Question

Solve the following quadratic equation:

\[x^2 - x + \left( 1 + i \right) = 0\]

Answer 1

\[ x^2 - x + \left( 1 + i \right) = 0\]

\[\text { Comparing the given equation with the general form } a x^2 + bx + c = 0, \text { we get }\]

\[a = 1, b = - 1 \text { and } c = \left( 1 + i \right)\]

\[x = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\]

\[ \Rightarrow x = \frac{1 \pm \sqrt{1 - 4\left( 1 + i \right)}}{2}\]

\[ \Rightarrow x = \frac{1 \pm \sqrt{- 3 - 4i}}{2} . . . \left( i \right)\]

\[\text { Let } x + iy = \sqrt{- 3 - 4i} . \text { Then }, \]

\[ \Rightarrow \left( x + iy \right)^2 = - 3 - 4i\]

\[ \Rightarrow x^2 - y^2 + 2ixy = - 3 - 4i \]

\[ \Rightarrow x^2 - y^2 = - 3 \text { and } 2xy = - 4 . . . \left( ii \right)\]

\[\text { Now }, \left( x^2 + y^2 \right)^2 = \left( x^2 - y^2 \right)^2 + 4 x^2 y^2 \]

\[ \Rightarrow \left( x^2 + y^2 \right)^2 = 9 + 16 = 25\]

\[ \Rightarrow x^2 + y^2 = 5 . . . \left( iii \right) \]

\[\text { From } \left( ii \right) \text { and } \left( iii \right)\]

\[ \Rightarrow x = \pm 1 \text { and } y = \pm 2\]

\[\text { As, xy is negative } \left[ \text { From } \left( ii \right) \right]\]

\[ \Rightarrow x = 1, y = - 2 \text { or,} x = - 1, y = 2\]

\[ \Rightarrow x + iy = 1 - 2i \text { or }- 1 + 2i\]

\[ \Rightarrow \sqrt{- 3 - 4i} = \pm \left( 1 - 2i \right)\]

\[\text{ Substituting these values in } \left( i \right), \text { we get }\]

\[ \Rightarrow x = \frac{1 \pm \left( 1 - 2i \right)}{2}\]

\[ \Rightarrow x = 1 - i, i\]

\[\text { So, the roots of the given quadratic equation are 1 - i and } i .\]