Answer
$$\sqrt{-128}=8\sqrt{2}i$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\sqrt{-128}& \xlongequal{ }\sqrt{128}\cdot i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}} \sqrt{ 64 \cdot 2 } i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}} \sqrt{ 64 } \cdot \sqrt{ 2 } i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}8\sqrt{2}i\end{aligned} $$ | |
| ① | Factor out the largest perfect square of 128. ( in this example we factored out $ 64 $ ) |
| ② | Rewrite $ \sqrt{ 64 \cdot 2 } $ as the product of two radicals. |
| ③ | The square root of $ 64 $ is $ 8 $. |
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Simplify Radical Expression