Answer
$$\sqrt{-17}\cdot\sqrt{-51}=-17\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\sqrt{-17}\cdot\sqrt{-51}& \xlongequal{ }\sqrt{17}\cdot i\sqrt{51}\cdot i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\sqrt{17}\cdot\sqrt{51}\cdot(-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-\sqrt{17}\cdot\sqrt{51} \xlongequal{ } \\[1 em] & \xlongequal{ }-\sqrt{867} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}- \, \sqrt{ 289 \cdot 3 } \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}- \, \sqrt{ 289 } \cdot \sqrt{ 3 } \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-17\sqrt{3}\end{aligned} $$ | |
| ① | $ i \cdot i = i^2 = -1 $ |
| ② | Put the minus sign in front of the result. |
| ③ | Factor out the largest perfect square of 867. ( in this example we factored out $ 289 $ ) |
| ④ | Rewrite $ \sqrt{ 289 \cdot 3 } $ as the product of two radicals. |
| ⑤ | The square root of $ 289 $ is $ 17 $. |
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