Question

Solve the following quadratic equation $i{x}^2-4x-4i=0$

Answer 1

given, $i{x}^2-4x-4i=0$

$\Rightarrow i(x^2+4xi-4)=0$

$\Rightarrow x^2+4xi-4=0$

$\Rightarrow x^2+2\times x\times 2i+{\left(2i\right)}^2=0$ where ${\left(2i\right)}^2=-4$

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

$\Rightarrow x+2i=0,x+2i=0$

$\therefore x=-2i,-2i$