Question

Solve the following quadratic equation:

\[2 x^2 - \left( 3 + 7i \right) x + \left( 9i - 3 \right) = 0\]

Answer 1

\[ 2 x^2 - \left( 3 + 7i \right) x + \left( 9i - 3 \right) = 0\]

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

\[a = 2, b = - \left( 3 + 7i \right) \text { and } c = \left( 9i - 3 \right)\]

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

\[ \Rightarrow x = \frac{\left( 3 + 7i \right) \pm \sqrt{\left( 3 + 7i \right)^2 - 8\left( 9i - 3 \right)}}{4}\]

\[ \Rightarrow x = \frac{\left( 3 + 7i \right) \pm \sqrt{- 16 - 30i}}{4} . . . \left( i \right)\]

\[\text { Let } x + iy = \sqrt{- 16 - 30i} . \text { Then }, \]

\[ \Rightarrow \left( x + iy \right)^2 = - 16 - 30i\]

\[ \Rightarrow x^2 - y^2 + 2ixy = - 16 - 30i \]

\[ \Rightarrow x^2 - y^2 = - 16 \text { and } 2xy = - 30 . . . \left( ii \right)\]

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

\[ \Rightarrow \left( x^2 + y^2 \right)^2 = 256 + 900 = 1156\]

\[ \Rightarrow x^2 + y^2 = 34 . . . \left( iii \right) \]

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

\[ \Rightarrow x = \pm 3 \text { and } y = \pm 5\]

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

\[ \Rightarrow x = - 3, y = 5 \text { or, } x = 3, y = - 5\]

\[ \Rightarrow x + iy = 3 - 5 i \text { or }, - 3 + 5 i\]

\[ \Rightarrow \sqrt{14 - 8\sqrt{2}i} = \pm \left( 3 - 5 i \right)\]

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

\[ \Rightarrow x = \frac{\left( 3 + 7i \right) \pm \left( 3 - 5 i \right)}{4}\]

\[ \Rightarrow x = \frac{3 + i}{2}, 3i\]

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