Finding root of complex equation

  • #1

Homework Statement


Good day,

I've been have having difficulties finding the roots of this:
Find the roots of 3ix^2 + 6x - i = 0
where i = complex number
i = sqrt(-1)

Homework Equations


quadratic formula (apologies for the large image)
quadratic-formula.jpg


The Attempt at a Solution



using the quadratic formula
[-6 (+-) sqrt (36 - 4(3i)i)] / 6i
= (-6/6i) + (sqrt(24)/6i)
multiplying by conjugate I get:
=i (+-) ((-6i * sqrt(24))/ 36)

I'm stuck here.
apparently the roots should have both real and imaginary parts, but I have 2 imaginary parts. ie x = Re + i Im
what exactly do I have to do next?

Thank you.
 

Answers and Replies

  • #2
528
33
You got ##i ^+_- \frac{\sqrt(24)}{6i}##
Now you can write it as ##i ^+_- \frac{2\sqrt(6)}{6i}##
now 1/i=-i
so ##i ^+_- (-i\frac{2\sqrt(6)}{6})## Now take a +/- B as a+b and a-b
 
  • #3
so, there is no real part for the left side of the answer?
or should I express the answer as:
(0 + i) + (−i * ((2√6)/6) ) and (0 + i) - (−i * ((2√6)/6) )
 
  • #4
oh, never mind.
I've finally got it.


Thank you for your time.
 

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