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

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Answer 1 · 2015-04-01T22:53:33

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#x^2+6x+2=0#

#(x^2+6x color(white)"ssssssss")+2=0#

#1/2 * 6 = 3#. and #3^2 = 9# so we need to add #9# to get a complete square:

#(x^2+6x +9-9)+2=0#

#(x^2+6x +9)-9+2=0#

#(x+3)^2-7=0#

#(x+3)^2=7#

#x+3= +-sqrt7#

#x= -3 +-sqrt7#

The solutions are: # -3 -sqrt7#, # -3 +sqrt7#