Question

Solve: $x^2-(3\surd 2-2i)x-\surd 2i=0$.

Answer 1

Solution of a quadratic equation by Shridhracharya rule:

Given quadratic equation is

$x^2-(3\surd 2-2i)x-\surd 2i=0$

Here, $a=1,b=-\left(3\sqrt{2}-2i\right),c=-\sqrt{2}i$

Substituting a and b and c in

We know:

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

$\Rightarrow x=\frac{\left(3\sqrt{2}-2i\right)\pm \sqrt{{\left(3\sqrt{2}-2i\right)}^2+4\sqrt{2}i}}{2}.........equation1$

Let $\sqrt{{\left(3\sqrt{2}-2i\right)}^2+4\sqrt{2}i}$ $=$$x+yi$

$$\begin{gathered} \Rightarrow {\left(3\sqrt{2}-2i\right)}^2+4\sqrt{2}i={\left(x+yi\right)}^2 \\ \Rightarrow 14-8\sqrt{2}i=x^2-y^2+2xyi \\ \Rightarrow x^2-y^2=142xy=-8\sqrt{2}............equation2 \end{gathered}$$

$$\begin{gathered} {\left(x^2+y^2\right)}^2={\left(x^2-y^2\right)}^2+4x^2{y}^2 \\ \Rightarrow {\left(x^2+y^2\right)}^2=196+128=324 \\ \Rightarrow x^2+y^2=18.......equation3 \end{gathered}$$

From eq.$2$ and eq.$3$,

$\Rightarrow x=-4,y=\sqrt{2}$or $x=4,y=-\sqrt{2}$

$\Rightarrow x+iy=4-\sqrt{2}i,$or $-4+\sqrt{2}i$

$\Rightarrow x+iy=\pm \left(4-\sqrt{2}i\right)$

Substituting these equations in eq.$1$$\Rightarrow x=\frac{\left(3\sqrt{2}-2i\right)\pm \left(4-\sqrt{2}i\right)}{2}$.