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Question
Solve the following quadratic equation:
\[x^2 - x + \left( 1 + i \right) = 0\]
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Solution
\[ x^2 - x + \left( 1 + i \right) = 0\]
\[\text { Comparing the given equation with the general form } a x^2 + bx + c = 0, \text { we get }\]
\[a = 1, b = - 1 \text { and } c = \left( 1 + i \right)\]
\[x = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[ \Rightarrow x = \frac{1 \pm \sqrt{1 - 4\left( 1 + i \right)}}{2}\]
\[ \Rightarrow x = \frac{1 \pm \sqrt{- 3 - 4i}}{2} . . . \left( i \right)\]
\[\text { Let } x + iy = \sqrt{- 3 - 4i} . \text { Then }, \]
\[ \Rightarrow \left( x + iy \right)^2 = - 3 - 4i\]
\[ \Rightarrow x^2 - y^2 + 2ixy = - 3 - 4i \]
\[ \Rightarrow x^2 - y^2 = - 3 \text { and } 2xy = - 4 . . . \left( ii \right)\]
\[\text { Now }, \left( x^2 + y^2 \right)^2 = \left( x^2 - y^2 \right)^2 + 4 x^2 y^2 \]
\[ \Rightarrow \left( x^2 + y^2 \right)^2 = 9 + 16 = 25\]
\[ \Rightarrow x^2 + y^2 = 5 . . . \left( iii \right) \]
\[\text { From } \left( ii \right) \text { and } \left( iii \right)\]
\[ \Rightarrow x = \pm 1 \text { and } y = \pm 2\]
\[\text { As, xy is negative } \left[ \text { From } \left( ii \right) \right]\]
\[ \Rightarrow x = 1, y = - 2 \text { or,} x = - 1, y = 2\]
\[ \Rightarrow x + iy = 1 - 2i \text { or }- 1 + 2i\]
\[ \Rightarrow \sqrt{- 3 - 4i} = \pm \left( 1 - 2i \right)\]
\[\text{ Substituting these values in } \left( i \right), \text { we get }\]
\[ \Rightarrow x = \frac{1 \pm \left( 1 - 2i \right)}{2}\]
\[ \Rightarrow x = 1 - i, i\]
\[\text { So, the roots of the given quadratic equation are 1 - i and } i .\]
