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Question
\[x^2 - x + 1 = 0\]
Solution
We have:
\[x^2 - x + 1 = 0\]
\[ \Rightarrow x^2 - x + \frac{1}{2} + \frac{3}{4} = 0\]
\[ \Rightarrow x^2 - 2 \times x \times \frac{1}{2} + \left( \frac{1}{2} \right)^2 - \frac{3}{4} i^2 = 0\]
\[ \Rightarrow \left( x - \frac{1}{2} \right)^2 - \left( \frac{i\sqrt{3}}{2} \right)^2 = 0\]
\[ \Rightarrow \left( x - \frac{1}{2} + \frac{i\sqrt{3}}{2} \right) \left( x - \frac{1}{2} - \frac{i\sqrt{3}}{2} \right) = 0\]
\[\Rightarrow \left( x - \frac{1}{2} + \frac{i\sqrt{3}}{2} \right) = 0\] or \[\left( x - \frac{1}{2} - \frac{i\sqrt{3}}{2} \right) = 0\]
\[\Rightarrow x = \frac{1}{2} - \frac{i\sqrt{3}}{2}\] or \[x = \frac{1}{2} + \frac{i\sqrt{3}}{2}\]
Hence, the roots of the equation are
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