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.

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