Question

If one root of the equation $x^2-(3\sqrt{2}-2i)x-6\sqrt{2}i=0$ is $-2i,$ then the other root of the equation is

• A. $2i$
• B. $3\sqrt{2}$
• C. $-3\sqrt{2}$
• D. $5\sqrt{2}$

Answer 1

Answer: (B)

Given, quadratic equation is
$x^2-(3\sqrt{2}-2i)x-6\sqrt{2}i=0$
Its one root is $-2i$
Let it's another root is $\alpha$
Then, sum of the roots
$\alpha -2i=3\sqrt{2}-2i$
$\therefore$ $\alpha =3\sqrt{2}$