How do you find the complex roots of #6c^3-3c^2-45c=0#?

1 Answer
mason m
Nov 27, 2016

Remove a common factor of #3c# from each term.

#6c^3-3c^2-45c=0#

#3c(2c^2-c-9)=0#

Checking the discriminant of the resulting quadratic, #Delta=b^2-4ac=(-1)^2-4(2)(-9)=73# we see that the quadratic will have two Real roots for a total of three Real roots (and no Complex roots).

The first zero comes from #3c=0#, or #c=0#.

The second two, through the quadratic formula for #2c^2-c-9=0# are:

#c=(1+-sqrt(1-4(2)(-9)))/4=(1+-sqrt73)/4#

We see that our three zeros are #c=0,(1+-sqrt73)/4# and that these are all three Real zeros.

Related questions
See all questions in Zeros
Impact of this question
1992 views around the world
You can reuse this answer
Creative Commons License