $- 72$ as $- 1 (72)$ .

$3 i \sqrt{2} + \sqrt{- 1 (72)}$

Rewrite $\sqrt{- 1 (72)}$ as $\sqrt{- 1} \cdot \sqrt{72}$ .

$3 i \sqrt{2} + \sqrt{- 1} \cdot \sqrt{72}$

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

$3 i \sqrt{2} + i \cdot \sqrt{72}$

Rewrite $72$ as $6^{2} \cdot 2$ .

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Factor $36$ out of $72$ .

$3 i \sqrt{2} + i \cdot \sqrt{36 (2)}$

Rewrite $36$ as $6^{2}$ .

$3 i \sqrt{2} + i \cdot \sqrt{6^{2} \cdot 2}$

Pull terms out from under the radical.

$3 i \sqrt{2} + i \cdot (6 \sqrt{2})$

Move $6$ to the left of $i$ .

$3 i \sqrt{2} + 6 i \sqrt{2}$

Add $3 i \sqrt{2}$ and $6 i \sqrt{2}$ .

$9 i \sqrt{2}$

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