How do you solve using the completing the square method #3x^2-7x-6=0#?

1 Answer
Mar 2, 2017

#color(green)(x=3)# or #color(green)(x=-2/3)#
#color(white)("XXX")#see below for solution method using "completing the square"

Explanation:

#3x^2-7x-6=0#
#rArrcolor(white)("XXX")color(blue)(x^2-7/3x)=color(blue)2#

If #x^2-7/3x# are the first two terms of a squared binomial:
#color(white)("XXX")(x+a)^2=x^2+2ax+a^2#
then
#color(white)("XXX")2a=-7/3color(white)("X")rarrcolor(white)("X")a=-7/6#
and to "complete the square" we will need to add
#color(white)("XXX")color(red)(a^2)=(-7/6)^2=color(red)(49/36)#

giving:
#color(white)("XXX")color(blue)(x^2-7/6x)color(red)(+49/36)=color(blue)2color(red)(+49/36)#

#rArrcolor(white)("XXX")(x-7/6)^2= (2xx36+49)/36=121/36= (11/6)^2#

#rArrcolor(white)("XXX")x-7/6=+-11/6#

#rArrcolor(white)("XXX")x=(7+-11)/6#

#rArrcolor(white)("XXX")x=18/6=3color(white)("X")orcolor(white)("X")x=-4/6=-2/3#