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

1 Answer
KillerBunny
Apr 5, 2015

We know that #(x+a)^2 = x^2+2ax+a^2#. Since your equation starts with #x^2-4x#, it means that we must find a value of #a# such that #2ax=-4x#, which means #a=-2#.

If #a=-2#, then we have #(x-2)^2 = x^2-4x+4#. You can see that your equation #x^2-4x+2# is very close: we only need to add #2#.

So, adding #2# to both sides, we get #x^2-4x+4=2#, which means

#(x-2)^2 = 2#. Extracting square roots at both sides, we get #x-2= \pm\sqrt(2)#, which leads us to the solutions

#x_{1,2} = 2 \pm \sqrt(2)#

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