Question

Solve the following quadratic equation:

$$x^2-(\sqrt{2}+i)x+\sqrt{2}i=0$$

Answer 1

$$x^2-(\sqrt{2}+i)x+\sqrt{2}i=0$$

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

$$a=1,b=-(\sqrt{2}+i)\text{ and }c=\sqrt{2}i$$

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

$$\Rightarrow x=\frac{(\sqrt{2}+i)\pm \sqrt{{(\sqrt{2}+i)}^2-4\sqrt{2}i}}{2}$$

$$\Rightarrow x=\frac{(\sqrt{2}+i)\pm \sqrt{1-2\sqrt{2}i}}{2}$$

$$\Rightarrow x=\frac{(\sqrt{2}+i)\pm \sqrt{{\left(\sqrt{2}\right)}^2-1^2-2\sqrt{2}i}}{2}$$

$$\Rightarrow x=\frac{(\sqrt{2}+i)\pm \sqrt{{(\sqrt{2}-i)}^2}}{2}$$

$$\Rightarrow x=\frac{(\sqrt{2}+i)\pm (\sqrt{2}-i)}{2}$$

$$\Rightarrow x=\sqrt{2},i$$

$$\text{ So, the roots of the given quadratic equation are }\sqrt{2}\text{ and }i.$$