How do you solve #16x^2-12x-1=0# by completing the square?

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Answer 1 · 2015-05-23T21:06:10

#(4x-3/2)^2 = 16x^2-12x+9/4#

So

#0 = 16x^2-12x-1 = 16x^2-12x+9/4-9/4-1#

#= (4x-3/2)^2-9/4-1#

#= (4x-3/2)^2 - 13/4#

Add #13/4# to both ends to get:

#(4x-3/2)^2 = 13/4 = (sqrt(13)/2)^2#

So #4x-3/2 = +-sqrt(13)/2#

Add #3/2# to both sides to get:

#4x = 3/2+-sqrt(13)/2#

Divide both sides by #4# to get

#x=3/8+-sqrt(13)/8#