How do you solve #2x^2 - 4x -1=0# by completing the square?

1 Answer
EZ as pi
Nov 19, 2016

#x = 2.22 " or " x = 0225#

Explanation:

#2x^2 -4x -1 =0#

Divide both sides by 2 to get just #x^2#

#x^2 -2x-1/2 =0" "larr #move #-1/2# to the other side
#x^2 -2x " "= 1/2" "larr# add #color(red)((-2/2)^2)# to both sides
#x^2 -2x " "color(red)(+1)= 1/2 color(red)(+1)#

#(x-1)^2" "=3/2" "larr# write as square of a binomial

#x-1" "=+-sqrt(3/2)" "larr# find square root of each side

#x = +sqrt(3/2)+1" "or x = -sqrt(3/2)+1" "larr# solve for x

#x = 2.22 " or " x = 0225#

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