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प्रश्न
x2 + x + 1 = 0
उत्तर
We have:
\[x^2 + x + 1 = 0\]
\[ \Rightarrow x^2 + x + \frac{1}{4} + \frac{3}{4} = 0\]
\[ \Rightarrow x^2 + \left( \frac{1}{2} \right)^2 + 2 \times x \times \frac{1}{2} - \left( \frac{\sqrt{3}i}{2} \right)^2 = 0\]
\[ \Rightarrow \left( x + \frac{1}{2} \right)^2 - \left( \frac{\sqrt{3}i}{2} \right)^2 = 0\]
\[ \Rightarrow \left( x + \frac{1}{2} + \frac{\sqrt{3}i}{2} \right) \left( x + \frac{1}{2} - \frac{\sqrt{3}i}{2} \right) = 0\]
\[\Rightarrow \left( x + \frac{1}{2} + \frac{\sqrt{3}i}{2} \right) = 0\] or, \[\left( x + \frac{1}{2} - \frac{\sqrt{3}i}{2} \right) = 0\]
\[\Rightarrow\] \[x = - \frac{1}{2} - \frac{\sqrt{3}i}{2}\] or, \[x = - \frac{1}{2} + \frac{\sqrt{3}i}{2}\]
Hence, the roots of the equation are \[- \frac{1}{2} - i\frac{\sqrt{3}}{2} \text { and } - \frac{1}{2} + i\frac{\sqrt{3}}{2}\].
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