How do you find the solution to the quadratic equation #2x^2 - 2x = 1#?

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Answer 1 · 2015-05-28T20:20:18

Here's how you solve it by completing the square:

#1 = 2x^2-2x = 2(x^2-x) = 2(x^2 -x +1/4 - 1/4)#

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

Divide both ends by #2# to get

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

Add #1/4# to both sides to get:

#(x-1/2)^2 = 3/4#

Take the square root of both sides, allowing for both positive and negative possibilities:

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

Add #1/2# to both sides to get:

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