How do you solve #3x^2-4x-2=0# by completing the square?

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Answer 1 · 2016-10-24T05:23:16

#x=2/3+sqrt(10)/3# , #x=2/3-sqrt(10)/3#

Firstly, we need to deal with the coefficient of #x^2# (the 3).

In order to remove the 3 we divide the whole expression by 3.

#(3x^2-4x-2)/3=0/3#

#x^2-4/3x-2/3=0#

next, we complete the square, here is the general formula

#x^2+ax=(x+a/2)^2-(a/2)^2#

so using this formula,

#x^2-4/3x-2/3=0#

goes to

#(x-4/6)^2-16/36-2/3=0#

simplifying down,

#(x-2/3)^2-10/9=0#

rearrange for x,

#(x-2/3)^2= 10/9#

#x-2/3= +-sqrt(10/9)#

#x=2/3+sqrt(10)/3# , #x=2/3-sqrt(10)/3#

these are the two solutions.