Question
Answer
$$(2x+i(x^2-1))^{0.5}=1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x+i(x^2-1))^{0.5}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(2x+ix^2-i)^{0.5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(1ix^2-i+2x)^{0.5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}1\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{i} $ by $ \left( x^2-1\right) $ $$ \color{blue}{i} \cdot \left( x^2-1\right) = ix^2-i $$ |
| ② | Combine like terms: $$ 2x+ix^2-i = ix^2-i+2x $$ |
| ③ | A non-zero polynomial raised to the power of 0 equals 1. |
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