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X 2 + X + 1 = 0 - Mathematics

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प्रश्न

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

उत्तर

We have:

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

\[ \Rightarrow x^2 + x + \frac{1}{4} + \frac{3}{4} = 0\]

\[ \Rightarrow \left( x + \frac{1}{2} \right)^2 - \frac{3}{4} i^2 = 0\]

\[ \Rightarrow \left( x + \frac{1}{2} \right)^2 - \left( \frac{i\sqrt{3}}{2} \right)^2 = 0\]

\[ \Rightarrow \left( x + \frac{1}{2} + \frac{i\sqrt{3}}{2} \right) \left( x + \frac{1}{2} - \frac{i\sqrt{3}}{2} \right) = 0\]

\[\Rightarrow \left( x + \frac{1}{2} + \frac{i\sqrt{3}}{2} \right) = 0\] 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 \[- \frac{1}{2} \pm \frac{i\sqrt{3}}{2} .\]

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अध्याय 14: Quadratic Equations - Exercise 14.1 [पृष्ठ ६]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 14 Quadratic Equations
Exercise 14.1 | Q 12 | पृष्ठ ६

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