How do you solve #2x^2-4x+8=0 # by completing the square?

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Answer 1 · 2016-06-05T15:13:44

no real solution

First, let's divide all terms by 2

#2/2x^2-4/2x+8/2=0#
#x^2-2x+4=0#

Let's add and subtract 1:
#x^2-2x+1-1+4=0#

so that the initial three terms represent the square of #(x-1)#

#(x-1)^2+3=0#

that's equivalent to:

#(x-1)^2=-3#

but this equation has no real solution, because a square is not negative