How do you solve #x^3-6x^2+6x=0#?

1 Answer
Dec 17, 2015

Answer:

I found:
#x_1=0#
#x_2=3+sqrt(3)#
#x_3=3-sqrt(3)#

Explanation:

We can try collecting #x#:
#x(x^2-6x+6)=0#
you get the first solution from the #x(...)=0# bit which gives you: #x_1=0#
the other two solutions are found solving the #x^2-6x+6=0# bit using the Quadratic Formula:
#x_(2,3)=(6+-sqrt(36-24))/2=(6+-sqrt(12))/2=(6+-2sqrt(3))/2=3+-sqrt(3)#
giving:
#x_2=3+sqrt(3)#
#x_3=3-sqrt(3)#