How do you solve by completing the square: #x^2 - 8x - 1 = 0#?

1 Answer
Alan P.
Mar 30, 2015

If #x^2-8x# are the first 2 terms of a squared expression of the form #(x+a)^2#
then the third term should be
#((-8)/2)^2 = 16#

If we are going to replace
#x^2-8x-1#
with
#x^2-8x+16#
we will need to add #+17# to both sides of the equation

#x^2-8x+16 = 17#

re-write the left-side as a square
#(x-4)^2 = 17#

then take the square root of both sides
#x-4 = +-sqrt(17)#

#rarr x = 4+sqrt(17)# or #x = 4 -sqrt(17)#

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