Question

Solve the following quadratic equation:

\[x^2 - \left( 2 + i \right) x - \left( 1 - 7i \right) = 0\]

Answer 1

\[ x^2 - \left( 2 + i \right) x - \left( 1 - 7i \right) = 0\]

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

\[a = 1, b = - \left( 2 + i \right) \text { and } c = - \left( 1 - 7i \right)\]

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

\[ \Rightarrow x = \frac{\left( 2 + i \right) \pm \sqrt{\left( 2 + i \right)^2 + 4\left( 1 - 7i \right)}}{2}\]

\[ \Rightarrow x = \frac{\left( 2 + i \right) \pm \sqrt{7 - 24i}}{2} . . . \left( i \right)\]

\[\text { Let  }x + iy = \sqrt{7 - 24i} . \text { Then }, \]

\[ \Rightarrow \left( x + iy \right)^2 = 7 - 24i\]

\[ \Rightarrow x^2 - y^2 + 2ixy = 7 - 24i \]

\[ \Rightarrow x^2 - y^2 = 7 \text { and } 2xy = - 24 . . . \left( ii \right)\]

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

\[ \Rightarrow \left( x^2 + y^2 \right)^2 = 49 + 576 = 625\]

\[ \Rightarrow x^2 + y^2 = 25 . . . \left( iii \right) \]

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

\[ \Rightarrow x = \pm 4 \text { and } y = \pm 3\]

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

\[ \Rightarrow x = - 4, y = 3 or, x = 4, y = - 3\]

\[ \Rightarrow x + iy = - 4 + 3i or, 4 - 3i\]

\[ \Rightarrow \sqrt{7 - 24i} = \pm 4 - 3i\]

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

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

\[ \Rightarrow x = 3 - i, - 1 + 2i\]

\[\text { So, the roots of the given quadratic equation are 3 - i and } - 1 + 2i . \]