Complex number and quadratic formula? ?

Question

? how would you solve the equation and find the roots of ix^2-3x-5i=0

Answer 1

i x^2 - 3x - 5i = 0 .. . . complete the square

simplify first, and remove the coefficient of the quadratic term

x^2 + 3i x - 5 = 0 ... (this is also the result if you multiply all terms by -i)

x^2 + 3i x = 5

x^2 + 3i x + (3i/2)^2 = 5 + (3i/2)

x^2 + 3i x - 9/4 = 5 - 9/4

x^2 + 3i x - 9/4 = 11/4

(x + 3i/2)^2 = 11/4

x + 3i/2 = ± √11 / 2

x = [ -3i ± √11 ] / 2

.. . .. .

Answer 2

The same way that you would solve for roots in a 'normal' quadratic equation: by using the quadratic formula,

- b +/- sqrt(b^2 - 4*a*c)

____________________

            2*a

Here, a = i , b = - 3, c = - 5i

so you get    

3 +/- sqrt (9 - 20)            

-------------------------   =

           2i

3 +/- sqrt(- 11)

----------------------

          2i

you can do a little more factoring here, if you like, but that's the idea.