How do you solve $x^{2} - 2 x + 6 = 0$ $x^2-2x+6=0$ by completing the square?

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How do you solve $x^{2} - 2 x + 6 = 0$ $x^2-2x+6=0$ by completing the square?

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Answer 1 · 2015-05-10T14:41:58

Write it as:
$x^{2} - 2 x = - 6$ $x^2-2x=-6$
Add and subtract $1$ $1$ ;
$x^{2} - 2 x + 1 - 1 = - 6$ $x^2-2x+1-1=-6$
$x^{2} - 2 x + 1 = - 6 + 1$ $x^2-2x+1=-6+1$
${(x - 1)}^{2} = - 5$ $(x-1)^2=-5$
So:
$x - 1 = \pm \sqrt{- 5}$ $x-1=+-sqrt(-5)$
$x = 1 \pm i \sqrt{5}$ $x=1+-isqrt(5)$
These are two Complex Solutions where $i = \sqrt{- 1}$ $i=sqrt(-1)$

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How do you solve $x^{2} - 2 x + 6 = 0$ $x^2-2x+6=0$ by completing the square?

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How do you solve $x^{2} - 2 x + 6 = 0$ $x^2-2x+6=0$ by completing the square?