Question

Solve the following quadratic equation:

\[\left( 2 + i \right) x^2 - \left( 5 - i \right) x + 2 \left( 1 - i \right) = 0\]

Answer 1

\[ \left( 2 + i \right) x^2 - \left( 5 - i \right) x + 2\left( 1 - i \right) = 0\]

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

\[a = \left( 2 + i \right), b = - \left( 5 - i \right) \text { and } c = 2\left( 1 - i \right)\]

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

\[ \Rightarrow x = \frac{\left( 5 - i \right) \pm \sqrt{\left( 5 - i \right)^2 - 4\left( 2 + i \right)2\left( 1 - i \right)}}{2\left( 2 + i \right)}\]

\[ \Rightarrow x = \frac{\left( 5 - i \right) \pm \sqrt{- 2i}}{2\left( 2 + i \right)}\]

\[ \Rightarrow x = \frac{\left( 5 - i \right) \pm \sqrt{- 2i}}{2\left( 2 + i \right)} . . . \left( i \right)\]

\[\text { Let } x + iy = \sqrt{- 2i} . \text { Then }, \]

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

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

\[ \Rightarrow x^2 - y^2 = 0 \text { and } 2xy = - 2 . . . \left( ii \right)\]

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

\[ \Rightarrow \left( x^2 + y^2 \right)^2 = 4\]

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

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

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

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

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

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

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

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

\[ \Rightarrow x = \frac{\left( 5 - i \right) \pm \left( 1 - i \right)}{2\left( 2 + i \right)}\]

\[ \Rightarrow x = 1 - i, \frac{4}{5} - \frac{2}{5}i\]

\[\text { So, the roots of the given quadratic equation are 1 - i and } \frac{4}{5} - \frac{2}{5}i . \]