How do you solve using the completing the square method # x^2-8x+7=0#?

1 Answer
Sep 20, 2016

Answer:

#x=7color(white)("XX")orcolor(white)("XX")x=1#

Explanation:

Given
#color(white)("XXX")x^2-8x+7=0color(white)("XX")#Note: I modified the question to include the #color(blue)(=0)#

Then
#color(white)("XXX")x^2-8x=-7#

Remembering the general relation:
#color(white)("XXX")x^2+2ax+a^2 = (x+a)^2#

If #x^2-8x# are to be the first two terms of #x^2+2ax+a^2#
then
#color(white)("XXX")a=-4#
and
#color(white)("XXX")a^2=16#

So we will need to add #16# (to both sides) to #underline("complete the square")#

#color(white)("XXX")x^2-8x+16 = -7+16#

#color(white)("XXX")(x-4)^2=9#

#color(white)("XXX")x-4=+-3#

#color(white)("XXX")x=4+-3#

#color(white)("XXX")x=7color(white)("XX")orcolor(white)("XX")x=1#