Question

Solve the following quadratic equation :

$2x^2-(3+7i)x+(9i-3)=0$

Answer 1

We have, $2x^2-(3+7i)x+(9i-3)=0$

On comparing with general form of quadratic
equation $a{x}^2+bx+c=0$

We get $a=1,b=-(3+7i),c=(9i-3)$

Roots of the equation are

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

$=\frac{(3+7i)\pm \sqrt{{(-(3+7i\left)\right)}^2-4\left(2\right)(9i-3)}}{2\times 2}$

$=\frac{(3+7i)\pm \sqrt{(9+2\times 3\times 78+49i^2)-8(9i-3)}}{4}$

$=\frac{(3+7i)\pm \sqrt{(9+42i-49-72i+24}}{4}$

$x=\frac{(3+7i)\pm \sqrt{-16-30i}}{4}$

$x=\frac{(3+7i)\pm \sqrt{9-2\times 3\times 5i-25}}{4}$

$x=\frac{(3+7i)\pm \sqrt{3^2-2\times 3\times 5i+(5i{)}^2}}{4}$

$=\frac{(3+7i)\pm \sqrt{(3-5i{)}^2}}{4}$

$=\frac{(3+7i)\pm (3-5i)}{4}$

$=\frac{(3+7i)+(3-5i)}{4},\frac{(3+7i)-(3-5i)}{4}$

$=\frac{(6+2i)}{4},\frac{\left(12i\right)}{4}$

$=\left(\frac{3+i}{2}\right),3i$

Hence, the roots are $\frac{3+i}{2}$ and $3i$