Question

Solve the following quadratic equation:

\[x^2 - \left( 3\sqrt{2} - 2i \right) x - \sqrt{2} i = 0\]

Answer 1

\[ x^2 - \left( 3\sqrt{2} - 2i \right) x - \sqrt{2}i = 0\]

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

\[a = 1, b = - \left( 3\sqrt{2} - 2i \right) \text { and } c = - \sqrt{2}i\]

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

⇒ x = `(-(-3sqrt2-2i)±sqrt((-(3sqrt2-2i))^2)-4(1)(-sqrt2i))/2(1)`

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

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

\[\text { Let } x + iy = \sqrt{14 - 8\sqrt{2}i} . \text { Then }, \]

\[ \Rightarrow \left( x + iy \right)^2 = 14 - 8\sqrt{2}i\]

\[ \Rightarrow x^2 - y^2 + 2ixy = 14 - 8\sqrt{2}i \]

\[ \Rightarrow x^2 - y^2 = 14 \text { and } 2xy = - 8\sqrt{2} . . . \left( ii \right)\]

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

\[ \Rightarrow \left( x^2 + y^2 \right)^2 = 196 + 128 = 324\]

\[ \Rightarrow x^2 + y^2 = 18 . . . \left( iii \right) \]

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

\[ \Rightarrow x = \pm 4 \text{ and }y = \pm \sqrt{2}\]

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

\[ \Rightarrow x = - 4, y = \sqrt{2}\text{  or }, x = 4, y = - \sqrt{2}\]

\[ \Rightarrow x + iy = 4 - \sqrt{2} i \text { or, } - 4 + \sqrt{2} i\]

\[ \Rightarrow \sqrt{14 - 8\sqrt{2}i} = \pm \left( 4 - \sqrt{2} i \right)\]

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

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

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

\[\text { So, the roots of the given quadratic equation are } \frac{\left( 3\sqrt{2} + 4 \right) - i\left( 2 + \sqrt{2} \right)}{2} \text{ and } \frac{\left( 3\sqrt{2} - 4 \right) - i\left( 2 - \sqrt{2} \right)}{2} .\]