How do you solve # y = x^2 - 8x + 7 # by completing the square?

1 Answer
May 23, 2015

Given #y=x^2-8x+7#

Note that if #x^2-8x# are the first two term of a squared expression #(x+a)^2#
then #a= 8/2 = 4# and #a^2 = 16#

#y= x^2 -8x+7#

#= x^2-8x + color(red)(4^2) + 7 color(red)(-16)#

#=(x-4)^2 -9#

Now, the real problem
What do you mean by solve and equation of the form #y = # and expression in terms of #x#?

I will assume you wish to determine the value(s) of #x# for which #y=0# (although this is not obvious).

If #(x-4)^2-9 = 0 (=y)#
then
#(x-4)^2 = 9#

#x-4 = +-sqrt(9) = +-3#

and
#x= 7# or #x=1#