How do you solve #2x^2-8x=-8#?

1 Answer
Alan P.
May 17, 2016

#x=2#

Explanation:

If
#color(white)("XXX")2x^3-8x=-8#
then
#color(white)("XXX")x^2-4x=-4# (after dividing by #2#)

#color(white)("XXX")x^2-4x+4=0# (after adding #4# to both sides)

#color(white)("XXX")(x-2)^2=0# (factoring)

#color(white)("XXX")(x-2)=0# (since if #a*b=0# then #a=0# or #b=0#)

#color(white)("XXX")x=2# (after adding #2# to both sides)

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