Question

Solve the following quadratic equation:
(i) $x^2-(3\sqrt{2}+2i)x+6\sqrt{2i}=0$

Answer 1

We have, $x^2-(3\sqrt{2}+2i)x+6\sqrt{2i}=0$
On comparing with general form of quadratic equation
$a{x}^2+bx+c=0$
We get $a=1,b=-(3\sqrt{2}+2i),c=6\sqrt{2}i$

Roots of the equation are

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

$=\frac{(3\sqrt{2}+2i)\pm \sqrt{{(-(3\sqrt{2}+2i))}^2-4\times 1\times 6\sqrt{2}i}}{2\times 1}$

$=\frac{(3\sqrt{2}+2i)\pm \sqrt{(18+12\sqrt{2}i-4)-24\sqrt{2}i}}{2}$

$=\frac{(3\sqrt{2}+2i)\pm \sqrt{(18-12\sqrt{2i}-4)}}{2}$

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

$=\frac{(3\sqrt{2}+2i)\pm \sqrt{{(3\sqrt{2}-2i)}^2}}{2}$

$=\frac{(3\sqrt{2}+2i)\pm (3\sqrt{2}-2i)}{2}$

$=\frac{(3\sqrt{2}+2i)+(3\sqrt{2}-2i)}{2},\frac{(3\sqrt{2}+2i)-(3\sqrt{2}-2i)}{2}$

$=3\sqrt{2},2i$

Hence, the roots are $3\sqrt{2}and2i$.