Question

Solve for x:
$2x^2+3ix+2=0,x\in C$

Answer 1

$\Rightarrow$ The give quadratic equation is $2x^2+3ix+2=0$, comparing it with $a{x}^2+bx+c=0$

$\Rightarrow$ We get, $a=2,b=3i,c=2$

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

$=\frac{-3i\pm \sqrt{(3i{)}^2-4(2\left)\right(2)}}{2\left(2\right)}$

$=\frac{-3i\pm \sqrt{9i^2-16}}{4}$

$=\frac{-3i\pm \sqrt{-9-16}}{4}$ [ Since, $i^2=-1$ ]

$=\frac{-3i\pm \sqrt{-25}}{4}$

$=\frac{-3i\pm 5i}{4}$

$\Rightarrow$ $x=\frac{-3i+5i}{4}$ and $x=\frac{-3i-5i}{4}$

$\therefore$ $x=\frac{i}{2}$ and $x=-2i$