How do you solve #9x² - 12x + 4 = -3#?

1 Answer
Nov 8, 2017

Answer:

#x=(2+-sqrt(3)i)/3#

Explanation:

Given
#color(white)("XXX")9x^2-12x+4=-3#

There are several ways to solve this.
I will demonstrate using a "completing the square method"

The given equation implies
#color(white)("XXX")9x^2-12x=-7# (after subtracting #4# from both sides)

#color(white)("XXX")9(x^2-4/3x)=-7#

#color(white)("XXX")9(x^2-4/3x+(2/3)^2)=-7+9 * (2/3)^2#

#color(white)("XXX")9(x-2/3)^2=-7+4#

#color(white)("XXX")(x-2/3)^2=-3/9=-1/3#

#color(white)("XXX")(x-2/3)=+-sqrt(-1/3)#

Note that for Real values #sqrt(-1/3)# is undefined,
but if we are allowed Complex values:
#color(white)("XXX")x-2/3=+-1/sqrt(3)i#
and
#color(white)("XXX")x=2/3+-1/sqrt(3)i=(2+-sqrt(3)i)/3#

Why is there not Real solution?
Note that the given equation is equivalent to #9x^2-12x+7=0#
and here is a graph of #9x^2-12x+7#
graph{9x^2-12x+7 [-12.41, 12.9, -4.42, 8.22]}
Notice that #9x^2-12x+7# never crosses the X-axis and therefore it is never equal to #0#