How do you factor #9x^2 + 6x + 1 = 12x#?

1 Answer
May 4, 2018

Answer:

#x = 1/3#

Explanation:

#9x^2 + 6x + 1 = 12x#

subtract #12x# from both sides

#9x^2 + 6x + 1 - 12x= 0#

#color(red)"6x - 12x"#

#9x^2 color(red)"- 6x" + 1 = 0#


then my way that I use when the coefficient of #x^2# not #1#
that I multiply the constant by the coefficient of #x^2# then I try to find two numbers that their product equals to the coefficient of #x^2# and their sum equals to the coefficient of #x#.


like here for example

#9x^2 - 6x + 1= 0#

#9*1=9#

so the two numbers are #-3# and #-3#

their product = 9 and their sum = -6

so then I put them in #color(red)"-6x"# place
#9x^2 color(red)"- 3x - 3x" + 1= 0#

the by grouping take the common factor

#color(red)"9x^2 - 3x" color(blue)" - 3x + 1= 0"#

#3x(3x-1)-(3x-1)=0#

so

#(3x-1)(3x-1) = 0#

#(3x-1)^2=0#

square root both sides

#3x-1 = 0#

add 1 to both sides

#3x = 1#

divide both sides by 3

#x = 1/3#