How do you complete the square to solve $x^{2} - 2 x - 3 = 0$ $x^2-2x-3=0$ ?

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Answer 1 · 2015-05-24T20:48:47

$0 = x^{2} - 2 x - 3 = x^{2} - 2 x + 1 - 1 - 3 = {(x - 1)}^{2} - 4$ $0 = x^2-2x-3 = x^2-2x+1-1-3 = (x-1)^2-4$

Adding $4$ $4$ to both ends we find:

${(x - 1)}^{2} = 4$ $(x-1)^2 = 4$

So $x - 1 = \pm \sqrt{4} = \pm 2$ $x-1 = +-sqrt(4) = +-2$

Adding $1$ $1$ to both sides, we get:

$x = 1 \pm 2$ $x = 1+-2$

That is $x = - 1$ $x = -1$ or $x = 3$ $x = 3$

How do you complete the square to solve x2−2x−3=0x^2-2x-3=0?