Question
Answer
$$-\sqrt{36}+\sqrt{200}+\sqrt{9}+\sqrt{-98}=-3+10\sqrt{2}+7\sqrt{2}i$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-\sqrt{36}+\sqrt{200}+\sqrt{9}+\sqrt{-98}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-6+10\sqrt{2}+3+\sqrt{-98} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-3+10\sqrt{2}+\sqrt{-98} \xlongequal{ } \\[1 em] & \xlongequal{ }-3+10\sqrt{2}+\sqrt{98}\cdot i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-3+10\sqrt{2} + \sqrt{ 49 \cdot 2 } i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-3+10\sqrt{2} + \sqrt{ 49 } \cdot \sqrt{ 2 } i \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}-3+10\sqrt{2}+7\sqrt{2}i\end{aligned} $$ | |
| ① | $$ - \sqrt{36} = -1 \cdot 6 = -6 $$ |
| ② | $$ \sqrt{200} =
\sqrt{ 10 ^2 \cdot 2 } =
\sqrt{ 10 ^2 } \, \sqrt{ 2 } =
10 \sqrt{ 2 }$$ |
| ③ | $$ \sqrt{9} = 3 $$ |
| ④ | Combine like terms |
| ⑤ | Factor out the largest perfect square of 98. ( in this example we factored out $ 49 $ ) |
| ⑥ | Rewrite $ \sqrt{ 49 \cdot 2 } $ as the product of two radicals. |
| ⑦ | The square root of $ 49 $ is $ 7 $. |
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