Question
Answer
$$(2+i)^2-4\cdot(2+i)+5=0$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2+i)^2-4\cdot(2+i)+5& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4+4i+i^2-4\cdot(2+i)+5 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4+4i-1-4\cdot(2+i)+5 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4i+3-4\cdot(2+i)+5 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}4i+3-(8+4i)+5 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}4i+3-8-4i+5 \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{4i}+3-8 -\cancel{4i}+5 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}0\end{aligned} $$ | |
| ① | Find $ \left(2+i\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2 } $ and $ B = \color{red}{ i }$. $$ \begin{aligned}\left(2+i\right)^2 = \color{blue}{2^2} +2 \cdot 2 \cdot i + \color{red}{i^2} = 4+4i+i^2\end{aligned} $$ |
| ② | $$ i^2 = -1 $$ |
| ③ | Combine like terms: $$ \color{blue}{4} +4i \color{blue}{-1} = 4i+ \color{blue}{3} $$ |
| ④ | Multiply $ \color{blue}{4} $ by $ \left( 2+i\right) $ $$ \color{blue}{4} \cdot \left( 2+i\right) = 8+4i $$ |
| ⑤ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 8+4i \right) = -8-4i $$ |
| ⑥ | Combine like terms: $$ \, \color{blue}{ \cancel{4i}} \,+ \color{green}{3} \color{orange}{-8} \, \color{blue}{ -\cancel{4i}} \,+ \color{orange}{5} = \color{orange}{0} $$ |
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