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} $