Question
Answer
$$-4+\dfrac{\sqrt{-128}}{8}=-4+\sqrt{2}\cdot i$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-4+\frac{\sqrt{-128}}{8}& \xlongequal{ }-4+\frac{\sqrt{128}}{8}i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-4 + \frac{ \sqrt{ 64 \cdot 2 } }{ 8 }i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-4 + \frac{ \sqrt{ 64 } \cdot \sqrt{ 2 } }{ 8 }i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-4+\frac{8\sqrt{2}}{8}i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-4 + \frac{ 8 \cdot \sqrt{ 2 } : \color{orangered}{ 8 }}{ 8 : \color{orangered}{ 8 }}i \xlongequal{ } \\[1 em] & \xlongequal{ }-4+\frac{\sqrt{2}}{1}i \xlongequal{ } \\[1 em] & \xlongequal{ }-4+\sqrt{2}\cdot i\end{aligned} $$ | |
| 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 $. |
| ④ | Divide numerator and denominator by $ \color{orangered}{ 8 } $. |
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