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Question
Solve the following quadratic equation:
\[x^2 - \left( 2 + i \right) x - \left( 1 - 7i \right) = 0\]
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Solution
\[ x^2 - \left( 2 + i \right) x - \left( 1 - 7i \right) = 0\]
\[\text { Comparing the given equation with the general form } a x^2 + bx + c = 0, \text { we get }\]
\[a = 1, b = - \left( 2 + i \right) \text { and } c = - \left( 1 - 7i \right)\]
\[x = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[ \Rightarrow x = \frac{\left( 2 + i \right) \pm \sqrt{\left( 2 + i \right)^2 + 4\left( 1 - 7i \right)}}{2}\]
\[ \Rightarrow x = \frac{\left( 2 + i \right) \pm \sqrt{7 - 24i}}{2} . . . \left( i \right)\]
\[\text { Let }x + iy = \sqrt{7 - 24i} . \text { Then }, \]
\[ \Rightarrow \left( x + iy \right)^2 = 7 - 24i\]
\[ \Rightarrow x^2 - y^2 + 2ixy = 7 - 24i \]
\[ \Rightarrow x^2 - y^2 = 7 \text { and } 2xy = - 24 . . . \left( ii \right)\]
\[\text { Now }, \left( x^2 + y^2 \right)^2 = \left( x^2 - y^2 \right)^2 + 4 x^2 y^2 \]
\[ \Rightarrow \left( x^2 + y^2 \right)^2 = 49 + 576 = 625\]
\[ \Rightarrow x^2 + y^2 = 25 . . . \left( iii \right) \]
\[\text { From } \left( ii \right) \text { and } \left( iii \right)\]
\[ \Rightarrow x = \pm 4 \text { and } y = \pm 3\]
\[\text { As, xy is negative } \left[ \text { From } \left( ii \right) \right]\]
\[ \Rightarrow x = - 4, y = 3 or, x = 4, y = - 3\]
\[ \Rightarrow x + iy = - 4 + 3i or, 4 - 3i\]
\[ \Rightarrow \sqrt{7 - 24i} = \pm 4 - 3i\]
\[\text { Substituting these values in } \left( i \right), \text { we get }\]
\[ \Rightarrow x = \frac{\left( 2 + i \right) \pm \left( 4 - 3i \right)}{2}\]
\[ \Rightarrow x = 3 - i, - 1 + 2i\]
\[\text { So, the roots of the given quadratic equation are 3 - i and } - 1 + 2i . \]
