How do you find the roots of #x^3-18x+27=0#?

1 Answer
Oct 31, 2016

Answer:

The roots are #(3, -3/2+3sqrt5/2, -3/2-3sqrt5/2)#

Explanation:

Let #f(x)=x^3-18x+27#
Then by trial and error, #f(3)=27-54+27=0#
so #x=3# is a root
Then we do a long division
#x^3##color(white)(aaaaa)##-18x+27##∣##x-3#
#x^3-3x^2##color(white)(aaaaaaaaa)##∣##x^2+3x-9#
#0+3x^2-18x#
#color(white)(aaaa)##0-9x#
#color(white)(aaaaaa)##-9x+27#
#color(white)(aaaaaa)##-9x+27#
#color(white)(aaaaaaaa)##0+0#
To find the other roots, we must solve
#x^x+3x-9=x^2+3x+9/4-9-9/4#
#(x+3/2)^2-45/4#
So #(x+3/2)^2=45/4##=>##x+3/2=+-sqrt45/2#
#x=-3/2+-3sqrt5/2#