Question

\[x^2 - x + 1 = 0\]

Answer 1

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 

\[\frac{1}{2} \pm \frac{i\sqrt{3}}{2} .\]