How do you solve #3x^3 - 2[ x - ( 3 - 2x ) ] = 3 ( x - 2 )^2#?

1 Answer
Jun 16, 2016

Answer:

Real solution: #x=1#

Complex solutions: # x in {1, sqrt(2)i, -sqrt(2)i}#

Explanation:

L.S. #=3x^3-2[x-(3-2x)]#

#color(white)("XXX")=3x^3-2[-3+3x]#

#color(white)("XXX")=3x^3-6x+6#

R.S. #=3(x-2)^2#

#color(white)("XXX")=3(x^2-4x+4)#

#color(white)("XXX")=3x^2-12x+12#

Since L.S. # = # R.S.
#3x^3+6x+6=3x^2-12x+12#

#rarr x^3+2x+2=x^2-4x+4#

#rarr color(green)(1)x^3-x^2+2x-color(blue)(2)=0#

In the hope of finding rational roots we try the factors of #color(blue)(2)/color(green)(1)#
enter image source here
and we find #color(red)(x=1)# is a root
#rarr (x-1)# is a factor.

#(x^3-x^2+2x-2) div (x-1) = x^2+2#

#(x^2+2)# has no Real roots (i.e. no Real solutions for #x#)

However, if we allow Complex solutions:
#color(white)("XXX")x=+-sqrt(2)i#