How do you solve #0=3x^2-11x+6# by completing the square?

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Answer 1 · 2015-06-05T19:05:00

#0=3x^2-11x+6#

#=3(x^2-11/3x+2)#

#=3(x^2-11/3x+121/36-121/36+2)#

#=3((x-11/6)^2-121/36+72/36)#

#=3((x-11/6)^2-49/36)#

Divide both sides by #3# to get:

#(x-11/6)^2-49/36 = 0#

Add #49/36# to both sides to get:

#(x-11/6)^2 = 49/36#

Take square root of both sides to get:

#x-11/6 = +-sqrt(49/36) = +-7/6#

Add #11/6# to both sides to get:

#x = 11/6+-7/6 = (11+-7)/6#

That is #x = 3# or #x = 2/3#