How do you solve by completing the square for #y= (2x^2) - 4x - 7#?

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Answer 1 · 2015-05-25T16:41:43

I will assume that you are trying to find the values of #x# for which the resulting #y=0#.

#0 = y = 2x^2-4x-7 = 2(x^2-2x+1-1)-7 = 2((x-1)^2-1) -7#

#=2(x-1)^2-2-7#

#=2(x-1)^2-9#

Add #9# to both ends to get:

#2(x-1)^2 = 9#

Divide both side by #2# to get:

#(x-1)^2 = 9/2 = 18/4#

So #x-1 = +-sqrt(18/4) = +-(3sqrt(2))/2#

Add #1# to both sides to get

#x = 1+-(3sqrt(2))/2#