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

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