Question
Answer
$$(x+3i)^2-(2x-3i)^2=18ix-3x^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+3i)^2-(2x-3i)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^2+6ix+9i^2-(4x^2-12ix+9i^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^2+6ix+9i^2-4x^2+12ix-9i^2 \xlongequal{ } \\[1 em] & \xlongequal{ }x^2+6ix+ \cancel{9i^2}-4x^2+12ix -\cancel{9i^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}18ix-3x^2\end{aligned} $$ | |
| ① | Find $ \left(x+3i\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 3i }$. $$ \begin{aligned}\left(x+3i\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 3i + \color{red}{\left( 3i \right)^2} = x^2+6ix+9i^2\end{aligned} $$Find $ \left(2x-3i\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2x } $ and $ B = \color{red}{ 3i }$. $$ \begin{aligned}\left(2x-3i\right)^2 = \color{blue}{\left( 2x \right)^2} -2 \cdot 2x \cdot 3i + \color{red}{\left( 3i \right)^2} = 4x^2-12ix+9i^2\end{aligned} $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 4x^2-12ix+9i^2 \right) = -4x^2+12ix-9i^2 $$ |
| ③ | Combine like terms: $$ \color{blue}{x^2} + \color{red}{6ix} + \, \color{green}{ \cancel{9i^2}} \, \color{blue}{-4x^2} + \color{red}{12ix} \, \color{green}{ -\cancel{9i^2}} \, = \color{red}{18ix} \color{blue}{-3x^2} $$ |
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