How do you solve by completing the square $x^{2} - 4 x - 11 = 0$ $x^2- 4x-11=0$ ?

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How do you solve by completing the square $x^{2} - 4 x - 11 = 0$ $x^2- 4x-11=0$ ?

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Answer 1 · 2015-04-01T17:38:53

$x = 2 - \sqrt{15}, 2 + \sqrt{15}$ $x=2-sqrt(15), 2+sqrt(15)$

We should reform this equation by a full square. So we have:
$x^{2} - 4 x + c = 0$ $x^2-4x+c=0$

I don't want to go deeper in this, lets conclude this part with $c = 4$ $c=4$

So:

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

Now, lets put ${(x - 2)}^{2}$ $(x-2)^2$ in the given equation.

$G i v e n :$ $Given:$ $1)$ $1)$ $x^{2} - 4 x - 11 = 0$ $x^2 -4x-11=0$
${(x - 2)}^{2} + n = x^{2} - 4 x - 11$ $(x-2)^2+n=x^2-4x-11$
$n = - 15$ $n=-15$

So:

${(x - 2)}^{2} - 15 = x^{2} - 4 x - 11$ $(x-2)^2-15=x^2-4x-11$
${(x - 2)}^{2} - 15 = 0$ $(x-2)^2-15=0$
$\sqrt{{(x - 2)}^{2}} = 15$ $sqrt((x-2)^2) = 15$
$| x - 2 | = \sqrt{15}$ $abs(x-2)=sqrt(15)$
$x - 2 = - \sqrt{15}, \sqrt{15}$ $x-2=-sqrt(15), sqrt(15)$
$x = 2 - \sqrt{15}, 2 + \sqrt{15}$ $x=2-sqrt(15),2+sqrt(15)$

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How do you solve by completing the square $x^{2} - 4 x - 11 = 0$ $x^2- 4x-11=0$ ?

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